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You have nothing in your shopping cart yet. Title: Solution manual of Engineering Economics 7th Ed Description: Solution manual of Engineering Economics 7th Ed by Leland Blank and Anthony Extracts from the notes are below, to see the PDF you'll receive please use the links aboveAE=mc 2ThiseBook is downloaded fromwww ... net∑1PlentyofeBooks ... For more Free eBooks and educationalmaterial visitwww ... netUploaded By$am$exy98theBooks Se v e nth Ed itionn thENGINEERINGECONOMY Se v e n th Ed itionENGINEERINGECONOMYLeland Blank, P ... Texas A & MUniversityAmerican University of Sharjah, United Arab EmiratesAnthony Tarquin, P ... University of Texas at El PasoTMTMENGINEERING ECONOMY: SEVENTH EDITIONPublished by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc ... Copyright © 2012 by The McGraw-HillCompanies, Inc ... Previous editions© 2005, 2002, and 1998 ... ,including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distancelearning ... This book is printed on recycled, acid-free paper containing 10% postconsumer waste ... MassarSenior MarketingManager: Curt ReynoldsDevelopment Editor: Lorraine K ... RolwesCover Designer: Studio Montage, St ... Library of Congress Cataloging-in-Publication DataBlank, Leland T ... — 7th ed ... cm ... ISBN-13: 978-0-07-337630-1 (alk ... Engineering economy ... Tarquin, Anthony J ... Title ... 4 ... 15—dc222010052297www ... comThis book is dedicated to Dr ... Sheppard, Jr ... M C GRAW-HILL DIGITAL OFFERINGSMcGraw-Hill Create™Craft your teaching resources to match the way you teach! With McGraw-Hill Create™, www ... com, you can easily rearrange chapters, combine material from othercontentsources, and quickly upload content you have written like your course syllabus or teaching notes ... Arrange your book to fit your teaching style ... Order a Create book and you’ll receive a complimentary print review copy in 3–5 business days or a complimentary electronic review copy (eComp) viaemail in minutes ... mcgrawhillcreate ... McGraw-Hill Higher Education and Blackboard HaveTeamed UpBlackboard, the Web-based course-management system, has partnered with McGraw-Hill to better allow students and faculty to use online materials and activities to complement face-to-faceteaching... You’ll transform your closeddoor classrooms into communities where students remain connected to their educational experience 24 hours a day ... McGraw-Hill and Blackboard cannow offer you easy access to industry leading technology and content, whether your campushosts it, or we do ...Electronic Textbook OptionsThis text is offered through CourseSmart for both instructors and students ... 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CONTENTSPreface to Seventh EditionLEARNINGSTAGE 1Chapter 1xiiiTHE FUNDAMENTALSFoundations of Engineering Economy1 ... 21 ... 41 ... 61 ... 81 ... 10Chapter 2Factors: How Time and Interest Affect MoneyPE2 ... 22 ... 42 ... 62 ... 13 ... 3Chapter 4EngineeringEconomics: Description andRole in Decision MakingPerforming an Engineering Economy StudyProfessional Ethics and Economic DecisionsInterest Rate and Rate of ReturnTerminology and SymbolsCash Flows: Estimation and DiagrammingEconomic EquivalenceSimple and Compound InterestMinimumAttractive Rate of ReturnIntroduction to Spreadsheet UseChapter SummaryProblemsAdditional Problems and FE Exam Review QuestionsCase Study—Renewable Energy Sources for Electricity GenerationCase Study—Refrigerator ShellsCalculations for Uniform Series That Are ShiftedCalculationsInvolving Uniform Series and Randomly Placed Single AmountsCalculations for Shifted GradientsChapter SummaryProblemsAdditional Problems and FE Exam Review QuestionsCase Study—Preserving Land for Public UseNominal and Effective Interest RatesPE4 ... 24 ... 44 ... 64 ... 84 ... 15 ... 35 ...5Advantages and Uses of Annual Worth AnalysisCalculation of Capital Recovery and AW ValuesEvaluating Alternatives by Annual Worth AnalysisAW of a Permanent InvestmentLife-Cycle Cost AnalysisChapter SummaryProblemsAdditional Problems and FE Exam Review QuestionsCase Study—TheChanging Scene of an Annual Worth AnalysisRate of Return Analysis: One Project7 ... 27 ... 47 ... 6Chapter 8Progressive Example—Water for Semiconductor Manufacturing CaseFormulating AlternativesPresent Worth Analysis of Equal-Life AlternativesPresent Worth Analysis of Different-LifeAlternativesFuture Worth AnalysisCapitalized Cost AnalysisChapter SummaryProblemsAdditional Problems and FE Exam Review QuestionsCase Study—Comparing Social Security BenefitsAnnual Worth Analysis6 ... 26 ... 46 ... 18 ... 38 ... 58 ... 7Chapter 9All-in-One Spreadsheet Analysis(Optional)Chapter SummaryProblemsAdditional Problems and FE Exam Review QuestionsCase Study—ROR Analysis with Estimated Lives That VaryCase Study—How a New Engineering Graduate Can Help His FatherBenefit/Cost Analysis and Public Sector EconomicsPE9 ... 29 ... 49 ... 6ProgressiveExample—Water Treatment Facility #3 CasePublic Sector ProjectsBenefit/Cost Analysis of a Single ProjectAlternative Selection Using Incremental B/C AnalysisIncremental B/C Analysis of Multiple, Mutually Exclusive AlternativesService Sector Projects and Cost-Effectiveness AnalysisEthicalConsiderations in the Public SectorChapter SummaryProblemsAdditional Problems and FE Exam Review QuestionsCase Study—Comparing B/C Analysis and CEA of Traffic Accident ReductionLEARNINGSTAGE 2218219220225226227228229230235238242246250251252258259EPILOGUE:SELECTING THE BASIC ANALYSIS TOOLLEARNINGSTAGE 3ixMAKING BETTER DECISIONSChapter 10Project Financing and Noneconomic Attributes10 ... 210 ... 410 ... 610 ... 111 ... 311 ... 511 ... 112 ... 312 ... 5Chapter 13Breakeven and Payback Analysis13 ... 213 ... 4LEARNINGSTAGE4Chapter 14Effects of InflationUnderstanding the Impact of InflationPresent Worth Calculations Adjusted for InflationFuture Worth Calculations Adjusted for InflationCapital Recovery Calculations Adjusted for InflationChapter SummaryProblemsAdditional Problems and FE Exam Review QuestionsCaseStudy—Inflation versus Stock and Bond InvestmentsCost Estimation and Indirect Cost Allocation15 ... 215 ... 415 ... 615 ... 8Chapter 16Breakeven Analysis for a Single ProjectBreakeven Analysis Between Two AlternativesPayback AnalysisMore Breakeven and Payback Analysis on SpreadsheetsChapterSummaryProblemsAdditional Problems and FE Exam Review QuestionsCase Study—Water Treatment Plant Process Costs322323325327329332334334338340341345348352355355361363ROUNDING OUT THE STUDY14 ... 214 ... 4Chapter 15An Overview of Capital Rationing amongProjectsCapital Rationing Using PW Analysis of Equal-Life ProjectsCapital Rationing Using PW Analysis of Unequal-Life ProjectsCapital Budgeting Problem Formulation Using Linear ProgrammingAdditional Project Ranking MeasuresChapter SummaryProblemsAdditional Problems and FE Exam ReviewQuestionsUnderstanding How Cost Estimation Is AccomplishedUnit MethodCost IndexesCost-Estimating Relationships: Cost-Capacity EquationsCost-Estimating Relationships: Factor MethodTraditional Indirect Cost Rates and AllocationActivity-Based Costing (ABC) for Indirect CostsMaking Estimatesand Maintaining Ethical PracticesChapter SummaryProblemsAdditional Problems and FE Exam Review QuestionsCase Study—Indirect Cost Analysis of Medical Equipment Manufacturing CostsCase Study—Deceptive Acts Can Get You in TroubleDepreciation Methods16 ... 216 ... 416 ... 6DepletionMethodsChapter SummaryAppendix16A ... 2 Switching between Depreciation Methods16A ... 117 ... 317 ... 517 ... 717 ... 9Chapter 18Sensitivity Analysis and Staged Decisions18 ... 218 ... 418 ... 6Chapter 19Determining Sensitivity to Parameter VariationSensitivity Analysis Using ThreeEstimatesEstimate Variability and the Expected ValueExpected Value Computations for AlternativesStaged Evaluation of Alternatives Using a Decision TreeReal Options in Engineering EconomicsChapter SummaryProblemsAdditional Problems and FE Exam Review QuestionsCase Study—Sensitivity tothe Economic EnvironmentCase Study—Sensitivity Analysis of Public Sector Projects—Water Supply PlansMore on Variation and Decision Making under Risk19 ... 219 ... 419 ... 1A ... 3A ... 5A ... 1B ... 3The Balance SheetIncome Statement and Cost of Goods Sold StatementBusinessRatios561561562563Appendix CCode of Ethics for Engineers566Appendix DAlternate Methods for Equivalence Calculations569D ... 2Appendix EGlossary of Concepts and TermsE ... 2Reference MaterialsFactor TablesPhoto CreditsIndexUsing Programmable CalculatorsUsing the Summation of aGeometric Series579581610611Important Concepts and GuidelinesSymbols and Terms569570573573576PREFACE TO SEVENTH EDITIONThis edition includes the time-tested approach and topics of previous editions and introduces significantly new print and electronic features useful to learningabout and successfully applying the exciting field of engineering economics ... Learning to understand, analyze, and manage the money side of anyproject is vital to its success ... This book is a great help to the learner and the instructor in accomplishing these goals by using easy-to-understand language,simple graphics, and online features ... Plus the supporting online materialsare new and updated to enhance the teaching and learning experience ... Studentsshould be at the sophomore level or above with a basic understanding of engineering conceptsand terminology ... Practitioners and professionalengineers who need a refresher in economic analysis and costestimation will find this book very useful as a reference document as well as a learning medium ... Each chapter starts with a statement of purpose and a specific learning outcome (ABET style) for each section ... The appendices are importantelements of learning for this text:Appendix AAppendix BAppendix CAppendix DAppendix ESpreadsheet layout and functions (Excel is featured)Accounting reports and business ratiosCode of Ethics for Engineers (from NSPE)Equivalence computations using calculators and geometric series; notablesConcepts, guidelines, terms, and symbols for engineering economicsThere is considerable flexibility in the sequencing of topics and chapters once the first sixchapters are covered, as shown in the progression graphic on the next page ... Foundations2 ... More Factors4 ... Present Worth6 ... Rate ofReturn8 ... Benefit/Cost10 ... Replacement12 ... Breakeven andPayback14 ... Estimation16 ... After-Tax18 ... Risk and Simulationafter Learning Stage 2 (Chapter 9) is completed ... The progression graphic can help in the design of the course content andtopic ordering ... This behavioral-basedapproachsensitizes the reader to whatis ahead, leading to improvedunderstanding and learning ... SECTIONTOPICLEARNING OUTCOME3 ... 3 ... 3 ... CONCEPTS AND GUIDELINESTime value of moneyIt is a well-known fact that money makes money ... This is the most important concept in engineeringeconomy ... Appendix E includes a briefdescription of each fundamental concept ... 6Numerous in-chapter examplesthroughout the book reinforce thebasic concepts and makeunderstanding easier ... A dot-com company plans to place money in a new venture capital fund that currently returns18% peryear, compounded daily ... 7], with r ϭ 0 ... 0 ... 716%Effective i% per year ϭ 1 ϩ ——365(b) Here r ϭ 0 ... 0 ... 415%Effective i% per 6 months ϭ 1 ϩ ——182()()PROGRESSIVE EXAMPLESPEWater for Semiconductor Manufacturing Case: The worldwide contribution ofsemiconductor sales is about $250billionper year, or about 10% of the world’sGDP (gross domestic product) ... Depending upon thetype and size of fabrication plant (fab),the need for ultrapure water (UPW) tomanufacture these tiny integrated circuitsis high, ranging from 500 to 2000 gpm(gallons per minute) ... Potable waterobtained frompurifying seawater orbrackish groundwater may cost from$2 to $3 per 1000 gallons, but to obtainUPW on-site for semiconductor manufacturing may cost an additional $1 to $3 per1000 gallons ... 5 billion toconstruct, with approximately 1% of thistotal, or $25 million, required to providethe ultrapure waterneeded, includingthe necessary wastewater and recyclingequipment ... It is fortunate toSeveral chapters include a progressiveexample—a more detailed problem statementintroduced at the beginning of the chapter andexpanded upon throughout the chapter inspecially marked examples ... have theoption of desalinated seawateror purified groundwater sources in thelocation chosen for its new fab ... SourceEquipment firstcost, $MSeawater(S)Ϫ20Groundwater(G)Ϫ22Ϫ0 ... 3Salvage value, % offirst cost510Cost of UPW, $ per1000 gallons45AOC, $M per yearAngular has made some initialestimatesfor the UPW system ... 2)PW analysis of different-life alternatives (Section 5 ... 5)Problems 5 ... 34bla76302_ch04_094-126 ... indd 212An icon in the margin indicates theavailability of an animated voice-over slidepresentation that summarizes the material inthe section and provides a briefexample forlearners who need a review or prefer videobased materials ... 3 ... In this case several methods can be used to find the equivalent present worth P ... • Use the F͞P factor to find the future worth of each disbursement in year 13, add them, andthen find the present worth of the total, using P ϭF(P͞F,i,13) ... • Use the P͞A factor to compute the “present worth” P3 ϭ A(P͞A,i,10) (which will be locatedin year 3, not year 0), and then find the present worth in year 0 by using the (P͞F,i,3) factor ... 8BreakevenChris and her father just purchased a ROR Ϸoffice building for $160,000 that is in need of alotIncremental small 17%of repairs, but is located in a prime commercial area of the city ... are $18,000 the first year, increasing by $1000 per year thereafter ... 4 ... SPREADSHEETSSolution12/11/10 6:52 PMFigure 13–11 shows the annual costs (column B) and the sales prices if the building is kept 2or3 years (columns C and E, respectively) ... These results bracket the paybackperiod for each retention period 7–12 price ... Figure and salesSpreadsheet between 3 and ROIC (column D) ... 6 ... (b) At a sales price of $370,000, the 8% return payback period is between 5 and 6 years (column F) ...bla76302_ch07_172-201 ... Cell tags or full cells detailbuilt-in functions and relations developedto solve a specific problem ... 8bla76302_ch13_340-364 ... indd 73FE EXAM AND COURSEREVIEWSEach chapter concludes with severalmultiple-choice, FE Exam–styleproblems that provide asimplifiedreview of chapter material ... 12/7/10 7:26 AMADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS8 ... 39 In comparing mutually exclusive alternatives bythe ROR method, you should:(a) Find the ROR of each alternative and pickthe one with the highest ROR(b) S lhlihil8 ... PEC hasa capacity of approximately 1300 MW (megawatts) of power, ofwhich 277 MW, or about 21%, is from renewable sources ... A constantquestion is how much of PEC’s generation capacity should befrom renewable sources, especially given the environmentalissues with coal-generated electricity and therising costs ofhydrocarbon fuels ... Consider yourself a member of the board of directors ofPEC ... As such, you do not represent a specific district within theentire service area; all other directors do represent a specificdistrict ... InformationHere are some data that you have obtained ... Electricitygeneration cost estimates are nationalin scope, not PEC-specific, and are provided in cents perkilowatt-hour (¢/kWh) ... 54 ... 14 ... 5Reasonable Average7 ... 68 ... 8National average cost of electricity to residential customers: 11¢/kWhPEC average cost to residential customers: 10 ... 92 ¢/kWh (renewablesources)Expected life of a generation facility: 20 to 40 years (it islikely closer to 20 than 40)Time to construct a facility: 2 to 5 yearsCapital cost to build a generation facility: $900 to $1500per kWYou have also learned that the PEC staff uses the wellrecognized levelized energy cost (LEC) method todeterminethe price of electricity that must be charged to customers tobreak even ... The LEC formula, expressedin dollars per kWh for (t ϭ 1, 2, ... If you wanted to know more about the new arrangement with the wind farm in south Texas for the additional 60 MW per year, what types of questionswouldyou ask of a staff member in your first meeting withhim or her?2 ... What about the ethical aspects of the government’s allowance for these plants to continue polluting the atmospherewith the emissions that may cause health problems forcitizens and further the effects of global warming? Whattypesof regulations, if any, should be developed for PEC(and other generators) to follow in the future?New and updated case studies at theend of most chapters present realworld, in-depth treatments andexercises in the engineeringprofession ... ACKNOWLEDGMENT OF CONTRIBUTORSIt takes the inputand efforts of many individuals to make significant improvements in a textbook ... Paul Askenasy, Texas Commission on Environmental QualityJack Beltran, Bristol-Myers SquibbRobert Lundquist, Ohio State UniversityWilliam Peet, Infrastructure Coordination, Government of NiueSallie Sheppard, TexasA&M UniversityWe thank the following individuals for their comments, feedback, and review of material to assistin making this edition a real success ... Dacquisto, Gonzaga UniversityHoushang Darabi, University of Illinois at ChicagoFreddie Davis, West Texas A&M UniversityEdward Lester Dollar,Southern Polytechnic State UniversityTed Eschenbach, University of AlaskaClara Fang, University of HartfordAbel Fernandez, University of the PacificDaniel A ... Fries, University of Central FloridaNathan Gartner, University of Massachusetts–LowellJohnny R ... Johnson, Valparaiso UniversityJustin W ...Madrid, New Mexico State UniversitySaeed Manafzadeh, University of Illinois at ChicagoQuamrul Mazumder, University of Michigan–FlintDeb McAvoy, Ohio UniversityGene McGinnis, Christian Brothers UniversityBruce V ... Rider, Ohio Northern UniversityJohn Ristroph, University of Louisiana atLafayetteSaeid L ... Song, Norfolk State UniversityJames Stevens, University of Colorado at Colorado SpringsJohn A ... Sutton, Purdue UniversityPete Weiss, Valparaiso UniversityxxAcknowledgment of ContributorsGreg Wiles, Southern Polytechnic State UniversityRichard Youchak, University ofPittsburgh at JohnstownWilliam A ... We hope you find the contents of this edition helpful inyour academic and professional activities ... comatarquin@utep ... When you have completed stage 1, you will beable to understand and work problems that account for thetime value of money, cash flowsoccurring at different times withdifferent amounts, and equivalence at different interest rates ... The factors commonly used in all engineering economy computations are introduced and applied here ... Also, after these chapters, youshould be comfortable using many of the spreadsheet functions ... Acheckmark icon in the margin indicates that a new concept or guidelineis introduced at this point ... SECTIONTOPICLEARNING OUTCOME1 ... 1 ... 1 ... 1 ... 1 ... 1 ... 1 ... 1 ... 1 ... 1 ... The need for engineering economy is primarily motivated by the work that engineersdo in performing analyses,synthesizing, and coming to a conclusion as they work onprojects of all sizes ... These decisions involve the fundamental elements of cash flows of money, time,and interest rates ... 1 ... Most decisions involve money, called capital or capital funds, which is usually limited inamount ... Engineers play a vitalrole in capital investment decisions based upon their ability and experienceto design, analyze, and synthesize ... Engineering economy deals with theeconomic factors ... Mathematical techniquessimplify the economic evaluation of alternatives ... Therefore, besides applications to projects in your futurejobs, what you learnfrom this book and in this course may well offer you an economic analysis tool for makingpersonal decisions such as car purchases, house purchases, major purchases on credit, e ... ,furniture, appliances, and electronics ... People make decisions; computers, mathematics, concepts,and guidelines assist people intheir decision-making process ... Therefore, the numbers used in engineering economy are best estimates of what is expected to occur ... In short, the variation between an amount or time estimated now and that observedin the future is caused by the stochastic (random)nature of all economic events ... Example 1 ... EXAMPLE 1 ... He found the average cost (tothe nearest dollar) to be $570 per repair from data taken over a 5-year period ... However, the last 3 years of costs are higher andmore consistent with an average of $631 ... If the analysis is to use the most recentdata and trends, a range of, say, Ϯ5% of $630 is recommended ... The criterion used to select an alternative in engineering economy for a specific set of estimatesis called a measure of worth ... This is theconcept of the time value of money ... The time value of money explains the changein the amount ofmoney over time for funds that are owned (invested) or owed (borrowed) ... The time value of money is very obvious in the world of economics ... If we borrow money today, in one form or another, we expect to return the originalamount plus some additional amount of money ... For example,assume youinvested $4975 exactly 3 years ago in 53 shares of IBM stock as traded on the NewYork Stock Exchange (NYSE) at $93 ... You expect to make 8% per year appreciation,not considering any dividends that IBM may declare ... 25 per share for a total of $6744 ... This increase in value represents arate ofreturn of 10 ... (These type of calculations are explained later ... 1 ... Implementing a structured procedure is the best approach to select the best solution to the problem ... 2 ... 4 ... Collect relevant, available data and define viable solution alternatives ... Identify an economic measure of worth criterion fordecision making ... 2Performing an Engineering Economy Study5 ... 6 ... 7 ... Technically, the last step is not part of the economy study, but it is, of course, a step needed tomeet the project objective ... Accordingly, steps 5 and 6 may result in selectionof an alternative different from the economically bestone ... This occurs when projects are independent of one another ... Figure 1–1 illustrates the steps above forone alternative ... Problem Description and Objective Statement A succinct statement of the problem andprimary objective(s) is very important to the formation of an alternative solution ... Theobjectives may be to generate the forecasted electricityStep instudy1ProblemdescriptionObjectivestatementAvailable data2Alternatives forsolution3Cash flows andother estimates4Measure of worthcriterion5Engineeringeconomic analysis6Best alternativeselection7One or more approachesto meetobjectiveExpected lifeRevenuesCostsTaxesProject financingImplementationand monitoringPW, ROR, B/C, etc ... New engineeringeconomy studybegins56Chapter 1Foundations of Engineering Economyneeded for 2015 and beyond, plus to not exceed all the projected emission allowances in thesefutureyears ... Words, pictures, graphs, equipment and service descriptions, simulations, etc ... The best estimates for parameters are also part of the alternative ... If changes in income (revenue)may occur, this parameter must be estimated ... For example, if two alternatives aredescribed and analyzed, onewill likely be selected and implementation initiated ... Cash Flows All cash flows are estimated for each alternative ... When cash flow estimates for specific parameters are expected tovary significantly from a point estimate made now, risk and sensitivity analyses (step 5) areneeded to improve thechances of selecting the best alternative ... Estimation ofcosts is discussed in Chapter 15, and the elements of variation (risk) and sensitivity analysis areincluded throughout the text ... The result of the analysis will be one or more numerical values; this can bein one of several terms, such as money, aninterest rate, number of years, or a probability ... Before an economic analysis technique is applied to the cash flows, some decisions aboutwhat to include in the analysis must be made ... Federal, state or provincial, county, and city taxes will impact the costs of everyalternative ... If taxes and inflation areexpected to impact all alternatives equally, theymay be disregarded in the analysis ... Also, if the impact of inflation over time is importantto the decision, an additional set of computations must be added to the analysis; Chapter 14covers the details ... For example, if alternative A has a rate of return (ROR)of15 ... 9% per year, B is better economically ... There are many possible noneconomic factors;some typical ones are•••••Market pressures, such as need for an increased international presenceAvailability of certain resources, e ... , skilled labor force, water, power, tax incentivesGovernment laws thatdictate safety, environmental, legal, or other aspectsCorporate management’s or the board of director’s interest in a particular alternativeGoodwill offered by an alternative toward a group: employees, union, county, etc ... At times, only one viable alternative is identified ... The do-nothing alternativemaintains the status quo ... 3Professional Ethics and Economic DecisionsWhether we are aware of it or not, we use criteria every day to choose between alternatives ... But how did youdefine best? Was the best route the safest, shortest, fastest, cheapest, most scenic, or what? Obviously, dependingupon which criterion or combination of criteria is used to identify the best, adifferent route might be selected each time ... Thus, when there areseveral ways of accomplishing a stated objective, the alternative with the lowest overall cost orhighest overall net income is selected ... 3 Professional Ethics andEconomic DecisionsMany of the fundamentals of engineering ethics are intertwined with the roles of money andeconomics-based decisions in the making of professionally ethical judgments ... For example, Chapter 9, Benefit/Cost Analysis and Public Sector Economics, includes material on the ethics ofpublic project contracts and public policy ... The terms morals and ethics are commonly used interchangeably, yet they have slightlydifferent interpretations ... Ethical practices can be evaluated byusing a code of morals or code of ethics that forms the standards to guide decisions andactions of individualsand organizations in a profession, for example, electrical, chemical,mechanical, industrial, or civil engineering ... Universal or common morals These are fundamental moral beliefs held by virtually all people ... It is possible for actions and intentions to come into conflict concerning a common moral ... Aftertheir collapse on September 11,2001, it was apparent that the design was not sufficient to withstand the heat generated by thefirestorm caused by the impact of an aircraft ... However, their designactions did not foresee this outcome as a measurable possibility ... These usually parallel the common moralsin that stealing, lying, murdering, etc ... It is quite possible that an individual strongly supports the common morals and has excellentpersonal morals, but these may conflict from time to time when decisions must be made ... If he or she does notknow how to work some test problems, but must make acertain minimum grade on the finalexam to graduate, the decision to cheat or not on the final exam is an exercise in following orviolating a personal moral ... The codestates the commonly accepted standards of honesty and integrity that each individual is expectedto demonstrate in her or his practice ...Although each engineering profession has its own code of ethics, the Code of Ethics forEngineers published by the National Society of Professional Engineers (NSPE) is very commonly used and quoted ... Here are three examples fromthe Code:“Engineers, in the fulfillment of their duties, shall holdparamount the safety, health, and welfare of the public ... 1)“Engineers shall not accept financial or other considerations, including free engineering designs, from material or equipment suppliers for specifying their product ... 5 ... ”(section III ... b)As with common and personal morals, conflicts can easilyrise in the mind of an engineerbetween his or her own ethics and that of the employing corporation ... Suppose the engineer has worked for years in a military defense contractor’sfacility and does the detailed cost estimations and economic evaluations of producing fighterjets for the Air Force ... Althoughthe employer and the engineer are not violating any ethics code,the engineer, as an individual, is stressed in this position ... Conflicts such as this can place individuals in real dilemmas with no or mostly unsatisfactoryalternatives ... Many money-related situations, such as those that follow, can haveethical dimensions ... • Family or personal connections with individuals in a company offer unfair or insider information that allows costs to be cut in strategic areas of a project ... While the system is operating:• Delayed or below-standard maintenance can be performed to save money when costoverrunsexist in other segments of a project ... • Safety margins are compromised because of cost, personal inconvenience to workers, tighttime schedules, etc ... 245–8) ... Some 500,000 persons were exposed to inhalation of this deadly gas that burns moist parts of the body ... Although Union Carbideowned the facility, the Indian governmenthad only Indian workers in the plant ... However, one ofthe surprising practices that caused unnecessary harm to workers was the fact that masks, gloves,and other protective gear were not worn by workers in close proximity to the tanks containingMIC ... Manyethical questions arise when corporations operate in international settings where thecorporate rules, worker incentives, cultural practices, and costs in the home country differ fromthose in the host country ... 3Professional Ethics and Economic Decisionsother cost-reducing factors ... It is important tounderstand that the translation from universal morals to personal morals andprofessional ethics does vary from one culture and country to another ... In some societies and cultures, corruption in the process of contract making is common andoften “overlooked” by the local authorities, who may also beinvolved in the affairs ... Find and punish theindividuals involved ... EXAMPLE 1 ... He has been a registered professional engineer (PE) for the last 15 years ... Carol offered to pay forhis time and talent, but Jamie saw no reason to take money for helping with data commonlyused by him in performing hisjob at Burris ... Yesterday, Jamie was called into his supervisor’s office and told that Burris had not receivedthe contract award in Sharpstown, where a metro system is to be installed ... This job wasgreatly needed by Burris, as the country and most municipalities were in a real economicslump, so muchso that Burris was considering furloughing several engineers if the Sharpstownbid was not accepted ... Jamie was astounded and angry ... The numbers used by the competitor to win the Sharpstown award were the same numbers that Jamie had prepared for Burrison this bid, and they closely matchedthe values that he gave Carol when he helped her ... As a result,Jamie was escorted out of his office and the building within one hour and told to not ask anyoneat Burris for a reference letter if he attempted to get another engineering job ... Refer to the NSPE Code of Ethics for Engineers (Appendix C) forspecific points of concern ... Some of these mistakes, oversights, and possible codeviolations are summarized here ... 1 ... ”910Chapter 1Foundations of Engineering EconomyCarol• Did not share the intended use of Jamie’s work• Did not seek information from Jamie concerning his employer’s intentionto bid on thesame project as her client• Misled Jamie in that she did not seek approval from Jamie to use and quote his informationand assistance• Did not inform her client that portions of her work originated from a source employed by apossible bid competitor• Likely violated, at least, Code of Ethics forEngineers section III ... a, which reads, “Engineers shall, whenever possible, name the person or persons who may be individually responsible for designs, inventions, writings, or other accomplishments ... 1 ... Computationally, interest is the differencebetween an ending amount of money and thebeginning amount ... There are always two perspectives to an amount of interest—interest paidand interest earned ... Interest is paid when a person or organization borrowed money (obtained a loan) and repays a larger amount over time ... The numerical values and formulas used are the same for bothperspectives, butthe interpretations are different ... 1]When interest paid over a specific time unit is expressed as a percentage of the principal, the result is called the interest rate ... 2]The time unit of the rate is called the interest period ... Shorter time periods can be used, such as 1% per month ... If onlythe rate is stated,for example, 8 ... LoanBankRepaymentϩ interestLoanBorrowerInvestorRepaymentϩ interest(a)Figure 1–2(a) Interest paid over time to lender ... (b)Corporation1 ... 3An employee at LaserKinetics ... Determine the interest amount and the interest rate paid ... Apply Equation [1 ... Interestpaid ϭ $10,700 Ϫ 10,000 ϭ $700Equation [1 ... $700Percent interest rate ϭ ———— ϫ 100% ϭ 7% per year$10,000EXAMPLE 1 ... , plans to borrow $20,000 from a bank for 1 year at 9% interest for newrecording equipment ... (b) Construct a column graph that shows the original loan amount and totalamount due after 1 yearused to compute the loan interest rate of 9% per year ... 2] for interest accrued ... 09) ϭ $1800The total amount due is the sum of principal and interest ... 2]: $1800 interest, $20,000 original loanprincipal, 1-year interest period ... Example 1 ... CommentNote that in part (a), the totalamount due may also be computed asTotal due ϭ principal(1 ϩ interest rate) ϭ $20,000(1 ... 1112Chapter 1Foundations of Engineering EconomyFrom the perspective of a saver, a lender, or an investor, interest earned (Figure 1–2b) is thefinal amount minus the initial amount, or principal ... 3]Interestearned over a specific period of time is expressed as a percentage of the original amountand is called rate of return (ROR) ... 4]The time unit for rate of return is called the interest period, just as for the borrower’s perspective ... The term return on investment (ROI) is used equivalently with ROR indifferent industries andsettings, especially where large capital funds are committed to engineering-oriented programs ... 2] and [1 ... EXAMPLE 1 ... (b) Calculate the amount of interest earned during this time period ... If X is the original deposit,Total accrued ϭ deposit ϩ deposit(interest rate)$1000 ϭ X ϩX(0 ... 05) ϭ 1 ... 381 ... 3] to determine the interest earned ... 38 ϭ $47 ... 3 to 1 ... When more than one interest period is involved, e ... , the amount of interest after 3 years, it is necessary to state whether the interest is accrued on a simple or compoundbasis from one period to the next ... Since inflationcan significantly increase an interest rate, some comments about the fundamentals of inflation are warranted at this early stage ... That is, $10 now will not purchase the same amount of gasolinefor your car (or most other things) as $10 did 10 years ago ... InflationIn simple terms, interest rates reflect twothings: a so-called real rate of return plus the expectedinflation rate ... The safest investments (such as government bonds) typically have a 3% to 4% real rate ofreturn built into their overall interest rates ... Clearly, inflation causes interest rates to rise ... And from the vantage point of the saver or investorin a fixed-interest account,1 ... Inflation means that cost and revenuecash flow estimates increase over time ... We see the effect of inflation in that moneypurchases less now than it did at a previous time ... Commonly, engineering economy studies assume that inflation affects all estimated valuesequally... However, if inflation were explicitly taken into account, and it was reducing the value of money at, say, an average of 4% per year,then it would be necessary to perform the economic analysis using an inflated interest rate ... 32% per year using the relations derived in Chapter 14 ... 5 Terminology andSymbolsThe equations and procedures of engineering economy utilize the following terms and symbols ... P ϭ value or amount of money at a time designated as the present or time 0 ... Also F is called future worth (FW)and future value (FV); dollarsA ϭ series of consecutive, equal, end-of-period amountsof money ... It should be clear that a present value P represents asingle sum of money at some time prior to a future value F or prior to the first occurrence of anequivalent series amount A ... e ... Both conditions mustexist before the series can be represented by A ... Unless stated otherwise, assume thatthe rate applies throughout the entire n years or interest periods ... All engineering economy problems involve the element of time expressed as n and interestrate i ... Additional symbols used in engineering economy are defined in Appendix E ... 6Today, Julie borrowed $5000 to purchase furniture for hernew house ... Determine the engineering economy symbols andtheir value for each option ... (b) One payment 3 years from now with interest based on 7% per year ... P ϭ $5000i ϭ 5% per yearn ϭ 5 yearsAϭ?(b) Repayment requires a single future amount F, which is unknown ... 7You plan to make alump-sum deposit of $5000 now into an investment account that pays 6%per year, and you plan to withdraw an equal end-of-year amount of $1000 for 5 years, startingnext year ... Define the engineering economy symbols involved ... P ϭ $5000A ϭ $1000 per year for 5 yearsF ϭ ? at end of year 6i ϭ 6%per yearn ϭ 5 years for the A series and 6 for the F valueEXAMPLE 1 ... (a) Identify the symbols, and(b) calculate the amount that had to be deposited exactly 1 year ago to earn $5000 in interestnow, if the rate of return is 6% per year ... Pϭ?i ϭ 6% per yearn ϭ 1 yearF ϭ P ϩ interest ϭ ? ϩ $5000(b) Let F ϭtotal amount now and P ϭ original amount ... Now we can determine P ... 1] through [1 ... F ϭ P ϩ PiThe $5000 interest can be expressed asInterest ϭ F – P ϭ (P ϩ Pi) – Pϭ Pi$5000 ϭ P(0 ... 330 ... 615Cash Flows: Estimation and Diagramming1 ... All cash flows occur during specific time periods, such as1 month, every 6 months, or 1 year ... Forexample, a payment of $10,000 once every year in December for 5 years is a series of 5 outgoingcash flows ... Engineering economy bases its computations on the timing, size, and direction of cash flows ... A plus sign indicates a cash inflow ... A negative orminus sign indicates a cash outflow ... Cash flowOf all the steps in Figure 1–1 that outline the engineering economy study, estimating cash flows(step 3) is the most difficult, primarily because it is an attempt to predict the future ... As you scan these, consider how the cash inflowor outflow may beestimated most accurately ... 2 million in capital expenditures for a water recycling unitAll of these are point estimates, that is, single-value estimates for cash flow elements of analternative, except for the last revenue and cost estimates listed above ... For the initial chapters, we will utilize point estimates... Once all cash inflows and outflows are estimated (or determined for a completed project), thenet cash flow for each time period is calculated ... 5][1 ... At the beginning of this section, the timing, size, and direction of cash flows were mentionedas important ... The end-of-period convention means that allcash inflows and all cash outflows are assumed totake place at the end of the interest period in which they actually occur ... End-of-period convention16Chapter 1Figure 1–4Foundations of Engineering EconomyYear 1Year 5A typical cash flow timescale for 5 years ... Cash flowi = 4% per year12345Year–In assuming end-of-period cash flows, it is important to understand that future (F) and uniformannual (A) amounts are located at the end of the interest period, which is not necessarilyDecember 31 ... 7 the lump-sum deposit took place on July 1, 2011, the withdrawals will take place on July 1 of eachsucceeding year for 6 years ... The cash flow diagram is a very important tool in an economic analysis, especially when thecash flow series is complex ... The diagram includes what is known, what is estimated, and what isneeded ... Cash flow diagram time t ϭ 0 is the present, and t ϭ 1 is the end of timeperiod 1 ... The time scale of Figure 1–4 is set up for 5 years ... While it is not necessary to use an exact scale on the cash flow diagram, you will probablyavoid errors if you make a neat diagram to approximate scale for both time and relative cash flowmagnitudes ... Avertical arrow pointing up indicates apositive cash flow ... We will use a bold, colored arrow to indicate what is unknownand to be determined ... The interest rate is also indicated on the diagram ... The arrowfor the unknown value is generally drawn in the opposite direction from the other cash flows;however, the engineering economycomputations will determine the actual sign on the F value ... Assume youborrow $8500 from a bank today to purchase an $8000 used car for cash next week, and you planto spend the remaining $500 on a new paint job for the car two weeks from now ... The cash flow signs and amounts fortheseperspectives are as follows ... 6Cash Flows: Estimation and Diagramming$850012Week0$500$8000Figure 1–6Cash flows from perspective of borrower for loan and purchases ... For your perspective,all three cash flows are involved and the diagram appears as shown in Figure 1–6 with a time scaleofweeks ... EXAMPLE 1 ... Carla Ramos, a lead engineer for Mexico and CentralAmerican operations, plans expenditures of $1 million now and each of the next 4 years justfor the improvement of field-based pressure-release valves ... SolutionFigure 1–7 indicates the uniform and negative cash flow series(expenditures) for five periods,and the unknown F value (positive cash flow equivalent) at exactly the same time as the fifthexpenditure ... Therefore, the last negative cash flow occurs at the end of the fourth year, when Falso occurs ... This addition demonstrates that year 0 is the end-of-periodpoint for theyear Ϫ1 ... 9 ... 10An electrical engineer wants to deposit an amount P now such that she can withdraw an equalannual amount of A1 ϭ $2000 per year for the first 5 years, starting 1 year after the deposit, anda different annual withdrawal of A2 ϭ $3000 per year for the following 3 years ... 5% per year?1718Chapter 1Foundations of Engineering EconomySolutionThe cash flows are shown in Figure 1–8 ... The withdrawals (positive cash inflow) for the A1 series occur at the end of years 1 through 5, and A2occurs in years 6 through 8 ... 5%P=?Figure 1–8Cash flow diagram with two different A series,Example 1 ... EXAMPLE 1 ... The annual rental incomefrom the compressor has been $750 ... The company plans to sell the compressor at the end of next year for$150 ... SolutionLet now be time t ϭ 0 ... 5] ... Present worth P is located at year 0 ... 11 ... 7Economic Equivalence191 ... Before we delveinto the economic aspects, think of the many types of equivalency wemay utilize daily by transferring from one scale to another ... 370 inches ϭ 1 meter100 centimeters ϭ 1 meter1000 meters ϭ 1 kilometer1 kilometer ϭ 0 ... Consider the equivalency of a speed of 110 kilometers per hour (kph) into milesper minute using conversions between distance and time scaleswith three-decimal accuracy ... 609 kilometers1 hour ϭ 60 minutes110 kph ϭ 68 ... 365 mph ϭ 1 ... Note that throughout these statements, the fundamental relations of 1 mile ϭ 1 ... If a fundamental relation changes, the entire equivalency isin error ... Economic equivalence is a combination of interest rate and time value of money to determine the different amounts of money at different points in time that are equal in economicvalue ... Amount accrued ϭ 100 ϩ 100(0 ... 06) ϭ $106If someone offered you a gift of $100 today or $106 one yearfrom today, it would make no difference which offer you accepted from an economic perspective ... However, the two sums of money are equivalent to each other only when theinterest rate is 6% per year ... In addition to future equivalence, we can apply the same logic to determine equivalenceforprevious years ... 06 ϭ $94 ... From these illustrations, we can state the following: $94 ... The fact that these sums are equivalent can be verified by computing the two interest rates for1-year interest periods ... 66$94 ... Economic equivalence20Chapter 1Foundations of Engineering Economyi = 6% peryearAmount, $10094 ... 00 interest$5 ... EXAMPLE 1 ... In general, batteries are stored throughout the year,and a 5% cost increase is added each year to cover the inventory carrying charge for the distributorship owner ... Make the calculationsnecessary to show which of the following statements are trueand which are false about batterycosts ... 60 one year from now ... (c) A $38 cost now is equivalent to $39 ... (d) A $3000 cost now is equivalent to $2887 ... (e) The carrying charge accumulated in 1 year on an investment of $20,000 worth ofbatteries is $1000 ... 05) ϭ $102 ... 60; therefore, it is false ...60͞1 ... 57 $98 ... 00͞1 ... 24 $200; therefore, it is false ... 05) ϭ $39 ... (d) Cost now is 2887 ... 05) ϭ $3031 ... (e) The charge is 5% per year interest, or $20,000(0 ... Comparison of alternative cash flow series requires the use of equivalence to determine whenthe series are economically equal or if one iseconomically preferable to another ... Example 1 ... EXAMPLE 1 ... He wants to borrow $10,000 now and repay itover the next 1 or 2 years ... Howard received2-year repayment options from banks A and B ... 8Simple and Compound InterestAmount to pay, $ per yearYearBank ABank B1−5,378 ...

002−5,378 ... 00Total paid−10,756 ... 00After reviewing these plans, Howard decided that he wants to repay the $10,000 after only1 year based on the expected increased revenue ... Now Howard has three options and wonders which one to take ... (This is determined by using computations that you willlearn in Chapter 2 ... The brother-in-law repayment plan requires a total of $600 in interest 1 year later plus theprincipal of $10,000, which makes the interest rate 6% per year ... Even though the sum of money repaid is smaller, the timing of the cash flowsand the interest rate make it less desirable ... Theinterestrate, timing, and economic equivalence must be considered ... 8 Simple and Compound InterestThe terms interest, interest period, and interest rate (introduced in Section 1 ... However, for more than one interest period, the terms simple interest and compound interest become important ... Thetotal simple interest over several periods is computed asSimple interest ϭ (principal)(number of periods)(interest rate)I ϭ Pni[1 ... EXAMPLE 1 ... The loan is for 3 years at 10% per year simple interest ... 10) ϭ $10,000Total interest for 3 years from Equation [1 ... 10) ϭ $30,0002122Chapter 1Foundations ofEngineering EconomyThe amount due after 3 years isTotal due ϭ $100,000 ϩ 30,000 ϭ $130,000The interest accrued in the first year and in the second year does not earn interest ... In most financial and economic analyses, we use compound interest calculations ... Thus,compound interest meansinterest on top of interest ... Now theinterest for one period is calculated asCompound interest ϭ (principal ϩ all accrued interest)(interest rate)[1 ... (jϭtϪ1It ϭ P ϩ⌺jϭ1)IJ (i)[1 ... 15Assume an engineering company borrows $100,000 at 10% per year compound interest andwill pay the principal and all theinterest after 3 years ... Graph the interest and total owed for each year, and compare withthe previous example that involved simple interest ... 8] ... 10) ϭ $10,000100,000 ϩ 10,000 ϭ $110,000110,000(0 ... 10) ϭ $12,100121,000 ϩ 12,100 ϭ $133,100The repayment plan requires no payment until year 3when all interest and the principal, a totalof $133,100, are due ... The differences due to compounding are clear ... Note that while simple interest due each year is constant, the compounded interest duegrows geometrically ... For example, if the loan is for 10 years, not 3, the extra paid for compoundinginterestmay be calculated to be $59,374 ... 823Simple and Compound Interest0Ϫ101I2I3I0YearϪ10I is constantϪ11123I increasesgeometricallyIIϪ12Amount owed (ϫ $1000)Amount owed (ϫ $1000)IϪ100Ϫ110Ϫ120Ϫ100Ϫ110Ϫ120GeometricincreaseArithmeticincreaseϪ130Ϫ130Ϫ140Ϫ140(a)(b)Figure 1–11Interest I owed and total amount owed for (a) simple interest (Example 1 ... 15) ... 15is to utilize the fact that compound interest increases geometrically ... In this case, the total amount due at the end of eachyear isYear 1: $100,000(1 ... 10)2 ϭ $121,000Year 3: $100,000(1 ... The generalform of theequation isTotal due after n years � principal(1 ؉ interest rate)n years� P(1 ؉ i)nYear[1 ... Equation [1 ... This fundamental relation will be used many times in the upcoming chapters ... This also shows that there aremany ways to take into account the time value of money ... 16Table 1–1 details fourdifferent loan repayment plans described below ... • Plan 1: Pay all at end ... Interest accumulates each year on the total of principal and all accrued interest ... The accrued interest is paid eachyear, and the entire principal is repaid at the end of year 5 ... The accrued interest and one-fifthof the principal (or$1000) are repaid each year ... • Plan 4: Pay equal amount of interest and principal ... Since the loan balance decreases at a rate slower than that in plan 3 due to the equalend-of-year payments, the interest decreases, but at a slower rate ... 00432 ... 56503 ... 20$5400 ... 006298 ... 447346 ... 64$5000 ...005832 ... 566802 ... 64Plan 2: Pay Interest Annually; Principal Repaid at End012345$400 ... 00400 ... 00400 ... 005400 ... 005400 ... 00Total$Ϫ400 ... 00Ϫ400 ... 00– 5400 ... 005000 ... 005000 ... 00$Ϫ7000 ... 00320 ... 00160 ... 00$5400 ... 003240 ... 001080 ... 00Ϫ1320 ... 00Ϫ1160 ... 00$5000 ... 003000... 001000 ... 00Plan 4: Pay Equal Annual Amount of Interest and Principal012345Total$400 ... 82258 ... 6592 ... 004479 ... 432411 ... 28$−1252 ... 28−1252 ... 28– 1252 ... 40$5000 ... 723227 ... 151159 ... 9Minimum Attractive Rate of Return(a) Make a statement about the equivalence of each plan at 8%compound interest ... Comment on the total amounts repaid for the two plans ... The difference in the total amounts repaid can be explained by the time value of money and by the partial repayment of principal prior toyear 5 ... 64 at the end of year 5Plan 2 $400 per year for 4 years and $5400 at the end ofyear 5Plan 3 Decreasing payments of interest and partial principal in years 1 ($1400)through 5 ($1080)Plan 4 $1252 ... This amount covers accrued interest and a partialamount of principal repayment ... Sincethe annual accrued interest of $400 is paid each year and the principal of $5000 is repaidin year5, the schedule is exactly the same as that for 8% per year compound interest, andthe total amount repaid is the same at $7000 ... Any deviation from this schedule willcause the two plans and amounts to differ ... 9 Minimum Attractive Rate of ReturnFor any investment to be profitable, the investor(corporate or individual) expects to receive moremoney than the amount of capital invested ... The definition of ROR in Equation [1 ... Engineering alternatives are evaluated upon the prognosis that a reasonable ROR can beexpected ... 2526Chapter 1Minimum Attractive Rateof Return (MARR)Cost ofcapitalFoundations of Engineering EconomyThe Minimum Attractive Rate of Return (MARR) is a reasonable rate of return establishedfor the evaluation and selection of alternatives ... MARR is also referred to as the hurdle rate,cutoff rate, benchmark rate, and minimum acceptable rate of return ... In theUnitedStates, the current U ... Treasury Bill return is sometimes used as the benchmark safe rate ... The MARR is not a rate that iscalculated as a ROR ... To develop a foundation-level understanding of how a MARR value is established and usedto make investment decisions, we return to the term capitalintroduced in Section 1 ... Althoughthe MARR is used as a criterion to decide on investing in a project, the size of MARR is fundamentally connected to how much it costs to obtain the needed capital funds ... The interest, expressed as a percentage rate peryear, is called the cost of capital ... Alternatively,you mightchoose to use your credit card and pay off the balance on a monthly basis ... Or, you could use funds from your savings account thatearns 5% per year and pay cash ... The 9%, 15%, and 5% rates are your cost of capital estimates to raise thecapital for the system by different methods of capitalfinancing ... Rate of return,percentExpected rate of return ona new proposalRange for the rate of return onaccepted proposals, if otherproposals were rejectedfor some reasonAll proposals must offerat least MARR tobe consideredMARRRate of return on“safe investment”Figure 1–12Size of MAAR relativeto other rate of return values ... 1027Introduction to Spreadsheet UseIn general, capital is developed in two ways—equity financing and debt financing ... Chapter 10 covers these in greater detail, buta snapshot description follows ... Individuals can use their own cash, savings, or investments ... Debtfinancing The corporation borrows from outside sources and repays the principal and interest according to some schedule, much like the plans in Table 1–1 ... Individuals, too, can utilize debtsources, such as the credit card (15% rate) and bank options (9% rate) described above ... If the HDTV ispurchased with 40% credit card money at 15% per year and 60% savingsaccount funds earning 5% per year, the weighted average cost of capital is 0 ... 6(5) ϭ9% per year ... So the inequalityROR Ն MARR Ͼ WACC[1 ... Exceptions may be government-regulated requirements(safety, security,environmental, legal, etc ... Often there are many alternatives that are expected to yield a ROR that exceeds the MARR asindicated in Figure 1–12, but there may not be sufficient capital available for all, or the project’srisk may be estimated as too high to take the investment chance ... The expected rate ofreturn on the unfunded project is called the opportunitycost ... Numerically, it is the largest rate of return of all the projects not accepted(forgone) due to the lack of capital funds or other resources ... e ... As an illustration of opportunity cost, refer to Figure 1–12 and assume a MARR of 12% peryear ...Meanwhile, proposal B has a ROR ϭ 14 ... Since proposal A is not undertaken due to the lack of capital, its estimated ROR of 13%is the opportunity cost; that is, the opportunity to make an additional 13% return is forgone ... 10 Introduction to Spreadsheet UseThe functions on a computer spreadsheetcan greatly reduce the amount of hand work for equivalency computations involving compound interest and the terms P, F, A, i, and n ... However, as cash flow series become more complex, the spreadsheetoffers a good alternative ... Appendix A is a primer on using spreadsheets and Excel ... AppendixA also includes a section on spreadsheet layout that is useful when theeconomic analysis is presented to someone else—a coworker, a boss, or a professor ... The functions are great supplemental tools, but they do not replace the understanding of engineering economy relations, assumptions, andtechniques ... To find the present value P: � PV(i%, n, A, F)To find the future value F: � FV(i%, n, A, P)To find the equal, periodic value A: � PMT(i%, n, P, F)To find the number of periods n: � NPER(i%, A, P, F)To find the compound interest rate i: � RATE( n, A, P, F)To find the compound interest rate i: � IRR(first_cell:last_cell)To find the present value P of any series: � NPV(i%, second_cell:last_cell) ؉ first_cellIf some of the parameters don’t apply to a particular problem, they can be omitted and zero isassumed ... If theparameter omitted is an interior one, the comma must be entered ... In all cases, thefunction must be preceded by an equals sign (ϭ) in thecell where the answer is to be displayed ... 6a, where theequivalent annual amount A is unknown, as indicated by A ϭ ? ... ) To find A using a spreadsheetfunction, simply enter the PMT function ϭ PMT(5%,5,5000) ... The answer ($1154 ... The answermay appear in red and in parentheses, or with a minus sign on your screento indicate a negative amount from the perspective of a reduction in the account balance ... 6b ... The FV function appears in the formula bar; and many examples throughout this text will include cell tags, as shown here, to indicatethe format ofimportant entries ... Once set up, the spreadsheet can beused to perform sensitivity analysis for estimates that are subject to change ... (Note: The spreadsheet examples maybe omitted, if spreadsheets are not used in the course ... )ϭ PMT(5%,5,5000)Figure 1–13Use of spreadsheetfunctions PMT and FV, Example 1 ... ϭ FV(7%,3,,5000)1 ... 17A Japan-based architectural firm has asked a United States–based software engineering groupto infuse GPS sensing capability via satellite into monitoring software for high-rise structuresin order to detect greater than expected horizontalmovements ... The inclusion of accurate GPS data is estimated to increase annual revenue overthat for the current software system by $200,000 for each of the next 2 years, and by $300,000for each of years 3 and 4 ... Develop spreadsheets to answer the questionsbelow ... (b) Repeat part (a) ifestimated revenue increases from $300,000 to $600,000 in years 3 and 4 ... This will decrease the real rateof return from 8% to 3 ... Solution by SpreadsheetRefer to Figure 1–14a to d for the solutions ... (Actually, all the questions can beanswered on one spreadsheet by changing the numbers ... )TheExcel functions are constructed with reference to the cells, not the values themselves, so that sensitivity analysis can be performed without function changes ... For example, the8% rate in cell B2 will be referenced in all functions as B2, not 8% ... See Appendix A for additional information about using cellreferencing and buildingspreadsheet relations ... As an illustration, for year 3 the interest I3 and revenue plus interest R3 areI3 ϭ (cumulative revenue through year 2)(rate of return)ϭ $416,000(0 ... Cell C8 relation for I3: ϭ F7*B2Cell E8 relation for CF3: ϭ B8 ϩ C8The equivalent amount after 4 years is$1,109,022, which is comprised of $1,000,000 intotal revenue and $109,022 in interest compounded at 8% per year ... (b) To determine the effect of increasing estimated revenue for years 3 and 4 to $600,000,use the same spreadsheet and change the entries in cells B8 and B9 as shown inFigure 1–14c... (c) Figure 1–14d shows the effect of changing the original i value from 8% to an inflationadjusted rate of 3 ... [Remember to return to the$300,000 revenue estimates for years 3 and 4 after working part (b) ... 2930Chapter 1Foundations of Engineering Economy(a) Total interest and revenue for basecase, year 4(b) Spreadsheet relations for base caseRevenue changed(c) Totals with increased revenue in years 3 and 4Rate ofreturnchanged(d) Totals with inflation of 4% per year consideredFigure 1–14Spreadsheet solutions with sensitivity analysis, Example 1 ... CommentLater we will learn how toutilize the NPV and FV Excel financial functions to obtain the sameanswers determined in Figure 1–14, where we developed each basic relation ... Problems31CHAPTER SUMMARYEngineering economy is the application of economic factors and criteria to evaluate alternatives,considering the time valueof money ... The concept of equivalence helps in understanding how different sums of money at differenttimes are equal in economic terms ... This power of compounding is very noticeable, especially overextended periods of time, and for larger sums of money ... The MARR is always higher than thereturn from a safe investmentand the cost to acquire needed capital ... 1 List the four essential elements involved in decision making in engineering economic analysis ... 2 What is meant by (a) limited capital funds and(b) sensitivity analysis?1 ... 1 ... Ethics1 ... Management wants to return some oftheengineering design work to the United Statesrather than export all of it to India, where the primary design work has been accomplished for thelast decade ... Stefanie and her design team were selected as atest case to determine the quality and speed of thedesign work they could demonstrate on amorefuel-efficient diesel locomotive ... One of her team members had agreat design idea on a key element that will improve fuel efficiency by approximately 15% ... S ... Although reluctant at first, Stefanie did go forwardwith a design that included the efficiency improvement, and no mention of the origin ofthe idea wasmade at the time of the oral presentation or documentation delivery ... Consult the NSPE Code of Ethics for Engineers(Appendix C) and identify sections that are pointsof concern about Stefanie’s decisions and actions ... 6 Consider the common moral precept that stealingis wrong ... One ofHector’s buddies takes ahigh-energy drink from a six-pack on the shelf,opens it, drinks it, and returns the empty can to thepackage, with no intention of paying for it ... Others do it all the time ... Personally, Hector believes this is a formof stealing ... 1 ... He has a strong belief in the universal moral that it iswrong to do serious harm toanother person ... 1 ... Although he had apassing score prior to the final, his final grade wasso low that he has now flunked the entire year andwill likely have to extend his graduation anothersemester or two ... He did realize during thesemester that he was doing something thatevenhe considered wrong morally and ethically ... The classroom was reconfigured for the final exam in a waythat he could not get any answers from classmates,and cell phones were collected prior to the exam,thus removing texting possibilities to friends outside the classroom who might help him onthefinal exam ... The question to Claude is,“What have you been doing throughout this yearto make passing scores repeatedly, but demonstrate such a poor command of Spanish on thefinal exam?”From an ethical viewpoint, what options doesClaude have in his answer to this question? Also,discusssome of the possible effects that this experience may have upon Claude’s future actions andmoral dilemmas ... 9 RKI Instruments borrowed $3,500,000 from a private equity firm for expansion of its manufacturingfacility for making carbon monoxide monitors/controllers ... Whatwas the interest rate on theloan?Foundations of Engineering Economy1 ... Theterms of the loan were such that the company couldpay interest only at the end of each year for up to 5years, after which the company would have to paythe entire amount due ... 11 Which of the following 1-year investments has thehighest rate ofreturn?(a) $12,500 that yields $1125 in interest,(b) $56,000 that yields $6160 in interest, or(c) $95,000 that yields $7600 in interest ... 12 A new engineering graduate who started a consulting business borrowed money for 1 year to furnishthe office ... However, because the new graduate had not built upacredit history, the bank made him buy loan-defaultinsurance that cost 5% of the loan amount ... What was the effective interest rate the engineerpaid for the loan?1 ... 14 The symbol P represents an amount of money at atime designated as present ... Explain what each symbol stands for: PW, PV, NPV,DCF, and CC ... 15 Identify the four engineering economy symbolsand their values from the following problem statement ... Thompson Mechanical Products is planning toset aside $150,000 now for possibly replacing itslarge synchronous refiner motors whenever it becomes necessary ... 16 Identify thefour engineering economy symbols andtheir values from the following problem statement ... 17 Identify the four engineering economy symbolsand their values from the following problem statement ... A green algae, Chlamydomonas reinhardtii,can produce hydrogen when temporarily deprivedof sulfur forup to 2 days at a time ... 4 million to commercialize the process ... 18 Identify the four engineering economy symbolsand their values from the following problem statement ... Vision Technologies, Inc ... The company expects to spend $100,000 per yearfor labor and $125,000 per year for supplies beforeaproduct can be marketed ... 21 Many credit unions use semiannual interest periodsto pay interest on customer savings accounts ... 5070—12020—15090——401101 ...MonthReceipts,$1000Disbursements,$1000JanFebMarAprMayJuneJulyAugSeptOctNovDec50080020012060090080070090050040018003005004004005006003003005004004007001 ... 1 ... 1 ... 20 Identify the following as cash inflows or outflowsto commercial air carriers: fuel cost, pensionplancontributions, fares, maintenance, freight revenue, cargo revenue, extra-bag charges, water andsodas, advertising, landing fees, seat preferencefees ... Atlas Long-Haul Transportation is considering installing Valutemp temperature loggers in allof its refrigerated trucks for monitoring temperaturesduring transit ... 25 Construct a cash flow diagram that represents theamount of money that will be accumulated in15 years from an investment of $40,000 now at aninterest rate of 8% per year ... 26 At an interest rate of 15% per year, an investmentof $100,000 one year ago is equivalent to howmuchnow?34Chapter 11 ... A companythat makes Ethernet adapters is planning to expandits production facility at a cost of $1,000,000 oneyear from now ... If the interest rate is 15% per year,how much of a discount is the company getting?1 ... One year ago, eachshare of stock was worth $40 ... Atwhatinterest rate would the firm’s offer be equivalent to the worth of the stock last year?1 ... The company said that although it could notgive bonuses this year, it would give each engineer two bonuses next year, the regular one of$8000 plus an amount equivalent to the $8000that each engineer should havegotten this year ... 30 University tuition and fees can be paid by usingone of two plans ... On-time: Pay total amount due when classes start ... (a) How much is paid in the early-bird plan?(b) What is the equivalent amount of the savingscompared to the on-time payment at the timethat the on-time paymentis made?Simple and Compound Interest1 ... 32 Iselt Welding has extra funds to invest for futurecapital expansion ... 33 To finance a new product line, a company thatmakes high-temperature ball bearings borrowed$1 ... If the com-Foundations of Engineering Economypany repaid the loan in a lump sumamount after2 years, what was (a) the amount of the paymentand (b) the amount of interest?1 ... e ... If the interest rate on the bondswas 9% per year, how much does the companyhave to pay the bond holders? The face value(principal) of the bonds is $6,000,000 ... 35 A solid waste disposal companyborrowed moneyat 10% per year interest to purchase new haulersand other equipment needed at the companyowned landfill site ... 36 If interest is compounded at 20% per year, howlong will it take for $50,000 to accumulate to$86,400?1 ... If aperson deposits $10,000 at 10% per year simpleinterest,what compound interest rate would yieldthe same amount of money in 3 years?MARR and Opportunity Cost1 ... 1 ... 1 ... 1 ... A staff member for the chief financial officerused key words to identify the projects and thenlisted them in order of projected rate of return asshown below ... 41913 ... 68 ... 42State the purpose for each of the following built-inspreadsheet functions ... 43 What are the values of the engineering economysymbols P, F, A, i, and n in the following functions?Use a question mark for the symbol that is to be35determined ... 44 Write the engineering economy symbol that correspondsto each of the following spreadsheetfunctions ... 45 In a built-in spreadsheet function, if a certain parameter is not present, (a) under what circumstances can it be left blank and (b) when must acomma be entered in its place?1 ... Sheryl receives simple interest andMarcelly gets compound interest ...Assume no withdrawals or further deposits are made during the 4 years ... 47 The concept that different sums of money at different points in time can be said to be equal to eachother is known as:(a) Evaluation criterion(b) Equivalence(c) Cash flow(d ) Intangible factors1 ... 49 All of the following areexamples of cash outflows,except:(a) Asset salvage value(b) Income taxes(c) Operating cost of asset(d ) First cost of asset1 ... 51 The following annual maintenance and operation(M&O) costs for a piece of equipment were collected over a 5-year period: $12,300, $8900,$9200, $11,000, and $12,100 ...In conducting a sensitivity analysis, themost reasonable range of costs to use (i ... , percentfrom the average) is:(a) Ϯ5% (b) Ϯ11% (c) Ϯ17% (d) Ϯ25%1 ... 53 Assume that you and your best friend each have$1000 to invest ... Yourfriend invests her money at a bank that pays 10%per year simple interest ...54 The time it would take for a given sum of money todouble at 4% per year simple interest is closest to:(a) 30 years (b) 25 years(c) 20 years (d) 10 years36Chapter 1Foundations of Engineering Economy1 ... 56 To finance a new project costing $30 million, acompany borrowed $21 million at 16% peryearinterest and used retained earnings valued at 12%per year for the remainder of the investment ... 5% (b) 13 ... 8% (d) 15 ... PEC has a capacity of approximately 1300 MW (megawatts) of power, ofwhich 277 MW, or about 21%, is from renewable sources ... A constantquestion is how much of PEC’sgeneration capacity should befrom renewable sources, especially given the environmentalissues with coal-generated electricity and the rising costs ofhydrocarbon fuels ... Consider yourself a member of the board of directors ofPEC ... As such, you do not represent a specific district within theentire servicearea; all other directors do represent a specificdistrict ... InformationHere are some data that you have obtained ... Electricity generation cost estimates are nationalin scope, not PEC-specific, and are provided in cents perkilowatt-hour (¢/kWh) ... 54 ... 14 ... 5Reasonable Average7 ... 68 ... 8Nationalaverage cost of electricity to residential customers: 11¢/kWhPEC average cost to residential customers: 10 ... 92 ¢/kWh (renewable sources)Expected life of a generation facility: 20 to 40 years (it islikely closer to 20 than 40)Time to construct a facility: 2 to 5 yearsCapital cost to build a generation facility:$900 to $1500per kWYou have also learned that the PEC staff uses the wellrecognized levelized energy cost (LEC) method to determinethe price of electricity that must be charged to customers tobreak even ... The LEC formula, expressedin dollars per kWh for (t ϭ 1, 2, ... If you wanted to know moreabout the new arrangement with the wind farm in south Texas for the additional 60 MW per year, what types of questions wouldyou ask of a staff member in your first meeting withhim or her?2 ... What about the ethical aspects of the government’s allowance for these plants to continue polluting theatmospherewith the emissions that may cause health problems forcitizens and further the effects of global warming? Whattypes of regulations, if any, should be developed for PEC(and other generators) to follow in the future?Case Study3 ... 27¢/kWh forthis year ... You did learn thefollowing:This is year tϭ 11 for LEC computation purposesn ϭ 25 yearsi ϭ 5% per yearE11 ϭ 5 ... 22 ¢/kWh (last year’s breakevencost to customers)From these sketchy data, can you determine the value of unknowns in the LEC relation for this year? Is it possible todetermine if the wind farm addition of 60 MW makesanydifference in the electricity rate charged to customers? If not,what additional information is necessary to determine theLEC with the wind source included?CASE STUDYREFRIGERATOR SHELLSBackgroundLarge refrigerator manufacturers such as Whirlpool, GeneralElectric, Frigidaire, and othersmay subcontract the molding oftheir plastic liners and door panels ... Because of improvements in mechanical properties, the molded plastic can sustain increased vertical and horizontal loading, thus significantly reducing the needfor attached metal anchors for some shelving ... The company presidentwants a recommendation on whetherInnovations should offer the new technology to the major manufacturers and an estimate of the necessary capital investmentto enter this market ... At this stage,you are not expected to perform a complete engineering economic analysis, for not enough information isavailable ... InformationSome information useful at this time is as follows:• The technology and equipment are expected to last about10 years before new methods are developed ... • The expected returns on capital investment used for thelast three new technology projects were compound rates of15%,5%, and 18% ... • Equity capital financing beyond $5 million is not possible ... • Annual operating costs have been averaging 8% of firstcost for major equipment ... 2 million ... You label these options as alternatives A and B ... Use the first four steps of the decision-making processto generally describe thealternatives and identify whateconomic-related estimates you will need to completean engineering economy analysis for the president ... Identify any noneconomic factors and criteria to be considered in making the alternative selection ... During your inquiries about alternative B from its manufacturer, youlearn that this company has already produceda prototype molding machine and has sold it to a companyin Germany for $3 million (U ... dollars) ... The company is willing to sell time on theequipment to Innovations immediately to produce its ownshells for U ... delivery ... Consider this as alternative C, anddevelop the estimates necessary to evaluate Cat the same time as alternatives A and B ... SECTIONTOPICLEARNING OUTCOME2 ... 2 ... 2 ... 2 ... 2 ... 2 ... 2 ... The cash flow is fundamental to every economic study ... This chapter develops derivations for all the commonly used engineering economyfactorsthat take the time value of money into account ... Spreadsheet functions are used in order to rapidly work with cash flow seriesand to perform sensitivity analysis ... PEThe Cement Factory Case: VotorantimCimentos North America, Inc ... The plant willbe called Houston American Cement, orHAC ...The plant investment, expectedto amount to $200 million, has beenplanned for 2012; however, it is currentlydelayed due to the economic downturnin construction ... Allanalysis will use a planning horizon of5 years commencing when the plantbegins operation ... 1)Uniform series factors (2 ... 3)Arithmeticgradient factors (2 ... 6)Determining unknown n values (2 ... 1 Single-Amount Factors (F͞P and P͞F )The most fundamental factor in engineering economy is the one that determines the amountof money F accumulated after n years (or periods) from a single present worth P, with interestcompounded onetime per year (or period) ... Therefore, if an amount P is invested at time t ϭ 0, the amount F1 accumulated1 year hence at an interest rate of i percent per year will beF1 ϭ P ϩ Piϭ P(1 ϩ i)where the interest rate is expressed in decimal form ... F2 ϭ F1 ϩ F1iϭ P(1 ϩ i) ϩ P(1 ϩ i)i[2 ... 1], will beF3 ϭ F2 ϩF2i40Chapter 2Factors: How Time and Interest Affect MoneyF = givenF=?i = given012n–2i = givenn–1n0P = given12n–2n–1nP=?(a)(b)Figure 2–1Cash flow diagrams for single-payment factors: (a) find F, given P, and (b) find P, given F ... To find F, given P,F � P(1 ؉ i)n[2 ... This is the conversion factorthat, when multiplied by P, yieldsthe future amount F of an initial amount P after n years at interest rate i ... Reverse the situation to determine the P value for a stated amount F that occurs n periodsin the future ... 2] for P ... 3]The expression (1 ؉ i)Ϫn is known as the single-payment present worth factor(SPPWF), or theP͞F factor ... The cash flow diagram is shown in Figure 2–1b ... A standard notation has been adopted for all factors ... It is always in the general form (X͞Y,i,n) ... For example, F͞Pmeans find F when given P ... Using this notation, (F͞P,6%,20) represents the factor that is used to calculate thefutureamount F accumulated in 20 periods if the interest rate is 6% per period ... Thestandard notation, simpler to use than formulas and factor names, will be used hereafter ... Thisinformation is also included inside the front cover ... 1Single-Amount Factors (F͞P and P͞F )To simplify routine engineeringeconomy calculations, tables of factor values have been prepared for interest rates from 0 ... These tables, found at the rear of the book, have a colored edge for easy identification ... The word discrete in the title of each table emphasizes that these tables utilize the end-of-periodconvention and thatinterest is compounded once each interest period ... Forexample, the value of the factor (P͞F,5%,10) is found in the P͞F column of Table 10 at period 10as 0 ... This value is determined by using Equation [2 ... 1(P͞F,5%,10) ϭ ————(1 ϩ i)n1ϭ ————(1 ... 61391 ... 4]A present amount P is determinedusing the PV function with the format� PV(i%,n,,F)[2 ... Refer to Appendix A or Excel online help for moreinformation on the use of FV and PV functions ... 1Sandy, a manufacturing engineer, just received a year-end bonus of $10,000 that will be investedimmediately ... Find the amount of funds that will beavailable in 20 years by using(a) hand solution by applying the factor formula and tabulated value and (b) a spreadsheet function ... The symbols and values areP ϭ $10,000Fϭ?i ϭ 8% per yearn ϭ 20 years(a) Factor formula: Apply Equation [2 ... Rounding to four decimals, we haveF ϭ P(1 ϩ i)n ϭ 10,000(1... 6610)ϭ $46,610Standard notation and tabulated value: Notation for the F͞P factor is (F͞P,i%,n) ... 6610)ϭ $46,610Table 13 provides the tabulated value ... (b) Spreadsheet: Use the FV function to find the amount 20 years in the future ... 4]; the numerical entry is ϭ FV(8%,20,,10000) ... 57) displayed ... )The FV functionhas performed the computation in part (a) and displayed the result ... 4142Chapter 2Factors: How Time and Interest Affect MoneyPEEXAMPLE 2 ... Delays beyond the anticipated implementation year of 2012 will require additional money to construct the factory ... (a) The equivalentinvestment needed if the plant is built in 2015 ... SolutionFigure 2–2 is a cash flow diagram showing the expected investment of $200 million ($200 M)in 2012, which we will identify as time t ϭ 0 ... Figure 2–2F3 = ?Cash flow diagram forExample 2 ... P−4 = ?−4−320082009−2−10122010201120122013201432015tYear$200 M(a) To find the equivalent investment required in 3 years, apply the F͞P factor ... F3 ϭ P(F͞P,i,n) ϭ 200(F͞P,10%,3) ϭ 200(1 ... 2 ($266,200,000)Now, use the FV function on a spreadsheet to find the same answer, F3 ϭ $266 ... (Refer to Figure2–3, left side ... 2 ... To determine theequivalent cost 4 years earlier, consider the $200 M in 2012 (t ϭ 0) as the future value Fand apply the P͞F factor for n ϭ 4 to find PϪ4 ... ) Table 15 suppliesthe tabulated value ... 6830)ϭ $136 ... This equivalence analysis indicates that at $136 ... 2 ... 2 Uniform SeriesPresent Worth Factor andCapital Recovery Factor (P͞A and A͞P)The equivalent present worth P of a uniform series A of end-of-period cash flows (investments)is shown in Figure 2–4a ... 3],and summing the results ... (1 ϩ i)1(1 ϩ i)2(1 ϩ i)311ϩ A ———— ϩ A ————(1 ϩ i)n(1 ϩ i)nϪ1[The terms inbrackets are the P͞F factors for years 1 through n, respectively ... [11111P ϭ A ———— ϩ ———— ϩ ———— ϩ ... 6]To simplify Equation [2 ... This results in Equation [2 ... Nowsubtract the two equations, [2 ... 7], and simplify to obtain the expression for P wheni 0 (Equation [2 ... [ (1 ϩ i)11111P——— ϭA ———— ϩ ———— ϩ ———— ϩ ... ϩ (1 ϩ1i)Ϫi P ϭ A1[ (1 ϩ1i) Ϫ (1 ϩ i) ]1ϩi(1 ϩ i)11111——— P ϭ A ———— ϩ ———— ϩ ... 7]]————1[1AP ϭ —— ———— Ϫ 1Ϫi (1 ϩ i)n][(1 ؉ i)n 1 ؊ P � A ——————i(1 ؉ i)n]i0[2 ... 8] is the conversion factor referred to as the uniform seriespresent worthfactor (USPWF) ... The cash flow diagram is Figure 2–4a ... 2n–2A=?(b)n–1n44Chapter 2Factors: How Time and Interest Affect MoneyTABLE 2–2Notation(P͞A,i,n)(A͞P,i,n)P͞A and A͞P Factors: Notation and EquationsFactorNameFactorFormulaUniform seriespresent worthCapitalrecoveryP͞AStandardNotation EquationExcelFunction(1 ϩ i)n Ϫ 1i(1 ϩ i)i(1 ϩ i)n—————(1 ϩ i)n − 1Find/GivenP ϭ A(P͞A,i,n)ϭ PV(i%,n,A)A ϭ P(A͞P,i,n)ϭ PMT(i%,n,P)—————nA͞PTo reverse the situation, the present worth P is known and the equivalent uniform seriesamount A is sought (Figure 2–4b)... Solve Equation [2 ... 9]The term in brackets is called the capital recovery factor (CRF), or A͞P factor ... Placement of PThe P͞A and A͞P factors are derived with the present worth P and the first uniform annualamount A one year (period) apart ... The factors and their use to find P and A are summarized inTable 2–2 and inside the front cover ... Tables at the end ofthe text include the factor values ... 4641 ... Spreadsheet functions can determine both P and A values in lieu of applying the P͞A and A͞Pfactors ... The format, is� PV(i%,n,A,F)[2 ... The format is� PMT(i%, n,P,F)[2 ... EXAMPLE 2 ... Thepresentworth isP ϭ 600(P͞A,16%,9) ϭ 600(4 ... 90The PV function ϭ PV(16%,9,600) entered into a single spreadsheet cell will display theanswer P ϭ ($2763 ... 2 ... 4 The Cement Factory CaseAs mentioned in the chapter introduction of this case, the Houston American Cement plantmay generate a revenue baseof $50 million per year ... With money worth10% per year, address the following question from the president: Will the initial investmentbe recovered over the 5-year horizon with the time value of money considered? If so, by howmuch extra in present worth funds? If not, what is the equivalent annualrevenue base requiredfor the recovery plus the 10% return on money? Use both tabulated factor values and spreadsheet functions ... The cash flow diagram is similar to Figure 2–4a, where thefirst A value occurs 1 year after P ... 7908)ϭ $189 ... To determine the minimum required to realize a 10% peryear return, use the A͞P factor ... A ϭ 200(A͞P,10%,5) ϭ 200(0 ... 76 per yearThe plant needs to generate $52,760,000 per year to realize a 10% per year return over5 years ... Figure 2–5 showsthe use of ϭ PV(i%,n,A,F) on the left side to find the present worth and the use ofϭ PMT(i%,n,P,F) on the right sideto determine the minimum A of $52,760,000 per year ... The minus sign placed beforeeach function name forces the answer to be positive, since these two functions always displaythe answer with the opposite sign entered on the estimated cash flows ... 4 ... 3 Sinking Fund Factor and Uniform SeriesCompoundAmount Factor (A͞F and F͞A)The simplest way to derive the A͞F factor is to substitute into factors already developed ... 3] is substituted into Equation [2 ... ][i(1 ϩ i)n1A ϭ F ———— —————n(1 ϩ i) (1 ϩ i)n − 1[[iA � F —————(1 ؉ i)n 12 ... 2 [[] 1 ؊ ] is the A͞F or sinking fund factor ... This isshown graphically in Figure 2–6a, where A is a uniform annual investment ... The last A value and F occur at the same time ... 12] can be rearranged to find F for a stated A series in periods 1 through n (Figure 2–6b) ... 13]The term in brackets is called the uniform series compound amount factor(USCAF), or F͞A factor ... It is important to remember that the future amount F occurs in the same period as the last A ... They are (F͞A,i,n) and(A͞F,i,n) ... As a matter of interest, the uniform series factors can be symbolically determined by using anabbreviated factor form ... Using the factor formulas, wehave(1 ϩ i)n Ϫ 1(1 ϩ i)n Ϫ 1(F͞A,i,n) ϭ [(1 ϩ i)n] —————— ϭ ——————nii(1 ϩ i)[]For solution by spreadsheet, the FV function calculates F for a stated A series over n years ... 14]The P may be omitted when no separate present worth value is given ... Theformat is� PMT(i%,n,P,F)[2 ... F = givenF=?i= giveni = given012n–2n–1nA=?(a)Figure 2–6Cash flow diagrams to (a) find A, given F, and (b) find F, given A ... 3Sinking Fund Factor and Uniform Series Compound Amount Factor (A͞F and F͞A)TABLE 2–3Notation(F͞A,i,n)(A͞F,i,n)47F͞A and A͞F Factors: Notation andEquationsFactorNameFind/GivenUniform seriescompound amountSinking fundFactorFormulaStandard NotationEquation(1 ϩ i)n Ϫ 1 F ϭ A(F͞A,i,n)ii————— A ϭ F(A͞F,i,n)(1 ϩ i)n − 1F͞A—————A͞FExcelFunctionsϭ FV(i%,n,A)ϭ PMT(i%,n,F)EXAMPLE 2 ... Ford capital earns at arate of 14% per year ...In $1000 units, the F value in year 8 is foundby using the F͞A factor ... 2328) ϭ $13,232 ... 5 ... 6 The Cement Factory CaseOnce again, consider the HAC case presented at the outset of this chapter, in which a projected$200 million investment can generate $50 million per year in revenue for 5 yearsstarting1 year after start-up ... Now the president would like the answers to a couple of new questionsabout the estimated annual revenues ... (a) What is the equivalent future worth of the estimated revenues after 5 years at 10% per year?(b) Assume that, due to the economic downturn, the presidentpredicts that the corporationwill earn only 4 ... What is the required amount of the annual revenue series over the 5-year period to be economically equivalent to the amount calculated in (a)?Solution(a) Figure 2–6b is the cash flow diagram with A ϭ $50 million ... We use tabulated values and thespreadsheetfunction to find F in year 5 ... In $1 million units, thefuture worth of the revenue series isF ϭ 50(F͞A,10%,5) ϭ 50(6 ... 255 ($305,255,000)PE48Chapter 2Factors: How Time and Interest Affect Moneyϭ ϪPMT(4 ... 5% for the cement factory case,Example 2 ... If the rate of return on the annualrevenues were 0%, the total amount after 5 years wouldbe $250,000,000 ... Spreadsheet: Apply the FV factor in the format ϭ ϪFV(10%,5,50) to determine F ϭ$305 ... Because there is no present amount in this computation, P is omittedfrom the factor ... (As before, the minus sign forces the FV functiontoresult in a positive value ... He wants the revenue stream to generate the equivalent that it would at a 10% per year return, that is, $305 ... 5% per year return is achievable ... Since the factortables do not include 4 ... In $1 million units,0 ... 255(A͞F,4 ... 255 —————— ϭ 305 ... 18279)(1 ... 798[]Theannual revenue requirement grows from $50 million to nearly $55,800,000 ... 6% each year ... 5% and F ϭ $305 ... We can use the cell reference method(described in Appendix A) for the future amount F ... 798 per year (in $1 million units) ... 4 Factor Values for Untabulated i or n ValuesOften it isnecessary to know the correct numerical value of a factor with an i or n value that isnot listed in the compound interest tables in the rear of the book ... • Use the formula listed in this chapter or the front cover of the book,• Use an Excel function with the corresponding P, F, or A value set to 1 ... When theformula is applied, the factor value is accurate since the specific i and n values areinput ... Additionally, the formulas become more complexwhen gradients are introduced, as you will see in the following sections ... For example, the P͞F factor is determined using the PV function with A omitted (or set to 0)andF ϭ 1, that is, PV(i%,n,,1) or PV(i%,n,0,1) ... Functions to determine the six common factors areas follows ... 4Factor Values for Untabulated i or n ValuesFactorTo Do ThisFind P, given F ... Find P, given A ... Find F, given A ... P͞FF͞PP͞AA͞PF͞AA͞FExcel Functionϭ ϪPV(i%,n,,1)ϭ ϪFV(i%,n,,1)ϭϪPV(i%,n,1)ϭ ϪPMT(i%,n,1)ϭ ϪFV(i%,n,1)ϭ ϪPMT(i%,n,,1)Figure 2–9 shows a spreadsheet developed explicitly to determine these factor values ... Thevalues for i ϭ 3 ... As we already know, these same functions willdetermine a final P, A, or F value when actual or estimated cash flow amounts areentered ... Also interpolation introduces somelevel of inaccuracy, depending upon the distance between the two boundary values selected fori or n, as the formulas themselves are nonlinear functions ... Refer to Figure 2–10 for a graphical description of the following explanation ... Second, find thecorresponding tabulated factor values(f1 and f2) ... Figure 2–9Enter requested i and nFigure 2–10Factor valueaxisf2cfLinear interpolation in factorvalue tables ... TableaKnownx1RequiredxbKnown i or nx2axis4950Chapter 2Factors: How Time and Interest Affect Money(x – x1)f ϭ f1 ϩ ———— (f2 – f1)(x2– x1)af ϭ f1 ϩ — c ϭ f1 ϩ db[2 ... 17]The value of d will be positive or negative if the factor is increasing or decreasing, respectively,in value between x1 and x2 ... 7Determine the P͞A factor value for i ϭ 7 ... SolutionFactor formula: Apply the formula from inside the front cover of the book for the P͞A factor ...0775)10 Ϫ 11 ... 75%,10) ϭ ————— ϭ ——————— ϭ ————ni(1 ϩ i)0 ... 0775)10 0 ... 78641Spreadsheet: Utilize the spreadsheet function in Figure 2–9, that is, ϭ ϪPV(7 ... 78641 ... Apply the Equation [2 ... 17] sequence, where x is the interest rate i, the bounding interest rates arei1 ϭ 7% and i2ϭ 8%, and the corresponding P͞A factor values are f1 ϭ (P͞A,7%,10) ϭ 7 ... 7101 ... 75 Ϫ 7)f ϭ f1 ϩ ——— (f2 – f1) ϭ 7 ... 7101 Ϫ 7 ... 0236 ϩ (0 ... 3135) ϭ 7 ... 2351ϭ 6 ... 2351 ... 75% and 10 years, plus it takes more calculations than using the formula or spreadsheet function ... 2 ... The cash flow series ofmaintenance costs involves aconstant gradient, which is $5000 per year ... The amount of change is called the gradient ... In thecase of a gradient, each year-end cash flow is different, so new formulas must be derived ... This is convenient because in actual applications, thebase amount is usuallysignificantly different in size compared to the gradient ... Assume these cost $2500; that is, $2500 is the baseamount ... 551Arithmetic Gradient Factors (P͞G and A͞G)01234nn–1TimeFigure 2–11Cash flow diagram of anarithmetic gradient series ... If you estimate that total costs will increase by $200 eachyear, the amount the secondyear is $2700, the third $2900, and so on to year n, when the total cost is 2500 ϩ (n Ϫ 1)200 ... Note that the gradient ($200) is first observed between year 1 and year 2, and the base amount ($2500 in year 1) is not equal to the gradient ... G ϭ constant arithmetic change incash flows from one time period to the next; G may be positiveor negative ... 18]It is important to realize that the base amount defines a uniform cash flow series of the size A thatoccurs eash time period ... If the base amount is ignored, a generalized arithmetic (increasing) gradient cash flow diagram is asshown in Figure 2–12 ... This is called a conventional gradient ... 8A local university has initiated a logo-licensing program with the clothier Holister, Inc ... Determine the gradient and construct a cash flow diagram that identifiesthe base amount and the gradient series ... 18], solved for G, determines thearithmetic gradient ... 52Chapter 2Factors: How Time and Interest Affect MoneyCF9 =$200,000$185,000$170,000$155,000$140,000$125,000$110,000G = $15,000CF1 = $95,000$80,0000123456789YearFigure 2–13Diagram for gradient series, Example 2 ... The cash flow diagram (Figure 2–13) showsthe base amount of $80,000 in years 1 through 9and the $15,000 gradient starting in year 2 and continuing through year 9 ... The addition of the two results in PT ... 19]where PA is the present worth of the uniform series only, PG is the present worth of the gradientseries only, and the ϩ or Ϫ sign is usedfor an increasing (ϩG) or decreasing (ϪG) gradient,respectively ... 20]Three factors are derived for arithmetic gradients: the P͞G factor for present worth, the A͞Gfactor for annual series, and the F͞G factor for future worth ... We use the single-payment present worth factor (P͞F,i,n), but the same result can beobtained by using the F͞P, F͞A, or P͞A factor ... P ϭ G(P͞F,i,2) ϩ 2G(P͞F,i,3) ϩ 3G(P͞F,i,4) ϩ ... [3nϪ2nϪ112P ϭ G ———— ϩ ———— ϩ ———— ϩ ... 21]Multiplying both sides of Equation [2 ... ϩ ———— ϩ ———— [2 ... 21] from Equation [2 ... [] [n1111iP ϭ G ———— ϩ ———— ϩ ... 23]The left bracketedexpression is the same as that contained in Equation [2 ... Substitute the closed-end form of the P͞A factor from Equation [2 ... 553Arithmetic Gradient Factors (P͞G and A͞G)PG = ?i = given01234nn–101234n–1nG2G3G(n – 2)G(n – 1)G(b)(a)Figure 2–14Conversion diagram from an arithmetic gradient to apresent worth ... 23] and simplify to solve for P G , the present worth of the gradient seriesonly ... 24]Equation [2 ... Figure 2–14a is converted into theequivalent cash flow in Figure 2–14b ... 25]Remember: The conventional arithmetic gradient starts in year 2, and P is located in year 0 ... 24] expressed asan engineering economy relation isPG ϭ G(P͞G,i,n)[2 ... 19] to calculate total present worth ... The equivalent uniform annual series AG for an arithmetic gradient G is found by multiplyingthe present worth in Equation [2 ... In standard notation form, theequivalent of algebraic cancellation of P can be used... 27]which is the rightmost term in Equation [2 ... The expression in brackets in Equation [2 ... This factorconverts Figure 2–15a into Figure 2–15b ... Factor valuesare tabulated in the two rightmost columns of factor values at the rear of this text ... There is no direct, single-cell spreadsheet function tocalculate PG or AG for an arithmeticgradient ... General formats for these functions are� NPV(i%, second_cell:last_cell) ؉ first_cell� PMT(i%, n, cell_with_PG)[2 ... 29]The word entries in italic are cell references, not the actual numerical values ... 2, for a description of cell reference formatting ... 10 ... Theresulting factor,(F͞G,i,n), in brackets, and engineering economy relation is[( ) () ]n1 (1 ϩ i) – 1FG ϭ G — ————— Ϫ niiEXAMPLE 2 ... At a recent meeting, the engineers estimated that a total of$500,000 will be deposited at the end of next year into an account for the repair of old andsafety-questionable

bridges throughout the area ... Determine theequivalent (a) present worth and (b) annual series amounts, if public funds earn at a rateof 5% per year ... According to Equation [2 ... The total present worth PToccurs in year 0 ... In $1000 units, the total present worth isPT ϭ 500(P͞A,5%,10) ϩ100(P͞G,5%,10)ϭ 500(7 ... 6520)ϭ $7026 ... 5Arithmetic Gradient Factors (P͞G and A͞G)012$5003$6004$7005$8006$9007$10008$11009$120010$1300$1400Figure 2–16Cash flow series with a conventional arithmetic gradient (in $1000 units),Example 2 ... PG = ?PA = ?A = $5001 29 10G = $1001 2910+$100BaseGradient $900PT = ?PT = PA + PG12$500$6003$7004$8005$9006$10007$11008$12009$130010$1400Figure 2–17Partitioned cash flow diagram (in $1000 units), Example 2 ... (b) Here, too, it is necessary to consider the gradient and the base amount separately ... 20] and occurs inyears 1 through 10 ... 0991)ϭ $909 ... Any other cash flows must be considered separately ... In this case, considering round-off error,AT ϭ PT (A͞P,5%,10) ϭ 7026 ... 12950)ϭ $909 ... 10 The Cement Factory CaseThe announcement of the HAC cement factory states that the $200 million (M) investmentisplanned for 2012 ... Further investigation may determine, for example, that the $200 M is a present worth in the year 2012 of anticipated investments during the next 4 years (2013 through 2016) ... As before, assume the timevalue of money for investment capital is 10% per year to answer the followingquestions usingtabulated factors and spreadsheet functions, as requested below ... (b) Given the planned investment series, what is the equivalent annual amount that will beinvested from 2013 to 2016? Use both tabulated factors and spreadsheet functions ... ) What must be the amount ofyearlyconstant decrease through 2016 to have a present worth of exactly $200 M in2012, provided $100 M is expended in 2013? Use a spreadsheet ... Figure 2–18 diagrams the cashflows with the shaded area showing the constantly declining investment each year ... Tabulated factors: Equation [2 ... Money isexpressed in $1 million units ... 30]ϭ 100(3 ... 3781)ϭ $207 ... 5 M ... Figure 2–19 shows the entries and functionNPV(i%,second_cell:last_cell) ... The result displayed in cell C9, $207 ... (Note that the NPV function does not consider two separate series ofcash flows as is necessary when using tabulatedfactors ... 5 M ... First, apply Equation [2 ... Both relations are illustrated here, in $1 million units,PT ϭ ?Figure 2–18Cash flow diagram for decreasing gradient in $1 million units, Example 2 ... i = 10% per year20130201420152016Year1234TimeBaseA ϭ $100$25$50$100GradientG ϭ $Ϫ252 ... 10a and b ...20]:AT ϭ 100 – 25(A͞G,10%,4) ϭ 100 Ϫ 25(1 ... 471($65,471,000 per year)Use PT :AT ϭ 207 ... 537(0 ... 471 per yearSpreadsheet: Apply the PMT function in Equation [2 ... 471 per year (Figure 2–19) ... It is an excellent tool to applywhen one cell entry must equal a specific value and only one other cellcan change ... This is the same as stating PT ϭ 200 in Equation [2 ... All other parameters retain their current value ... When OK is clicked, the solution is displayed; G ϭ $Ϫ26 ... Refer to Figure 2–20 again ... 721 M, the equivalent total present worth invested over the 4 yearswill be exactly $200 M ... 10c ...721 to makepresent worth exactly $20058Chapter 2Factors: How Time and Interest Affect Money2 ... This change occursevery year on top of a starting amount in the first year of the project ... A geometric gradient series is a cash flow series that either increases or decreases by a constantpercentageeach period ... g ϭ constant rate of change, in decimal form, by which cash flow values increase or decreasefrom one period to the next ... A1 ϭ initial cash flow in year 1 of the geometric seriesP g ϭ present worth of the entire geometric gradient series, including the initial amountA1Note that the initial cashflow A1 is not considered separately when working with geometricgradients ... The relation to determine the total presentworth Pg for the entire cash flow series may be derived by multiplying each cash flow in Figure 2–21a by the P͞F factor 1͞(1 ϩ i)n ... ϩ ——————(1 ϩ i)n(1 ϩ i)1(1 ϩ i)2(1 ϩ i)3[(1 ϩg)2(1 ϩ g)nϪ11ϩ g1ϭ A1 ——— ϩ ———— ϩ ———— ϩ ... 31]Multiply both sides by (1 ϩ g)͞(1 ϩ i), subtract Equation [2 ... [()1ϩg n1 Ϫ ———1ϩiPg ϭ A1 ———————iϪg]gi[2 ... 32] is the (P͞A,g,i,n) or geometric gradient series presentworth factor for values of g not equal to the interest rate i ... 31] andobserve that the term 1/(1 + i) appears n times ... nA1(1 – g)n – 12 ... ϩ ———(1 ϩ i) (1 ϩ i) (1 ϩ i)(1 ϩ i)nA1Pg ϭ ———(1 ϩ i))[2 ... The equation for Pg and the (P͞A,g,i,n) factor formula arePg ϭ A1(P͞A,g,i,n)()1؉g n1 ؉1 ؊ ——— i(P͞A,g,i,n) � —————— i−gn———1؉i[2 ... 35]g� iIt is possible to derivefactors for the equivalent A and F values; however, it is easier to determinethe Pg amount and then multiply by the A͞P or F͞P factor ... Once the cash flows are entered, P and A are determined using the NPV and PMTfunctions, respectively ... 11A coal-fired power plant has upgraded an emission controlvalve ... The maintenance cost is expected to be high at $1700 the first year, increasing by 11% per year thereafter ... Solution by HandThe cash flow diagram (Figure 2–22) shows the salvage value as a positive cash flow and allcosts as negative ... 35] for g i to calculate Pg ... PT ϭ Ϫ8000 Ϫ Pg ϩ200(P͞F,8%,6)1 Ϫ (1 ... 08)6ϭ Ϫ8000 Ϫ 1700 ——————— ϩ 200(P͞F,8%,6)0 ... 11ϭ Ϫ8000 Ϫ 1700(5 ... 11 ... 11)$1700(1 ... 11)3$1700(1 ... 11)5Placement ofGradient Pg60Chapter 2Factors: How Time and Interest Affect MoneySolution by SpreadsheetFigure 2–23 details the spreadsheet operations tofind the geometric gradient present worth Pgand total present worth PT ... Cell tags detail the relations for thesecond and third components; the first cost occurs at time 0 ... If this factor is used repeatedly, it is worthwhile using cell referenceformatting so that A1, i, g, and n values can be changed and thecorrect value is always obtained ... Present worth of salvageϭ ϪPV(8%,6,,200)Present worth of maintenance costs, Eq ... 35]ϭ Ϫ1700* ((1-((1 ... 08))^6)/(0 ... 11))Figure 2–23Geometric gradient and total present worth calculated via spreadsheet, Example 2 ... PEEXAMPLE 2 ... The revenueseries estimateof $50 million annually is quite optimistic, especially since there are manyother cement product plants operating in Florida and Georgia on the same limestone deposit ... ) Therefore, it is important to be sensitive in our analysis to possibly declining and increasing revenue series, depending upon thelonger-term success of theplant’s marketing, quality, and reputation ... Determine the present worth and future worth equivalents of all revenues during this 5-year timeframe at the same rate used previously, that is, 10% per year ... In year 1, A1 ϭ $50 M and revenues decrease in year 5 toA1(1 Ϫ g)nϪ1 ϭ50 M(1 Ϫ 0 ... 88)4 ϭ $29 ... [2 ... 10 and g ϭ Ϫ0 ... In $1 million units,[()]0 ... 10Pg ϭ 50 ——————— ϭ 50[3 ... 10 Ϫ (Ϫ0 ... 80F ϭ 152 ... 80(1 ... 08This means that the decreasing revenue stream has a 5-year future equivalent worth of$246 ... If you look back to Example 2 ... 7Determining i or n forKnown Cash Flow Valuesuniform revenue series of $50 M annually is $305 ... In conclusion, the 12% declininggeometric gradient has lowered the future worth of revenue by $59 ... 2 ... An example for which i is soughtmay be stated as follows: A company invested money to develop a new product ...There are several ways to find an unknown i or n value, depending upon thenature of the cash flow series and the method chosen to find the unknown ... The mostdifficult and complex involves finding i or n for irregular cash flows mixed with uniform andgradient series utilizing solution by hand andcalculator ... Single Amounts—P and F OnlyHand or Calculator Solution Set up the equivalence relation and (1) solve for the variableusing the factor formula, or (2) find the factor value and interpolate in the tables ... (See below and Appendix A for details ... Spreadsheet Solution Use the IRR or RATEfunction to find i or the NPER function to find n ... Spreadsheet Solution Use the IRR or RATE function to find i or the NPER function to findn ... )Besides the PV, FV, and NPV functions, other spreadsheet functions useful in determining iare IRR (internal rate of return) and RATE, and NPER (number ofperiods) to find n ... In all threeof these functions, at least one cash flow entry must have a sign opposite that of others in orderto find a solution ... 36]To use IRR to find i, enter all cash flows into contiguous cells, including zero values ... 37]The single-cell RATE function finds i when an A series and singleP and/or F values are involved ... [2 ... 13If Laurel made a $30,000 investment in a friend’s business and received $50,000 5 years later,determine the rate of return ... 1P ϭ F(P͞F,i,n) ϭ F ————(1 ϩ i)n130,000 ϭ 50,000 ————(1 ϩ i)510 ... 2i ϭ —— Ϫ 1 ϭ 0 ... 76%)0 ... ( )P ϭ F(P͞F,i,n)30,000 ϭ50,000(P͞F,i,5)(P͞F,i,5) ϭ 0 ... 6000 for n ϭ 5 lies between 10% and 11% ... 76% ... 14Pyramid Energy requires that for each of its offshore wind power generators $5000 per yearbe placed into a capital reserve fund to cover unexpected major rework on field equipment ... What rate of return did this practiceprovide to the company? Solve by hand andspreadsheet ... Either the A͞F or F͞A factor can be used ... 0500From the A͞F interest tables for 15 years, the value 0 ... By interpolation, i ϭ 3 ... F = $100,000 Figure 2–24Diagram to determine the rateof return, Example 2 ... i= ?012345678A =$500091011121314 152 ... A single-cell solution using the RATE function can be applied since A ϭ $Ϫ5000 occurseach year and F ϭ $100,000 takes place in the last year of the series ... 98% ... The IRR function is much better for answering “what if ” questions ... In any cell enter the IRR function ... 98%is displayed ... The IRR function does not need these numbers, but it makes the cash flow entry activity easier and moreaccurate ... Figure 2–25Use of RATE and IRRfunctions to determinei value for a uniformseries, Example 2 ... i using RATE functionϭ RATE(15,-5000,,100000)i using IRR functionϭIRR(E2:E17))EXAMPLE 2 ... All analysis thus far has taken place at 10% per year; however, the parent company has made it clear that its other international plants are able to show a 20% per year returnon the initial investment ... SolutionIf hand solution is utilized, the present worth relation can beestablished and the n valuesinterpolated in the tables for each of the three rate of return values ... 00This is a good opportunity to utilize a spreadsheet and repeated NPER functions fromEquation [2 ... Figure 2–26 shows the single-cellϭ NPER(i%,50,Ϫ200) function for each rate of return ... 15 ...Capability in using these formulas and theirstandard notation manually and with spreadsheets is critical to complete an engineering economystudy ... Additionally, you can solvefor rate of return i or time n ... 1 Look up the numerical value for the following factors from the interest tables ... (P͞F,6%,8)2 ...(A͞G,15%,20)4 ... (P͞G,35%,15)Determination of F, P, and A2 ... , affordto spend now on an energy management system ifthe software will save the company $21,300 peryear for the next 5 years? Use an interest rate of10% per year ... Thecost is $985,000, and a $100,000 deposit will holdone of the first100 “cars ... At an interest rate of 10% peryear, what is the effective total cost of the PAV inyear 3?2 ... If the fund earnedinterest at 6% per year, how much was in the account 14 years after it was started?2 ... The distributor of theinclinometers is temporarily overstocked and is offering them at a 40%discount from the regular costof $142 ... Assume the interest rate is10% per year ... 6 One of the biggest vulnerabilities in a control system is network devices, such as Ethernet-basednetwork switches that are located in unsecuredlocations and accessible to everyone ... The company is consideringexpanding its manufacturinglines now or doing it in 3 years ... 9 million, what equivalent amountcould the company afford to spend in 3 years? Theinterest rate is 15% per year ... 4 The Moller Skycar M400 is a flying car knownas a personal air vehicle (PAV) that is expected2 ... If the new equipment willcost $220,000 to purchase and install,how much must the company save each year for3 years in order to justify the investment, if theinterest rate is 10% per year?2 ... of Carnegie, Pennsylvania, makes acontrol pinch valve that provides accurate, repeatable control of abrasive and corrosive slurries,outlasting gate, plug, ball, and even satellite coatedvalves ... 2 ... 3 million for its corporate headquarters, whatmust the building be worth in 10 years? The company expects all expenditures to earn a rate of return of at least 18% per year ... 10 CGK Rheosystems makes high-performance rotationalviscometers capable of steady shear andyield stress testing in a rugged, compact footprint ... 11 Five years ago a consulting engineer purchased abuilding for company offices constructed of bricksthat were not properly fired ... Because of the problem with thebricks, the selling price of the building was25%below the price of comparable, structurally soundbuildings ... This resulted in restoring the building to its fair market value ... 12 Metso Automation, which manufactures addressable quarter-turn electric actuators, is planning toset aside $100,000 now and $150,000 one yearfrom now for possiblereplacement of the heatingand cooling systems in three of its larger manufacturing plants ... 13 Syringe pumps often fail because reagents adhereto the ceramic piston and deteriorate the seal ... One of Trident’s customers expects to reducedowntime by 30% as a result of the new seal design ... 14 Chinaspends an estimated $100,000 per year oncloud seeding efforts, which includes using antiaircraft guns and rocket launchers to fill the skywith silver iodide ... If the yields of cash cropswill increase by 4% each year for the next 3 yearsbecause of extra irrigation water captured behinddams during cloudseeding, what is the maximumamount the farmers should spend now on the cloudseeding activity? The value of the cash crops without the extra irrigation water would be $600,000per year ... 2 ... 07 million to improve(i ... , deepen) a retention basin and reconstruct thespillway that was severely damagedin a flood2 years ago ... If the projects are assumed tohave a 20-year life, what is the annual worth of thesavings at an interest rate of 6% per year?2 ... 5 miles per gallon for cars and light trucks bythe year 2016 ... If a person purchases a new car in2012 and keeps it for 5 years, how much must besavedin fuel costs each year to justify the extracost? Use an interest rate of 8% per year ... 17 In an effort to reduce childhood obesity by reducing the consumption of sugared beverages,some states have imposed taxes on soda andother soft drinks ... However, if taxes were increased to 18 cents onthe dollar,Sturm calculated they would make a66Chapter 2Factors: How Time and Interest Affect Moneysignificant difference ... Use aninterest rate of 6% per year ... 18 The Texas Tomorrow Fund (TTF) is a programstarted in 1996 in Texas wherein parents could prepay their child's college tuition when the childwasyoung ... Later, the Texaslegislature allowed universities to set their own tuition rates; tuition costs jumped dramatically ... If the TTF fund grew at a rate of 4% per year,while tuition costs increased at 7% per year, determine the state’s shortfall when a newborn enterscollege 18 years later ... 19 HenryMueller Supply Co ... e ... Annual cash flows areshown in the table below ... YearIncome, $1000Cost, $100012345678200 200 200 200 200 200 200 20090 90 90 90 90 90 90 902 ... If the new units will cost $350,000, howmuch should the company set aside each year, ifthe account earns 10% per year?calculated values, assuming the formula-calculatedvalue is the correct one ... 25 Profits from recycling paper, cardboard, aluminum, and glass at a liberal arts college have increased at a constant rate of $1100 in each of thelast 3 years ... 26 A report by the Government Accountability Office (GAO) showsthat the GAO expects theU ... Postal Service to lose a record $7 billion atthe end of this year, and if the business model isnot changed, the losses will total $241 billion bythe end of year 10 ... 27 Rolled ball screws are suitable for high-precisionapplications such as water jet cutting ... Determine theequivalent annual cost at aninterest rate of 8% per year ... 21 Find the numerical value of the following factorsusing (a) interpolation and (b) the formula ... (A͞P,13%,15)2 ... 22 Find the numerical value of the following factorsusing (a) interpolation, (b) the formula, and (c) aspreadsheet function ...(F͞P,14%,62)2 ... 23 For the factor (F͞P,10%,43), find the percent difference between the interpolated and formulacalculated values, assuming the formula-calculatedvalue is the correct one ... 24 For the factor (F͞A,15%,52), find the percent difference between the interpolated and formula-2 ... 7-m-diametermilling head thatemits low vibration and processes stress-relievedaluminum panels measuring up to 6000 mm long ... If the company offers to repay the loan with $60,000 in year 1 andamounts increasing by $10,000 each year throughyear 5, how much can the company borrow at aninterest rate of 10%per year?2 ... (a) What is the amount of the cash flow in year 3?(b) What is the future worth of the entire cashflow series in year 10? Let i ϭ10% per year ... 30 For the cash flows below, determine the amount in year 1, if the annual worth in years 1 through 9 is $601 ... Year12345Cost, $1000AA ϩ 30A ϩ60A ϩ 90A ϩ 1202 ... 1 billionavailable 5 years from now to finance production of a handheld “electronic brain” that, basedon your behavior, will learn how to controlnearly all the electronic devices in your home,such as the thermostat, coffee pot, TV, andsprinkler system ... If the amount setaside at the endof year 1 is $50 million, howmuch will the constant increase G have to beeach year? Assume the investment accountgrows at a rate of 18% per year ... 32 Tacozza Electric, which manufactures brush dcservomotors, budgeted $75,000 per year to pay forcertain components over the next 5 years ...Geometric Gradient2 ... Calculate the firsttwo annual worth factor values, that is, A values forn ϭ 1 and 2, that would be in a 10% interest tablefor a growth rate of 4% per year ... 34 Determine the present worth of a geometric gradient series with a cash flow of $50,000 in year 1 andincreases of 6% eachyear through year 8 ... 2 ... Contract 1 has a cost of $10,000 in year 1; costswill escalate at a rate of 4% per year for 10 years ... 2 ... A new contractbetween the two entities resulted in a reduction infuture price increases in the cost of the water from67A ϩ 150 A ϩ 18089A ϩ 210A ϩ 2408% per year to 4%per year for the next 20 years ... 2 ... Usean interest rate of 6% per year ... 38 Gesky Industrial Products manufactures brushlessblowers for boilers, food service equipment, kilns,and fuel cells ... What was the interest rate on the loan? Usehand and spreadsheet solutions ... 39 If the value of Jane’sretirement portfolio increased from $170,000 to $813,000 over a 15-yearperiod, with no deposits made to the account overthat period, what annual rate of return did shemake?2 ... According toConsumer Credit Counseling Service, a homeowner with a $100,000 mortgage and a 520 creditscore will pay$110,325 more in interest chargesover the life of a 30-year loan than a homeownerwith the same mortgage and a credit score of 720 ... 41 During a period when the real estate market inPhoenix, Arizona, was undergoing a significantdownturn, CSM Consulting Engineers made anagreement with adistressed seller to purchase anoffice building under the following terms: totalprice of $1 ... CSM was able to make this deal becauseof poor market conditions at the time of purchase,and, at the same time, planning to sell the buildingin 4 years (when market conditions would probably be better) and moveto a larger office buildingin Scottsdale, Arizona ... 9 million, whatrate of return per year did the company make onthe investment?2 ... The contract required the company to repay the investors through an innovative mechanism called faux dividends, a series ofuniform annual payments over a fixed periodoftime ... 43 Bessimer Electronics manufactures addressableactuators in one of its Maquiladora plants inMexico ... If the company doesmake the annual investments, what rate of returnwill it realize?2 ... Your boss saw a report submitted by the chief financial officer (CFO)that said the equivalent annualworth of maintaining the equipment used in producing the resins was$48,436 over the last 5 years ... Your bossthought $48,436 was too high, so she asked you todetermine what interest rate the CFO used in making the calculations ... 45 Acme Bricks, a masonry products company, wantsto have$600,000 on hand before it invests in newconveyors, trucks, and other equipment ... 46 An engineer who was contemplating retirementhad $1 ... However, a severe recession caused his portfolio to decrease to only 55% of the original amount, so hekept working ... 6 million value?Factors: How Time andInterest Affect Money2 ... If you invest $200,000 of the company’s money ina natural gas well that is expected to provide income of $29,000 per year, how long must the wellproduce at that rate in order to get the money backplus a rate of return of 10% per year?2 ... She invested ina stock fund thataveraged a 12% rate of return overthat period ... 49 A mechanical engineering graduate who wantedto have his own business borrowed $350,000from his father as start-up money ... If the engineer was able to pay his father$15,000 in year 1, $36,700 in year 2, and amountsincreasing by $21,700 eachyear, how manyyears did it take for the engineer to repay theloan?2 ... The cost at theend of the next year (year 1) is expected to be$13,000 ... 51 In cleaning out some files that were left behind bythe engineer who preceded you in your current job,you found an old report that had a calculation forthepresent worth of certain maintenance costs forstate highways ... 03)͞(1 ϩ 0 ... 06 Ϫ 0 ... What is its value?2 ... If the cash flow inyear 1 is $35,000 and the gradient amount is$19,000, what is the value of n at an interest rate of10% per year?2 ... If the cash flow inyear 1 is $25,000 and the gradient increaseis18% per year, what is the value of n? The interestrate is 10% per year ... 54 The amount of money that Diamond Systems canspend now for improving productivity in lieu ofspending $30,000 three years from now at an interest rate of 12% per year is closest to:(a) $15,700(b) $17,800(c) $19,300(d )$21,3502 ... If the conveyor belt resulted in costsavings of $4200 per year, the length of time itwould take for the company to recover its investment at 8% per year is closest to:(a) Less than 9 years(b) 9 to 10 years(c) 11 to 12 years(d ) Over 12 years2 ... S ... If U ... Garment’s operating cost per machineis $22,000 for year 1 and increases by aconstant $1000 per year through year 5, what isthe equivalent uniform annual cost per machinefor the 5 years at an interest rate of 8% per year?(a) $23,850(b) $24,650(c) $25,930(d ) Over $26,0002 ... 58 At i ϭ 4% per year, A for years 1 through 6 of thecash flowsshown below is closest to:(a) $300(b) $560(c) $800(d ) $10400123456$300$400Years2 ... 60 A small construction company is considering thepurchase of a used bulldozer for $61,000 ... 61 The cost of lighting and maintaining the tallestsmokestack in the United States (at a shutteredASARCO refinery) is$90,000 per year ... 62 An enthusiastic new engineering graduate plans tostart a consulting firm by borrowing $100,000 at10% per year interest ... 63 An engineer who believed in “save now and playlater” wanted to retire in 20 years with $1 ... At 10% per year interest, to reach the $1 ... 64 The cost of aborder fence is $3 million permile ... 6 million(b) $4 ... 9 million(d ) Over $5 ... 65 An investment of $75,000 in equipment thatwill reduce the time for machining self-lockingfasteners will save $20,000 per year ... 66 The number of years required for an account to accumulate $650,000 if Ralph deposits$50,000 eachyear and the account earns interest at a rate of 6%per year is closest to:(a) 13 years(b) 12 years(c) 11 years(d) 10 years2 ... , manufactures cleaning nozzlesfor reverse-pulse jet dust collectors ... The rate of return per year on the investment is closest to:(a) 20%(b) 18%(c) 16%(d) Less than15%Factors: How Time and Interest Affect Money2 ... If the cost inyear 1 was $26,000 and it increased by $2000 peryear through year 5, the present worth of the costsat an interest rate of 10% per year is closest to:(a) $102,900(b) $112,300(c) $122,100(d) $195,8002 ... Ifyour investments earn 10% peryear, the amountyou will have at the end of year 20 is closest to:(a) $242,568(b) $355,407(c) $597,975(d) $659,1252 ... If income in year 1 was $300,000 and it decreased by$30,000 per year through year 4, the annual worthof the income at 10% per year is closest to:(a) $310,500(b) $258,600(c)$203,900(d) $164,8002 ... She has gotten interested in the major effects thattime and interest rates have on the amount of money necessary to do things and the significant growth in the amount ofmoney when a large number of years are considered ... The four situations are described here ... ManhattanIsland was purchased in 1626 for $24 ... B ... Case StudyC ... Sundara estimated the annual rate of return must bequite good, especially given that she is lucky to earn 4%per year on her own investments these days ... A friend who is not good with money, went to a pawnshop and borrowed $200 for oneweek and paid $30 ininterest ... However, she didnot know whether the interest was simple or compounded monthly, and how much may be owed werethis loan not paid off for 1 year ... What is the annual interest rate for each situation? Include both the annual simple and the compound ratesfor situationD ... Calculate and observe the total amount of money involved in each situation at the end of the time periodscompared to the starting amount ... Think of a situation for yourself that may be similar toany of those above ... 71CHAPTER 3CombiningFactors andSpreadsheetFunctionsL E A R N I N GO U TC O M E SPurpose: Use multiple factors and spreadsheet functions to find equivalent amounts for cash flows that have nonstandard placement ... 1Shifted series• Determine the P, F or A values of a seriesstarting at a time other than period 1 ... 2Shifted series and single cashflows• Determine the P, F, orA values of a shifted seriesand randomly placed single cash flows ... 3Shifted gradients• Make equivalence calculations for shiftedarithmetic or geometric gradient series thatincrease or decrease in size of cash flows ... Fora given sequence of cash flows, there are usually several correct ways to determinethe equivalent present worth P, future worth F, or annual worth A ... 3 ... In this case several methods can be used to find the equivalent present worth P ... • Use the F͞P factor to find the future worth of each disbursement in year 13, add them, andthen find the present worth of the total, using P ϭ F(P͞F,i,13)... • Use the P͞A factor to compute the “present worth” P3 ϭ A(P͞A,i,10) (which will be locatedin year 3, not year 0), and then find the present worth in year 0 by using the (P͞F,i,3) factor ... For Figure 3–1, the “present worth” obtained using the P͞A factor is located in year 3 ... Note that a P value is alwayslocated1 year or period prior to the beginning of the first series amount ... The most commonmistake made in working problems of this type is improper placement of P ... Placement of PTo determine a future worth or F value, recall that the F͞A factor derived in Section 2 ... Figure 3–3 shows the locationofthe future worth when F͞A is used for Figure 3–1 cash flows ... Placement of FIt is also important to remember that the number of periods n in the P͞A or F͞A factor is equalto the number of uniform series values ... Figures 3–2 and 3–3 show Figure 3–1 renumbered to determine n ϭ 10 ... A = $50Figure 3–2P3 = ?01234152637485A = $50961011127891310YearnLocation of present worthand renumbering for n forthe shifted uniform seriesin Figure 3–1 ... F=?0123415267348591012 13Year761189n10A = $50As stated above, several methods can be used to solve problems containing a uniform seriesthatis shifted ... Specific steps should be followed to avoid errors:1 ... 3 ... 5 ... Locate the present worth or future worth of each series on the cash flow diagram ... Draw another cash flow diagram representing the desired equivalent cash flow ... These steps are illustrated below ... 1The offshore design groupat Bechtel just purchased upgraded CAD software for $5000 nowand annual payments of $500 per year for 6 years starting 3 years from now for annual upgrades ... The symbol PA is used throughout this chapterto represent the present worth of a uniform annual series A, and PA represents the presentworth'at a time other than period 0 ... Thecorrect placement of PA and the diagram renumbering to obtain n are also indicated ... Also, n ϭ 6, not 8, for the P͞A factor ... PA ϭ $500(P͞A,8%,6)'Since PA is located in year 2, now find PA in year 0 ... 1 ... PT ϭ P0 ϩ PAϭ 5000 ϩ 500(P͞A,8%,6)(P͞F,8%,2)ϭ 5000 ϩ500(4 ... 8573)ϭ $6981 ... 1Calculations for Uniform Series That Are ShiftedThe more complex that cash flow series become, the more useful are the spreadsheet functions ... The NPV function, like the PV function, determinesthe P values, but NPV can handle any combination of cash flows directly fromthe cells ... Use the formatNPV(i%, second_cell:last_cell) ؉ first_cellFirst_cell contains the cash flow for year 0 and must be listed separately for NPV to correctlyaccount for the time value of money ... The easiest way to find an equivalent A over n years for a shifted series is with the PMT function, wherethe P value is from the NPV function above ... PMT(i%, n, cell_with_P,F)Alternatively, the same technique can be used when an F value was obtained using the FV function ... ”It is very fortunate that any parameter in a spreadsheet function can itself be a function ... The format isPMT(i%, n,NPV(i%,second_cell:last_cell) ؉ first_cell,F)[3 ... All three of these functions are illustrated in Example 3 ... EXAMPLE 3 ... If the machine will be recalibrated for each of 6 years starting 3 years after purchase, calculate the 8-year equivalentuniform series at 16% per year ... Solution by HandFigure 3–5aand b shows the original cash flows and the desired equivalent diagram ... Then either the A͞P factor or theA͞F factor can be used ... Present worth method ... ) Calculate PA for the shifted series in year 2,'followed by PT in year 0 ... P'A ϭ 8000(P͞A,16%,6)PT ϭ PA(P͞F,16%,2) ϭ 8000(P͞A,16%,6)(P͞F,16%,2)'ϭ8000(3 ... 7432) ϭ $21,907 ... A' ϭ PT (A͞P,16%,8) ϭ $5043 ... (Refer to Figure 3–5a ... F ϭ 8000(F͞A,16%,6) ϭ $71,820The A͞F factor is now used to obtain A' over all 8 years ... 207576Chapter 3PT = ?0Combining Factors and Spreadsheet FunctionsЈPA = ?1 2F=?34567 8i = 16% per year01234A =$8000678AЈ = ?(a)5(b)ϭ ϪPMT(16%,8,B12)ϭ ϪPMT(16%,8,NPV(16%,B4:B11) ϩ B3ϭ NPV(16%,B4:B11) ϩ B3(c)Figure 3–5(a) Original and (b) equivalent cash flow diagrams; and (c) spreadsheet functions to determineP and A, Example 3 ... Solution by Spreadsheet(Refer to Figure 3–5c ... Use the NPVfunction to display P ϭ $21,906 ... There are two ways to obtain the equivalent A over 8 years ... (1) Enter the PMT function making direct reference tothe P value (see cell tag for D͞E5), or (2) use Equation [3 ... 3 ... 1 are applied to the uniform series and the single-amount formulas are applied to the one-time cash flows ... 3 and 3 ... For spreadsheet solutions, it is necessary to enter the net cashflows before using the NPV and other functions ... 3An engineering company in Wyoming that owns 50 hectares of valuable land has decided tolease the mineral rights to a mining company ... Theengineeringcompany makes a proposal to the mining company that it pay $20,000 per year for 20 yearsbeginning 1 year from now, plus $10,000 six years from now and $15,000 sixteen years fromnow ... 2Calculations Involving Uniform Series and Randomly Placed Single AmountsSolutionThe cash flowdiagram is shown in Figure 3–6 from the owner’s perspective ... PT ϭ 20,000(P͞A,16%,20) ϩ 10,000(P͞F,16%,6) ϩ 15,000(P͞F,16%,16)ϭ $124,075Note that the $20,000 uniform series starts at the end of year 1, so the P͞A factor determines thepresent worth at year 0 ... 3 ... Then you obtain the A value bymultiplying P or F by the appropriate A͞P or A͞F factor ... 4 illustrates this procedure ... 4A design-build-operate engineering company in Texas that owns a sizable amount of landplans to lease the drilling rights (oil and gas only) to a mining and exploration company ... e ... Utilize engineering economyrelations byhand and by spreadsheet to determine the five equivalent values listed below at 16% per year ... 2 ... 4 ... Total present worth PT in year 0Future worth F in year 22Annual series over all 22 yearsAnnual series over the first 10 yearsAnnual series over the last 12 yearsSolution by HandFigure 3–7 presents the cash flows with equivalent P and F values indicated in the correct yearsfor the P͞A, P͞F, and F͞A factors ... PT in year 0: First determine P' of the series in year 2 ... 7778Chapter 3Combining Factors and Spreadsheet FunctionsP'A ϭ 20,000(P͞A,16%,20)PT ϭ P'A(P͞F,16%,2) ϩ10,000(P͞F,16%,6) ϩ 15,000(P͞F,16%,16)ϭ 20,000(P͞A,16%,20)(P͞F,16%,2) ϩ 10,000(P͞F,16%,6)ϩ 15,000(P͞F,16%,16)ϭ $93,625[3 ... F in year 22: To determine F in year 22 from the original cash flows (Figure 3–7), findF for the 20-year series and add the two F values for the two single amounts ... F ϭ20,000(F͞A,16%,20) ϩ 10,000(F͞P,16%,16) ϩ 15,000(F͞P,16%,6)[3 ... A over 22 years: Multiply PT ϭ $93,625 from (1) above by the A͞P factor to determinean equivalent 22-year A series, referred to as A1–22 here ... 16635) ϭ $15,575[3 ... In thiscase, the computation is A1–22 ϭ F(A͞F,16%,22) ϭ $15,575 ... 4... A over years 1 to 10: This and (5), which follows, are special cases that often occur inengineering economy studies ... This occurs when a defined studyperiod or planning horizon is preset for the analysis ... ) To determine the equivalent A series for years 1 through 10 only (call it A1–10),the PT valuemust be used with the A͞P factor for n ϭ 10 ... A1–10 ϭ PT (A͞P,16%,10) ϭ 93,625(0 ... 5]5 ... This transforms Figure 3–7 into the 12-year series A11–22 in Figure 3–8b ... 03241) ϭ $79,457[3 ... This is another demonstration of the time value of money ... The $20,000 series andthe two single amounts havebeen entered into separate columns, B and C ... 279Calculations Involving Uniform Series and Randomly Placed Single AmountsFigure 3–8A1–10 = ?12345678910112122 Year2021 22Cash flows of Figure 3–7converted to equivalentuniform series for(a) years 1 to 10 and(b) years 11 to 22 ... Example 3... values are all entered so that the functions will work correctly ... To prepare for sensitivityanalysis, the functions are developed using cell reference format or global variables, as indicated in the column E function ... 1 ... The sum of these in F14 is PT ϭ $93,622, whichcorresponds to the value inEquation [3 ... Alternatively, PT can be determined directlyvia the sum of two NPV functions, shown in row 15 ... The FV function in row 18 uses the P value in F14 (preceded by a minus sign) to determine F twenty-two years later ... 3] ... To find the 22-year A series starting in year 1, the PMT function inrow 21 referencesthe P value in cell F14 ... 4] toobtain A1–22 ... The cell reference format isϭ PMT(D1,22,−(NPV(D1,B6:B27)ϩB5 ϩ NPV(D1,C6:C27)ϩC5)) ... and 5 ... These are both true for theseries requested here ... 5] and [3 ... CommentRemember that round-off error will always be present whencomparing hand and spreadsheetresults ... Also, be very careful when constructing spreadsheet functions ... Always checkyour function entries carefully before touching ... 3 Calculations for Shifted GradientsIn Section 2 ... The P͞G factor, Equation [2 ... The present worth of an arithmetic gradient willalways be located two periods before thegradient starts ... The relation A ϭ G(A͞G,i,n) was also derived in Section 2 ... The A͞G factor in Equation [2 ... Recall that the base amount must be treated separately ... A conventional gradient series starts between periods 1 and 2 of the cash flow sequence ... Then value in the P͞G and A͞Gfactors for a shifted gradient is determined by renumbering the time scale ... The n value for the gradient factor is determined bythe renumbered period where the last gradient increase occurs ... Example 3 ... EXAMPLE 3 ... has tracked the average inspection cost on a roboticsmanufacturing line for8 years ... Analyze the gradient increase, using the P͞G factor ... Figure 3–10b and c partitions these twoseries ... It isclear that n ϭ 5 for the P͞G factor ... 3 ... 5 ... 19] ... PT ϭ PA ϩ PG ϭ 100(P͞A,i,8) ϩ 50(P͞G,i,5)(P͞F,i,3)It is important to note that the A͞G factor cannot be used to find anequivalent A value in periods 1 through n for cash flows involving a shifted gradient ... To find the equivalent annual series in years 1 through 10 for the gradient seriesonly, first find the present worth PG of the gradient in actual year 5, take this present worth backto year 0, and annualize the present worthfor 10 years with the A͞P factor ... To find the equivalent A series of a shifted gradient through all the n periods, first find thepresent worth of the gradient at actual time 0, then apply the (A͞P, i, n) factor ... 8182Chapter 3Combining Factors and Spreadsheet FunctionsEXAMPLE 3 ...0123$50$504567$50$70$90$110$130Figure 3–12Diagram of a shifted gradient, Example 3 ... SolutionThe base amount annual series is AB ϭ $50 for all 7 years (Figure 3–13) ... The gradient n is 5 ... P0 ϭ PG(P͞F,i,2) ϭ 20(P͞G,i,5)(P͞F,i,2)Annualize the gradient present worth from year 1 through year 7 toobtain AG ... AT ϭ 20(P͞G,i,5)(P͞F,i,2)(A͞P,i,7) ϩ 50For a spreadsheet solution, enter the original cash flows into adjacent cells, say, B3 throughB10, and use an embedded NPV function in the PMT function ... P0 = ?0PG = ?12AT = ?30$501$50453674Year$50AB = $5025Gradient n$70$90G =$20$110$130Figure 3–13Diagram used to determine A for a shifted gradient, Example 3 ... In Section 2 ... The factor was derived to find thepresent worth in year 0, with A1 in year 1 and the first gradient appearing in year 2 ... Referto Figure 2–21 as a refresher for the cash flows ... 35] is the formula usedfor the factor ... 3 ... 7Chemical engineers at a Coleman Industries plant in the Midwest have determined that asmall amount of a newly available chemical additive will increase the water repellency ofColeman’s tent fabric by 20% ... He expectsthe annual price to increase by 12% per year thereafter for thenext 8 years ... Use i ϭ 15% per year to determine the equivalent total presentworth for all these cash flows ... The total present worth PT is found using g ϭ 0 ... 15 ... 34] and [2 ... PT ϭ 35,000 ϩ A(P͞A,15%,4) ϩ A1(P͞A,12%,15%,9)(P͞F,15%,4)1 Ϫ (1 ... 15)9ϭ 35,000 ϩ 7000(2 ... 5718)0 ... 12ϭ 35,000 ϩ19,985 ϩ 28,247[]ϭ $83,232Note that n ϭ 4 in the (P͞A,15%,4) factor because the $7000 in year 5 is the cash flow of theinitial amount A1 ... If cells B1 through B14are used, the function to find P ϭ $83,230 isNPV(15%,B2:B14)ϩB1The fastest way to enter the geometric series is to enter $7000 for year 5(into cell B6) and setup each succeeding cell multiplied by 1 ... PT = ?0Pg = ?123 450$7000i = 15% per year678910 11 12 13123456789YearGeometricgradient n$7840$35,000$17,33112% increaseper yearFigure 3–14Cash flow diagram including a geometric gradient with g ϭ 12%, Example 3 ...Decreasing arithmetic and geometric gradients are common, and they are often shifted gradient series ... Equivalence computations for present worth P and annual worth A are basically thesame as discussed in Chapter 2, except for the following ... For shifted, decreasing gradients:• The base amount A(arithmetic) or initial amount A1 (geometric) is the largest amount in thefirst year of the series ... • The amount used in the factors is –G for arithmetic and –g for geometric gradient series ... Figure 3–15 partitions a decreasing arithmetic gradient series with G ϭ $−100 that is shifted1 year forward in time ...PT ϭ $800(P͞F,i,1) ϩ 800(P͞A,i,5)(P͞F,i,1) Ϫ 100(P͞G,i,5)(P͞F,i,1)EXAMPLE 3 ... This type of new-technology glass uses electrochrome coating to allow rapid adjustment to sun and dark in building glass, as well as assistingwith internal heating and cooling cost reduction ... All cash flow estimates arein $1000units, and the interest rate expectation is 8% per year ... Years 6 through 10: No new investment and no withdrawals ... 3 ... If the withdrawal series is over- or underfunded, what is the exact amount available inyear 11, provided all other estimates remain the same?Solution by HandFigure 3–16 presentsthe cash flow diagram and the placement of the equivalent P values usedin the solution ... Pg,10 = ?Pg,0 = ?$20,000A1 = $20,000g = Ϫ0 ... 8 ... The present worth in year 0 isPG ϭ Ϫ[7000(P͞A,8%,5) Ϫ 1000(P͞G,8%,5)]ϭ $Ϫ20,577Withdrawal series: Decreasing, shifted geometric series starting in year 12with A1 ϭ $20,000,g ϭ Ϫ0 ... If the present worth in year 10 is identified as Pg,10, thepresent worth in year 0 is Pg,0 ... 35] for the (P͞A,Ϫ20%,8%,5) factor ... 7]}1 ϩ (Ϫ0 ... 08ϭ 20,000 ———————— (0 ... 08 Ϫ (Ϫ0 ... 7750)(0 ... Either additional fundsmust be invested or less must be withdrawn to makethe series equivalent at 8% per year ... 7] and set Pg,0 ϭ ϪPG ϭ 20,577 ... 7750)(0 ... 8586Chapter 3Combining Factors and Spreadsheet Functionsϭ NPV(8%,B4:B18) + B3(a)(b)Figure 3–17Spreadsheet solution, Example 3 ... (a) Cash flows and NPV function and (b) Goal Seek to determine initialwithdrawal amount in year 11 ... To determine if the investment series will cover the withdrawal series, enterthe cash flows (in column B using the functions shown in column C) and apply the NPV function shown in the cell tag to display PT ϭ $ϩ5130 directly ... The Goal Seek tool is very handy indetermining the initial withdrawal amount that resultsin PT ϭ 0 (cell B19) ... Each succeeding withdrawal is 80% of the previous one ... However, nowestablish the entry in B4 as the changing cell ... CHAPTER SUMMARYIn Chapter 2, we derived the equations to calculate the present, future, or annualworth of specificcash flow series ... For example, when auniform series does not begin in period 1, we still use the P͞A factor to find the “present worth”of the series, except the P value is located one period ahead of the first A value, not at time 0 ... With this information, it is possible to solve for P, A, or Ffor any conceivable cash flow series ... Thoughspreadsheet solutions are fast, they do remove some of the understanding of how the time valueof money and the factors change the equivalent value of money ... 1 Industrial Electric Services has a contract with theU ... Embassy in Mexico to providemaintenancefor scanners and other devices in the building ... e ... 2 Civil engineering consulting firms that provide services to outlying communities are vulnerable to anumber of factors that affect the financial conditionof the communities, such as bond issues and realestate developments ... At the end ofthat time, amild recession slowed the development, so the parties signed another contract for $190,000 per yearfor 2 more years ... 3 ... Determine thepresent worth at an interest rate of 12% per year ... Show (a) hand and(b) spreadsheet solutions ... 4 Standby power for water utility pumps andotherelectrical devices is provided by diesel-poweredgenerators ... The utility estimates that by switching to gas,it will save $22,000 per year, starting 3 years fromnow ... 3 ... i = 10% per year012$20045678$90$2003$90Year$90$2003 ... 56 per 1000 gallons ... 28 per 1000 gallons ... ) If FortBliss uses 2billion gallons of water each year,what is the present worth of the discount for a 20year period at an interest rate of 6% per year?3 ... One of the latest schemesfor big-time athletics is the “sports mortgage ... In return, the seats themselveswill stay locked in at current-year prices ... A fan plans to purchasethe sports mortgage alongwith a current-season ticket and pay for both now,then buy a ticket each year for the next 30 years ... 8 The cash flow associated with making self-lockingfasteners is shown below ... 89Income, $1000 20 20 20 20 30 30 30 30 30Cost, $10008 8 8 8 12 12 12 1212Year0123456730253 ... In an effort to pay off the loan quickly, thecompany made four payments in years 1 through4, with each payment being twice as large as thepreceding one ... 10 Revenue from the sale of ergonomic hand toolswas $300,000 in years 1 through 4 and $465,000 inyears 5 through 9... 3 ... They believe that they will need$2,000,000 in year 20 ... They alreadyhave $25,000 in their investment account ... 12 Costs associated with the manufacture of miniature high-sensitivity piezoresistive pressure transducers are $73,000 per year ... Using an interest rate of 10% per year, determine(a)the equivalent annual cost of the manufacturingoperations and (b) the equivalent annual savings inyears 1 through 5 ... 13 Calculate the equivalent annual cost in years 1through 9 of the following series of disbursements ... Show (a)hand and (b) spreadsheet solutions ... 14 For the cash flows below, findthe value of x thatmakes the equivalent annual worth in years 1through 7 equal to $300 per year ... Show solutions (a) by handand (b) using the Goal Seek tool ... 15 Precision Instruments, Inc ... The company borrowed$10,000,000 with the understanding that it wouldmake a $2,000,000 payment at theend of year 1and then make equal annual payments in years 2through 5 to pay off the loan ... 16 A construction management company is examiningits cash flow requirements for the next 7 years ... Specifically, the companyexpects to spend $6000 one year from now, $9000three years from now, and$10,000 six years fromnow ... 17 Find the equivalent annual worth for the cashflows shown, using an interest rate of 12% peryear ... YearCash Flow, $1234567892020202060606060603 ... YearIncome, $͞YearExpense, $͞Year01–45–1007002000Ϫ2500Ϫ200Ϫ3003 ... The clubwill pay $1000 per year plusmake $350,000 inimprovements at the park ... M ... If the clubmakes $100,000 worth of improvements nowand then $50,000 worth each year for the next5 years, what is the equivalent annual cost ofthe lease to the club at an interest rate of 10%per year?3 ... In return,season ticket prices stay locked in atcurrentyear prices, and the package can be sold in thesecondary market, while taking a tax write-offfor donating to a school ... The firstpayment is made now (i ... , beginning-of-yearpayment), and an additional nine payments areto be made at the end of each year for the next9 years ... What is the totalamount of the paymenteach year in years 0 through 9? Use an interestrate of 10% per year ... 21 The expansion plans of Acme Granite, Stone &Brick call for the company to add capacity for anew product in 5 years ... If the company sets aside $55,000 nowand $90,000 in year 2, what uniformannualamount will it have to put in an account in years 3through 5 to have the $360,000? Assume the account earns interest at 8% per year ... 22 For the cash flows shown, calculate the futureworth in year 8 using i ϭ 10% per year ... 24 New actuator element technology enables engineers to simulatecomplex computer-controlledmovements in any direction ... 25 Austin Utilities is planning to install solar panelsto provide some of the electricity for its groundwater desalting plant ... The first phase will cost $4 million inyear 1 and $5 million in year 2 ... Let i ϭ10% per year ... 3 ... i = 10% peryear12345xxx6782x2xYearx2x3003 ... If you make annualdeposits of a uniform amount A into the accountthat earns interest at a rate of 7% per year, howmany years from now will it be until the value ofthe account is equal to 10 times the value of asingle deposit?026100Cash Flow,

$11xYear0345678Random Single Amounts and Uniform Series3 ... The replacement process will cost thecompany $50,000 three years from now ... 3 ... If thecompany spent only $42,000 in year 1, what uniform annual amount should the company expect tospend in each of the next 4 years to expendtheentire budget? Assume the company uses an interest rate of 10% per year ... 30 A recently hired chief executive officer wants toreduce future production costs to improve the company’s earnings, thereby increasing the value of thecompany’s stock ... By how much must annualcosts decrease in years3 through 7 to recover theinvestment plus a return of 12% per year?3 ... Use i ϭ 10% per year ... 27 For the cash flows shown in the diagram, determine the value of x that will make the future worthin year 8 equal to $Ϫ70,000 ... 32 For the following series of income and expenses,find the equivalent valuein year 9 at an interest rateof 12% per year ... 3 ... Year1910YearsIncome, $Expense, $Cash Flow, $13 13 13 13 16 19 22 25 283101–67–910–1609,00028,00038,000Ϫ70,000Ϫ13,000Ϫ14,000Ϫ19,0003 ... The contractprice is $1 ... The payment plan is Z dollarsnow, 2Z dollars in year 2, and 3Z dollars inyears 3through 5 ... 34 A foursome of entrepreneurial electrical engineering graduates has a plan to start a new solar powerequipment company based on STE (solar thermalelectric) technology ... Within the agreement, the loan isto be repaid by allocating 80% of the company’sprofits each year for thefirst 4 years to the investors ... The company’s business plan indicates that they expect to make noprofit for the first year, but in years 2 through 4, theyanticipate profits to be $1 ... If theinvestors accept the deal at an interest rate of 15%per year, and the business plan works to perfection,what is theexpected amount of the last loan payment (in year 5)?Shifted Gradients3 ... Let i ϭ 10% per year ... 372345678A low-cost noncontact temperature measuring toolmay be able to identify railroad car wheels that are inneed of repair long before a costly structural failureoccurs ... 3 ... Year12345678910CashFlow, $90 90 90 85 80 75 70 65 60 553 ... There will be no income in years 1 and 2, but inyear 3 income will be $250,000, and thereafter itwill increase according to an arithmetic gradientthrough year 15 ... 3 ... Use an interest rate of 10% per year ... 41 The cost associated with manufacturinghighperformance lubricants closely follows the cost ofcrude oil ... 4 million inyears 1 through 3, after which the cost increased by3% per year through this year ... e ... Show both (a) hand and (b) spreadsheet solutions ... 42 Find the future worth in year 10 of $50,000 inyear 0 and amounts increasing by15% per yearthrough year 10 at an interest rate of 10% per year ... 43 The cost of tuition at public universities has beensteadily increasing for many years ... 8 million for the first 3 years ... 77 million in year 4,$1 ... What is theequivalent annual worth of the pumping costs overthe 9 years at an interest rateof 12% per year?4 years for all students who finished in the top 3%of their class ... If the tuition for the first 4 years will be$7200 per year and it increases by 5% per year forthe next 5 years, what is the present worth of thetuition cost at an interest rate of 8% per year?3 ... The deal was structured such thattheprivate equity firm received $3 million immediately after the deal was closed (in year 0) throughthe sale of some assets ... 36 million, and it is projected to increaseby 12% each year through expansion of the customer base ... 3 ... Assumei ϭ 10% per year ... 49 Income from the mining of mineraldeposits usually decreases as the resource becomes more difficult to extract ... Use an interest rate of 18% per year ... 3 ... i = 10% per year012345678Year$220Shifted Decreasing Gradients$270$3203 ... Year0CashFlow, $1230 8000 8000 8000ϪG458000Ϫ2G8000Ϫ3G3 ... YearCash Flow, $YearCashFlow, $012385080075070045676506005505003 ... 51 Income from the sale of application software (apps)is usually constant for several years and then decreases quite rapidly as the market gets close tosaturation ... Determine the equivalent annual income in years 1through 7, using an interest rate of10% per year ... 52 Determine the future worth in year 10 of a cashflow series that starts in year 0 at $100,000 anddecreases by 12% per year ... 92Chapter 3Combining Factors and Spreadsheet FunctionsADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS3 ... If the companyplans todeposit money each year, starting now, theequation that represents the deposit each year at8% per year interest is:(a) 1,900,000(A͞F,8%,3)(b) 1,900,000(A͞F,8%,4)(c) 1,900,000 ϩ 1,900,000(A͞F,8%,3)(d) 1,900,000 ϩ 1,900,000(A͞F,8%,2)3 ... 55 For the diagram shown, the respective values of nfor thefollowing equation are:P0 ϭ 100(P͞A,10%,n)(P͞F,10%,n)(a) 6 and 1(b) 6 and 2(c) 7 and 1(d) 7 and 23 ... Assume the interestrate is 8% per year ... 59 The net present worth in year 0 of the followingseries of incomes and expenses at 8% per year isclosest to:YearsIncome, $(a)(c)2345673 ... If the companyplans to deposit money into an account each year for 4 yearsbeginning 2 years from now (first deposit is in year 2)to pay for the expansion, the equation that representsthe amount of the deposit at 9% per year interest is:(a) A ϭ 10,000,000(A͞F,9%,5)(b) A ϭ 10,000,000(A͞F,9%,4)(c) A ϭ10,000,000(A͞P,9%,4)(d) A ϭ 10,000,000(A͞F,9%,4)(P͞F,9%,1)3 ... 60 For the cash flows shown, the equivalent annualworth in periods 1 through 5 at an interest rate of10% per year is closest to:(a)(c)$100Ϫ1000Ϫ100Ϫ200$14,300$16,100i = 10% per year112,00070090001–67–11P0 = ?0Expenses,$$1120$1350(b)(d)$1240$14903 ... 62 In order to have cash available for unforeseenemergencies, Baring Systems, a military contractor, wants to have $2,000,000 in a contingencyfund 4 years from now ... Its mission is toensure the preservation of the natural resources, while providing necessary, butminimal, development for recreationaluse by the public ... A southern U ... state, which has long-term groundwaterproblems, has asked the TPL to manage the purchase of10,000 acres of aquifer recharge land and the development ofthree parks of different use types on the land ... Totalannual purchaseamounts are expected to decrease 25% eachyear through year 5 and then cease for this particular project ... 5 million citizens immediately to the southeast of this acreage relies heavily on the aquifer’s water ... The bond interest rate is an effective 7% per year ... Increases in construction costs areexpected to be $100,000 each year throughyear 6 ... Use the bondissue funds to assist with this purchase ... • Raise the remaining project funds over the next 2 years inequal annual amounts ... Case Study Exercises1 ... If the TPL did agree to fund all costs except the $3 million bond proceeds nowavailable, determine the equivalent annual amount that must be raised in years 4through 6 to supply all remaining project funds ... 3 ... tpl ... Identify someeconomic and noneconomic factors that you believemust be considered when the TPL is deciding to purchase land to protect it from real estatedevelopment ... SECTIONTOPICLEARNING OUTCOME4 ... 4 ... 4 ... 4 ... 4 ... 4 ... 4 ... 4 ... 4 ... In all engineering economy relations developed thus far, the interest rate has beena constant, annual value ... In fact,weekly, daily, and even continuous compounding may be experienced in some projectevaluations ... —have interest rates compounded for a time period shorter than1 year ... This chapter explains how to understand and use nominal and effective interest ratesin engineering practice and in daily life situations ... PEThe Credit Card Offer Case: Today, Davereceived a special offer of a newcreditcard from Chase Bank linked with themajor airline that he flies frequently ... The bonus packageincludes extra airline points (once thefirst purchase is made), priority airportcheck-in services (for 1 year), severalfree checked-bag allowances (for upto 10 check-ins), extra frequent-flyerpoints on theairline, access to airlinelounges (provided he uses the cardon a set time basis), plus several otherrewards (rental car discounts, cruise tripamenities, and floral order discounts) ... Dave has a credit card currently witha bank that he is planning to leave dueto its poor customer service and highmonthly fees... In the page that accompanies theoffer letter, “pricing information” isincluded ... A summary ofseveral of these rates and fees follows ... 24% per year (sum of the current U ... Government prime rate of 3 ... 99%, which is the APR added to determinethe balance transfer APR for Chase Bank)19 ... 99%per year (maximum penalty APR)$85; free the first year$5 or 3% of each transfer, whichever is greater$10 or 3% of each advance, whichever is greater$39 each occurrence, if balance exceeds $250$39 each occurrence$39 each occurrence*All APR rates are variable, based on a current 3 ... 99% addedtodetermine purchase͞balance transfer APR; with 15 ... 99% added to determine penalty APR ... If no minimum payment is receivedwithin 60 days, the penalty APR applies to all outstanding balances and all future transactionson the account ... 1)Effective annual interest rates (4 ... 6)4 ... Here we discussnominal and effective interest rates, whichhave the same basic relationship ... For example, if aninterest rate is expressed as 1% per month, the terms nominal and effective interest rates mustbe considered ... The interest amounts for loans, mortgages, bonds, and stocksare commonly based uponinterest rates compounded more frequently than annually ... In our own personal finances, we managemost cash disbursements and receipts on a nonannual time basis ... First, consider a nominal interest rate ... By definition,r � interest rate per time period � number of periods[4 ... 1] ... 5% per month isthe same as each of the following nominal rates ... 1]What This Is24 months12 months6 months3 months1 ... 5 ϫ 12 ϭ 18%1 ... 5 ϫ 3 ϭ 4 ... ” These nominal rates are calculated in the same way that simple ratesare calculated using Equation [1 ... After the nominal rate has been calculated, thecompounding period (CP) must be included in the interest rate statement ... 5%per month ... 5% per quarter, compounded monthly ... Effective interest rate iAn effective interest rate i is a rate wherein the compounding of interest is taken intoaccount ... The most common form of interest rate statementwhen compounding occurs over time periodsshorter than 1 year is “% per time period, compounded CP-ly,” for example, 10% per year, compounded monthly, or 12% per year, compounded weekly ... If the CP is not mentioned, it is understood to4 ... For example, an interest rate of“1 ... Anequivalenteffective rate statement, therefore, is 1 ... All of the following are effective interest rate statements because either they state they areeffective or the compounding period is not mentioned ... StatementCPWhat This Isi ϭ 10% per yeari ϭ effective 10% per year,compounded monthly1i ϭ 1_% per month2CPnot stated; CP ϭ yearCP stated; CP ϭ monthEffective rate per yearEffective rate per year1i ϭ effective 1_% per month,2compounded monthlyi ϭ effective 3% per quarter,compounded dailyCP not stated; CP ϭ monthEffective rate per monthCP stated; CP ϭ monthEffective rate per month; termseffectiveand compounded monthly are redundantEffective rate per quarterCP stated; CP ϭ dayAll nominal interest rates can be converted to effective rates ... All interest formulas, factors, tabulated values, and spreadsheet functions must use an effectiveinterest rate to properly account for the time valueof money ... This is the same as the nominal rate ... 25% on a monthly basis ... This is the same as an effective rate ... As we will learn in the following sections, the effectiverate is always greater than or equal to the nominal rate, and similarly APY Ն APR ... Interest period (t)—The period of time overwhich the interest is expressed ... The time unit of 1 year isby far the most common ... Compounding period (CP)—The shortest time unit over which interest is charged or earned ... If CP is not stated, it is assumed to be the same as the interestperiod ... If the compounding period CP and the time period tare the same, the compounding frequency is 1, for example, 1% per month, compounded monthly ... It has an interest period t of1 year, a compounding period CP of 1 month, and a compounding frequency m of 12 times peryear ... In previous chapters, all interest rates had t and CP values of 1 year, sothe compounding frequency was always m ϭ 1 ... Now, it will be necessary to express a nominal rate as aneffective rate on the same time base as the compounding period ... 2]As an illustration, assume r ϭ 9% per year, compounded monthly; then m ϭ 12 ... 2] is used to obtain the effective rate of 9%͞12 ϭ0 ... 9798Chapter 4Nominal and Effective Interest RatesNote that changing the interest period t does not alter the compounding period, which is 1 monthin this illustration ... 5% per 6 months,compounded monthly, are two expression of the same interest rate ... 1Three different bank loan rates for electricgeneration equipment are listed below ... (a) 9% per year, compounded quarterly ... (c) 4 ... SolutionApply Equation [4 ... The graphic in Figure 4–1 indicates the effective rate per CP and how the interest rate is distributed over time ... 25%Quarter(b) 9% peryearCompoundingFrequency (m)Month()Distribution over Time Period t4 ... 75%0 ... 75%112 ... 75%3 ... 75%1 ... 25% ... 25% ... 25% ... 25% ... 173%(c) 4 ... 173%112 14 1626 WeekFigure 4–1Relations between interest period t, compounding period CP, and effective interest rate per CP ... Basicallythere are three ways to express interestrates, as detailed in Table 4–1 ... For the first format, a nominal interest rate is given and thecompounding period is stated ... In the second format, the stated rate is identified as effective (or APY could also be used),so the rate is used directly in computations ... This rateis effective over a compoundingperiod equal to the stated interest period of 1 year in this case ... TABLE 4–1Various Ways to Express Nominal and Effective Interest RatesFormat of Rate StatementExample of StatementWhat about the Effective Rate?Nominal rate stated,compounding period stated8% per year,compoundedquarterlyFind effective rate for any timeperiod (next two sections)Effective rate statedEffective 8 ... 243% per yeardirectly for annual cash flowsInterest rate stated, nocompounding period stated8% per yearRate is effective for CP equal to statedinterest period of 1 year; find effective rate forall other time periods4 ... 2 The Credit Card Offer CaseAs described in the introduction to this case, Dave has been offered what is described as a creditcard deal that should not be refused—at least that is what the Chase Bank offer letter implies ... 24% is an annual rate, with no compounding periodmentioned ... Therefore, we should conclude that the CP is 1 year, the same as the annual interestperiod of the APR ... (a) First, determine the effective interest rates for compounding periods of 1 year and1 month so Dave knows some effective rates he might be paying when he transfers the$1000balance from his current card ... What is the amount of the total balance he owes?Now, Dave looks a little closer at the fine print of the “pricing information” sheet and discoversa small-print statement that Chase Bank uses the daily balance method (including new transactions) to determine the balanceused to calculate the interest due at payment time ... Solution(a) The interest period is 1 year ... 2] for both CP values of 1 year (m ϭ 1compounding period per year) and 1 month (m ϭ 12 compounding periods per year) ... 24͞1 ϭ 14 ... 24͞12 ϭ 1 ... Amount owed after 1 month ϭ 1000 ϩ 1000(0 ... 03(1000)ϭ1000 ϩ 11 ... 87Including the $30 fee, this represents an interest rate of (41 ... 187%for only the 1-month period ... 2], now with m ϭ 365 compounding periods per year ... 24͞365 ϭ 0 ... 2 Effective Annual Interest RatesIn this section, effective annual interest rates are calculated ... For example,we will learnthat a nominal 18% per year, compounded quarterly is the same as an effective rate of19 ... The symbols used for nominal and effective interest rates arer ϭ nominal interest rate per yearCP ϭ time period for each compoundingm ϭ number of compounding periods per yeari ϭ effective interest rate percompounding period ϭ r͞mia ϭ effective interest rate per yearThe relation i ϭ r͞m is exactly the same as Equation [4 ... 99PE100Chapter 4Nominal and Effective Interest RatesFigure 4–2P(1 + i)m = P(1 + ia)Future worth calculationat a rate i, compounded mtimes in a year ... Like compound interest, aneffective interest rate at any pointduring the year includes (compounds) the interest rate for all previous compounding periodsduring the year ... We set P ϭ $1 forsimplification ... Since interest may be compounded several times during the year, use the effective annualrate symbol ia to write the relation forF with P ϭ $1 ... The effective rate i per CP must be compounded through all m periodsto obtain the total effect of compounding by the end of the year ... The effective annual interest rate formulafor ia isia � ) 1 ؉ i)m 3 ... 4]1 ؊ ] calculates the effective annual interest rate ia for any number ofcompoundingperiods per year when i is the rate for one compounding period ... 3] can besolved for i to determine the effective interest rate per compounding period ... 4]As an illustration, Table 4–2 utilizes the nominal rate of 18% per year for different compounding periods (year to week) to determine theeffective annual interest rate ... Table 4–3 summarizes the effective annualrate for frequently quoted nominal rates using Equation [4 ... A standard of 52 weeks and365 days per year is used throughout ... 8 ... 346151 ... 5918Rate perCompoundPeriod, i%9%24 ... 34615% in each51 ... 5%944 ...5%29%118%Distribution of i over the Year ofCompounding Periodsr ϭ 18% per year, compounded CP-lyEffective Annual Interest Rates Using Equation [4 ... 0034615)52 Ϫ 1 ϭ 19 ... 015)12 Ϫ 1 ϭ 19 ... 045)4 Ϫ 1 ϭ 19 ... 09)2 Ϫ 1 ϭ 18 ... 18)1 Ϫ 1 ϭ 18%Effective AnnualRate, ia � ) 1 ؉ i)m 1021 ؊ Chapter4Nominal and Effective Interest RatesTABLE 4–3 Effective Annual Interest Rates for Selected Nominal RatesNominalRate r %Semiannually(m � 2(Quarterly(m � 4(Monthly(m � 21(Weekly(m � 25(Daily(m � 563( ... 1238 ... 0636 ... 0234 ... 5062 ... 5011 ... 50234567891012151820253040500 ... 501 ... 009328 ... 15819 ... 50612 ... 3229 ... 1807 ... 0795 ... 0203 ... 0051 ... 2500 ... 21363 ... 07334 ... 56221 ... 68316 ... 38110 ... 2298 ... 1166 ... 0424 ... 5102 ... 5011 ... 1810 ... 54746 ... 55127 ... 86519 ... 38112 ... 2439 ... 1367 ... 0605 ... 0153 ... 0041 ... 2500 ... 00056 ... 56332 ... 81021 ... 36015 ... 20310872 ... 98649 ... 14028 ... 18319 ... 51712 ... 3299 ... 1847 ... 0815 ... 0203 ... 0051 ... 2500 ... 15064 ... 39034 ... 71422 ... 74516 ... 41710 ... 2478 ... 1266 ... 0454 ... 5112 ... 5011 ... 4790 ... 86948 ... EXAMPLE 4 ... She purchased Southwest stock for $6 ... 14 per share ... Help Janice understand exactlywhat she earned in terms of (a) effective annual rate and(b) effective rate for quarterly compounding, and for monthly compounding ... Solution(a) The effective annual rate of return ia has a compounding period of 1 year, since thestock purchase and sales dates are exactly 1 year apart ... 90 per shareand using the definition of interest rate in Equation [1 ... 24ia ϭ ———————————— ϫ 100% ϭ —— ϫ 100% ϭ 90 ... 90(b) For the effective annual rates of 90 ... 43%,compounded monthly, apply Equation [4 ... Quarter:Month:m ϭ 4 times per year1͞4i ϭ (1 ... 17472 Ϫ 1 ϭ 0 ... 472% per quarter,compounded quarterly ... 9043) Ϫ 1 ϭ 1 ... 05514This is 5 ... CommentNote that these quarterly and monthly rates are less than the effective annual rate divided bythe number of quarters or months per year ... 43%͞12 ϭ7 ... This computation is incorrect because it neglects the fact that compoundingtakesplace 12 times during the year to result in the effective annual rate of 90 ... The spreadsheet function that displays the result of Equation [4 ... The format is4 ... 5]Note that the rate entered in the EFFECT function is the nominal annual rate r% per year, notthe effective rate i% per compounding period ... 3]... 25% per year, compoundedquarterly, and you want to find the effective annual rate ia ... 25%,4)to display ia ϭ 5 ... This is the spreadsheet equivalent of Equation [4 ... 25͞4 ϭ 1 ... ia ϭ (1 ϩ 0 ... 05354(5 ... The NOMINAL spreadsheet function finds the nominal annual rate r ... 6]This function is designedto display only nominal annual rates ... For example, if the effective annual rateis 10 ... 381%,4) to display r ϭ 10% per year, compounded quarterly ... 1] ... 5% ... EXAMPLE 4 ... 24% per year, or 1 ... He will transfera balance of $1000 and plans to pay it and the transfer fee of $30, due at the end of thefirstmonth ... Dave accepts the employment offer, and in his hurried, excited departure, he forgets to send the creditcard service company a change of address ... 2 to be $1041 ... (a) If this situation continues for a total of 12 months, determine the total due after 12 monthsand the effective annual rate ofinterest Dave has accumulated ... 99% per year after one latepayment of the minimum payment amount, plus a late payment fee of $39 per occurrence ... Solution(a) Because Dave did not pay the first month’s amount, the new balance of $1041 ... 99%͞12 ϭ 2 ... The first3 months and last 2 months aredetailed below ... PEChapter 4Nominal and Effective Interest Ratesϭ SUM(B6:D6)ϭ J4/12ϭ K4/12Month 1: transfer feeMonths 2-12: late-payment feeInterest rate per monthMonth 1: 1 ... 499%Figure 4–3Monthly amounts due for a credit card, Example 4 ... Month 1:Month 2:Month 3:1000 ϩ 1000(0 ...871041 ... 87(0 ... 911106 ... 91(0 ... 57 ... 25 ϩ 1689 ... 02499) ϩ 39 ϭ $1770 ... 46 ϩ 1770 ... 02499) ϩ 39 ϭ $1853 ... 71 after 12 periods ... 71 ϭ 1000(F͞P, i,12) ϭ 1000(1 ϩ i)1͞121 ϩ i ϭ (1 ... 05278i ϭ 5 ... 3] to determine the effectiveannual rate of 85 ... 12ia ϭ (1 ϩ i )m Ϫ 1 ϭ (1 ... 85375(85 ... 24% (or1 ...207% per year, compounded monthly ... 3], with a smallrounding error included,ia ϭ (1 ϩ i)m Ϫ 1ϭ (1 ... 15207First, Dave will not pay at the stated rate of 14 ... 207% ... 99% and (2) the monthlyfees of $39 for not making a payment ... 99% per year ... 207% to 85 ... CommentThis is but one illustration ofwhy the best advice to an individual or company in debt is tospend down the debt ... 4 ... 3] is applied to find ia the result is usually not an integer ... There arealternative ways to find the factor value ... • Use the spreadsheet function with ia (as discussed in Section 2 ... • Linearly interpolate between twotabulated rates (as discussed in Section 2 ... 4 ... 3] in Section 4 ... We can generalize this equation to determine the effective interestrate for any time period (shorter or longer than 1 year) ... 7]where i � effective rate for specified time period (say, semiannual)r � nominal interest rate for same time period(semiannual)m � number of times interest is compounded per stated time period (times per6 months)The term r͞m is always the effective interest rate over a compounding period CP, and m isalways the number of times that interest is compounded per the time period on the left of theequals sign inEquation [4 ... Instead of ia, this general expression uses i as the symbol for theeffective interest rate, which conforms to the use of i throughout the remainder of this text ... 5 and 4 ... EXAMPLE 4 ... An engineer is on aTesla committee to evaluate bids for new-generation coordinate-measuring machineryto bedirectly linked to the automated manufacturing of high-precision vehicle components ... To get a clear understanding of finance costs, Tesla management asked the engineer to determine the effectivesemiannual and annual interest rates for each bid ... 8% per year, compounded monthly(a)Determine the effective rate for each bid on the basis of semiannual periods ... (c) Which bid has the lowest effective annual rate?Solution(a) Convert the nominal rates to a semiannual basis, determine m, then use Equation [4 ... For bid 1,r ϭ 9% per year ϭ 4 ... 045 2Effective i% per 6 months ϭ 1 ϩ——— Ϫ 1 ϭ 1 ... 55%2Table 4–4 (left section) summarizes the effective semiannual rates for all three bids ... 5Semiannual RatesAnnual RatesBidNominal r per6 Months, %CP per6 Months, mEquation [4 ... 7],Effective i, %1234 ... 04 ... 556 ... 489128 ... 3112 ... 16(b) For the effective annual rate, thetime basis in Equation [4 ... For bid 1,r ϭ 9% per yearm ϭ 4 quarters per year()0 ... 0931 Ϫ 1 ϭ 9 ... (c) Bid 3 includes the lowest effective annual rate of 9 ... 48% when interest is compounded monthly ... 6A dot-com company plans to place money in a new venture capital fund that currently returns18% peryear, compounded daily ... 7], with r ϭ 0 ... 0 ... 716%Effective i% per year ϭ 1 ϩ ——365(b) Here r ϭ 0 ... 0 ... 415%Effective i% per 6 months ϭ 1 ϩ ——182()()4 ... The payment period (PP) is the length of time between cash flows (inflows or outflows) ... It is important to determine if PP ϭ CP, PP Ͼ CP, orPP Ͻ CP ... These time periods areshown in Figure 4–4 ... r = nominal 8% per year, compounded semiannuallyCP6 monthsFigure 4–4One-year cash flow diagram for a monthly payment period (PP) andsemiannual compoundingperiod (CP) ... 5TA BLE 4–5 Section References for EquivalenceCalculations Based onPayment Period and Compounding Period ComparisonLengthof TimeInvolvesSingle Amounts(P and F Only)Involves Uniform Seriesor Gradient Series(A, G, or g)PP ϭ CPPP > CPPP < CPSection 4 ... 5Section 4 ... 6Section 4 ... 7As we learned earlier, to correctly performequivalence calculations, an effective interest rateis needed in the factors and spreadsheet functions ... The next three sections (4 ... 7) describe procedures to determine correct i and n values for engineering economy factorsand spreadsheet functions ... Table 4–5 provides the section reference ... 5 and4 ... A general principle to remember throughout these equivalence computations is that whencash actually flows, it is necessary to account for the time value of money ... After 3months there is no cash flow and no need to determine the effect of quarterly compounding ... 4 ... In virtuallyall situations, PPwill be equal to or greater than CP ... If the rate is 8% per year, for example, PP ϭ CP ϭ 1 year ... The procedures to perform equivalence computations are the same forboth situations, as explained here ... Method 1 is easier to apply, because the interest tablesin the back of the text can usually provide thefactor value ... For spreadsheets, either method is acceptable; however, method 1 is usually easier ... The relations to calculate P andF areP ϭ F(P͞F, effective i% per CP, total number of periods n)[4 ... 9]For example, assume that the stated credit card rate is nominal 15% per year, compoundedmonthly ...To find P or F over a 2-year span, calculate the effective monthlyrate of 15%͞12 ϭ 1 ... Then 1 ... Any time period can be used to determine the effective interest rate; however, the interest ratethat is associated with the CP is typically the best because it is usually a whole number ... Method 2: Determine theeffective interest rate for the time period t of the nominal rate, andset n equal to the total number of periods, using this same time period ... 8] and [4 ... For a credit card rate of 15% per year, compounded monthly, thetime period t is 1 year ... 15Effective i% per year ϭ 1 ϩ ——12)12Ϫ 1 ϭ 16 ... 25%,24) ϭ 0... 076%,2) ϭ 0 ... EXAMPLE 4 ... The company sells compost produced by garbage-to-compost plants inthe United States and Vietnam ... Find theamount in the account now (after 10 years) at an interest rate of 12% per year, compoundedsemiannually ... Both methods are illustrated to find F in year 10... There are n ϭ (2)(number of years) semiannual periods for each cash flow ... 9] isF ϭ 1000(F͞P,6%,20) ϩ 3000(F͞P,6%,12) ϩ 1500(F͞P,6%,8)ϭ 1000(3 ... 0122) ϩ 1500(1 ... 634 million)Express the effective annual rate, based on semiannual compounding ... 12 2Effective i% per year ϭ 1 ϩ —— Ϫ 1 ϭ 12 ...Use the factor formula (F͞P,i,n) ϭ (1 ... 9] to obtain the same answer as above ... 36%,10) ϩ 3000(F͞P,12 ... 36%,4)ϭ 1000(3 ... 0122) ϩ 1500(1 ... 634 million)F=?01234$100056$1500$3000Figure 4–5Cash flow diagram, Example 4 ... 78910YearEquivalence Relations: Series with PP Ն CP4 ... 03%,24)CashFlow SeriesInterest Rate$500 semiannuallyfor 5 years$75 monthly for3 years$180 quarterly for15 years$25 per monthincrease for4 years$5000 per quarterfor 6 years4 ... This also establishes the time unit of the effective interest rate ... Then value is the total number of quarters ... This is a directapplication of the following general guideline:When cash flows involve a series (i ... , A, G, g) and the payment period equals or exceeds thecompounding period in length:••Find the effective i per payment period ... In performing equivalence computations for series, only these values of i and n can be usedininterest tables, factor formulas, and spreadsheet functions ... Table 4–6 shows the correct formulation for several cash flow series and interest rates ... EXAMPLE 4 ... What is the equivalent total amount after the last payment, if these funds are takenfrom a pool that has been returning 8% per year,compounded quarterly?SolutionThe cash flow diagram is shown in Figure 4–6 ... Applying the guideline, we need to determine an effective semiannual interest rate ... 7] with r ϭ 4% per 6-monthperiod and m ϭ 2 quarters per semiannual period ... 04 2Effective i% per 6 months ϭ 1 ϩ —— Ϫ 1 ϭ 4 ... 04% ...04% seems reasonable, since we expect the effective rate to be slightlyhigher than the nominal rate of 4% per 6-month period ... The relation for F isF ϭ A(F͞A,4 ... 3422)ϭ $9171 ... 3422 using a spreadsheet, enter the FV function fromFigure 2–9, that is, ϭ ϪFV(4 ... Alternatively, the final answer of $9171... 04%,14,500) ... 8 ... 9 Credit Card Offer CaseIn our continuing credit card saga of Dave and his job transfer to Africa, let’s assume he didremember that the total balance is $1030, including the $30 balance transfer fee, and he wantsto set up a monthly automatic checking account transfer to pay off theentire amount in 2 years ... 99% per year ... SolutionThe monthly A series is needed for a total of n ϭ 2(12) ϭ 24 payments ... 24%͞12 ϭ 1 ... Solution by hand:Use a calculator or hand computation to determine the A͞P factor value ... 187%,24) ϭ 1030(0 ... 57 per month for 24 monthsSolution bySpreadsheet: Use the function ϭ ϪPMT(1 ... 04813 to determine A for n ϭ 24 payments ... 187%,24,1030) to directly display the required monthly payment of A ϭ $Ϫ49 ... The effective annual interest rate or APY is computed using Equation [4 ... 24%per year, compounded monthly, and m ϭ 12 times peryear ... 1424Effective i per year ϭ 1 ϩ ———12)12– 1 ϭ 1 ... 207% per yearThis is the same effective annual rate ia determined in Example 4 ... Equivalence Relations: Series with PP Ն CP4 ... 10The Scott and White Health Plan (SWHP) has purchased a robotized prescription fulfillmentsystem for fasterand more accurate delivery to patients with stable, pill-form medication forchronic health problems, such as diabetes, thyroid, and high blood pressure ... The expected life is 10 years ... Find this semiannual A valueboth by hand and by spreadsheet, if capital funds are evaluated at 8% per year, using twodifferent compounding periods:Rate 1 ... Rate 2 ... SolutionFigure 4–7 shows the cash flow diagram ... This pattern makes the solution by hand quite involved if the P͞F factor, not the P͞Afactor, is used to find P for the 10 annual $200,000 costs ... Solution by hand—rate 1: Steps to find the semiannual Avalue are summarized below:PP ϭ CP at 6 months; find the effective rate per semiannual period ... Number of semiannual periods n ϭ 2(10) ϭ 20 ... , 20 periods because the costs are annual,not semiannual ... [⌺20P ϭ 3,000,000 ϩ 200,000(P͞F,4%,k)kϭ2,4]ϭ 3,000,000 ϩ 200,000(6 ... 24681234A per 6months = ?101214161820891006 months567$200,000 per yeari1 = 8%, compounded semiannuallyi2 = 8%, compounded monthlyP = $3 millionFigure 4–7Cash flow diagram with two different compounding periods, Example 4 ... Years111112Chapter 4Nominal and Effective Interest RatesSolution byhand—rate 2: The PP is 6 months, but the CP is now monthly; therefore, PP Ͼ CP ... 7] is applied withr ϭ 4% and m ϭ 6 months per semiannual period ... 04Effective semiannual i ϭ 1 ϩ ———6P ϭ 3,000,000 ϩ 200,000[)6Ϫ 1 ϭ 4 ... 067%,k)kϭ2,4]ϭ 3,000,000 ϩ 200,000(6 ... 067%,20) ϭ $320,064Now,$320,064, or $1286 more semiannually, is required to cover the more frequent compounding of the 8% per year interest ... 067% ... Solution by spreadsheet—rates 1 and 2: Figure 4–8 presents a general solution for the problemat both rates ... They continue thecash flow pattern of $200,000 every other 6months through cell B32 ... This allows somesensitivity analysis to be performed for different PP and CP values ... This technique works well for spreadsheets once PP and CP are entered in the time unit of the CP ... The final A values in D14 ($318,784) and F14 ($320,069) arethe same (except forrounding) as those above ... 10 ... 7 Equivalence Relations: Single Amountsand Series with PP Ͻ CPIf a person deposits money each month into a savings account where interest is compoundedquarterly, do all the monthly deposits earn interest before the next quarterly compounding time?If a person'scredit card payment is due with interest on the 15th of the month, and if the full payment is made on the 1st, does the financial institution reduce the interest owed, based on earlypayment? The usual answers are no ... 7compounded, bank loan were made early by a large corporation, the corporatefinancial officerwould likely insist that the bank reduce the amount of interest due, based on early payment ... The timing of cash flow transactions between compoundingpoints introduces the question of how interperiod compounding is handled ... For a no-interperiod-interest policy, negative cash flows(deposits or payments, depending on theperspective used for cash flows) are all regarded as made at the end of the compounding period,and positive cash flows (receipts or withdrawals) are all regarded as made at the beginning ... This procedure can significantly alter the distribution of cashflows beforethe effective quarterly rate is applied to find P, F, or A ... 5 and 4 ... Example 4 ... Of course, noneconomic factors may be present ... 11Last year AllStar Venture Capital agreed to invest funds in Clean Air Now (CAN), a start-upcompany in Las Vegas that is an outgrowth of research conducted inmechanical engineering atthe University of Nevada–Las Vegas ... The venture fund managergenerated the cash flow diagram in Figure 4–9a in $1000 units from AllStar’s perspective ... The receipts were unexpected this first year; however, the product has great promise,and advance orders have comefrom eastern U ... plants anxious to become zero-emissioncoal-fueled plants ... How much is AllStar in the “red” at the end of the year?Receipts from CAN$90$90$120$45100123456$758910$501112Month$100$1507YearPayments toCAN$200(a)$180010123$165245637894101112$50QuarterMonth$150$175$200F=?(b)Figure 4–9(a) Actual and (b) moved cash flows (in $1000) for quarterly compounding periods using no interperiodinterest, Example 4 ... 113114Chapter 4Nominal and Effective Interest RatesSolutionWith nointerperiod interest considered, Figure 4–9b reflects the moved cash flows ... Calculatethe F value at 12%͞4 ϭ 3% per quarter ... If PP Ͻ CP and interperiod compounding is earned, then the cash flows are not moved, and theequivalent P, F, or A values are determined using the effective interest rate perpayment period ... The effective interest rate formula will have an m value less than 1, because there isonly a fractional part of the CP within one PP ... When the nominal rate is 12% per year, compoundedquarterly (the same as 3% per quarter, compounded quarterly), the effective rate per PP isEffectiveweekly i% ϭ (1 ... 228% per week4 ... Continuous compounding is present when the duration of CP, the compounding period, becomesinfinitely small and m, the number of times interest is compounded per period, becomes infinite ... As m approaches infinity, the effective interest rate Equation [4 ... First,recall the definition of the natural logarithm base ... 71828ϩhϱh[4 ... 7] as m approaches infinity is found by using r͞m ϭ 1͞h, which makesm ϭ hr ... 11]Equation [4 ... As an illustration, if the nominal annual r ϭ 15% per year, the effectivecontinuous rate per year isi% ϭ e0 ... 183%For convenience, Table 4–3includes effective continuous rates for the nominal rates listed ... 5] or [4 ... A value of 10,000 or higher provides sufficientaccuracy ... 12 ... 8Effective Interest Rate for Continuous CompoundingEXAMPLE 4 ... (b) An investor requires an effective return of at least 15% ... 5%, or 0 ... By Equation [4 ... 015 −1 ϭ 1 ... 18 per year isi% per year ϭ er − 1 ϭ e0 ... 722%(b) Solve Equation [4 ... er − 1 ϭ 0 ... 15ln er ϭ ln 1 ... 13976Therefore, a rate of 13 ... The general formula to find the nominal rate, given the effectivecontinuous rate i, is r ϭ ln(1 ϩ i) ... 5% and annual rater ϭ 18% with a large m to display effective ivalues ... 5%,10000)ϭ EFFECT(18%,10000)effective i ϭ 1 ... 722% per year(b) Use the function in Equation [4 ... 976% per year, compounded continuously ... 13Engineers Marci and Suzanne both invest $5000 for 10 years at 10% per year ... SolutionMarci: For annual compounding the future worth isF ϭP(F͞P,10%,10) ϭ 5000(2 ... 11], first find the effective i per year for use in the F͞P factor ... 10 Ϫ 1 ϭ 10 ... 517%,10) ϭ 5000(2 ... For comparison, daily compounding yields an effective rate of 10 ... 517% for continuous compounding ... Examples of costs are energy and water costs, inventory costs, andlabor costs ... In these cases, the economicanalysis can be performed for continuous cash flow (also called continuous funds flow) and thecontinuous compounding of interest as discussed above ... In fact, the monetary differences for continuous cash flows relativeto the discrete cash flow and discretecompounding assumptions are usually not large ... 4 ... Loan rates may increase from one year to another ... The mortgage rate is slightly adjusted annually to reflect the age of the loan, the current cost of mortgagemoney, etc ... If the variation in i is large, the equivalent valueswill vary considerably fromthose calculated using the constant rate ... To determine the P value for future cash flow values (Ft) at different i values (it) for each year t,we will assume annual compounding ... Using standard notation and the P͞F factor,P � F1(P͞F,i1,1) ؉ F2(P͞F,i1,1)(P͞F,i2,1) ؉ ... (P͞F,in,1)[4 ... 12] is the expression forthe present worth of the future cash flow ... (P͞F,in,1)[4 ... Since the equivalent P has beendetermined numerically using the varying rates, this new equation will have only one unknown,namely, A ... 14 illustrates this procedure ... 14CE, Inc ... The net profit from the equipment for each ofthe last 4 yearshas been decreasing, as shown below ... The return has been increasing ... Take the annual variation of rates of return intoaccount ... Equation [4 ... Since for both years 1 and 2 the net profit is $70,000 and theannual rate is 7%, the P͞A factor can be used for these 2 years only ... 8080) ϩ 35(0 ... 7284)](1000)ϭ $172,816[4 ... 14 ... 14], set it equal to P ϭ $172,816, and solve for A ... See Figure 4–10 for the cash flow diagramtransformation ... 8080) ϩ (0 ... 7284)] ϭ A[3 ... 25% is used, the result is A ϭ $52,467 ... When there is a cash flow in year 0 and interest rates vary annually, this cash flow mustbeincluded to determine P ... This is accomplished byinserting the factor value for (P͞F,i0,0) into the relation for A ... 00 ... In this case, theA value is determined using the F͞P factor, and the cash flow in year n is accounted for by including the factor (F͞P,in,0) ϭ 1 ... CHAPTER SUMMARYSince many real-world situations involve cash flow frequencies and compounding periods otherthan 1 year, it is necessary to use nominal and effective interest rates ... r mEffective i � ) 1 ؊ ( — ؉ 1mYear118Chapter 4Nominal and Effective Interest RatesThe m is the number of compounding periods (CP) per interestperiod ... All engineering economy factors require the use of an effective interest rate ... If only single amounts (P and F)are present, there are several ways to perform equivalence calculations using the factors ... This requires that the relative lengths of PPand CP be considered as i and n are determined... From one year (or interest period) to the next, interest rates will vary ... PROBLEMSNominal and Effective Rates4 ... 4 ... 4 ... 4 ... 4 ... 2% per year, compounded monthly; (d) effective3 ... Given Interest RateDesired Interest Rate1% per month3% per quarter2% per quarter0 ... 1% per 6 monthsNominalrate per yearNominal rate per 6 monthsNominal rate per yearNominal rate per quarterNominal rate per 2 years4 ... 5% per year, compounded monthly?4 ... 4 ... Determine the APY ... 4 ... 87% compounded quarterly, determine (a) the effectivequarterly rate and (b) the nominal annual rate ... 6 Identify thefollowing interest rate statements aseither nominal or effective: (a) 1 ... 6% per month, compounded weekly; (e) 0 ... 4 ... A less well-known company inthe chip business has been growing fast enoughthat the company uses a minimum attractive rate ofreturn of 60% per year ... 4 ... (Assume 4weeks͞month ... 13 An interest rate of 21% per year, compoundedevery 4 months, is equivalent to what effective rateper year? Show hand and spreadsheet solutions ... 14 An interest rate of 8% per 6 months, compoundedmonthly, is equivalent to what effective rate perquarter?4 ... The owner wasconfused by the terminology and asked you to help ... 16 In ‘N Out Payday Loans advertises that for a fee ofonly $10, you can immediately borrow up to $200for one month ... 17 A government-required truth-in-lending documentshowed that the APR was 21% and the APY was22 ... Determine thecompounding frequency atwhich the two rates are equivalent ... 18 Julie has a low credit rating, plus she was furloughed from her job 2 months ago ... Since she is a littleshort on money to pay her rent, she decided to borrow $100 from a loan company, which will chargeher only $10 interest if the $110 ispaid no morethan 1 week after the loan is made ... 19 Assume you deposit 25% of your monthly checkof $5500 into a savings account at a credit unionthat compounds interest semiannually ... 20 Interest is compounded quarterly, and singlepayment cash flows (that is, F and P) are separatedby 5 years ...21 When interest is compounded quarterly and auniform series cash flow of $4000 occurs every6 months, what time periods on i and n must beused?4 ... If the replacement is expected to take119place in 3 years, how much will the company havein its investment set-aside account? Assume thecompanycan achieve a rate of return of 12% peryear, compounded quarterly ... 23 Wheeling-Pittsburgh Steel is investigating whetherit should replace some of its basic oxygen furnaceequipment now or wait to do it later ... e ... How much can the company afford tospend now, if its minimum attractive rate ofreturnis 1 ... 24 How much can Wells Fargo lend to a developerwho will repay the loan by selling 6 view lots at$190,000 each 2 years from now? Assume thebank will lend at a nominal 14% per year, compounded semiannually ... 25 How much will be in a high-yield account at theNational Bank of Arizona12 years from now ifyou deposit $5000 now and $7000 five years fromnow? The account earns interest at a rate of 8% peryear, compounded quarterly ... 26 Loadstar Sensors is a company that makes load͞force sensors based on capacitive sensing technology ... If the companyhas already set aside $12million in an investmentaccount for the expansion, how much more mustthe company add to the account next year (i ... ,1 year from now) so that it will have the $28 million 4 years from now? The account earns interestat 12% per year, compounded quarterly ... 27 A structural engineering consultingcompany isexamining its cash flow requirements for the next6 years ... Specifically, the company expects to spend $21,000 twoyears from now, $24,000 three years from now,and $10,000 five years from now ... 28 Irvin Aerospace of Santa Ana, California, wasawarded a 5-year contract to develop anadvancedspace capsule airbag landing attenuation systemfor NASA’s Langley Research Center ... What is the present worth of thecontract at 16% per year, compounded quarterly, ifthe quarterly cost in years 1 through 5 is $2 millionper quarter?120Chapter 44 ... (A drillcollar is the heavy tubularconnection between adrill pipe and a drill bit ... 30 In 2010, the National Highway Traffic Safety Administration raised the average fuel efficiency standard to 35 ... The rules will costconsumers an average of $926 extra per vehicle inthe 2016 model year ... How much will the monthly savings in thecost ofgasoline have to be to recover Yolanda’sextra cost? Use an interest rate of 0 ... 4 ... S ... 3 million of the $87 million capital cost for adesalting plant constructed and operated by El PasoWater Utilities (EPWU) ... 85 per 1000 gallonsfor 20 years ... 3 million capital costis amortized at an interest rate of 6%per year,compounded monthly ... 32 Beginning in 2011, city hall, administrative offices, and municipal courts in the city of El Paso,Texas, will go on a 10 hour͞day, 4-day workweekfrom the beginning of May through the end of September ... If this workschedule continues for the next 10 years, what isthefuture worth of the savings at the end of thattime (i ... , end of year 2020)? Use an interest rateof 0 ... 4 ... Prices ranged from $999 for steel models such asDad Remembered to $3199 for the Sienna Bronzecasket ... An individual purchased a SiennaBronze casket and made 12 equal monthly payments(in months 1 through 12) at no interest ... 34 NRG Energy plans to construct a giant solar plantin Santa Teresa, New Mexico, to supply electricityto West Texas Electric ... It will provide enoughpower for 30,000 homes ... 35 Many college students have Visa credit cards thatcarry an interest rate of “simple24% per year”(that is, 2% per month) ... 25 ... 25 per month andadds no other charges to the card?4 ... S ... After working for a year or so,he found himself in financial trouble, and he borrowed $500 from a friend in the finance department at his office ... The twogot separated doing different jobs, and 1year wentby ... (a) What does Bart now owe his friend? (b) Whateffective annual interest rate did Bart pay on this$500 loan?4 ... Thefee will be reduced by $10 each month of the2-year contract ... Assume AT&T pays $499 for aniPhone that it sells for $199 ... e ... 5% per month on its $300 investment in acustomer who terminates the contractafter 12 months?4 ... 5 years from now at aninterest rate of 2% per month?Problems4 ... Thecompany plans to borrow $3 ... Howmuch will the company have to get in a lump-sumpayment when the project is over in order to earn24% per year, compounded quarterly,on its investment?4 ... Year12345Cash Flow, $200,0000350,0000400,0004 ... F=?i = 1% per month012345$30$30$30$30$306$50$508 Years7$30$50121Lee antiscalant) for use at its nanofiltration waterconditioning plant ... If the chemical cost is $11 per day,determine the equivalent cost per month atan interest rate of 12% per year, compounded monthly ... 4 ... S ... What isthe future worth of the income (after the 2½ years)at an interest rate of 6% per year, compoundedquarterly? Assume there is no interperiod compounding ... 47 The Autocar E3 refuse truck has an energy recovery systemdeveloped by Parker HannifinLLC that is expected to reduce fuel consumptionby 50% ... (The truck recharges the accumulators when itbrakes ... How much can a private wastehauling company afford to spend now on therecovery system, if it wants to recover its investment in 3 years plus a return of 14%

per year,compounded semiannually? Assume no interperiod compounding ... 42 According to the Government Accountability Office (GAO), if the U ... Postal Service does notchange its business model, it will lose $480 millionnext month and $500 million the month after that,and the losses will increase by$20 million permonth for the next 10 years ... 25% per month, what is the equivalent uniformamount per month of the losses through year 10?Continuous Compounding4 ... What is the present worth ofthe maintenance costs at an interest rate of 10%per year, compounded quarterly?4 ... 3% per month,compounded continuously?Equivalence When PP Ͻ CP4 ... 4 ... 48 What effective interest rate per year is equal to1 ... 4 ... 6% per month, compounded continuously?4 ... If the company wantsto make an effective 25% per year, compoundedcontinuously, what nominal daily rate of return hasto berealized? Assume 365 days per year ... 52 U ... Steel is planning a plant expansion that is expected to cost $13 million ... 122Chapter 4Nominal and Effective Interest Rates4 ... What is the present worth of the costs atan interest rate of 10% per year, compoundedcontinuously?nanowires can be extrudedby blasting the carbonnanotubes with an electron beam ... 7 million in year 1,$2 ... 4 million in year 3 todevelop the technology, determine the presentworth of the investments in year 0, if the interestrate in year 1 is 10% and in years 2 and 3 it is 12%per year ... 54 Many small companies use accountsreceivable ascollateral to borrow money for continuing operations and meeting payrolls ... 25% per month after4 months, how much will the company owe at theend of 1 year?4 ... P=?i = 10%4 ... If the interest rate was 8%per year for the first 3 years and then it increased to10% in years 4 and 5, what isthe equivalent futureworth (in year 5) of the maintenance cost? Showhand and spreadsheet solutions ... 56 By filling carbon nanotubes with miniscule wiresmade of iron and iron carbide, incredibly thin6$160ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS4 ... 59 An interest rate is aneffective rate under all of thefollowing conditions, except when:(a) The compounding period is not stated(b) The interest period and compounding periodare the same(c) The interest statement says that the interestrate is effective(d) The interest period is shorter than the compounding period4 ... 61 Aninterest rate of 1 ... 5% per month(b) 4 ... 0% per quarter, compounded continuously9% per 6 months4 ... 68% per year, compounded monthly,is the closest to:(a) 12% per year(b) 12% per year, compounded annually(c) 1% per month(d) 1% per month, compounded annually4 ... 02%(b) 12 ... 31%(d) 12 ...64 If you make quarterly deposits for 3 years into anaccount that compounds interest at 1% per month,the value of n in the F͞A factor that will determineF at the end of the 3-year period is:(a) 3(b) 12(c) 36(d) None of these123Additional Problems and FE Exam Review Questions4 ... 1 ... 5% per quarter,compoundedmonthly(a) Both are nominal rates ... (c) Rate 1 is effective and rate 2 is nominal ... 4 ... e ... (b) The compounding period is equal to the payment period ... (d) The compounding period is longer than thepayment period ... 67 An engineer who is saving for her retirement plansto deposit $500every quarter, starting one quarterfrom now, into an investment account ... 68 A company that makes flange-mount, motorized rotary potentiometers expects to spend $50,000 for acertain machine 4 years from now ... 69 For the cash flow diagram shown, the unit of thepayment period (PP) is:(a)Months(b) Quarters(c) Semiannual (d) YearsF=?i = 1% per month012345678191011212 Quarter3Year$400 $400$420$440$460$480$5004 ... At an interest rate of effective10% per year, compounded semiannually, the equation that represents the equivalent annual worth A inyears 1 through 5 is:(a) A ϭ10,000(P͞F,10%,2)(A͞P,10%,5)ϩ 10,000(A͞F,10%,5)(b) A ϭ 10,000(A͞P,10%,4) ϩ 10,000(A͞F,10%,5)(c) A ϭ 10,000(P͞F,5%,2)(A͞P,5%,10)ϩ 10,000(A͞F,5%,10)(d) A ϭ [10,000(F͞P,10%,5)ϩ 10,000](A͞F,10%,5)4 ... If youwant to know the total after 4 years, the value of nyou should use in the F͞A factor is:(a) 2(b)4(c) 8(d) 124 ... At an interest rate of 14% per year,compounded semiannually, the uniform amountthat must be deposited into a sinking fund every6 months is closest to:(a) $21,335(b) $24,825(c) $27,950(d) $97,995124Chapter 4Nominal and Effective Interest RatesCASE STUDYIS OWNING A HOME ANET GAIN OR NET LOSS OVER TIME?BackgroundThe Carroltons are deliberating whether to purchase a houseor continue to rent for the next 10 years ... Plus, the high schoolthat their children attend is very good for their college prepeducation, and they all like the neighborhood where theylive now ...If the Carroltons do not buy a house, they will continueto rent the house they currently occupy for $2700 permonth ... Additionally, they will add to this investment at theend of each year the same amount as the monthly 15-yearmortgage payments ... InformationTwo financing plans using fixed-ratemortgages are currentlyavailable ... PlanDescriptionA30-year fixed rate of 5 ... 0% per year interest;10% down paymentBOther information:• Price of the house is $330,000 ... • Up-front fees (origination fee, survey fee, attorney’s fee,etc ... Any money not spent on the down payment or monthly payment willbe invested and return at a rate of 6% per year(0 ... The Carroltons anticipate selling the house after 10 yearsand plan for a 10% increase in price, that is, $363,000 (afterall selling expenses are paid)Case Study Exercises1 ... No taxes are considered on proceeds from thesavings or investments ... TheCarroltonsdecided to use the largest future worth after 10 years toselect the best of the plans ... Plan A analysis: 30-year fixed-rate loanAmount of money required for closing costs:Down payment (10% of $330,000)Up-front fees (origination fee, attorney’sfee, survey, filing fee, etc ... 25%͞12 ϭ 0 ... A ϭ297,000(A͞P,0 ... 005522)ϭ $1640Add the T&I of $500 for a total monthly payment ofPaymentA ϭ $2140 per monthThe future worth of plan A is the sum of three futureworth components: remainder of the $40,000 availablefor the closing costs (F1A); left-over money from thatavailable for monthly payments(F2A); and increase inthe house value when it is sold after 10 years (F3A) ... F1A ϭ (40,000 Ϫ 36,000)(F͞P,0 ... 5%,120)ϭ $116,354Case StudyNet money from the sale in 10 years (F3A) is the difference between the net selling price ($363,000) and theremaining balance on the loan ... 4375%,120)Ϫ1640(F͞A,0 ... 6885) Ϫ 1640(157 ... Perform this analysis if all estimates remain the same,except that when the house sells 10 years after purchase,the bottom has fallen out of the housing market and thenet selling price is only 70% of the purchase price, thatis, $231,000 ... An engineeringeconomicanalysis evaluates cash flow estimates for parameters such as initialcost, annual costs and revenues, nonrecurring costs, and possiblesalvage value over an estimated useful life of the product; process,or service ... After completing these chapters, you will be able to evaluatemost engineeringproject proposals using a well-accepted economicanalysis technique, such as present worth, future worth, capitalizedcost, life-cycle costing, annual worth, rate of return, or benefit /costanalysis ... Important note: If depreciation and͞or after-tax analysis is to beconsidered along with the evaluation methodsin Chapters 5 through9, Chapter 16 and/or Chapter 17 should be covered, preferably afterChapter 6 ... SECTIONTOPICLEARNING OUTCOME5 ... 5 ... 5 ... 5 ... 5 ... Afuture amount of money converted to its equivalent value now has a present worth(PW) that is always less than that of the future cashflow, because all P͞F factors havea value less than 1 ... For this reason, presentworth values are often referred to as discounted cash flows (DCF), and the interest rate is referred to as the discount rate ... Up to this point, present worth computations have beenmade for one project or alternative ... Twoadditional applications are covered here—future worth and capitalized cost ... To understand how to organize an economic analysis, this chapter begins with a description of independent and mutually exclusive projects as well as revenue and costalternatives ... This industry produces the microchips usedin manyof the communication, entertainment,transportation, and computing deviceswe use every day ... Ultrapure water isobtained by special processes that commonly include reverse osmosis͞deionizingresin bed technologies ... A fab costs upward of $2 ... A newcomer to the industry,AngularEnterprises, has estimated the cost profiles for two options to supply its anticipated fab with water ... The initial cost estimates for the UPW system aregiven below ... 5Ϫ0 ... Life of UPW equipment 10 yearsUPW needs1500 gpmOperating time16 hours perday for 250 daysper yearThis case is usedin the following topics(Sections) and problems of this chapter:PW analysis of equal-life alternatives(Section 5 ... 3)Capitalized cost analysis (Section 5 ... 20 and 5 ... 1 Formulating AlternativesThe evaluation and selection of economic proposals require cash flow estimates over a statedperiod of time,mathematical techniques to calculate the measure of worth (review Example 1 ... From all theChapter 5Present Worth AnalysisMandatesntioeasmaorInfId130ExperienceEstimatesPlansProposalsNot viable1BNot viable2DmCEEitherof theseMutuallyexclusivealternatives12m+DNNatureofproposalsIndependentprojectsSelectalljustifiedDN12……Selectonlyone…AViableDN = do nothingmTypes of cash flow estimates* Costs only* Revenues and costsRevenue alternativeCost alternativePerform evaluation and make selectionFigure 5–1Progression from proposals to economic evaluation toselection ... This progression is detailed in Figure 5–1 ... Once the obviously nonviable ideasare eliminated, the remaining viable proposals are fleshed out to form the alternatives to beevaluated ... The nature of the economic proposals is always one of two types:Mutually exclusive alternatives: Only one ofthe proposals can be selected ... Independent projects: More than one proposal can be selected ... The do-nothing (DN) proposal is usually understood to be an option when the evaluation isperformed ... 2131Present Worth Analysis of Equal-Life AlternativesThe DN alternative or project means that thecurrent approach is maintained; nothing newis initiated ... Do nothingIf it is absolutely required that one or more of the defined alternatives be selected, do nothing isnot considered ... Mutually exclusive alternatives and independent projects are selected in completely different ways ... Only one is chosen,and therest are rejected ... For independent projects one, two or more, infact, all of the projects that are economically justified can be accepted, provided capital fundsare available ... Independent projects are evaluated one at a time and compete only with the DN project ... When performed correctly asdescribed in each chapter, anyof the techniques will reach the same conclusion of which alternative or alternatives to select ... A parallel can be developed between independent and mutually exclusive evaluation ... Zero, one, two, or more may be selected ... This number includes the DN alternative, asshown in Figure 5–1 ... Commonly, in real-world applications, thereare restrictions, such as an upper budgetary limit, that eliminate many of the 2m alternatives ... Chapter 12 treats independent projects with a budget limitation; this is called capital budgeting ... Cash flow estimates determine whether thealternatives are revenue- or cost-based ... Definitions for these types follow:Revenue: Each alternative generates cost (cash outflow) and revenue (cash inflow) estimates,and possibly savings, also considered cash inflows ... Cost: Each alternative has only cost cash flow estimates ... These arealsoreferred to as service alternatives ... Differences in evaluation methodology aredetailed in each chapter ... 2 Present Worth Analysis of Equal-Life AlternativesThe PW comparison of alternatives with equal lives is straightforward ... The present worth method is quite popular in industry because allfuturecosts and revenues are transformed to equivalent monetary units NOW; that is, all futurecash flows are converted (discounted) to present amounts (e ... , dollars) at a specific rate of return,which is the MARR ... The required conditions and evaluation procedure are as follows:If the alternatives have thesame capacities for the same time period (life), the equal-servicerequirement is met ... Equal-servicerequirement132Chapter 5Present Worth AnalysisFor mutually exclusive (ME) alternatives, whether they are revenue or cost alternatives, the following guidelines are applied to justify a single project or toselect one from several alternatives ... Two or more alternatives: Select the alternative with the PW that is numerically largest,that is, less negative or more positive ... Note that the guideline to select one alternative with the lowest cost or highest revenue usesthe criterion of numerically largest ... Theselections below correctly apply the guideline for two alternatives A and B ... The selection guideline is as follows:Independent projectselectionOne or more independent projects: Select all projects with PW Ն 0 at the MARR ... All PW analyses require a MARR for use as the i value in the PW relations ...EXAMPLE 5 ... During lab research, three equal-service machines need to be evaluatedeconomically ... The MARR is10% per year ... The salvage values are considered a “negative” cost, so a ϩ signprecedes them ... )The PW of each machine is calculated at i ϭ 10% for n ϭ 8 years ... PWE ϭ Ϫ4500 Ϫ900(P͞A,10%,8) ϩ 200(P͞F,10%,8) ϭ $Ϫ9208PWG ϭ Ϫ3500 Ϫ 700(P͞A,10%,8) ϩ 350(P͞F,10%,8) ϭ $Ϫ7071PWS ϭ Ϫ6000 Ϫ 50(P͞A,10%,8) ϩ 100(P͞F,10%,8) ϭ $Ϫ6220The solar-powered machine is selected since the PW of its costs is the lowest; it has thenumerically largest PW value ... 3133Present WorthAnalysis of Different-Life AlternativesEXAMPLE 5 ... With the options of seawater or groundwater sources,it is a good idea to determine if one system is more economical than the other ... SolutionAn important first calculation is the cost of UPW per year ... 44 M per yearGroundwater: (5͞1000)(1500)(60)(16)(250) ϭ $1 ... In $1 million units:PW relation: PW ϭ first cost Ϫ PW of AOC Ϫ PW of UPW ϩ PW of salvage valuePWS ϭ Ϫ20 Ϫ 0 ... 44(P͞A,12%,10) ϩ 0 ... 5(5 ... 44(5 ... 3220)ϭ $Ϫ30 ... 3(P͞A,12%,10) Ϫ 1 ... 10(22)(P͞F,12%,10)ϭ Ϫ22 Ϫ 0 ... 6502) Ϫ 1 ... 6502) ϩ 2 ... 3220)ϭ $Ϫ33 ... 52 M ... 3 PresentWorth Analysis of Different-LifeAlternativesWhen the present worth method is used to compare mutually exclusive alternatives that havedifferent lives, the equal-service requirement must be met ... 2 is followed, with one exception:The PW of the alternatives must be compared over the same number ofyears and must endat the same time to satisfy the equal-service requirement ... A fair comparison requires that PW values representcash flows associated with equal service ... The equal-service requirement issatisfied by using either of two approaches:LCM: Compare the PW of alternatives over a periodof time equal to the least commonmultiple (LCM) of their estimated lives ... This approach does not necessarily consider the useful life of an alternative ... For either approach, calculate the PW at the MARR and use the same selection guideline asthat for equal-life alternatives ... For example, lives of 3and 4 years are compared over a 12-year period ... Additionally, the LCM approach requires that some assumptions be made aboutsubsequent life cycles ... The service provided will be needed over the entire LCM years or more ... The selected alternative can be repeated over each life cycle of the LCMin exactly the samemanner ... Cash flow estimates are the same for each life cycle ... If the cash flows are expected to change by any other rate, then the PW analysis mustbe conducted using constant-value dollars, which considers inflation (Chapter 14) ... For the study period approach, a time horizon ischosen overwhich the economic analysis is conducted, and only those cash flows which occur during that timeperiod are considered relevant to the analysis ... An estimated market value at the end of the study period must be made ... Thestudy period approach is often used in replacement analysis(Chapter 11) ... EXAMPLE 5 ... , plans to purchase new cut-and-finish equipment ... Vendor AFirst cost, $Annual M&O cost, $ per yearSalvage value, $Life, yearsVendor BϪ15,000Ϫ3,5001,0006Ϫ18,000Ϫ3,1002,0009(a) Determine which vendor should be selected on the basis of a present worthcomparison,if the MARR is 15% per year ... If a study period of 5 years is used and the salvage values are not expected tochange, which vendor should be selected?Solution(a) Since the equipment has different lives, compare them over the LCM of 18 years ... These are years 6 and 12 for vendor A andyear 9 for B ... Calculate PW at 15% over 18 years ... 3Present Worth Analysis of Different-Life Alternatives135PWA = ?$100012$1000612$10001617 18Year$3500$15,000$15,000$15,000Vendor APWB = ?$200012$200016917 18Year$3100$18,000$18,000Vendor BFigure 5–2Cash flow diagram fordifferent-life alternatives, Example 5 ... Vendor B is selected, since it costs less in PW terms; that is, the PWB value is numerically larger than PWA ... The PW analysis isPWA ϭ Ϫ15,000 Ϫ 3500(P͞A,15%,5) ϩ 1000(P͞F,15%,5)ϭ $Ϫ26,236PWB ϭ Ϫ18,000 Ϫ 3100(P͞A,15%,5) ϩ 2000(P͞F,15%,5)ϭ$Ϫ27,397Vendor A is now selected based on its smaller PW value ... In situations such asthis, the standard practice of using a fixed study period should be carefully examined toensure that the appropriate approach, that is, LCM or fixed study period, is used to satisfythe equal-service requirement ... 4Water for Semiconductor Manufacturing CaseWhen we discussed this case in the introduction, we learned that the initial estimates of equipment life were 10 years for both options of UPW (ultrapure water)—seawater and groundwater ... However, it is expected that, instead of complete replacement, atotal refurbishment of the equipment for $10 M after 5 years will extend the life throughthe anticipated 10th year of service ... 2 ... PE136Chapter 5Present Worth AnalysisSeawater optionϭ ⟨OC ϩ UPW cost/yearϭ Ϫ0 ... 44Groundwater optionϭ ⟨OC ϩ UPW cost/yearϭ Ϫ0 ... 80Both optionsSalvagevaluesincluded hereSalvage values are thesame after 5 and 10 yearsFigure 5–3PW analyses using LCM and study period approaches for water for semiconductor manufacturing case,Example 5 ... SolutionA spreadsheet and the NPV function are a quick and easy way to perform this dual analysis ...LCM of 10 years: In the top part of the spreadsheet, the LCM of 10 years is necessary to satisfy the equal-service requirement; however, the first cost in year 5 is the refurbishment cost of$Ϫ10 M, not the $Ϫ20 M expended in year 0 ... 94 M in year 5 accounts for the continuing AOC and annual UPW costof$Ϫ1 ... The NPV functions shown on the spreadsheetdetermine the 12% per year PW values in $1 million units ... 31PWG ϭ $Ϫ33 ... Study period of 5 years: The lower portion of Figure 5–3 details a PW analysis using thesecond approach to evaluating different-life alternatives, that is, a specific studyperiod, whichis 5 years in this case study ... Again the economic decision is reversed as the 12% per year PW values favor the seawateroption ... 43PWG ϭ $Ϫ28 ... Both are correct answers given the decision of how the equal-service requirement is met ... 5 ... The PW values areOption S:Option G:n ϭ 5years, PWS ϭ $Ϫ26 ... n ϭ 10 years, PWG ϭ $Ϫ33 ... For independent projects, use of the LCM approach is unnecessary since each project is compared to the do-nothing alternative, not to each other, and satisfying the equal-service requirement is not a problem ... 5 ... The n value in the F͞Pfactor is eitherthe LCM value or a specified study period ... Future worth analysis over a specified study period is often utilized if the asset (equipment, abuilding, etc ... Supposean entrepreneur is planning to buy a company and expects to trade it within 3 years ... Example 5 ... Another excellent application of FWanalysis is for projects that will comeonline at the end of a multiyear investment period, such as electric generation facilities, toll roads,airports, and the like ... The selection guidelines for FW analysis are the same as for PW analysis; FW Ն 0 means theMARR is met or exceeded ... EXAMPLE 5 ... Therewas a net loss of £10 million at the end of year 1 of ownership ... This means that breakevennet cash flow was achieved this year ... (a) The British conglomerate has just been offered £159 ... Use FW analysis to determine if the MARR will berealized at this selling price ... 5 million ... FW3 ϭϪ75(F͞P,25%,3) Ϫ 10(F͞P,25%,2) Ϫ 5(F͞P,25%,1) ϩ 159 ... 36 ϩ 159 ... 86 millionNo, the MARR of 25% will not be realized if the £159 ... ME alternativeselection138Chapter 5Present Worth AnalysisFW = ?£159 ... 5 ... (b) Determine the future worth 5 years from now at 25% per year ... The A͞G and F͞Afactors are applied to the arithmetic gradient ... 81 millionThe offer must be for at least £246 ... This is approximately3 ... 5 ... A perpetualor infinite life is the effective planning horizon ... The economic worth of these types of projects orendowments is evaluated using the present worth of the cash flows ...The formula to calculate CC is derived from the PW relation P ϭ A(P͞A,i%,n), where n ϭ ϱtime periods ... We replace the symbols P and PW withCC as a reminder that this is a capitalized cost equivalence ... 1]Solving for A or AW, the amount of new money that is generated each year by a capitalizationofan amount CC isAW � CC (i)[5 ... Equation [5 ... If $20,000 is invested now (this is the5 ... This leaves theoriginal $20,000 to earn interest so that another $2000 will be accumulated the next year ... An annual operating cost of $50,000and a rework cost estimated at $40,000 every 12 years are examples ofrecurring cash flows ... The procedure to determine the CC for an infinite sequence of cash flows is as follows:1 ... 2 ... This is their CC value ... Find the A value through one life cycle of all recurring amounts ... ) Add this to all other uniformamounts (A) occurring in years 1 through infinity ... 4 ... This isanapplication of Equation [5 ... 5 ... Drawing the cash flow diagram (step 1) is more important in CC calculations than elsewhere,because it helps separate nonrecurring and recurring amounts ... EXAMPLE 5 ... The director wants to know the total equivalent cost of all future costs incurred to purchase thesoftware system ... The system has an installed cost of $150,000 and an additional cost of $50,000 after10 years ... In addition, there is expected to be a recurring major upgrade cost of $15,000every 13 years ... Solution(a) The five-step procedure to find CC now is applied ... Draw a cash flow diagramfor two cycles (Figure 5–5) ... Find the present worth of the nonrecurring costs of $150,000 now and $50,000 in year10 at i ϭ 5% ... CC1 ϭ Ϫ150,000 Ϫ 50,000(P͞F,5%,10) ϭ $Ϫ180,695024681012142026$5000$8000$15,000i = 5% per year$50,000$150,000Figure 5–5Cash flows for two cycles of recurringcosts and all nonrecurring amounts, Example 5 ... $15,000Year139140Chapter 5Present Worth Analysis3 and 4 ... 1] ... 05 ϭ $Ϫ16,940There are several ways to convert the annual software maintenance cost series to Aand CC values ... 05 ϭ $Ϫ100,000Second, convert the step-up maintenance costseries of $−3000 to a capitalized costCC4 in year 4, and find the present worth in year 0 ... )Ϫ3,000CC4 ϭ ———— (P͞F,5%,4) ϭ $Ϫ49,3620 ... The total capitalized cost CCT for Haverty County Transportation Authority is the sumof the four component CC values ... 2] determines the AW value forever ... 05)ϭ $17,350Correctly interpreted, this means Haverty County officials have committed the equivalentof $17,350 forever to operate and maintain the toll management software ... Since the capitalized cost represents thetotal present worth of financing and maintaining a given alternative forever, thealternatives willautomatically be compared for the same number of years (i ... , infinity) ... This evaluation is illustrated inExample 5 ... EXAMPLE 5 ... 4) to the point that the life of the seawater optioncan be extended to 10 years with a major refurbishment cost after 5 years ... In $1 million units,theestimates and PW values (from Figure 5–3) are as follows:Seawater: PS ϭ $Ϫ20; AOCS ϭ $Ϫ1 ... 05(20) ϭ $1 ... 31Groundwater: PG ϭ $Ϫ22; AOCG ϭ $Ϫ2 ... 10(22) ϭ $2 ... 16If we assume that the UPW (ultrapure water) requirement will continue for the foreseeablefuture, a good number to know is thepresent worth of the long-term options at the selectedMARR of 12% per year ... Select the option with the lower CC ... 5 ... 31(0 ... 43CCS ϭ Ϫ6 ... 12 ϭ $Ϫ53 ... 16(0 ... 87CCG ϭ Ϫ5 ... 12 ϭ $Ϫ48 ... CommentIf the seawater-life extension is not considered a viable option, the original alternative of5 yearscould be used in this analysis ... AS,5 years ϭ Ϫ20(A͞P,12%,5) Ϫ 1 ... 05(20)(A͞F,12%,5)ϭ $Ϫ7 ... 33/0 ... 08Now, the economic advantage of the groundwater option is even larger ... To determine capitalized cost for the finite life alternative,calculate the equivalent A value for one life cycle and divide by theinterest rate (Equation [5 ... This procedure is illustrated in Example 5 ... EXAMPLE 5 ... The goal is zero landfill by 2020 ... The interest rate for state-mandated projectsis 5% per year ... No contract period is stated; thus the contract and services are offered for as long as the State needs them ... Expectedlife of the equipment is 5 yearswith no salvage value ... (b) Determine the maximum number of sites at which the equipment can be purchased andstill have a capitalized cost less than that of the contractor option ... The contract, as proposed, has a long life ... The annual charge of A ϭ $25,000is dividedby i ϭ 0 ... Summing the two values results inCCC ϭ $Ϫ8 ... For the finite, 5-year purchase alternative, column B shows the first cost ($Ϫ275,000 persite), AOC ($Ϫ12,000), and equivalent A value of $−755,181, which is determined viathe PMT function (cell tag) ... 1 million ... (b) A quick way to find themaximum number of sites for which CCP Ͻ CCC is to use Excel’s Goal Seek tool, introduced in Chapter 2, Example 2 ... (See Appendix A for detailson how to use this tool ... The result, shown in column C, indicates that 5 ... Since the number of sites141142Chapter 5Present Worth AnalysisϭPMT($B$1,B13,ϪB12*B4) ϩ B14*B4Figure 5–6Spreadsheet solution of Example 5 ... must be an integer, 5 or fewer sites will favor purchasing the equipment and 6 or moresites will favor contracting the separation services ... By the way, another way to determine the number of sites is by trialand error ...CHAPTER SUMMARYThe present worth method of comparing alternatives involves converting all cash flows to presentdollars at the MARR ... When the alternatives have different lives, the comparison must be made for equal-serviceperiods ... Both approaches compare alternatives in accordance withthe equal-service requirement ... If the life of the alternatives is considered to be very long or infinite, capitalized cost is thecomparison method ... PROBLEMSTypes of Projects5 ... 2 (a)(b)What is meant by the do-nothing alternative?When is the do-nothing alternative not anoption?5 ... 5 ... 115 ... 6 Whattwo approaches can be used to satisfy theequal-service requirement?Alternative Comparison—Equal Lives5 ... Which one should be selected on thebasis of a present worth analysis at 10% per year?In-houseFirst cost, $Annual cost, $ per yearAnnual income, $ per yearSalvage value, $Life,yearsContractϪ30Ϫ514250Ϫ23 ... 8 The manager of a canned food processing plantmust decide between two different labeling machines ... Machine B will cost $51,000 to buyand will have an annual operating cost of $17,000during its 4-year life ... 9 A metallurgical engineer is considering two materialsfor use in a space vehicle ... (a) Which should be selected on the basis ofa present worth comparison at an interest rate of12% per year? (b) At what first cost for the material not selected above will it become the moreeconomic alternative?Material XFirst cost, $Maintenance cost, $ per yearSalvage value,$Life, yearsMaterial YϪ15,000Ϫ9,0002,0005Ϫ35,000Ϫ7,00020,00055 ... In one particular year, the company offered 1000 shares of eitherclass A or class B stock ... Class B stock wasselling for $20 per share, but its price was expectedto increase by 12% per year ... Plan A involves remodeling the firestationson Alameda Avenue and Trowbridge Boulevard thatare 57 and 61 years old, respectively ... ) Thecost for remodeling the Alameda station is estimated at $952,000 while the cost of redoing theTrowbridge station is $1 ... Plan B calls forbuying 5 acres of land somewhere between the twostations,building a new fire station, and selling theland and structures at the previous sites ... The size of the new fire station would be 9000square feet with a construction cost of $151 ... Contractor fees for overhead, profit,etc ... (Assume all of the costs for plan Boccur at time 0 ... Under plan B, the sale of the oldsites is anticipated to net a positive $500,000 five years in thefuture ... 5 ... The company needs to analyze the economic feasibility of rainwater drains in a 60-acrearea that it plans to develop ... Ifno drains are installed, the cost of refilling andgrading the washed out area is expected to be$1500 perthunderstorm ... The cost of thepipe will be $3 per foot for the total length of7000 feet required ... Assuming that thunderstorms occur regularly at3-month intervals, starting 3 months from now,which alternative should be selected on the basis ofa present worth comparison using an interest rate of4% perquarter?5 ... A250-mm line will have an initial cost of $155,000,whereas a 300-mm line will cost $210,000 ... If the lines are144Chapter 5Present Worth Analysisexpected to last for 30 years, which size should beselected on the basis of a present worth analysisusing an interest rate of 10% per year?5 ... Ifgaseous chlorine is added, a chlorinatorwill be required that has an initial cost of $8000and a useful life of 5 years ... Alternatively, dry chlorine can be added manually at a cost of $1000 per year for chlorine and$1900 per year for labor ... 15 Anion, an environmental engineering consultingfirm, is trying tobe eco-friendly in acquiring an automobile for general office use ... The hybrid under considerationis GM’s Volt, which will cost $35,000 and have arange of 40 miles on the electric battery and severalhundred more miles when the gasoline enginekicks in ... The Leaf’s relatively limited range creates apsychological effect known as range anxiety ... The Leafcould be leased for $349 per month (end-of-monthpayments) for 5 years after an initial $1500 downpayment for “account activation ... 75% per month? Assume the operating cost will be the same for both vehicles ... 16 A pipeline engineer working inKuwait for the oilgiant BP wants to perform a present worth analysis onalternative pipeline routings—the first predominatelyby land and the second primarily undersea ... Perform the analysis for the engineer at 15% per year ... 17 An electric switch manufacturing company has tochoose one of threedifferent assembly methods ... Method B will cost $80,000 to buy andwill have an annual operating cost of $6000 overits 4-year service life ... Methods A and B will have nosalvage value, but method C will have some equipment worth an estimated $12,000 ... 5 ... The company recently settled a lawsuitbyagreeing to pay $60 million in mitigation costs related to acid rain ... The question of how to distribute the money overtime has been posed ... 5 millionin each of years 1 through 10) ... Determine which plan is more economical on thebasis of a present worth analysis over a 10-yearperiod at an interestrate of 10% per year ... 19 Machines that have the following costs are underconsideration for a robotized welding process ... Show (a) hand and(b) spreadsheet solutions ... 20 Water for Semiconductor Manufacturing CasePEThroughout the present worth analyses, the decision between seawater andgroundwater switchedmultiple times in Examples 5 ... 4 ... 145ProblemsSeawater (S)Life n,years1055(studyperiod)Groundwater (G)PW at12%, $SelectedLife n,yearsϪ20Ϫ30 ... 31Ϫ20Ϫ26 ... 16NoNo10Ϫ22Ϫ33 ... 32NoThe confusion about the recommended source forUPW has not gone unnoticed by thegeneral manager ... The study period is set by the manageras 10 years, simply because that is the time period onthe lease agreement for the building where the fabwill be located ... What is themaximum first cost that Angular Enterprises shouldpay for the seawater option?5 ... In a field application,physical constraints compromise the pipe layout, so the engineeris considering installing the airflow probes in anelbow, knowing that flow measurement will be lessaccurate but good enough for process control ... This plan will have a first cost of $26,000 with anannual maintenance cost estimated at$5000 ... The stainlesssteel probe can be installed in a drop pipe with thetransmitter located in a waterproof enclosure on thehandrail ... Its maintenance cost is estimated to be $1400 per year plus$2500 in year 3 for replacement of signal processing software ... At an interest rate of 10% per year,whichone should be selected on the basis of a presentworth comparison?5 ... A plastic linerwill cost $0 ... This removal will cost $500,000 ... 20 per square foot ... 23A sports mortgage is the brainchild of Stadium Capital Financing Group, a company headquartered inChicago, Illinois ... In California, thelocked-in priceperiod is 50 years ... (a) Which fan madethe better deal if the interest rate is 8% per year?(b) What should fan X be willing to pay up front forthe mortgage to make the two plans exactly equivalent economically? (Assume he has no reason togive extra money to UCLA at this point ... 24 Achemical processing corporation is consideringthree methods to dispose of a non-hazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract ... Determine which has theleast cost on the basis of a present worth comparison at 10% per year for the followingscenarios:(a) The estimates as shown(b) The contract award cost increases by 20%every 2-year renewalLand Application IncinerationFirst cost, $Annual operatingcost, $ per yearSalvage value, $Life, yearsContractϪ130,000Ϫ95,000Ϫ900,000Ϫ60,0000Ϫ120,00025,0003300,000602146Chapter 5PresentWorth Analysis5 ... The resultsin the table use a MARR of 14% per year ... ILife n, yearsPW over n years, $PW over 6 years, $PW over 12 years, $316 ... 9439 ... 1215 ... 45KL126Ϫ257 ... 46Ϫ653 ... 46Ϫ257 ... 46Future Worth Comparison5 ... Robot X will have a first cost of $80,000, anannualmaintenance and operation (M&O) cost of$30,000, and a $40,000 salvage value ... Which should be selected on the basis of a futureworth comparison at an interest rate of 15% peryear? Use a 3-year study period ... 27 Two processes can be used for producing a polymer that reduces friction loss inengines ... Process W will have a first costof $1,350,000, an operating cost of $25,000 peryear, and a $120,000 salvage value after its 4-yearlife ... Which processshould be selected on the basis of a future worthanalysis at an interest rate of 12% per year?5 ... PFirst cost, $Annual operating cost, $ peryearSalvage value, $Life, yearsQϪ23,000 Ϫ30,000Ϫ4,000 Ϫ2,5003,0001,000365 ... St ... System K will have a first cost of$1,600,000, an operating cost of $70,000 per year,and a salvage value of $400,000 after its 4-year life ... Which systemshould be selected on the basis of a future worthanalysis at aninterest rate of 12% per year?5 ... 7 million over 10 years ... However, due to a real estate–induced recession in theUnited States, the developer sought and was granteda new contract ... Assume that the cost for razing the existingbuildings is $1 ... Determine the difference inthe future worth cost in year 7of the two contractsat an interest rate of 10% per year ... 31 A wealthy businessman wants to start a permanentfund for supporting research directed toward sustainability ... e ... If the fund earns interest at a rate of 8% peryear, how much money must be donated each time?5 ... Determine the capitalizedcost of $10,000every 5 years forever, starting 5 years from nowat an interest rate of (a) 3% and (b) 8% per year ... 5 ... Use an interest rateof 12% per year ... 34 Water for Semiconductor Manufacturing CasePEIt is anticipated that the needs for UPW (ultrapurewater) at the new Angular Enterprises sitewillcontinue for a long time, as long as 50 years ... Thesecosts were determined (Example 5 ... 58 million and CCG ϭ $Ϫ48 ... Groundwater is the clear economic choice ... It would mean a dependence upon a contractor to supply the water, but theequipment, treatment, and other costly activities toobtainUPW on-site would be eliminated ... (b) The annual cost starts at $5 million for the firstyear only, and then it increases 2% per year ... )5 ... Calculate the capitalized cost of the maintenanceusing an interest rate of 10% per year ... 35 Compare the alternatives shown on the basis oftheir capitalized costsusing an interest rate of10% per year ... 38 A patriotic group of firefighters is raising moneyto erect a permanent (i ... , infinite life) monumentin New York City to honor those killed in the lineof duty ... There will be an additional one-time costof $20,000 in 2 years to add names of those whowere missedinitially ... 37 Because you are thankful for what you learned inengineering economy, you plan to start a permanent scholarship fund in the name of the professorwho taught the course ... The interest that is accumulated between now and year 12 is to be added tothe principal of the endowment ... If youwant the amount of thescholarships to be $40,000 per year, how muchmust you donate now if the fund earns interest at arate of 8% per year?ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS5 ... (b) The alternatives will be used only throughthe life of the longest-lived alternative ... (d) Atleast one of the alternatives will have afinite life ... 40 When only one alternative can be selected fromtwo or more, the alternatives are said to be:(a) Mutually exclusive(b) Independent alternatives(c) Cost alternatives(d) Revenue alternatives5 ... 42 The alternatives shown are to be compared on thebasisof their present worth values ... 43 The value of the future worth for alternative P at aninterest rate of 8% per year is closest to:PϪ23,000Ϫ4,000(a)(b)(c)(d)Ϫ30,000Ϫ2,5003,0003First cost, $Annual operating cost,$ per yearSalvage value, $Life, yearsQ1,0006FWP ϭ $Ϫ88,036FWP ϭ $Ϫ86,026FWP ϭ$Ϫ81,274FWP ϭ $Ϫ70,1785 ... 45 A donor (you) wishes to start an endowment thatwill provide scholarship money of $40,000 per yearbeginning in year 5 and continuing indefinitely ... 46 through 5 ... Alternative IInitial cost, $Annual income, $ per yearAnnual expenses, $ per yearSalvage value, $Life,yearsAlternative JϪ150,00020,000Ϫ9,00025,0003Ϫ250,00040,000Ϫ14,00035,0006The interest rate is 15% per year ... 46 In comparing alternatives I and J by the presentworth method, the value of n that must be used in11,000(P͞A,i,n) for alternative I is:(a) 3(b) 6(c) 18(d) 365 ... 48 In comparingalternatives I and J by the presentworth method, the equation that yields the presentworth of alternative I is:(a) PWI ϭ Ϫ150,000 ϩ 11,000(P͞A,15%,3) ϩ25,000(P͞F,15%,3)(b) PWI ϭ Ϫ150,000 ϩ 11,000(P͞A,15%,6) ϩ25,000(P͞F,15%,6)(c) PWI ϭ Ϫ150,000 ϩ 11,000(P͞A,15%,6) ϩ175,000(P͞F,15%,3) ϩ25,000(P͞F,15%,6)(d) PWI ϭ Ϫ150,000 ϩ 11,000(P͞A,15%,6) Ϫ125,000(P͞F,15%,3) ϩ 25,000(P͞F,15%,6)Problems 5 ... 50 are based on the followinginformation ... 5 ... 50 In comparing the machines on a present worthbasis, the present worth of machine Y is closest to:(a) $Ϫ112,320(b) $Ϫ122,060(c)$Ϫ163,040(d) $Ϫ175,980Case Study1495 ... 51 The capitalized cost of $10,000 every 5 yearsforever, starting now at an interest rate of 10% peryear, is closest to:(a) $Ϫ13,520(b) $Ϫ16,380(c) $Ϫ26,380(d) $Ϫ32,590CASE STUDYCOMPARING SOCIAL SECURITY BENEFITSBackgroundWhen Sherylgraduated from Northeastern University in2000 and went to work for BAE Systems, she did not paymuch attention to the monthly payroll deduction for socialsecurity ... However, this was so far in the future that shefully expected this government retirement benefit system tobe broke and gone by the timeshe could reap any benefitsfrom her years of contributions ... Recently, they both received notices from theSocial Security Administration of their potential retirementamounts, were they to retire and start social security benefits at preset ages ... InformationThey found that their projected benefits aresubstantially thesame, which makes sense since their salaries are very close toeach other ... At age 62, your payment wouldbe aboutAt you full retirement age (67years), your payment would beaboutAt age 70, your payment wouldbe about$1400 per month$2000 per month$2480 per monthThesenumbers represent a reduction of 30% for early retirement (age 62) and an increase of 24% for delayed retirement (age 70) ... In other words, if Sheryl starts her $2000 benefit at age 67, Brad can receive a benefit equal to 50% of hers ... In the meantime, his benefits will have increased by 24% ... Allthese options led them to define four alternative plans ... B: Each takes full benefits at full retirement age of 67and receives $2000 per month ... D: One person takes full benefits of $2000 per month atage 67, and the other person receives spousal benefits($1000 per month at age 67) and switches todelayedbenefits of $2480 at age 70 ... Case Study ExercisesBrad and Sheryl are the same age ... Withthis as the interest rate, the analysis for the four alternatives ispossible ... Can you please helpthem? (Do the analysis for one person at a time, not the couple, and stop at the age of 85 ... How much intotal (without the time value of money considered) will each plan A through D pay through age 85?2 ... Plot the future worth values for all four plans on onespreadsheet graph ... Economically, what is the best combination of plans forBrad and Sheryl, assuming they both live to be 85 yearsold?5 ... Answerthe question ... SECTIONTOPICLEARNING OUTCOME6 ... 6 ... 6 ... 6 ... 6 ... In this chapter, we add to our repertoire of alternative comparison tools ... Here we learn the equivalent annual worth, orAW, method ... Annual worth is also known by other titles ... The alternative selected by the AW methodwill always be the same as that selectedby the PW method, and all other alternative evaluation methods, provided they are performed correctly ... Thismethod considers all costs of a product, process, or system from concept to phaseout ... 1 Advantages and Uses of Annual Worth AnalysisFor manyengineering economic studies, the AW method is the best to use, when compared toPW, FW, and rate of return (Chapters 7 and 8) ... The AW value, which has the same interpretation as A used thus far, isthe economic equivalent of the PW and FW values at the MARR for n years ... 1]The n in the factorsis the number of years for equal-service comparison ... When all cash flow estimates are converted to an AW value, this value applies for every yearof the life cycle and for each additional life cycle ... The AW value determined overone life cycle is the AW for all future life cycles ... As with the PW method,there are three fundamental assumptions of the AW method that shouldbe understood ... The services provided are needed for at least the LCM of the lives of the alternatives ... The selected alternative will be repeated for succeeding life cycles in exactly the same manner as for the first life cycle ... Allcash flows will have the same estimated values in every life cycle ... If, in a particular evaluation, the first two assumptions are not reasonable, a study period must be established for the analysis ... In the third assumption, allcash flows are expected to change exactly with the inflation (or deflation) rate ...AW analysis for a stated study period is discussed in Section 6 ... EXAMPLE 6 ... 3, National Homebuilders, Inc ... The PW analysis used the LCM of 18 years ... The diagram in Figure 6–1 shows the cash flows for all three lifecycles (first cost $Ϫ15,000; annual M&O costs $Ϫ3500; salvage value $1000) ...In Example 5 ... Equal-servicerequirement and LCM152Figure 6–1Chapter 6Annual Worth AnalysisPW = $45,036PW and AW valuesfor three life cycles,Example 6 ... 3 life cycles0123456789101112131415161718$10000123456Life cycle 1i = 15%$1000$35000123456$15,000Life cycle2$1000$35000123456$15,000Life cycle 3$3500$15,00001234567891011121314151617181920•••continuesAW = $7349SolutionCalculate the equivalent uniform annual worth value for all cash flows in the first life cycle ... Now Equation [6 ... AW ϭ Ϫ45,036(A͞P,15%,18) ϭ $Ϫ7349The one-life-cycle AWvalue and the AW value based on 18 years are equal ... The AW method is especially useful in certain types of studies: asset replacement andretention time studies to minimize overall annual costs (both covered in Chapter 11), breakevenstudies and make-or-buy decisions (Chapter 13), and all studiesdealing with production or manufacturing costs where a cost /unit or profit /unit measure is the focus ... 2Calculation of Capital Recovery and AW Values153If income taxes are considered, a slightly different approach to the AW method is used by somelarge corporations and financial institutions ... Thisapproach,covered in Chapter 17, concentrates upon the wealth-increasing potential that an alternative offers acorporation ... 6 ... This is the total first cost of all assets and services required to initiatethe alternative ... Use this amount as P ... This is the terminal estimated value of assets at the end of theiruseful life ... For study periods shorter than the useful life, S is the estimated market value ortrade-in value at the end of the study period ... This is the equivalent annual amount (costs only for cost alternatives;costs and receipts for revenue alternatives) ... Salvage/market valueThe annual worth (AW)value for an alternative is comprised of two components: capitalrecovery for the initial investment P at a stated interest rate (usually the MARR) and the equivalentannual amount A ... In equation form,AW � CR ؉ A[6 ... The total annual amount A is determined from uniform recurringcosts (and possiblyreceipts) and nonrecurring amounts ... 2] ... )The recovery of an amount of capital P committed to an asset, plus the time value of thecapital at a particular interest rate, is a fundamental principle of economic analysis ... Any expected salvage value is considered in the computation of CR ... If there is someanticipatedpositive salvage value S at the end of the asset’s useful life, its equivalent annual value is recovered using the A͞F factor ... Accordingly, CR is calculated asCR � ؊ P(A͞P,i,n) ؉ S(A͞F,i,n)[6 ... 2Lockheed Martin is increasing its booster thrust power in order to win more satellite launch contractsfrom European companies interested in opening up new global communications markets ... Annual operating costs for the system are expected to start the first year and continue at$0 ... The useful life of the tracker is 8 years with a salvage value of $0 ... Calculate the CR and AW values for the system, ifthe corporate MARR is 12% per year ... 3] to calculate the capital recovery ... 46Capital recovery154Chapter 6Annual Worth AnalysisCR ϭ Ϫ12 ... 5(A͞F,12%,8)ϭ Ϫ12 ... 20130) ϩ 0 ... 08130)ϭ $Ϫ2 ... It means thateach and every year for 8 years, the equivalent total net revenue from the tracker must beatleast $2,470,000 just to recover the initial present worth investment plus the required return of12% per year ... 9 million each year ... Since CR ϭ $Ϫ2 ... AW ϭ Ϫ2 ... 9 ϭ $Ϫ3 ... Figure 6–2$0 ... 967 80123456AW = ?78(a) Cash flow diagramfor satellite trackercosts, and (b) conversion to an equivalentAW

(in $1 million),Example 6 ... $5 ... 0(a)(b)There is a second, equally correct way to determine CR ... There is a relation between the A͞P and A͞F factors ... 3] ... Subtracting S from the initial investment P before applyingthe A͞P factor recognizes that the salvage value will be recovered ... However, the factthat S is not recovered until year n of ownership iscompensated for by charging the annual interest S(i) against the CR ... The first method,Equation [6 ... For solution by spreadsheet, use the PMT function to determine CR only in a single spreadsheet cell ... The format is� PMT(i%,n,P,؊S)[6 ... 2 ... 46million ... 46,Ϫ0 ... The answer of $Ϫ2 ... As we learned in Section 3 ... In the case of Example 6 ... The one-cellPMT function, with the PV function embedded (in bold), can be written as ϭ PMT(12%,8,8؉PV(12%,1,5؊),Ϫ0 ... 47 ... 3155Evaluating Alternatives by Annual Worth Analysis6 ... The AW iscalculated over the respective life of each alternative, and the selection guidelines are the same as those used for the PW method ... Two or more alternatives: Select the alternative with the AW that is numerically largest,that is, less negative or more positive ... If any of the three assumptions in Section 6... Then the cash flow estimates over the study period are converted to AWamounts ... EXAMPLE 6 ... Many students at the area universities and community colleges work part-timedelivering orders made via the web ... The systems provide a link between the web order-placement software and the On-Star systemfor satellite-generated directions to any address in the area ... Each system costs $4600, has a 5-year useful life, and may be salvaged for an estimated$300 ... The MARR is 10% ... Perform the solution by hand and by spreadsheet ... Is this projectfinancially viable at the MARR?(c) Based onthe answer in part (b), determine how much new net income Heavenly Pizzamust have to economically justify the project ... Solution by Hand(a) The capital recovery amount calculated by Equation [6 ... CR ϭ Ϫ5[4600(A͞P,10%,5)] ϩ 5[300(A͞F,10%,5)]ϭ Ϫ5[4600(0 ... 16380)]ϭ $Ϫ5822The five systems mustgenerate an equivalent annual new revenue of $5822 to recover theinitial investment plus a 10% per year return ... The annual operating cost series,combined with the estimated $6000 annual income, forms an arithmetic gradient serieswith a base amount of $5000 and G ϭ $Ϫ100 ... Apply Equation [6 ...AW ϭ CR ϩ A ϭ Ϫ5822 ϩ 5000 Ϫ 100(A͞G,10%,5)ϭ $Ϫ1003The system is not financially justified at the net income level of $6000 per year ... 3 ... 0 ϭ Ϫ5822 ϩ (R Ϫ 1000) Ϫ 100(A͞G,10%,5)R ϭ Ϫ5822 Ϫ 1000 Ϫ 100(1 ... Figure 6–4Spreadsheet solutionof Example 6 ... (a) Capital recoveryin cell B16, (b) AWincell E17, and (c) GoalSeek template andoutcome in cell B5 ... 4AW of a Permanent Investment157Cell references are used in the spreadsheet functions to accommodate changes in estimated values ... (b) The annual worth AW ϭ $–1003 is displayed in column E using the PMT functionshown ... (c) Theminimum required income is determined in the lower part of Figure 6–4 ... EXAMPLE 6 ... Since the meals areprepared in one central location and distributed by trucks throughout the city, the equipmentthat keeps food and drink cold and hot is very important ... Use the cost estimates below to select themoreeconomic unit at a MARR of 8% per year ... AWH ϭ annual equivalent of P Ϫ annual M&O ϩ annual equivalent of Sϭ Ϫ15,000(A͞P,8%,4) Ϫ 6000 ϩ 0 ... 30192) Ϫ 6000 ϩ 3000(0 ... 4(20,000)(A͞F,8%,12)ϭ Ϫ20,000(0 ... 7350 ϩ 0 ... 13270) ϩ 8000(0 ... If the projects are independent, the AW at the MARRis calculated ... Independent projectselection6 ... 5 ... For this type of analysis,the annual worth (and capital recovery amount) of the initial investment is the perpetual annualinterest on the initial investment, that is, A ϭ Pi ϭ (CC) i ... 2] ... Thisautomatically annualizes them for each succeeding life cycle ...2] ... 5The U ... Bureau of Reclamation is considering three proposals for increasing the capacity of themain drainage canal in an agricultural region of Nebraska ... The capacity of the canal will have to be maintained in the future near its design peak flow because of increased water demand ... Theequipment is expected to have a 10-year life with a $17,000salvage value ... To control weeds in thecanal itself and along the banks, environmentally safe herbicides will be sprayed during the irrigation season ... Proposal B is to line the canal with concrete at an initial cost of $4 million ... In addition, liningrepairs will have to be made every 5 years at a cost of $30,000 ... Estimates are an initial costof $6 million, annual maintenance of $3000 for right-of-way, and a life of 50 years ... SolutionSince this is an investment for a permanent project, compute the AW for one cycle of all recurring costs ... 3], with nA ϭ10and nC ϭ 50, respectively ... Proposal ACR of dredging equipment:Ϫ650,000(A͞P,5%,10) ϩ 17,000(A͞F,5%,10)Annual cost of dredgingAnnual cost of weed controlProposal BCR of initial investment: Ϫ4,000,000(0 ... CommentNote the use of the A͞F factor for the lining repair cost in proposal B ... If the 50-year life of proposal C is considered infinite, CR ϭ P(i) ϭ $Ϫ300,000, instead of$Ϫ328,680 for n ϭ 50 ... How long lives of 40 or moreyears are treated economically is a matter of “local” practice ... 4159AW of a Permanent InvestmentEXAMPLE 6 ... Bart just received hisbonus in the amount of $8530 ...Bart’s long-term plans are to quit the Coop job some years in the future when he isstill young enough to start his own business ... (a) Use a spreadsheet to determine the amount of annual year-end withdrawal that he can anticipate (starting 1 year after he quits) that will continue forever ... (b) Determinethe amount Bart must accumulate after 15 and 20 years to generate $3000 peryear forever ... The accumulated amount after n ϭ 15 years is indicatedas Fafter 15 ϭ ? and the withdrawal series starts at the end of year 16 ... The spreadsheet in Figure 6–6 shows the functions and answers for n ϭ 15 yearsincolumns C and D ... The perpetual withdrawal is determined by viewing this accumulated amountas a P value and by applying the formulaA ϭ P(i) ϭ 23,535(0 ... Answers for n ϭ 20 years are displayed in column E ... A=?i = 7%01234131415∞16171819202122 …Figure 6–5P = $8530Fafter 15 = ?Diagram for a perpetualseries starting after 15years of accumulation,Example 6 ... Figure 6–6Spreadsheet solution,Example 6 ... 160Chapter 6Annual Worth Analysis(b) To obtain a perpetual annual withdrawal of $3000, it is necessary to determine how muchmust be accumulated 1 year before the firstwithdrawal of $3000 ... 07This P value is independent of how long Bart works at the Coop, because he must accumulatethis amount to achieve his goal ... Note thatthe number of years n does not enter into the function ϭ 3000͞B13 ... The row 15 function indicates that Bart will have to work at the Coop forjust under 24 additional years ... 5 Life-Cycle Cost AnalysisThe PW and AW analysis techniques discussed thus far have concentrated on estimates for firstcost P, annual operating and maintenance costs (AOC or M&O), salvage value S, and predictableperiodic repair and upgrade costs, plus anyrevenue estimates that may favor one alternative overanother ... A life-cycle cost analysis includes these additional estimates to the extent that theycan be reliably determined ... Estimates will cover the entire life span from the earlyconceptual stage, through the design and development stages,throughout the operating stage, andeven the phaseout and disposal stages ... Some typical LCC applications are life-span analysis for military and commercial aircraft,new manufacturing plants, new automobile models, new and expanded product lines, and government systems at federal and state levels... S ... Most commonly the LCC analysis includes costs, and the AW method is used for the analysis,especially if only one alternative is evaluated ... Public sector projects are usuallyevaluated using a benefit/cost analysis (Chapter 9), rather than LCC analysis, because estimatesto the citizenry aredifficult to make with much accuracy ... Some examples of indirect cost components are taxes, management, legal,warranty, quality, human resources, insurance, software, purchasing, etc ... LCC analysis is most effectively applied when a substantial percentage of the life span (postpurchase) costs,relative to the initial investment, will be expended in direct and indirect operating, maintenance, and similar costs once the system is operational ... However, let’s assume thatExxon-Mobil wants to evaluate the design, construction, operation, and support of a new typeand style of tanker that can transportoil over long distances of ocean ... To understand how a LCC analysis works, first we must understand the phases and stages ofsystems engineering or systems development ... Generally, the LCC estimates may be categorized into a simplified6 ... Acquisition phase: all activities prior to the delivery ofproducts and services ... • Preliminary design stage—Includes feasibility study, conceptual, and early-stage plans;final go–no go decision is probably made here ... ; there is some acquisition of assets, if economically justifiable ... • Construction and implementation stage—Includes purchases,construction, and implementation of system components; testing; preparation, etc ... Phaseout and disposal phase: covers all activities to transition to a new system; removal/recycling/disposal of old system ... 7In the 1860s, General Mills Inc ... both started in the flour business in the TwinCities ofMinneapolis–St ... In the decade of 2000 to 2010, General Mills purchased Pillsbury for a combination cash and stock deal worth more than $10 billion and integrated the product lines ... At this point only costestimates have been addressed—no revenues or profits ... Use LCC analysis at the industryMARR of 18% to determine the size of the commitment in AW terms ... Since all estimates are for costs, they are not preceded bya minus sign ... 5 million0 ... 5 million1 ... 0 millionEquipment acquisition (years 1 and 2)Current equipment upgrades (year 2)New equipment purchases (years 4 and 8)$2 ...75 million2 ... 0 million5 ... 2 millionper year thereafter3 ... 0 million each year161162Chapter 6Annual Worth AnalysisSolutionLCC analysis can get complicated rapidly due to the number of elements involved ... Values are in$1 million units ... 5Preliminary design: product and equipmentPW ϭ 1 ...187Detailed design: product and test marketing, and equipmentPW ϭ 1 ... 0(P͞F,18%,2) ϭ $3 ... 0(P͞A,18%,2) ϩ 1 ... 0(P͞F,18%,4) ϩ 2 ... 04 81 Ϫ ——1 ... 2 ————— (P͞F,18%,2) ϭ $6 ... 14Use: marketingPW ϭ 8 ... 0(P͞A,18%,8) Ϫ 0 ... 0(P͞F,18%,5)ϭ $20 ... 0 million in year 3΄()΅1 ... 18PW ϭ 4 ... 4120 ...0(P͞A,18%,2)(P͞F,18%,8) ϭ $0 ... 238 million ... AW � 01$ ... 832 million per yearThis is the LCC estimate of the equivalent annual commitment to the two proposedproducts ... For example, if one alternative will produce 20 million units per year and a second alternative will operate at 35 million per year, theAW values should be compared on a currency unit/unit produced basis, such as dollar/unit or euro/hour operated ... Forsome systems, typically defense systems, operating and maintenance costs rise fast after acquisition and remain high until phaseout occurs ... It is not unusualto have 75% to 85% of theentire life span LCC committed during the preliminary and detailLife-Cycle Cost AnalysisCosts6 ... D%%BBReducedtotal LCCCommittedCumulative LCCCumulative LCCCommittedActual#1#1CEF#2AATimeAcquisitionphaseOperationphaseTimeAcquisitionphase(a)Operationphase(b)Figure 6–8LCCenvelopes for committed and actual costs: (a) design 1, (b) improved design 2 ... As shown in Figure 6–8a, the actual or observed LCC (bottom curve AB) will trailthe committed LCC throughout the life span (unless some major design flaw increases the totalLCC of design #1 above point B) ... Amoreeffective design and more efficient equipment can reposition the envelope to design #2 inFigure 6–8b ... It is this lower envelope #2 we seek ... Even though an effective LCC envelope may be established early in the acquisition phase,it is not uncommon that unplanned cost-saving measures areintroduced during the acquisitionphase and early operation phase ... This style of ad hoc cost savings, often imposed by management early in the design stage and/or construction stage, can substantially increase costs later,especially in the after-sale portion of the use stage ... 163164Chapter 6AnnualWorth AnalysisCHAPTER SUMMARYThe annual worth method of comparing alternatives is often preferred to the present worth method,because the AW comparison is performed for only one life cycle ... AW for the first life cycle is the AW for the second, third, andall succeeding life cycles, under certainassumptions ... For infinite-life (perpetual) alternatives, the initial cost is annualized simply by multiplying Pby i ... Life-cycle cost analysis is appropriate for systems that have a large percentage of costs inoperating and maintenance ... PROBLEMSafter its 4-year life ... Annual Worth Calculations6 ... 6 Asports mortgage is an innovative way to financecash-strapped sports programs by allowing fans tosign up to pay a “mortgage” for the right to buygood seats at football games for several decadeswith season tickets locked in at current prices ... If a fan pays a $130,000 “mortgage” fee now (i ... ,in year 0)when season tickets are selling for $290each, what is the equivalent annual cost of thefootball tickets over the 50-year period at an interest rate of 8% per year?6 ... 6 ... 4 James developed the two cash flow diagrams shownat the bottom of this page ... Calculate the annual worth value of each overtherespective life cycles to demonstrate that they arethe same ... 6 ... 5 An asset with a first cost of $20,000 has an annualoperating cost of $12,000 and a $4000 salvage valueAlternative AAlternative B$1000012$25$253i = 10% peryearYear0$10002345$25$251$25$25$25$25$4000$5000$50006$25YearProblems6 ... The companysold it today for $45,000 ... A completeoverhaul at the end of year 4 cost an extra $3600 ... 6 ... The cost ofperiodic maintenance has been $800 every 2 years ... Capital Recovery6 ... He anticipated asalvage value of $50,000 after 10years ... (a) Did he recover his investmentand a 12% per year return? (b) If the annual M&Ocost was $10,000 the first year and increased by aconstant $1000 per year, was the AW positive ornegative at 12% per year? Assume the $50,000 salvage was realized ...11Sylvia has received a $500,000 inheritance fromher favorite, recently deceased aunt in Hawaii ... She hopes to make 8%per year on this purchase over an ownership periodof 20 years ... No annual M&O costs are considered in the analysis ... 12 Humana Hospital Corporation installed a newMRImachine at a cost of $750,000 this year in its medical professional clinic in Cedar Park ... Humana uses a returnrequirement of 24% per year for all of its medicaldiagnostic equipment ... Also, you are asked todraw two cash flow diagrams, one showing theMRI purchase and sale cash flow and asecond depicting the required capital recovery each year ... 13 Polypropylene wall caps, used for covering exterior vents for kitchen cooktops, bathroom fans,dryers, and other building air exhausts, can bemade by two different methods ... Method Y will have a first cost of $140,000,an operating cost of$24,000 per year, and a$19,000 salvage value after its 4-year life ... 14 Nissan’s all-electric car, the Leaf, has a base priceof $32,780 in the United States, but it is eligiblefor a $7500 federal tax credit ... Thecost for leasing the vehicle will be $4200 per year(payable at the end of each year) after aninitialization charge of $2500 paid now ... If the company expectsto be able to sell the car and charging station for40% of the base price of the car alone at the end of3 years, should the company purchase or lease thecar? Use an interest rate of 10% per year and annual worth analysis ... 15 A newstructural design software package is available for analyzing and designing three-sided guyedtowers and three- and four-sided self-supportingtowers ... A site license has a one-time cost of $22,000 ... Determine whichstrategy should be adopted at an interest rate of10% per year for a 4-year planningperiod usingthe annual worth method of evaluation ... 16 The city council in a certain southwestern city isconsidering whether to construct permanent restrooms in 22 of its smaller parks (i ... , parks of less166Chapter 6Annual Worth Analysisthan 12 acres) or pay for portable toilets on a yearround basis ...8 million ... The service life of a permanent restroom is 20 years ... 6 ... Solar cells will cost $16,600 to install andwill have a useful life of 5 years with no salvagevalue ... ,are expected to be $2400 ... Since the air sampling projectwill end in 5 years, the salvage value of the line isconsidered to be zero ... 18The cash flows for two small raw water treatmentsystems are shown ... MFFirst cost, $Annual cost, $ per yearSalvage value, $Life, yearsUF–33,000–8,0004,0003–51,000–3,50011,00066 ... Twoequivalent robot instruments are available ... Robot Watcheye will have afirst cost of $125,000, annual M&Ocosts of$27,000, and a $33,000 salvage value after its5-year life ... (a) Which robot is the better economic option?(b) Using the spreadsheet Goal Seek tool, determine the first cost of the robot not selected in(a) so that it will be the economic selection ... 20 TT Racing and Performance MotorCorporationwishes to evaluate two alternative CNC machinesfor NHRA engine building ... Machine RFirst cost, $Annual operating cost,$ per yearSalvage value, $Life, yearsMachine SϪ250,000Ϫ40,000Ϫ370,500–50,00020,000330,00056 ... For a loader that has a first cost of $39,000 andfirst-year M&Ocost of $17,000, compare theequivalent annual worth of a loader kept for 4 yearswith one kept for 5 years at an interest rate of12% per year ... 6 ... A managerasked you to determine which of the followingtwo machines will have the lower (a) capital recovery and (b) equivalent annual total cost ... MachineAuto1 has a first cost of $62,000 and anoperating cost of $21,000 in year 1, increasingby 8% per year through year 5, after which timeit will have a scavenge value of $2000 ... Permanent Investments6 ... The first cost is $200,000now, and an update budget of $100,000 every7 years forever is requested... 6 ... Use an interest rate of 10% peryear ... 25 A Pennsylvania coal mining operation has installed an in-shaft monitoring system for oxygentank and gear readiness for emergencies ... Maintenance costs are expected to be $150,000 inyear 3, $175,000 in year 4, and amounts increasing by $25,000 peryear through year 6 and remain constant thereafter for the expected 10-yearlife of the system ... 6 ... CondiϪ25,000Ϫ9,0003,0003First cost, $Annual cost, $ per yearSalvage value, $Life, yearsTorroϪ130,000Ϫ2,500150,000ϱin usage for crop irrigation, is considering thepurchase of one of the Blantonsystems ... The estimated costs and associated cash flow diagrams over a 10-year period are summarized below and on the next page,respectively, for each of the five alternatives ... For alternatives A and B, there is an extra cost of $15,000 perinstallation year to maintain the manual system inplace now... Alternative Cis a retrofit of the current manual system with nodesign or development costs, and there is no level1 option ... AlternativeXFirst cost, $Annual cost,$ per yearOverhaul every10 years, $Salvage value, $Life, yearsYZϪ90,000Ϫ40,000Ϫ400,000Ϫ20,000Ϫ650,000Ϫ13,000——Ϫ80,0007,000325,00010200,000ϱLife-Cycle Cost6 ... Alarge farming corporation in India, where depletion is occurring at an alarming rate of 1 ... 27 A new bridge across the Allegheny River inPittsburgh is expected to be permanent and willhave an initial cost of $30 million ... The annual inspection andoperating costs are estimated to be $50,000 ... 6 ... (a) Determine the alternative that is economically best ... Use aspreadsheet to answer this question ... 30 The Pentagon asked a defense contractor to estimate the life-cycle cost for a proposed light-dutysupport vehicle ... Use the cost estimates (shownin $1 million) for the 20-year lifecycle to calculate the annual LCC at an interest rateof 7% per year ... 53 ... 50 ... 15 ... 510 ... 33 ... 26 ... 20 ... 41 ... 73 ... 53 ... 37 ... 5Po͞D2 ... 296 ... Reportson the disposition of each service will also beentered by field personnel, then filed and archived by the system ...The system is expected to be widely usedover time for other aircraft maintenance scheduling ... The engineer, who mustmake a presentation next week of the best estimates of costs over a 20-year life period, has decided to use the life-cycle cost approach of costestimations ... 169Additional Problemsand FE Exam Review QuestionsCost in Year ($ millions)Cost Category123456 on 10 18Field study0 ... 1 1 ... 5Software design0 ... 9Hardware purchases5 ... 1 0 ... 1 0 ... 2 0 ... 06developmentSystem implementation1 ... 7Field hardware0 ... 0 2 ... 3 2 ... 5 0 ... 6 3 ... 76 ... S ... Proposal A involves an off-the-shelf“bare-bones” design and standard grade constructionof walls, windows, doors, and other features ... The initial cost forA will be $750,000 ... Minor remodeling willbe required in years 5, 10, and 15 at a cost of $150,000each time in order to render the units usable for20 years ... Proposal B willinclude tailored design andconstruction costs of $1 ... There will be no salvage value at the end of the20-year life ... 33 A medium-size municipality plans to develop asoftware system to assist in project selection during the next 10 years ... There are three alternativesunder consideration, identified as M, N,and O ... Use an annuallife-cycle cost approach to identify the best alternative at 8% per year ... 34 All of the following are fundamental assumptionsfor the annual worth method of analysis except:(a) The alternatives will be needed for only onelife cycle ... (c) The selected alternative will be repeated forthesucceeding life cycles in exactly the samemanner as for the first life cycle ... 6 ... (b) Find the AW of each over the life of theshortest-lived alternative ... (d) Find the AW of each alternative over its lifewithout considering the life of the otheralternatives ... 36 The annual worth of an alternative can becalculated from the alternative’s:(a) Present worth by multiplying by (A/P,i,n)(b) Future worth by multiplying by (F/A,i,n)(c) Either (a) or (b)(d) Neither (a) nor (b)6 ... At an interest rate of 10%per year, the values of n that you could use in the(A͞P,i,n) factors to make a correct comparison bythe annual worthmethod are:AFirst cost, $Annual cost, $ per yearSalvage value, $Life, years(a)(b)(c)(d)BϪ50,000Ϫ10,00013,0003Ϫ90,000Ϫ4,00015,0006n ϭ 3 years for A and 3 years for Bn ϭ 3 years for A and 6 years for BEither (a) or (b)Neither (a) nor (b)170Chapter 6Annual Worth Analysis6 ... e ... At an interest rate of10% per year, theequation that represents the perpetual AW of X1 is:X1First cost, $Annual cost, $ per yearSalvage value, $Life, years(a)(b)(c)(d)Y1Ϫ50,000Ϫ10,00013,0003Ϫ90,000Ϫ4,00015,0006AWX1 ϭ Ϫ50,000(0 ... 10)AWX1 ϭ Ϫ50,000(0 ... 10) Ϫ 10,000Ϫ 37,000(P/F,10%,3)(0 ... 10)AWX1 ϭϪ50,000(A/P,10%,3) Ϫ 10,000ϩ 13,000(A/F,10%,3)6 ... (b) Multiply $10,000 by (A/F,i,10) ... (d) Multiply $10,000 by (A/F,i,n) and thenmultiply by i ... 40 through 6 ... The alternatives are mutually exclusive and the MARR is6% per year ... 40 The annual worth of vendor 2 cash flow estimatesis closest to:(a)$Ϫ63,370(b) $43,370(c) $Ϫ43,370(d) $63,3706 ... 42 The AW values for the alternatives are listedbelow ... 43 The capital recovery amount for vendor 3 is:(a) $40,000 per year(b) $60,000 per year(c) $43,370 per year(d) $100,000 per year6 ... (b) The estimates are wrong somewhere ... (d) The alternativeshould have a longer life sorevenues will exceed costs ... 45 Estimates for one of two process upgrades are asfollows: first cost of $40,000, annual cost of $5000per year, market value that decreases by $2000 peryear to the salvage value of $20,000 after the expected life of 10 years ... 46 The perpetualannual worth of investing $50,000now and $20,000 per year starting in year 16 andcontinuing forever at 12% per year is closest to:(a) $Ϫ4200(b) $Ϫ8650(c) $Ϫ9655(d) $Ϫ10,655Case Study6 ... (b) A monetary estimate of new capital funds required each year for the life of the alternative ... Does notconsider the salvage value, since itis returned at the end of the alternative’s life ... The estimates used and theannual worth analysis at MARR ϭ 15% are summarizedbelow ... PowrUpCost and installation, $Annual maintenance cost,$ per yearSalvage value, $Equipment repair savings, $Useful life,yearsLloyd’sϪ26,000Ϫ800Ϫ36,000Ϫ3002,00025,00063,00035,00010The spreadsheet in Figure 6–9 is the one Harry used to makethe decision ... The Lloyd’s protectors were installed ... In fact, the maintenance contract cost (which includes quarterly inspection) is going from $300 to $1200 peryear nextyear and will then increase 10% per year for the next10 years ... He believes savings will decrease by $2000 per yearhereafter ... Case Study Exercises1 ... 2 ... If theseestimates had been made 3 years ago, would Lloyd’sstill have been the economic choice?3 ... CHAPTER 7Rate of ReturnAnalysis:OneProjectL E A R N I N GO U T C O M E SPurpose: Understand the meaning of rate of return and perform an ROR evaluation of a single project ... 1Definition• State and understand the meaning of rate ofreturn ... 2Calculate ROR• Use a PW or AW relation to determine theROR of a series of cash flows... 3Cautions• State the difficulties of using the ROR method,relative to the PW and AW methods ... 4Multiple RORs• Determine the maximum number of possibleROR values and their values for a specific cashflow series ... 5Calculate EROR• Determine the external rate of return usingthe techniques ofmodified ROR and return oninvested capital ... 6Bonds• Calculate the nominal and effective rate ofreturn for a bond investment ... Whether it is an engineering project with cashflow estimates or an investment in a stock or bond, the rate of return is a wellaccepted way of determining if the project orinvestment is economically acceptable ... Correct procedures to calculate a rate of return using a PW orAW relation are explained here, as are some cautions necessary when the ROR technique isapplied to a single project’s cash flows ... We will discuss the computation ofROI in the latter part of thischapter ... This chapter describes how to recognize this possibility and an approach to find the multiple values ... Two of the techniques are covered: the modified ROR technique and the ROIC (return on invested capital) technique ... Finally, the rate of return for a bond investment is discussed ... 1Interpretation of a Rate of Return ValueFrom the perspective of someone who has borrowed money, the interest rate is applied to theunpaid balance so that the total loan amount and interest are paid in full exactly with the last loanpayment ... The interest rate is the return on this unrecovered balance sothat the total amount lent andthe interest are recovered exactly with the last receipt ... Rate of return (ROR) is the rate paid on the unpaid balance of borrowed money, or the rateearned on the unrecovered balance of an investment, so that the final payment or receiptbrings the balance to exactly zero withinterest considered ... It isstated as a positive percentage; the fact that interest paid on a loan is actually a negative rate ofreturn from the borrower’s perspective is not considered ... In terms of an investment, a return of i ϭ Ϫ100%means the entire amount is lost ... Example 7 ... EXAMPLE 7 ... From thelender’sperspective, the investment in this young engineer is expected to produce an equivalent netcash flow of $315 ... A ϭ $1000(A͞P,10%,4) ϭ $315 ... Compute the amountof the unrecovered investment for each of the 4 years using (a) the rate of return on the unrecovered balance (the correct basis)and (b) the return on the initial $1000 investment ... Solution(a) Table 7–1 shows the unrecovered balance at the end of each year in column 6 using the10% rate on the unrecovered balance at the beginning of the year ... Rate of returnChapter 7Rate of Return Analysis: One ProjectTA BLE 7–1Unrecovered Balances Using a Rate of Return of 10% on the UnrecoveredBalance(1)(2)(3) 00Ϫ784 ... 51Ϫ286 ... 0078 ... 7528 ... 88$Ϫ1000 ... 47ϩ315 ... 47ϩ315 ... 47237 ... 72286 ... 00TA BLE 7–2$Ϫ1000 ... 53Ϫ547 ... 790Unrecovered Balances Using a 10% Return on the Initial Amount(1) ... 01 � ) 2(00Ϫ784 ... 06Ϫ353 ... 00ϩ315 ... 47ϩ315 ... 47—$215 ... 47215 ... 47$861 ... 00Ϫ784 ... 06Ϫ353 ... 12(b) Table 7–2 shows the unrecovered balance if the 10% return is always figured on the initial $1000 ... 12,because only $861 ... (c) As shown in column 3, a total of $400 in interest must ... 01 � ) 2( ( (2)(3be earned if the 10% return each year isbased on the initial amount of $1000 ... 88 in interest must be earned if a10% return on the unrecovered balance is used ... Figure 7–1 illustrates the correct interpretation of rate of return in Table 7–1 ... 001000 ... 45$215 ... 53Loan balance, $174$237 ... 75547 ...72$28 ... 79$286 ... Year7 ... 47 receipt represents 10% interest on the unrecovered balance in column 2plus the recovered amount in column 5 ... Clearly, an interest rate applied only to the principal represents a higher rate than is stated ... This is sometimes referred to as the installment financingproblem ... One popularexample is a “no-interest program” offered by retail stores on the sale of major appliances, audioand video equipment, furniture, and other consumer items ... Further,the program’s fine print may stipulate that the purchaser use a credit card issued by the retailcompany, which oftenhas a higher interest rate than that of a regular credit card, for example,24% per year compared to 15% per year ... Usually, the correct definition of i as interest onthe unpaid balance does not apply directly; i has often been manipulated to the financial disadvantage of the purchaser ... 4 using the CreditCard Case inChapter 4 ... 2 Rate of Return Calculation Using a PWor AW RelationThe ROR value is determined in a generically different way compared to the PW or AW value fora series of cash flows ... Using the MARR, which is established independent of any particular project’s cash flows,amathematical relation determines the PW value in actual monetary units, say, dollars or euros ... Therefore, ROR may beconsidered a relative measure, while PW and AW are absolute measures ... Another definition of rate of return isbased on our previous interpretations of PW and AW ... To determine� AWor[7 ... It is the root of the RORrelation ... Rate of return176Chapter 7Rate of Return Analysis: One � PWor[7 ... 0 the rate of return, develop the ROR equation using either a PW or AW relation, setit equal to 0, and solve for the interest rate ... That is, solve for i using either of the relations0ProjectFigure 7–2$1500Cash flow for whicha value of i is to bedetermined ... If i* Ͻ MARR, the project is not economically viable ... In rate of return calculations, the objective is to find the interest rate i* at whichthe cash flows are equivalent ... For example, if you deposit $1000 now and are promisedpayments of $500 three years from now and $1500 five years from now, the rate of return relationusing PW factors and Equation [7 ... 3]The value of i* that makes the equality correct is to be determined (see Figure 7–2) ... 3], we have the form 0 ϭ PW ... 9% by hand using trial and error or using aspreadsheet function ... Using i* ϭ 16 ... It will show that the unrecovered balances eachyear, starting with $Ϫ1000 in year 1, are exactly recovered by the $500 and $1500 receipts inyears 3 and 5 ... That is, if the above interest rate is known to be 16 ... 9%,3) ϩ 1500(P͞F,16 ... The only differences are whatis given and what is sought ... Thespreadsheet is faster; the first helps in understanding how ROR computations work ... 2 ... i* Using Trial and ErrorThe general procedure of using a PW-based equation is as follows:1 ... 2 ... 1] ... Select values of i by trial and error until the equation is balanced ... If thecash flows are combined in such a mannerthat the income and disbursements can be represented by a single factor such as P͞F or P͞A, it is7 ... The problem, then, is to combine the cash flows into the format of only one of the factors ... Convert all disbursements into either single amounts (P or F) oruniform amounts (A) byneglecting the time value of money ... The scheme selected for movementof cash flows should be the one that minimizes the error caused by neglecting the time valueof money ... 2 ... 3 ... The rate obtained is a good estimate for the first trial ... The procedure is illustrated inExample 7 ... i* by Spreadsheet The fastest way to determine an i* value when there is a series of equalcash flows (A series) is to apply the RATE function ... The format is� RATE(n,A,P,F)[7 ... The format is� IRR(first_cell:last_cell,guess)[7 ... The PW-based procedure for sensitivity analysis and agraphical estimation of the i* value isas follows:1 ... 3 ... 5 ... Set up the ROR relation in the form of Equation [7 ... Enter the cash flows onto the spreadsheet in contiguous cells ... Use the NPV function to develop a PW graph (PW versus i values) ... EXAMPLE 7 ... Engineers with Monarch Paints haverecommended to management an investment of $200,000 now in novel methods that will reduce the amount of wastewater, packagingmaterials, and other solid waste in their consumer paint manufacturing facility ... Determine the rate of returnusing hand and spreadsheet solutions ... 1 ... 2 ... 1] format forthe ROR equation ... 6]177178Chapter 7Rate of Return Analysis: One Project$300,000i* = ?$15,000012345678910$200,000Figure 7–3Cash flow diagram, Example 7 ... 3 ... All income will be regardedas a single F in year 10 so that the P͞F factor can be used ... Only for the firstestimate of i, define P ϭ$200,000, n ϭ 10, and F ϭ 10(15,000) ϩ 300,000 ϭ $450,000 ... 444The roughly estimated i is between 8% and 9% ... 0 ϭ Ϫ200,000 ϩ 15,000(P͞A,9%,10) ϩ 300,000(P͞F,9%,10)0 Ͻ $22,986The result is positive, indicating that the return is more than 9% ... 0 ϭ Ϫ200,000 ϩ 15,000(P͞A,11%,10) ϩ300,000(P͞F,11%,10)0 Ͼ $Ϫ6002Since the interest rate of 11% is too high, linearly interpolate between 9% and 11% ... 00 ϩ ———————— (2 ... 00 ϩ 1 ... 58%Solution by SpreadsheetThe fastest way to find i* is to use the RATE function (Equation [7 ... The entryϭ RATE(10,15000,Ϫ200000,300000)displays i* ϭ 10 ... It is equally correct touse the IRR function ... For a complete spreadsheet analysis, use the procedure outlined above ... 2 ... 4 ... Figure 7–3 shows cash flows ... 6] is the ROR relation ... The IRR function in cell B14 displays i* ϭ 10 ... To graphically observe i * ϭ 10 ... The NPV functionis used repeatedly to calculate PW for the xy scatterchart ... 3Special Considerations When Using the ROR Methodϭ NPV(C4,$B$3:$B$12) + $B$2ϭ IRR(B2:B12)Figure 7–4Spreadsheet to determine i* and develop a PW graph, Example 7 ... Just as i* can be found using a PW equation, it mayequivalently be determined using an AWrelation ... Solution byhand is the same as the procedure for a PW-based relation, except Equation [7 ... In thecase of Example 7 ... 55% is determined using the AW-based relation ... Internally, IRR calculates the NPV function at different i values until NPV ϭ 0 ... )7... As mentioned earlier, an ROR analysis is performed using a different basis than PW and AW analyses ... As a result, there are some assumptions and special considerations with ROR analysis that must be made when calculating i* and in interpreting its real-world meaning ... • Multiple i* values ... Thispossibility is discussed in Section 7 ... • Reinvestment at i* ... e ... However, the ROR method assumes reinvestment at the i* rate ... g ... In such cases, the i* value is not a good basis for decision making ... 5 ... To correctly use the ROR methodto choose from two or more mutually exclusive alternativesrequires an incremental analysisprocedure that is significantly more involved than PW and AW analysis ... If possible, from an engineering economic study perspective, the AW or PW method at astated MARR should be used in lieu of the ROR method ... And it is easy tocompare a proposed project’sreturn with that of in-place projects ... As an illustration, if a project is evaluated at MARR ϭ 15% and has PW Ͻ 0, there is no need tocalculate i*, because i* Ͻ 15% ... 7 ... 2 a unique rate of return i* was determined ... This is called a conventional (or simple) cash flow series ... Such a series is callednonconventional (nonsimple) ... Relatively large net cash flow (NCF) changes in amount and sign canoccur in projects that require significant spending at the end of the expected life ... The cash flow diagram will appearsimilar to Figure 7–5a ... TABLE 7–3 Examples of Conventional and NonconventionalNet Cash Flow fora 6-year ProjectSign on Net Cash Flow by YearType of Series0123456ConventionalConventionalConventionalNonconventionalNonconventionalNonconventionalϪϪϩϪϩϪϩϪϩϩϩϩϩϪϩϩϪϪϩϩϩϩϪϪϩϩϩϪϪϩϩϩϪϪϩϩϩϩϪϪϩϩPositiveNCF01Number ofSignChangesPositiveNCFPositiveNCFnϪ1 n2Year01111223nϪ12ni* = ?PhaseoutcostsInitialinvestmentInitialinvestment(a)Midlifeinvestmentsi* = ?(b)Figure 7–5Typical cash flow diagrams for projects with (a) large restoration or remediation costs, and (b) upgrade or refurbishment costs ... 4Multiple Rate ofReturn ValuesWhen there is more than one sign change in the net cash flows, it is possible that there will bemultiple i* values in the Ϫ100% to plus infinity range ... Test 1: (Descartes’) rule of signs states that the total number of real-number roots is always lessthan or equal to the number of sign changesin the series ... 1] or [7 ... (It is possible that imaginary values or infinity may also satisfy theequation ... Zero values in the series are neglected when applying Norstrom’s criterion ... There may benegative roots that satisfy the ROR relation, but these are not useful i* values ... , Sn ... With the results ofthese two tests, the ROR relation is solved for either the unique i* or themultiple i* values, using trial and error by hand, using a programmable calculator, or by spreadsheet using an IRR function that incorporates the “guess” option ... Examples 7 ... 4 illustrate the testsand solution for i* ... 3Sept-ÎlesAluminum Company operates a bauxite mine to supply its aluminum smelter located about 2 km from the current open pit ... Thelease for the land will cost $400,000 immediately ... Thisis expected to cost $300,000 ... Perform an ROR analysis that will provide the following information:(a) Type of cashflow series and possible number of ROR values(b) PW graph showing all i* values(c) Actual i* values determined using the ROR relation and spreadsheet function(d) Conclusions that can be drawn about the correct rate of return from this analysisSolution(a) The net cash flows will appear like those inFigure 7–5a with an initial investment of$Ϫ400,000, annual net cash flow (NCF) of $75,000 for years 1 through 10, and a phaseout cost of $Ϫ300,000 in year 10 ... Theseries is nonconventional based on the sign changes throughout the series ... Test #2: There is one sign change in the cumulative NCFseries, which indicates a uniquepositive root or one positive i* value ... 3 ... There are two times thatthe parabolic-shaped curve crosses the PW ϭ 0 line; these are approximately i1* ϭ Ϫ18%and i2*ϭ 5% ... 7]i* values by hand If hand solution is chosen, the same procedure used in Example 7 ... However,the technique to estimate the initial i value will not workas well in this case since the majority of the cash flows do not fit either the P͞F or theF͞P factor ... 25% ... 7] with various i values will approximate thecorrect answer of about 4 ... This complies with the test results of one positive i* value ... Enteringdifferent values in theoptional “guess” field will force the function to find multiple i* values, if they exist ... i1* ϭ Ϫ18 ... 53%This result does not conflict with test results, as there is one positive value, but a negativevalue also balances the ROR equation ... 53% is accepted as the correct internal rate ofreturn (IROR) for theproject ... EXAMPLE 7 ... does contract-based work forautomobile manufacturers throughout the world ... Year0123Cash Flow ($1000)ϩ2000Ϫ500Ϫ8100ϩ68007 ... (b) Write the PW equation and approximate the i* value(s) by plotting PW vs i ... Since there are twosign changes in thecash flow sequence, the rule of signs indicates a maximum of two i*values ... As many as two i* values can be found ... 4YearCash Flow($1000)SequenceNumber0123ϩ2000Ϫ500Ϫ8100ϩ6800Cumulative Cash Flow($1000)S0S1S2S3ϩ2000ϩ1500Ϫ6600ϩ200(b) The PW relation isPW ϭ 2000 Ϫ500(P͞F,i,1) Ϫ 8100(P͞F,i,2) ϩ 6800(P͞F,i,3)The PW values are shown below and plotted in Figure 7–7 for several i values ... i%PW ($1000)51020304050ϩ51 ... 55Ϫ106 ... 01Ϫ11 ... 85Figure 7–7100Present worth of cashflows at several interestrates, Example 7 ... 75PW (ϫ $1000)5025i%01020304050Ϫ25Ϫ50Ϫ75Ϫ100Ϫ125Figure 7–8 presents the spreadsheet PW graph with the PW curve crossing the x axis atPW ϭ 0 two times ... The values arei1* ϭ 7 ... 35%(c) Since both i* values are positive, they are not of much value, because neither can beconsidered the true ROR of the cash flow series ...4 ... This problem is a good example of when anapproach discussed in the next section should be taken ... The function will find the one ROR closest to 10% that satisfies the PW relation ... Often,the results are unbelievable or unacceptable values that are rejected ... Assume there are two i* values for aparticular cash flow series ... Both i* Ͼ 0Discard both values ... If both i* values are discarded, proceed to the approach discussed in the next section to determineone rate of return value for the project ... Always determine the PW or AW at the MARR first for a reliable measure of economic justification ...This recommendation is not to dissuade you from using the ROR method ... 7 ... 5Techniques to Remove Multiple Rates of Return• More than one positive i* value or all negative i* values are obtained when the PW graph andIRR function are developed, and• A single, reliable rate of return value isrequired by management or engineers to make a cleareconomic decision ... The selected approach dependsupon what estimates are the most reliable for the project being evaluated ... The result of follow-up analysis to obtain a single ROR value when multiple, nonuseful i* valuesare present does notdetermine the internal rate of return (IROR) for nonconventional netcash flow series ... We will refer to the resulting value as the external rate of return (EROR) as a reminder thatit is different from the IROR obtained in all previous sections ... Take the following view: You arethe project manager and theproject generates cash flows each year ... We will call thisthe investment rate ii ... Other years, the net cash flowwill be negative and you must borrow funds from some source to continue ... Each year, you must consider the time value of money, which must utilize either theinvestment rate or the borrowingrate, depending upon the sign on the NCF of the preceding year ... The resulting ROR value will not be the same for each method, because slightlydifferent additional information is necessary and the cash flows are treated in slightly differentfashions from the time value of money viewpoint ... However, theinvestment and borrowing rates must be reliably estimated, since the results may be quite sensitive to them ... Return on Invested Capital (ROIC) Approach Though more mathematically rigorous,this technique provides a more reliable estimate of the EROR and it requires only the investmentrate ii ...Before covering the techniques, it would be good to review the material in Section 7 ... 1 ... Modified ROR ApproachThe technique requires that two rates external to the project net cash flows be estimated ... This applies to all positive annual NCF ... • Borrowing rate ib is the rate at which funds areborrowed from an external source to providefunds to the project ... The weighted average cost ofcapital (WACC) can be used for this rate ... However, this is nota good idea as it implies that the company is willing to borrow funds and invest in projects at thesame rate ... Commonly MARR Ͼ WACC, sousually ii Ͼ ib ... 9 for a quick reviewof MARR and WACC and Chapter 10 for a more detailed discussion of WACC ... Figure 7–9 is a reference diagram that has multiple i* values, since the net cash flows changesign multiple times ... 1 ... 2 ... 3 ... FWn ϭ PW0 (F͞P,iЈ%,n)[7 ... 9]4 ... If iЈ Ͼ MARR, theproject is economically justified ... As in other situations, on the rare occasion that iЈ ϭ MARR, there is indifference to the project's economic acceptability; however, acceptance is the usual decision ... 5The cash flows experienced by Honda Motors in Example 7 ... There aretwo positive i* values thatsatisfy the PW relation, 7 ... 35% per year ... Studies indicate that Honda has a WACC of8 ... Due to the nature of this contract business, any excess funds generated are expectedto earn at a rate of 12% per year ... 5Techniques to Remove Multiple Rates of ReturnSolution by SpreadsheetUsing theinformation in the problem statement, the rate estimates are as follows:MARR:Investment rate, ii:Borrowing rate, ib:9% per year12% per year8 ... Figure 7–10 shows the result of i' ϭ 9 ... Since 9 ... ϭ MIRR(B2:B5,E2,E3)Figure 7–10Spreadsheet application of MIRR function, Example 7 ... It is vital that theinterpretation be correct ... 39% is not the internal rate of return(IROR); it is the external ROR (EROR) based on the two external rates for investing and borrowing money ... Step 1 ... 5% ... 5%,1) Ϫ 8100(P͞F,8 ... Find FW3 of all positive NCF at ii ϭ 12% ... Find the rate i' at which the PW and FW areequivalent ... 939 (9 ... Since i' Ͼ MARR of 9%, the project is economically justified using this ERORapproach ... Return on invested capital (ROIC) is a rate-of-return measure of how effectively a project utilizesthe funds invested in it, that is, funds that remain internal to the project ... 187188Chapter 7Rateof Return Analysis: One ProjectThe technique requires that the investment rate ii be estimated for excess funds generated in anyyear that they are not needed by the project ... It involves developing a series of futureworth (F) relations moving forward 1 year at a time ... Usually, ii is set equal to the MARRϾ 0(extra 1؊� net cash flow in year tiif Ft � future worth in year t based on previous year and time value of moneyNCFt ... The ROIC method uses the following procedure to determine a single external rate of return iЉ and to evaluate the economic viability of theproject ... 1 ... , n years) ... 10]where FtϽ 0(project uses all available funds){2 ... The iЉ value is the ROIC for the specified investment rate ii ... Fortunately, the Goal Seek spreadsheet tool can assist in the determination of iЉ because there is only one unknown in the Fn relation and the target value is zero ... 6 ... 1؊funds available)k� iiЉif FtThe guideline for economic decision making is the same as above, namely,If ROIC Ն MARR, the project is economically justified ... It is important to remember that the ROIC is an external rate of return dependent uponthe investment rate choice ... This is a separate technique to find a single rate for theproject ... 6Once again, we will use the cash flows experienced by Honda Motors in Example 7 ... Use the ROIC method to determine the EROR value ... Year0Net Cash Flow ($1000) ϩ20001Ϫ5002Ϫ81003ϩ6800Solution by HandThe hand solution is presented first to provide the logic of the ROIC method... Figure 7–11details the cash flows and tracks the progress as each Ft is developed ... 10] is applied to develop each Ft ... Year 0:Year 1:F0 ϭ $ϩ2000Since F0 Ͼ 0, externally invest in year 1 at ii ϭ 12% ... 12) Ϫ 500 ϭ $ϩ1740Since F1 Ͼ 0, use ii ϭ 12% for year 2 ... 5189Techniques to Remove Multiple

a)(b)(c)(d)Figure 7–11Application of ROIC method at ii ϭ 12% per year: (a) original cash flow; equivalent form in (b) year 1, (c) year 2, and (d) year 3 ... 12) Ϫ 8100 ϭ)$8100$8100� ؊$ 1516 50023012301230123YearF2$01 ؆ = ? � 0471$F3 = 0at i $6800F1� 0002$ Rates of Return$6800$6800F0$Ϫ6151Now F2 Ͻ 0, use iЉ for year 3, according to Equation [7 ... Year 3:F3 ϭ Ϫ6151(1 ϩ iЉ) ϩ 6800This is the last year ... Go to step 2 ... Solve for iЉ ϭ ROIC from F3 ϭ 0 ... 1055 (10 ... Since ROIC > MARR ϭ 9%, the project is economically justified ... The future worth values F1 through F3 aredetermined by the conditional IF statements in rows 3 through 5 ... In each year, Equation [7 ... If there are surplus funds generated by the project,Ft–1 Ͼ 0 and the investment rate ii (in cell E7) is used to find Ft ... Figure 7–12Spreadsheet application of ROIC method using Goal Seek, Example 7 ...190Chapter 7Rate of Return Analysis: One ProjectThe Goal Seek template sets the F3 value to zero by changing the ROIC value (cell E8) ... 55% per year ... 55% Ͼ 9%, the MARR, the projectis economically justified ... 55%) is different than the MIRR rate (9 ... Alsothese are both different than themultiple rates determined earlier (7 ... 35%) ... Now that we have learned two techniques to remove multiple i* values, there are some connections between the multiple i* values, the external rate estimates, and the resulting externalrates (iЈ and iЉ) obtained by the two methods ... That is, all fourparameters have the same value ... Finally, it is very important to remember the following fact ... When the MARR is established, this is, in effect, fixing the i* value ... 7 ... One very common form of IOU is a bond—a longterm note issued by a corporation or a government entity (the borrower) to financemajor projects ... Bonds are usually issued in face value amounts of $1000, $5000,or $10,000 ... The bond dividend is paid c times peryear ... The amount of interest isdetermined using the stated dividend or interest rate, called the bond coupon rate b ... 11]There are many types or classifications of bonds... For example, Treasury securities are issued in different monetary amounts ($1000 andup) with varying periods of time to the maturity date (Bills up to 1 year; Notes for 2 to 10 years) ... S ... ” The safe investment rate indicatedin Figure 1Ϫ6 as the lowest level for establishing a MARR is the coupon rateon a U ... Treasurysecurity ... 7 ... 7General Electric just released $10 million worth of $10,000 ten-year bonds ... (a) Determine the amount a purchaser will receive each6 months and after 10 years ... What are the dividend amounts and the final payment amount at the maturity date?Solution(a) UseEquation [7 ... 10,000 (0 ... (b) Purchasing the bond at a discount from face value does not change the dividend or final repayment amounts ... The cash flow series for a bond investment is conventional and has one unique i*, which isbest determined by solving a PW-based rate of return equation in theform of Equation [7 ... EXAMPLE 7 ... It is offering small-denomination bonds at a discount price of $800 for a 4% $1000 bondthat matures in 20 years with a dividend payable semiannually ... The PW-based equation for calculating the rate of return is0 ϭ Ϫ800 ϩ 20(P͞A,i*,40) ϩ 1000(P͞F,i*,40)Solve by theIRR function or by hand to obtain i* ϭ 2 ... The nominal interest rate per year is computed by multiplying i* by 2 ... 8435)(2) ϭ 5 ... 5], the effective annual rate isia ϭ (1 ... 7678%191192Chapter 7Rate of Return Analysis: One ProjectEXAMPLE 7 ... He took a financial risk and bought a bond from acorporation thathad defaulted on its interest payments ... The bond paid no interest for the first 3 years after Gerry bought it ... Perform hand and spreadsheet analysis ... 08)I ϭ —————— ϭ $200 per quarter4The effective rate of return per quarter can be determined by solving the PW equationdeveloped on a per quarter basis ... 1% per quarter, which is a nominal 16 ... Solution by SpreadsheetOnce all the cash flows are entered into contiguous cells, the function ϭ IRR(B2:B42) is usedin Figure 7–13, row 43, to display the answer of a nominal rate of return of 4 ... (Note that many of the rowentries have been hidden to conserve space ... 10%(4) ϭ 16 ... Figure 7–13Spreadsheet solution fora bond investment,Example 7 ... If a bond investment is being considered and a required rate of return is stated, the samePW-based relation used to find i* can be used to determine the maximum amountto pay forthe bond now to ensure that the rate is realized ... As an illustration, in the lastexample, if 12% per year, compounded quarterly, is the target MARR, the PW relation isused to find the maximum that Gerry should pay now; P is determined to be $6004 ... 0 ϭ ϪP ϩ 200(P͞A,3%,28)(P͞F,3%,12) ϩ11,000(P͞F,3%,40)P ϭ $6004Problems193CHAPTER SUMMARYThe rate of return of a cash flow series is determined by setting a PW-based or AW-based relationequal to zero and solving for the value of i* ... Most people, however, can have considerable difficulty in calculating a rate of returncorrectly foranything other than a conventional cash flow series ... The maximum number of i* values is equal to the number ofchanges in the sign of the net cash flow series (Descartes’ rule of signs) ... When multiple i* values are indicated, either of the two techniques covered in this chapter canbe applied to find asingle, reliable rate for the nonconventional net cash flow series ... Usually, the investment rate is setequal to the MARR, and the borrowing rate takes on the historical WACC rate ... If an exact ROR is not necessary, it is strongly recommended that the PW or AW method atthe MARR be used to decideupon economic justification ... 1 Under what circumstances would the rate of returnbe (a) Ϫ100%, and (b) infinite?7 ... How much extra interest didthe company pay?7 ... 6 In 2010, the city of Houston, Texas, collected$24,112,054 in fines from motorists because oftraffic violations caught by red-lightcameras ... Thenet profit, that is, profit after operating costs, issplit equally (that is, 50% each) between the cityand the operator of the camera system ... 4 Assume you borrow $50,000 at 10% per year interest and you agree to repay the loan in five equalannual payments ... 7 P&G sold its prescriptiondrug business toWarner-Chilcott, Ltd ... 1 billion ... 5 International Potash got a $50 million loan amortized over a 10-year period at 10% per year interest ... (a) What is the amount of each payment?(b) What is the total amount of interest paid?How does the total interest paid comparewith the principal of theloan?7 ... As a result of the claimant payouts,insurance companies raised homeowners' insurance rates by an average of $59 per year for eachof the 160,000 households in the affected city ... 9 Determine the rate of return for the cash flows shownin the diagram ... )$7000i=?01234$200$200567$90$908$200Year$90$3000to 1,694,247 in 2015 ... 14 U ... Census Bureau statistics show that the annualearnings for persons with a high school diplomaare $35,220 versus $57,925 for someone with abachelor’s degree ... (Hint: The investment inyears 1 through 4 is the cost ofcollege plus theforegone earnings, and the income in years 5through 35 is the difference in income between ahigh school diploma and a bachelor’s degree ... 10 The Office of Naval Research sponsors a contestfor college students to build underwater robots thatcan perform a series of tasks withouthuman intervention ... If theteam spent $2000 for parts (at time 0) and the project took 2 years, what annual rate of return did theteam make?7 ... The bonds carried a 5 ... The U ... economy wasin a recession at that time, so as part of the federalstimulus program, the Utility gets a 35% reimbursement onthe dividend it pays ... 11 For the cash flows shown, determine the rate of(b) What is the total dollar amount the Utility willreturn ... 12 In an effort to avoid foreclosure proceedings onstruggling mortgage customers, Bank of Americaproposed an allowance that a jobless customermake no payment on theirmortgage for up to9 months ... The bank would give them$2000 for moving expenses ... If the banksaved $40,000 in foreclosure costs, what rate ofreturn per month did the bank make on the allowance? Assume the first payment that was skippedwas due at the end of month 1 and the $40,000foreclosuresavings and $2000 moving expenseoccurred at the end of the 9-month forbearanceperiod ... 13 The Closing the Gaps initiative by the TexasHigher Education Coordinating Board establishedthe goal of increasing the number of students inhigher education in Texas from 1,064,247 in 20007 ... The contractalso required the company to provide buyout packagesfor 400 workers ... 7 ... Rubbersidewalks, Inc ... The District of Columbia spent $60,000 fora rubber sidewalk to replace broken concrete in aresidential neighborhood lined with towering willow oaks ... 18 Efficient light jets (ELJs) are smaller aircraftthatmay revolutionize the way people travel by plane ... 5 and $3 million, seat 5 to 7people, and can fly up to 1100 miles at cruisingspeeds approaching 425 mph ... The company invested $500 million(time 0) and began taking orders 2 years later ... 8 million, what rate of returnwill the company make overa 10-year planningperiod? Assume 500 of the planes are deliveredeach year in years 6 through 10 and that the company’s M&O costs average $10 million per year inyears 1 through 10 ... )7 ... E-sports entertainment inNew York City purchased five machines for $6000each and took in an average of$600 total per weekin sales ... 7 ... The projecttook 10 years of planning and cost $4 million ... What rate of return does theventure represent, if increased fishing and recreation activities are valued at $270,000 per yearbeginning in year 11 and they continue in perpetuity? (If assigned by your instructor,show bothhand and spreadsheet solutions ... 24 According to Descartes’ rule of signs, what is themaximum number of real-number values that willbalance a rate of return equation?7 ... 26 According to Norstrom’s criterion, there are tworequirements regarding the cumulative cash flowsthat must besatisfied to ensure that there is onlyone positive root in a rate of return equation ... 27 According to Descartes’ rule of signs, how manypossible i* values are there for the cash flowsshown?Year123456Net Cash ϩ4100 Ϫ2000 Ϫ7000 ϩ12,000 Ϫ700 ϩ800Flow, $7 ... 29 According to Descartes’ rule andNorstrom’s criterion, how many i* values are possible for the cashflow (CF ) sequence shown?Year12345Net Cashϩ16,000 Ϫ32,000 Ϫ25,000 ϩ50,000 Ϫ8,000Flow, $Cumulative ϩ16,000 Ϫ16,000 Ϫ41,000 ϩ9,000 ϩ1,000CF, $7 ... Multiple ROR ValuesYear7 ... 22 Explain at least three types of projects inwhichlarge net cash flow changes may cause sign changesduring the life of the project, thus indicating thepossible presence of multiple ROR values ... 31 Stan-Rite Corp of Manitowoc, Wisconsin, is a B toB company that manufactures many types of industrial products, including portable measuringarmswith absolute encoders, designed to perform3D inspections of industrial parts ... 7 ... 0123425,000 15,0004,00018,000Ϫ30,000 Ϫ7,000 Ϫ6,000 Ϫ12,000196Chapter 7Rate of Return Analysis: One ProjectYearExpensesRevenues0123456$Ϫ30Ϫ20Ϫ25Ϫ15Ϫ22Ϫ20Ϫ30$01819365238707 ... At the end ofFebruary, she spent this$50 and an additional $150 to buy clothes ... Herconclusion was that over the 4 months, she had received $25 more than she spent ... If so, determine the multiple rates and comment on their validity ... 33 Veggie Burger Boy sells franchises to individualswho want to start small inthe sandwiches-forvegeterians business and grow in net cash flowover the years ... He was allowed to borrow atthe end of his first year from the corporation’s capital incentive fund with a promise to repay the loanin addition to the annual share that the corporationcontractually receives from annual sales... Year01NCF, $ 5000 Ϫ10,100250034562000 2000 2000 2000The corporate chief financial officer (CFO) hassome questions concerning this NCF series ... (a) Plot the PW versus i graph and estimate therate of return for this franchise ... (c) Basing your conclusions on Descartes’ andNorstrom’s rules,provide the CFO withsome advice on what ROR value is the mostreliable for this franchise over the 6-year period ... based on an older technology to produce meatproducts ... In 2012, prior to the sale of thefacility and property, Vaught spent $1 million tomake the site environmentally acceptable to apotential buyer ... Use a spreadsheet to dothe following ... (b) Find all rates that are real numbers betweenϪ25% and ϩ50%, and calculate the PWvalue for interest rates in this range ... YearϪ5Ϫ6 Ϫ107 ... The service was not well receivedafter the first year and was removed from the market ... Now, inyear5, VistaCare has spent a large sum on research tobroaden the application of this service ... NCF valuesare in $1 million units ... 36 In calculating the external rate of return by themodified rate of return approach, it is necessary touse two different rates of return, the investment rateii and the borrowingrate ib ... 37 In the modified rate of return approach for determining a single interest rate from net cash flows,state which interest rate is usually higher, the investment rate ii or the borrowing rate ib ... 7 ... Year0123Net Cash Flow, $ ϩ16,000 Ϫ32,000 Ϫ25,000 ϩ70,0007 ... 34 In 2011, Vaught Industriesclosed its plant inMarionsville following labor, environmental, andsafety problems ... The cash flows shown are those he recorded for thefirst 6 years as his own boss ... (After using the procedure, use the MIRRfunction to confirm your answer ... 40 Samara, an engineer working for GE, invested herbonusmoney each year in company stock ... e ... At the endof year 7, she sold the stock for $52,000 to buy acondo; she purchased no stock that year ... Samarasold all of the remaining stock for $28,000 immediately after the investment at the end of year 10 ... (b) Determine the external rate of return byhand,using the modified rate of return approach with an investment rate of 12% peryear and a borrowing rate of 8% ... (d) Enter the cash flows into a spreadsheet, and usethe IRR function to find the i* value ... Explain why this is so, given that the investmentrate is 12% per year ... )7 ... of Solon, Ohio, makesvariable areaflowmeters (VAFs) that measure liquid and gas flowrates by means of a tapered tube and float ... The revenue was $160,000 per year inyears 1 through 10 ... 7 ... (a)(b)Cash Flow, $100001234Calculate the external rate of return using thereturn on invested capital (ROIC) approachwith aninvestment rate of 15% per year ... )6NCF, $ Ϫ9000 ϩ4100 Ϫ2000 Ϫ7000 ϩ12,000 ϩ700 ϩ800Year(c)Ϫ653084Ϫ10Ϫ12Determine the number of positive roots to therate of return relation ... 7 ... YearCash Flow, $012343000Ϫ20001000Ϫ600038007 ... Theproduct did poorly after only 1 year on the market ...New development funds have been expended this year(year 5) at a cost of $1 ... Determine theexternal rate of return using the ROIC approachand an investment rate of 15% per year ... 1% per year ... 45 What is the bond coupon rate on a $25,000 mortgage bond that has semiannual interest paymentsof$1250 and a 20-year maturity date?7 ... What are the amount and frequency of the dividend payments?7 ... 48 What is the present worth of a $50,000 debenturebond that has a bond coupon rate of 8% per year,payable quarterly? The bond matures in 15 years ... 198Chapter 7Rate of Return Analysis:One Project7 ... If thebond maturity date is 20 years from the date theywere issued and the interest rate in the marketplaceis now 12% per year, compounded semiannually,what is the present worth (now) of one bond?issued 30-year bonds with a face value of $25 million ... Because the marketinterestrate increased immediately before the bonds weresold, the city received only $23 ... What was the semiannual interest ratewhen the bonds were sold?7 ... 125 million to improve the VanBuren dam in central El Paso and to finance threeother drainage projects ... If the bonddividend rate would havebeen 4% per year, payable quarterly, with a bond maturity date 18 yearsafter issuance, what is the present worth of thedividend savings to EPWU rate payers? Assumethe market interest rate is 6% per year ... 55 An investor who purchased a $10,000 mortgagebond today paid only $6000 for it ...Because the bond is in default, it will pay no dividend for the next 2 years ... 51 A recently issued industrial bond with a face valueof $10,000 has a coupon rate of 8% per year, payableannually ... Jeremy is interested in buying one bond ... 52 Due to a significant troop buildup at the local military base, aschool district issued $10,000,000 inbonds to build new schools ... If an investor is able to purchase one of the bonds that has a face value of$5000 for $4800, what rate of return per 6 monthswill the investor realize? Assume the bond is keptto maturity ... 53 As the name implies, a zero-coupon bond paysnodividend, only the face value when it matures ... 54 To provide infrastructure in the outlying areas ofMorgantown, West Virginia, the city council7 ... The bonds have a call date of this year ifGSI decides to take advantage of it ... If the company buysthe bonds back now for $11 million, determine therateof return that the company will make (a) perquarter and (b) per year (nominal) ... )7 ... Market interest ratesdropped, and the company called the bonds (i ... ,paid them off in advance) at a 10% premium on theface value ... 5 million to retire the bonds ... 58 All of the following mean the same as rate ofreturnexcept:(a) Internal rate of return(b) Time for return of capital(c) Interest rate(d) Return on investment7 ... 60 The internal rate of return on an investment refersto the interest rate earned on the:(a) Initial investment(b) Unrecovered balance of the investment(c) Money recovered from an investment(d)Income from an investment7 ... (b) The interest rate you get is a simple interestrate ... (d) The total of the cumulative cash flows isequal to 0 ... 62 According to Descartes’ rule of signs, for a netcash flow sequence of Ϫ ϪϩϩϪϩ, the number ofpossible i values is:(a) 2(b) 3(c) 4(d) 57 ... (b) The cumulativecash flow must start outnegatively ... (d) The net cash flow must start out positively ... 64 According to Descartes’ rule and Norstrom’s criterion, the number of positive i* values for thefollowing cash flow sequence is:Year1Revenue, $Cost, $(a)(b)(c)(d)23425,00030,00015,0007,0004,0006,00018,00012,00012347 ... 66 A company that uses a minimum attractive rate ofreturn of 10% per year is evaluating new processesto improve operational efficiency ... Alternative IAlternative JϪ40,000Ϫ15,0005,0003First cost, $Annual cost, $ per yearSalvagevalue, $Life, yearsϪ50,000Ϫ12,0005,0006The statement that is most correct is:(a) The alternatives are revenue alternatives ... (c) The alternatives are revenue alternatives andDN is an option ... 7 ... uses a MARR of 8% peryear ... The estimate associated with the process follows ... 68 When one is usingthe modified ROR method to remove multiple ROR values, an additional estimateneeded besides the cash flows and their timings is:(a) The ROIC value(b) External rate of return(c) Investment rate(d) Internal rate of return7 ... The correctcomputation for the present worth in year 0 is:Year1NCF, $(a)(b)23Ϫ10,0000045Ϫ19,000 ϩ25,000Ϫ10,000 Ϫ 19,000(P͞F,12%,4)Ϫ10,000 Ϫ 19,000(P͞F,12%,4) ϩ25,000(P͞F,10%,5)200Chapter 7(c)(d)25,000(P͞F,10%,5)Ϫ10,000 Ϫ 19,000(P͞F,10%,4)7 ... If the future worth computation in year t isFt Ͻ 0, the ROIC rate iЉ is used ... (b) The resulting external rate of returnwill bepositive ... (d) The sequence has nonremovable negativeROR values ... 71 The meaning of return on invested capital for acorporation is best stated as:(a) A rate-of-return measure that equates the internal and external ROR(b) A measure of how effectively the corporationuses capital funds investedin it(c) The value at which borrowing ROR and investing ROR are equal(d) The external rate of return value is based ontotal capital invested7 ... The bond matures20 years from now ... 73 A $20,000 mortgage bond that is due in 1 yearpays interest of $500 every 3 months ... 5% per year, payablequarterly(b) 5% per year, payable quarterly(c) 5% per year, payable semiannually(d) 10% per year, payable quarterly7 ... If an investor purchases the bond now for $9000 and holds it to maturity, the rate of return received can be determinedby the following equation:(a) 0 ϭ Ϫ9000 ϩ 400(P͞A,i,10)ϩ10,000(P͞F,i,10)(b) 0 ϭ Ϫ9000 ϩ 400(P͞A,i,20)ϩ 10,000(P͞F,i,20)(c) 0 ϭ Ϫ10,000 ϩ 400(P͞A,i,20)ϩ 10,000(P͞F,i,20)(d) 0 ϭ Ϫ9000 ϩ 800(P͞A,i,10)ϩ 10,000(P͞F,i,10)7 ... The bond is for salenow for $4500 ... Over time they decided to work on solar energy production ideas ... For residential applications, thecollector could be mounted along side aTV dish and be programmed to track the sun ... The system serves as a supplement to the electricity providedby the local power company ... This was great news for lowincome dwellers on government subsidy that are required topay their own utility bills ... Net cashflow after all expenses, loan repayment,and taxes for the first 4 years was acceptable; $55,000 at theCase Studyend of the first year, increasing by 5% each year thereafter ... However, after serious discussion replaced the initialexcitement of the sales offer, the trio decided to not sell at thistime ... Duringthe next year, the fifth year of the partnership, theengineer who had received the patents upon which the collector and generator designs were based became very displeasedwith the partnering arrangements and left the trio to go intopartnership with an international firm in the energy business ... Net cashflow dropped to $40,000 in year 5 and continued todecrease by $5000 per year ... Thiswas considered too much of a loss, so the two owners did notaccept ... It is now 12 years since the system was publicly launched ... 201Case Study ExercisesIt is now 12 years after the products were developed, andtheengineers invested most of their savings in an innovativeidea ... To help with the analysis, determine the following:1 ... 2 ... 3 ... 4 ... Is thereany indication that multiple rates of return may be present? If so, use the spreadsheet already developed tosearch for ROR values in the range Ϯ100% otherthanthe one determined in exercise 3 above ... Assume you are an investor with a large amount ofready cash, looking for an innovative solar energy product ... Explain your logic foroffering this amount ... SECTIONTOPICLEARNING OUTCOME8 ... 8 ... 8 ... 8 ... 8 ... 8 ... 8 ... This chapter presents themethods by which two or more alternatives can be evaluated using a rate of return (ROR) comparison based on the methods of the previouschapter ... The ROR analysis evaluates the increments between two alternatives in pairwisecomparisons ... 8 ... As we have learned, the PW and AW techniquescan be used to do so, and are the recommended methods ... Let’s assume that a company uses a MARR of 16% per year, that the company has $90,000available for investment, and that two alternatives (A and B) are being evaluated ... AlAternative B requires $85,000 and has an i* of 29% per year ...However, this is notnecessarily so ... What happens to the investment capital that isleft over? It is generally assumed that excess funds will be invested at the company’s MARR,as we learned in previous chapters ... If alternative A is selected, $50,000 will return35% per year ... The rateof return on thetotal capital available, then, will be the weighted average ... 35) ϩ 40,000(0 ... 6%90,000If alternative B is selected, $85,000 will be invested at 29% per year, and the remaining $5000will earn 16% per year ... 29) ϩ 5000(0 ... 3%90,000These calculations show that even though the i* for alternative A ishigher, alternative B presentsthe better overall ROR for the $90,000 ... This simple example illustrates a major fact about the rate of return method for ranking andcomparing alternatives:Under some circumstances, project ROR values do not provide the same ranking of alternativesas do PW and AWanalyses ... When independent projects are evaluated, no incremental analysis is necessary betweenprojects ... Therefore, the only comparison is with the do-nothing alternative for each project ... 8 ... Based upon the equivalence relations (PW and AW), RORevaluation makes the equal-serviceassumption ... Therefore,the LCM (least common multiple) of lives for each pairwise comparison must be used ... A format for hand or spreadsheet solutions is helpful (Table 8–1) ... At the end of each life cycle, the salvage value and initial investment for the next cyclemust be included for the LCM case ...Allincremental cash flows outside the period are neglected ... Only for the purpose of simplification, use the convention that between two alternatives, theone with the larger initial investment will be regarded as alternative B ... 1]The initial investment and annual cash flows for each alternative (excludingthe salvage value)are one of the types identified in Chapter 5:Revenue alternative, where there are both negative and positive cash flowsCost alternative, where all cash flow estimates are negativeRevenue or costalternativeIn either case, Equation [8 ... EXAMPLE 8 ... The company has the opportunity tobuy aslightly used machine for $15,000 or a new one for $21,000 ... Each machine is expected to have a 25-year lifewith a 5% salvage value ... SolutionIncremental cash flow is tabulated in Table 8–2 ... The salvage values in year 25 are separatedfrom ordinary cash flow for clarity ... However, rememberthat several years were combined when performingthe analysis ... 8 ... 1YearUsed PressNew PressIncrementalCash Flow(New – Used)01–2525$Ϫ15,000Ϫ8,200ϩ750$Ϫ21,000Ϫ7,000ϩ1,050$Ϫ6,000ϩ1,200ϩ300Cash FlowEXAMPLE 8 ... Type A has an initial cost of $70,000 anda life of 8 years ... Theannual operating cost for type A is expected to be $9000, while the AOC for type B isexpected to be $7000 ... Solution by HandThe LCM of 8 and 12 is 24 years ... Solution by SpreadsheetFigure 8–1 shows the incremental cash flows for the LCM of 24 years ... The incrementalvalues in column D are theA)$ Ϫ70,000Ϫ9,000Ϫ70,000Ϫ9,000ϩ5,000Ϫ9,000$ ؊ result of subtractions of column B from C ... The total incremental cash flow shouldagree in both the column D total and the subtraction C29 Ϫ B29 ... This possible dilemma is discussed later in the chapter ... 2Type AType BIncrementalCash Flow(BϪ95,000–7,000$Ϫ25,000ϩ2,000–7,000ϩ67,000–7,000–95,000–7,000ϩ10,000–7,000ϩ2,000Ϫ83,000–7,000ϩ67,000–7,000–7,000ϩ10,000$Ϫ338,000ϩ2,000Cash FlowYear01–789–111213–151617–2324Ϫ9,000Ϫ9,000Ϫ70,000Ϫ9,000ϩ5,000Ϫ9,000Ϫ9,000ϩ5,000$Ϫ411,000ϩ2,000ϩ7,000$ϩ73,000205206Chapter 8Rate of Return Analysis: Multiple AlternativesFigure 8–1Spreadsheet computation of incrementalcash flows forunequal-life alternatives, Example 8 ... Starting new life cycle for Aϭ initial costϩ AOC ϩ salvage� ؊ 000,07 ؊ 000,9 ؉ 000,5Check on summationsIncremental column shouldequal difference of columns8 ... This is important in an incrementalROR analysis in order to determine the ROR earned on the extra funds expended for the largerinvestment alternative ... In Example 8 ... If thenew machine is purchased, there will be a “savings” of $1200 per yearfor 25 years, plus an extra $300 in year 25 ... If theequivalent worth of the savings is greater than the equivalent worth of the extra investment at theMARR, the extra investment should be made (i ... , the larger first-cost proposalshouldbe accepted) ... It is important to recognize that the rationale for making the selection decision is the same asif only one alternative were under consideration, that alternative being the one represented by theincremental cash flow series ... As further clarification of this extra investment rationale,consider the following: Therate of return attainable through the incremental cash flow is an alternative to investing at theMARR ... 1 states that any excess funds not invested in the alternative are assumed tobe invested at the MARR ... 8 ... Accordingly,prior to performing an incremental ROR analysis, it isadvisable to determine the internal rate ofreturn i* for each alternative ... The guideline is as follows:For multiple revenue alternatives, calculate the internal rate of return i* for each alternative,and eliminate all alternatives that have an i* Ͻ MARR ... As an illustration, if the MARR ϭ 15% and twoalternatives have i* values of 12% and 21%,the 12% alternative can be eliminated from further consideration ... If both alternatives have i* Ͻ MARR, no alternative isjustified and the do-nothing alternative is the best economically ... Alternatives that cannot meet the MARR may be eliminated fromfurtherevaluation using this option ... The IRR function applied to each alternative’s cash flow estimates can quickly indicate unacceptable alternatives, as demonstrated in Section 8 ... When independent projects are evaluated, there is no comparison on the extra investment ... For example, assumeMARR ϭ 10%, and three independent projects are availablewith ROR values of*iA ϭ 12%i* ϭ 9%B*iC ϭ 23%Projects A and C are selected, but B is not because i* Ͻ MARR ... 4 Rate of Return Evaluation Using PW:Incremental and BreakevenIn this section we discuss the primary approach to makingmutually exclusive alternative selections by the incremental ROR method ... Use hand solution or spreadsheet functions to find ⌬i* , the internalB–A*ROR for the series ... (⌬i* may replace ⌬i* when only two alternatives are present ... Because of the reinvestment requirement for PW analysis for different-lifeassets, the incremental cash flow series may contain several sign changes, indicating multiple ⌬i*values ... The correctapproach is to follow one of the techniques of Section 7 ... This means that the single externalROR (⌬iЈ or ⌬iЈЈ) for the incremental cash flow series is determined ... As statedearlier, it is always possible, and generally advisable, to use a PW orAW analysis at an established MARR in lieu of the ROR method when multiple rates are indicated ... Order the alternatives by initial investment or cost, starting with the smaller one, calledA ... 2 ... 3 ... Independentprojectselection208Chapter 8Rate of Return Analysis: Multiple Alternatives4 ... If necessary, use Norstrom’s criterion to determine if a single positive root exists ... Set up the PW ϭ 0 equation and determine ⌬i* ... Select the economically better alternative as follows:ME alternativeselectionIf ⌬i* Ͻ MARR,select alternative A ... BϪAIf ⌬i* is exactly equal to or very near the MARR, noneconomic considerations help inthe selection of the “better” alternative ... For example, if the MARR is 15% per year and you have established that⌬i* is in the 15% to 20% range, an exact value is not necessary to accept Bsince you alreadyBϪAknow that ⌬i* Ն MARR ... Multiple guessvalues can be input to find multiple roots in the range Ϫ100% to ϱ for a nonconventional series,as illustrated in Example 7 ... If this is not the case, to be correct, the indication of multiple rootsin step 4 requires that one of the techniques ofSection 7 ... EXAMPLE 8 ... Ford and its suppliers are seeking additional sources forlight, long-life transmissions ... Two United States–based vendors make the required dies ... Show both hand and spreadsheet solutions ... Use the procedure described above todetermine ⌬i* ... Alternatives A and B arecorrectly ordered with the higher first-cost alternative in column 2of Table 8–4 ... The cash flows for the LCM of 10 years are tabulated ... 3Year01–5Cash Flow A(1)$ Ϫ8,000Ϫ3,5005—6–1010Ϫ3,500—$Ϫ43,000Cash Flow B(2)$Ϫ13,000Ϫ1,600ϩ2,000Ϫ13,000Ϫ1,600ϩ2,000$Ϫ38,000IncrementalCashϪ5,000ϩ1,900Ϫ11,000ϩ1,900ϩ2,000$ ϩ5,0008 ... 3 ... The incremental cash flow diagram is shown in Figure 8–2 ... There are three sign changes in the incremental cash flow series, indicating as many asthree roots ... 5 ... 2]In order to resolve any multiple-root problem, we can $ � ) 2 ) ؊ ( 1( ( Flow(3assume that the investment rate ii in theROIC technique will equal the ⌬i* found by trial and error ... 2] for thefirst root discovered results in ⌬i* between 12% and 15% ... 65% ... Since the rate of return of 12 ... CommentIn step 4, the presence of up to three i* values is indicated ... 65% ... 65%, weassume that anypositive net cash flows are reinvested at 12 ... If this is not a reasonable assumption, theROIC or modified ROR technique (Section 7 ... The other two roots are very large positive and negative numbers, as the IRR function ofExcel reveals ... Solution by SpreadsheetSteps 1 through 4 arethe same as above ... Figure 8–3 includes the same incremental net cash flows from Table 8–4 calculated incolumn D ... 65% using the IRR function ... 3 ... Since the rate of return on the extra investment is greater than the 12% MARR, the highercost vendor B is selected ... Forexample, row 17 uses theNPV function to verify that the present worth is positive at MARRϭ12% ... The rate of return determined for the incremental cash flow series or the actual cash flows canbe interpreted as a breakeven rate of return value ... Equivalently, the breakeven ROR is the ivalue, i*, at which the PW (or AW) values oftwo alternatives’ actual cash flows are exactlyequal to each other ... For example, if the PW versus ⌬i graph for the incremental cash flows inTable 8–4 (and spreadsheet Figure 8–3) is plotted for various interest rates, the graph shown inFigure 8–4 is obtained ... 65% ... 65%, the extra investment for B isjustified ... 65%, the opposite is true—the extra investment in B should not be made,and vendor A is selected ... 65%, the alternatives are equally attractive ... 3, provides the same results ... 65%1800For MARRin this range,select B1600For MARRin this range,select A14001200PW of incremental cashflows, $Breakeven ROR100080060040020006789101112141516 ⌬i%– 200– 400– 600Vendor BVendor A– 800Figure 8–4Plot of present worth of incremental cash flows for Example 8 ... 8 ... Now, the same conclusions are reached using thefollowing logic:• If MARR Ͻ 12 ... • If MARR Ͼ 12 ... • If MARRis exactly 12 ... Example 8 ... More of breakeven analysis is covered in Chapter 13 ... 4New filtration systems for commercial airliners are available that use an electric field to remove upto 99 ... This is vitally important, as manyof the flu germs, viruses, and other contagious diseases are transmittedthrough the systems thatrecirculate aircraft air many times per hour ... , can also be sizable ... • Plot two graphs: PW versus i values for both alternatives’ cash flows and PW versus ⌬ivalues for incremental cash flows ... Air Cleanser(Filter 1)Initial cost per aircraft, $Estimated savings, $ per yearEstimatedlife, yearsPurely Heaven(Filter 2)Ϫ1000375Ϫ1500700 in year 1, decreasing by100 per year thereafter55Solution by SpreadsheetRefer to Figure 8–6 as the solution is explained ... The cash flow sign tests for each filter indicate no multiple rates ... The PW values of filter 1 and filter 2 cash flows areplottedon the right side for i values ranging from 0% to 60% ... The higher-cost filter 2(Purely Heaven) is selected ... As expected, the curve crosses the PW ϭ 0 line at approximately 17%, indicating the same economic conclusion of filter 2 ... 3 cash flows(not incremental) ... 4 ... 5213Rate of ReturnEvaluation Using AWFigure 8–6 provides an excellent opportunity to see why the ROR method can result in selecting the wrong alternative when only i* values are used to select between two alternatives ... The inconsistencyoccurs when the MARR is set less than the breakeven rate between two revenuealternatives ... In Figure 8–6 the incremental breakevenrate is 16 ... The MARR is lower than breakeven; therefore, the incremental ROR analysis results in correctly selecting filter 2 ... 41% > 23 ... Thiserror occurs because the rate of return method assumes reinvestment at the alternative’s RORvalue,while PW and AW analyses use the MARR as the reinvestment rate ... 8 ... However, for the AW-based technique, there are two equivalent ways to perform theevaluation: (1) using the incremental cash flows over the LCM of alternative lives, just as for thePW-based relation (Section 8 ... There is nodifference betweenthe two approaches if the alternative lives are equal ... Since the ROR method requires comparison for equal service, the incremental cash flowsmust be evaluated over the LCM of lives ... The same six-step procedure of the previous section (forPW-based calculation) is used, except instep 5 the AW-based relation is developed ... Whether the lives are equal or unequal, set up the AW relation for the cash flows of eachalternative, form the relation below, and solve for i* ... 3]For both methods, all equivalent values are on an AW basis, so the i* that results from Equation [8 ... Example 8 ...EXAMPLE 8 ... 3, using an AW-basedincremental ROR method and the same MARR of 12% per year ... 2], for the incremental cash flow inExample 8 ... 65% ... Write an AW-basedrelation on the incremental cash flow series over the LCM of 10 years, or write Equation [8 ... For the incremental method,the AW equation is0 ϭ Ϫ5000(A͞P,⌬i*,10) Ϫ 11,000(P͞F,⌬i*,5)(A͞P,⌬i*,10) ϩ 2000(A͞F,⌬i*,10) ϩ 1900Equal-servicerequirement214Chapter 8Rate of Return Analysis: Multiple AlternativesIt is easy to enter the incremental cash flows onto a spreadsheet, as in Figure 8–3, column D,and use the ϭ IRR(D4:D14)function to display ⌬i* ϭ 12 ... For the second method, the ROR is found using the actual cash flows and the respectivelives of 10 years for A and 5 years for B ... 0 ϭ Ϫ13,000(A͞P,i*,5) ϩ 2000(A͞F,i*,5) ϩ 8000(A͞P,i*,10) ϩ 1900Solution again yields i* ϭ 12 ... CommentIt is very important to remember thatwhen an incremental ROR analysis using an AW-basedequation is made on the incremental cash flows, the LCM must be used ... 6 Incremental ROR Analysis of Multiple AlternativesThis section treats selection from multiple alternatives that are mutually exclusive, using the incremental ROR method ...The analysis is based upon PW (or AW) relations for incremental cash flows betweentwo alternatives at a time ... When the i* values on several alternatives exceed the MARR, incremental ROR evaluationis required ... ) For all alternatives (revenue or cost), the incremental investment must be separatelyjustified ... 1 ... ME alternativeselectionSelect the one alternativeThat requires the largest investment, andIndicates that the extra investment over another acceptable alternative is justified ... The incremental ROR evaluation procedure for multiple, equal-life alternatives is summarized below ... The termsdefender and challenger are dynamic in that they refer, respectively, to the alternative that is currently selected (the defender)and the one that is challenging it for acceptance based on ⌬i* ... The steps for solution by hand or by spreadsheet are as follows:1 ... Record the annual cashflow estimates foreach equal-life alternative ... Revenue alternatives only: Calculate i* for the first alternative ... If i* Ͻ MARR, eliminate the alternative andgo to the next one ... The next alternative is now the challenger ... (Note: This is wheresolution by spreadsheet can be a quick assist ... Label it thedefender and go tostep 3 ... Determine the incremental cash flow between the challenger and defender, using the relationIncremental cash flow ϭ challenger cash flow Ϫ defender cash flowSet up the ROR relation ... 6Incremental ROR Analysis of Multiple Alternatives4 ... (PWis most commonly used ... If ⌬i* Ն MARR, thechallenger becomes the defender and the previous defender is eliminated ... 6 ... It is the selected one ... It is vital that the correct alternatives be compared, or the wrong alternative may be selected ... 6Caterpillar Corporation wants to build a spare parts storage facility in the Phoenix, Arizona,vicinity ...The initial cost of earthwork and prefab building and the annual net cash flow estimates are detailed in Table 8–5 ... If the MARR is 10%, use incremental ROR analysis to select the one economically best location ... 6AInitial cost, $Annual cash flow, $ per yearLife,yearsϪ200,000ϩ22,00030BCϪ275,000ϩ35,00030Ϫ190,000ϩ19,50030DϪ350,000ϩ42,00030SolutionAll sites have a 30-year life, and they are revenue alternatives ... 1 ... 2 ... The ROR relation includes only theP͞A factor ... 7436 and⌬ic* ϭ 9 ... Since 9 ... Now the comparison is A to*DN, and column 2shows that ⌬iA ϭ 10 ... This eliminates the do-nothing alternative;the defender is now A and the challenger is B ... 6C(1)Initial cost, $Cash flow, $ per yearAlternatives comparedIncremental cost, $Incremental cash flow, $Calculated (P͞A,⌬i*,30)⌬i*,%Increment justified?AlternativeselectedϪ190,000ϩ19,500C to DNϪ190,000ϩ19,5009 ... 63NoDNA(2)Ϫ200,000ϩ22,000A to DNϪ200,000ϩ22,0009 ... 49YesAB(3)Ϫ275,000ϩ35,000B to AϪ75,000ϩ13,0005 ... 28YesBD(4)Ϫ350,000ϩ42,000D to BϪ75,000ϩ7,00010 ... 55NoB215216Chapter 8Rate of Return Analysis: Multiple Alternatives3... From the interest tables, look up the P͞A factor at the MARR, which is (P͞A,10%,30) ϭ9 ... Now, any P͞A value greater than 9 ... The P͞A factor is 5 ... For referencepurposes, ⌬i* ϭ 17 ... 5 ... 6 ... 7143 (⌬i* ϭ 8 ... Location D is eliminated,and only alternative B remains; it is selected ... Since C was notjustified inthis example, location A was not compared with C ... To demonstrate how important it is to apply the ROR method correctly, consider the following ... 6310 ... 3511 ... Location D is selected ... In fact,it will earn only 8 ... This is another example of the ranking inconsistency problem of theRORmethod mentioned in Section 8 ... For cost alternatives, the incremental cash flow is the difference between costs for two alternatives ... Therefore, thelowest-investment alternative is the initial defender against the next-lowest investment (challenger) ... 7 using a spreadsheet solution ... 7The completefailure of an offshore platform and the resulting spillage of up to 800,000 to1,000,000 gallons per day into the Gulf of Mexico in the spring of 2010 have made major oil producers and transporters very conscious of the harm done to people’s livelihood and all forms ofaquatic life by spills of this magnitude ...The Sierra Club, Greenpeace, and other international environmental interest groups arein favor of the initiative ... Annual cost estimates are expected to behigh to ensure readiness at any time ... 5% ... 8 ... 7Machine 1First cost, $Annual operating cost, $Salvage value, $Life, yearsMachine 2Machine3Machine 4Ϫ5,000Ϫ3,500ϩ5008Ϫ6,500Ϫ3,200ϩ9008Ϫ10,000Ϫ3,000ϩ7008Ϫ15,000Ϫ1,400ϩ1,0008Solution by SpreadsheetFollow the procedure for incremental ROR analysis ... 1 ... 2 ... 3 ... 4 ... 57% by applying the IRR function ... This return exceeds MARR ϭ 13 ... The comparison continues for 3-to-2in column E, where the return is negative at ⌬i* ϭϪ18 ... Finally the 4-to-2 comparison has an incremental ROR of 13 ... 5% ... ϭ IRR(D6:D14)Figure 8–7Spreadsheet solution to select from multiple cost alternatives, Example 8 ... CommentAs mentioned earlier, it is not possible to generate a PW versus igraph for each cost alternativebecause all cash flows are negative ... The curves willcross the PW ϭ 0 line at the ⌬i* values determined by the IRR functions ... This is another application of the principle of equal-service comparison ... It is always possible to rely on PW or AW analysis of the incrementalcash flows at the MARRto make the selection ... However, it is still necessary to make the comparison over the LCMnumber of years for an incremental analysis to be performed correctly ... 7 All-in-One Spreadsheet Analysis (Optional)For professors and students who like to pack a spreadsheet, Example8 ... Now that the IRR, NPV, and PV functions aremastered, it is possible to perform a wide variety of evaluations for multiple alternatives on asingle spreadsheet ... A nonconventional cash flow seriesfor which multiple ROR values may be found, and selection from both mutually exclusive alternatives andindependent projects, are included in this example ... 8In-flight texting, phone, and Internet connections provided at airline passenger seats are anexpected service by many customers ... Four optionaldata handling features that build upon one another are available from the manufacturer, but atan addedcost per unit ... g ... All four options are expected to boost annual revenues by varying amounts ... (a) Using MARR ϭ 15%, perform ROR, PW, and AW evaluations to select the one level ofoptions that is the most promising economically ... If no budget limitations are considered at this time, whichoptionsare acceptable if the MARR is increased to 20% when more than one option may beimplemented?Figure 8–8Spreadsheet analysis using ROR, PW, and AW methods for unequal-life, revenue alternatives, Example 8 ... Chapter SummarySolution by Spreadsheet(a) The spreadsheet (Figure 8–8) isdivided into six sections:Section 1 (rows 1, 2): MARR value and the alternative names (A through D) are in increasing order of initial cost ... These arerevenue alternatives with unequal lives ... Section 4 (rows 21, 22): Because these are all revenue alternatives, i* values are determined by the IRR function... Columns F and H were inserted to make space for theincremental evaluations ... Section 5 (rows 23 to 25): The IRR functions display the ⌬i* values in columns F and H ... Since ⌬i* ϭ 19 ... Thefinal comparison of D to C over 12 years results in ⌬i* ϭ 11 ... Alternative C is the chosen one ... The AWvalue over the lifeof each alternative is calculated using the PMT function at the MARR with an embeddedNPV function ... For both measures, alternative C has the numerically largest value, as expected ... (b) Since each option is independent of the others, and there is no budget limitation at thistime,each i* value in row 21 of Figure 8–8 is compared to MARR ϭ 20% ... Of the four, options B and C havei* Ͼ 20% ... CommentIn part (a), we should have applied the two multiple-root sign tests to the incremental cash flowseries for the C-to-B comparison ... Therefore, up to three realnumber roots mayexist ... 42%CϪBwithout using a supplemental (Section 7 ... This means that the investment assumption of 19 ... If the MARR ϭ 15%, or some otherearning rate were more appropriate, the ROIC procedure could be applied to determine a singlerate, which would be different from 19 ... Depending upon theinvestment rate chosen,alternative C may or may not be incrementally justified against B ... CHAPTER SUMMARYJust as present worth and annual worth methods find the best alternative from among several,incremental rate of return calculations can be used for the same purpose ... The incrementalinvestment evaluation is conducted between only twoalternatives at a time, beginning with the lowest initial investment alternative ... Rate of return values have a natural appeal to management, but the ROR analysis is oftenmore difficult to set up and complete than the PW or AW analysis using anestablished MARR ... If there is no budget limitation when independent projects are evaluated, the ROR value ofeach project is compared to the MARR ... 219220Chapter 8Rate of Return Analysis: Multiple AlternativesPROBLEMSUnderstanding Incremental ROR8 ... 2 If the rate of return on theincremental cash flowbetween two alternatives is less than the minimumattractive rate of return, which alternative shouldbe selected, if any?8 ... There are revenue andcost cash flow estimates ... The first one is 3 ... 2% below the MARR, andthe third is 2 ... Which alternatives, if any, must he include inthe incrementalROR analysis?8 ... Alternative Y requiresa larger investment than alternative X ... 5 Victoria is comparing two mutually exclusivealternatives, A and B ... (a) What is known about the ROR on the increment between A and B?(b) Which alternative should be selected?8 ... Only one can be

selected ... A more expensive microwave model will yield a rate of return of 22% per year ... 7 If $80,000 is invested at 30% and another $50,000is invested at 20% per year, what is the overall rateof return on the entire $130,000?8 ... If the overall rateof return on the $100,000 was 30% and the rateofreturn on the $30,000 invested in Z1 was 15%,what was the rate of return on Z2?8 ... , which manufactures rigid shaft couplings, has $600,000 to invest ... Project XProject YProject ZiX ϭ 24%iY ϭ 18%iZ ϭ 30%The initial investment required for each project is$100,000, $300,000, and $200,000,respectively ... 10 Two options are available for setting up a wirelessmeter scanner and controller ... A more permanent system has a higherfirst cost of $73,000, but it has an estimated life of6 years and a salvage value of $15,000 ... Ifthe two options are compared using an incrementalrate of return, whatare the incremental cash flowsin (a) year 0 and (b) year 2?8 ... Machine XFirst cost, $Annual operating cost, $ per yearSalvage value, $Life, yearsMachine YϪ35,000Ϫ31,60002Ϫ90,000Ϫ19,4008,00048 ... Alternative P Alternative QFirst cost, $Annual operating cost, $ per yearAnnual revenue, $ peryearSalvage value, $Life, yearsϪ50,000Ϫ8,60022,0003,0003Ϫ85,000Ϫ2,00045,0008,00068 ... Alternative A has a 3-year life and alternative B a 6-year life ... 14 Standby power for pumps at water distributionbooster stations can be provided by either gasolineor diesel-powered engines ... 15 Several high-value parts for NASA’s reusablespace exploration vehicle can be either anodizedor powder-coated ... Determine (a) the rate of return on theincremental cash flows and (b) which one should beselected if the company’s MARR is 25% per year ... Powder Coat?Ϫ21,000?3Ϫ65,000?6,0003The incrementalAW cash flow equation associated with (powder coat – anodize) is0 ϭ –14,000(A͞P,i,3) ϩ 5000 ϩ 2000(A͞F,i,3)What is (a) the first cost for anodizing, (b) the annual cost for powder coating, and (c) the resale(salvage) value of the anodized parts?Incremental ROR Comparison (Two Alternatives)8 ... TheP3), $100001–910$Ϫ4600110020008 ... This situation can be dealt with by drilling a new well at a cost of $1,000,000 or by installing a tank and self-cleaning screen ahead of ؊ incremental cash flow associated with two alternatives for chemical storage and handling systemsIncremental Cash Flow(X3thedesalting equipment ... A new well will have a pump that ismore efficient than the old one, and it will requirealmost no maintenance, so its operating cost will beonly $18,000 per year ... 8 ... The costs associated with producing chemicallytreated vinyl rollers and fiber-impregnated rubberrollers areshown below ... Assume the company’s MARR is 21% per year ... )TreatedFirst cost, $Annual cost, $ per yearSalvage value, $Life, yearsImpregnatedϪ50,000Ϫ100,0005,0003Ϫ95,000Ϫ85,00011,00068 ... Design 2B willcost $3 million to install and $135,000 per year tomaintain ... 7 million to install and$70,000 per year to maintain ... 222Chapter 8Rate of Return Analysis: Multiple Alternatives8 ... Additive A has a first cost of$110,000 and an annual operating cost of $60,000 ... If the companyuses a 3-year recovery period for paint productsand a MARR of 20% per year, which process iseconomicallyfavored? Use an incremental RORanalysis ... 21 The manager of Liquid Sleeve, Inc ... The costsassociated with each are shown below ... Do the followingusing a PW-based rate of return analysis and aspreadsheet:(a) Determine which nanoparticle type the company should select using the ⌬i* value ...Indicatethe breakeven i* value and the MARR valueon the plot ... Is the answer the same as in part (a)?Type FeFirst cost, $Annual operating cost, $ per yearSalvage value, $Life, yearsType AlϪ150,000 Ϫ280,000Ϫ92,000 Ϫ74,00030,00070,000248 ... The incremental cashflows between the twoalternatives, J and S, have anincremental rate of return that is less than 40%,which is the MARR of the company ... She believes the company can implement costcontrols to reduce the annual cost of the moreexpensive process ... 23 The incremental cash flows for two alternativeelectrode setups areNPT), $01Ϫ89Ϫ56,000ϩ8,900ϩ12,0008 ... It must decide between two machines for a finishing operation ... The company’s MARR is 18% per year ... 25 A manufacturer of ؊ shown ... (a) Determine which should be selected using anAW-based rate of return analysis ... YearIncremental Cash Flow(Drylochydraulic equipment is trying todetermine whether it should use monoflange double block and bleed (DBB) valves or a multi-valvesystem (MVS) for chemical injection ... Use an AW-based rate of returnanalysis and a MARR of 18% per year to determine the better of the two options ... 26 Poly-ChemPlastics is considering two types of injection molding machines—hydraulic and electric ... Electric machinetechnology (EMT) will have a first cost of $800,000,annual costs of $150,000, and a salvage value of$130,000 after 5 years ... 223Problems(b)(c)Determine which machine the companyshould select,if the MARR ϭ 16% per year ... 8 ... He calculated iA ϭ 34 ... 2% and recommended acceptance of Asince its rate of return exceeded the establishedMARR of 30% by a greater amount than project B ... Do the following to helpEduardo better understand the rate of return methodand what this reduction inMARR means ... (b) Perform the correct analysis using eachMARR value ... Alternative AFirst cost, $Annual operating cost,$ per yearAnnual revenue, $ per yearSalvage value, $Life, yearsi*, %a MARR of 10% per year to determine whichalternative is best using an incremental rate ofreturn analysis ...245,00020,000629 ... 28 Four mutually exclusive revenue alternativesare under consideration to automate a bakingand packaging process at Able Bakery Products ... The rate ofreturn on each increment of investment was lessthan the MARR ... 29 A WiMAX wireless network integrated with asatellitenetwork can provide connectivity toany location within 10 km of the base station ... An independent cable operator is considering three bandwidth alternatives ... 30 Xerox’s iGenX high-speed commercial printerscost $1 ... The machines cost$500,000 to $750,000 depending on what optionsthe clientselects ... Theoperating costs and revenues generated are relatedto a large extent to the speed and other capabilitiesof the copier ... The company uses a3-year planning period and a MARR of 15% peryear ... CopierInitialInvestment,$OperatingCost,$ per YearAnnualRevenue,$ per YearSalvageValue,$iGen-1iGen-2iGen-3iGen-4–500,000–600,000–650,000–750,000–350,000–300,000–275,000–200,000ϩ450,000ϩ460,000ϩ480,000ϩ510,000ϩ70,000ϩ85,000ϩ95,000ϩ120,000Alternative BϪ40,000Ϫ5,500–40,000–46,000–61,000OperatingAnnualCost,Income,$1000 per Year $1000 per Year8 ... hasdetermined that any one offive machines can be used in one phase of its chilicanning operation ... If the minimum attractiverate of return is 20% per year, determine whichmachine should be selected on the basis of a rate ofreturn analysis ... 32 Five revenue projects are under consideration byGeneralDynamics for improving material flowthrough an assembly line ... From the calculations, determine which project, ifany, should be undertaken if the company’s MARRis (a) 11 ... 5% per year ... ComparisonIncremental Rate of Return, %B vs DNA vs BD vs DNE vs BE vs DE vs AC vs DNC vs AE vs DNAvs DNE vs CD vs CD vs B13%19%11%15%24%21%7%19%12%10%33%33%29%ABCDE–25,000–35,000–40,000–60,000–75,000(a)(b)(c)9 ... 113 ... 420 ... 3 9 ... 3 25 ... 5 24 ... 5 27 ... 8—If the alternatives are mutually exclusive andthe MARR is 26% per year, which alternative should be selected?Ifthe alternatives are mutually exclusive andthe MARR is 15% per year, which alternative should be selected?If the alternatives are independent and theMARR is 15% per year, which alternative(s)should be selected?8 ... The initial costs and cash flows of eachproject are shown ... 9% per year,whichalternative should be selected?DEϪ80,000Ϫ60,000Ϫ40,000Ϫ30,000Ϫ20,000Overall Incremental Rate of Return, %ROR, % ABCDE141617128—1211172412—1423211114—3529172335—1724212917—8 ... An engineer performed the following analysis to select the bestmachine, all of which have a 10-year life ... 35 The plant manager at Automaton Robotics is looking at the summarized incremental rate of returninformation shown below for five mutually exclusive alternatives, one of which must be chosen ... Which alternative is best if the minimum attractive rate of returnis (a) 15% per year and (b) 12%per year?ABCDEIncrementalRate of Return, %BInitial cost, $Ϫ7,000 Ϫ23,000 Ϫ9,000 Ϫ3,000 Ϫ16,000Cash flow, $ per1,0003,500 1,4005002,200yearRate of return14 ... 215 ... 713 ... 33 The five alternatives shown here are being evaluated by the rate of return method ... 623 ... 120 ... 372 to 13 to 24 to3Ϫ16,000 Ϫ12,000 Ϫ26,000ϩ6,000ϩ3,000ϩ5,00035 ... 414 ... S ... At a MARR of7 ... Project IDFirstCost, $1000AnnualIncome, $1000Rate ofReturn, %ABCDEϪ20,000Ϫ10,000Ϫ15,000Ϫ70,000Ϫ50,000ϩ2000ϩ1300ϩ1000ϩ4000ϩ260010 ... 06 ... 75 ... 38 When conducting a rate of return (ROR)analysisinvolving multiple mutually exclusive alternatives, the first step is to:(a) Rank the alternatives according to decreasing initial investment cost(b) Rank the alternatives according to increasinginitial investment cost(c) Calculate the present worth of each alternative using the MARR(d) Find the LCMbetween all of the alternatives8 ... 40 When comparing independent projects by theROR method, you should:(a) Find the ROR of each project and pick theones with the highest ROR(b) Select all projects that have an overall ROR ՆMARR(c) Select the project with an overall ROR ՆMARR that involvesthe lowest initialinvestment cost(d) Select the project with the largest initial investment that has been incrementally justified8 ... The only scenario thatrequires an incremental investment analysis toselect an alternative is that:(a) X has an overall ROR of 22% per year, and Yhas an overall ROR of 24% peryear(b) X has an overall ROR of 19% per year, and Yhas an overall ROR of 23% per year(c) X has an overall ROR of 18% per year, and Yhas an overall ROR of 19% per year(d) X has an overall ROR of 28% per year, and Yhas an overall ROR of 26% per year8 ... 43 For these alternatives, the sum ofthe incrementalcash flows is:YearAB012345Ϫ10,000ϩ2,500ϩ2,500ϩ2,500ϩ2,500ϩ2,500Ϫ14,000ϩ4,000ϩ4,000ϩ4,000ϩ4,000ϩ4,000(a)(b)(c)(d)$2500$3500$6000$80008 ... uses a minimum attractiverate of return of 8% per year, compounded annually ... The cash flow estimatesassociated with eachprocess are shown below ... 45 For the four independent projects shown, the one orones to select using a MARR of 14% per year are:ProjectABCD(a)(b)(c)(d)Rate of Return,% per Year14121510Only COnly A and COnly ACan’t tell; need to conduct incrementalanalysis226Chapter 8Rate of ReturnAnalysis: Multiple AlternativesProblems 8 ... 48 are based on the followinginformation ... InitialOverall RORAlternative Investment, $ versus DN, %ABCDEϪ25,000Ϫ35,000Ϫ40,000Ϫ60,000Ϫ75,0009 ... 113 ... 420 ... 3 9 ... 3 25 ... 5 24 ... 5 27 ... 8—8 ... 48 If the projects are independent, instead ofmutuallyexclusive, the one or ones to select at an MARR of18% per year are:(a) B and C(b) B, D, and E(c) D and E(d) B, C, and E8 ... It will be able to translate digital versions ofthree-dimensional computer models, containing a wide varietyof part shapes with machined and highly finished (ultrasmooth)surfaces ... Additionally, Make-to-Specs will build the code for superfinefinishing of surfaces with continuous control of the finishingmachines ... The server first costand estimated contribution to annual net cash flow aresummarized below ... They haveasked that, at this stage of the project, all analyses beperformedusing both life estimates for each system ... If the MARR ϭ 12%, which server should be selected?Use the PW or AW method to make the selection ... Use incremental ROR analysis to decide between theservers at MARR ϭ 12% ... Use any method of economic analysis to display onthespreadsheet the value of the incremental ROR betweenserver 2 with a life estimate of 5 years and a life estimateof 8 years ... ButI don’t think we can keep doing the same thing for many moreyears ... Elmerwas sharing thoughts on Gulf Coast Wholesale Auto Parts, acompany he has owned andoperated for 25 years on the southern outskirts of Houston, Texas ... Additionally, Gulf Coast operates a rebuild shop serving these sameretailers for major automobile components, such as carburetors,transmissions, and air conditioning compressors ... Part of his job atEnergcon Industries is to performbasic rate of return andpresent worth analyses on energy management proposals ... John summarizedall the estimates over a 10-year horizon ... Option 1: Remove rebuild ... The removal of the rebuild operations and theswitch to an “all-parts house” are expected to cost$750,000 in the first year ...Expenses are projectedat $0 ... Option 2: Contract rebuild operations ... If expenses stay the same for 5 years, they will average$1 ... Elmer thinks revenues under a contract arrangementcan be $1 ... Option 3: Maintain status quo and sell out after 5 years(Elmer’s personal favorite) ... Projections are $1 ...15 million per year in revenue ... Elmer’s wish is to sell out completely after 5 moreyears at this price, and to make a deal that the newowner pay $500,000 per year at the end of year 5 (saletime) and the same amount for the next 3 years ... Elmer has a close friend in theantique auto parts business who ismaking a “killing,”so he says, with e-commerce ... The trade-out would cost an estimated $1 million for Elmer immediately ... Expenses areestimated at $3 million per year and revenues at$3 ... Option 5: Lease arrangement ... The first-cut estimatesfor this option are $1 ... Case Study ExercisesHelp Johnwith the analysis by doing the following:1 ... 2 ... Findany multiple rates in the range of 0% to 100% ... If John’s father insists that he make 25% per year or moreon the selected option over the next 10 years, what shouldhe do? Use all the methods of economic analysis youhave learned so far (PW, AW,ROR) so John’s father canunderstand the recommendation in one way or another ... Prepare plots of the PW versus i for each of the five options ... 5 ... Stewart, Consultant, Communications and High Tech Solutions Engineering, Accenture LLP ... SECTIONTOPICLEARNING OUTCOME9 ... 9 ... 9 ... 9... 9 ... 9 ... The evaluation methods of previous chapters are usually applied to alternativesin the private sector, that is, for-profit and not-for-profit corporations and businesses ... In the case of public projects, the owners and users (beneficiaries) are the citizens and residents of a government unit—city,county, state, province,or nation ... Public-private partnerships have become increasingly common, especially for large infrastructure projects such as major highways, power generation plants, water resource developments, and the like ... The different formats of B/C analysis, and associated disbenefits ofan alternative, are discussed here ... Performed correctly, the benefit/cost method will always select the same alternative as PW, AW, and RORanalyses ... Finally, there is a discussion on professionalethics and ethical dilemmas in the public sector ... An expectation ofover 100,000 new residents in thenextseveral years and 500,000 by 2040prompted the development of the plantstarting in 2012 ... The project istermed WTF3, and its initial capital investment is $540 million for the treatment plant and two large steel-pipetransmission mains (84- and 48-inch)that will be installed via tunneling approximately100 to 120 feet under suburban areas of the city to reach currentreservoirs ... Besides thetreatment plant construction on the 95acre site, there must be at least threelarge vertical shafts (25 to 50 feet in diameter) bored along each transmissionmain to gain underground access forequipment and debrisremoval duringthe tunneling operations ... There are major long-term benefits forthe new facility ... • The new treatment plant is at ahigher elevation than the currenttwo plants, allowing gravity flow toreplenish reservoirs, thereby usinglittle or no electric pumping ... • It will provide a water quality thatis moreconsistent due to the location of the raw water intakes ... The disbenefits are mostly short-termduring the construction of WTF3 andtransmission mains ... • Large amounts of dust and smokewill enter the atmosphere in aresidential area during the 3½ yearsof construction, tunneling, andtransmission maincompletion ... • Natural landscape in plant and tunnel shaft sites will be destroyed ... • There may be delays in fire andambulance services in emergencies,since many neighborhood streets arecountry-road width and offer onlysingle ingress/egress streets for neighborhoods along the indicated routes ... •Newly generated revenues will beused to pay off the capital fundingbonds approved for the plant’s construction ... Public and elected official intervention has now caused some of theconclusions using the criteria mentionedabove to be questioned by the generalmanager of Allen Water Utilities ...1)Incremental B/C analysis, two alternatives (Section 9 ... 4)9 ... (Notable exceptions are the long-life alternatives discussed in Chapters 5 (PW) and 6 (AW) where capitalizedcost analysis was applied ... These are called public sector projects ... The primary purpose is to provide service to the citizenry forthepublic good at no profit ... Upon reflection, it is surprising how much of what we use on a daily or as-needed basis ispublicly owned and financed to serve us—the citizenry ... 1Public Sector ProjectsPolice and fire protectionCourts and prisonsFood stamp and rent relief programsJob trainingPublichousingEmergency reliefCodes and standardsThere are significant differences in the characteristics of private and public sector alternatives ... CharacteristicPublic sectorPrivate sectorSize of investmentLargeSome large; more medium to smallOften alternatives developed to serve public needs requireyears)Shorter (2–25 years)The long lives of public projects often prompt the use of the capitalized cost method, whereinfinity is used for n and annual costs are ؉ large initial investments, possiblydistributed over several years ... CharacteristicPublic sectorPrivate sectorLife estimatesLonger (30–50calculated as A ϭ P(i) ... For example, at i ϭ 7%,there will be a very small difference in 30 and 50 years, because (A/P,7%,30) ϭ 0 ... 07246 ... Public sector projectsoften have undesirable consequences, as interpreted by some sectors of the public ... The economic analysis shouldconsider theseconsequences in monetary terms to the degree estimable ... ) To perform a benefit/cost economic analysis of public alternatives, the costs (initial andannual), the benefits, and the disbenefits, if considered, must be estimated as accurately aspossible in monetary units ... Benefits—advantages to beexperienced by the owners, the public ... Disbenefits may be indirect economic disadvantages of the alternative ... For example, assume a short bypass around a congested area in town isrecommended ... Relative to revenue cash flow estimates in the private sector, benefit estimates are much harder tomake, and vary more widely around uncertain averages ... 5 ... In fact, the disbenefit itself may not be known at the time the evaluation is performed ... Taxes are collected from those who are the owners—the citizens (e ... , federal gasolinetaxes for highways are paid by all gasoline users, and health carecosts are covered by insurancepremiums) ... Bonds are often issued:U ... Treasury bonds, municipal bond issues, and special-purpose bonds, such as utility districtbonds ... Also, private donors may provide fundingfor museums, memorials, parks, and garden areas through gifts ... Governmentagenciesare exempt from taxes levied by higher-level units ... (Private corporations and individual citizens do pay taxes ... This results in interest rates in the 4% to 8% range ... As a matter of standardization, directives to use a specific interest rate are beneficial becausedifferent government agencies are able toobtain varying types of funding at different rates ... Standardized rates tend to increase the consistency of economic decisions and to reduce gamesmanship ... The public sector interest rate is identifiedas i; however, it is referred to by other names to distinguish it from the private sector rate ...CharacteristicAlternative selectioncriteriaPublic sectorPrivate sectorMultiple criteriaPrimarily based on rateof returnMultiple categories of users, economic as well as noneconomic interests, and special-interestpolitical and citizen groups make the selection of one alternative over another much more difficultin public sector economics ... It is important to describe and itemize the criteria and selectionmethod prior to the analysis ... Viewpoint is discussed below ... Elected officials commonly assist with the selection, especially when pressure is brought to bear by voters, developers, environmentalists, andothers... The viewpoint of the public sector analysis must be determined before cost, benefit, and disbenefit estimates are made and before the evaluation is formulated and performed ... 9 ... In general, the viewpoint of the analysis should be as broadly defined as thosewho will bear the costs of the project andreap its benefits ... 1 ... 1 Water Treatment Facility #3 CaseThe situation with the location and construction of the new WTF3 and associated transmissionmains described in the chapter’s introduction has reached a serious level because of recentquestions posed by some city council members and citizengroups ... The lead consultant, Joel Whiterson,took engineering economy as a part of his B ... education and has previously worked oneconomic studies in the government sector, but never as the lead person ... He realized thatno viewpoint of the study was defined, and, in fact, the estimates were neverclassified as costs,benefits, or disbenefits ... Joel defined two viewpoints: a citizen of Allen and the Allen Water Utilities budget ... Please help with this classification ... Cost of water: 10% annual increase to AllenhouseholdsAverage of $29 ... Bonds: Annual debt service at 3% per year on$540 million$16 ...2 million (year 20)3 ... Property values: Loss in value, sales price,and property taxes$4 million (years 1–5)5 ... M&O: Annual maintenance and operationscosts$300,000 plus 4% per year increase (years 1–20)7 ... (How this classification is done will vary depending upon who does the analysis ...)Viewpoint 1: Citizen of the city of Allen ... Costs: 1, 2, 4, 6Benefits: 5, 7Disbenefits: 3Viewpoint 2: Allen Water Utilities budget ... Costs: 2, 3, 6Benefits: 1, 5, 7Disbenefits: 4Citizens view costs in a different light than a city budget employee does ... Similarly, the Allen Water Utilities budget interpretsestimate3 (payment for use of land to Parks and Recreation) as a real cost; but a citizen might interpretthis as merely a movement of funds between two municipal budgets—therefore, it is a disbenefit, not a real cost ... However, agreement onthe disbenefits and their monetary estimates is difficult (toimpossible) to develop, often resulting in the exclusion of any disbenefits from the economic analysis ... e ... Most of the large public sector projects are developed through public-private partnerships(PPPs) ... Fullfunding by the government unit may not be possible using traditional means—fees, taxes,andbonds ... The government unit cannot make a profit, but the corporation(s)involved can realize a reasonable profit; in fact, the profit margin is usually written into the contract that governs the design, construction, and operation of the project ... In these formats, a government unit took responsibilityforfunding and possibly some of the design elements, and later all operation activities, while thecontractor did not share in the risks involved—liability, natural disasters, funding shortfalls, etc ... Commonly these are called design-build contracts, under which contractors take on more andmore of thefunctions from design to operation ... The most reliance is placed upona contractor or contractors with a DBOMF contract, as described below ... It requires the contractor(s) to perform all the DBOMF activities with collaboration and approval of the owner (the government unit) ... Although acontractor mayassist in some instances, the funding (obtaining the capital funds) remains thegovernment’s responsibility through bonding, commercial loans, taxation, grants, and gifts ... In virtually all cases, some forms of design-buildarrangements for public projects are made because they offer several advantages tothe government and citizens served:• Cost and time savings in the design, build, and operate phases• Earlier and more reliable (less variable) cost estimates9 ... There are, of course, disadvantages to this arrangement ... Another risk is that a reasonable profit may not be realizedby the private corporationdue to low usage of the facility during the operate phase ... The subsidy may cover costs plus (contractually agreed-to) profit if usageis lower than a specified level ... 9 ... The B/C analysis was developed to introduce greater objectivity into public sector economics,and as one response to the U ...Congress approving the Flood Control Act of 1936 ... All cost andbenefit estimates must be converted to a common equivalent monetary unit (PW, AW, or FW) atthe discount rate (interest rate) ... 1]Present worth and annual worth equivalencies are preferred to future worth values ... Salvage valuesandadditional revenues to the government, when they are estimated, are subtracted from costs inthe denominator ... Most commonly, disbenefits are subtracted from benefits and placed in the numerator ... The decision guideline is simple:If B/C Ն 1 ... If B/C Ͻ 1 ... If the B/C value is exactly or very near 1 ...� —————————— � ——— costsC[9 ... 2] disbenefits are subtracted from benefits, not added to costs ... For example, if the numbers10, 8, and 5 are used to represent the PW of DB/C ؊ disbenefits B ؊ The conventional B/C ratio, probably the most widely used, is calculated as follows:benefitsbenefits, disbenefits, and costs, respectively, thecorrect procedure results in B/C ϭ (10 Ϫ 8)͞5 ϭ 0 ... The incorrect placement of disbenefitsin the denominator results in B/C ϭ 10͞(8 ϩ 5) ϭ 0 ... 40 ... However, regardless of whether disbenefits are (correctly) subtracted from the numerator or (incorrectly)added to costs in the denominator, a B/C ratio ofless than 1 ... 0 by the secondmethod, and vice versa ... Maintenance and operation (M&O) costs are placed in the numerator and treated in a mannerProject evaluation236Chapter 9Benefit/Cost Analysis and Public Sector Economicssimilar to disbenefits ...� ———————————————— initial investment[9 ... The modified B/C ratiowill obviously yield a different value than the conventional B/C method ... M&O costsModified B/C ؊ disbenefits ؊ Once all amounts areexpressed in PW, AW, or FW terms, the modified B/C ratio is calculated asbenefitsThe benefit and cost difference measure of worth, which does not involve a ratio, is based onthe difference between the PW, AW, or FW of benefits and costs, that is, B − C ... This method has the advantage of eliminating the discrepancies notedabove when disbenefits are regarded as costs, because Brepresents net benefits ... Subtracting disbenefits from benefits:Adding disbenefits to costs:B Ϫ C ϭ (10 Ϫ 8) Ϫ 5 ϭ Ϫ3B Ϫ C ϭ 10 Ϫ (8 ϩ 5) ϭ Ϫ3Before calculating the B/C ratio by any formula, check whether the alternative with the largerAW or PW of costs also yields a larger AW or PW of benefits ... By thevery nature of benefits and especially disbenefits, monetary estimates are difficult tomake and will vary over a wide range ... This approach assists in determining the economic and public acceptance risk associated with a defined project ... EXAMPLE 9 ... In aproposal for the foundation’s board ofdirectors to construct a new hospital and medical cliniccomplex in a deprived central African country, the project manager has developed some estimates ... Award amount:Annual costs:Benefits:Disbenefits:$20 million (end of) first year, decreasing by $5 million per year for 3additional years; localgovernment will fund during the first year only$2 million per year for 10 years, as proposedReduction of $8 million per year in health-related expenses for citizens$0 ... 6 million per year for removal of arable land and commercialdistrictsUse the conventional and modified B/C methods to determine if thisgrant proposal is economically justified over a 10-year study period ... SolutionInitially, determine the AW for each parameter over 10 years ... 864 per year$2 per year$8 per yearUse $0 ... 2Benefit/Cost Analysis of a Single Project237The conventional B/C analysis applies Equation [9 ... 8 ... 6B/C ϭ—————— ϭ 0 ... 864 ϩ 2 ... 3] ... 0 Ϫ 0 ... 0Modified B/C ϭ ——————— ϭ 0 ... 864The proposal is not justified economically since both measures are less than 1 ... If the lowdisbenefits estimate of $0 ... It is possible to develop a direct formula connection between the B/C of a public sector andB/C ofa private sector project that is a revenue alternative; that is, both revenues and costs areestimated ... 3] and the PW method we have used repeatedly ... ) Let’s neglect the initial investment in year 0 for a moment, andconcentrate on the cash flows of the project for year 1 through its expected life ... 3] may� ——————————— PW be written asPW of (B Ϫ D) Ϫ PW of CModified B/C ϭ ————————————PW of initial investmentThis relation can be slightly rewritten to form the profitability index (PI), which can be used toevaluate revenue projects in the public or private sector ... , n,PW of NCFtPIof initial investment[9 ... The PI measure ofworth provides a sense of getting the most for the investment dollar (euro, yen, etc ... This is a “bang for the buck”measure ... The evaluation guideline for a single project using the PI is the same as for the conventionalB/C or modified B/C ... 0, the project iseconomically acceptable at the discount rate ... 0, the project is not economically acceptable at the discount rate ... The PI has another name: the present worth index(PWI) ... This application is discussed in Chapter 12, Section 12 ... EXAMPLE 9 ... 51-mile toll road onthe outskirts of Atlanta’s suburbanarea ... Highway construction is expected to requireProject evaluation238Chapter 9Benefit/Cost Analysis and Public Sector Economics5 years at an average cost of $3 ... The discount rate is 4% per year, and thestudy period is 30 years ... Initial investment: $88 million distributed over 5 years; $4 millionnow and in year 5 and$20 million in each of years 1 through 4 ... Annual revenue/benefits: Include tolls and retail/commercial growth; start at $2 million inyear 1, increasing by a constant $0 ... Estimable disbenefits: Include loss of business income, taxes, and property value in surrounding areas; start at$10 million in year 1, decrease by $0 ... SolutionThe PW values in year 0 for all estimates must be developed initially usually by hand, calculator, or spreadsheet computations ... All values arepositive because of the sign convention for B/C and PI measures ... 89PW of costs ϭ $26 ... 41PW of disbenefitsϭ $80 ... 3] ... 41 Ϫ 80 ... 87Modified B/C ϭ —————————— ϭ 0 ... 89The toll road proposal is not economically acceptable, since B/C Ͻ 1 ... (b) From the private corporation viewpoint, Young Construction will apply Equation [9 ... 167 ... 87PI ϭ ——————— ϭ 1 ... 89The proposal is clearly justifiedwithout the disbenefits, since PI > 1 ... The private projectperspective predicts that every investment dollar will return an equivalent of $1 ... CommentThe obvious question that arises concerns the correct measure to use ... The public project setting will commonly use some form of the B/C ratio withthedisbenefit considered ... Then the numerical dilemma presented above should not occur ... 3 Alternative Selection Using IncrementalB/C AnalysisThe technique to compare two mutually exclusive alternatives using benefit/cost analysis is virtually the same as that for incremental ROR in Chapter 8 ... Thehigher-costalternative is justified if ⌬B/C is equal to or larger than 1 ... The selection rule is as follows:ME alternativeselectionIf ⌬B/C Ն 1 ... If ⌬B/C Ͻ 1 ... 9 ... This same rulewas used for incremental ROR analysis ... We already know the first, all costs have a positive sign in the B/Cratio ... Alternativesare ordered by increasing equivalent total costs, that is, PW or AW of all costestimates that will be utilized in the denominator of the B/C ratio ... If two alternatives, A and B, have equal initial investments and lives, but B has a larger equivalentannual cost, then B must be incrementally justified against A ...4below ... Follow these steps to correctly perform a conventional B/C ratio analysis of two alternatives ... 1 ... 2 ... Calculate the incremental cost (⌬C) for the larger-cost alternative ... 3 ... Calculate the incremental benefits (⌬B) for the larger-cost alternative ... 4 ... 2], (B Ϫ D)/C ... Use the selectionguideline to select the higher-cost alternative if ⌬B/C Ն 1 ... When the B/C ratio is determined for the lower-cost alternative, it is a comparison with the donothing (DN) alternative ... 0, then DN should be selected and compared to the secondalternative ... In public sector analysis, the DN alternative isusually the current condition ... 4The city of Garden Ridge, Florida, has received designs for a new patient room wing to themunicipal hospital from two architectural consultants ... The costs and benefits are the same in mostcategories, but the city financial manager decided that the estimates belowshould be considered to determine which design to recommend at the city council meeting next week and topresent to the citizenry in preparation for an upcoming bond referendum next month ... The discount rate is 5%, and the life of the building isestimated at 30 years ... (b) Once the two designs werepublicized, the privately owned hospital in the directly adjacentcity of Forest Glen lodged a complaint that design A will reduce its own municipal hospital’s income by an estimated $500,000 per year because some of the day-surgery featuresof design A duplicate its services ... The cityfinancial managerstated that these concerns would be entered into the evaluation as disbenefits of the respective designs ... Solution(a) Since most of the cash flows are already annualized, the incremental B/C ratio will useAW values ... Follow the steps of the procedureabove:1 ... AWA ϭ 10,000,000(A/P,5%,30) ϩ 35,000ϭ $685,500AWB ϭ 15,000,000(A/P,5%,30) ϩ 55,000 ϭ $1,030,7502 ... The incremental cost is⌬C ϭ AWB Ϫ AWA ϭ $345,250 per year3 ... The benefits for the ⌬B/C analysis are not the estimates themselves, but the difference if design B is selected ... ⌬B ϭ usageA Ϫ usageB ϭ $450,000 Ϫ $200,000 ϭ$250,000 per year4 ... 2] ... 72$345,2505 ... 0, indicating that the extra costs associated with design Bare not justified ... (b) The revenue loss estimates are considered disbenefits ... Now$350,000⌬B͞C ϭ ———— ϭ 1 ... In this case the inclusion of disbenefits has reversed the previous economic decision... New disbenefits will surely be claimed in the near future by other special-interest groups ... Usually, the expected useful life of a public project is long (25 or 30 or more years), soalternatives generally have equal lives ... As with ROR analysis of two alternatives, this is an excellent opportunityto use theAW equivalency of estimated (not incremental) costs and benefits, if the implied assumption that the project could be repeated is reasonable ... EXAMPLE 9 ... Thetwo options for constructing this main were open trench (OT) for the entire 6 ... 3 miles ... 9 ... He stated the equivalent annual costs inaninternal e-mail some months ago, based on the expected construction periods of 24 and36 months, respectively, as equivalent toAWOT ϭ $1 ... 37 million per yearThis analysis indicated that the open-trench option was economically better, at that time ... Use the estimates below that Joel has unearthedto perform a correct incrementalB/C analysis and comment on the results ... Open trench (OT)Trench-tunnel (TT)6 ... 3Trench for 2 ... 3 miles: 210036175,000150,000140,00020,00020,00060,0005,000Distance, milesFirst cost, $ per footTime to complete, monthsConstruction support costs, $ permonthAncillary expenses, $ per month:EnvironmentalSafetyCommunity interfaceSolutionOne of the alternatives must be selected, and the construction lives are unequal ... However, the study period of 50 years is a reasonable evaluation time frame, since the mains are considered permanent installations... PWOT ϭ PW of construction ϩ PW of construction support costsϭ 700(6 ... 20 million per yearPWTT ϭ [700(2 ... 3)](5280) ϩ 175,000(12)(P͞A,3%,3)ϭ $61,010,460AWTT ϭ 61,010,460(A͞P,3%,50)ϭ $2 ... The incremental cost is⌬C ϭ AWTT Ϫ AWOT ϭ 2 ... 20 ϭ $1 ... PWOT-anc ϭ 310,000(12)(P͞A,3%,2)ϭ$7,118,220AWOT-anc ϭ 7,118,220(A͞P,3%,50)ϭ $276,685 per year241242Chapter 9Benefit/Cost Analysis and Public Sector EconomicsPWTT-anc ϭ 85,000(12)(P͞A,3%,3)ϭ $2,885,172AWTT-anc ϭ 2,885,172(A͞P,3%,50)ϭ $112,147 per year⌬B ϭ AWOT-anc Ϫ AWTT-anc ϭ 276,685 Ϫ 112,147 ϭ $164,538per year ($0 ... ⌬B/C ϭ 0 ... 17 ϭ 0 ... 0, the trench-tunnel option is not economically justified ... 9 ... 3 ... 6 ... 0 when this selected alternative has been compared with another justified alternative ... The previous two examples (9 ... 5) are good illustrations ofthe second type of implied benefit estimation ... Atleast one alternative must have B/C Ն 1 ... If all alternatives are unacceptable, the DN alternative is the choice ... 6 ... )As in the previous section when comparing two alternatives, selection from multiple alternatives by incremental B/C ratio utilizes equivalent total costs to initially order alternativesfromsmallest to largest ... Also, remember that all costs areconsidered positive in B/C calculations ... The procedure for incremental B/C analysis ofmultiple alternatives is as follows:Determine the equivalent total cost for all alternatives ... Order the alternatives by equivalent total cost, smallest first ...Direct benefits estimation only: Calculate the B/C for the first ordered alternative ... 0, eliminate it ... 0 ... 0 becomes the defender and the nexthigher-cost alternative is the challenger in the next step ... )5 ... 2 ... 4 ... 5]9 ... Calculate the ⌬B/C for the first challenger compared to the defender ... 6]If ⌬B/C Ն1 ... 6], the challenger becomes the defender and the previous defender is eliminated ... 0, remove the challenger and the defenderremains against the next challenger ... Repeat steps 5 and 6 until only one alternative remains ... In all the steps above, incremental disbenefits may be considered byreplacing ⌬B with ⌬(B Ϫ D) ... 6Schlitterbahn Waterparks of Texas, a very popular water and entertainment park headquartered in New Braunfels, has been asked by four different cities outside of Texas to consider building a park in their area ... The annual M&O costsare expected to be the same for alllocations ... SolutionThe viewpoint is that of Schlitterbahn, and the benefits are direct estimates ... The results are presented inTable 9–1 ... AW of total costs and an example for city 1 are determined in $1 million units ... 5(0 ... 5ϭ $6 ... The four alternatives are correctly ordered by increasing equivalenttotal cost inTable 9–1 ... 6City 1City 2City 3City 4First cost, $ millionEntrance fee costs, $/yearAnnual revenue, $ million/yearInitial cash incentive, $Property tax reduction, $/yearSales tax sharing, $/year38 ... 0250,00025,000310,00040 ... 2350,00035,000320,00045 ... 0500,00050,000320,00060 ...4800,00080,000340,000AW of total costs, $ million/yearAW of total benefits, $ million/yearOverall B/CAlternatives comparedIncremental costs ⌬C, $/yearIncremental benefits ⌬B, $/year⌬B/CIncrement justified?City selected6 ... 3771 ... 9487 ... 06Yes17 ... 6140 ... 08 ... 4541 ... 1643 ... 64Yes310 ...9541 ... 2360 ... 22No3Eliminated243244Chapter 9Benefit/Cost Analysis and Public Sector Economics3 ... AW of benefits ϭ revenue ϩ initial incentive(A͞P,7%,8)ϩ property tax reduction ϩ sales tax sharingϭ 7 ... 25(0 ... 025 ϩ 0 ... 377 ($7,377,000 per year)4 ... 1] ... 377͞6 ... 06City 2 is eliminated with B/C2ϭ 0 ... 5 ... 6 ... 06, using Equation [9 ... City 1 is economicallyjustified and becomes the defender ... Repeat steps 5 and 6 ... 112 Ϫ 6 ... 164⌬B ϭ 10 ... 377 ϭ 3 ... 077͞1 ... 64City 3 is well justified and becomes the defender against city 4 ... 22for the 4-to-3 comparison ... Note that the DN alternative couldhave been selected had no proposal met the B/C or⌬B/C requirements ... The only comparison is betweeneach project separately with the do-nothing alternative ... 0 are accepted ... When a budget limitation is imposed, the capital budgeting procedure discussed inChapter 12 must be applied ... Equation[5 ... Example 9 ... EXAMPLE 9 ... 3 million; however, when it was publicized, influential people around Allen spoke strongly againstthe location ... Some of the plant design had already been completedwhen the general manager announced that this site was not the best choice anyway, and that itwould besold and a different, better site (location 2) would be purchased for $28 ... This was well over the budget amount of $22 ... Asit turns out, there was a third site (location 3) available for $35 ... In his review and after much resistance from Allen Water Utilities staff, the consultant, Joel,received a copy of theestimated costs and benefits for the three plant location options ... Using the assumption of a very long life forthe WTF3 facility and the established discount rate of 3% per year, determine what Joel discovered when he did the B/C analysis ... 4Incremental B/C Analysis of Multiple, Mutually ExclusiveAlternativesLocation 1Total benefits, $ per yearLocation 319 ... 028 ... 035 ... 05126310608623Land cost, $ millionFacility first cost, $ millionBenefits, $ per year:Pumping cost savingsSales to area communitiesAdded revenue from AllenLocation 21914SolutionA spreadsheet can be very useful whenperforming an incremental B/C analysis of three ormore alternatives ... Figure 9–1b details all the functionsused in the analysis ... In $1 million units,AW of costs ϭ A of land cost ϩ A of facility first costϭ (19 ... 0)(0 ... 379 per yearAW of benefits ϭ $23Location 2:AW of costs ϭ $14 ... 430AW of benefits ϭ$14Location 1:Though the AW of cost values are close to one another, the increasing order is locations 2, 1,and 3 to determine ⌬B/C values ... 97) is not economically justified at the outset ... Location 2 is justifiedagainst the DN alternative (B/C2 ϭ 1 ... Location 1 is a clear winner with ⌬B/C ϭ 27 ... Inconclusion, Joel has learned that location 1 is indeed the best and that, from the economicperspective, the general manager was incorrect in stating that location 2 was better ... (a)(b)Figure 9–1Incremental B/C analysis for WTF3 case: (a) numerical results and (b) functions developed for the analysis ...Location 1 was initially purchasedand planned for WTF3 ... 9 ... A large percentage of service sector projects are generated by and dependent upon the private sector (corporations, businesses,and other for-profit institutions) ... A service sector project is a process or system that provides services toindividuals, businesses,or government units ... Manufacturing and construction activities are commonly not considered a service sector project, thoughthey may support the theme of the service provided ... The intangible and intellectual work done by engineers and other professionals is often a part ofaservice sector project ... In otherwords, undue risk may be introduced into the decision because of poor monetary estimates ... This is a public and a service project, but its (economic) benefits are quite difficultto estimate ... In all but the last case, benefits in monetaryterms will be poor estimates ... Inservice and public sector projects, as expected, it is the benefits that are the more difficultto estimate ... The CEA approach utilizes a costeffectiveness measure or the cost-effectiveness ratio (CER) as a basis of ranking projects andselecting the best of independent projects or mutually exclusivealternatives ... 7]In the red-light camera example, the effectiveness measure (the benefit) may be one of the samples mentioned earlier, accidents averted or deaths prevented ... (The reciprocal of Equation [9 ... ) With costs in the numerator, smaller ratio valuesare more desirable for the same value of thedenominator, since smaller ratio values indicate alower cost for the same level of effectiveness ... 5Service Sector Projects and Cost-Effectiveness Analysis247alternative selection ... The ordering criteria are as follows:Independent projects: Initially rank projects by CER value ... Return again to thepublic/service project of red-light cameras ... If the projects are mutually exclusive, “total accidents averted ” is the correct ranking basisand an incremental analysis is necessary ... To select some from several (independent) projects, a budget limit, termed b, is inherently necessary once ordering iscomplete ... The procedures and examples follow ... Determine the equivalent total cost C and effectiveness measure E, and calculate the CERmeasure for each project ... Order projects from the smallest to the largest CER value ... Determine cumulative cost for each project and compare with the budgetlimit b ... The selection criterion is to fund all projects such that b is not exceeded ... 8Recent research indicates that corporations throughout the world need employees who demonstrate creativity and innovation for new processes and products ... RollingsFoundation for Innovative Thinking has allocated$1 million in grant funds to award to corporations that enroll their top R&D personnel in a 1- to 2-month professional training program intheir home state that has a historically proven track record over the last 5 years in helpingindividuals earn patents ... Columns 2 and 3give the proposed number ofattendees and cost per person, respectively, and column 4 provides the historical track record of program graduates in patents per year ... SolutionWe assume that across all programs and all patent awards there is equal quality ... TA BLE 9–2Data for Programs to Increase Patents Used forCEAProgram(1)Total Personnel(2)Cost/Person, $(3)5-Year History,Patents/Graduate/Year(4)1234565035572412875000450080002500550038000 ... 11 ... 12 ... 6Independent projectselection248Chapter 9Benefit/Cost Analysis and Public Sector Economics1 ... 7], the effectiveness measure E is patentsper year, and the CER isprogram cost per person CCER ϭ —————————— ϭ —Epatents per graduateThe program cost C is a PW value, and the E values are obtained from the proposals ... The CER values are shown in Table 9–3 in increasing order, column 5 ... Cost per course, column 6, and)3 )/( 4( ( cumulative costs, column 7, are determined ... Programs 4, 2, 5, 3, and 6 (68 of the 87 people) are selected to not exceed $1 million ... 8Program(1)425361TotalCost/PersonPersonnelC, $(2)(3)2435125787502,5004,5005,5008,0003,8005,000Patentsper Year E(4)CER, $ perPatent(5� ⌺(6)2 ... 12 ... 90 ... 51,1901,4521,8974,2116,33310,00060,000157,50066,000456,000330,600250,00060,000217,500283,500739,5001,070,1001,320,100CommentThis is the first time that a budget limit has been imposed for the selection among ( � ) 2 )( 3(CumulativeCost, $(7 ( � ProgramCost, $(6independentprojects ... For mutually exclusive alternatives and no budget limit, the alternative with the highest effectiveness measure E is selected without further analysis ... The analysis isbased on the incremental ratio ⌬C/E, and the procedure is similar to that we have applied forincremental ROR andB/C, except now the concept of dominance is utilized ... For mutually exclusive alternatives, the selection procedure is as follows:1 ... Record the cost foreach alternative ... Calculate the CER measure for the first alternative ... This CER is a baseline for the next incremental comparison, and the first

alternative becomes the new defender ... Calculate incremental costs (⌬C) and effectiveness (⌬E) and the incremental measure ⌬C/Efor the new challenger using the relationcost of challenger Ϫ cost of defender⌬C⌬C/E ϭ ——————————————————————— ϭ ——effectiveness ofchallenger Ϫ effectiveness of defender ⌬E4 ... Otherwise, no dominance is present and bothalternatives are retained for the next incremental evaluation ... Dominance present: Repeat steps 3 and 4 to compare the next ordered alternative (challenger) and new defender ... 9 ... Repeat steps 3 and 4 tocompare the new challenger and newdefender ... 6 ... 7 ... EXAMPLE 9 ... 8 decided to fund its 50 R&Dpersonnel to attend one of the innovation and creativity programs at its own expense ... 0 patentsper year ... TABLE 9–4Mutually Exclusive Alternatives Evaluated by Cost-EffectivenessAnalysis,Example 9 ... 12 ... 1119018971452125,000275,000225,000SolutionFrom Table 9–2, three programs—2, 4, and 5—have a historical record of at least two patentsper graduate per year ... Use the procedure to perform the incremental analysis ... The alternatives are ranked by increasing patents per yearin Table 9–4, column 4 ... The CER measure for program 4 is compared to the DN alternative ... 1patents per graduate3 ... ⌬C 5500 Ϫ 25005-to-4 comparison: ⌬C/E ϭ —— ϭ —————— ϭ 37502 ... 1⌬E4 ... Program 5 is more expensive for more patents; however, clear dominance is notpresent; bothprograms are retained for further evaluation ... Dominance not present: Program 5 becomes the new defender, and program 2 is the newchallenger ... ⌬C 4500 Ϫ 55002-to-5 comparison: ⌬C/E ϭ —— ϭ —————— ϭ Ϫ50003 ... 9⌬ECompared to C/E5 ϭ 1897, this increment is much cheaper—morepatents for less moneyper person ... 6 ... ⌬C 4500 Ϫ 25002-to-4 comparison: ⌬C/E ϭ —— ϭ —————— ϭ 20003 ... 1⌬EThis does not represent dominance of program 2 over 4 ... This occurs when there is not lower cost and higher effectiveness of one alternativeover another; that is, one alternativedoes not dominate all the others ... Now the budget and other considerations (probably noneconomic) are brought to bear tomake the final decision ... 249250Chapter 9Benefit/Cost Analysis and Public Sector EconomicsCost-effectiveness analysis is a form of multiattribute decision-making in whicheconomicand noneconomic dimensions are integrated to evaluate alternatives from several perspectives bydifferent decision makers ... 9 ... Above these is the long-standing expectation that public servants have integrity ... Impartiality, consideration of a wide range of circumstances, and the use ofrealistic assumptions are but three ofthe foundation elements upon which engineers should base their recommendations to decisionmakers ... Engineers are routinely involved in two of the major aspects of public sector activities:Public policy making—the development of strategy for public service,behavior, fairness,and justice ... An example is transportation management ... Public officials use these findings to establish public transportation policy ... Consider traffic control,where the use and placement of traffic control signs, signals, speed limits, parking restrictions, etc ... (In effect, this issystemsengineering, that is, an application of the life-cycle phases and stages explained inSection 6 ... )Whether in the arena of policy making or public planning, engineers can find ethical compromisea possibility when working with the public sector ... • Use of technology Many public projects involve the use ofnew technology ... It is common and expectedthat engineers make every attempt to apply the latest technology while ensuring that the publicis not exposed to undue risk ... These restrictions may be based on financialreasons, politically-charged topics, client-favored options, or a wide variety of otherreasons ... • Negative community impact It is inevitable that public projects will adversely affect somegroups of people, or the environment, or businesses ... Engineers who find (stumble onto) such negative impacts may be pressured by clients, managers, or public figures to overlook them, thoughChapterSummaryTABLE 9–5Some Ethical Considerations When Performing B/C and CEA AnalysisWhat the Study Includes Ethical DimensionExampleAudience for studyIs it ethical to select a specificgroup of people affected by theproject and neglect possibleeffects on other groups?Construct children’s healthcare clinicsfor city dwellers, but neglect rural families with poor transportation means ... Greater good for community as a wholeVulnerable minority groups, especially economically deprivedones, may be disproportionallyaffected ... Reliance on economicmeasures onlyIs it acceptable to reduce allcostsand benefits to monetaryestimates for a decision, thensubjectively impute nonquantified factors in the final decision?Softening of building codes can improve the financial outlook for homebuilders; however, increased risks offire loss, storm and water damage tostructures, and reduced future resalevaluesare considered only in passing asa new subdivision is approved by theplanning and zoning committee ... the Code of Ethics for Engineers dictates a full and fair analysis and report ... Considering this outcome in the recommendation to the transportation department should be a goal of the analyzingengineers, yet pressure to bias the results may be quite high ... As discussed earlier,estimations for benefits, disbenefits, effectiveness measures, and costs can be difficult and inaccurate, but these analysis tools are often the best available to structure a study ... CHAPTER SUMMARYThe benefit/costmethod is used primarily to evaluate alternatives in the public sector ... 0 for the incremental equivalent total cost to be economically justified ... For independent projects, no incremental B/C analysis is necessary ... 0 are selected provided there is no budget limitation ... The characteristics of public sectorprojects are substantially different from those of the private sector: initial costs are larger; expectedlife is longer; additional sources of capital funds include taxation, user fees, and governmentgrants; and interest (discount) rates are lower ... Evaluation by B/Canalysis can be difficult with no good way tomake monetary estimates of benefits ... The concept of dominance is incorporated into the procedure for comparing mutually exclusive alternatives ... Examples are included ... 1 What is the difference between disbenefits andcosts?9 ... 9 ... 9 ... (a) Loss of income to local businesses becauseof a newfreeway(b) Less travel time because of a loop bypass(c) $400,000 annual income to local businessesbecause of tourism created by a national park(d) Cost of fish from a hatchery to stock a lake atthe state park(e) Less tire wear because of smoother roadsurfaces(f) Decrease in property values due to theclosure of a government research lab(g) School overcrowding because of a militarybase expansion(h) Revenue to local motels because of an extended weekend holidaypublic-private partnership between the sheriff’soffice and a private security company ... Thecategories chosen will vary depending upona person’s viewpoint ... (a) Select the top two viewpoints (in your opinion) for each of the following individuals asthey would categorize estimates as a cost,benefit, or disbenefit ... An industrial plant manager in the county2 ... County commissioner (elected office)4 ... Project B/C Value9 ... 5 What is afundamental difference between DBOMand DBOMF contracts?9 ... Theroad will have to be maintained at a cost of $25,000per year ... Theimproved accessibility has led to a 150% increasein the property values along the road ... 9 ... Enforcement is proposed to be a9 ... Since it has been linkedto253Problemscancer of the bladder, kidney, and other internalorgans, the EPA has lowered the arsenic standardfor drinking water from 0 ... 010 parts per million (10 parts per billion) ... If it is estimated that there are 90 millionhouseholds in the United States and that the lowerstandard can save 50 livesper year valued at$4,000,000 per life, what is the benefit/cost ratio ofthe regulation?9 ... Use aninterest rate of 8% per year ... 11 A project to extend irrigation canals into an areathat was recently cleared of mesquite trees (a nuisance tree in Texas) and large weeds is projected tohave a capital cost of$2,000,000 ... Annual favorable consequences to the general public of $820,000 per year will be offset tosome extent by annual adverse consequences of$400,000 to a portion of the general public ... 12 Calculate the B/C ratio for the following cash flowestimates at a discount rate of 7% per year ... 13 Thebenefits associated with a nuclear power plantcooling water filtration project located on the OhioRiver are $10,000 per year forever, starting inyear 1 ... Calculate the B/C ratio at i ϭ10% per year ... 14 A privately funded wind-based electric powergeneration company in the southern part of thecountry hasdeveloped the following estimates(in $1000) for a new turbine farm ... Calculate (a) the profitability index and (b) the modified B/C ratio ... 15 For the values shown, calculate the conventionalB/C ratio at i ϭ 10% per year ... 16 A proposal to reduce traffic congestion on I-5 hasa B/C ratio of 1 ... The annualworth of benefitsminus disbenefits is $560,000 ... 17 Oil spills in the Gulf of Mexico have been knownto cause extensive damage to both public and private oyster grounds along the Louisiana and Mississippi shores ... This procedure inevitably results in death tosome of the saltwater shellfish whilepreventingmore widespread destruction to public reefs ... If theFish and Wildlife Service spent $110 million inyear 0 and $50 million in years 1 and 2 to minimize environmental damage from one particularoil spill, what is the benefit-to-cost ratio providedthe efforts resulted in saving 3000 jobs valued atatotal of $175 million per year? Assume disbenefitsassociated with oyster deaths amounted to $30 million in year 0 ... 9 ... If the operatingcost is $600,000 per year and the public healthbenefits are assumed to be $800,000 per year, whatinitial investment in the GFH system is necessaryto guarantee amodified B/C ratio of at least 1 ... 254Chapter 9Benefit/Cost Analysis and Public Sector Economics9 ... 3 million ... Benefits of $340,000 per year and disbenefits of $40,000 per year have also been identified ... 9 ... To thePeopleBenefits:$300,000 now and$100,000 peryear thereafterDisbenefits: $40,000per yearTo theGovernmentCosts:$1 ... 21 In 2010, Brazil began construction of the BeloMonte hydroelectric dam on the Xingu River(which feeds the Amazon River) ... It will begin producingelectricity in 2015 ... They say it willdevastate wildlife and the livelihoods of 40,000people who live in the area to beflooded ... 2 billion per year ... Assume that thedisbenefits will occur evenly through the 5-yearconstruction period and anticipated benefits willbegin at the end of 2015 and continue indefinitely ... 0 ... 22 In the United States, the average number of airplanes in the sky on an average morning is 4000 ...Aerospace company Rockwell Collins developedwhat it calls a digital parachute—a panic-buttontechnology that will land any plane in a pinch atthe closest airport, no matter what the weather orgeography and without the help of a pilot ... Assumethat the cost of retrofitting 20,000 commercial air-planes is$100,000 each and the plane stays in service for 15 years ... 9 ... This has required homeowners living in valley areas near theriver to purchase flood insurance costing between$145 and $2766 per year ... As a result, 13,000 propertieswere freed of the federal mandate to purchaseflood insurance ... If theaverage cost of floodinsurance is $460 per household per year, calculate the benefit-to-cost ratio of the levee-raisingproject ... 9 ... Benefits:$20,000 in year 0 and$30,000 in year 5Disbenefits:$7000 in year 3Savings (to government): $25,000 in years 1–4Cost:$100,000 in year 0Project life:5 years9 ... 9 ...Determine the profitability index for thefinancial results listed below using a MARR of8% per year ... 2 millionYear 5$Ϫ3 ... 2 million per year$2 ... 27 A project had a staged investment distributed overthe 6-year contract period ... 255ProblemsYear0Investment, $1000NCF, $1000 peryear123Ϫ25005Ϫ10709456selected at an interest rate of 8% per year and a5-year study period ... 28 In comparing two alternatives by the B/C method,if the overall B/C ratio for both alternatives is calculated to be exactly 1 ... 29 In comparing alternatives X and Y by the B/Cmethod, if B/CX ϭ 1 ... 8, whatisknown about the B/C ratio on the increment of investment between X and Y?9 ... Themountain site (MS) will use injection wells that cost$4 ... This site will be able to accommodate150 million gallons per year ... At thissite, 890 million gallons can be injected each year ... 00 per 1000 gallons, whichalternative, if either, should be selectedaccording to the B/C ratio method? Use an interestrate of 8% per year and a 20-year study period ... 31 The estimates shown are for a bridge under consideration for a river crossing in Wheeling, WestVirginia ... East LocationWest Location11 ϫ10100,000990,000120,000ϱ27 ϫ 10690,0002,400,000100,000ϱInitial cost, $Annual M&O, $/yearBenefits, $/yearDisbenefits, $/yearLife, years69 ... Use a B/C analysis and an interestrate of 8% per year ... 33 Conventional and solar alternatives are availablefor providing energy at a remote radar site ... 34The two alternatives shown are under considerationfor improving security at a county jail in TravisCounty, New York ... Extra Cameras(EC)New Sensors(NS)38,00049,000110,00026,00087,00064,000160,000—First cost, $Annual M&O, $/yearBenefits, $/yearDisbenefits, $/year9 ... S ... Separatecontractors proposed two methods ... For this method, the costs will be$14,100 for concrete, $6000 for metal decking,$4300 for joists, and $2600 for beams ... Specialadditives will be included in the lightweight concretethat will improve the heat-transfer properties of thefloor ... 9 ... Determinewhich projectshould be selected on the basis of aB/C analysis at i ϭ 8% per year and a 20-yearstudy period ... 37 Two routes are under consideration for a new interstate highway ... The short transmountain route256Chapter 9Benefit/Cost Analysis and Public Sector Economicswould be 10 kilometers long and wouldhave an initial cost of $45 million ... Regardless ofwhich route is selected, the volume of traffic is expected to be 400,000 vehicles per year ... 30 per kilometer, determine which route should be selected onthe basis of (a) conventional B/C analysis and(b) modified B/C analysis ... 9 ... Location E wouldrequire an investment of $3 million and $50,000 per year to maintain ... The operating cost of location W will be$65,000 per year ... The disbenefits associatedwith each location are $30,000 per year for locationE and $40,000 per year for location W ... Use an interestrate of 12% per year to determinewhich location, ifeither, should be selected on the basis of (a) theB/C method and (b) the modified B/C method ... 39 Three engineers made the estimates shown belowfor two optional methods by which new constructiontechnology would be implemented at a site for public housing ... Set up aspreadsheetfor B/C sensitivity analysis and determine if option1, option 2 or the do-nothing option is selected byeach of the three engineers ... Engineer BobEngineer JudyEngineer ChenAlternativeABCDEFPW of capital, $PW of benefits, $8070505572764352898581849 ... One must be accepted ...Comparison⌬B/CRatioA versus BB versus CC versus DA versus CA versus DB versus DC versus ED versus E0 ... 41 ... 10 ... 91 ... 99 ... The future worth ofcosts, benefits, disbenefits, and cost savings isshown ... Determine which of theprojects, if any, should be selected, if the projectsare (a)independent and (b) mutually exclusive ... 43 From the data shown below for six mutuallyexclusive projects, determine which project, ifany, should be selected ... 40 A group of engineers responsible for developingadvanced missile detection and tracking technologies, such as shortwave infrared, thermalinfrareddetection, target tracking radar, etc ... The present worth (in $ billions) of the capital requirementsand benefits is shown for each alternative in thetable ... Annual cost, $per yearAnnual benefits,$ per yearB/C ratio (alternative vs ... 231 ... 870 ... 711 ... 07A versus C ϭ 0 ... 02B versus D ϭ 0 ... 00Bversus F ϭ 1 ... 06C versus F ϭ 1 ... 44 Four mutually exclusive revenue alternatives arebeing compared using the B/C method ... 2023 ... 7510 ... 6515 ... DN AABCD303852810 ... 181 ... 16BCD— 2 ... 30 1 ... 58 1 ... 45—9 ... One accepted measure of effectiveness ofa program is the percentage ofenrollees quitting ... 9 ... Louis, Missouri, is considering various proposals regarding the disposal of used tires ... An incremental B/Canalysis was initiated but never completed ... (b) Determine which alternative should be selected ... 051 ... 34IncrementalB/C WhenCompared withAlternativeJKLM— ? ? ?—? ?— ?—The Cancer Society provides annual cost-offsetfunding to cancer patients so more people can affordthese programs ... Louis has thecapacity to treat each year the number of peopleshown ... (b) Offer programs using as many techniques aspossible to treat up to 1300 people per yearusing themost cost-effective techniques ... 46 In an effort to improve productivity in a largesemiconductor manufacturing plant, the plantmanager decided to undertake on a trial basis a series of actions directed toward improving employee morale ... Periodically, the company surveyed the employees to measurethe change in morale ... The per-employee cost of each strategy(identified as A through F) and the resultant measurement score are shown in the next column ... Determine which strategies are the best to implement ... )Cost, $/EnrolleeTechniqueAcupunctureSubliminal messageAversiontherapyOutpatient clinicIn-patient clinicNicotine replacementtherapy (NRT)% QuittingTreatmentCapacity per Year700150170025001800130091103941202505002004005501009 ... (a) Calculate the cost effectiveness ratiofor each alternative, and (b) use the CER to identify the best alternative ... 49 Anengineering student has only 30 minutes beforethe final exam in statics and dynamics ... There is time for using onlyone method of assistance before the exam; he mustselect well ... The method and estimates follow ... 50 During the design and specifications developmentstages of a remote meter readingsystem for residential electricity use (a system that allows monthlyusage to be transmitted via phone lines with noneed to physically view meters), the two engineersworking on the project for the city of Forest Ridgenoted something different from what they expected ... The second designer,anindustrial/systems engineer, further noted that allthe hardware specifications provided to them bythis same liaison came from the same distributor,namely, Delsey Enterprises ... Upon review,they learned that Don is the son-in-law of the cityliaison and Susan is his stepdaughter ... 51 Explain thedifference between public policy making and public planning ... 52 Since transportation via automobile was introduced,drivers throughout the country of Yalturia in easternEurope have driven on the left side of the road ... This is a major policy change for the country and will require significant publicplanning andproject development to implement successfully andsafely ... Identify six of the projects you deem necessary ... ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS9 ... 54 All of the following are usually associated withpublic sector projects except:(a) Funding from taxes(b)Profit(c) Disbenefits(d) Infinite life9 ... com company(c) Dam with irrigation canals(d) Mass transit system9 ... 57 In a conventional B/C ratio:(a) Disbenefits and M&O costs are subtractedfrom benefits ... (c) Disbenefits and M&O costs are added tocosts ... 9 ... (b) Benefits are subtracted from costs ... (d)M&O costs are put in the numerator ... 59 If two mutually exclusive alternatives have B/Cratios of 1 ... 5 for the lower- and higher-costalternatives, respectively, the following is correct:Case Study(a)(b)(c)(d)9 ... 5 ... 4 and 1 ... The B/C ratio on the increment betweenthem is less than 1 ... The higher-costalternative is the better oneeconomically ... 1,0 ... 3 ... 61 An alternative has the following cash flows:Benefits of $50,000 per yearDisbenefits of $27,000 per yearInitial cost of $250,000M&O costs of $10,000 per yearIf the alternative has an infinite life and the interestrate is 10% per year, the B/C ratio isclosest to:(a) 0 ... 66(c) 0 ... 169 ... 65(b) 0 ... 80(d) 1 ... 63 In evaluating three mutually exclusive alternativesby the B/C method, the alternatives are ranked A,B, and C, respectively, in terms of increasing cost,and the following results are obtained for theoverall B/C ratios: 1 ... 9, and 1 ... On the basisofthese results, you should:(a) Select A(b) Select C(c)(d)259Select A and CCompare A and C incrementally9 ... 5 ... The first cost of the alternative at an interestrate of 10% per year is closest to:(a) $23,300(b) $85,400(c) $146,100(d) $233,0009 ... (b) CEA expresses outcomes in natural unitsrather than incurrency units ... (d) CEA is more time-consuming and resourceintensive ... 66 Several private colleges claim to have programsthat are very effective at teaching enrollees howto become entrepreneurs ... If the total cost ofthe programs is $25,000 and $33,000, respectively, the incremental cost-effectiveness ratio isclosest to:(a) 6250(b) 5500(c) 4000(d) 13339 ... 68 Of the following, the word not related to ethics is:(a) Virtuous(b) Honest(c) Lucrative(d) Proper9 ... Poor highway lighting may be one reason that proportionately more traffic accidents occur at night ... For example, an accident with afatality is valued at260Chapter 9Benefit/Cost Analysis and Public Sector Economicsapproximately $4 million, while an accident in which thereis property damage (to the car and contents) is valued at$6000 ... Observed reductions in accidents seemingly caused by too lowlighting can be translated intoeither monetary estimates ofthe benefits B of lighting or used as the effectivenessmeasure E of lighting ... Theproperty damage category is commonly the largest based onthe accident rate ... Number of Accidents Recorded1UnlightedLightedAccident TypeDayNightDayNightPropertydamage3791992069839The ratios of night to day accidents involving property damage for the unlighted and lighted freeway sections are 199/379 ϭ0 ... 406, respectively ... To quantify the benefit, theaccident rate ratio from the unlighted section will be applied tothe lighted section ... Thus, there wouldhave been (2069)(0 ... This is a difference of 247 accidents ... To determine the cost of the lighting, it will be assumed thatthe light poles are center poles 67 meters apart with 2 bulbseach ... Since these data were collected over 87 ... 8Installation cost ϭ $3500 ———0 ... 8/0 ... 10 per kWh ... 4kilowatt/bulb)ϫ (12 hours/day)(365 days/year)ϫ ($0 ... Therefore, theannualized cost C at i ϭ 6% per year isTotal annual cost ϭ $4,585,000(A/P,6%,5)ϩ 459,024ϭ $1,547,503If a benefit/cost analysis is the basis for a decision on additional lighting, the B/C ratio is1,482,000B/C ϭ ————— ϭ 0 ... 0, thelighting is not justified ... If a cost-effectiveness analysis (CEA) isapplied, due to a judgment that the monetary estimates forlighting’s benefit is not accurate, the C/E ratio is1,547,503C/E ϭ ————— ϭ 6265247This can serve as a base ratio for comparison when an incremental CEA is performed foradditional accident reduction proposals ... Install poles at twice the distance apart (134 meters) ... Install cheaper poles and surrounding safety guards,plus slightly lowered lumen bulbs (350 watts) at acost of $2500 per pole; place the poles 67 metersapart ... Install cheaper equipment for $2500 per polewith350-watt lightbulbs and place them 134 meters apart ... Case Study ExercisesDetermine if a definitive decision on lighting can be determined by doing the following:1 ... 2 ... From an understanding viewpoint, consider the following:3 ... What would the lighted, night-to-day accident ratiohave to be tomake alternative Z economically justifiedby the B/C ratio?5 ... Doesone seem more appropriate in this type of situation thanthe other? Why? Can you think of other bases that mightbe better for decisions for public projects such as this one?Portion of data reported in Michael Griffin, “Comparison of theSafety of Lighting on Urban Freeways,” Public Roads, vol ... 8–15, 1994 ... Anymethod—PW, AW, FW, ROR, or B/C—can be used to select one alternative from two or more andobtain the same, correct answer ... Yet different information about an alternative is available with each different method ...Table LS2–1 gives a recommended evaluation method for different situations, if it is not specifiedby the instructor in a course or by corporate practice in professional work ... Interpretation of the entries in eachcolumn follows ... Public sector projects arecommonly evaluated using the B/C ratio and usuallyhave long lives that may be considered infinite for economic computation purposes ... For cost alternatives, the revenue cash flow series is assumed to be equal for all alternatives ... Service sector projects for which benefits are estimated using a nonmonetary effectiveness measure are usually evaluatedwith a method such as cost-effectiveness analysis ... Recommended method: Whether an analysis is performed by hand, calculator, or spreadsheet, themethod(s) recommended in Table LS2–1 will correctly select one alternative from two or more asTABLE LS2–1Recommended Method to CompareMutually Exclusive Alternatives,Provided the Method Is Not PreselectedEvaluation PeriodType of AlternativesRecommended MethodSeries toEvaluateEqual lives ofalternativesRevenue or costAW or PWCash flowsPublic sectorB/C, based on AW or PWIncrementalcash flowsRevenue or costAWCashflowsPublic sectorB/C, based on AWIncrementalcash flowsRevenue or costAW or PWPublic sectorB/C, based on AW or PWUpdated cashflowsUpdated incrementalcash flowsRevenue or costPublic sectorAW or PWB/C, based on AWUnequal lives ofalternativesStudy periodLong to infiniteCashflowsIncrementalcash flows262Chapter 9Benefit/Cost Analysis and Public Sector Economicsrapidly as possible ... For example, if lives are unequal and therate of return is needed, it is best to first apply the AW method at the MARR and then determine the selected alternative’s i* using the same AWrelation with i as the unknown ... For spreadsheet analyses, this means that the NPV or PV functions (for presentworth) or the PMT function (for annual worth) is applied ... Once the evaluation method is selected, a specific procedure must be followed ... Table LS2–2 summarizes the importantTABLELS2–2Characteristics of an Economic Analysis of Mutually Exclusive Alternatives Once theEvaluation Method Is DeterminedSeries toEvaluateRate ofReturn;Interest RateDecisionGuideline:Select1EqualLivesCash flowsMARRUnequalLCMCash flowsMARRPWStudy periodStudy periodLongtoinfiniteInfinityUpdatedcash flowsCash flowsMARRCCFuture worthTimePeriod forAnalysisPWPresent worthLives ofAlternativesPWEvaluationMethodNumericallylargest PWNumericallylargest PWNumericallylargest PWNumericallylargest CCEquivalenceRelationFWAWAnnual worthAWMARRSame aspresent worth for equal lives, unequal lives,and study periodEqual orunequalStudy periodLivesCash flowsMARRStudy periodUpdatedcash flowsCash flowsMARRIncrementalcash flowsIncrementalcash flowsCash flowsFind ⌬i*AWEqualLivesPW or AWUnequalLCM of pairAWUnequalLivesPW orAWStudy periodStudy periodUpdatedincrementalcash flowsFind ⌬i*PWBenefit/costInfinityPW or AWRate of returnLong toinfiniteEqual orunequalEqual orunequalLong toinfiniteLCM of pairsIncrementalcash flowsIncrementalcash flowsIncrementalcash flowsDiscount rateAWAW or PW1Lowest equivalentcost or largest equivalent income ... 0Last ⌬B/C Ն1 ... 0Learning Stage 2: Epilogueelements of the procedure for each method—PW, AW, ROR, and B/C ... The meaning of the entries in Table LS2–2 follows ... The capitalized cost (CC) relation is a PW relation for infinite life, andthe FW relation is likelydetermined from the PW equivalent value ... Lives of alternatives and time period for analysis The length of time for an evaluation (then value) will always be one of the following: equal lives of the alternatives, LCM of unequallives, specified study period, or infinity because the lives are very long ... •Incremental ROR and B/C methods require the LCM of the two alternatives being compared ... • CC analysis has an infinite time line and uses the relation P ϭ A͞i ... The LCM of the two alternatives compared must be used ... Bothapproaches find the incremental rate of return ⌬i* ... Rate of return (interestrate) The MARR value must be stated to complete the PW, FW,or AW method ... The ROR method requires that the incremental rate be found in orderto select one alternative ... Decision guideline The selection of one alternative is accomplished using the general guideline in the rightmost column ... Thisis correct for both revenue and cost alternatives ... Thismeans that the ⌬i* exceeds MARR, or the ⌬B/C exceeds 1 ... EXAMPLE LS2–1Read through the problem statement of the following examples, neglecting the evaluationmethod used in the example ... Is this the method used in the example? (a) 8 ...5, (c) 5 ... 4 ... (a) Example 8 ... Use the AW or PW valueat the MARR of 10% ... (b) Example 6 ... The B/C ratio of AW values is the bestchoice ... 263264Chapter 9Benefit/Cost Analysis and Public Sector Economics(c) Since Example 5 ... Since one life is long, capitalized cost, based on P ϭ A/i, is best inthiscase ... (d) Example 5 ... It involves 5-year and 10-year costalternatives ... The PW method for the LCMof 10 years and a study period of 5 years were both presented in the example ... Thechapters in this stage introduce information-gathering andtechniques that make decisions better ... The future iscertainly not exact ... After completing these chapters, you will be able to go beyondthe basic alternative analysis tools of the previous chapters ... Important note: If asset depreciation and taxes are to be considered by an after-tax analysis, Chapters 16 and 17 should be coveredbefore or in conjunctionwith these chapters ... SECTIONTOPICLEARNING OUTCOME10 ... 10 ... 10 ... 10 ... 10 ... 10 ... 10 ... This chapter discusses the different ways to finance a project through debt andequity sources and explains how the MARR is established ... Some ofthe parameters specified earlier are unspecifiedhere, and in future chapters ... Until now, only one dimension—the economic one—has been the basis for judging theeconomic viability of one project, or the selection basis from two or more alternatives ... 10 ... In Chapter 1 the MARR was described relative to the weighted costs of debt andequity capital... To form the basis for a realistic MARR, the types and cost of each source of project financingshould be understood and estimated ... There are several terms and relationships important to the understanding of project financing and the MARR that is specified to evaluate projects using PW, AW,FW, orB/C methods ... 9 will complement the following material ... The MARR is then set relative to the cost of capital ... MARR values changeover time due to changing circumstances ... This rate is determined by finding the ROR (i*) value of theproject cash flows ... Before we discuss cost of capital, we reviewthe two primary sources of capital ... Debt financing includes borrowing viabonds, loans, and mortgages ... The amount of outstanding debt financing is indicated in the liabilities section of the corporatebalance sheet ... Owners’ funds are further classified as common and preferred stock proceeds orowners’ capital for a private (non-stock-issuing) company ... The amount of equity is indicated in the net worthsection of the corporate balance sheet ... Therefore, the cost of debt capitalis 8% as shown in Figure 10–1 ... Management may increasethis MARR in increments that reflect its desire for addedreturn and its perception of risk ... Suppose thisamount is 2% ... Also, if the risk associated with the investment is considered substantial enough to warrant an additional 1% return requirement, the final MARR is 11% ... The recommended approach does not follow the logic presented above ... Then the i*value is determined from theestimated net cash flows ... Now, additional return requirements and risk factors are considered to determine if 3%above the MARR of 8% is sufficient to justify the capital investment ... This is the opportunity cost discussed previously—the unfunded project i* has establishedthe effective MARR for emissioncontrol system alternatives at 11%, not 8% ... The debt and equitycapital mix changes over time and between projects ... It is altered for different opportunities and types of projects ... The effective MARR varies from one project to another and through time because offactorssuch as the following:Project risk ... This is encouraged by the higher cost of debtcapital for projects considered risky ... Investment opportunity ... This common reaction to investment opportunity can create havoc when theguidelines for setting a MARR are too strictly applied ... Governmentintervention ... This may occur through price limits, subsidies,import tariffs, and limitation on availability ... Examples are steel imports,foreign capital investment, car imports, and agricultural product exports ... , thus tending to move the MARR up or down ... If corporate taxes are rising (due to increasedprofits, capital gains, localtaxes, etc ... Use of after-tax analysis may assist ineliminating this reason for a fluctuating MARR, since accompanying business expenses willtend to decrease taxes and after-tax costs ... As debt and equity capital become limited, the MARR is increased ... Theopportunity costhas a large role in determining the MARR actually used ... 2Debt-Equity Mix and Weighted Average Cost of CapitalMarket rates at other corporations ... These variations areoften based on changes in interest rates for loans, which directly impact the cost of capital ... 1]The total or effective tax rate,including federal, state, and local taxes, for most corporations is inthe range of 30% to 50% ... 35) ϭ 15 ... EXAMPLE 10 ... Carl, anarchitect, has worked in home design with Bulte Homes since graduation ... They both reside inRichmond, Virginia ... Carl and Christy want to expand into a regional e-business corporation ... Identifysome factors that might cause the loan rate to vary when BA provides the quote ... SolutionIn all cases the direction of the loan rate and the MARR will be the same ... Investment opportunity: The rate could increase if other companies offering similar services have alreadyapplied for a loan at other BA branches regionally or nationwide ... The intervention maybe designed to boost the housing economic sector in an effort to offset a significantslowdown in new home construction ... Capital limitation: Assume the computer equipment and software rights held by CarlandChristy were bought with their own funds and there are no outstanding loans ... Market loan rates: The local BA branch probably obtains its development loan moneyfrom a large national pool ... ”10 ... A company with a 40–60 D-E mix has 40% of its capital originating from debt capitalsources (bonds,loans, and mortgages) and 60% derived from equity sources (stocks and retained269270Chapter 10Project Financing and Noneconomic AttributesFigure 10–2General shape of differentcost of capital curves ... Most projects are funded with a combination of debt and equity capital madedebt fraction)(cost of debt capital)[10 ... (؉ ) � ) equity fraction)(cost of equity capital availablespecifically for the project or taken from a corporate pool of capital ... If known exactly, these fractions are used to estimate WACC; otherwise the historical fractionsfor each source are used in the relationWACCSince virtually all corporations have a mixture of capital sources, the WACC is a value between the debt and equity costs of capital ... 2] is expanded ... 3]Figure 10–2 indicates the usual shape of cost of capital curves ... There is virtually always a mixture of capital sources involved for any capitalizationprogram ... Most firms operate over a range of D-E mixes ... However, another company may be considered “risky” with only 20% debt capital ... EXAMPLE 10 ... In an effortto develop sustainable and renewable vegetable sources, a new commercial vertical crop technology is being installed through apublic-private partnership with Valcent Products ... ,” www ... net, June 16, 2010 news release ... 3Determination of the Cost of Debt Capitalillustration purposes, assume that the present worth of the total system cost is $20 million withfinancing sources and costs as follows ... 8% per year$4 million at 5 ...9% per yearThere are three existing international vertical farming projects with capitalization and WACCvalues as follows:Project 1:Project 2:Project 3:$5 million with WACC1 ϭ 7 ... 2%$7 million with WACC3 ϭ 4 ... SolutionTo apply Equation [10 ... These are 0 ... 2 for retained earnings, and 0 ... WACCHKϭ 0 ... 9%) ϩ 0 ... 2%) ϩ 0 ... 8%) ϭ 6 ... 119; project 2 has0 ... 167 ... WACCW ϭ 0 ... 9%) ϩ 0 ... 2%) ϩ 0 ... 8%) ϭ 9 ... The WACC value can be computed using before-tax or after-tax values for cost of capital ... 3 below ... 4]The effective tax rate is a combination of federal, state, and local tax rates ...Equation [10 ... 2] for an after-tax WACC rate ... 10 ... (We learned about bonds inSection 7 ... ) In most industrialized countries, bond dividends and loan interest payments are taxdeductible as a corporate expense ... The cost of debt capital is, therefore, reducedbecause there is an annual tax savings ofTe)[10 ... 6]To find the cost of debt capital, develop a PW- or AW-based relation of the net cash flow(NCF) ؊ � expenses (1 tax savings ؊ � expenses � expenses (Te)Net cash flow ( � ) expenses) (effective tax rate the expense cash flow times the effective tax rate Te ... In formula form,Tax savingsseries with i* as the unknown ... This is the cost of debt capital used in the WACC computation, Equation [10 ... EXAMPLE 10 ... If the effective tax rate of the company is 30% and the bonds are discounted2%, compute the cost of debt capital (a) before taxes and (b) after taxes from thecompanyperspective ... Solution by Hand(a) The annual bond dividend is $1000(0 ... Using the company perspective, find the i* in the PW relation0 ϭ 980 Ϫ 80(P/A, i*,10) Ϫ 1000(P/F, i*,10)i* ϭ 8 ... 3%, which is slightly higher than the8% bond interest rate, because of the 2% sales discount ... 5]shows atax savings of $80(0 ... The bond dividend amount for the PWrelation is now $80 Ϫ 24 ϭ $56 ... 87% ... The after-tax net cash flow is calculated using Equation [10 ... 3 ... 08Bond dividend after taxesϭ (Ϫ1000*0 ... 3)ϭ IRR(C3:C13)Figure 10–3Use of IRR function to determine cost of debt capital beforetaxes and after taxes, Example10 ... EXAMPLE 10 ... Company managershave decided to put $10,000 down now from retained earnings and borrow $10,000 at aninterest rate of 6% ... (a) What is the after-tax cost of debt capital ifthe effective tax rate is 42%? (b) How are the interest rate and cost of debtcapital used tocalculate WACC?10 ... 42) ϭ $348 by Equation [10 ... The loan repayment is $10,000 in year 10 ... 48% ... Nor is 3 ... Thecost of the $10,000 equity capital is needed to determine the WACC ... 4 Determination of the Cost of EquityCapital and the MARREquity capital is usually obtained fromthe following sources:Sale of preferred stockSale of common stockUse of retained earningsUse of owner’s private capitalThe cost of each type of financing is estimated separately and entered into the WACC computation ... One additional method for estimating the cost of equity capital via commonstockis presented ... Issuance of preferred stock carries with it a commitment to pay a stated dividend annually ... Preferred stock may be sold at a discount to speed the sale, inwhich case the actual proceeds from the stock should be used as the denominator ... 53% ... The dividends paidare not a true� ؉ ———————— expected dividend growth rateprice of stockDV1� ؉ —— gP[10 ... Stated another way, it is the compound growth rate on dividends that the indication of what the stock issue will actually cost in the future ... If Re is the cost of equity capital (indecimal form),first-year dividendRecompany believes isrequired to attract stockholders ... S ... If a 5% or $1 dividend is planned for the first year and anappreciation of 4% per year is anticipated for future dividends, the cost of capital for this common stock issue from Equation [10 ... 1Re ϭ —— ϩ 0 ... 0920The retained earnings and owner’sfunds cost of equity capital is usually set equal to thecommon stock cost, since it is the shareholders and owners who will realize any returns fromprojects in which these funds are invested ... MarketsecuritylineRe␤Ͼ1Premiumincreasesfor morerisky securitiesRm –RfPremiumRmRfSelectedmarketportfolio01 ... 3] ... Because of the fluctuations in stock prices and the higher return demandedby some corporations’ stocks compared to others, this valuation technique is commonly applied ... 8]where ␤ ϭ volatility of a company’s stock relative to other stocks in themarket (␤ ϭ 1 ... S ... The coefficient ␤(beta) indicates how the stock is expected to vary compared to a selected portfolio of stocks inthe same general market area, often the Standard and Poor’s 500 stock index ... 0, thestock is less volatile, so the resulting premium can be smaller; when ␤ Ͼ 1 ...Security is a word that identifies a stock, bond, or any other instrument used to develop capital ... This is a plot of a market security line, which is a linear fit by regression analysis to indicate the expected return for different␤ values ... As ␤ increases, thepremium return requirement grows ... Oncecomplete, this estimated cost of common stock equity capital can be includedin the WACC computation in Equation [10 ... EXAMPLE 10 ... It is envisioned that processes for prepared meats can be completed more safely and faster using thisautomated control software ... SafeSoft, which has a historicalbeta value of 1 ... The securitymarket line indicates that a 5% premium above the risk-free rate is desirable ... S ... 10 ... 8] ... 0 ϩ 1 ... 0) ϭ 7 ... In theory, a correctly performed engineering economy study uses a MARR equal to the costof the capital committed to the specific alternatives in the study ... Fora combination of debt and equity capital, the calculated WACC sets the minimum forthe MARR ... The risks associated with an alternative should be treated separatelyfrom the MARR determination, as stated earlier ... Unfortunately, the MARR is often set above the WACC because management doeswant to account for risk by increasing the MARR ... 6The Engineering Products Division of 4M Corporation has two mutually exclusive alternativesA and B with ROR values of i* ϭ 9 ... 9% ... 5% and the remainder from the same equity funds mentioned above ... Make the economic decision on alternativeA versus B under each financingscenario ... SolutionThe capital is available for one of the two mutually exclusive alternatives ... Only alternative A is acceptable; alternative B is not since the estimated return of 5 ... Under financing plan 2, with a D-E mix of 25–75,WACC ϭ 0 ... 5) ϩ 0 ... 0) ϭ 9 ... 625% ...10 ... 2 ... The leverage offered by larger debt capital percentages increases the riskiness of projects undertaken by the company ... This is sometimes referred to as a highlyleveraged corporation ... Thus, a reasonable balance between debt and equity financing is important for275276Chapter 10ProjectFinancing and Noneconomic Attributesthe financial health of a corporation ... 7 illustrates the disadvantages of unbalancedD-E mixes ... 7Three auto parts manufacturing companies have the following debt and equity capital amountsand D-E mixes ... Amount of CapitalCompanyDebt($ in Millions)Equity($in Millions)D-E Mix (%–%)ABC10204040201020−8050−5080−20Assume the annual revenue is $15 million for each one and that after interest on debt is considered, the net incomes are $14 ... 4, and $10 ... Compute the return oncommon stock for each company, and comment on the return relative tothe D-E mixes ... Inmillion dollars,14 ... 36 (36%)4013 ... 67 (67%)ReturnB ϭ ——2010 ... 00 (100%)ReturnC ϭ ——10As expected, the return is by far the largest for highly leveraged C, where only 20% of thecompany is in the hands of the ownership ... The use of large percentages of debt financinggreatly increases the risk taken by lenders andstock owners ... The leverage of large D-E mixes does increase the return on equity capital, as shown in previous examples; but it can also work against the owners and investors ... Example 10 ... EXAMPLE 10 ... S ... As a consequence, the D-E mixes ofthe so-called traditional companies (American, United, Delta, and others) have become larger on the debt side than is historically acceptable ... In an effort to reduce costs, assume thatthree airlines joined forces to cooperate on a range of services (baggage handling, onboardfood preparation, ticketservices, and software development) by forming a new company calledFullServe, Inc ... Table 10–1 summarizes the D-E mixes and the total equity capitalization for each airlineafter its share of $5 B was removed from available equity funds ... For10 ... 8AirlineCompanyCorporate D-EMix, %AmountBorrowed,$BEquity CapitalAvailable, $ BNationalGlobalPanAm30–7065–3591–91 ... 254 ... 03 ... 7example, National had 30% of its capitalization in debt capital; therefore, 30% of $5 B wasborrowed, and 70% was provided from National’s equity fund ... 0 billion, only 20% of its original value ... 0 billion inequity capital was returned to each airline ... Assuming the loan and equity amounts are thesame as shown in Table 10–1, determine the resulting equity capital situation for each airlineafter it pays off the loan from its own equity funds ... SolutionDetermine the level of post-FullServe equity capital usingthe following relation, in $ billions ... 0 ϩ 1 ... 50 ϭ $4 ... 7 ϩ 1 ... 25 ϭ $1 ... 7 ϩ 1 ... 55 ϭ $3 ... The debt capital to fund the failed FullServe effort has affected Nationalairlines the least, in large part due to its low D-E mix of 30%–70% ... The same principles discussed above for corporations are applicableto individuals ... As an example, assume two engineers each have a take-home amount of$40,000 after all income tax, social security, and insurance premiums are deducted from theirannual salaries ... If Jamal has a total debt of $25,000 and Barry owes $100,000, the remaining amount ofthe annualtake-home pay may be calculated as follows:Amount Paid, $ per YearPersonTotalDebt, $Cost ofDebt at 15%, $Repayment of Debt, $Amount Remainingfrom $40,000, $JamalBarry25,000100,0003,75015,0001,2505,00035,00020,000Jamal has 87 ... 277278Chapter 10Project Financing and NoneconomicAttributes10 ... Thedecision-making process explained in that chapter (Figure 1–1) included the seven steps listed onthe right side of Figure 10–5 ... In all prior evaluations, only one attribute—the economicone—has been identified and used to select the best alternative ... As we are all aware, mostevaluations do and should take into account multiple attributes indecision making ... However,these noneconomic dimensions tend to be intangible and often difficult, if not impossible, toquantify with economic and other scales ... This section and the next describe some of the techniques thataccommodatemultiple attributes in an engineering study ... Public and service sector projects are excellent examples of multiple-attribute problem solving ... High levels of complexity are introduced into the selection process by the multiple attributes thoughtto be important in selecting an alternative for the

dam’s location, design, environmental impact, etc ... The discussion below concentrates on the expanded step 4 and the next section focuses on the evaluationmeasure and alternative selection of step 5 ... To seek input from individuals other than the analyst isimportant; it helps focus the study on keyattributes ... ••••Comparison with similar studies that include multiple attributesInput from experts with relevant past experienceSurveys of constituencies (customers, employees, managers) impacted by the alternativesSmall group discussions using approaches such as focus groups, brainstorming, ornominalgroup technique• Delphi method, which is a progressive procedure to develop reasoned consensus from differentperspectives and opinionsFigure 10–5Consider multiple attributesExpansion of the decisionmaking process to includemultiple attributes ... Understand the problem; define the objective... Collect relevant information; define alternatives ... Make estimates ... Identify the attributes for decisionmaking ... Determine the relativeimportance (weights) of attributes ... For each alternative, determineeach attribute’s value rating ... Evaluate each alternative usinga multiple-attribute technique ... 4 ...5 ... 6 ... 7 ... 10 ... There are approximately 8000 options for each planethat must be decided upon by Delta’s engineering, purchasing, maintenance, and marketing personnel before the order to Boeing is placed ... An economic study based on the equivalent AW of the estimated passenger income pertrip has determined that 150 of these options areclearly advantageous ... A Delphi study was performed using input from 25 individuals ... From these two studies it was determined that there are 10 key attributes for options selection ... • Safety: mean time to failure (MTTF) of flight-critical components ...(Basically, this is the attribute evaluated bythe economic study already performed ... The economic attribute of extra revenue may be considered an indirect measure of customer satisfaction, one that is more quantitative than customer opinion/satisfaction survey results ... However, the point is that theeconomic studymay directly address only one or a few of the key attributes vital to alternative decision making ... Risk is a possible variation in a parameter from an expected, desired, or predicted value thatmay be detrimental to observing the intended outcome(s) of the product, process, or system ...Risk is present when there are two ormore observable values of a parameter and it is possible to assume or estimate the chance thateach value may occur ... Considerations of variation, probabilistic estimates, etc ... Formalized sensitivity analysis, expected values, simulation,and decision trees are someof the techniques useful in handling risk ... Theweight, a number between 0 and 1, is based upon the experienced opinion of one individual or agroup of persons familiar with the attributes, and possibly the alternatives ... Otherwise,some averaging technique must be applied to arrive at one weight valuefor each attribute ... Weights Wi for each attribute are entered on the left side ... Attribute weights are usually normalized such that their sum over all the alternatives is 1 ... Thisnormalizing implies that each attribute’s importance score is divided by the sum S over all attributes ... , m) aremNormalized� ————————— � ———————— i [10 ... nWmValue ratings VijOf the many procedures developed to assign weights to an attribute, an analyst is likely to relyupon one that is relatively simple, such as equal weights:͚ W ϭ 1 ... 9]iϭ1importance scoreiimportance scoreWeight calculation: Wiweighting, rank order, or weighted rank order ... Each is briefly presented below ... This is the default approach ... 10] ... In this case, the final evaluation measure for an alternative will be the sum overall attributes ... By Equation [10 ... , m͞S ... Weighted Rank Order The m attributes are again placed in the� —— m[10 ... 11] automatically normalizes the weights ... If repair time is only one-half as important as safety, and order of increasing importance ... The most important attribute is assigned a score, usually 100, and all other attributes are scored relative to it between 100and 0 ... 10] takes the formsiWithelast two attributes are each one-half as important as repair time, the scores and weights areas follows ... 5050/200 ϭ 0 ... 12525/200 ϭ 0 ... 000Pairwise Comparison Each attribute is compared to each other attribute in a pairwise fashion using a rating scale that indicates the importance of one attribute� Environment—210—111—30 ... � Constructability3 � Cost2 over the other ... 6TABLE 10–3Attribute iCostConstructabilityEnvironmentSum of scoresWeight WiMultiple Attribute Analysis: Identification and Importance of Each AttributePairwise Comparison of Three Attributes to Determine Weights116720 ... Define the importance comparison scale as follows:0 if attribute is less important than one compared to1 if attribute is equally important as one compared to2 if attribute is more important than one compared toSet up a table listing attributes across the top and down the side, and perform thepairwise comparison for each column attribute with each row attribute ... The arrow to the right of the table indicates the direction of comparison, i ... , column with row attribute ... The complement score of 0 is placed in the reverse comparison ofconstructability with cost ... 11], where the sum for eachcolumn is si ... s1 ϭ 3si ϭ 3 ϩ 1 ϩ 2 ϭ 6Cost weight W1 ϭ 3͞6 ϭ 0 ... 167 and W3 ϭ 2͞6 ϭ 0 ... There are other attribute weighting techniques, especially for group processes, such as utilityfunctions, and the Dunn-Rankin procedure ... If this consistency is important in thatseveral decision makers with diverseopinions about attribute importance are involved in a study, amore sophisticated technique may be warranted ... 4-3 Value Rating of Each Alternative by Attribute This is the final step prior to calculating the evaluation measure ... These are the entries within the cells in Table 10–2 ... The scale for the valuerating can vary depending upon what is easiest to understand for thosewho do the valuation ... However,the most popular is a scale of 4 or 5 gradations about the perceived ability of an alternative toaccomplish the intent of the attribute ... g ... The last two scales can give a negative impact to theevaluationmeasure for poor alternatives ... 281282Chapter 10Project Financing and Noneconomic AttributesTA BLE 10–4Completed Layout for Four Attributes and ThreeAlternatives for Multiple Attribute EvaluationAlternativesAttributesWeights123SafetyRepairCrew needsEconomic0 ... 250 ...125695543698167If we now build upon the aircraft purchase illustration to include value ratings, the cells arefilled with ratings awarded by a decision maker ... Initially, there will be one such table for each decision maker ... Determination of this evaluation measure is discussed below ... 7 EvaluationMeasure for Multiple AttributesThe need for an evaluation measure that accommodates multiple attributes is indicated in step 5of Figure 10–5 ... This section introduces a single-dimension measurethat is widely accepted ... The resulting evaluation measure is a formula that calculates an aggregatedmeasure for use in selecting from two or morealternatives ... This reduction process removes much of the complexity of trying to balance the different attributes;however, it also eliminates much of the robust information captured by the process of ranking attributes for their importance and rating eachalternative’s performance against each attribute ... The most used additive model is the weighted attribute method,also called the additive weight technique ... 12]i�1The Wi numbers are the attribute importance weights, and Vij is the value rating by attribute i foreach alternative j ... 10] ... (If an equalweight of Wi ϭ 1 ... )The selection guideline is as follows:ME alternative selectionChoose the alternative with the largest Rj value ... Sensitivity analysis for any score, weight, or value rating is used to determine sensitivity of the decision to it ... Chapter SummaryEXAMPLE 10 ... niueisland ... Thespreadsheet in Figure 10–6, left two columns, presents the attributes and normalizedweights Wi published in the RFP for use in selecting one of the tenders presenting proposals ... The next four columns (C through F ) include value ratingsbetween 0 and 100 developed by a group of decision makerswhen the details of each proposal wereevaluated against each attribute ... Use these weights and ratings to determine which proposal to pursue first ... Equation [10 ... As an illustration, for proposal 3,R3 ϭ 0 ... 20(60) ϩ 0 ... 35(85) ϩ 0 ... 5 ϩ 12 ... 5 ϩ 29 ... 0ϭ 84 ... CommentAny economic measure canbe incorporated into a multiple attribute evaluation using this method ... Figure 10–6Attributes, weights, ratings, and evaluation measure for Niue workboat proposals, Example 10 ... 1Used with permission of Government of Niue, Infrastructure Department, “Request for Proposal: Supplyof Workboat,”released March 30, 2010, www ... nu/Documents/workboattender3022 ... CHAPTER SUMMARYThe interest rate at which the MARR is established depends principally upon the cost of capitaland the mix between debt and equity financing ... Risk, profit, and other factors can be considered after theAW,PW, or ROR analysis is completed and prior to final alternative selection ... If multiple attributes, which include more than the economic dimension of a study, are to beconsidered in making the alternative decision, first the attributes must be identified and theirrelative importance assessed ...Theevaluation measure is determined using a model such as the weighted attribute method, where themeasure is calculated by Equation [10 ... The largest value indicates the best alternative ... Use the results below to determine the opportunity cost for each measure ... 1 List at least three factors thataffect the MARR,and discuss how each one affects it ... 2 State whether each of the following involves debtfinancing or equity financing ... 3 million(c) Short-term loan of $75,000 from a local bank(d ) Issuance of $3 million worth of 20-yearbonds(e) Del Engineering buyback of $8 million of itsown stockusing internal funds10 ... uses an after-tax MARR of12% per year ... 10 ... Heestimates that his cost to complete the project willbe $7 ... He wants to bid an amount thatwill give him an after-tax rate of return of 15% peryear if he gets the job, but he doesn’t know howmuch he should bid on a before-taxbasis ... (a) The expression for determining the overalleffective tax rate isstate rate ϩ (1 Ϫ state rate)(federal rate)What should his before-tax MARR be inorder for him to make an after-tax MARRof 15% per year?(b) How much should he bid on the job?10 ... Immediatelyafter the investment was made,another investment opportunity came up for which the investorsdidn’t have enough capital ... If the group’s effective taxrate is 32%, what after-tax rate of return wouldthe forgone project have yielded?10 ... 57,13812 ... 41,0519 ...09368,00015,0008,0008,0005,000AECBD8,00013,00021,00036,00044,000ACEDB8,00016,00021,00029,00044,00010 ... He expects a return of 4% per year on all of his investments ... Tomhas determined that the proposal’s “risk factor”will require an additional 3% per year return forhim to accept it ... (b)Determine the effective MARR for his business if Tom turns down the proposal ... 8 Electrical generators produce not only electricity,but also heat from conductor resistance and fromfriction losses in bearings ... If $18 million came from mortgages andbond sales, what was the total amount of thefinancing?10 ... 10 ... The business student has$30,000 in student loans that come due at graduation ... Theengineering senior owes $50,000, 50% from hisparents with no interest due and 50% from a285Problemscredit union loan ... (a) What is the D-E mix for each student?(b) If their grandparents pay the loans infull atgraduation, what are the amounts on thechecks they write for each graduate?(c) When grandparents pay the full amount atgraduation, what percent of the principaldoes the interest represent?dividends at a rate of 5% per year, and the remaining 60% from retained earnings, which currentlyearn 9%per year ... 10 ... The balance sheet for First Engineering indicates a totaldebt of $87 million, and that of Midwest Development indicates a net worth of $62 million ... 10 ... 1% for the year in itsreport to stockholders ... The annual report also mentions that projects within the corporation are 75%funded byits own capital ... 10 ... invested $50 million ... What is thereturn on the company’s equity, if the net income is$5 million on a revenue base of $6 million?10 ... Thefinancing profile, with interest rates, is as follows:$3 million in stock sales at 15% per year, $4 million in bonds at 9%, and $6 million inretainedearnings at 7% per year ... 14 Growth Transgenics Enterprises (GTE) is contemplating the purchase of its rival ... He learned there are twoplans being considered ... Plan 2 requires only 20% equityfunds with the balance borrowed at a higher rate of8% per year ... 2% will not be exceeded,what isthe maximum cost of debt capital allowed for each plan? Are these rates higheror lower than the current estimates?10 ... The current plan is 60%equity capital and 40% debt financing ... 17 BASF will invest $14 million this year to upgradeits ethylene glycol processes ... Equity capital costs 14 ... Debtcapital costs 10%per year before taxes ... (a) Determine the amount of annual revenueafter taxes that is consumed in covering theinterest on the project’s initial cost ... Determine theamount of annual revenue needed to coverthe interest with this plan, and explain the effect it may have on thecorporation’s abilityto borrow in the future ... 18 A couple planning for their child’s college education can fund part of or all the expected $100,000tuition cost from their own funds (through an education IRA) or borrow all or part of it ... Use a spreadsheet togenerate a plot of the WACC curve with theestimated loan interest rates below and determine thebest D-E mix for the couple ... 06 ... 09 ... 013 ... 19 Over the last few years, Carol’s Fashion Store, astatewide franchise, has experienced the D-Emixes and costs of debt and equity capital on several projects summarized below ... (b) Determine whatmix of debt and equity capital provided the lowest WACC ... 5%13 ... 011 ... 912 ... 8%7 ... 99 ... 512 ... 20 For Problem 10 ... 0% increasesto 14 ... Cost of Debt Capital10 ... If the company’s effective tax rate is 33%, determine the company’s costof debt capital (a) before taxes and (b) after taxes ... 22 Acompany that makes several different types ofskateboards, Jennings Outdoors, incurred interestexpenses of $1,200,000 per year from varioustypes of debt financing ... If the company’s effective tax rate is29%, what was the company’s cost of debt capital(a) before taxes and (b) after taxes?10 ... , amanufacturer of cable assemblies forpolycrystalline photovoltaic solar modules, requires $3 ... The companyplans to sell 15-year bonds that carry a dividend of6% per year, payable semiannually ... Determine(a) the nominal annual after-tax cost of debt capital and (b) the effective annual after-tax costofdebt capital ... 24 Tri-States Gas Producers expects to borrow$800,000 for field engineering improvements ... The company will pay an effective 8% per year to the bankfor 8 years ... Thebond issue will be for 800 10-year bonds of $1000each that require a 6% per year dividend payment ... 25 TheSullivan Family Partnership plans to purchasea refurbished condo in their hometown for investment purposes ... 5%per year after all relevant income taxes are paid ... If the effectivetax rate is 22% per year, based only on these data,answer the following ... 26 Determine the cost of equity capital to Hy-LokUSA if the company sells 500,000 shares of itspreferred stock at a 5% discount from its price of$130 ... 10 ... 92 per share ... 2% per year, determine the cost of equity capitalfor the stock offering ... 28 The cost of debt capital is lower after taxes thanbefore taxes ... Whyare the after-tax rates not thesame for both typesof financing?10 ... The beta value for its stockis 1 ... Use the capital asset pricing model and a3 ... 2% ... 30 Management at Hirschman Engineering has askedyou to determine the cost of equity capital basedon the company’s common stock ... Last year, the first year fordividends, thestock paid $0 ... 50 on the New York Stock Exchange ... Hirschman Engineering stockhas a volatility that is higher than the norm at 1 ... If safe investments are returning 5 ... 10 ... 93 per share on an average price of $18 ... The company expectsto grow the dividend rate at a maximum of 1 ... The stockvolatility is 1 ... 95% per year dividend ... S Treasury bills arereturning 4 ... Determine the company’s cost ofequity capital last year using (a) the dividendmethod and (b) the CAPM ... 32 Last year a Japanese engineering materials corporation, Yamachi Inc ... S ... Now, Euro bonds are being purchasedwith a realized average return of 3 ... The volatilityfactor of Yamachi stock last year was 1 ... 18 ... 1% dividends per year ... 10 ... Sheasked the finance manager for the corporateMARR ... Funds SourceAmount, $ Average Cost, %Retained earningsStock salesLong-termloans4,000,0006,000,0005,000,0007 ... 89 ... The study is after taxes and part (a) providedthe before-tax MARR ... Different D-E Mixes10 ... 35 In a leveraged buyout of one company by another,the purchasing company usually obtains borrowedmoney and inserts as little of its own equity as possible intothe purchase ... 10 ... Financing will be equally splitbetween common stock ($250,000) and a loanwith an 8% after-tax interest rate ... The effective tax rate is 50% ... Recent analysis shows that ithas a volatility rating of 0 ... Nationally, the safest investment is currently paying 3% per year ... 37 FairmontIndustries primarily relies on 100% equity financing to fund projects ... The Fairmont owner can supply the money frompersonal investments that currently earn an average of 7 ... The annual net cash flowfrom the project is estimated at $30,000 for thenext 15 years ... If the MARR is the WACC, determinewhichplan, if either, should be undertaken ... 10 ... has an opportunity to invest $10,000,000 in a new engineering remotecontrol system for offshore drilling platforms ... Omega’s share of the annual net cash flowis estimated to be $1 ... Omega is about to initiate CAPM as288Chapter 10Project Financingand Noneconomic Attributesits common stock evaluation model ... 22 andis paying a premium of 5% on its common stockdividend ... S ... Is the venture financially attractive if the MARR equals (a) the cost of equitycapital and (b) the WACC?10 ... The phase Iinstalled price for the dies and machineryis$2,000,000 ... The WACC over thelast 5 years has averaged 10% per year ... The first requires an investment of40% equity funds at 9% and a loan for thebalance at an interest rate of 10% per year ... 5% per year ... With this restriction, what isthe maximum loan interest rate that can be incurred for eachof the financing alternatives?10 ... Utilize aspreadsheet solution to (a) select any combinationof the projects if the MARR is equal to the aftertax WACC and (b) determine if the same projectsshould be selected if the risk factors are enough torequire an additional 2% per year for the investment to be made... 5% per year ... The effective tax rate is 30% per year ... 41 Two friends each invested $20,000 of their own(equity) funds ... Theresa, being a risk taker, leveraged the$20,000 and purchased a $100,000 condo for rentalproperty ... (a) Determine the year-end values of their equityfunds if there was a 10%increase in thevalue of the stocks and the condo ... (c) Use your results to explain why leverage canbe financially risky ... 42 In multiple attribute analysis, if three different alternatives are to be evaluated on the basis of eightattributes that are considered of equal importance,what is the weight of eachattribute?10 ... Determine the weight of each attributeusing the importance scores ... Safety2 ... Impact4 ... AcceptabilityImportance Score604080302010 ... , and J ... 10 ... Use the statementsto determine the normalized weights if assignedscores are between 0 and 100 ... Flexibility (F)2 ... Uptime (U)4 ...46 Different types and capacities of crawler hoes arebeing considered for use in a major excavation ona pipe-laying project ... For the information that follows,determine the weighted rank order, using a 0-to-10scale and the normalized weights ... Truck versus hoe height 90% as important as trenchingspeed2 ... Type of subsoil30% as important as trenching speed4 ... Hoe trenching speedMost important attribute6 ... 47 John, who works at Swatch, has decided to use theweighted attribute method to compare three systems for manufacturing a watchband ... John’sratings for each alternative are asfollows:AlternativeAttributeEconomic return Ͼ MARRHigh throughputLow scrap rate123501001007060401003050Use the weights below to evaluate the alternatives ... 48 The Athlete’s Shop has evaluated two proposalsfor weight lifting and exercise equipment ... In addition tothis economic measure,three more attributeswere independently assigned a relative importance score from 0 to 100 by the shop managerand the lead trainer ... 0 as shown in the following table ... AttributeEconomicsDurabilityFlexibilityMaintainabilityProposal AProposal B1 ... 351 ... 250 ... 000 ... 00Select the better proposalusing each of the following methods ... 49 The term opportunity cost refers to:(a) The first cost of an alternative that has beenaccepted for funding(b) The total cost of an alternative that has beenaccepted for funding(c) The rate of return or profit available on thenext-best alternative that had to beforgonedue to lack of capital funds(d) The cost of an alternative that was not recognized as an alternative that actually represented a good opportunity10 ... 51 All of the following are examples of debt capitalexcept:(a) Retained earnings(b) Long-term bonds(c) Loan from a local bank(d) Purchase ofequipment using a credit card10 ... 53 If a public utility expands its capacity to generateelectricity by obtaining $41 million from retainedearnings and $30 million from municipal bondsales, the utilities’ debt-to-equity mix is closest to:(a) 58% debt and 42% equity(b) 73% debt and 27% equity(c) 27% debtand 73% equity(d) 42% debt and 58% equityIf Medzyme does not want to exceed its historicalweighted average cost of capital (WACC), and it isforced to go to a D-E mix of 75–25, the maximumacceptable cost of equity capital is closest to:(a) 7 ... 2%(c) 9 ... 9%10 ... financed a new product as follows:$5million in stock sales at 13 ... 9% per year, and$3 million through convertible bonds at 7 ... The company’s WACC is closest to:(a) 9% per year(b) 10% per year(c) 11% per year(d) 12% per year10 ... The weight to assign to attribute1 is:(a) 0 ... 20(c) 0 ... 2510 ... 2% and the corporate effective tax rateis39%, the approximated before-tax rate of return isclosest to:(a) 6 ... 4%(c) 18 ... 7%10 ... Equity capitalhas cost 11%; however, debt capital that historically cost 9% has now increased by 20% per year ... 58 For eight attributes rank-ordered in terms of increasing importance, the weighting of the sixthattribute is closest to:(a) 0 ... 14(c) 0 ... 03CASE STUDYWHICH IS BETTER—DEBT OR EQUITY FINANCING?BackgroundInformationPizza Hut Corporation has decided to enter the catering business in three states within its Southeastern U ... Division,using the name Pizza Hut At-Your-Place ... 5 million... A feasibility study completed last year indicated that theAt-Your-Place business venture could realize an estimatedannual net cash flow of $300,000 before taxes in the threestates ... An engineer with Pizza Hut’s Distribution Division hasworked with the corporate finance office to determine how tobestdevelop the $1 ... There are two viable financing plans ... (A simplifying assumption that $75,000 of the principal is repaid with eachannual payment can be made ... The financial manager informed theengineer that stock is paying $0 ... This dividend pattern is expected to continue,based on the currentfinancial environment ... What values of MARR should the engineer use to determine the better financing plan?Case Study2 ... He does not knowhow to consider all the tax angles for the debt financingin plan A ... Is A or Bthe better plan?3 ... Plot the WACC curve and compareits shape with that of Figure10–2 ... SECTIONTOPICLEARNING OUTCOME11 ... 11 ... 11 ... 11 ... 11 ... 11 ... One of the most common and important issues in industrial practice is that of replacement or retention of an asset, process, or system that is currently installed ... The fundamental question answered by a replacementstudy (also called a replacement͞retentionstudy) about a currently installed system is, Should it be replaced now or later? When anasset is currently in use and its function is needed in the future, it will be replaced at sometime ... A replacement study is usually designed to first make the economic decisionto retain orreplace now ... If the decision is to retain,the cost estimates and decision can be revisited each year to ensure that the decision toretain is still economically correct ... A replacement study is an application of the AW method of comparing unequal-life alternatives, first introduced in Chapter 6 ... Ifa study period is specified, the replacement study procedure is different from that usedwhen no study period is set ... After-taxreplacement analysis is included in Chapter 17 ... The products are sold to a wide range ofindustries from the nuclear and solar powerindustry to sports equipmentmanufacturersof specialty golf and tennis gear, where kilntemperatures up to approximately 1700°Care needed ... Two are in usecurrently at plant locations on each coastof the country; one kiln is 10 years old, andthe second was purchased only 2 years agoand serves, primarily, the ceramicsindustryneeds on the west coast ... During the last two or three quarterlymaintenance visits, the Harper teamleader and the head of B&T quality havediscussed the ceramic and metal industryneeds for higher temperatures ... Thesemay find use in hypersonic vehicles,engines, plasma arc electrodes,cuttingtools, and high-temperature shielding ... This unit will have loweroperating costs and significantly greaterfurnace efficiency in heat time, transit,and other crucial parameters ... For identification, letPT identify the currently installedpusher-plate tunnel kiln (defender)GH identify the proposed newgraphite hearth kiln (challenger)Relevant estimates follow in $ millionsfor monetary units ... 2; starts at $3 ... 4increases10%͞yearLife, years6 (remaining)Heatingelement,$M—$38; with notrade-in12 (estimated)$2 ... 2)Replacement study (Section 11 ... 5)Replacement value (Section 11 ... 18 and 11 ... 1Basics of a Replacement StudyThe need for a replacement study can develop from several sources:Reduced performance ... This usuallyresults in increased costs of operation, higher scrap and rework costs, lost sales, reduced quality, diminished safety, and larger maintenance expenses ... Newrequirements of accuracy, speed, or other specifications cannotbe met by the existing equipment or system ... Obsolescence ... The ever-decreasing development cycle time to bring new products to market isoften the reason for premature replacement studies, that is, studies performed beforetheestimated useful or economic life is reached ... Defender and challenger are the names for two mutually exclusive alternatives ... Areplacement study compares these two alternatives ... (This is thesame terminology used earlier for incremental ROR and B͞C analysis, but both alternativeswere new) ...Also called trade-in value, this estimate is obtained from professional appraisers, resellers, or liquidators familiar with the industry ... In replacement analysis,the salvage value at the end of one year is used as the market value at the beginning of thenext year ... The term equivalent uniform annual cost(EUAC) may be used in lieu ofAW, because often only costs are included in the evaluation; revenues generated by thedefender or challenger are assumed to be equal ... ) Therefore, all values will be negative when onlycosts are involved ... Economic service lifeEconomic service life (ESL) for analternative is the number of years at which the lowest AWof cost occurs ... The next sectionexplains how to find the ESL ... The currentmarket value (MV) is the correct estimate to use for P for the defender in a replacementstudy ... It isincorrect to use the following as MV for the defender first cost: trade-invalue that doesnot represent a fair market value, or the depreciated book value taken from accountingrecords ... ), this cost is added to the MV to obtain the estimateddefender first cost ... This amount is almost always equal to P, the first costof the challenger ... 1295Basics of a Replacement StudyIf anunrealistically high trade-in value is offered for the defender compared to its fairmarket value, the net cash flow required for the challenger is reduced, and this fact should beconsidered in the analysis ... In equation form, this is P Ϫ (TIV Ϫ MV) ... e ... e ... Of course, when the trade-in and market values arethe same, the challenger Pvalue is used in all computations ... Sometimes, an analyst or manager will attempt to increase this first cost by an amount equal to theunrecovered capital remaining in the defender, as shown on the accounting records for the asset ... This leads us to identify two additionalcharacteristics ofreplacement analysis, in fact, of any economic analysis: sunk costs and nonowner’s viewpoint ... The replacement alternative for an asset, system, or process that hasincurred a nonrecoverable cost should not include this cost in any direct fashion; sunk costsshould be handled in arealistic way using tax laws and write-off allowances ... For example, assume an asset costing $100,000two years ago has a depreciated value of $80,000 on the corporate books ... If the replacement alternative (challenger) hasa first cost of $150,000, the $80,000 from the current asset is a sunk costwere the challengerpurchased ... The second characteristic is the perspective taken when conducting a replacement study ... The nonowner’s viewpoint, also called the outsider’s viewpoint or consultant’s viewpoint, provides the greatest objectivity in a replacement study ... Additionally, it assumes theservices provided by the defender can be purchased now by making an “initialinvestment” equal to the market value of the defender ... As mentioned in the introduction, a replacement study is an application of the annual worthmethod ... If the planning horizon is unlimited, that is, a study period is notspecified, the assumptions are as follows:1 ... 2 ... When this challenger replaces the defender (now or later), it will be repeated for succeedinglife cycles ... Cost estimates for every life cycle of the defender and challenger will be the same as in theirfirst cycle ... We discussed this previously for theAWmethod (and the PW method) ... The replacement procedure discussed in Section 11 ... When theplanning horizon is limited to a specified study period, the assumptions above do not hold ... 5 discusses replacement analysis over a fixed study period ... 1Only 2 years ago, Techtron purchased for$275,000 a fully loaded SCADA (supervisory control and data acquisition) system including hardware and software for a processing plant operating on the Houston ship channel ... Actual M&O costs have been $25,000 per year, and the book valueis $187,000 ... Given thesefactors, the system is likelyworth nothing if kept in use for the final 3 years of its anticipateduseful life ... A 5-year life, salvagevalue of 15% of stated first cost or $60,000, and an M&O cost of $15,000 per year are goodestimates for the new system ... Using the above values as the best possible today, state the correct defender andchallengerestimates for P, M&O, S, and n in a replacement study to be performed today ... Allothers—original cost of $275,000, book value of $187,000, and trade-in value of$100,000—are irrelevant to a replacement study conducted today ... First costP ϭ $Ϫ400,000M&O costA ϭ $Ϫ15,000 peryearExpected lifen ϭ 5 yearsSalvage valueS ϭ $60,000CommentIf the replacement study is conducted next week when estimates will have changed, the defender’s first cost will be $80,000, the new market value according to the appraiser ... 11 ... In reality, the best lifeestimate to use in the economicanalysis is not known initially ... The best life estimate is called the economic service life ... 11 ... ToWtal AAWof co stsOCof AC a p it a l re c o v ery0YearsEconomicservice lifeThe ESL is also referred to as the economic life or minimum cost life ... When n years have passed, the ESL indicates that theasset should be replacedto minimize overall costs ... The ESL is determined by calculating the total AW of costs if the asset is in service 1 year,2 years, 3 years, and so on, up to the last year the asset is considered useful ... 1]The ESL is the n value for the smallest total AW of costs ... Therefore, $–200 isa lower cost than $Ϫ500 ... The CR component of totalAW decreases, while the AOC component increases, thus forming the concave shape ... Decreasing cost of capital recovery ... Capital recovery is calculated by Equation [6 ... The salvage value S, which usually decreases with time, is theestimatedmarket value (MV) in that year ... 2]Increasing cost of AW of AOC ... To calculate the AW of the AOC series for 1, 2, 3, ... The complete equation for total AW of costs over k years (k ϭ 1, 2, 3, ... 3]j�1P ϭ initial investment or current market valueSk ϭ salvage value or market value after kyearsAOCj ϭ annual operating cost for year j ( j ϭ 1 to k)Capital recovery298Chapter 11Replacement and Retention DecisionsThe current MV is used for P when the asset is the defender, and the estimated future MV valuesare substituted for the S values in years 1, 2, 3, ... To determine ESL byspreadsheet, the PMT function (with embedded NPV functions asneeded) is used repeatedly for each year to calculate capital recovery and the AW of AOC ... The PMT function formats for the capital recoveryand AOC components for each year k (k ϭ 1, 2, 3, ...When the spreadsheet is developed, it is recommended that the PMT functions in year 1 be developed using cell-reference format; then drag down the function through each column ... Augmenting the table with an Excel xyscatter chart(0 (؉ PMT(i%,years,NPV(i%,year_1_AOC:year_k_AOC [؊ 4graphically displays the cost curves in the general form of Figure 11–1, and the ESLis easily identified ... 2 illustrates ESL determination by hand and by spreadsheet ... 2A 3-year-old backup power system is being considered for early replacement ... Estimated future market values and annual operatingcosts for the next5 years are given in Table 11–1, columns 2 and 3 ... Solution by HandEquation [11 ... , 5 ... Column 5gives the equivalent AW of AOC for k years ... 3] isTotal AW3 ϭ ϪP(A͞P,i,3) ϩ MV3(A͞F,i,3) Ϫ [PW of AOC1,AOC2, and AOC3](A͞P,i,3)ϭ Ϫ20,000(A͞P,10%,3) ϩ 6000(A͞F,10%,3) Ϫ[5000(P͞F,10%,1)ϩ 6500(P͞F,10%,2) ϩ 8000(P͞F,10%,3)](A͞P,10%,3)ϭ Ϫ6230 Ϫ 6405 ϭ $Ϫ12,635A similar computation is performed for each year 1 through 5 ... Therefore, the defender ESL is n ϭ 3 years,and the AW value is $Ϫ12,635 ... Solution by SpreadsheetSee Figure 11–2 for the spreadsheet screenshot and chart that shows the ESL is n ϭ 3 yearsand AW ϭ $Ϫ12,634 ... 2Economic Service LifeCurrent market valueϭ PMT($B$1,$A9,$B$2,Ϫ$B9)ϭ ϪPMT($B$1,$A9,NPV($B$1,$C$5:$C9)ϩ0)ESL of defenderTotal AW is minimumat n ϭ 3 yearsTotal AWcurveCapitalrecoverycurveFigure 11–2Determination of ESL and plot of curves, Example 11 ... ) Contents of columns D and E are described below ... 4] ... The $ symbols are included for absolute cell referencing, neededwhen the entry is dragged down through the column ... For example, in actual numbers, thecell-reference PMT function inyear 5 shown on the spreadsheet reads ϭ PMT (10%,5,20000,Ϫ0), resulting in $Ϫ5276 ... Column E: The NPV function embedded in the PMT function obtains the present worth inyear 0 of all AOC estimates through year k ... For example, in year 5, the PMT in numbers is ϭϪPMT(10%,5,NPV(10%,C5:C9)ϩ0) ... The graph plots the AW ofAOC curve, which constantly increases in cost because the AOC estimates increaseeach year ... If the same MV were estimated for eachyear, the curve would appear like Figure 11–1 ... This indicates that the ESL is relatively insensitive tocosts ... Previously we had a specific life estimated to be n yearswith associated other estimates: first cost in year 0, possibly a salvage value in year n, and anAOC that remained constant or varied each year ... This is the economic service life when nis fixed ... Therefore,we can conclude thefollowing:When the expected life n is known and specified for the challenger or defender, no ESL computations are necessary ... This AW value is the correct oneto use in the replacement study ... First the market/salvage series is needed ... For example, an asset with afirst cost of P can lose market valueof, say, 20% per year, so the market value series for years0, 1, 2, ... 8P, 0 ... , respectively ... Marginal costs (MC) are year-by-year estimates of the costs to own and operate an asset for thatyear ... Thesum of the AW values of the first two of these components is the capital recovery amount ... 3] ... AW of� total AW of costs[11 ... The ESL analysis presented in Example 11 ... This is demonstrated in Example 11 ... EXAMPLE 11 ... When the current tunnel kiln was purchased 2 years ago for $25 million, an ESL study indicated that the minimum cost life was between 3 and 5 years of the marginal costsexpected 8-year life ... Now, the same type of question arises for the proposed graphite hearthmodel that costs $38 million new: What are the ESL and the estimated total AW of costs? TheManager of Critical Equipment at B&T estimates that the market value after only 1 year willdrop to $25 million andthen retain 75% of the previous year’s value over the 12-year expectedlife ... SolutionFigure 11–3 is a spreadsheet screen shot of the two analyses in $ million units ... A brief description of each analysis follows ... 2Economic Service LifeESL analysisTwo AW seriesare identicalMarginal costanalysisFigure 11–3Comparison of annual worth series resulting from ESL analysis and marginal cost analysis, Example 11 ... ESL analysis: Equation [11 ... , 12 years (columns C,D, and E) in the top of Figure 11–3 ... The result in column F is the total AW series that is of interest now ... Row 33 detailsthe functionsfor year 12 ... The two AW series are identical, thus demonstrating that Equation [11 ... Therefore, either an ESL or a marginal cost analysis will provide the same information for a replacement study ... 32 million at its full 12-year life ... These conclusions are based on the extent to whichdetailed annual estimates aremade for the market value ... Year-by-year market value estimates are made ... These are the best n and AWvalues for the replacement study ... Yearly market value estimates are not available ... Use it to calculate the AW over n years ... 301302Chapter 11Replacement andRetention DecisionsUpon completion of the ESL analysis (item 1 above), the replacement study procedure in Section 11 ... 3 Performing a Replacement StudyReplacement studies are performed in one of two ways: without a study period specified or withone defined ... The procedure discussed in thissection applies when no study period (planning horizon) is specified ... 5 is applied ... The complete study is finished if the challenger (C) is selected to replace the defender (D) now ... Use the annual worth and lifevalues for C and D determined in the ESL analysis in the following procedure ... Thereplacement study procedure is:New replacement study:1 ... When thechallenger is selected, replace the defender now, and expect to keep the challenger for nCyears ... If the defender is selected, plan to retain it for upto nD more years ... ) Next year, perform thefollowing steps ... Determine if allestimates are still current for both alternatives, especially first cost, marketvalue, and AOC ... If yes and this is year nD, replace the defender ... This stepmay be repeated several times ... Whenever the estimates have changed, update them and determine new AWC and AWD values ... Figure 11–4Replacement studyOverview of replacementstudy approaches ... 3Performing a Replacement StudyIf the defender is selected initially (step 1), estimates may need updating after 1 year of retention (step 2) ... Either significantchanges in defender estimates or availability of a new challenger indicates thata new replacement study is to be performed ... Example 11 ... The planning horizon is unspecified in thisexample ... 4Two years ago, Toshiba Electronics made a $15 million investment in new assembly linemachinery ... The equipment sorts, tests, and performs insertion-order kitting onelectroniccomponents in preparation for special-purpose circuit boards ... Due to the new standards, coupled with rapidly changing technology, anew system is challenging the retention of these 2-year-old machines ... The i is 10% and theestimates are below ... (b) Perform the replacement study now ... Thechallenger is making largeinroads to the market for electronic components assembly equipment, especially with thenew international standards features built in ... Also, this prematurely outdatedequipment is more costly to keep serviced, so the estimated AOC next year has been increased from $8000 to$12,000 and to $16,000 two years out ... Solution(a) The results of the ESL analysis, shown in Figure 11–5, include all the MV and AOC estimates in columns B and C ... The total AW of costs is for each year, should the challenger be placed into servicefor that number of years ... 3], where the A͞G factoraccommodates the arithmetic gradientseries in the AOC ... 4 ... 4] ... The lowest AW cost (numerically largest) values for the replacement study are as follows:Challenger:Defender:AWC ϭ $Ϫ19,123AWD ϭ $Ϫ17,307for nC ϭ 4 yearsfor nD ϭ 3 yearsThe challenger total AW of cost curve (Figure 11–5) isclassically shaped and relatively flatbetween years 3 and 6; there is virtually no difference in the total AW for years 4 and 5 ... (b) To perform the replacement study now, apply only the first step of the procedure ... Prepare to perform the one-year-later analysis 1 year from now ... Apply the steps for theone-year-later analysis:2 ... Go to step 3 toperform a new ESL analysis for the defender ... The defender estimates in Figure 11–5 are updated below for the ESL analysis ... 3] ... Year kMarketValue, $AOC, $Total AWIf Retained k More Years, $01212,0002,0000—Ϫ12,000Ϫ16,000—Ϫ23,200Ϫ20,81911 ...Therefore, replace the defendernow, not 2 years from now ... PEEXAMPLE 11 ... Amarketing study revealed that the improving business activity on the west coast implies thatthe revenue profile between the installed kiln (PT) and the proposed new one (GH) would bethe same, with the new kiln possibly� ؊$ bringing in new revenue within the next couple of years ... Assume you are the leadengineer and that you previously completed the ESL analysis on the challenger (Example 11 ... It indicates that for the GH system the ESL is its expected useful life ... 21millionThe president asked you to completethe replacement study, stipulating that, due to the rapidlyrising annual operating costs (AOC), the defender would be retained a maximum of 6 years ... SolutionAfter some data collection, you have good evidence that the market value for the PT systemwill stay high, but that the increasing AOC isexpected to continue rising about $1 ... The best estimates for the next 6 years in $ million units are these:Year123456Market Value, $ MAOC, $ M per year22 ... 222 ... 422 ... 620 ... 818 ... 018 ... 2You developed a spreadsheet and performed the analysis in Figure 11–6 ... 0(A͞P,15%,3) ϩ 22 ...Since AWPT ϭ$Ϫ8 ... 32 million, you should recommend keeping the current kiln only 1 more year and doing another study during the yearto ... 5 ... 05 � 1 year with total equivalent annual cost AWPT 2(P͞F,15%,1) ϩ 6 ... 6(P͞F,15%,3)](A͞P,15%,3)ϭ –9 ... 34 Ϫ[14 ... 43798)ϭ $Ϫ9 ... Defender: ESL nPTdetermine if the current estimates are still reliable ... A comparison of Figure 11–3 (top), column F, and Figure 11–6, column F, shows us that the largesttotal AW of the current system ($–11 ... 32 M for 12 years) ... It would take some significant estimatechanges to justify the challenger ... 4 AdditionalConsiderations in aReplacement StudyThere are several additional aspects of a replacement study that may be introduced ... • Future-year replacement decisions at the time of the initial replacement study• Opportunity cost versus cash flow approaches to alternative comparison• Anticipation of improvedfuture challengersIn most cases when management initiates a replacement study, the question is best framed as,“Replace now, 1 year from now, 2 years from now, etc ... In other words,at the time it is performed, step 1 of the procedure does answer the replacement question formultiple years ... The firstcosts (P values) for the challenger and defender have been correctly taken as theinitial investment for the challenger C and current market value for the defender D ... This approach, also called the conventionalapproach, is correct for every replacement study ... Use ofthe cash flow approach is stronglydiscouraged for at least two reasons: possible violation of theequal-service requirement and incorrect capital recovery value for C ... Therefore, the cash flow approach can work only when challenger and defender lives are exactly equal ... If this equalservice comparison reason is not enough to avoid thecash flow approach, consider what happensto the challenger’s capital recovery amount when its first cost is decreased by the market value ofthe defender ... 3] will decrease, resulting in afalsely low value of CR for the challenger, were it selected ... The conclusion is simple:Use the initial investment of Cand the market value of D as the first costs in the ESL analysis andin the replacement study ... The expectation of ever-improving challengers can offer strong encouragement to retain the defender untilsome situational elements—technology, costs, market fluctuations, contract negotiations, etc ...5307Replacement Study over a Specified Study Periodstabilize ... A large expenditure on equipment whenthe standards changed soon after purchase forced an early replacement consideration and a large lossof invested capital ... It isimportant to understand trends, new advances, and competitivepressures that can complement theeconomic outcome of a good replacement study ... Adding needed features to a currently installed defender may prolong its useful life and productivity until challenger choices are more appealing ... If taxes should be considered, proceed now, or after the next section, toChapter 17and the after-tax replacement analysis in Section 17 ... 11 ... The AW values for the challenger and for the remaining life of the defender are not based on theeconomic service life; the AW is calculated over the study period only ... This means that the defender or challenger is not neededbeyond the study period ... 1—service needed for indefinite future, best challenger available now, and estimates will be identical for future life cycles ... This is especially important forthe defender ... When the defender’s remaining life is shorter than the study period, the cost of providingthe defender’sservices from the end of its expected remaining life to the end of the studyperiod must be estimated as accurately as possible and included in the replacement study ... 1 ... Develop all the viable ways to use the defender and challenger during the study period ... The AW values for the challenger

anddefender cash flows are used to build the equivalent cash flow values for each option ... Selection of the best option ... Select the option with the lowest cost, or highest income if revenues are estimated ... )The following examples use this procedure and illustrate the importance of making costestimates for the defender alternative when its remaining life is less than the study period ... 6Claudia works with Lockheed-Martin (LMCO) in the aircraft maintenance division ... S ... A key piece of equipment for maintenanceoperations is an avionics circuit diagnostics system ... It has no capital recoverycosts remaining, and the following arereliable estimates: current market value ϭ $70,000, remaining life of 3 more years, no salvagevalue, and AOC ϭ $30,000 per year ... Claudia has found that there is only one good challenger system ... Study period308Chapter 11Replacement and RetentionDecisionsRealizing the importance of accurate defender alternative cost estimates, Claudia asked thedivision chief what system would be a logical follow-on to the current one 3 years hence, ifLMCO wins the contract ... The company would keep itfor the entire 10 additional years for use on an extensionof this contract or some other application that could recover the remaining 3 years of invested capital ... Claudia’s estimate of the first cost of this same system 3 years from now is$900,000 ... The division chief mentioned any study had to be conducted using the interest rate of 10%,as mandated by the U... Office of Management and Budget (OMB) ... SolutionThe study period is fixed at 10 years, so the intent of the replacement study assumptions is notpresent ... Further, any analyses to determine the ESL values are unnecessary since alternative lives arealready set and no projected annual marketvalues are available ... Since the defender will be replaced now or in3 years, there are only two options:1 ... 2 ... Cash flows are diagrammed in Figure 11–7 ... Equation [11 ... 6 ... 5Replacement Study over a Specified Study PeriodThe second option has more complex cost estimates ... Added to this isthe capital recovery for the defender follow-on for thenext 7 years ... (It is notunusual for the recovery of invested capital to be moved between projects, especially for contractwork ... The final cash flows are shown in Figure 11–7b ... The CR and AW for all 10 years areCRDF ϭ Ϫ900,000(A͞P,10%,10) ϭ$Ϫ146,475AWDF ϭ (Ϫ146,475 Ϫ 50,000)(F͞A,10%,7)(A͞F,10%,10) ϭ $Ϫ116,966[11 ... This is the AWfor option 2 ... Retain the defender now and expectto purchase the follow-on system 3 years hence ... If this assumption were not made, its capital recovery cost wouldbe calculated over 7 years, not 10, inEquation [11 ... This raises theannual worth to AWD ϭ $Ϫ163,357 ... EXAMPLE 11 ... Because of flight increases, new fire-fighting capacity is needed once again ... Estimates are presented below ... Presently OwnedFirst cost P, $AOC, $Market value, $Salvage value, $Life, yearsNew PurchaseDoubleCapacityϪ151,000 (3 years ago)Ϫ1,50070,00010% of P12Ϫ175,000Ϫ1,500—12% of P12Ϫ190,000Ϫ2,500—10% of P12SolutionIdentify option 1 as retention of the presently owned truck and augmentation with a new samecapacity vehicle ... Option 1Option 2Presently OwnedP, $AOC, $S, $n,yearsAugmentationDouble CapacityϪ70,000Ϫ1,50015,1009Ϫ175,000Ϫ1,50021,00012Ϫ190,000Ϫ2,50019,00012309310Chapter 11Replacement and Retention Decisions(a) For a full-life 12-year study period of option 1,AW1 ϭ (AW of presently owned) ϩ (AW of augmentation)ϭ [Ϫ70,000(A͞P,12%,9) ϩ15,100(A͞F,12%,9) Ϫ 1500]ϩ [Ϫ175,000(A͞P,12%,12) ϩ 21,000(A͞F,12%,12) Ϫ 1500]ϭ Ϫ13,616 Ϫ 28,882ϭ $Ϫ42,498This computation assumes the equivalent services provided by the current fire truck can bepurchased at $−13,616 per year for years 10 through 12 ... (b) The analysis for an abbreviated 9-year study period is identical, except that n ϭ 9 in eachfactor; that is, 3 fewer years are allowed for the augmentation and double-capacity trucksto recover the capital investment plus a 12% per year return ... AW1 ϭ $Ϫ46,539AW2 ϭ $Ϫ36,873Option 2 is again selected ... It involves the capital recoveryamount for the challenger, when the strict definition of a study period is applied ... Highly abbreviated study periods tend to disadvantage the challengerbecause no consideration of time beyond the end of the study period is made in calculating thechallenger’s capital recovery amount ... For example, if thestudy period is 5 years andthe defender will remain in service 1 year, or 2 years, or 3 years, cost estimates must be made todetermine AW values for each defender retention period ... OptionDefenderRetained, YearsChallengerServes, YearsWXYZ32102345The respective AW values for defenderretention and challenger use define the cash flows foreach option ... 8 illustrates the procedure using the progressive example ... 8 Keep or Replace the Kiln CasePEWe have progressed to the point that the replacement study between the defender PT andchallenger GH was completed (Example 11 ...The defender was the clear choice with amuch smaller AW value ($Ϫ8 ... 32 M) ... They know the current tunnel kiln is much cheaperthan the new graphite hearth, but the prospect of future new business should not be dismissed ... 5Replacement Study over a Specified Study PeriodTA BLE 11–2Replacement Study Options and Total AW Values, Example 11 ... 50Ϫ9 ... 59Ϫ10 ... 16Ϫ11 ... 21Ϫ15 ... 31Ϫ18 ... 10—The president asked, “Is it possible to determine when it is economically the cheapest topurchase the new kiln, provided the current one is kept at least 1 year, but no more than 6years,its remaining expected life?” The chief financial officer answered, yes, of course ... (b) Discuss the next step in the analysis based onthe conclusion reached here ... We know the MARR is 15% per year, the study period has been established at 6 years, and the defender PT will stay in place between 1and 6 years ... The total AW valueswere determined for the defender in Example 11 ... 3 (Figure 11–3) ... Use theprocedure for a replacement study with a fixed study period ... There are six options in this case; the defenderis retained from 1 to 6 years while the challenger is installed from 0 to 5 years ...Figure 11–8 presents the options and the AW series for each optionfrom Table 11–2 ... Step 2: Selection of the best option ... The conclusion is clearly to keep thedefender in place for 6 more years ... If the analysis is to be carried further, the possibility of increased revenue based onservices of thechallenger’s high-temperature and operating efficiency should be considered next ... A revenue increase for the challenger will reduce its AW of costs and possibly make it moreeconomically viable ... 8 ... 6 Replacement ValueOften it is helpful to know the minimum market value of the defender necessaryto make thechallenger economically attractive ... This is a breakeven value between AWC and AWD; it is referred to as the replacement value(RV) ... The AWC is known, so RV can be determined ... Determination of the RV for a defender is an excellent opportunity to utilize the Goal Seek toolin Excel ...Example 11 ... EXAMPLE 11 ... This is basedon the ESL analysis that concluded the following (Examples 11 ... 5):Defender:Challenger:ESL nPT ϭ 1 year with AWPT ϭ $Ϫ8 ... 32 millionThe original defender price was $25 million, and a current market value of $22 million wasestimated earlier (Figure 11–6) ... What will you discover RV to be? The MARR is 15% per year ... 32and solve for RV ... 20MV ϭ $22 ... 32 ϭ ϪRV(A͞P,15%,1) ϩ 22 ... 201 ... 32 ϩ 22 ... 20RV ϭ $25 ... CommentTo find RV using a spreadsheet, return to Figure 11–6 ... 50), and the required value is $Ϫ12 ... The “changing” cell is thecurrent market value (cell F2), currently $22 ... When “OK” is touched, $25 ... This is the RV ... Best(economic) challenger is described as the one with the lowest annual worth (AW) of costs forsome number of years ... However, if reasonable estimates of the expected market value (MV) andAOCProblems313for each year of ownership can be made, these year-by-year (marginal) costs help determine thebest challenger ... The resulting nC and AWC values are usedin the replacement study ... Replacement studies in which no study period (planning horizon) is specified utilize the annual worthmethod of comparing two unequal-life alternatives ... When a study period is specified for the replacement study, it is vital that the market value andcost estimates for the defender be as accurate as possible ... All the viable options for using the defender and challenger are enumerated, and theirAWequivalent cash flows are determined ... This option determines how long the defender is retained before replacement ... 1 In a replacement study, what is meant by “takingthe nonowner’s viewpoint”?11 ... The company thought the asset would last5 years and that its book value would decreaseby$20,000 each year and, therefore, be worthless atthe end of year 5 ... If the replacement ispurchased immediately at a first cost of $75,000and if it will have a lower annual worth, what is theamount of the sunk cost? Assume the company’sMARR is 15% per year ... 3 As a muscle car aficionado, a friendof yours likesto restore cars of the 60s and 70s and sell them fora profit ... Another opportunity hascome up (a 1969 Dodge Charger) that he is thinking of buying because he believes he could sell itfor a profit of $60,000 after it is completely restored ... He thought that thecompletely restored Shelby wouldbe worth$195,000, resulting in a tidy profit of $22,000, butin its half-restored condition, the most he could getnow is $115,000 ... (a) What is wrong with this thinking?(b) What is his sunk cost in the Shelby?11 ... 11 ... At that time, it was expected to be used for 10 years and then traded infor its salvagevalue of $10,000 ... The company estimates that the old cranecan be used, if necessary, for another 3 years, atwhich time it would have a $23,000 estimatedmarket value ... Determine the values ofP, n, S, and AOC that should be used for the existing crane in a replacement analysis ... 6 Equipment thatwas purchased by Newport Corporation for making pneumatic vibration isolatorscost $90,000 two years ago ... Experience with this type of equipment has shownthat the operating cost for the first 4 years is$65,000 per year, after which it increases by $6300per year ... Determine the values of P, S, andAOCif a replacement study is done (a) now and(b) 1 year from now ... 7 A piece of equipment that was purchased 2 yearsago by Toshiba Imaging for $50,000 was expectedto have a useful life of 5 years with a $5000 salvage314Chapter 11Replacement and Retention Decisionsvalue ... Increaseddemand now requires that the equipmentbe upgraded again for another $17,000 so that itcan be used for 3 more years ... Alternatively,it can be replaced with new equipment priced at$65,000 with operating costs of $14,000 per yearand a salvage value of $23,000 after 6 years ... Determine the values ofP, S, AOC, and nfor the defender in a replacement study ... 8 For equipment that has a first cost of $10,000 andthe estimated operating costs and year-end salvagevalues shown below, determine the economic service life at i ϭ 10% per year ... 9 To improve package tracking at a UPS transfer facility,conveyor equipment was upgraded withRFID sensors at a cost of $345,000 ... Thesalvage value of the equipment is expected to be$140,000 for the first 3 years, but due to obsolescence, it won’t have a significant value after that ... 10 Economic service life calculations for an asset areshown below ...YearsRetainedAW of FirstCost, $AW of OperatingCost, $ per YearAW ofSalvage Value, $12345Ϫ51,700Ϫ27,091Ϫ18,899Ϫ14,827Ϫ12,398Ϫ15,000Ϫ17,000Ϫ19,000Ϫ21,000Ϫ23,00035,00013,8106,6484,3092,45711 ... Maintenance can bedone at 1-, 2-, 3-, or 4-year intervals, but the longerthe intervalbetween servicing, the higher the cost ... What interval shouldbe scheduled for maintenance to minimize theoverall equivalent annual cost? The interest rate is8% per year ... 12 A construction company bought a 180,000 metricton earth sifter at a cost of $65,000 ... The operating cost is expected to followtheseries described by 40,000 ϩ 10,000k, where k isthe number of years since it was purchased (k ϭ 1,2, ... The salvage value is estimated to be$30,000 for years 1 and 2 and $20,000 for years 3through 7 ... 11 ... [Note that the numbers are annual worth values associated withvarious years of retention;that is, if the equipment is kept for, say, 3 years, the AW (years 1through 3) of the first cost is $32,169, the AW ofthe operating cost is $51,000, and the AW of thesalvage value is $6042 ... From the information available, determine the following:(a) The interest rate used in the ESL calculations ... Use theinterest rate determinedin part (a) ... 14 A large, standby electricity generator in a hospitaloperating room has a first cost of $70,000 and maybe used for a maximum of 6 years ... 15)n,where n is the number of years after purchase ... At an interest rate of 12% peryear, what are the economic service lifeand theassociated AW value?315Problems11 ... The salvage value cannot go below zero ... The interest rate is 15% per year ... 11 ... Use an interest rate of14% per year and hand solution ... 17 Use the annual marginal costs to find the economicservice life for Problem 11 ... Assume the salvage valuesare the best estimates of future market value ... 11 ... 3, the market value (salvage value)series of the proposed $38 million replacement kiln(GH) dropped to $25 million in only 1 year and thenretained 75% of the previous year’s market valuethrough the remainder of its 12-year expected life ... Additionally,the heating element replacement in year 6 will probably cost $4 million, not$2 million ... Starting inyear 5, the AOC is expected to increase by 25% peryear, not 10% as predicted earlier ... 3) ... (b) In percentage changes, estimate how muchthese new cost estimates may affect theminimum-cost life andAW of cost estimate ... 19 In a one-year-later analysis, what action should betaken if (a) all estimates are still current and theyear is nD, (b) all estimates are still current and theyear is not nD, and (c) the estimates have changed?11 ... If Retained ThisNumber of YearsAW Value,$ perYear12345Ϫ62,000Ϫ51,000Ϫ49,000Ϫ53,000Ϫ70,000A challenger has ESL ϭ 2 years and AWC ϭ$Ϫ48,000 per year ... (b) When should the next replacement evaluationtake place?11 ... The presently owned ones were purchased 4years ago for $250,000 ... Alternatively, new controlled-environmentroomscould be purchased at a cost of $270,000 ... Determine whether thecompany should upgrade or replace ... 11 ... Increased demand necessitated an upgrade costing$30,000 one year ago ... Its annual operating cost will be $47,000, andit will have a $22,000 salvage after 3 years ... If replaced now,the existing equipment willbe sold for $9000 ... 316Chapter 11Replacement and Retention Decisions11 ... A new car will cost $26,000 and haveannual operation and maintenance costs of $1200 peryear with an $8000 salvage value in 5 years (which isits estimated economic service life) ... Its operatingcost is expected to be $1900this year, with costs increasing by $200 per year ... Assuming used cars likethe one presently owned will always be available,should the presently owned car be sold now, 1 yearfrom now, 2 years from now, or 3 years from now?Use annual worth calculations at i ϭ 10% peryearand show your work ... 24 A pulp and paper company is evaluating whether itshould retain the current bleaching process thatuses chlorine dioxide or replace it with a proprietary “oxypure” process ... Use an interest rate of15% per year to perform the replacement study ... 27 The data associated withoperating and maintaining an asset are shown below ... e ... At an interest rate of 10%per year, estimate the AW of keeping the machinefrom year 1 to year 2 ... 26 A crushing machine that is a basic component of ametal recycling operation is wearing out faster thanexpected ... At that time, the buyerthought themachine would serve its needs for at least 5 years,at which time the machine would be sold to asmaller independent recycler for $80,000 ... If it iskept, the machine’s operating cost will be $37,000per year for the next 2 years, after which it will bescrapped for $1000 ... Alternatively, thecompany can outsource the process nowfor a fixed cost of $56,000 per year ... 25 A machine that is critical to the Phelps-Dodge copper refining operation was purchased 7 years agofor $160,000 ... The situation has changed ... If kept inservice, it can be minimally upgraded at a cost of$43,000, which willmake it usable for up to2 more years ... Alternatively, the company can purchase a newsystem that will have an equivalent annual worthof $Ϫ47,063 per year over its ESL ... Calculate the relevant annual worth values, and determine when thecompany should replace the machine ... 28 A machine that cost$120,000 three years ago canbe sold now for $54,000 ... Itsoperating cost was $18,000 for the first 3 years ofits life, but the M&O cost is expected to be$23,000 for the next 2 years ... At an interest rate of 10%per year, determine if the presently owned machine should be replaced now, 1 year from now,or2 years from now ... 29 The projected market value and M&O costs associated with a presently owned machine are shown(next page) ... If the presentlyowned machine is replaced now, the cost of thefixed-price contract will be $33,000 for each of thenext 3 years ... Determine ifand when the defendershould be replaced with theoutside vendor using an interest rate of 10% peryear ... YearMarket Value, $M&O Cost,$ per Year0123430,00025,00014,00010,0008,000—Ϫ24,000Ϫ25,000Ϫ26,000—11 ... It is appraised at a current market value of only $50,000 ... The challenger,which can be purchased for$300,000, has an expected life of 10 years and a $50,000 salvagevalue ... Assume the AOC estimates are the same for bothalternatives ... 31 For the estimates in Problem 11 ... Is this a maximum or minimum for the upgrade, if the current system is to be retained?11 ... It is expected to have the marketvalues andannual operating costs shown below for its remaining useful life of up to 3 years ... Richter should retain the present plotter ... 11 ... 34 State what is meant by the cash flow approach toreplacement analysis, and list two reasons why it isnot a good idea to use this method ... 35 ABBCommunications is considering replacingequipment that had a first cost of $300,000 fiveyears ago ... Since the present equipment or the proposedequipment can be used for any or all of the 3-yearperiod, one of the company’s industrial engineersproduced AW cost information for the defenderandchallenger as shown below ... Determinewhen the defender should be replaced to minimizethe cost to ABB for the 3-year study period usingan interest rate of 10% per year ... 36 The table below shows present worth calculationsof the costs associated with using a presentlyowned machine (defender) anda possible replacement (challenger) for different numbers of years ... Show solutions (a) by hand and(b) by spreadsheet ... At an interest rateof 15% per year, determine how many more years11 ... Accordingly, a seniorengineer has recommended that a 2-year-old piece318Chapter 11Replacement andRetention Decisionsof precision measurement equipment be replacedimmediately ... (a) Perform the replacement analysis using theannual worth method for a specified 3-yearstudy period ... Comment on the effectmade by the 3-year study period ... 38 An industrial engineer at a fiber-optic manufacturingcompany is considering two robots to reducecosts in a production line ... Robot Y will have a first costof $97,000, an annual M&O cost of $27,000, andsalvage values of $60,000, $51,000, and $42,000after 1, 2, and 3 years, respectively ... 39 A 3-year-old machine purchased for $140,000 is notable tomeet today’s market demands ... The current machine will have an annual operating cost of $85,000per year and a $30,000 salvage value in 3 years ... The replacement machine,which will serve the company now and for at least8 years, will cost $220,000 ... It will have an estimated operating cost of$65,000 per year ... (a) Should the company replace the presentlyowned machine now, or do it 3 years from now?(b) Compare the capital recovery requirements forthe replacement machine (challenger) over thestudy period and an expected life of 8 years ... 40 Two processes can be used for producing apolymer that reduces friction loss in engines ... Process L, the challenger, will have afirst cost of $230,000, an operating cost of $65,000per year, and salvage values of $100,000 after 1year, $70,000 after 2 years, $45,000 after 3 years,and $26,000 after its maximum expected 4-yearlife ... Youhave beenasked to determine which process to select when (a) a 2-year study period is used and(b) a 3-year study period is used ... 41 Keep or Replace the Kiln CaseIn Example 11 ... This is a significantly shortenedperiod compared to the expected 12-year life ofthe challenger ... 11 ... The plantmanager, who isdedicated to cutting costs but notsacrificing quality and hygiene, has the projecteddata shown in the table below if the current systemwere retained for up to its maximum expected lifeof 5 years ... 0 million per yearif Nabisco signs on for 4 to 10 years, and $5 ... Years RetainedAW, $ per YearClose-DownExpense, $012345—Ϫ2,300,000Ϫ2,300,000Ϫ3,000,000Ϫ3,000,000Ϫ3,500,000Ϫ3,000,000Ϫ2,500,000Ϫ2,000,000Ϫ1,000,000Ϫ1,000,000Ϫ500,000(a)At a MARR ϭ 8% per year, perform a replacement study for the plant manager with afixed study period of 5 years, when it is anticipated that the plant will beshut down dueto the age of the facility and projected technological obsolescence ... (Hint:Calculate AW values for all combinations ofdefender͞challenger options ... 43 In 2008, Amphenol Industrial purchased a newquality inspection system for $550,000 ... Currently the expected remaining life is 3yearswith an AOC of $27,000 per year and an estimatedsalvage value of $30,000 ... If theMARR for the corporation is 12% per year, findthe minimum trade-in value now necessary tomake the president’s replacement economicallyadvantageous ... 44 A CNC milling machine purchased by Proto ToolandDie 10 years ago for $75,000 can be used for 3more years ... A challenger will cost $130,000 with an economic life of6 years and an operating cost of $32,000 per year ... On the basis of319these estimates, what market value for the existingasset will render the challenger equally attractive?Use aninterest rate of 12% per year ... 45 Hydrochloric acid, which fumes at room temperatures, creates a very corrosive work environment,causing steel tools to rust and equipment to failprematurely ... Its operating cost is $75,000 per year ... It is expected to have a $30,000 salvagevalue after its 6-year ESL ...46 Engine oil purifier machines can effectively remove acid, pitch, particles, water, and gas fromused oil ... Its operating cost is higher than expected,so if it is not replaced now, it will likely be used foronly 3 more years ... A more efficient challenger, purifier B, will cost $150,000with a $50,000 salvagevalue after its 8-year ESL ... What is the market value for A that will make thetwo purifiers equally attractive at an interest rate of12% per year?ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS11 ... 48 A sunk cost is the difference between:(a) The first cost and the salvage value(b) Themarket value and the salvage value(c) The first cost and the market value(d) The book value and the market value11 ... The operating costsare $9000 per year, and it is expected to last 4 moreyears with a $5000 salvage value ... The value that shouldbe used as P for the presently owned vehicle inareplacement study is:(a) $45,000(b) $5000(c) $50,000(d) $24,00011 ... 51 In looking for ways to cut costs and increaseprofit (to make the company’s stock go up), oneof the company’s industrial engineers (IEs) determined that the equivalent annual worth of anexisting machine over its remaining usefullife of3 years will be $–70,000 per year ... If the engineer uses a3-year study period and an interest rate of 15%per year, she should recommend that the existingmachine be:(a) Replaced now(b) Replaced 1 year from now(c) Replaced 2 years from now(d) Not replaced11 ... It can be replaced now or laterwith amachine that will have an AW of $–90,000 peryear if it is kept for 2 years or less, –$65,000 if itis kept between 3 and 5 years, and $–110,000 if itis kept for 6 to 8 years ... The replacement must be made now or 3 years fromnow, according to the department supervisor ... 53 The cost characteristicsof a CO testing machine atDytran Instruments are shown below ... The equation for determining the AW of keeping the tester for 2 years is:(a) AW ϭ Ϫ100,000(A͞P,i,8) Ϫ [42,000(P͞F,i,1)ϩ 47,000(P͞F,i,2)](A͞P,i,8) ϩ 40,000(A͞F,i,8)(b) AW ϭ Ϫ100,000(A͞P,i,2) Ϫ [42,000(P͞F,i,1)ϩ 47,000(P͞F,i,2)](A͞P,i,2) ϩ40,000(A͞F,i,2)(c)(d)AW ϭ Ϫ100,000(A͞P,i,2) Ϫ 47,000ϩ 40,000(A͞F,i,2)AW ϭ Ϫ100,000(A͞P,i,2) Ϫ 42,000ϩ 40,000(A͞F,i,2)Machine Age,YearsM&O Costs,$ per YearSalvage Valueat End of Year,$12345678Ϫ42,000Ϫ47,000Ϫ49,000Ϫ50,000Ϫ52,000Ϫ54,000Ϫ63,000Ϫ67,00060,00040,00031,00024,00015,00010,00010,00010,00011 ... RetentionPeriod, YearsAW Value,$ per Year12345Ϫ92,000Ϫ81,000Ϫ87,000Ϫ89,000Ϫ95,000A challenger has an economic service life of7 years with an AW of $–86,000 per year ... If all future costs remain as estimated for the analysis, the company should purchase the challenger:(a) Now(b) After 2 years(c) After 3 years(d) Never11 ... The defender should be replaced:(a) Now(b) 1 year from now(c) 2 years from now(d) 3 years from nowAW Value, $ perYearNumber ofYears RetainedDefenderChallenger12345Ϫ14,000Ϫ13,700Ϫ16,900Ϫ17,000Ϫ18,000Ϫ21,000Ϫ18,000Ϫ13,100Ϫ15,600Ϫ17,500Case Study321CASE STUDYWILL THE CORRECT ESL PLEASE STAND?BackgroundNew pumper system equipment is under consideration by aGulf Coastchemical processing plant ... Because of the variable quality of the rawchemical and the high pressures imposed on the pump chassisand impellers, a close log is maintained on the number ofhours per year that the pump operates ... As currently planned, rebuild andM&O cost estimates are increasedaccordingly when cumulative operating time reaches the 6000-hour mark ... Estimates made forthis pump are as follows:First cost:Rebuild cost:M&O costs:MARR:$Ϫ800,000$Ϫ150,000 whenever 6000 cumulative hours are logged ... A maximum of3 rebuilds is allowed ... Determine the economic servicelife of the pump ... The plant superintendent told you, the safety engineer,that only one rebuild should be planned for, becausethese types of pumps usually have their minimum-costlife before the second rebuild ... Comment on the practicality of ESL ϭ 6 years, giventhe MV calculated ... In a separateconversation, the line manager told youto not plan for a rebuild after 6000 hours, becausethe pump will be replaced after a total of 10,000hours of operation ... He also told you to assume now that the 15%growth rate applies from year 1 forward ... What do you think of these suggestions from theplantsuperintendent and the line manager?CHAPTER 12IndependentProjects withBudgetLimitationL E A R N I N GO U T C O M E SPurpose: Select independent projects for funding when there is a limitation on the amount of capital available for investment ... 1Capital rationing• Explain how a capitalbudgeting problem isapproached ... 2Equal-life projects• Use PW-based capital budgeting to select fromseveral equal-life independent projects ... 3Unequal-life projects• Use PW-based capital budgeting to select fromseveral unequal-life independent projects ... 4Linear programming• Set up the linearprogramming model and use theSolver spreadsheet tool to select projects ... 5Ranking options• Use the internal rate of return (IROR) andprofitability index (PI) to rank and select fromindependent projects ... If the projects are not mutually exclusive,they are categorized as independent of one another, asdiscussed at the beginning of Chapter 5 ... Itis possible to select any number of projects from none (do nothing) to all viable projects ... This limit is considered as each independent project is economicallyevaluated ... They determine the economically best rationing of initial investmentcapital amongindependent projects based upon different measures, such as PW, ROR, andthe profitability index ... 12 ... When a corporation hasseveral options for placing investment capital, a “reject or accept” decision must be made foreach project ... Selection of one project does not impact the selection decisionfor any other project ... The term project is used to identify each independent option ... The term mutually exclusive alternative continuesto identify a project when only one may be selected from several ... Two examples of contingent projects A andB are as follows: A cannot be accepted unless B isaccepted; and A can be accepted in lieu of B,but both are not needed ... For example, B must be accepted if both A and C are accepted ... A capital budgeting study has the following characteristics:• Several independent projects are identified, and net cash flow estimates are available ... • A statedbudgetary constraint restricts the total amount available for investment ... This investment limitis identified by the symbol b ... By nature, independent projects are usually quite different from one another ... In the private sector, sample projects maybe a new warehousing facility, expanded product base,improved quality program, upgraded information system, automation, and acquisition of another firm ... The typicalcapital budgeting problem is illustrated in Figure 12–1 ... Present worth analysis using the capital budgeting process is the recommended method toselect projects ... IndependentprojectselectionLimited budget324Chapter 12Figure 12–1Independent Projects with Budget LimitationIndependentprojectsBasic characteristics of acapital budgeting study ... LifeCapitalinvestmentlimitInvestmentB ... LifeSelect 0 to all 3 projectsObjective: Maximize PWvalue of selectionwithin capitallimitEqual-service requirementThis guideline is not different from that used for selection in previous chapters for independentprojects ... The primary difference now is that the amount ofmoney available to invest is limited, thus the title capital budgeting or rationing ... Previously, PW analysis had therequirement of equal service between alternatives ... Rather, the selection guideline has the following implied assumption ... Opportunity costThis fundamental assumption is demonstrated to be correct at the end of Section 12 ... Another dilemma of capital rationing among independent projects concernsthe flexibility of thecapital investment limit b ... For example, assume project A has a positive PW value at the MARR ... However, in the examples here, we will not exceed a stated investment limit ... 9 and 10 ... The lack of capital to fund the next project defines the ROR level thatis forgone ... 12 ... Twoare the internal rate of return (IROR), discussed in Chapter 7, and theprofitability index (PI), also called the present worth index (PWI), introduced in Chapter 9 ... The capital budgeting process,covered in the next 3 sections, does find the optimal solution for PW values ... Application of these two measuresis presented in Section 12 ... 12 ... Eachfeasible bundle must have a total investment that does not exceed b ... The total number of bundles for m projects is 2m ... For m ϭ 4, there are 24 ϭ 16 bundles, and for m ϭ 16, 216 ϭ 65,536 bundles ... The bundle with the largest PW valueis selected ...ProjectInitial Investment, $ABCDϪ10,000Ϫ5,000Ϫ8,000Ϫ15,000If the investment limit is b ϭ $25,000, of the 16 bundles, there are 12 feasible ones to evaluate ... Theviable bundles are shown below ... Develop all mutually exclusive bundles with a total initial investment that does not exceedthe capital limitNCFjtj0[12 ... Select the bundle with the (numerically) largest PWj value ... Any bundle with PWj Ͻ 0 is discarded, because it does not ( ؊ NCF (P͞F,i,t ͚ initial investmentt� nj� ؊ � PW of bundle net cash flows b ... Sum the net cash flows NCFjt for all projects in each bundle j ( j ϭ 1, 2, ... , nj) ... 3 ... PWjproduce a return of at least the MARR ... 1The projects review committee of Microsoft has $20 million to allocate next year to new software product development ... Allamounts are in $1000 units ... Select the project(s)if a 15% return is expected ... Remember the units are in $1000 ... There are 25 ϭ 32possible bundles ... The $21,000 investment for E eliminates it from all bundles ... The bundle net cash flows, column 4, are the sum of individual project net cashflows ... Use Equation [12 ... Since the annual NCFand life estimates are the same for a bundle, PWj reduces toPWj ϭ NCFj(P͞A,15%,9) ϪNCFj04 ... Bundle 2 does not return15%, since PW2 Ͻ 0 ... This leaves $6 million uncommitted ... 12 ... 3 Capital Rationing Using PW Analysisof Unequal-Life ProjectsUsually independent projects do not have the same expected life ... 1, thePW method for solution of the capital budgeting problemassumes that each project will last forthe period of the longest-lived project nL ... Therefore, use of the LCM of lives is notnecessary, and it is correct to use Equation [12 ... EXAMPLE 12 ... Solve by hand and by spreadsheet ... Of the 24 ϭ 16 bundles, 8 are economically feasible ... 1] are summarized inTable 12–3 ... TA BLE 12–3Present Worth Analysis for Unequal-Life Independent Projects,Example 12 ... It is necessaryto initially develop the mutually exclusive bundles manually and total net cash flows each yearusing each project’s NCF ... Bundle 5 (projects A and C) has the largest PW value (row16) ... 2 ... 1] is correct ... Refer toFigure 12–3, which uses the general layout of a two-project bundle ... The P͞A factor is used for PW computation ... At the end of the shorter-lived project, the bundle has a total future worthof NCFj(F͞A,MARR,nj) as determined for each project ... The assumption of thereturn at the MARRis important; this PW approach does not necessarily select the correct projects if the return is notat the MARR ... Finally,Figure 12–3Representative cash flowsused to compute PW for abundle of two independent unequal-life projectsby Equation [12 ... FWBPWBnB = nLProjectBInvestmentfor BFWAPWAFuture worthPeriod ofreinvestmentat MARRnAProject AInvestmentfor ABundle PW = PWA + PWBnL12 ... 2 ... This is the bundle PW ϭ PWA ϩ PWB ... 2]Substitute the symbol i for the MARR, and use the factor formulas to simplify ... 3]ϭ NCFj(P͞A,i,nj)Since the bracketedexpression in Equation [12 ... To demonstrate numerically, consider bundle j ϭ 7 in Example 12 ... The evaluation is inTable 12–3, and the net cash flow is pictured in Figure 12–4 ... FW ϭ 5220(F͞A,15%,4)(F͞P,15%,5) ϩ 2680(F͞P,15%,4) ϭ $57,111The present worth at the initial investment time isPW ϭϪ16,000 ϩ 57,111(P͞F,15%,9) ϭ $235The PW value is the same as PW7 in Table 12–3 and Figure 12–2 ... If this assumption is not realistic, the PW analysismust be conducted using the LCM of all project lives ... 4 Capital Budgeting Problem FormulationUsing Linear ProgrammingThe procedurediscussed above requires the development of mutually exclusive bundles oneproject at a time, two projects at a time, etc ... As the number of independent projects increases, this processbecomes prohibitively cumbersome and unworkable ... The problem is formulated using theinteger linear programming(ILP) model, which means simply that all relations are linear and thatthe variable x can take on only integer values ... The formulation inwords follows ... Constraints:• Capital investment constraint is that the sum of initial investments must not exceed a specified limit ... For the math formulation, define b asthe capital investment limit, and let xk (k ϭ 1 to m projects)be the variables to be determined ... Note that the subscript k represents each independent project, not a mutuallyexclusive bundle ... 4]Յbkϭ1xk ϭ 0 or 1for k ϭ 1, 2, ... 1] at MARR ϭ i ... 5]t�1Computer solution is accomplished by a linearprogramming software package which treats the ILPmodel ... The Solver tool is similar to Goal Seek with significantly more capabilities ... This means that thefunction Z in Equation [12 ... Also, multiple changing cells can be identified, sothe 0 or 1 value of the unknowns can be determined ... 4] canbeaccommodated ... 3 illustrates its use ... 3Review Example 12 ... (a) Formulate the capital budgeting problem using the math programming model presented in Equation [12 ... (b) Select the projects using Solver ... The capital investment limit is b ϭ $20,000 in Equation [12 ... kϭ4͚ PW xMaximize:kkϭZkϭ1kϭ4͚ NCFConstraints:k0 xkՅ 20,000kϭ1xk ϭ 0 or 1for k ϭ 1, 2, 3, 4Now, substitute the PWk and NCFk0 values from Table 12–3 into the model ... We have the complete 0-or-1 ILP formulation ... 2 is written asx1 ϭ 1x2 ϭ 0x3 ϭ 1x4 ϭ 012 ... 3 ... The spreadsheettemplate can be expanded in eitherdirection if needed ... The descriptions below and the cell tag identify the contents of the rows and cellsin Figure 12–5, and their linkage to Solver parameters ... Cell I5 is the expression for Z, the sum of the PW values forthe projects ... Rows 6 to 18: These are initial investments and net cash flowestimates for eachproject ... Row 19: The entry in each cell is 1 for a selected project and 0 if not selected ... Since each entry must be 0 or 1, a binary constraint is placed on all row 19 cells in Solver, as shown in Figure 12–5 ... Solver will find the solution to maximize Z ... TheNPV functions aredeveloped for any project with a life up to 12 years at theMARR entered in cell B1 ... Row 22: This row shows the initial investment for the selected projects ... This cell has the budget limitation placed on it by the constraint in Solver ... To solve the example, set all values in row 19 to 0, set up the Solverparameters as describedabove, and click on Solve ... ) If needed, further directions on saving the solution,making changes, etc ... 5, and on the Excel help function ... 331332Chapter 12Independent Projects with Budget Limitation12 ... 2 to 2 ... However, it isvery common in industrial, professional, andgovernment settings to learn that the rate of returnis the basis for ranking projects ... 2, is determined by setting a PW or AW relation equal to zero and solving for i* — theIROR ... 6]tϭ1This is the same as Equation [12 ... The spreadsheetfunction RATE or IRR will provide the same answer ... If there is nobudget limit, select all projects that have IROR Ն MARR ... This can occur because IROR ranking maximizes the overall rate of return, notnecessarily the PW value ... Another common ranking method is the profitability index (PI) that we learned inSection 9 ... This is a “bang for the buck” measure that� ———————— PW of initial investmentͦ NCFj0 ͦ[12 ... The numerator has only cash flows ——————————— provides a sense of getting the most forthe investment dollar over the life of the project ... 2 for more details ... The PI measure is defined ast�nj͚ NCF (P͞F, i, t)jtPW of net cash flowst�1that result from the project for years 1 through its lifenj ... Similar to theprevious case, the selection guideline is as follows:Independent projectselectionOnce the project ranking by PI is complete, select all projects in order without exceeding theinvestment limit b ... 0 ... Example 12 ... Noneof these resultsare incorrect; they simply maximize different measures, as you will see ... Additionally, greater complexity is introduced when dependent and contingent projects are involved ... 4Georgia works as a financial analyst in the Management Science Group of General Electronics ... 5333Additional ProjectRanking MeasuresIROR, PI, and PW Values for Five Projects, Example 12 ... 51 ... 80 ... 41 ... 60 ... 01 ... She has confirmed the computations and is ready to do the ranking and make theselection ... Use the PI measure to rank and select projects ... Compare the selected projects by the three methodsand determine which one will maximize the overall ROR value of the $18 million budget ... (a) Ranking by overall IROR values indicates that projects 1 and 5 should be selected with$13 million of the $18 million budget expended ... (b) As an example, the PI for project 1 is calculated using Equation [12 ...4000(P͞A,15%,6)PI1 ϭ ————————|Ϫ8000|ϭ 15,138͞8,000ϭ 1 ... Again, the remaining $5 million is assumed to generate a return of MARR ϭ 15% per year ... Projects 1 and 3, rather than 1 and 5, are selected for a total PW ϭ $8 ... The remaining $2 million is assumed to earn 15% per year ... PWranking results in theselection of projects 1 and 3 ... In $1000 units,Projects 1 and 5NCF, year 0:NCF, years 1–3:NCF, years 4–6:$Ϫ13,000$6,600$4,0000 ϭ Ϫ13,000 ϩ 6600(P͞A,i,3) ϩ 4000(P͞A,i,3)(P͞F,i,3)Ranking of Projects by Different Measures, Example 12 ... 526 ... 891 ... 412 ... 632421,0001 ... 920 ...1% ... 1(13,000) ϩ 15 ... 4%Projects 1 and 3NCF, year 0:NCF, years 1–5:NCF, year 6:$Ϫ16,000$6,700$4,0000 ϭ Ϫ16,000 ϩ 6700(P͞A,i,5) ϩ 4000(P͞F,i,6)By IRR function, the rate of return is 33 ... The overall return on the entire budget isROR ϭ [33 ... 0(2000)]͞18,000ϭ 31 ... 4% ... 189 (i ... , 7 ... 051)million, as determinedfrom Table 12–5, column 7 ... Capital budgeting involves proposed projects, eachwith an initial investment and net cash flows estimated over the life of the project ... ••••Selection is made from among independent projects ... Maximizing the present worth of the net cash flows is theobjective ... The present worth method is used for evaluation ... There are a maximum of 2m bundles for m projects ... Reinvestment of net positive cash flows at theMARR is assumed for all projects with lives shorter than that of the longest-lived project ... Excel’s Solver tool solves this problembyspreadsheet ... Two measures are theinternal rate of return (IROR) and the profitability index (PI), also called the PW index ... When there are a large number of projects, the IROR basis is commonly applied inindustrial settings ... 1 Define the following terms: bundle, contingentproject, dependentproject ... 2 State two assumptions made when doing capital rationing using a PW analysis for unequal-life projects ... 3 For independent projects identified as A, B, C, D,E, F, and G, how many mutually exclusive bundlescan be formed?12 ... Projects X and Y perform the same function withdifferentprocesses; both should not be selected ... 5 Five projects have been identified for possibleimplementation by a company that makes dry iceblasters—machines that propel tiny dry ice pellets at supersonic speeds so they flash-freeze andthen lift grime, paint, rust, mold, asphalt, and335Problemsothercontaminants off in-place machines and awide range of surfaces ... Determine which bundles are possible,provided the budget limitation is (a) $34,000 and(b) $45,000 ... The project costs and 18% per year PWvalues are as shown ... 7 Develop all acceptable mutually exclusive bundles for the fourindependent projects describedbelow if the investment limit is $400 and the following project selection restriction applies: Project 1 can be selected only if both projects 3 and 4are selected ... 6 Four independent projects (1, 2, 3, and 4) are proposed for investment by Perfect Manufacturing,Inc ... Projects1 and 4 should not both be selected;they are essentially duplicates ... 10 The capital fund for research project investment atSummaCorp is limited to $100,000 for next year ... InitialProject Investment, $IIIIIIϪ25,000Ϫ30,000Ϫ50,000ABC12 ... Select the best bundle if the capital budget limit is $45,000 andthe MARR is the cost ofcapital, which is 9% per year ... 9 The general manager for Woodslome ApplianceCompany Plant #A14 in Mexico City has fourindependent projects that she can fund this year to6,0009,00015,000Life, SalvageYears Value, $4444,000Ϫ1,00020,00012 ... Use (a) hand and (b)spreadsheet-basedPW analysis and a 15% per year return requirementto help this engineer make the best decision from apurely economic perspective ... 12 Dwayne has four independent vendor proposals tocontract the nationwide oil recycling services forFord Corporation manufacturing plants ...Revenue sharing ofrecycled oil sales with Ford is a part of the requirement ... The corporate MARRis 10% per year ... (b) A larger budget of $5 ... (c) There is no limit on spending ... 5 millionϪ3 ... 8 millionϪ2 ... 14 Charlie’s Garage has $70,000 to spend on newequipment that may increase revenue for hiscarrepair shop ... All are expected to last 3 years ... 17 The independent project estimates below havebeen developed by the engineering and financemanagers ... Select the economically best projects using the PWmethod and (a) hand solution and (b) spreadsheetsolution ... 5Ϫ3 ... 8Ϫ2 ... 18 Use the PWmethod to evaluate four independentprojects ... The MARR is 12% per year, and up to$16,000 in capital investment funds are available ... 13 Use the PW method at 8% per year to select up tothree projects from the four available ones if nomore than $20,000 can be invested ... Project12 ... Use the PWmethod toevaluate mutually exclusive bundles tomake the selection ... )234Ϫ50005Ϫ8,0005Ϫ9,0003Ϫ10,0004Year12345NCF Estimates, $ per Year1000170024003000380050050050050010,50050005000200000017,00012 ... 18 using a spreadsheet ... 12 ... 12 ... 5% per year ... 0million can be invested... (b) If the life of project 3 can be increased from5 to 10 years for the same $1 million investment, use Goal Seek to determine the NCF inyear 1 for project 3 alone to have the samePW as the best bundle in part (a) ... With these newNCF and life estimates, what are the bestprojects forinvestment?337ProblemsEstimated NCF, $ per YearProjectInvestment,$MLife,YearsYear 1Ϫ0 ... 1Ϫ1 ... 21 Formulate the linear programming model, developa spreadsheet, and solve the capital rationing problem in Example 12 ... 12 ... 17 ... 25 Using the estimates in Problem 12 ... Other RankingMeasures12 ... (a) Determine the IROR, PI, and PW values ifthe MARR is 15% per year ... 27 An engineer at Delphi Systems is considering theprojects below, all of which can be considered tolast indefinitely ... (a) Determine which projects should be selectedon the basis of IROR if the budget limitationis$39,000 ... 019 ... 314 ... 012 ... The company’s MARR is 15% per year ... (Solve by hand orspreadsheet as instructed ... 12 ... 20(a), using the linear programming model anda spreadsheet ... 24 Johnson and Johnson is expanding its first-aidproducts line for individuals allergic to latex ... 18 ... The MARRis 12%per year, and the budget limit is $16 million ... First Cost, $ABCDEGradientafter Year 1123ProjectAnnualProjectProject First Cost, $ Income, $ per Year Life, YearsLANDTϪ30,000Ϫ15,000Ϫ45,000Ϫ70,000Ϫ40,0009,0004,90011,1009,00010,000101010101012 ... The American Society of CivilEngineers (ASCE) has teamedwith municipalities, counties, and several excavationcompanies to develop robots that can travel throughmains, detect leaks, and repair many of them immediately ... There is a $100 million limit on capital funding,and the MARR is established at 12% per year ... Solve byspreadsheet, unless instructed to use hand solution ... Is it economically justified?(c) Determine the overall rate of return for the$100 million with the projects selected inpart (a) ... ProjectFirstCost, $ MEstimated AnnualSavings, $M per YearProjectLife, YearsWXYZϪ12Ϫ25Ϫ45Ϫ605 ... 312 ...03468338Chapter 12Independent Projects with Budget Limitation12 ... 12 ... All projects have a 10-year life ... (c) Are different projects selected using the twomethods?ProjectFirst Cost, $Net Income,$ per YearABCDEϪ18,000Ϫ15,000Ϫ35,000Ϫ60,000Ϫ50,0004,0002,80012,60013,0008,00012 ... The

company always has more projects to engage in than it hascapital to fund projects ... Since all projects are considered long-term ventures, the company uses an infinite period for their life ... ProjectIROR, %18 ... 334 ... 39 ... 33 A project that has a condition associated with itsacceptance or rejection isknown as:(a) A mutually exclusive alternative(b) A contingent project(c) A dependent project(d) Both (b) and (c)12 ... 35 All of the following are correct when a capitalbudgeting problem is solved using the 0-1 integerlinear programming model except:(a) Partial investment in a project is acceptable ... (c)Budget constraints may be present for thefirst year only or for several years ... 12 ... A bundle may consist of only one project ... A bundle may include contingent and dependent projects ... 37 When there are 5 projects involved in a capitalbudgeting study, the maximum number of bundlesthat can beformulated is:(a) 6(b) 10(c) 31(d) 3212 ... 39 The independent projects shown below are underconsideration for possible implementation byRenishaw Inc ... If the company’s MARR is 14% per year and ituses the IROR method of capital budgeting, theprojects it should select under a budget limitationof$105,000 are:(a) A, B, and C(b) A, B, and D(c) B, C, and D(d) A, C, and DProjectFirst Cost, $AnnualIncome, $ per YearRate ofReturn, %ABCDEϪ20,000Ϫ10,000Ϫ15,000Ϫ70,000Ϫ50,0004,0001,9002,60010,0006,00020 ... 017 ... 312 ... 40 For a project that requires an initial investmentof $26,000 andyields $10,000 per year for4 years, the PI at an interest rate of 10% per yearis closest to:(a) 1 ... 22(c) 1 ... 56CHAPTER 13Breakeven andPaybackAnalysisL E A R N I N GO U T C O M E SPurpose: Determine the breakeven for one or two alternatives and calculate the payback period with and withoutareturn required ... 1Breakeven point• Determine the breakeven point for oneparameter ... 2Two alternatives• Calculate the breakeven point of a parameterand use it to select between two alternatives ... 3Payback period• Determine the payback period of a project ati ϭ 0% and i Ͼ 0% ... 13 ... Breakevenanalysis is performed to determine the value of a variable or parameter of a project or alternative that makes two elements equal, for example, thesales volume that will equate revenues and costs ... Breakevenanalysis is commonly applied in make-or-buy decisions when a decision is needed aboutthesource for manufactured components, services, etc ... There are two types of payback: return (i Ͼ 0%) and no return(i ϭ 0%) ... These aspects are discussed in depth in this chapter ... If the variable of interest is allowed to vary, theapproaches of sensitivity analysis (Chapter 18) should be used ... 13 ...This form of breakeven analysis has been used many times so far ... Methodsused to determine the quantity includeDirect solution by hand if only one factor is present (say, P͞A) or only single amounts areestimated (for example, P and F)Trial and error by hand or calculator when multiple factors arepresentSpreadsheet when cash flow and other estimates are entered into cells and used in residentfunctions (PV, FV, RATE, IRR, NPV, PMT, and NPER) or tools (Goal Seek and Solver) ... For example, the variable may be a design element to minimize cost or theproduction level needed to realizerevenues that exceed costs by 10% ... The breakeven point QBE is determined from mathematical relations, e ... , product revenue and costs ormaterials supply and demand or other parameters that involve the parameter Q ... The unit of the parameter Q may vary widely: units per year, cost perkilogram, hours per month,percentage of full plant capacity, etc ... A linear revenuerelation is commonly assumed, but a nonlinear relation is often more realistic ... Costs, which may be linear or nonlinear, usually include two components—fixed andvariable—as indicated in Figure 13–1b ... These includecosts such as buildings, insurance, fixed overhead, someminimum level of labor, equipment capital recovery, and information systems ... These include costs such as direct labor, materials, indirect costs, contractors, marketing, advertisement, and warranty ... Even if nounits are produced, fixed costs areincurred at some threshold level ... R, revenue per yearNonlinear(2)LinearQ, units per year(a) Revenue relations—(1) increasing and(2) decreasing revenue per unitTC ϭ FC ϩ VCTotalcost, TCVariable, VCCost per yearCost per yearTC ϭ FC ϩ VCTCVCFCFixed, FCQ, units per year(b) Linear costrelationsQ, units per year(c) Nonlinear cost relationscannot last long before the plant must shut down to reduce fixed costs ... Variable costs change with production level, workforce size, and other parameters ... When FC and VC are added, they form the total cost relation TC ... Figure 13–1c shows aVC)[13 ... R ϭ TCrQ ϭ FC ϩ vQwherer ϭ revenue per ؉ FC ؊ ) TC� R ؊ total cost� R ؊ � revenue general TC curve for anonlinear VC in which unit variable costs decrease as the quantity level rises ... If Q Ͼ QBE, there is apredictable profit; but if Q Ͻ QBE, there is a loss ... Profit is calculated asProfitunitv ϭ variable cost per unitSolve for the breakeven quantity Q ϭ QBE for linear R and TC functions ... 2]13 ... $TCRProfitmaximizedProfit rangeLossLossQBEQPQBEQ, units per yearFigure 13–3Breakeven points and maximum profit point for a nonlinear analysis ... For example, if the variable cost perunit is reduced, then theTC line has a smaller slope (Figure13–2) and the breakeven point will decrease ... A similaranalysis is possible for fixed VC and increased levels of production, as shown in the next example ... Figure 13–3presents this situation for two breakeven points ... 343344Chapter13Breakeven and Payback AnalysisOf course, no static R and TC relations—linear or nonlinear—are able to estimate exactly therevenue and cost amounts over an extended period of time ... EXAMPLE 13 ... Normal production level is 60 diverter systems per month, butdue to significantly improvedeconomic conditions in Asia, production is at 72 per month ... Fixed costsVariable cost per unitRevenue per unitFC ϭ $2 ... 2] to determine the breakeven number of units ... FCQBE ϭ ———rϪv2400ϭ ———— ϭ 60 units per month75 Ϫ 35Figure 13–4 is a plot of R and TC lines ... Theincreased productionlevel of 72 units is above the breakeven value ... 1 ... 2345Breakeven Analysis Between Two Alternatives(b) To estimate profit (in $1000 units) at Q ϭ 72 units per month, use Equation [13 ... Profit ϭ R Ϫ TC ϭ rQ Ϫ (FC ϩ vQ)ϭ (r Ϫ v)Q Ϫ FCϭ (75 Ϫ 35)72 Ϫ 2400ϭ $480[13 ... (c) To determine the requireddifference r Ϫ v, use Equation [13 ... 4 million ... 33 per unit45The spread between r and v must be $53,330 ... e ... In some circumstances, breakeven analysis performed on a per unit basis is more meaningful ... 2], but the TC relation is divided by Q toobtain an expression for cost per unit, also termedaverage cost per unit Cu ... 4]Cu ϭ —— ϭ ———— ϭ —— ϩ vQQQAt the breakeven quantity Q ϭ QBE, the revenue per unit is exactly equal to the cost per unit ... 4] takes on the shape of a hyperbola ... This is the same as setting profit equal to zero in Equation [13 ... It may be necessary to perform somedimensional analysis initially to obtain the correct revenueand total cost relations in order to use the same dimension for both relations, for example, $ perunit, miles per month, or units per year ... 2 Breakeven Analysis Between Two AlternativesNow we consider breakeven analysis between two mutuallyexclusive alternatives ... Equating the two PW or AW relations determines the breakeven point ... The parameter can be the interest rate i, first cost P, annual operating cost (AOC), or any parameter ... For example, the incremental ROR value (⌬i*) is the breakeven rate between alternatives ... InSection11 ... If the market value islarger than RV, the decision should favor the challenger ... Figure 13–5 illustratesthe breakeven concept for two alternatives with linear cost relations ... However, alternative 2 has a smaller variable cost,as indicated by its lower slope ... Thus, if the number of units of thecommonvariable is greater than the breakeven amount, alternative 2 is selected, since the total costwill be lower ... Breakeven346Chapter 13Breakeven and Payback AnalysisFigure 13–5Alt ... Alt ... 1 FCAlt ... The AW is preferred when the variable units areexpressed on a yearly basis, and AWcalculations are simpler for alternatives with unequal lives ... 1 ... 2 ... 3 ... Selection between alternatives is based on this guideline:If the anticipated level of the common variable is below the breakeven value, select the alternative with the higher variable cost (larger slope) ... (Refer to Figure 13–5 ... 2Asmall aerospace company is evaluating two alternatives: the purchase of an automatic feedmachine and a manual feed machine for a finishing process ... One person will operate the machine at a rate of $12 per hour ... Annual maintenance and operating cost is expected to be $3500 ... However, threeworkers will be required at $8 perhour each ... Allprojects are expected to generate a return of 10% per year ... 1 ... 2 ... 5xThe VC is developed in dollars per year ... 5xϭ $–6992 Ϫ 1 ... 2Breakeven Analysis Between Two AlternativesSimilarly, the annual variable cost and AW for the manual feed machineare$81 hour x tonsAnnual VC ϭ —— (3 operators) ——— ———hour6 tons yearϭ 4xAWmanual ϭ Ϫ8000(A͞P,10%,5) Ϫ 1500 Ϫ 4xϭ $Ϫ3610 Ϫ 4x3 ... AWauto ϭ AWmanualϪ6992 Ϫ 1 ... 5 is smaller than the manual feed VC slope of 4 ... This meansthe company contracts to buy the product or service fromthe outside, or makes it within the company ... Where the two cost relations cross is the make-buy decision quantity ... EXAMPLE 13 ... It is faced witha make-or-buy decision ... The steel arm of the lift can be purchased internationally for $3 ... If manufactured on site, two machines will be required ...Machine A will require an overhaul after 3 years costing $3000 ... A total of fouroperators will be required for the two machines at a rate of $12 ... In anormal 8-hour period, the operators and two machines can produce parts sufficient to manufacture 1000 units ... (a) Number of units to manufacture eachyear to justify the in-house (make) option ... The company expects to produce 10,000 unitsper year ... 1 ... 2 ... Annual VC ϭ (cost per unit)(units per year)4 operators $12 ... 4xThe annual fixed costs for machines A and B are the AW amounts ... 347Chapter 13Breakeven and Payback Analysis3 ... 50x)and the make option yieldsϪ3 ... 4xϪ3 ... 5]A minimum of 6565 lifts must be produced each year to justify the make option, whichhas the lower variable cost of 0 ... (b) Substitute 10,000 for x and PA for the to-be-determined first cost of machine A (currently$18,000) in Equation [13 ... Solution yields PA ϭ$58,295 ... Even though the preceding examples treat only two alternatives, the same type of analysis canbe performed for three or more alternatives ... The results are the ranges through which each alternative ismore economical ... Between 40 and 60, alternative 2 is more economical; and above60,alternative 3 is favored ... If the costs increaseor decrease uniformly, mathematical expressions that allow direct determination of the breakeven point can be developed ... 3 Payback AnalysisPayback analysis is another use of the present worth technique ... Payback is allied with breakeven analysis;this is illustrated later in the section ... Alternative 1Alternative 2Alternative 3Total cost, $/year348Breakevenpoints4060Output, units/hourFigure 13–6Breakeven points for three alternatives ... 3349Payback AnalysisThe payback period np is an estimated time for the revenues, savings, and any othermonetary benefits to completely recover the initial investment plus a stated rate of return i ... No return; iϭ0%: Also called simple payback, this is the recovery of only the initial investment ... An example application of payback may be a corporate senior manager who insists that everyproposal return theinitial cost and some stated return within 3 years ... The payback period should be determined using a required i Ͼ 0% ... After the formulas are presented, a couple of cautions about payback usage are provided ... For both types, the terminology is P for the initial investment in the asset, project, contract,NCF (P͞F, i, t)t[13 ... 9]After np years, the cash flows will recover the investment in year 0 plus the required ؉͚ � ؊ P NCFt[13 ... 7]t� npDiscounted, i Ͼ 0%; NCFt varies annually:0 ؉͚ � ؊ P � %0 ; NCFt varies annually:0 etc ... Using Equation [1 ... Note that np is usually not an integer ... , np,t� npNo return, ireturn ofi% ... If the estimated life is less than np years, there is not enough time to recover theinvestment and i% return ... Consequently, it is preferable to use payback as aninitial screening method or supplemental tool rather than as the primary means to select analternative ... • Either type of paybackdisregards all cash flows occurring after the payback period ... Payback analysis utilizes a significantly different approach to alternative evaluation than theprimary methods of PW, AW, ROR, and B͞C ... However, the information obtained from discounted payback analysis performed at an appropriate i Ͼ0% can be very useful in that a sense of the riskinvolved in undertaking an alternative is provided ... Even here, the 6-year payback is considered supplemental information and does notreplace a complete economic analysis ... 4The board of directors of Halliburton International has just approved an $18million worldwide engineering construction design contract ... The contract has a potentially lucrative repayment clausePayback period350Chapter 13Breakeven and Payback Analysisto Halliburton of $3 million at any time that the contract is canceled by either party during the10 years of the contractperiod ... (b) Determinethe no-return payback period and compare it with the answer for i ϭ 15% ... Show both hand and spreadsheet solutions ... The single $3 million payment (call it CV forcancellation value) could be received at any time within the 10-year contract period ... 9] is altered to include CV ...3 years, found by trail and error ... (b) If Halliburton requires absolutely no return on its $18 million investment, Equation [13 ... 0 ϭ Ϫ18 ϩ 5(3) ϩ 3There is a very significant difference in np for 15% and 0% ... 3 years, while the no-return payback period requires only 5 years ... Solution bySpreadsheetEnter the function ϭ NPER(15%,3,Ϫ18,3) to display 15 ... Change the rate from 15% to0% to display the no-return payback period of 5 years ... When cash flows that occurafter np are neglected, it is possible to favor short-lived assets even when longer-lived assetsproduce a higher return ...Comparison of short- and long-lived assets in Example 13 ... EXAMPLE 13 ... Machine 2 is expected to be versatile and technologically advanced enoughto provide net income longer than machine 1 ... 8] and [13 ... 57 years at i ϭ 15% ... 13 ... 57$12,000Cash flows neglectedby payback analysis$3000per year$1000 per year01234567Machine 2891011121314np = 9 ... 5 ... 57 years, which is less than the 7-year life ... 52 years, which is less than the 14-year life ... Now, use a 15% PW analysis to compare the machines and comment on any difference in therecommendation ... Compare them over theLCM of 14 years ... This result is the opposite of the payback period decision ... As illustrated in Figure 13–7 (for one lifecycle for each machine), payback analysis neglects all cash flow amounts that may occur afterthe payback time has been reached ... Often a shorter-lived alternative evaluated bypayback analysis mayappear to be more attractive, when the longer-lived alternative has cash flows later in its lifethat make it more economically attractive ... They can be used in conjunction to determine the payback period when a desired level of breakeven is specified ... By working together inthisfashion, better economic decisions can be made ... 6 illustrates the second of thesituations mentioned above ... 6The president of a local company expects a product to have a profitable life of between 1 and5 years ... The cost and revenue estimates are as follows:Fixed costs: Initial investment of$80,000 with $1000 annual operating cost ... Revenue: Twice the variable cost for the first 5 years and 50% of the variable cost thereafter ... Since values of XBE aresought for np ϭ 1, 2, 3, 4, 5, solve for breakeven by substituting each payback period ... 80,000——— ϩ 1000Fixed cost, FCnpRevenue perunit, r$16Variable cost per unit, v(years 1 through 5 only)$8The breakeven relation from Equation [13 ... [13 ... Equation [13 ... The breakeven values are the same as those above,e ... , sell 5125 units per year to pay back in 2 years ... -Figure 13–8Breakeven number of units for different payback periods,Example 13 ... 13 ... Examples 13 ... 8 demonstrate the use of Goal Seek for both types of problems ... 4More Breakeven and Payback Analysis on SpreadsheetsEXAMPLE 13 ... autoblog ... The design is based on the fact that a person naturally steps downward on his or her foot when surprised,shocked, or struck with a medicalemergency ... Assume that for the manufacture of pedal components, two equally qualified machines have been identified and estimates made ... However, the automated controls, safety features, and ergonomic design of machine 2 make it a better choice forthe plant inthe opinion of the project engineer ... The parameters to concentrate on are (a) first cost, (b) net cash flow, and (c) life of machine 2,if all other estimates remain the same ... 7 ... (a) Figure 13–10a: By forcing the AW for machine 2 to equal $193, Goal Seek finds a breakeven of $96,669 ... (b) Figure 13–10b: (Remember to reset the first cost to $ –110,000 on the spreadsheet ... Therefore, if the NCF estimate can realistically beincreased from $22,000 to $25,061, again machine 2 will be economically equivalent ... The easiest approach is to use the NPER function to find the payback period ... 13 years ...353354Chapter 13Breakeven and Payback Analysis(a) First cost(b) NCFFigure 13–10Breakeven values for (a) first cost and (b) annual net cash flow using Goal Seek, Example 13 ... EXAMPLE 13 ... The estimated costs each yearfor repairs, insurance, etc ... At an expected 8% per year return, usespreadsheet analysis to determine the payback periodif the building is (a) kept for 2 years and sold for $290,000 sometime beyond year 2 or (b) keptfor 3 years and sold for $370,000 sometime beyond 3 years ... The NPV function is applied (columns D and F) todetermine when the PW changes sign fromplus to minus ... When PW Ͼ 0, the 8% return is exceeded ... If the building is soldafter exactly 3 years for $290,000, the payback period was not exceeded; but after 4 yearsit is exceeded ... If the building is sold after 4 or 5 years, the payback is not exceeded; however, asale after 6 years is beyond the8%-return payback period ... 8355ProblemsCHAPTER SUMMARYThe breakeven point for a variable for one project is expressed in terms such as units per year orhours per month ... Use the following decision guideline:Single Project(Refer to Figure 13–2 ... Use thefollowing guideline to select analternative:Two Alternatives(Refer to Figure 13–5 ... This is a supplemental analysis technique used primarily for initial screening prior to a full evaluation by PW or some other method ... PROBLEMSBreakeven Analysis for a Project13 ... Banner Engineering’s QT50R radar-basedsensor featuresfrequency-modulated technology toaccurately monitor or detect objects up to 15 milesaway while resisting rain, wind, humidity, and extreme temperatures ... (a) What could the company’s fixed cost peryear be in order for Banner to break evenwith sales of 9000 units per year?(b) If Banner’s fixed cost isactually $750,000per year, what is the profit at a sales level of7000 units per year?13 ... The fixed costs associatedwith manufacturing are $800,000 per year ... 3 A metallurgical engineer has estimated that thecapital investment cost for recovering valuablemetals (nickel, silver, platinum, gold, etc ... Theequipment will have a useful lifeof 15 years with no salvage value ... 9,where E is the efficiency of the metal recovery operation (in decimal form) ... What must theaverage selling price per pound be for the precious metals that are recovered and sold in orderfor the company to break even at its MARRof15% per year?Problems 13 ... 7 are based on the followinginformation ... Use the following costand revenue figures, quoted in U ... dollars per hundredweight (cwt), recorded for this year to calculate the answers for each plant ... 502 ... 4 Determine the breakeven point for each plant ... 5 Estimate theminimum revenue per hundredweightrequired for next year if breakeven values andvariable costs remain constant, but fixed costsincrease by 10% ... 6 During this year, the French plant sold 950 units inEurope, and the U ... plant sold 850 units ... 13 ... Determine the decreases in dollaramounts andpercentages in variable cost necessary to meet this goal, if the number of units sold isthe same as this year ... 8 The National Highway Traffic Safety Administration raised the average fuel efficiency standard to35 ... The rules will cost consumers an average of $926 extra per vehicle in the2016 model year... 5 mpg and keeps itfor 5 years ... 75% per month?13 ... S ... K ... The fixed cost of the center is $775,000with an average variable cost of $1 and revenue of$2 ... (a) Find the percentage of the call capacity thatmust be placed each year to break even ... This is expected toincrease the center’s fixed costto $900,000 ofwhich 50% will be allocated to the new product line ... How does this required revenue compare withthe current center revenue of $2 ... 10 The addition of a turbocharger to a small V-6 engine that gets 18 miles per gallon of gasoline canboost its power to that of a V-8 engine andincreasefuel efficiency at the same time ... 25 per gallon and the interest rate is 1%per month ... 11 Transporting extremely heavy patients (peoplewho weigh more than 500 pounds) is much moredifficult than transporting normal-weight patients ... The extra feesare justified by the ambulance companieson thebasis of the specialty equipment required and theextra personnel involved ... 13 ... A number of companies make devicesthat they claim will significantly increase a vehicle’s fuel efficiency ... Assuming the device works as claimed, for avehicle that currently gets 20 miles per gallon(mpg), how manymiles would the owner have todrive each year to break even in 5 years? Assumethe cost of gasoline is $3 ... 13 ... Assume you had been buying gasoline for$2 ... 98 pergallon at the station where you usually go ... 13 ... The initial cost for equipment conversionwill be $200 million with a 20% salvagevalueanytime within a 5-year period ... Theproduction capacity for the first year will be4000 units ... 15 For the last 2 years, The Health Company hasexperienced a fixed cost of $850,000 per yearand an (r Ϫ v) value of $1 ... Internationalcompetition has become severe enough thatsome financial changesmust be made to keepmarket share at the current level ... (b) If fixed costs and revenue per unit remain attheir current values, what type of changemust take place to make the breakeven pointgo down?13 ... 15) Expandthe analysis performed in Problem 13 ... The financial manager estimates that fixedcosts will fall to $750,000when the required production rate to break even isat or below 600,000 units ... 17 Providing restrooms at parks, zoos, and other cityowned recreation facilities is a considerable expense for municipal governments ... Thecost of renting and servicing a portable restroomis $7500per year ... He remarked that the rather high costis due to the necessity to use expensive materialsand construction techniques that are tailored tominimize damage from vandalism that often occurs in unattended public facilities ... 13 ... A geosynthetic bentonite clay liner(GCL) will cost $1 ...Alternatively,357a high-density polyethylene (HDPE) geomembrane can be installed that will have a useful life of12 years ... 19 An irrigation canal contractor wants to determinewhether he should purchase a used Caterpillarmini excavator or a Toro powered rotary tiller forservicing irrigation ditches in anagricultural areaof California ... Fixed costs for insurance, license, etc ... The excavator will requireone operator at $15 per hour and maintenance at$1 per hour ... 15 mile of ditch can beprepared ... The tiller costs $1200 and has a useful life of5 years with no salvage value ... 20 per hour, and with thetiller,the two workers can prepare 0 ... The contractor’s MARR is 10% per year ... 13 ... Such a system involves recirculation of the partially treated water back into the feed tank, causing the water to heat up ... The single-pass system, good for 3 years, requiresa small chiller costing $920 plus stainlesssteeltubing, connectors, valves, etc ... Thecost of water, treatment charges, electricity, etc ... 10 per hour ... 28 per hour to operate ... 13 ... Ahigh-use component (expected usage is 5000 unitsper year) can be purchased for $25 per unit withdelivery promised within a week ... Labor358Chapter 13and otheroperating costs are estimated to be$35,000 per year over the study period of 5 years ... Neglect the element of availability (a) to determine the breakeven quantityand (b) to recommend making or buying at the expected usage level ... 22 A partner in a medium-size A/E (architectural͞engineering) designfirm is evaluating two alternativesfor improving the exterior appearance of the building they occupy ... The paint is expected toremain attractive for 4 years, at which time repainting will be necessary ... Alternatively, the building can be sandblasted now and every 6 years at a cost 40% greaterthan theprevious time ... 23 A junior mechanical engineering student is cooping this semester at Regency Aircraft, whichcustomizes the interiors of private and corporatejets ... The first cost isnot easy to estimate due to many options, but theannual revenue and M&O costs should net out at$ϩ15,000 per year overa 10-year life ... Determine thebreakeven first cost of the machine to just recoverits first cost and a return of 8% per year under twoscenarios:I: No outside revenue will be developed bythe machine ... Solve using (a) hand and (b) spreadsheet solutions ... 24 Ascarate Fishing Club (a nonprofitorganizationdedicated to teaching kids how to fish) is considering two options for providing a heavily stockedpond for kids who have never caught a fish before ... The purchase price will be $400 ... Breakeven and Payback AnalysisOption 2 is an in-ground pond that will be excavated by club members atno cost and lined withfabric that costs $1 per square foot ... Assume 300 ft2of liner will be purchased ... Maintenance inside thefence is expected to cost $20 per year ... At an interest rate of 6%per year, how long would the above-ground poolhave to last to break even?13 ... Alternative 1 is a gravel baseand pavement with an initial cost of $500,000 that will last for 15 yearsand has an annual upkeep cost of $100 per mile ... Annual reapplication of themix is required ... 05 mile ... (b) A drive in a pickup indicates atotal of 12 ... Which is the moreeconomical alternative?13 ... When the lagoon isfull, it isnecessary to remove the sludge to a sitelocated 8 ... Currently, when the lagoon is full, the sludge is removed by pump into a tank truck and hauled away ... Thecompany pays a contract individual to operate thepump and oversee environmental and safety factors at a rate of $100 per day, plus the truckanddriver must be rented for $200 per day ... The pump would have an initialcost of $1600 and a life of 10 years and will cost$3 per day to operate ... (a) If the pipeline will cost $12 per meter to construct and will have a 10-year life, how manydays per year must the lagoon require pumping to justifyconstruction of the pipeline?(b) If the company expects to pump the lagoononce per week every week of the year, howmuch money can it afford to spend now onthe 10-year life pipeline to just break even?13 ... (a) Use an AWrelation to determine the minimum number ofhours per year to operate thepumps that will justify the Auto Green system, if the MARR is 10%per year ... (c) Write the spreadsheet functions to display the payback period forboth 0% and 3% per month ... 31 (a)Nutra Jet (N) Auto Green (A)Initial cost, $Life, yearsRebuild cost, $Time before rebuild, annuallyor minimum hoursCost tooperate, $ per hourϪ4,0003Ϫ1,0002,000Ϫ10,3006Ϫ2,2008,0001 ... If the asset will be in service for 12 years,should it be purchased?0 ... 28 An engineering practitioner can lease a fullyequipped computer and color printer system for$800 per month or purchase one for $8500 now andpay a $75 per monthmaintenance fee ... Show both (a) hand and (b) spreadsheet solutions ... 29 The office manager of an environmental engineering consulting firm was instructed to make an ecofriendly decision in acquiring an automobile forgeneral office use ... The hybrid under consideration isGM’s Volt, which will cost$35,000, will have asalvage value of $15,000 after 5 years, and willhave a range of 40 miles on the electric battery,plus several hundred more miles when the gasoline engine kicks in ... The Leaf’s relatively limitedrange creates a psychological effect known asrange anxiety (RA), which has the companyleaning toward purchasing the Volt ... The accountant for the consulting firm told theoffice manager that the Leaf is the better economicoption based on an evaluation she performed earlier ... 75% per month ... 30 How long will you have to sell a product that hasan income of $5000 per month andexpenses of(b)13 ... 15 million ... Determine thenumber of years the equipment must be used toobtain payback at MARR values of (a) 0% and 8%per year and (b) 15% and 16% per year ... 13 ... The year index is k ϭ 1, 2, 3, ... For a preliminary conclusion, should theequipment be purchased if the actualusefullife is 7 years?13 ... She is considering the purchase of a three-bedroomlodge in upper Montana that will cost $250,000 ... IfClarisa spends an average of $500 per month forutilities and the investment increases at a rate of2% per month, how long would it be before shecould sell the property for$100,000 more than shehas invested in it?13 ... When she completedher engineering management degree, she sold thebusiness and her grandparents told her to keep the360Chapter 13Breakeven and Payback Analysismoney as a graduation present ... 36 Buhler Tractor sold a tractor for $45,000 toTomEdwards 10 years ago ... 37 National Parcel Service has historically owned andmaintained its own delivery trucks ... The study period is nomore than 24 months for either alternative ... Usethe first cost and net cash flow estimates to determine the payback in months with a nominal 9%per year returnfor the (a) purchase option and(b) lease option ... 38 Julian Browne, owner of Clear Interior Environments, purchased an air scrubber, HEPA vacuum,and other equipment for mold removal for $15,000eight months ago ... For the last 4 months, acontract generated a net $6000 per month ... Determine (a)the no-return payback period and(b) the nominal 18%-per-year payback period ... 39 Explain why payback analysis may favor an alternative with a shorter payback period when it is notthe better choice economically ... 40 When comparing two alternatives, why is it best touse no-return payback analysisas a preliminaryscreening tool prior to conducting a complete PWor AW evaluation?Spreadsheet Problems13 ... 007Q2 ϩ 32QTC ϭ 0 ... 2Q ϩ 8(a) Plot R and TC ... Estimate the amount of profit at this quantity ... The equations areProfit ϭ aQ2 ϩ bQ ϩ cϪbQp ϭ ——2aϪb2 ϩ cMaximum profit ϭ ——4aUsethese relations to confirm the graphicalestimates you made in (a) ... )13 ... Revenuefor the first year was $50,000 ... Problems 13 ... 44 are based on the followinginformation ... The current system has a fixed cost of $300,000per year and a variable cost of $10 per unit ... A newly proposed process will addonboard features that allow the revenue to increase to $16 perunit, but the fixed cost will now be $500,000 per year ... 2 hour required to produce each unit ... 43 Determine the annual breakeven quantity for(a) the current system and (b) the new system ... 44 Plot the two profit relations and estimategraphically the breakeven quantity between the twoalternatives ... 45 through 13 ... 13 ... Mid-Valley Industrial Extension Service, a state-sponsoredagency, provides water quality sampling services to allbusiness and industrial firms in a 10-county region ... Now an outsourcing agency has offered to takeover this function on aper sample basis ... The MARR for government projects is 5%per year, and a study period of 8 years is chosen ... 46 Use a spreadsheet to graph the AW curves for bothoptions for test loads between 0 and 4000 per yearin increments of 1000 tests ... Sample costs average $25 each... Outsourced: Contractors quote sample cost averagesof $100 for the first 5 years, increasing to$125 per sample for years 6 through 8 ... 47 The service director has asked the outsource company to reduce the per sample costs by 25% acrossthe board over the 8-year study period ... 46 beforeanswering ... 48 Assume the Extension Service can reduce its annual salaries from $175,000 to $100,000 peryear and the per sample cost from $25 to $20 ... ) What isthe new annual breakeven test quantity?ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS13 ... 52 AW1 ϭϪ23,000(A͞P,10%,10)ϩ4000(A͞F,10%,10)Ϫ 3000 Ϫ 3xAW2 ϭ Ϫ8,000(A͞P,10%,4) Ϫ 2000 − 6xFor these two AW relations, the breakeven point x,in miles per year, is closest to:(a) 1130(b) 1224(c) 1590(d) 65513 ... Alternative A willhave fixed costs of $42,000 per year and will require 2 workers at $48 per dayeach ... Alternative B will have fixed costs of$56,000 per year, but with this alternative, 3 workers will generate 200 units of product ... 53 To make an item in-house, equipment costing$250,000 must be purchased ... Buying the item externally will cost $100 per unit ... 51 When the variable cost is reducedfor linear totalcost and revenue lines, the breakeven point decreases ... (b) The two lines will now cross at zero ... (d) The total cost line becomes nonlinear ... 54 A procedure at Mercy Hospital has fixed costs of$10,000 per year and variable costs of $50 per test ... The number of tests that must beperformed each year for the two operations to breakeven is closest to:(a) 290 (b) 455 (c) 750 (d) Over 800362Chapter 1313 ... Alternative X has fixedcosts of $10,000 per year with a variable cost of$50 per unit ... The number of units that must be produced eachyear in order for alternative Y to be favoredisclosest to:(a) Y will be favored for any level of production(b) 125(c) 375(d) X will be favored for any level of production13 ... Material C will cost$100,000 per mile and last for 10 years ... Material D will cost $30,000 per mileand last for 5 years ... 57 A construction company can purchase a pieceofequipment for $50,000 and spend $100 per day inoperating costs ... Alternatively, the company can lease the equipment for $400 per day ... 58 A tractor has a first cost of $40,000, a monthly operating cost of $1500, and a salvage value of$12,000 in 10 years ... An identical tractor can be rented for$3200 permonth (operating cost not included) ... 4%,10) Ϫ 1500nϩ 12,000(A͞F,11 ... 59 An anticorrosive coating for a chemical storagetank will cost $5000 and last 5 years if touchedup at the end of 3 years at a cost of $1000 ... 60 The price of a car is $50,000 today ... You nowhave $25,000 in aninvestment that is earning 20%per year ... 61 Process A has a fixed cost of $16,000 per year anda variable cost of $40 per unit ... If the company’s MARR is 10% per year, the fixedcost of process B that will make the two alternatives have the same annual cost at a productionrate of 1000 units per year isclosest to:(a) Less than $10,000(b) $18,000(c) $27,000(d) Over $30,00013 ... 63 Two methods of weed control in an irrigation canalare under consideration ... The lining will last 20 years ... Method B involves sprayinga chemical that costs $40 per gallon ... In determining the number ofmiles per year thatwould result in breakeven, thevariable cost for method B is closest to:(a) $5 per mile(b) $15 per mile(c) $20 per mile(d) $40 per mile13 ... Aeration is used primarily for the physical removal ofgases or volatile compounds, while sludge recirculation canbe beneficial for turbidity removal and hardnessreduction ... With the huge increases inelectricity cost that have occurred in some localities, however, it became necessary to review the cost-effectiveness ofall water treatment processes that consume significantamounts of energy ... there was neither aeration nor recirculation ... The reduction was 18%when neither aeration norrecirculation was used ... With sludge recirculation alone, the turbidity reduction was only 6%, meaning thatsludge recirculation alone actually resulted in an increase inturbidity—the difference between 18% and 6% ... The calculationsare based on the following data:Aerator motorϭ 40 hpAerator motor efficiency ϭ 90%Sludge recirculation motor ϭ 5 hpRecirculation pump efficiency ϭ 90%InformationThis study was conducted at a 106 m3 per minute watertreatment plant where, under normal operating circumstances, sludge from the secondary clarifiers is returned tothe aerator andsubsequently removed in the primary clarifiers ... To evaluate the effect of sludge recirculation, the sludgepump was turned off, but aeration was continued ... Finally, both processes were discontinued ... The results obtained from the four operating modesshowed that the hardness decreased by 4 ... e ...When only sludge was recirculated, the reduction was3 ... There was no reduction due to aeration only, or whenChemicaladditionsElectricity cost ϭ 9 ¢͞kWh (previous analysis)Lime cost ϭ 7 ... 62 mg͞L per mg͞L hardnessCoagulant cost ϭ 16 ... 5As a first step, the costs associated with aeration andsludgerecirculation were calculated ... Aeration cost:40 hp ϫ 0 ... 09 $͞kWh ϫ 24 h͞dayϬ 0 ... 75 kW͞hp ϫ 0 ... 90 ϭ $9 per day or $275 per monthFigure 13–12FlashmixFlocculationPrimaryclarifierSecondaryclarifierAeratorFilterSchematic of watertreatment plant ... I ... AlternativeDescription1Aeration� ) 3 ) ؊ ( 6(Normal operating conditionSludge recirculationand aeration23Aeration(1)Extra Cost forRemoval of—2196275— ( � ) 4 ) ؉ ( 5(NetSavings(7 ( � ) 1 ) ؉ ( 2(Hardness(4)Turbidity(5)TotalExtra Cost(6 ( onlySludge recirculationonlyNeither aeration norsludge recirculationRecirculation(2)4Total Savings(32752196138026246984518491107Ϫ1574ϩ10892196275247113804691849ϩ622The estimates appear in columns 1 and 2 of the cost summaryin Table 13–1 ... The calculations below are based on a design flow of53 m3͞minute ... The extra turbidity reaching theflocculators could require further additionsof the coagulatingchemical ... Since the average dosage before discontinuation of aeration was 10 mg͞L,the incremental chemical cost incurred because of the increased turbidity in the clarifier effluent would be(10 ϫ 0 ... 165 $͞kg ϫ 60 min͞hϫ 24 h͞day ϭ $27 ... e ... Changes in hardness affect chemical costsby virtue of thedirect effect on the amount of lime required for water softening ... 1 mg͞L (that is, 258 mg͞L ϫ 4 ... However, with sludge recirculation only, the reduction was9 ... 3 mg͞L attributed toaeration ... 3 mg͞L ϫ 0 ... 079 $͞kgϫ60 min͞h ϫ 24 h͞day ϭ $8 ... The total savings and total costs associated withchangesin plant operating conditions are tabulated in columns 3 and6 of Table 13–1, respectively, with the net savings shown incolumn 7 ... ” This condition would result ina net savings of $1089 per month, compared to a net savingsof $622 per month when both processes are discontinued anda net costof $1574 per month for aeration only ... In summary, the commonly applied water treatment practices of sludge recirculation and aeration can significantly affect the removal of some compounds in the primary clarifier ... Case Study Exercises1 ... Does a decrease in the efficiency of the aerator motormakethe selected alternative of sludge recirculation onlymore attractive, less attractive, or the same as before?3 ... If the efficiency of the sludge recirculation pump werereduced from 90% to 70%, would the net savings difference between alternatives 3 and 4 increase, decrease, orstay the same?5 ... If the costof electricity decreased to 8 ¢͞kWh, whichalternative would be the most cost-effective?7 ... L E A R N I N G S TA G E 4Rounding Out the StudyLEARNING STAGE 4Rounding Outthe StudyCHAPTER14Effects of InflationCHAPTER15Cost Estimation andIndirect CostAllocationCHAPTER16DepreciationMethodsCHAPTER17After-Tax EconomicAnalysisCHAPTER18Sensitivity Analysisand Staged DecisionsCHAPTER19More on Variationand Decision Makingunder RiskThis stage includes topics to enhance your ability to perform athorough engineering economic study of one project or several alternatives... Techniques of costestimation to better predict cash flows are treated in order to basealternative selection on more accurate estimates ... An expanded version of sensitivity analysis isdeveloped to examine parameters that vary over a predictable rangeof values ... Finally, the elements of risk andprobabilityare explicitly considered using expected values, probabilistic analysis,and spreadsheet-based Monte Carlo simulation ... Use the chart in the Preface todetermine appropriate points at which to introduce the material inLearning Stage 4 ... SECTIONTOPICLEARNING OUTCOME14 ... 14 ... 14 ...14 ... This chapter concentrates upon understanding and calculating the effects of inflation in time value of money computations ... The annual inflation rate is closely watched and historically analyzed by governmentunits, businesses, and industrial corporations ... In the first decade of the 21st century,inflation has notbeen a major concern in the United States or most industrialized nations ... Factors such as the costof energy, interest rates, availability and cost of skilled people, scarcity of materials, political stability, and other, less tangible factors have short-term and long-term impacts on theinflationrate ... The basic techniques to do so are covered here ... 1 Understanding the Impact of InflationWe are all very well aware that $20 now does not purchase the same amount as $20 did in 2005and purchases significantly less than in 2000 ... Inflation is an increase in the amount of money necessary toobtain the same amount ofgoods or services before the inflated price was present ... Inflation decreasesthe purchasing ability of money in that less goods or services can be purchased for the sameone unit of money ... Thevalue of money has decreased, and as a result, it takes more money for the sameamount ofgoods or services ... To make comparisons between monetary amountsthat occur in different time periods, the different-valued money first must be converted to constant-value money in order to represent the same purchasing power over time ... Money in one period of time t1 can be brought to� ————————————— inflation rate between t1 and t2[14 ... Dollars in period t2 are called future dollars or then-current dollars and have inflationtaken into account ... 1] isfuture the same value as money in another periodof time t2 by using the equationamount in period t2Amount in period t1f )n[14 ... 3]We can express future dollars in terms of constant-value dollars, and vice versa, by applying thelast two equations ... As an illustration, use the price of a cheese pizza ... 99March 2011If inflation ؉ � constant-value dollars(1 f )nFuture dollars ؉ 1)� —————— dollarsConstant-value dollarson food prices averaged 5% during the last year, in constant-value 2010 dollars, thiscost is last year’s equivalent of$8 ... 05 ϭ $8 ... 3], is$8 ... 05) ϭ $9 ... 44 in 2012 buys exactly the same cheese pizza as $8 ... If inflationaverages 5% per year over the next 10 years, Equation [14 ... $8 ... 05)10 ϭ $13 ... Insome areas of the world, hyperinflation may average 50% peryear ... 99 to $518 ... Placed into an industrial or business context, at a reasonably low inflation rate averaging 4%per year, equipment or services with a first cost of $209,000 will increase by 48% to $309,000over a 10-year span ... Make nomistake: Inflation is a formidableforce in our economy ... Only the first two are interest rates ... This is the rate at which interest is earned when theeffects of changes in the value of currency (inflation) have been removed ... (The equation used to calculate i, with theinfluence of inflation removed, is derivedlater in Section 14 ... ) The real rate of return thatgenerally applies for individuals is approximately 3 ... This is the “safe investment”rate ... Inflation-adjusted or market interest rate if ... This is the interest rate we hear everyday ... It is also known as the inflated interest rate ... The determinationof this valueis discussed in Section 14 ... Inflation rate f ... Deflation is the opposite of inflation in that when deflation is present, the purchasingpower of the monetary unit is greater in the future than at present ... Inflationoccurs much more commonly than deflation, especially at the national economy level ... Temporaryprice deflation may occur in specific sectors of the economy due to the introduction of improved products, cheaper technology, or imported materials or products thatforce current prices down ... However, deflation over a short time in a specific sector of an economy can beorchestrated through dumping ...The prices will godown for the consumer, thus forcing domestic manufacturers to reduce their prices in orderto compete for business ... Prices may then return to normal levels and, in fact, become inflated over time, if competition has been significantlyreduced ... However, if deflation occurs at a moregeneral14 ... Another result is that individuals and families have less money to spend due to fewer jobs, lesscredit, and fewer loans available; an overall “tighter” money situation prevails ... Inthe extreme case, this can evolve over time into a deflationary spiral that disrupts the entireeconomy ...Engineering economy computations that consider deflation use the same relations as those forinflation ... 2] and [14 ... For example, if deflation is estimated to be 2% per year, an asset that costs $10,000 today would have a first cost 5 years fromnow determined by Equation [14 ... 10,000(1 Ϫ f )n ϭ10,000(0 ... 9039) ϭ $903914 ... The calculations involved in this procedure are illustrated in Table14–1, where the inflation rate is 4% per year ... Column 3 shows the cost in future dollars,and column 4 verifies the cost in constant-value dollars via Equation [14 ... When the futuredollars of column 3 areconverted to constant-value dollars (column 4), the cost is always$5000, the same as the cost at the start ... The actual cost (in inflated dollars) of theitem 4 years from now will be $5849, but in constant-value dollars the cost in 4 years will stillamount to $5000 ... Two conclusions can be drawn ...)And$5000 four years in the future has a PW of only $3415 now in constant-value dollars at a realinterest rate of 10% per year ... The effect of compounded inflation and interest rates can be large, as youcan see by the shaded area ... Consider the P͞Fformula, where i is the real interest rate ... 1͞)3ϭ (4)(P͞F,10%,n)012345000(0 ... 04) ϭ 2085408(0 ... 04) ϭ 2255000520054085624584950005200͞(1 ... 04)2 ϭ 50005624͞(1 ... 04)4 ϭ 500050004545413237573415369370Chapter 14Effects of InflationFigure 14–1Comparison of constantvalue dollars, future dollars, and ( 5)� %01 $, � nPresent Worthat Realiif )n[14 ... 6]where i ϭ real interest ratef ϭ inflation rateFor a real interest rate of 10% per year and an inflation rate of 4% per year, Equation [14 ... 4% ... 10 ϩ 0 ... ؉ � F(P͞F,if ,n)(1 ———— � F their presentworth values ... 2] ... 4](1 ϩ i ϩ f ϩ if )nIf the term i ϩ f ϩ if is defined as if, the equation becomes1P10(0 ... 144Table 14–2 illustrates the use of if ϭ 14 ... As shown in column 4, the present worth for each yearis the same as column 5 of Table 14–1 ... That is, either i or if is introduced into the P͞A, P͞G, or Pg factors, depending upon whether the cash flow is expressed in constant-value (today’s) dollars

orfuture dollars, respectively ... If the cash flow is expressed in future dollars, the PW value is obtained using if ... 2] and then find the PW at the real interest rate i ... 2Present Worth Calculations Adjusted for InflationTABLE 14–2Present Worth Calculation Using an Inflated Interest RateYearn(1)CostinFuture Dollars, $(2)(P͞F,14 ... 87410 ... 66790 ... 1Glyphosate is the active ingredient in the herbicide Roundup® marketed by Monsanto Co ... Contributions to Monsanto’srevenue have been reduced significantly by international dumping of generic glyphosate, asannounced in mid-2010 ... Assume whenthe price was set at $16 per gallon,there was a prediction that in 5 years the price would inflate to $19 per gallon ... (a) Determine the annual rate of inflation over 5 years to increase the price from $16 to $19 ... Compare this resultwith $10 per gallon that Monsanto predicted would be the longer-term price... (d) Determine the market interest rate that must be used in economic equivalencecomputations, if inflation is considered and an 8% per year real return is expectedby Monsanto ... (a) Solve Equation [14 ... 1916 ϭ 19(P͞F,f,5) ϭ ————(1 ϩ f )51 ϩ f ϭ (1 ... 2f ϭ 0 ... 5% per year)(b) If the price deflationrate is 3 ... F ϭ P(F͞P,Ϫ3 ... 035)5ϭ 12(0 ... 04The price will fall to exactly $10 per gallon after 5 years, as Monsanto predicted ... Kilman and I ... 371372Chapter 14Effects of Inflation(c) Five years in the future, at 3 ... 5%,5) ϭ 12(1 ... 1877)ϭ $14 ... 25 versus $16 per gallon) ... 5% per year and a real return of8% per year, Equation [14 ... 78% per year ... 08 ϩ 0 ... 08)(0 ... 1178 (11 ... 2A 15-year $50,000 bond that has a dividend rate of 10% per year, payable semiannually, is currently for sale ... 5% each 6-month period, what is thebond worth now (a) without an adjustment for inflation and (b) when inflation isconsidered?Show both hand and spreadsheet solutions ... 10)]͞2 ϭ $2500 ... if ϭ 0 ... 025 ϩ (0 ... 025) ϭ 0 ... 6%,30) ϩ 50,000(P͞F,6 ... 9244) ϩ 50,000(0 ... Withoutan inflation adjustment, the PV function is developed at the nominal 4% rate for 30 periods;with inflation considered the rate is if ϭ 6 ...CommentThe $18,985 difference in PW values illustrates the tremendous negative impact made by only2 ... 06% per year) ... Yet, this is worth only$39,660 in constant-value dollars ... 2 ... 2Present Worth Calculations Adjusted for InflationEXAMPLE 14 ... She wishes to calculate a project’s PWwithestimated costs of $35,000 now and $7000 per year for 5 years beginning 1 year from nowwith increases of 12% per year thereafter for the next 8 years ... Solution(a) Figure 14–3 presents the cash flows ... 34] and [2 ... PW ϭ Ϫ35,000 Ϫ 7000(P͞A,15%,4)1 ... 15Ϫ ———————— (P͞F,15%,4)0 ... 12{[ ( )]}ϭ Ϫ35,000 Ϫ 19,985 Ϫ 28,247ϭ $Ϫ83,232In the P͞A factor, n ϭ 4 because the $7000 cost in year 5 is the A1 term in Equation [2 ... PW = ?01PWg = ?24530$7000i = 15% per year61273849 10 11 12 1356789YearGeometric series year$7840$35,000$17,33112% increaseper yearFigure 14–3Cash flowdiagram, Example 14 ... (b) To adjust for inflation, calculate the inflated interest rate by Equation [14 ... if ϭ 0 ... 11 ϩ (0 ... 11) ϭ 0 ... 65%,4)1 ... 2765Ϫ ————————— (P͞F,27 ... 2765 Ϫ 0 ... 2545) Ϫ 30,945(0 ... The present value of future inflateddollars is significantly less when the inflation adjustmentis included ... 373374Chapter 14Effects of InflationExamples 14 ... 3 above add credence to the “buy now, pay later” philosophy ... If cash is not readily available at that time, the debts cannot be repaid ... In the longer term, thisbuy now, pay later approach must be tempered with sound financial practicesnow, and in the future ... 3 Future Worth CalculationsAdjusted for InflationIn future worth calculations, a future amount F can have any one of four different interpretations:Case 1 ... Case 2 ... Case 3 ... Case 4 ... Depending upon which interpretation is intended, the F value is calculated differently, if)n P (F͞P, if ,n)F ؉ asdescribed below ... Case 1: Actual Amount Accumulated It should be clear that F, the actual amount of moneyaccumulated, is obtained using the inflation-adjusted (market) interest rate ... 7]For example, when we quote a market rate of 10%, the inflation rate is included ... P(1f )n[14 ... The percentage loss in purchasing power is a measure of how much less ... In 7 years, the purchasing power has risen, but only to $1481 ... 3159(1 ... Therefore, weconclude that 4% inflation over 7 years reduces the purchasing power of money by 24% ؉ f) n(1 ؉ 1)� ———— � —————— f[14 ... 3Future Worth Calculations Adjusted for InflationThe real interest rate i represents the rate at which today’s dollars expand with their same purchasing power into equivalent future dollars ... The use of this؉1� ——— fi ؊ ... This real interest rate is the i in Equation [14 ... if ϭ i ϩ f ϩ ifϭ i(1 ϩ f ) ϩ fifinterest rate is appropriate for calculating the future worth of an investment (such as a savings account or money market fund) whenthe effect of inflation must be removed ... 9]0 ... 04i ϭ ————— ϭ 0 ... 04(5 ... 77%,7) ϭ $1481The market interest rate of 10% per year has been reduced to a real rate thatis less than 6% peryear because of the erosive effects of 4% per year inflation ... Simply put, future dollars are worth less, so more are needed ... This is the situation if someone asks, “Howmuch will a car cost in 5 years if its current cost is $20,000 and its price will increase by the inflation rate of 6% peryear?” (The answer is $26,765 ... To find the future cost, substitute f for the interest rate in the F͞P factor ... 10]Reconsider the $1000 used previously ... Maintaining purchasing power and earning interest must account for both increasingprices (case 3) and the time value of money ... Thus, to make arealrate of return of 5 ... For the same $1000 amount,if ϭ 0 ... 04 ϩ 0 ... 04) ϭ 0 ... 77% per year and inflation of f ϭ 4% per year ... The calculations made in this section explain the following:••••The amount of $1000 now at a market rate of 10% per year will accumulate to $1948 in 7 years ... An item with acost of $1000 now will cost $1316 in 7 years at an inflation rate of 4% per year ... 77%with inflation considered at 4% per year ... 5% mentioned earlier ... Define the symbol MARRf as the inflation-adjusted or marketMARR, which is calculated in a fashion similar to if ... 11]375376Chapter 14TA BLE 14–%01 , � 7,if � 0001$ , n 3Effects of InflationCalculation Methods for Various Future Worth InterpretationsFuture WorthDesiredCase 1: Actual dollarsaccumulatedUse stated marketrate if in equivalenceformulasCase 2: Purchasing powerof accumulated dollars interms of constant-valuedollarsExample forP� %4Method ofCalculationUse market rate if inequivalence anddivide by (1 ϩ f )norUse real iF ϭ 1000(F͞P,10%,7)1000 (F͞P,10%,7)F ϭ ————————(1 ... 77%,7)Case 3: Dollars required forsame purchasing powerUse f in place of i inequivalenceformulasF ϭ 1000(F͞P,4%,7)Case 4: Future dollars � ftomaintain purchasingpower and to earn a returnCalculate if and use inequivalenceformulasF ϭ 1000(F͞P,10%,7)The real rate of return i used here is the required rate for the corporation relative to its cost ofcapital ... Theinflation-adjusted MARR is calculated by including the inflation rate of, say, 4% peryear ... 11] ... 13 ϩ 0 ... 13(0 ... 1752(17 ... 4Abbott Mining Systems wants to determine whether it should upgrade a piece of equipmentused in deep mining operations in one of its international operations now or later ... However, if the company selects plan I, the purchase will be deferred for 3 years whenthe cost is expected to riseto $300,000 ... The inflation ratein the country has averaged 3% per year ... Solution(a) Inflation not considered: The real rate, or MARR, is i ϭ 12% per year ... Calculate the FW value for plan A three years from now andselect the lower cost ... (b) Inflation considered: This iscase 4; the real rate (12%), and inflation of 3% must beaccounted for ... 11] ... 12 ϩ 0 ... 12(0 ... 153614 ... FWA ϭ Ϫ200,000(F͞P,15 ... The inflation rate of 3% per year has raised the equivalent future worth of costs by 9 ... This is the same as an increase of 3% per year, compoundedover 3 years, or (1 ...3% ... , are present ... In these cases, the government may take drastic actions: redefine the currency in terms ofthe currency of another country, control banks and corporations, and control the flow of capitalinto and out of the country in order to decrease inflation ... To appreciatethe disastrous effect ofhyperinflation on a company’s ability to keep up, we can rework Example 14 ... The FWA amount skyrockets and plan I is aclear choice ... Good economic decisions in a hyperinflated economy are very difficult to make using traditional engineering economy methods, since the estimated future values aretotally unreliableand the future availability of capital is uncertain ... 4 Capital Recovery CalculationsAdjusted for InflationIt is particularly important in capital recovery (CR) calculations used for AW analysis to includeinflation because current capital dollars must be recovered with future inflated dollars ...This suggests the use of the inflated interest rate in theA͞P formula ... 8%,5) ϭ $325 ... This suggests the use of a higher interest rate, that is, the if rate, to produce a lower A value in theA͞F formula ... 8%,5) ϭ $137 ... 80 ... 59 versus $163 ... 377378Chapter 14Effects of InflationEXAMPLE 14 ... 58 today, ifthe market interest rate is 10% per year and inflation is8% per year?SolutionFirst, find the actual number of inflated dollars required 5 years in the future that is equivalentto $680 ... This is case 3; Equation [14 ... F ϭ (present purchasing power)(1 ϩ f )5 ϭ 680 ... 08)5 ϭ $1000The actual amount of theannual deposit is calculated using the market interest rate of 10% ... A ϭ 1000(A͞F,10%,5) ϭ $163 ... 85% as determined using Equation [14 ... To put these calculations into perspective, if the inflation rate is zero when the real interest rate is 1 ... 58 today is obviously $680 ... Then the annual amountrequired to accumulate this future amount in 5 years is A ϭ680 ... 85%,5) ϭ $131 ... This is $32 ... 80 calculated above forf ϭ 8% ... To make up thepurchasing power difference, more higher-value dollars are required ... 63 per year is required ... People tend to pay off less of their incurred debt at eachpayment because theyuse any excess money to purchase additional items before the price is further inflated ... All this is due to the spiraling effect of increasing inflation ... CHAPTER SUMMARYInflation, treated computationally as an interest rate, makes the cost of the same product or service increaseover time due to the decreased value of money ... Some important relations are the following:Inflated interest rate: if ϭ i ϩ f ϩ ifReal interest rate: i ϭ (if − f )͞(1 ϩ f)PW of a future amount with inflation considered: P ϭ F(P͞F, if, n)Future worth in constant-value dollars of a present amount with the samepurchasing power:F ϭ P(F͞P,i,n)Future amount to cover a current amount with inflation only: F ϭ P(F͞P, f, n)Future amount to cover a current amount with inflation and interest: F ϭ P(F͞P,if , n)Annual equivalent of a future amount: A ϭ F(A͞F,if , n)Annual equivalent of a present amount in future dollars: A ϭP(A͞P,if , n)Hyperinflation implies very high f values ... This can, and usually does, cause a national financial disaster when it continues over extended periods of time ... 1 What is the difference between today’s dollars andconstant-value dollars (a) when using today as thereference point in time and (b)when using 2 yearsago as the reference point?14 ... 14 ... 4 Determine the inflation-adjusted interest rate for agrowth company that wants to earn a real rate ofreturn of 20% per year when the inflation rate is5% per year ... 5 For a high-growth company that wants to make areal rate of return of 30% peryear, compoundedmonthly, determine the inflation-adjusted nominalinterest rate per year ... 5% per month ... 6 A high-tech company whose stock trades on theNASDAQ stock exchange uses a MARR of 35% peryear ... 7 Calculate the inflation-adjusted interest rate perquarter when the real interest rate is4% per quarterand the inflation rate is 1% per quarter ... 8 Calculate the real interest rate per month if thenominal inflation-adjusted interest rate per year,compounded monthly, is 18% and the inflation rateper month is 0 ... 14 ... The contract price is fixed at$45,000 per year for 4 years ... 14 ... 11 Assumethat you want to retire 30 years from nowwith an amount of money that will have the samevalue (same purchasing power) as $1 ... If you estimate the inflation rate will be 4%per year, how many future (then-current) dollarswill you need?Adjusting for Inflation14 ... 13 The inflation rate in a Central Americancountry is6% per year ... 14 During periods of hyperinflation, prices increaserapidly over short periods of time ... 6 billion ... 4 billion ... 15 A trust was set up by your grandfather that statesyou are to receive $250,000 exactly 5 years fromtoday ... 14 ... Determine (a) the number of constant-value dollars 5years in the future that is equivalent to$30,000 now and (b) the number of future dollarsthat will be equivalent to $30,000 now ... 17 The Pell Grant program of the federal governmentprovides financial aid to needy college students ... 18 Ford Motor Company announced that the price ofits F-150 pickuptrucks is going to increase by onlythe inflation rate for the next 3 years ... 1% per year, what is theexpected price of a comparably equipped trucknext year? 3 years from now?14 ... The cost of the dampeners today is $120,000,but the company has to wait until a permit is approved for its bidirectional port-to-plant productpipeline ... Because of intense foreign competition, the manufacturer plans to increase the price only by the inflation rate each year ... 8% peryear and the company’s MARR is 20% per year,estimate the cost of the dampeners in 2 years interms of (a) today’s dollars and (b) future dollars ...20 A machine currently under consideration byHolzmann Industries has a cost of $45,000 ... When the purchasing managerchecked the invoice for the machine he purchased5 years ago, he saw that the price was $29,000 ... 21 A report by the National Center for Public Policyand Higher Education statedthat tuition and fees(T&F) at public colleges and universities increasedby 439% over the last 25 years ... When the report was written, T&F at a4-year public university constituted 28% of theMFI of $52,000 (tuition and fees at a private university constituted 76% of MFI) ... 22 The headline on a Chronicle ofHigher Educationarticle reads “College Costs Rise Faster than Inflation ... (a) What was the average annual percentage increase over that period of time? (b) If thereal increase in tuition (i ... , without inflation) was5% per year, what was the inflation rate per year?14 ... ” At the University ofKansas,Jayhawks fans can sign up to pay $105,000 over10 years for the right to buy top seats for footballduring the next 30 years ... Season tickets in tier 1 are currently selling for $350 ... What is the dollar amount ofthe savings on the tickets (with no interest considered), if ticket prices rise at a rate of3% per yearfor the next 30 years?Present Worth Calculations with Inflation14 ... What are they?14 ... At a real interest rate of10% per year and inflation rate of 4% per year, whatis the present worth of the cost of the equipment?14 ... The supplier quoted a price of$125,000 if the unit is purchased within thenext3 years ... Assuming the tower will not be purchased for 3 years, calculate the present worth atan interest rate of 10% per year and an inflationrate of 4% per year ... 27 How much can the manufacturer of superconducting magnetic energy storage systems afford tospend now on new equipment in lieuof spending$75,000 four years from now? The company’s realMARR is 12% per year, and the inflation rate is3% per year ... 28 Find the present worth of the cash flows shown ... Assume a realinterest rate of 8% per year and an inflation rate of6% per year ... 29 A doctor is on contract to a medium-sizedoil company to provide medical services at remotely located, widely separated refineries ... The doctorcan buy a used Lear jet now or wait for a new verylight jet (VLJ) that will be available 3 years fromnow ... 9 million, payable when the plane is delivered in 3 years ... If theMARR is 15% per year and theinflation rate is projected to be 3% per year, what is the present worth ofthe VLJ with inflation considered?14 ... The HDD machine will include an innovative pipe loader design andmaneuverable undercarriage system ... At a real MARR of10% per year and an inflation rate of 5% per year,determine if thecompany should buy now or later(a) without any adjustment for inflation and(b) with inflation considered ... 31 An engineer must recommend one of two rapidprototyping machines for integration into an upgraded manufacturing line ... SalesmanA gave her the estimates in constant-value(today’s) dollars,while saleswoman B providedthe estimates in future (then-current) dollars ... Use PW analysis to determine whichmachine the engineer should recommend ... 32 A salesman from Industrial Water Services (IWS),who is trying to get his foot in the door for a largeaccount in Fremont, offered waterchlorinationequipment for $2 ... This is $400,000 morethan the price offered by a competing saleswomanfrom AG Enterprises ... If the equipment has381a 2-year warranty, determine which offer is better ... 14 ... A small pipelinewill cost less to purchase (including valves andother appurtenances) but willhave a high head lossand, therefore, a higher pumping cost ... Determine present worth values if futureworth values are FWS ϭ $2 ... 5 million ... 4% permonth ... 14 ... 1 million 5 years fromnow, or (3) pay an amount of money 5 years fromnow that will have the same purchasing power as$850,000 now ...35 If the inflation rate is 6% per year and a personwants to earn a true (real) interest rate of 10% peryear, determine the number of future dollars shehas to receive 10 years from now if the present investment is $10,000 ... 36 How many future dollars would you need 5 yearsfrom now just to have the samebuying power as$50,000 now, if the deflation rate is 3% per year?14 ... 38 Harmony Corporation plans to set aside $60,000per year beginning 1 year from now for replacingequipment 5 years from now ... 39 The strategic plan of a solar energy company thatmanufactures high-efficiency solar cellsincludesan expansion of its physical plant in 4 years ... If the company sets aside$7,000,000 now into an account that earns interestat 7% per year, what will the inflation rate have tobe in order for the company to have exactly theright amount of money for the expansion?14 ... The account is now worth$25,000 ... 41 A Toyota Tundra can be purchased today for$32,350 ... If the price of thetruck increases exactly in accordance with an estimated inflation rate of 3 ... 14 ... In part, his will stated: “Thecapital shall be invested by my executors in safesecurities and shall constitute a fund, the intereston whichshall be annually distributed in the formof prizes to those who, during the preceding year,shall have conferred the greatest benefit on mankind ... In addition to agold medal and a diploma, each recipient receivesa substantial sum of money that depends on theFoundation’s income that year ... In 1996, theaward was $653,000; it was$1 ... (a) If the increase between 1996 and 2009 wasstrictly due to inflation, what was the averageinflation rate per year during that 13-yearperiod?(b) If the Foundation expects to invest moneywith a return of 5% above the inflation rate,how much will a laureate receive in2020,provided the inflation rate averages 3% peryear between 2009 and 2020?Effects of Inflation14 ... The companyis planning to add larger-capacity robotic arms toone of its assembly lines 3 years from now ... 4 million ... 8% per year ... 44 The data below show two patterns of inflation thatare exactlythe opposite of each other over a20-year time period ... 45 Factors that increase costs and prices—especiallyfor materials and manufacturing costs sensitive tomarket, technology, and labor availability—can beconsidered separately using the real interest rate i,the inflation rate f, and additional increasesthatgrow at a geometric rate g ... Thegeometric rate is the same one used in the geometric series (Chapter 2) ... This is over and above the inflation rate ... 46 An electric utility is considering two alternativesfor satisfying state regulations regarding pollution control for one of its generating stations ... S ...The station iscurrently producing excess VOCs and oxides ofnitrogen ... Plan A involves replacingthe burners and switching from fuel oil to naturalgas ... Plan B involves going to the foreign cityand running gas lines to many of the “backyard”brick-making sites that now use wood, tires, andothercombustible waste materials for firing thebricks ... S ... The initial cost of plan B will be$1 ... Additionally, the electric company would subsidizethe cost of gas for the brick makers to the extentof $200,000 per year ... For a 10-year project periodand no salvage value for either plan, which oneshould beselected on the basis of an annualworth analysis at a real interest rate of 7% peryear and an inflation rate of 4% per year?14 ... Her goal is to save enoughmoney over the next 3 years so that when she begins her trip, the amount she has accumulated willhave the same buying power as $72,000 today ...48 An entrepreneur engaged in wildcat oil well drilling is seeking investors who will put up $500,000for an opportunity to reap high returns if the venture is successful ... How much will the investors have toreceive each year to recover their money if aninflation rate of 5% per year is to be included inthecalculation?14 ... manufactures in situ calibrationverification systems that confirm flow measurement accuracies without removing the meters ... If the company’s real MARR is 15%per year, which process has the lower annual costwhen inflation of 5% per year is considered?Process XFirst cost,$Operating cost, $ per yearSalvage value, $Life, yearsProcess YϪ65,000Ϫ40,00005Ϫ90,000Ϫ34,00010,000514 ... The company is considering a newannealing-drawing process to reduce costs ... 7 million now, how muchmust be saved each year to recover the investmentin 5 years if the company’sMARR is a real 12%per year and the inflation rate is 3% per year?14 ... Determine the equivalent cost of the system if the realinterest rate is 10% per year and the inflation rateis 4% per year ... 52 Maintenance costs for pollution control equipmenton a pulverized coal cyclone furnace are expectedto be$180,000 now and another $70,000 threeyears from now ... If the companyuses a real interest rate of 9% per year and the inflation rate averages 3% per year, what is theequivalent annual cost of the equipment?14 ... 5 million for 5 years to finance start-up costs for a new project involvingsite reclamation... The averageinflation rate is 5% per year ... (b) Determine the capital recovery if the company is satisfied with accumulating $2 ... (c) Determine the capital recovery in part (b)without considering inflation ... 54 Inflation occurs when:(a) Productivity increases(b) The value of the currency decreases(c) Thevalue of the currency increases(d) The price of gold decreases14 ... 56 To calculate how much something will cost if youexpect its cost to increase by exactly the inflationrate, you should:(a) Multiply by (1 ϩ f )n(b) Multiply by [(1 ϩ f )n͞(1 ϩ i)n](c) Divide by (1 ϩ f )n(d) Multiply by (1 ϩ if )n14 ... 4%(b) 7%(c)9%(d) 15 ... 58 The market interest rate if is 12% per year, compounded semiannually ... 59 Construction equipment has a cost today of$40,000 ... 60 The amount of money that would be accumulatednow from an investment of $1000 25 years ago ata market rate of 5% per year and an inflationrateaveraging 2% per year over that time period isclosest to:(a) $1640(b) $3385(c) $5430(d) Over $550014 ... 62 If you expect to receive a gift of $50,000 six yearsfrom now, the present worth of the gift at a realinterest rate of 4% per year and an inflation rate of3% per year is closest to:(a) $27,600(b)$29,800(c) $33,100(d) $37,20014 ... For a real interest rate of 5%per year and an inflation rate of 4% per year, theannual capital recovery requirement for the equipment (in future dollars) is determined by:(a) AW ϭ Ϫ30,000(A͞P,4%,20) Ϫ 7000ϩ 5000(A͞F,4%,20)(b) AW ϭ Ϫ30,000(A͞P,5%,20) Ϫ 7000ϩ5000(A͞F,5%,20)(c) AW ϭ Ϫ30,000(A͞P,9%,20) Ϫ 7000ϩ 5000(A͞F,9%,20)(d) AW ϭ Ϫ30,000(A͞P,9 ... 2%,20)14 ... Wheninflation is moderately high, bonds offer a low return relativeto stocks, because the potential for market growth is not present with bonds ... However, bonds do offer a steady income thatmay beimportant to an individual, and they serve to preserve theprincipal invested in the bond, because the face value is returned at maturity ... He has a collection of stocks in hisretirement portfolio, but no bonds ... ” He can choose additionalstocks or bonds, but has decided to not split the $50,000between the two forms of investments ... He assumes the effects of federal and state income taxes willbe the same for both forms of investment ... Bond purchase: If he purchased a bond, he would have apredictable income of 5% per year and the $50,000 facevalue after the 12-year maturity period ...Can you answer them for him for both choices?1 ... If he decided to sell the stock or bond immediately afterthe fifth annual dividend, what is his minimum sellingprice to realize a 7% real return? Include an adjustmentof 4% per year for inflation ... If Earl needed some money in the future, say, immediatelyafter the fifth dividend payment, what would bethe minimum selling price in future dollars, if he wereonly interested in recovering an amount that maintainedthe purchasing power of the original price?4 ... Earl plans to keep the stocks or bonds for 12 years, thatis, until the bond matures ... For what amountmust hesell the stocks after 12 years, or buy the bonds now toensure he realizes this return? Do these amounts seemreasonable to you, given your knowledge of the waythat stocks and bonds are bought and sold?CHAPTER 15CostEstimation andIndirect CostAllocationL E A R N I N GO U T C O M ESPurpose: Make cost estimates using different methods; demonstrate the allocation of indirect costs using traditional andactivity-based costing rates ... 1Approach• Explain the bottom-up and design-to-cost (topdown) approaches to cost estimation ... 2Unit method• Use the unit method to make apreliminary costestimate ... 3Cost index• Use a cost index to estimate a present cost basedon historical data ... 4Cost capacity• Use a cost-capacity equation to estimatecomponent, system, or plant costs ... 5Factor method• Estimate total plant cost using the factormethod ... 6Indirect cost rates• Allocateindirect costs using traditional indirectcost rates ... 7ABC allocation• Use the Activity-Based Costing (ABC) method toallocate indirect costs ... 8Ethics and profit• Describe how biased estimation can become anethical dilemma ... In reality, they are not; they must be estimated ... Cost estimation is importantin all aspects of a project, but especially in the stages of project conception, preliminary design, detailed design, and economic analysis ... Inengineering practice, the estimation of costs receives much more attention than revenueestimation; costs are the topic of this chapter ... Therefore, allocation ofindirect costs for functionssuch as utilities, safety, management and administration, purchasing, and quality is madeusing some rational basis ... 15 ... In general, most cost estimates are developed for either aproject or a system; however, combinations of these are very common ... A system is usually anoperational design that involves processes, services, software, and other nonphysical items ... Of course, many projects will have major elements that are not physical, so estimates of both types must be developed ... There would be no operational system if only the costs of computer hardware plus wireandwireless connectors were estimated; it is equally important to estimate the software, personnel, and maintenance costs ... In real-world practice, the cash flows forcosts and revenues must be estimated prior to the evaluation of a project or comparison of alternatives ... Revenue estimates utilized byengineers are usually developed in marketing,sales, and other departments ... Normally direct costs are estimated with some detail, then the indirect costs are added using standard rates and factors ... Accordingly, many industrial settings require some estimating for indirect costsas well ... Primarily, directcosts are discussed here ... ••••What cost components must be estimated?What approach to cost estimation will be applied?How accurate should the estimates be?What estimation techniques will be utilized?Costs to Estimate If a project revolves around a single piece of equipment, for example,anindustrial robot, the cost components will be significantly simpler and fewer than the componentsfor a complete system such as the manufacturing and testing line for a new product ... Examples of costcomponents are the first cost P and the annual operating cost (AOC), also called the M&OcostsDirect͞Indirect costs388Chapter 15Cost Estimation and Indirect Cost Allocation(maintenance and operating) of equipment ... Listed below are sample elements of the first cost and AOC components ... AOC component (part of the equivalent annual cost A):Elements: Direct labor cost for operatingpersonnelDirect materialsMaintenance costs (daily, periodic, repairs, etc ... When costs for an entire system must be estimated, the number of cost components and elements is likely to be in the hundreds ... For familiar projects (houses, office buildings, highways, and some chemical plants) therearestandard cost estimation software packages available ... ) and estimates costs with time-proven, built-in relations ... However, there are no “canned” softwarepackages for a large percentage of industrial, business, service and public sector projects ... For a simple rendition of this approach, seeFigure15–1 (left side) ... Theprice is then determined by adding indirect costs and the profit margin, which is usually a percentage of the total cost ... This approach works well when competition is not a dominant factor in pricing the product or service ... The competitive price establishes the target cost ... Thisapproach is useful in encouraging innovation,new design, manufacturing process improvement, and efficiency ... This approach places greater emphasis on the accuracy of the price estimation activity ... The designto-cost approach is best applied in the early stages of a new or enhanced product design... Usually, the resulting approach is some combination of these two philosophies ... Historically, the bottom-upapproach was more predominant in Western engineering cultures, especially in North America ... 1389Understanding How Cost Estimation Is AccomplishedFigure 15–1Top-downapproachRequiredpriceCompetitivepriceDesiredprofitϩAllowedprofitϪTotal cost ϭϭ Target costIndirectcostsϩϩIndirectcostsDirectcostsMaintenanceand operationsϩϩMaintenanceand operationsDirectlaborϩϩDirectlaborDirectmaterialsϩϩDirectmaterialsEquipment andϩcapital recoveryϩEquipmentcapitalrecoveryCostcomponentestimatesAt design stageBefore design stageBottom-up approachCostcomponentestimatesDesign-to-cost approachglobalization of engineering design has speeded the adoption of the design-to-cost approachworldwide ... The accuracy required increases as the projectprogresses from preliminary design to detailed design and on toeconomic evaluation ... When utilized at early and conceptual design stages, estimates are referred to as order-ofmagnitude estimates and generally range within Ϯ20% of actual cost ... Every project setting has its own characteristics, but arange of Ϯ5% of actualcosts is expected at the detailed design stage ... Obviously, the desire for better accuracy has to be balanced against the cost of obtaining it ... Simplified cost estimationprocesses for bottom-upand top-down approaches ... The use of the unit method and cost indexes basethepresent estimate on past cost experiences, with inflation considered ... They are called cost estimating relationships (CERs) ... Most cost estimates made in a professional setting are accomplished in part or wholly usingsoftware packages linked to updated databases that contain cost indexes and rates forthe locations, products, or processes being studied ... Corporations usually standardize on one or two packages to ensure consistency over time and projects ... 2 Unit MethodThe unit method is a popular preliminary estimation technique applicable to virtually all professions ... CT ϭ u ϫ N[15 ... Somesample unit cost factors (and values) areTotal average cost of operating an automobile (52¢ per mile)Cost to bury fiber cable in a suburban area ($30,000 per mile)Cost to construct a parking space in a parking garage ($4500 per space)Cost of constructing interstate highway ($6 ... If house constructioncostsaverage $225 per square foot, a preliminary cost estimate for an 1800-square-foot house, usingEquation [15 ... Similarly, a 200-mile trip should cost about $104 for the car onlyat 52¢ per mile ... This is illustrated in Example 15 ... EXAMPLE 15 ... Since a Ϯ20% estimate is acceptable at thisN, $MaterialsMachinery, toolingLabor, castingLabor, � preliminary stage, a unitmethod estimate is sufficient ... Materials: 3000 tons at $45 ... 1] to each of the five areas and sum the results to obtain the total costestimate of $566,700 ... 15 ... 1ResourceAmount NUnit Cost Factor u, $Cost Estimate, ufinishingLabor, indirect3000 tons1500 hours3000 hours1200 hours400 hours45 ... 3 Cost IndexesThis section explains indexes and their use in cost estimation ... A preliminary cost estimate is oftenbased on a cost index ... The index isdimensionless and measures relative cost change over time ... ”Timely updating of the index is very important ... This index includes such items as rent, food, transportation, and certain services ... Table 15–2 is a listing of some of the more common indexes ... S ... S ... 1381 ... 5389 ... 6394 ... 3395 ... 7444 ... 2499 ... 4575 ... 9555 ... 51039 ... 81061 ... 31089 ... 91104... 61178 ... 51302 ... 31449 ... 61461 ... 2]where Ct ϭ estimated cost at present time tC0 ϭ cost at previous time t0It ϭ index value at time tI0 ϭ index value at time t0Generally, the indexes for equipment and materials are made up of a mix of components thatare assigned certain weights, with thecomponents sometimes further subdivided into more basicitems ... These subcomponents, in turn, are built up from even more basic items such as pressure pipe, black pipe, and galvanized pipe ... The base period of 1957 to1959 is assigned a value of 100 for the CEPCI, 1913 ϭ 100 for the ENR index,and 1926 ϭ 100for the M&S equipment cost index ... For example, the CE plant cost index is available at www ... com͞pci ... construction ... This latter site offers a comprehensiveseries of construction-related resources, including several ENR cost indexes and cost estimationsystems ... eng-tips ...EXAMPLE 15 ... The engineer finds that a project of similar complexity and magnitude was completed 5 years ago at a skilled labor cost of $360,000 ... What is the estimated skilled labor cost forthe new project?15 ... Using Equation [15 ... The cost index will vary, perhaps with the region of the country,the type of product or service,and many other factors ... The development of the cost index requires the actualcost at different times for a prescribed quantity and quality of the item ... The index each year (period) is determined as the cost divided by the base-year cost and multiplied by 100 (or 1) ...EXAMPLE 15 ... He has decided to expand into newareas and wants to make cost estimates for three of the more significant labor costs involvedin making these types of films ... (a) Make 2008 the base year, and determine the cost indexes using a basis of 1 ... Commenton the trend of each index overthe years ... The cost in 2010 was $78 per hour;assume a worst-case scenario is that the graphics index continues the same arithmetic trendit had from 2010 to 2011 ... Solution(a) For each type of service, calculate It͞I0 where t ϭ 2005, 2006, ... Table 15–5 presents the indexes ... Stuntmen labor cost:Rising until 2009, then stable ... TA BLE 15–4Average Hourly Costs for Three Services, Example 15 ... 3Index It ͞I0Type of Service2005200620072008200920102011GraphicsStuntmenActors0 ... 710 ... 840 ... 890 ... 861 ... 001 ... 001 ... 240 ... 161 ... 831 ... 210 ... 16 ... 18; the index value in 2014 will be1... 18) ϭ 1 ... Equation [15 ... C2014 ϭ C2010(I2014͞I2010) ϭ 78(1 ... 16)ϭ 78(1 ... 4 Cost-Estimating Relationships:Cost-Capacity EquationsDesign variables (speed, weight, thrust, physical size, etc ... Cost-estimating relationships (CERs) use thesedesign variables to predict costs ... One of the most widelyused CER models is a cost-capacity equation ... This is also knownas the power law and sizing model ... 3]where C1 ϭ cost at capacity Q1C2 ϭ cost at capacity Q2x ϭ correlating exponentThe value of the exponent for various components, systems, or entire plants can be obtained orderived from anumber of sources, including Plant Design and Economics for Chemical Engineers, Preliminary Plant Design in Chemical Engineering, Chemical Engineers’ Handbook,technical journals (especially Chemical Engineering), the U ... Environmental Protection Agency,professional or trade organizations,consulting firms, handbooks, and equipment companies ... When an exponent value for a particular unit is not known, it is common practice to use the value of x ϭ 0 ... Infact, in the chemical processing industry, Equation [15 ... The exponent x in the cost-capacity equation is commonly in the range 0 Ͻ xՅ 1 ... If x ϭ 1, a linear relationship is present ... It is especially powerful to combine the time adjustment of the cost index (It͞I0) from Equation [15 ... If the index isembedded into the cost-capacity computation in Equation [15 ... C2,t ϭ (cost at time 0 of level 2) ϫ (time adjustment cost index)[ ( ) ] ( II )Q2ϭC1,0 ——Q1xt—0This is commonly expressed without the time subscripts ... [15 ... 5TABLE 15–6Cost-Estimating Relationships: Factor MethodSample Exponent Values for Cost-Capacity EquationsSize RangeExponent1–100 MGD0 ... 1–100 MGD5–300 hp200–2100 hp20–8000 ft3͞min15–400 ft20 ... 05–50 MGD0 ... 04–5 MGD0 ... 2 MGD100–2000 gal0 ... 140 ... 710 ... 980 ... 320 ... 710 ... 550 ... 021 ... 690 ... 350 ... 67Component͞System͞PlantActivated sludge plantAerobic digesterBlowerCentrifugeChlorine plantClarifierCompressor, reciprocating (air service)CompressorCyclone separatorDryerFilter,sandHeat exchangerHydrogen plantLaboratoryLagoon, aeratedPump, centrifugalReactorSludge drying bedsStabilization pondTank, stainlessNote: MGD ϭ million gallons per day; hp ϭ horsepower; scfd ϭ standard cubic feet per day ... 4The total design and construction cost for a digester to handle a flowrate of 0 ... 7 million in 2010 ... 0 MGD ... 2 to 40 is 0 ... The cost index in 2010of 131 has been updated to 225 for this year ... 3] can estimate the cost of the larger system in 2010, but it must be updated bythe cost index to today’s dollars ... 4] performs both operations at once ... 0 0 ... 5ϭ 1,700,000(1 ...718) ϭ $3 ... 5 Cost-Estimating Relationships:Factor MethodAnother widely used model for preliminary cost estimates of process plants is called the factormethod ... The method is based on the premise that fairly reliable total plant costs can beobtained by multiplying the cost of the major equipment by� hCE[15 ... This follows the cost estimation approaches certain factors ... These factors are commonly referred to as Lang factors after Hans J ... 395396Chapter 15Cost Estimation and Indirect Cost AllocationIn its simplest form, the factor method is expressed in the same form as the unit methodCTpresented in Figure 15–1 ... 10;solid-fluid process plants, 3 ... 74 ... EXAMPLE 15 ... 08 million ... 63, estimate the plant’s total cost ... 5] ... 63(2,080,000)ϭ $7,550,400Subsequent refinements of the factor method have led to the development of separate factorsfor direct and indirect cost components ... 1are specificallyidentifiable with a product, function, or process ... Examples of indirect costs are general administration, computer services, quality, safety,taxes, security, and a variety of support functions ... 1 ... For indirect costs, some of the factors apply to equipment costs only, while others apply tofI)[15 ... 6Traditional Indirect Cost Rates and AllocationEXAMPLE 15 ... If the direct cost factor is 1 ... 25, determine the total plant cost ... 6] ... 61 ϩ 0 ... 86The total plant cost isCT ϭ 2 ؉ thetotal direct cost ... The overall cost factor h can bewritten asnhϭ1ϩ͚f[15 ... Therefore, Equation [15 ... 6 and 15 ... )](1... 7A new container-handling crane at the Port of Singapore is expected to have a deliveredequipment cost of $875,000 ... , is 0 ... The construction factor is 0 ... 21 ... Solution(a) Total equipment cost is $875,000 ... 6] ish ϭ 1 ϩ 0 ... 53 ϩ 0 ... 23The total cost isCT ϭ 2 ... 7] is used to estimate totalcost ...49 ϩ 0 ... 02iiϭ1CT ϭ [875,000(2 ... 21) ϭ $2,138,675CommentNote the decrease in estimated cost when the indirect cost is applied to the equipment cost onlyin part (a) ... 15 ... For the manufacturing environment, it can be stated generally that thestatement of cost of goods sold (discussed in Appendix B)is one end product of this system ... All costs incurred in one departmentor process line are collected under a cost center title, for example, Department 3X ... Of course, this in itself is no easy chore, and the cost of the trackingsystem may prohibit collection of all direct cost data to the level of detail desired... 397398Chapter 15Cost Estimation and Indirect Cost AllocationTA BLE 15–7Sample Indirect Cost Allocation BasesCost CategoryTaxesHeat, lightPowerReceiving, purchasingPersonnel, machine shopBuilding maintenanceSoftwareQuality controlIndirect costsPossible Allocation BasisSpaceoccupiedSpace, usage, number of outletsSpace, direct labor hours, horsepower-hours, machine hoursCost of materials, number of orders, number of itemsDirect labor hours, direct labor costSpace occupied, direct labor costNumber of accessesNumber of inspectionsIndirect costs are costs associatedwith property taxes, service and maintenance departments, personnel, legal, quality, supervision, purchasing, utilities, software development,etc ... Detailed collection of these data iscost-prohibitive and often impossible; thus, allocation schemes are utilized to distribute theexpenses on a reasonable basis� ————————————— estimated basis level[15 ... For example, if a division hastwo producing ... Historically, common bases havebeen direct labor cost, direct labor hours, machine-hours, number of employees, space, and directmaterials ... estimated total indirect costsIndirect cost ratedepartments, the total indirect cost allocated to a department is used as the numerator in Equation [15 ... Example 15 ... EXAMPLE 15 ... The following information was obtained from last year’s budget for the three machines ... Cost SourceAllocation BasisMachine 1Machine 2Machine 3Direct laborcostDirect labor hoursDirect material costEstimated Activity Level$100,0002000 hours$250,000SolutionApplying Equation [15 ... 50 per direct labor dollarindirect budget50,000Machine 2 rate ϭ ———————— ϭ ———2000direct labor hoursϭ $25 per direct labor hourindirect budget50,000Machine 3rate ϭ ——————— ϭ ————material cost250,000ϭ $0 ... 6Traditional Indirect Cost Rates and AllocationNow the actual direct labor costs and hours and material costs are determined for thisyear, and each dollar of direct labor cost spent on machine 1 implies that $0 ... Similarly, indirect costs areadded for machines 2 and 3 ... For example, if direct materials are the basis for allocation tofour separate processing lines, the blanket rate istotal indirect costsIndirect cost rate ϭ ———————————total direct materials costIf $500,000 in indirect costs and $3 million in materials are estimated for thefour lines, the blanket indirect rate is 500,000͞3,000,000 ϭ $0 ... Blanket rates are easierto calculate and apply, but they do not account for differences in the type of activities accomplished in each cost center ... For example, light machinery may contribute less per hour than heavy, more expensivemachinery ... The use of blanket rates in these cases is not recommended, as theindirect cost will be incorrectly allocated ... The approach should be the application of different bases fordifferent machines, activities, etc ... 8 ... Realization that morethan one basis should be normally used in allocatingindirect costs has led to the use of activitybased costing methods, as discussed in the next section ... This results in the totalcost of production, which is called the cost of goods sold, or factory cost ... If the total indirect cost budget is correct, the indirect costs charged to all cost centers for theperiod of timeshould equal this budget amount ... Experience in indirect cost estimation assists in reducing the variance at theend of the accounting period ... 9Since the manager determined indirect cost rates for BestWay (Example 15 ... Perform the computations using the data inTable 15–8 ... SolutionStart with thecost of goods sold (factory cost) relation given by Equation [B ... 8 are applied:Machine 1 indirect ϭ (labor cost)(rate) ϭ 2500(0 ... 00) ϭ $18,750Machine 3 indirect ϭ (material cost)(rate) ϭ 19,550(0 ... Based on the annual indirect cost budget of $150,000, one month represents 1͞12 of thetotalor150,000Monthly budget ϭ ————12ϭ $12,500The allocation variance for total indirect cost isVariance ϭ 12,500 Ϫ 23,910 ϭ $Ϫ11,410This is a large budget underallocation, since much more was actually charged than allocated ... 3% underallocation of indirectcosts ... Once estimates of indirect costsare determined, it is possible to perform an economic analysis ofthe present operation versus a proposed operation ... 10 ... 10For several years, Cuisinart Corporation has purchased the carafe assembly of its major coffeemaker line at an annual cost of $2 ... The suggestion to make the component in-househas been made ... The allocated hours column is the timenecessary to produce the carafes for a year ... Perform an economic analysis for the make alternative,assuming that a market rate of 15% per year is the MARR ... Use the data of Table 15–9 to calculate the indirect cost allocation ...10Indirect CostsDepartmentBasis,HoursRateper Hour, $AllocatedHoursMaterialCost, $DirectLaborCost, $ABCLaborMachineLabor1051525,00025,00010,000200,00050,00050,000200,000200,000100,000300,000500,00015 ... 15 ... Whereonce as much as 35% to 45% of the final product cost wasrepresented in labor, now thelabor component is commonly 5% to 15% of total manufacturing cost ... The use ofbases, such as direct labor hours, to allocate indirect cost is not accurate enough for automated and technologically advanced environments ... 8] ... A product that by traditional methods mayhave contributed a large portion to profit may actually be a loser when indirect costs are allocated more correctly ... This may indicate that they areprofitable, when in actuality they are losing money ... Itis designed to identify cost centers, activities, and cost drivers ... Cost centers: The final products orservices of the corporation are called cost centers or costpools ... Activities: These are usually support departments (purchasing, quality, IT, maintenance, engineering, supervision) that generate the indirect costs which are then distributed to the costcenters ... Examples are the number of purchase� ———————————— total volume of cost driver[15 ... Use the rate to allocate indirect cost to cost centers for each activity ... The orders, cost of engineering change orders,number of machine setups, number of safety violations, and the like ... 1 ... 2 ... 3 ... total cost of activityABC indirect cost rateallocationfor costs generated by the purchasing department, for example, is based on the number of purchase orders (step 2) to support laser production ... 401402Chapter 15Cost Estimation and Indirect Cost AllocationEXAMPLE 15 ... However, accounts such as business travelhave historically been

allocated on the basis of the number of employees at the plants in France,Italy, Germany, and Spain ... The ABC system is chosen to augment the traditional method to more preciselyallocate travel costs to major product lines at each plant ... If total employment of 29,100 isdistributed as follows, allocatethe $500,000 ... The ABC method will beapplied to allocate travel costs to major product lines ... 5% of $1 million30% of $500,000Further, the study indicates that in 1 year a total of 500 travel vouchers were processed bythe management of the major five product lines produced at the four plants ...Product lines—1, 3, and 5; vouchers—80 for line 1, 30 for 3, 30 for 5 ... Product line—5; vouchers—140 for line 5 ... Solution(a) In this case, Equation [15 ... travel budgetIndirect cost rate ϭ ———————total workforce$500,000ϭ ———— ϭ $17 ... ParisFlorenceHamburgBarcelona$17 ... Use the 4-stepprocedure to allocate travel costs to the five products ... The total amount to be allocated is determined from the percentages of eachplant’s support budget devoted to travel ... 05(2,000,000) ϩ ... 30(500,000) ϭ $500,00015 ... 11ProductLine1ParisFlorenceHamburgBarcelonaTotal2508010025$230$5034302553020$30$20140$170Total75140145140$500Step 2 ... The allocation will be tothe products directly, not to the plants ... Step 3 ... 9] determines an ABC allocation rate ... Table 15–10 summarizes the vouchers and allocation byproduct line and by city ... Comparison of the by-plant totals in Table 15–10 (far rightcolumn) with the respective totals in part (a) indicates a substantial difference inthe amounts allocated, especially to Paris, Hamburg, and Barcelona ... CommentLet’s assume that product 1 has been produced in smalllots at the Hamburg plant for a numberof years ... In the ABC analysis, Hamburg has a total of $145,000 travel dollarsallocated, $100,000 from product 1 ... This indicates to management the need to examine the manufacturing lot size practices at Hamburg andpossibly other plants, especially when aproduct is currently manufactured at more than one plant ... This is not a good approach, since ABC isnot a complete cost system ... The two systemswork well together with the traditional methods allocating costs that have identifiable directbases, for example, direct labor ... 15 ... Public agencies,privatecorporations, and not-for-profit businesses all make economic decisions based on these403404Chapter 15Cost Estimation and Indirect Cost Allocationestimates, most of which are made by employees of the organizations or by outside consultantshired to perform specific activities under contract ... Thepersonal morals and adherence to codes of professional ethics discussed in Section 1 ... The NSPE Code of Ethics for Engineers (Appendix C) referenced previously starts with a listof six Fundamental Canons ... ”Acts that bias the results from experimental samples, previous cost data, or survey resultsfor thepurposes of personal gain, increased profits, or favoritism are examples of unethical behavior ... Use accepted theory and techniques in taking statistical samples, building budget elements,and drawing conclusions that are included in proposals, applications, and recommendations ... The secondcase study at the end of this chapter presents an example of some ethical challengespresent when preparing estimates and proposals for contract work ... There are bottom-up andtop-down approaches; each treats price and cost estimates differently ... The Consumer Price Index (CPI) is an often-quotedexample of cost indexing ... Two of them areCost-capacity equation—good for estimating costs from design variables for equipment, materials, and constructionFactor method—good for estimating total plant costTraditional cost allocation uses an indirect cost rate determined for a machine, department,product line, etc ... With increased automation and information technology, different techniques of indirect cost allocation have been developed ... The ABC method allocates indirect costs on the rationale that purchase orders, inspections,machine setups, reworks, etc ... Improved understanding of howthe companyor plant accumulates indirect costs is a major by-product of implementing the ABC method ... 1 Rank the following estimate types in terms of timespent to carry out the estimate (most time to leasttime): partially designed, design 60% to 100%complete, order of magnitude,scoping͞feasibility,detailed estimate ... 2 Classify the following cost elements as first cost(FC) components or annual operating cost (AOC)components for a piece of equipment on theshop floor: supplies, insurance, equipment cost,utility cost, installation, delivery charges, laborcost ... 3 State whether actual(A) or estimated (E) costs aremore likely to be used to carry out the followingactivities: calculate taxes, make bids, pay bonuses,determine profit or loss, predict sales, set prices,405Problemsevaluate proposals, distribute resources, plan production, and set goals ... 4 Identify the output and input variablesin both thebottom-up and top-down approaches to cost estimating ... 5 Classify the following costs as typically direct (D)or indirect (I):Project staffUtilitiesRaw materialsProject suppliesAdministrative staffAudit and legalRentEquipment trainingLaborMiscellaneous office supplies15 ... 15 ... 8 Use the unitcost method to determine the preliminary cost of a guardrail for a small bridge if a totalof 120 linear feet will be required at a cost of$58 ... 15 ... Prepare a preliminarycost estimate for the garage if the cost per parkingspace is $4700 ... 10 The Office of the Undersecretary of Defense periodically releasesunit cost data for use in militaryconstruction programs ... 11 The Department of Defense uses area cost factors(ACFs) to adjust for differences in construction costsin different parts of the country (and world) ... 70 while the ACF for Rapid City, South Dakota, is0 ... If a cold storage processing warehousecosts$1,350,000 in Rapid City, estimate the cost forAndros Island ... 12 Preliminary cost estimates for jails can be madeusing costs based on either unit area (square feet)or unit volume (cubic feet) ... 13 The unit area and unit volume total project costsfor a library are $114 per square foot and $7 ...Based on these numbers,what is the average height of library rooms?15 ... If the vibrators cost $76 per day and the concrete pump costs$580 per day, estimate (a) the cost of the equipmentper cubic yard of concrete and (b) the equipmentcost for placing 56 cubic yards of concrete ... 15 A labor crew forplacing concrete consists of1 labor foreman at $25 ... 60 per hour, 5 laborers at $23 ... 45 perhour ... Determine(a) the cost per day of labor for the C20 crew, (b) thecost of the C20 crew per cubic yard of concrete, and(c) the cost to place 250 cubic yards of concrete ... 16 Site work activities associated withconstructing asmall bridge are shown in the table below ... Use the data to determine (a) the total cost for structural excavation, (b)the total cost for the pile-driving rig, and (c) thetotal labor cost for the site work ... 351 ... 315 ... 781 ... 132 ... 57Pile-driving rigPiling, steel,drivingLegend: cy ϭ cubic yard; ls ϭlump sum; lf ϭ linear footCost Indexes15 ... If the state index for Texas is76 ... 5, estimate thetotal construction cost of a middle school for 800students in each state ... 18 A consulting engineering firm is preparing a preliminary cost estimate for a design-construct project of a coal processing plant ... Usethevalues in Table 15–3 to estimate the cost of construction for a similar-size plant in mid-2010 ... 19 If the editors at ENR decide to redo the construction cost index so that the year 2000 has a basevalue of 100, determine the value for the year (a)1995 and (b) 2009 ... 20 (a) Estimate the value of the ENRconstructioncost index by using the average (compounded)percentage change in its value between 1995and 2005 to predict the value in 2009 ... 21 An engineer who owns a construction companythat specializes in large commercial projects noticed that material costs increased at a rate of 1%per monthover the past 12 months ... 15 ... 35 ... 38 ... 23 Electropneumatic general-purpose pressure transducers convert supply pressure to regulated outputpressure in direct proportion to an electrical inputsignal ... 6 ... 3?15 ... 40 in February 2007 ... 32 and 4874 ... If a generalcontractor in Atlanta wonconstruction jobs totaling $54 ... Cost Estimation and Indirect Cost Allocation15 ... 37 when the base yearwas 1913 with a value of 100 ... 0, the CCI for August 2010would be 8 ... In this case, what is the CCIvalue for 1967 when the base year is 1913?15 ... 51 in August 2010 ... 55 per ton ... 27 Acontractor purchased equipment costing $40,000in 2010 when the M&S equipment cost index wasat 1461 ... He remembers purchasing the sameequipment for $21,771 many years ago, but he doesnot remember the year that he did so ... 68% per yearover that time period and the equipment increasedinprice exactly in proportion to the index, (a) inwhat year did he purchase the equipment and (b)what was the value of the index in that year?Cost-Capacity and Factor Methods15 ... 29 Use the exponent values in Table 15–6 to estimatethe cost for the following equipment to be placedon an offshore drillingplatform ... (b) The cost of a 1700-gallon stainless steel tankif a 900-gallon tank costs $4100 ... 30 A high-pressure stainless steel pump (1000 psi)with a variable-frequency drive (VFD) is installedin a seawater reverse-osmosis pilot plant that is recovering water from membrane concentrate at arate of 4gallons per minute (gpm) ... Because of favorable resultsfrom the pilot study, the city utility wants to go witha full-scale system that will produce 500 gpm ... 37 ... 31 A 0 ... Estimatethe cost of a 2-MGD tower if the exponent in thecost-capacity equation is 0 ... Problems15 ... Make the best cost estimatepossible forthe VFD for a 100-hp motor ... 15 ... 52 times the cost of the 30 m2 unit ... 34 Reinforced concrete pipe (RCP) that is 12 inchesin diameter had a cost of $12 ... The cost for 24-inch RCPwas $27 ... If the cross-sectional area ofthe pipe is considered the “capacity” in the costcapacity equation,determine the value of the exponent in the cost-capacity equation that exactlyrelates the two pipe sizes ... 35 A 100,000 barrel per day (bpd) fractionation towercost $1 ... 3 ... 8, provided the exponent in thecost-capacity equation is 0 ... 36 A mini wind tunnel for calibrating vane or hotwire anemometerscost $3750 in 2002 when theM&S equipment index value was 1104 ... If theindex value is now 1620 ... The cost-capacity equationexponent is 0 ... 15 ... The engineer used the M&S equipment index forthe years 1998 and 2008 and the cost-capacityequation with an exponent value of 0 ... If theoriginalequipment had only one-fourth the capacity of the new equipment, what was the cost of theoriginal equipment in 1998?15 ... For example, a SRR of 2 ... 942 ... 942C1 when Q2͞Q1 ϭ 2 ... 15 ... 8 million ... 25, what is the total plantcost expected to be?40715 ... )while maximizing orifice life and machineperformance ... 32 million, what is the overallcost factor for the system?15 ... If the direct cost factor is 1 ... 45 (applies to equipmentonly), determine the expected cost of the laboratory ... 42 The equipment cost for a 10 gallon per minutefarm-scale ethanol fuel production plant is$243,000 ... 28 and forinstallation is 0 ... The indirect costfactor for licenses, insurance, etc ... 84 (appliedto total direct cost) ... 15 ... 3 million ... 35 and the indirect cost factor is0 ... 15 ... If thedirect cost factor is 3 ... 38, what is the estimated total plant cost? Theindirect cost factor applies to the total direct cost ... 45 Nicole is anengineer on temporary assignment ata refinery operation in Seaside ... The equipment itself is estimated at $250,000 witha construction cost factor of 0 ... 30 ... (a) If the indirect cost factorshould be 0 ... 40 indirectcost factor is used ... 46 The company you work for currently allocates insurance costs onthe basis of cost per direct laborhour ... If the direct labor hours for408Chapter 15Cost Estimation and Indirect Cost Allocationdepartments A, B, and C are expected to be 2000,8000, and 5000, respectively, this year, determinethe allocation to each department ... However, the accountant advises themanager to notbe concerned because the allocation rates havedecreased each month ... 47 The director of public works needs to distribute theindirect cost allocation of $1 ... The recorded amountsfor this year are as follows:Indirect Cost, $MonthBranchMiles DrivenDirectLaborHours275,000247,000395,00038,00031,00055,500NorthSouthMidtownBranchDirectLaborHoursBasisIndirect CostAllocationLast Year, $NorthSouthMidtown350,000200,000500,00040,00020,00064,000MilesLaborLabor300,000200,000450,000(a)(b)Determine the rates for this year for eachbranch ...What percentage of this year’sbudget is now distributed?15 ... Because of the nature and use of three ofthese stations, each is considered a separate costcenter for indirect cost allocation ... Machineoperating hours are used as the allocation basis forall machines ... Use the data collectedthis year todetermine the indirect cost rate foreach center ... 401 ... 371 ... 9228003400350036006000260038003500During the evaluation, the following additionalinformation from departmental and accountingrecords is obtained ... This information follows:MilesDrivenAllocatedFebruaryMarchAprilMayJuneRecordsfor This YearRateDirect LaborHoursMonthFebruaryMarchAprilMayJune(a)(b)Cost, $MaterialCost, $DepartmentalSpace, ft2640640640640800256025602560272033205400460057006300650020002000350035003500With this information determine the allocation basis used each month ... 15 ... 9millionthis year ... Your manager asksthat both direct and indirect costs be includedwhen in-house manufacturing (make alternative)is evaluated ... 5 million and a lifeof 6 years ... 5 million per year ... Perform the AW evaluation atMARR ϭ 12% per year over a 6-year study period ... Department15 ... Themanager has obtained records of allocation rates and actual charges forthe prior 3 months and estimates for this month(May) and next month (see the table) ... 40 per $$0 ... 51 The municipal water and desalinization utility in aCalifornia city currently allocates some costs formaintenance shop workers topumping stationsbased on the number of pumps at each station ... Information about the stations is below ... (a) Allocate the budget to each station based onthe number of service trips ... Station IDNo ... 52 Factory Direct manufactures and sells manufacturedhomes ... Each plant builds different modelsand floorplans ... Because of these advances, the CFO plans to use build-time per unit asthe new basis ... The data shown represents average costs and times ... PlantTexasOklahoma20,00012,70018,600400415355Allocation, $DFWYYZMEX55,00020,83315,0006910330,000187,500150,000The airline’sbaggage management director suggests thatan allocation on the basis of baggage traffic, not flights,will better represent the distribution, primarily based onthe fact that the high fees now charged to passengers tocheck luggage have significantly changed the number ofbags handled at the major hubs ...15 ... 54 Using the baggage-traffic basis, determine the allocation rate using last year’s total of $667,500,and distribute this amount to the hubs this year ... 55 What are the percentage changes in allocation ateach hub using the two different bases?15 ... For this year, in roundnumbers, the budgets andallocation of $1 millionadvertising indirect costs are as follows:SiteABudget, $Allocation, $(a)BCD2 million200,0003 million300,0004 million400,0001 million100,000Determine the allocation if the ABC method isused with a new basis ... The costdriver is the number of guests during the year ... The averagenumber of lodgingnights for guests at each site is as follows:SiteAProblems 15 ... 55 use the followinginformation ... Last year $667,500 was distributed as follows:Length of stay, nights(c)BCD3 ... 51 ... 75Comment on the distribution of advertisingcosts using the two methods ... 410Chapter 15CostEstimation and Indirect Cost AllocationADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS15 ... (b) Cost estimates are an output variable ... (d) Both (a) and (b) are correct ... 58 A ratio of the cost of something today to its cost atsome time in the past is called a:(a) Cost-capacity index(b)Cost index(c) Buyer’s guide(d) Bluebook index15 ... 60 A 50-hp turbine pump was purchased for $2100 ... 76, a 200-hp turbine pump could be expected to cost about:(a) $6020(b) $5320(c) $4890(d) $4260100,000 units per day was $3 million, the value ofthe exponent in the cost-capacity equation is:(a) 0... 39(c) 0 ... 6015 ... If the overall cost factorfor the complete system is 2 ... 65 The delivered-equipment cost for setting up aproduction and assembly line for high-sensitivity,gas-damped accelerometers is $650,000 ... 82 and0 ... 61 The city built a recreation park in 1980 for$500,000 ... 17 at that time ...16, the estimatedcost is closest to:(a) $695,800(b) $750,700(c) $820,300(d) $910,50015 ... An allocation basis that may not bereasonable is:(a) Miles of toll road monitored(b) Average number of cars patrolling per hour(c) Amount of car traffic per section of tollroad(d) Cost to operate a patrol car15 ... 9 ...32 and the M&S indexvalue was 1449 ... 67 The IT department allocates indirect costs to userdepartments on the basis of CPU time at the rate of$2000 per second ... If the IT indirect budget for theyear is $8 ... 5%(c) 55%(d) Not enough information to determine15 ... If the cost for a plant with a capacityof15 ... 2 ... (a)(b)(c)(d)15 ... Cost of engineering changes processedSize of the workforceAdministrative cost to process a changeorder1231 and 32 ... (a)(b)(c)(d)It is an excellent replacement for a traditionalcost accounting system ... It can help explain the economic impact ofmanagement decisions ... ,initiated the manufacture andsales of a portable sterilization unit (Quik-Sterz) that can beplaced in the hospital room of a patient ... This new unitmakes the instruments available at the point and time of usefor burn and severe wound patients who are in a regular patient room ... The standardversion sellsfor $10 ... 75 ... InformationMedical Dynamics has historically used an indirect cost allocation system based upon direct hours to manufacture forTA BLE 15–11all its other product lines ... However, Arnie, the person who performedthe indirect cost analysis and set the sales price, is no longerat thecompany, and the detailed analysis is no longer available ... A search ofthese files revealed the manufacturing and cost information inTable 15–11 ... 73 per unit for the standard model and $27 ... Last year management decided to place the entire plant onthe ABC system of indirect cost allocation ... Fiveactivities and their cost drivers were identified for the Medical Dynamics manufacturing operations(Table 15–12) ... Historical Records of Direct and Indirect Cost Analyses for Quik-SterzQuik-Sterz Direct Cost (DC) EvaluationModelStandardPremiumDirect Labor,$͞Unit1Direct Material,$͞UnitDirectLabor,Hours͞UnitTotal DirectLabor Hours5 ... 002 ... 750 ... 50187,500125,000AllocatedIDC, $Sales,Units/Year1 ... 33 million250,000Quik-Sterz Indirect Cost (IDC) EvaluationModelDirect Labor,Hours͞UnitStandard0 ... 501Average direct labor rate is $20 per hour ... Thefirst impression of the productionmanager is that the new system will show that indirect costs for Quik-Sterz are about thesame as they have been for other products over the last severalyears when a standard model and an upgrade (premium) modelwere sold ... Fundamentally, there are two reasonswhy the production manager does notlike to produce premiumversions: They are less profitable for the company, and they require significantly more time and operations to manufacture ... Use traditional indirect cost allocation to verify Arnie’scost and price estimates ... Use the ABC method to estimate the indirect cost allocation and total costfor each model ... If the prices and number of units sold are the samenext year (750,000 standard and 250,000 premium),and all other costs remain constant, compare theprofit from Quik-Sterz under the ABC method withthe profit using the traditional indirect cost allocation method ... What prices shouldMedical Dynamics charge next yearbased on the ABC method and a 10% markup over cost?What is the total profit from Quik-Sterz predicted to beif sales hold steady?5 ... CASE STUDYDECEPTIVE ACTS CAN GET YOU IN TROUBLEContributed by Dr ... The law requires that land disturbed by thesetypes of miningactivities be returned to a productive capacity that is as goodas or better than its productive capacity before mining ... To this end,mining companies must sample the different soils found inthe areas to be disturbed by mining activities ... Soils in a low pH range (pH values Ͻ 5) are indicativeoflow fertility ... To meet the baseline for pH, the acreage of the mine soilswith low pH should not exceed the acreage of the unminedsoils with low pH ... The different soilswithin the 600 acres were depicted in the County Soil Surveywhere the mining activities were to take place ... Assessment of thedataindicated that 30% of the area (180 acres) occupied by thesoils in the area to be disturbed had pH values between 4 ... 9 ... Six years later, 450 acres had been mined and the terrainsurface had been leveled to reestablish premine slopes ... 0 and4 ... The president of Yucatan indicated that thecompanywould submit a revised soil baseline based on new samplingin the remaining 150 acres of unmined soils because, in hisopinion, the first soil baseline was biased ... The company quickly hireda consultant to develop the new baseline, and subsequentlyYucatan submitted the final report from itsconsultant to theState Department of Mining and Reclamation ... 0 and 4 ... Comparative results between the old and new samples can be expressed as follows:Percent and Acreage of AreaSoil BaselinesOld Soil BaselineRevised Soil BaselinepH: 4 ... 930%180 acres45%270 acresA rough statisticalcheck between the old and revised soilbaselines indicated that the results were mixed ... Contained in the submitted new-sample packagewas a letter from the Yucatan consultant ... 413The State Department of Mining and Reclamation staffconcluded that the revised soil pH sample data had beencarefully “screened” to reduce the amount of remediation workthat Yucatan Mining would have to complete ... It was also notedthat should Yucatan Mining disagree with this response, thecase would be filed with the legal staff as a contested case ... Case Study Questions1 ... Whatactions would you directyour staff to take concerningthis situation?2 ... What type of“audit” procedures might you want implemented toidentify these possibly unethical activities?3 ... When the 30 lowest pH samples were used to establish the new baseline, were the samples still random,according to experimental designstandards? If so, why?If not, why not?4 ... You have golfed togetherseveral times, your and his children are on the same soccer team at school, and your families are members ofthe same community swimming pool club ... As a matter of principle and practice, do you believethere is some amount of data-altering or bias-makingthat is allowed before an application (such as the onedescribed here) should be considered the result of professionally unethical acts? How would you define sucha threshold limit?CHAPTER 16DepreciationMethodsL E A R N I N GO U T C O M E SPurpose: Use depreciation ordepletion methods to reduce the book value of a capital investment in an asset and naturalresource ... 1Terminology• Define and use the basic terms of assetdepreciation ... 2Straight line• Apply the straight line (SL) method ofdepreciation ... 3Declining balance• Apply the declining balance (DB) anddoubledeclining balance (DDB) methods ofdepreciation ... 4MACRS• Apply the modified accelerated cost recoverysystem for tax depreciation purposes for U ... based corporations ... 5Recovery period• Select the asset recovery period for MACRSdepreciation ... 6Depletion• Explain depletion; apply costdepletion andpercentage depletion methods ... 1Historical methods• Apply the sum-of-years-digits (SYD) and unitof-production (UOP) methods of depreciation ... 2Switching• Switch between classical depreciation methods;explain how MACRS provides for switching ... 3MACRS and switching• CalculateMACRS rates using switching betweenclassical methods and MACRS rules ... Although the depreciation amount isnot an actual cash flow, the process of depreciating an asset on the books of the corporation accounts for the decrease in an asset’s value because of age, wear, and obsolescence ... Anintroduction to depreciation types, terminology, and classical methods is followed by a discussionof the Modified Accelerated Cost Recovery System (MACRS), which is the standard in theUnited States for tax purposes ... Why is depreciation important to engineering economy? Depreciation is a tax-alloweddeduction included in tax calculations in virtually all industrialized countries ... This chapter concludes with an introduction to two methods of depletion, which are usedto recover capital investments in deposits of natural resources such as oil, gas, minerals, ores,and timber ... Additionally, theappendix includes an in-depthderivation of the MACRS depreciation rates from the straight line and declining balancerates ... 16 ... Most descriptions are applicable to corporations as well as individuals who own depreciable assets ... The method used to depreciate an asset is a way to account for thedecreasing value of the assetto the owner and to represent the diminishing value (amount) of the capital funds invested in it ... Though the term amortization is sometimes used interchangeably with the term depreciation,they are different ... In addition, the term capital recovery is sometimes used toidentify depreciation ... The term depreciation is usedthroughout this book ... Tax depreciation Used by a corporation or business to determine taxes due based on current tax laws of the government entity (country, state, province, etc ... The methods applied for these two purposes may or may not utilizethe same formulas ... There are classical, internationally accepted depreciation methods used to determine book depreciation: straight line, declining balance, and the historical416Chapter 16Depreciation Methodssum-of-years-digits method ... In most industrialized countries, the annual tax depreciation istax-deductible; that is, it is subtracted from income when calculating the amount of taxes due each year ... Tax depreciation may be calculated and referred to differently in countries outside the UnitedStates ... Where allowed, tax depreciation is usually based on an accelerated method, wherebythedepreciation for the first years of use is larger than that for later years ... In effect, accelerated methods defersome of the income tax burden to later in the asset’s life; they do not reduce the total taxburden ... First cost P or unadjusted basis B is the delivered and installed cost of the assetincludingpurchase price, delivery and installation fees, and other depreciable direct costs incurred toprepare the asset for use ... When thefirst cost has no added, depreciable costs, the basis is the first cost, that is, P ϭ B ... The bookvalue is determined at the end of each year t (t ϭ 1, 2, ... Recoveryperiod n is the depreciable life of the asset in years ... Both of these values may be different from the asset’sestimated productive life ... Because of the structure of depreciation laws,the book value and market value may be substantially different ... However, a computer workstation may have a marketvalue much lowerthan its book value due to rapidly changing technology ... The salvage value, expressed as an estimated dollar amount or as a percentage of the first cost,may be positive, zero, or negative due to dismantling and carry-away costs ... This rate may be the same each year, which is calledthe straight line rate d, ordifferent for each year of the recovery period ... Included is most manufacturing and service industry property—vehicles, manufacturingequipment, materials handling devices, computers and networking equipment, communications equipment, office furniture, refining processequipment, construction assets, andmuch more ... Land itself is considered real property, but it is not depreciable ... This convention is utilized inthis text and in most U ... -approved tax depreciation methods ... 16 ... Book value, $SLMACRSDBEstimatedsalvagevalueTime, yearsRecovery periodAsmentioned before, there are several models for depreciating assets ... Accelerated models, such as the declining balance (DB) method, decrease the book value to zero (or to the salvage value) more rapidly thanthe straight line method, as shown by the general book value curves in Figure 16–1 ... Eachfunction is introduced and illustrated as the method is explained ... Onethat may be of interest to a U ... -based small or medium-sized business performing an economicanalysis is the Section 179 Deduction ... Up to aspecified amount, the entire basis of an asset is treated as a business expense in theyear of purchase ... The limit changes with time; it was$24,000 in 2002; $102,000 in 2004; $125,000 in 2007; and $250,000 in 2008–2010 ... Investments above these limits must be depreciated using MACRS ... S ... In 1981, all classical methods, including straight line, declining balance, and sum-of-yearsdigits depreciation, were disallowed as tax deductible and replaced by the Accelerated Cost Recovery System (ACRS) ... To this date, the following is the law in the United States ... MACRS has the DB and SL methods, in slightly different forms, embedded in it, but these twomethods cannot be useddirectly if the annual depreciation is to be tax deductible ... S ... Most other countries still recognize the classical methods of straight line and declining balance for tax or book purposes ... Appendix Section 16A ... 418Chapter 16Depreciation MethodsTax law revisions occur often, and depreciation rules arechanged from time to time in theUnited States and other countries ... S ... irs ... Pertinent publications can be downloaded ... MACRS and most corporate tax depreciation laws are discussed in it ... 2 Straight Line (SL) DepreciationStraight line depreciation derives its name from the fact that the book valuedecreases linearlywith time ... Straight line depreciation is considered the standard against which any depreciation model iscompared ... For tax depreciation, as mentioned earlier, it is not used directly in the United States, but it is commonly used in most countries for tax purposes ... S ... 5) ... In equationS� ——— n[16 ... , n)Dt ϭ annual depreciation chargeB ϭ first cost or unadjusted basisS ϭ estimated salvage valuen ϭ recovery perioddt ϭ depreciation rate ϭ 1͞nBook ValueBSince the asset is depreciated by the same amount each year, the book value after t years of service,؊S)dtB ؊ � ) B form,Dtdenoted by BVt, will be equal to the first cost B minus the annual depreciation times t ... 2]Earlier we defined dt as a depreciation rate for a specific year t ... 3]d ϭ dt ϭ —nThe format for the spreadsheet function to display the annual depreciation Dt in a single-celloperation is� SLN(B, S, n)[16 ... 1If anasset has a first cost of $50,000 with a $10,000 estimated salvage value after 5 years,(a) calculate the annual depreciation and (b) calculate and plot the book value of the asset aftereach year, using straight line depreciation ... 1] ... 16 ... 1 ... 2] ... For years 1 and 5, for example,BV1 ϭ 50,000 Ϫ 1(8000) ϭIf BVtϪ1 is not known, the depreciation in year t can be calculated using B and d ... 8]Book value in[7 ... 16]1؊� ) d)BVt $42,000BV5 ϭ 50,000 Ϫ 5(8000) ϭ $10,000 ϭ S16 ... 5]In this case the method is called double declining balance (DDB) ... 2; so 20% of the book value is removed annually ... 5͞n ... Dtyear t is determined in one of two ways: by using the rate d and basis B or bysubtracting the current depreciation charge from the previous book value ... 9][16 ... Like theSL method, DB is embedded in the MACRS method, but the DB method itself cannot be used todetermine the annual tax-deductibledepreciation in the United States ... Declining balance is also known as the fixed percentage or uniform percentage method ... If d ϭ 0 ... Therefore, the depreciation amount decreases each year ... The implied salvage value after nyears is the BVn amount, that is,Implied S ϭ BVn ϭ B(1 Ϫ d)n[16 ...However, if the implied S Ͻ estimated S, it isnecessary to stop charging further depreciation when the book value is at or below the estimatedsalvage value ... (Thisguideline is important when the DB method can be used directly for tax depreciation purposes ... The range for d is 0 Ͻ d Ͻ 2͞n ... 12]Thespreadsheet functions DDB and DB are used to display depreciation amounts for specificyears ... For the double declining balance method, the format is� DDB(B, S, n, t, d)[16 ... If omitted, this optionalentry is assumed to be 2 for DDB ... 5 makes the DDB function display 150%declining balance methodamounts ... No further depreciation is charged when thisoccurs ... 11] ... 1] ... Caution is needed when using this function ... 12] ... Therefore, if the depreciation rate is known, always use the DDB function to ensurecorrect results ... 2 and 16 ... EXAMPLE 16 ... Theequipment will be DDB depreciated overan expected life of 12 years ... (a) Calculate the depreciation and book value foryears 1 and 4 ... (b) Calculate the implied salvage value after 12 years ... 1667 per year ... 8]and [16 ... Year 1:Year 4:D1 ϭ (0 ... 1667)1Ϫ1 ϭ $4167BV1 ϭ 25,000(1 Ϫ 0 ... 1667)(25,000)(1 Ϫ 0 ... 1667)4 ϭ $12,054The DDBfunctions for D1 and D4 are, respectively, ϭ DDB(25000,2500,12,1) andϭ DDB(25000,2500,12,4) ... 3Declining Balance (DB) and Double Declining Balance (DDB) Depreciation(b) From Equation [16 ... 1667)12 ϭ $2803Since the estimated S ϭ $2500 is less than $2803, the asset is not fully depreciatedwhenits 12-year expected life is reached ... 3Freeport-McMoRan Copper and Gold has purchased a new ore grading unit for $80,000 ... Use the DB and DDBmethods to compare the schedule of depreciation and book values for each year ... Solution by HandAn implied DB depreciation rate is determinedby Equation [16 ... (10,000d ϭ 1 Ϫ ———80,000)1͞10ϭ 0 ... 1877 Ͻ 2͞n ϭ 0 ... Table 16–1 presents the Dt values using Equation [16 ... 10] rounded to the nearest dollar ... 1877(64,984) ϭ $12,197BV2 ϭ 64,984 Ϫ 12,197 ϭ $52,787Because we round off to even dollars, $2312 is calculated for depreciation inyear 10, but$2318 is deducted to make BV10 ϭ S ϭ $10,000 exactly ... 2 result in the depreciation and book value series in Table 16–1 ... 3Declining Balance, $Double Declining Balance, $Year tDtBVtDtBVt012345678910—15,01612,1979,9088,0486,5385,3114,3143,5042,8462,31880,00064,98452,78742,87934,83128,29322,98218,66815,16412,31810,000—16,00012,80010,2408,1926,5545,2434,1943,3552,68473780,00064,00051,20040,96032,76826,21420,97216,77713,42210,73710,000Solution by SpreadsheetThespreadsheet in Figure 16–3 displays the results for the DB and DDB methods ... Since the fixed rates are close—0 ... 2 forDDB—the annual depreciation and book value series are approximately the same for the twomethods ... 12], but note in the cell tagsthat the DDB function is used in both columns Band D to determine annual depreciation ... 12] and maintains it to only three significant digits ... 1ϭ DDB(B$2,B$3,B$4,$A18,10*$B$5)ϭ DDB(B$2,B$3,B$4,$A18)Figure 16–3Annual depreciation and book value using DB and DDB methods, Example 16 ... were used in column B (Figure 16–3), the fixedrate applied would be 0 ... The resultingDt and BVt values for years 8, 9, and 10 would be as follows:tD t, $BVt, $89103,5012,8422,30815,12012,2779,969Also noteworthy is the fact that the DB function uses the implied rate without a check to haltthe book value at the estimated salvage value ... However,the DDB function uses a relation different from that of the DB function to determine annual depreciation—one that correctly stops depreciating at the estimatedsalvage value, as shown in Figure 16–3, cells E17–E18 ... 4 Modified Accelerated Cost RecoverySystem (MACRS)In the 1980s, the United Statesintroduced MACRS as the required tax depreciation method forall depreciable assets ... Corporations are free to applyany of the classical methods for book depreciation ... 1Many aspects of MACRS deal with the specific depreciation accounting aspects of tax law ... Additional information on how the DDB,DB, and SL methods are embedded into MACRS and howto derive the MACRS depreciation rates is presented and illustrated in the chapter appendix,Sections 16A ... 3 ... 14]R ... 16 ... As for other methods, the book valuein year t is determined by subtracting the depreciation amount from the previousD[16 ... The basis B (or first cost P) is completely depreciated; salvage is always assumed to be zero,or S ϭ $0 ... g ... 5 or 39 yearsfor real property (e ... , rental property or structures)Depreciation rates؊͚ sum of accumulated depreciationj� t� B ؊ � first cost year’s bookvalueBVt ϭ BVtϪ1 Ϫ Dt[16 ... BVtprovide accelerated write-off by incorporating switching between classicalmethods ... 5 explains how to determine an allowable MACRS recovery period ... 14] are included in Table 16–2 ... The MACRS rates start with the DDB rate or the 150% DB rate and switch when the SL methodoffers faster write-off... 3344 ... 817 ... 0032 ... 2011 ... 5214 ... 4917 ... 498 ... 0018 ... 4011 ... 225 ... 508 ... 706 ... 757 ... 686 ... 715 ... 928 ... 467 ... 556 ... 566 ... 235 ... 905 ... 905 ... 894 ... 464 ... 285 ... 905 ... 905 ... 464 ... 464 ... 462 ... 464 ... 2367891011121314151617–2021423424Chapter 16Depreciation MethodsForreal property, MACRS utilizes the SL method for n ϭ 39 throughout the recovery period ... 02564 ... The MACRS real property rates in percentage amounts areYear 1Years 2–39Year 40100d1 ϭ 1 ... 564%100d40 ϭ 1 ... 5 years, which applies only to residential rental property,uses the SL method in asimilar fashion ... Also note that the extra-year rate is one-half of the previous year’s rate ... This convention assumes that all property is placed in service at the midpoint of the tax year of installation ... This removes some ofthe accelerated depreciation advantage and requires that one-half year ofwhere B[17 ... 0؊depreciation be taken inyear n ϩ 1 ... However, thevariable declining balance (VDB) function, which is used to determine when to switch betweenclassical methods, can be adapted to display MACRS deprecation for each year ... 2 of this chapter and Appendix A of the text ... 5), MIN(n, tϭ first cost0 ϭ salvage value of S ϭ 0n ϭ recovery periodif MACRS n ϭ 3, 5, 7, or 10dϭ 21 ... EXAMPLE 16 ... This chemical is a resin used in plastic pipe, retail bags, blowmolding, and injection molding ... The chief engineer asked the finance director to provide an analysis of the difference between (1) theDDB method, which is the internal book depreciation and book value method used at the plant, and (2) the requiredMACRS tax depreciation and its book value ... Use hand and spreadsheetsolutions to do the following:(a) Determine which method offers the larger total depreciation after 2 years ... Solutionby HandThe basis is B ϭ $400,000 and the estimated S ϭ 0 ... The MACRS ratesfor n ϭ 3 are taken from Table 16–2, and the depreciation rate for DDB is dmax ϭ 2͞3 ϭ0 ... Table 16–3 presents the depreciation and book values ... 6667) ϭ $29,629, except this would make BV3 Ͻ $20,000 ... 16 ...4MACRSDDBYearRateTaxDepreciation, $BookValue, $BookDepreciation, $012340 ... 44450 ... 0741133,320177,80059,24029,640400,000266,68088,88029,6400266,66788,88924,444BookValue, $400,000133,33344,44420,000(a) The 2-year accumulated depreciation values from Table 16–3areMACRS:DDB:D1 ϩ D2 ϭ $133,320 ϩ 177,800 ϭ $311,120D1 ϩ D2 ϭ $266,667 ϩ 88,889 ϭ $355,556The DDB depreciation is larger ... )(b) After 2 years the book value for DDB at $44,444 is 50% of the MACRS book value of $88,880 ... Thisoccurs because MACRS always removes the entire first cost,regardless of the estimated salvage value ... 4) ... (a) The 2-year accumulated depreciation values areMACRS (add cells B6 ϩ B7):DDB (add cells D6 ϩ D7):$133,333 ϩ 177,778 ϭ $311,111$266,667 ϩ 88,889 ϭ $355,556(b) Book values after 2 years areMACRS (cell C7):DDB (cellE7):$88,889$44,444The book values are plotted in Figure 16–4 ... CommentIt is advisable to set up a spreadsheet template for use with depreciation problems in this andfuture chapters ... Figure 16–4Spreadsheet screen shot of MACRS and DDB depreciation and book value, Example 16 ...425426Chapter 16Depreciation MethodsMACRS simplifies depreciation computations, but it removes much of the flexibility ofmethod selection for a business or corporation ... 16 ... For book depreciation the n value should be theexpected useful life ... There are tables that assist in determining the life andrecovery periodfor tax purposes ... The U ... government requires that all depreciable property be classified into a property classwhich identifies its MACRS-allowed recovery period ... Virtually any propertyconsidered in an economic analysis has a MACRS n value of 3, 5, 7, 10, 15, or 20 years ... The firstis the general depreciation system (GDS) value, which we use in examples and problems ... Therates utilize the DDB method or the 150% DB method with a switch to SL depreciation ... The far right column of Table 16–4 lists the alternative depreciation system (ADS) recovery period range ... 5Officefurniture; some manufacturing equipment;railroad cars, engines, tracks; agricultural machinery; petroleum and natural gas equipment; allproperty not in another class710–15Equipment for water transportation, petroleum refining, agriculture product processing, durable-goodsmanufacturing,shipbuilding1015–19Land improvements, docks, roads, drainage, bridges,landscaping, pipelines, nuclear power productionequipment, telephone distribution1520–24Municipal sewers, farm buildings, telephone switching buildings, power production equipment (steamand hydraulic), water utilities2025–50Residential rental property (house, mobile home)27 ... 6Depletion Methodsrecovery period than the GDS ... The use of ADS is generally a choice left to a company, but it is requiredfor some special asset situations ... This electable SL option is, however, sometimes chosenby businesses that are youngand do not need the tax benefit of accelerated depreciation during thefirst years of operation and asset ownership ... 16 ... We now turn to irreplaceable natural resources and the equivalent of depreciation, whichis called depletion ... The two methods of depletion for book or tax purposes areused to writeoff the first cost, or value of the estimated quantity, of resources in mines, wells,quarries, geothermal deposits, forests, and the like ... Details forU ... taxes on depletion are found in IRS Publication 535, Business Expenses ... Cost depletion may be applied to most types ofnatural resources and must be� ———————— resource capacity[16 ... The total cost depletioncannot exceed the first cost of the resource ... Percentage depletion This is a special consideration given for natural resources ... The depletion amount for year t is calculated asPercentage applied to timber production ... first costCDtGIt[16 ... TheU ... government does not generally allow percentage depletion to be applied to oil and gas wells(except small independent producers) ... S ... DepositSulfur, uranium, lead, nickel, zinc,and some other ores and � gross income from property� PD �� percentage depletion rate depletion tmineralsGold, silver, copper, iron ore, andsome oil shaleOil and natural gas wells (varies)Coal, lignite, sodium chlorideGravel, sand, peat, some stonesMost other minerals, metallic oresPercentage ofGross Income, PD221515–2210514427428Chapter 16Depreciation MethodsEXAMPLE 16 ... Anestimated 350 million board feet of lumber is harvestable ... (b) After 2 years the total recoverable board feet was reestimated upward to be 450 million fromthe time the rights were purchased ... Solution(a) Use Equation [16 ... 700,000CDt ϭ ———— ϭ $2000 per million board feet350Multiply CDt by theannual harvest to obtain depletion of $30,000 in year 1 and $44,000in year 2 ... (b) After 2 years, a total of $74,000 has been depleted ... Additionally, with thenew estimate of 450 million board feet, a total of 450 Ϫ 15 Ϫ 22 ϭ 413 million board feetremains ... , the cost depletion factor is626,000CDt ϭ———— ϭ $1516 per million board feet413EXAMPLE 16 ... It has an anticipated gross income of $5 ... 0 million per year after year 5 ... Compute annual depletion amounts for the mine ... 15 ... 15(5 ... 15(3 ... 75 million is written off in 5 years, and the remaining $6 ... The total number of years is$6 ... 9 ϭ18 ... In many of the natural resource depletion situations, the tax law allows the larger of the twodepletion amounts to be claimed each year ... Therefore, it is wise to calculate both depletionamounts and select the larger ... CDAt ϭ cost depletion amountPDAt ϭ percentage depletion amountTIt ϭ taxableincomeThe guideline for the tax-allowed depletion amount for year t is{Depletion ϭ max[CDAt, PDAt]max[CDAt, 50% of TIt]if PDAt Յ 50% of TItif PDAt Ͼ 50% of TItChapter SummaryFor example, assume a medium-sized quarry owner calculates the following for 1 year ... For tax purposes, apply theguideline above and use the cost depletion of $275,000, since it is larger than50% of TI ... In the United States, the MACRS method is the onlyone allowed for tax depreciation ... Depreciation does not result in cashflow directly ... The annual depreciation amount is tax deductible, which can result in actualcashflow changes ... Common relations for each method are summarized in Table 16–5 ... • The estimated salvage value is always considered ... Declining Balance (DB)• The method accelerates depreciation compared to the straight line method ... • The most used rate is twice the SL rate, which is calleddouble declining balance (DDB) ... • It is not an approved tax depreciation method in the United States ... Modified Accelerated Cost Recovery System (MACRS)• It is the only approved tax depreciation system in the United States ... • It always depreciates to zero; that is, it assumes S ϭ 0 ... • Depreciationrates are tabulated ... • MACRS straight line depreciation is an option, but recovery periods are longer than those forregular MACRS ... The annualcost depletion factor is applied to the amount of resource removed ... Percentage depletion, which can recover morethan the initial investment, reduces theinvestment value by a constant percentage of gross income each year ... 1 Sum-of-Years-Digits (SYD) and Unit-ofProduction (UOP) DepreciationThe SYD method is a historical accelerated depreciation technique that removes much of thebasis in the first one-third of the recovery period; however, write-S)SUM[16A ... ؊ � ) ————— B 1Dt؉t؊salvage value)sum of years digitsn ؊ � ) ————————————— basis off is not as rapid as for DDBor MACRS ... The mechanics of the method involve the sum of the year’s digits from 1 through the recoveryperiod n ... depreciable years remainingDtThe rate of depreciation decreases each year and equals the multiplier in Equation [16A ... nϪtϩ1dt ϭ ————SUM[16A ... The function format isϭ SYD(B,S,n,t)EXAMPLE 16A ... SolutionThe sum of the[2 ... 0 ؉ t͞2 ؊ jϭnSUM ϭn(n ϩ 1)͚ j ϭ ————2jϭ1The book value for any year t is calculated ast(nyear’s digits is 36, and the depreciation amount for the second year by Equation [16A ... Figure 16A–1 is a plot of the book values for an $80,000 asset with S ϭ $10,000 and n ϭ10 years using the four depreciation methods that we have learned ... 16A ... A second depreciation method that is not allowed

for tax purposes, but useful in some situations is the unit-of-production (UOP) method ... Suppose a highway contractorhas a series of state highway department contracts that will last several years and that earth moving equipment is purchased for use on all contracts ... For year t, UOPdeprecation iscalculated asactual usage for year tDt ϭ —————————— (basis Ϫ salvage)total lifetime usage[16A ... 2Zachry Contractors purchased an $80,000 mixer for use during the next 10 years for contractwork on IH-10 in San Antonio ... Use the actual usage peryear shown in Table 16A–1 and the unit-of-production method to determine annual depreciation ... 4] to determine the annual depreciation based on the estimated total lifetime amount of material, 2 million m3 in thiscase ... If the mixeris continued in service after the 2 million m3 is processed, no further depreciation is allowed ... 2YeartActualUsage, 1000 m3AnnualDepreciation Dt , $CumulativeDepreciation, $12–89–1040020010016,0008,0004,00016,00072,00080,000Total200080,000431432Chapter 16Depreciation Methods16A ... It also maximizes the present value of accumulated and total depreciation over the recovery period ...The approach below is an inherent part of MACRS ... General rules of switching are summarized here ... Switching is recommended when the depreciation for year t by the currently used method isless than that for a new method ... 2 ... 3 ... When switching from a DB method, the estimated salvageD (P͞F, i, t)t[16A ... Switching is most advantageous from a rapid ͚ � value,not the DB-implied salvage value, is used to compute the depreciation for the new method;we assume S ϭ 0 in all cases ... )4 ... In all situations, the criterion is to maximize the present worth of the total depreciation PWD ... t� nPWDwrite-off method such as DDB to the SL model ... 11] exceeds the salvage value estimated at purchase time; that is, switch ifBVn ϭ B(1 Ϫ d)n Ͼ estimated S[16A ... Depending upon the values of d and n, theswitch may be best in the later years or last year of the recovery period, which removes the impliedS inherent to the DDB model ... For each year t, compute the two depreciation charges ... 7][16A ... Select the larger depreciation value ... 9]3 ... 5] ... This approach is usually not taken, but the switching technique will workcorrectly for all depreciation methods ... Once these are understood, the mechanicsof the switching can be speeded up by applying the spreadsheet function16A ... This is a quite powerful function that determines the depreciation for 1 year or the total over several years for the DB-to-SL switch ... 10]Appendix A explains all the fields in detail, but for simple applications, where the DDBand SLannual Dt values are needed, the following are correct entries:start_t is the year (tϪ1)end_t is year td is optional; 2 for DDB is assumed, the same as in the DDB functionno_switch is an optional logical value:FALSE or omitted—switch to SL occurs, if advantageousTRUE—DDB or DB method isapplied with no switching to SL depreciation considered ... This is discussed in Example 16A ... You may notice that the VDB function is the same one used to calculate annual MACRS depreciation ... 3The Outback Steakhouse main office has purchased a $100,000 online document imaging system withan estimated useful life of 8 years and a tax depreciation recovery period of 5 years ... (d) Perform the DDB-to-SL switch using a spreadsheet and plotthe book values ... Solution by HandThe MACRS method is not involved in this solution ... 1] determines the annual SL depreciation ... PWD ϭ20,000(P͞A,15%,5) ϭ 20,000(3 ... 40 ... The value PWD ϭ$69,915 exceeds $67,044 for SL depreciation ... (c) Use the DDB-to-SL switching procedure ... The DDB values for Dt in Table 16A–2 are repeated in Table 16A–3 for comparisonwith the DSL values from Equation [16A ... The DSL values changeeach year becauseBVtϪ1 is different ... For illustration, compute DSL values for years 2 and 4 ... The column “Larger Dt” indicates a switch in year 4 with D4 ϭ $10,800 ... Total depreciationwith switching is $100,000 compared to the DDB amount of $92,224 ... With switching, PWD ϭ $73,943, which is anincrease over both the SL and DDBmethods ... 3bYeartDt , $BVt , $(P͞F,15%,t)Present Worthof Dt , $01234540,00024,00014,4008,6405,184100,00060,00036,00021,60012,9607,7760 ... 75610 ... 57180 ... 3cDDB Method, $DDDBBVt01234*5—40,00024,00014,4008,6405,184Totals100,00060,00036,00021,60012,9607,776SL MethodDSL, $LargerDt , $P͞FFactorPresentWorth ofDt, $20,00015,00012,00010,80012,96040,00024,00014,40010,80010,8000 ... 75610 ... 57180 ... Solution by Spreadsheet(d) In Figure 16A–2, column D entries are theVDB functions to determine that the DDB-toSL switch should take place in year 4 ... If TRUE were entered, the decliningbalance model would be maintained throughout the recovery period, and the annual depreciation amounts would be equal to those in column B ... The terminal book value in year 5 forthe DDBmethod is BV5 ϭ $7776, while the DDB-to-SL switch reduces the book value to zero ... The results here are thesame as in parts (b) and (c) above ... ϭ DDB($C$3,0,5,$A7)ϭ VDB($E$3,0,5,$A6,$A7,2,FALSE)Figure 16A–2Depreciation for DDB-to-SL switch using the VDB function, Example 16A ...16A ... When the switch to SL takes place, which is usually in the last 1 to3 years of the recovery period, any remaining basis is charged off in year n ϩ 1 so that the bookvalue reaches zero ... For recovery periods of 15 and 20 years, 150% DB with the half-year convention and theswitch to SL apply ...Only the MACRS rates for the GDS recovery periods (Table 16–4) utilize the DDBto-SL switch ... EXAMPLE 16A ... 3, parts (c) and (d), the DDB-to-SL switching method was applied to a$100,000, n ϭ 5 years asset resulting in PWD ϭ $73,943 at i ϭ 15% ... SolutionTable 16A–4 summarizes thecomputations for depreciation (using Table 16–2 rates), bookvalue, and present worth of depreciation ... This is so, in part, because the half-yearconvention disallows 50% of the first-year DDB depreciation (which amounts to 20% of thebasis) ... TA BLE 16A–4Depreciation and Book Value UsingMACRS,Example 16A ... 200 ... 1920 ... 11520 ... 000tϭ6PWD ϭ͚ D (P͞F,15%,t) ϭ $69,016ttϭ116A ... In the first year, some adjustments have been made to compute theMACRS rate ... The half-year convention is always imposed, and any remaining book value in year n is removed in year n ϩ 1 ... Sincedifferent DB depreciation rates apply for different n values, the following summary maybe used to determine Dt and BVt values ... 435436Chapter 16Depreciation MethodsFor n ϭ 3, 5, 7, and 10 Use DDB depreciation with the half-year convention, switching to SLdepreciation in year t when DSL Ն DDB ...2, and add onehalf year when computing DSL to account for the half-year convention ... [16A ... 12]tϭ12 nBVtϪ1—————n Ϫ t ϩ 1 ... 13]t ϭ 2, 3, ... For n ϭ 15 and 20 Use 150% DB with the half-year convention and the switch to SL whenDSL Ն DDB ... 7]DDB ϭ dt(BVtϪ1)wheredt ϭ{0 ... 50——nt ϭ 2, 3,... 14]EXAMPLE 16A ... (a) Use Equations [16A ... 13] to obtain the annualdepreciation and book value ... Solution(a) With n ϭ 5 and the half-year convention, use the DDB-to-SL switching procedure to obtain the results in Table 16A–5 ... 4(2880) ϭ $11522880DSL ϭ ————— ϭ $11525 Ϫ 4 ϩ 1 ... 13] ...(b) The actual rates are computed by dividing the “Larger Dt” column values by the first costof $10,000 ... t123456dt0 ... 320 ... 11520 ... 057616A ... 5TABLE 16A–5YearsDDBtdtDDB, $SL DepreciationDSL, $LargerDt, $BVt, $0123456—0 ... 40 ... 40 ... But the logic behind the MACRSrates is describedhere for those interested ... The subscripts DB and SL have been inserted along withthe year t ... , t) on d ... , n are(iϭtϪ1dDB,t ϭ d 1 Ϫ(͚diiϭ1iϭtϪ11Ϫ͚di))[16A ... 5[16A ... dSL,nϩ1 ϭ 0 ... 17]The DB and SL rates are compared each year to determine which is larger and when the switchto SL depreciationshould occur ... 6Verify the MACRS rates in Table 16–2 for a 3-year recovery period ... 33, 44 ... 81, and 7 ... SolutionThe fixed rate for DDB with n ϭ 3 is d ϭ 2͞3 ϭ 0 ... Using the half-year convention in year 1and Equations [16A ... 17], the results are as follows:d1:dDB,1 ϭ 0 ... 5(0 ... 3333d2: Cumulativedepreciation rate is 0 ... dDB,2 ϭ 0 ... 3333) ϭ 0 ... 3333dSL,2 ϭ ————— ϭ 0 ... 5(larger value)437438Chapter 16Depreciation Methodsd3: Cumulative depreciation rate is 0 ... 4445 ϭ 0 ... dDB,3 ϭ 0 ... 7778) ϭ 0 ... 7778dSL,2 ϭ ————— ϭ 0 ... 5Both values are the same; switch to straight linedepreciation ... d4 ϭ 0 ... 5(0 ... 0741PROBLEMSFundamentals of Depreciation16 ... 2 What is the difference between book value andmarket value?16 ... 16 ... 16 ... 16 ... S ... irs ... (a) What is the definition of depreciation according to the IRS?(b) What is the description of the term salvagevalue?(c) Whatare the two depreciation systems withinMACRS, and what are the major differencesbetween them?(d) What are the properties listed that cannot bedepreciated under MACRS?(e) When does depreciation begin and end?(f ) What is a Section 179 deduction?Salvage value ϭ 10% of purchasepriceOperating cost (with technician) ϭ $185,000per yearThe manager of the department asked the newesthire to enter the appropriate data in the tax accounting program ... 8 Stahmann Products paid $350,000 for a numericalcontroller during the last month of 2007 and had itinstalled at a cost of $50,000... Stahmann soldthe system at the end of 2011 for $45,000 ... 16 ... The annual depreciation was1͞n using the relevant life value ... Straight Line Depreciation16 ... 16 ... 11 Pneumatics Engineering purchased a machine thathad a first cost of $40,000, an expected useful lifeof 8 years, a recovery period of10 years, and asalvage value of $10,000 ... Theinflation rate is 6% per year and the company’sProblemsMARR is 11% per year ... 16 ... Determine (a) the first cost of the asset and (b) theassumed salvage value ... 13 Lee Company of Westbrook, Connecticut, manufactures pressure relief inserts forthermal reliefand low-flow hydraulic pressure relief applicationswhere zero leakage is required ... If the book value at the end of year 3 is $30,000 andthe company assumed that the machine would beworthless at the end of its 5-year useful life,(a) what is the book depreciation charge each yearand (b)what was the first cost of the machine?16 ... Write a single-cell spreadsheet function todisplay the book value after 5 years of straight linedepreciation ... 16 ... The companyplanned to use the machine for 10 years; however,due to rapid obsolescence it will be retired afteronly 4 years in 2012 ... (a) What isthe amount of capital investment remaining when the asset is prematurely retired?(b) If the asset is sold at the end of 4 years for$175,000, what is the amount of capital investment lost based on straight line depreciation?(c) If the new-technology machine has an estimated cost of $300,000, how manymoreyears should the company retain and depreciate the currently owned machine to make itsbook value and the first cost of the new machine equal to each other?16 ... Tabulate the values for SLdepreciation, accumulated depreciation, and bookvalue for each year if (a) S ϭ 0 and (b) S ϭ $16,000 ...43916 ... S ... It has B ϭ$2,000,000 and a salvage value of 20% of B ... The general managers of the two plants want to know thedifference in (a) the depreciation amount for year5 and (b) the book value after 5 years ... Declining Balance Depreciation16 ... Explain the differences between these rates ...19 Equipment for immersion cooling of electroniccomponents has an installed value of $182,000with an estimated trade-in value of $40,000 after15 years ... 16 ... However, the station will be book-depreciated tozero over a recovery period of 30 years ... (c) What is the impliedsalvage value for DDB? (d)Use a spreadsheet tobuild the depreciation and book value schedulesfor both methods to verify your answers ... 21 A video recording system was purchased 3 yearsago at a cost of $30,000 ... The system is to be replaced this year witha trade-in value of $5000 ... 22 An engineer with Accenture MiddleEast BV inDubai was asked by her client to help him understandthe difference between 150% DB and DDB depreciation ... (a) What are the book values after 12 years forboth methods?(b) How do the estimated salvage and these bookvalues compare in value after 12 years?(c) Which of the two methods,when calculatedcorrectly considering S ϭ $30,000, writes offmore of the first cost over 12 years?440Chapter 16Depreciation Methods16 ... For taxdepreciation, the SL method with n ϭ10 yearswas used, but for book depreciation, Boyditch applied the DDB method with n ϭ 7 years and neglected thesalvage estimate ... (a) Compare the sales price today with the bookvalues using the SL and DDB methods ... 16 ... Therefore, hekeeps two sets of books, one for tax purposes(MACRS) and one for equipment managementpurposes (SL) ... 16 ... The instructor asked her tographically compare the totalpercent of first costdepreciated for an asset costing B dollars over alife of n ϭ 5 years for DDB and 125% DB depreciation ... Use a spreadsheetunless otherwise instructed ... 30 The manager of a Glidden Paint manufacturingplant is aware that MACRS and DDB are both accelerated depreciation methods;however, out of curiosity, she wants to determine which one providesthe faster write-off in the first 3 years for a recentlypurchased mixer that has a first cost of $300,000, a5-year recovery period, and a $60,000 salvagevalue ... The annual MACRS depreciation rates are 20%, 32%, and19 ... MACRSDepreciation16 ... Give an example of each ... 26 What was one of the prime reasons that MACRSdepreciation was initiated in the mid-1980s?16 ... Since the robot isunique in its capabilities, the company expects tobe able to sell it in 4 years for $95,000 ... (b) Determine the book value of the robot attheend of year 2 ... 28 Animatics Corp ... The company purchased an asset 2 yearsago that has a 5-year recovery period ... (a) What was the first cost of the asset?(b) How much was the depreciation charge inyear 1?(c) Develop the complete MACRS depreciationand book value schedule using theVDBfunction ... 31 Railroad cars used to transport coal from Wyomingmines to Texas power plants cost $1 ... Develop the depreciation and book value schedules forthe GDS MACRS method by using two methods ona spreadsheet—the VDB function and the MACRSrates ... 32 A 120-metric-tontelescoping crane that cost$320,000 is owned by Upper State Power ... (a) Compare book valuesfor MACRS and standard SL depreciation over a7-year recovery period ... 16 ... Estimated salvageis $150,000 for any year after 5 years of use ... (a) GDS MACRS where a recovery period of10 years isallowed(b) Double declining balance with a recoveryperiod of 15 years(c) ADS straight line as an alternative to MACRS,with a recovery period of 15 years16 ... com has installed $100,000 worth ofdepreciable software and equipment that represents the latest in Internet teaming and basketcompetition,intended to allow anyone to enjoy441Problemsthe sport on the Web or in the alley ... The company can depreciateusing MACRS for a 5-year recovery period or optfor the ADS alternate system over 10 years usingthe straight line method ... (a) Construct the book value curves for bothmethods on onegraph ... (b) After 3 years of use, what percentage of the$100,000 basis is removed for each method?Compare the two percentages ... 35 A company has purchased special-purpose equipment for the manufacture of rubber products(asset class 30 ... In this case, the ADS alternative toMACRS is requiredfor tax depreciation purposes ... Using a recovery period of 3 years, except for the ADS alternative, which requires a 4-year recovery withhalf-year convention included, prepare a singlegraph showing the annual recovery rates (inpercent) for the three methods ... 38 A sand and gravel pit purchased for$900,000 isexpected to yield 50,000 tons of gravel and 80,000tons of sand per year ... (a) Determine the depletion charge according tothe percentage depletion method ... (b) If taxable income is $100,000 for the year, isthis depletion charge allowed? If not, howmuch is allowed?16 ... The cost of thepurchase was $9 million ... If the companysold 20,000 tons in year 1 and 30,000 tons in year2, what would the depletion charges be each yearaccording to the cost depletion method?16 ... TheColorado mine has the taxable income and salesresults summarized below ... YearTaxableIncome,$Sales,OuncesAverageSales Price,$ per Ounce1231,500,0002,000,000800,000180055002300111512211246Depletion16 ... During the past 3 years the amount of coal removed was 21,000, 18,000, and 20,000 tons,respectively ... Determine (a) the cost depletion allowance foreach year and (b) thepercentage of the purchaseprice depleted thus far ... 37 NA Forest Resources purchased forest acreagefor $500,000 from which an estimated 200 million board feet of lumber is recoverable ... 10 per boardfoot ... In years 3 to 10, however, the company expectsto remove 20 million board feet per year ...(a) Determine the depletion amount in year 2 bythe cost depletion method ... 16 ... During the last 5 years, the amount extracted each year was 60,000, 50,000, 58,00060,000, and 65,000 tons ... 5 million tons of usable stonesand gravel ... 2 million ... (a) Determine the depletion charge each year,usingthe larger of the values for the two depletion methods ... (b) Compute the percent of the initial cost thathas been written off in these 5 years, usingthe depletion charges in part (a) ... 5 million tons remaining,rework parts (a) and (b) ... 42 All of the following types of real property are depreciable except:(a)Warehouses(b) Land(c) Office buildings(d) Test facilities16 ... (b) The taxpayer must use the property in anincome-producing activity ... (d) The property must have a determinable useful life of more than 1 year ... 44 A machine with a 5-year life has a first cost of$20,000 and a $2000 salvage value ...According to theclassical straight line method, the depreciationcharge in year 2 is nearest to:(a) $2800(b) $3600(c) $4500(d) $530016 ... The machine has a first cost of $40,000with a $5000 salvage value ... 00%, 18 ... 40%, respectively ... 46 A coal mine purchased for $5 million has enoughcoal tooperate for 10 years ... The coal isexpected to sell for $150 per ton, with annual production expected to be 10,000 tons ... The depletioncharge for year 6 according to the percentage depletion method would be closest to:(a) $75,000(b) $100,000(c) $125,000(d) $150,00016 ... If the firstcost was $20,000,the salvage value used in the depreciation calculation was closest to:(a) $0(b) $2500(c) $5000(d) $750016 ... If the first cost of the asset was$80,000, the salvage value that was used in the depreciation calculation was closest to:(b) $2000(a) $0(c) $5000(d) $800016 ... 00%, 18 ... 40%, 11 ... 22%,respectively):(a) $58,700(b) $62,400(c) $53,900(d) $46,10016 ... If the MACRS depreciation rates foryears 1, 2, and 3, were 0 ... 32, and 0 ... 51 A lumber company purchased a tract of land for$70,000 that contained an estimated 25,000 usabletrees ... In the first year of operation, the lumbercompany cutdown 5000 trees ... 52 An asset with a first cost of $50,000 and an estimated salvage value of $10,000 is depreciated bythe MACRS method ... 53 Under the General Depreciation System (GDS) ofasset classification, any asset that is not in a statedclass is automatically assigned a recovery period of:(a) 5years(b) 7 years(c) 10 years(d) 15 years16 ... (b) Salvage value is neglected ... (d) The straight line method is required ... Sum-of-Years-Digits Depreciation16A ... 216A ... Use the SYD methodto tabulate annual depreciation and book value ... The salvage value is expected to be 10% of thefirst cost ... If Bϭ $12,000, n ϭ 6 years, and S is estimatedat 15% of B, use the SYD method to determine(a) the book value after 3 years and (b) the rateof depreciation and the depreciation amount inyear 4 ... 416A ... Volvo Motors decided to use the unit-of-production depreciationmethod because the number of testcrashes peryear in which the robot would be involved wasnot estimable ... A new hybrid car was purchased by PedernalesElectric Cooperative as a courier vehicle totransport items between its 12 city offices ... Alternatively, it could have been retained for100,000 miles ... Five-yearDDB depreciation wasapplied ... Use the actual annual miles driven,listed below, to plot the book values for bothmethods ... Show hand orspreadsheet solution, as instructed ... 6An asset has a first cost of $45,000, a recoveryperiod of 5 years, and a $3000 salvage value ... 7If B ϭ $45,000, S ϭ $3000, and n ϭ 5-yearrecoveryperiod, use a spreadsheet and i ϭ18% per year to maximize the present worth ofdepreciation, using the following methods:DDB-to-SL switching (this was determined inProblem 16A ... Given thatMACRS is the required depreciation system inthe United States, comment on the results ... 8HempsteadIndustries has a new milling machinewith B ϭ $110,000, n ϭ 10 years, and S ϭ $10,000 ... Use a spreadsheet to solve this problem ... 9Reliant Electric Company has erected a largeportable building with a first cost of $155,000and an anticipated salvage of $50,000 after25 years ... 10 Verify the 5-yearrecovery period rates forMACRS given in Table 16–2 ... 16A ... A 5-year recoveryperiod and MACRS depreciation have been usedto write off the basis ... Determine the MACRS depreciation, using theswitching rules to find the difference between thebook value and the trade-in value after 3 years ... 12Use the computations in Equations [16A ... 13] to determine the MACRS annual depreciation for the following asset data:B ϭ $50,000 and a recovery period of 7 years ... 13 The 3-year MACRS recovery rates are 33 ... 45%, 14 ... 41%, respectively ... CHAPTER 17After-TaxEconomicAnalysisL E A R N I NGO U T C O M E SPurpose: Perform an after-tax economic evaluation considering the impact of pertinent tax regulations, income taxes,and depreciation ... 1Terminology and rates• Know the fundamental terms and relations of after-taxanalysis; use a marginal tax rate table ... 2CFBT and CFAT•Determine cash flow series before taxes and after taxes ... 3Taxes and depreciation• Demonstrate the tax advantage of accelerateddepreciation and shortened recovery periods ... 4Depreciation recapture• Calculate the tax impact of DR; explain the use ofcapital gains and capital losses ... 5After-taxanalysis• Evaluate one project or multiple alternatives usingafter-tax PW, AW, and ROR analysis ... 6After-tax replacement• Evaluate a defender and challenger in an after-taxreplacement study ... 7EVA analysis• Evaluate an alternative using after-tax economic valueadded analysis; compare to CFATanalysis ... 8Taxes outside the United States• Understand the fundamental practices for depreciationand tax rates in international settings ... 9VAT• Demonstrate the use and computation of a valueadded tax on manufactured products ... The change from estimating cash flow before taxes (CFBT) to cashflow after taxes (CFAT) involves a consideration of significant tax effects that may alter the final decision, and estimates the magnitude of the effect on cash flow over the life of the alternative that taxes may have ... Replacement studies are discussedwith tax effects that occur at the time that a defender isreplaced ... Allthese methods use the procedures learned in earlier chapters, except now with tax effectsconsidered ... Templates for tabulation of cash flow after taxes by hand and by spreadsheetare developed ... S ... irs ... Publications 542, Corporations, and 544, Sales and OtherDispositions of Assets,are especially applicable to this chapter ... 17 ... The perspective of a financial study is that of the corporation and how the tax structure and laws affectprofitability ... There are many types of taxes levied upon corporations and individuals in all countries, including the United States ... Federal governmentsrely on income taxes for asignificant portion of their annual revenue ... for the citizenry ... Income tax is the amount of the payment (taxes) on income or profit that must be delivered to afederal (or lower-level) government unit ... Corporate income taxesare usually submitted quarterly, and the last paymentof the year is submitted with the annual taxreturn ... S ... The website www ... gov provides information on tax laws, rates,publications, etc ... Though the formulas are much more complex when applied to a specific situation, two fundamental relations form the basis for income tax computations ... Taxableincome ϭ revenue Ϫ operating expenses Ϫ depreciationThese terms and relations for corporations are now described ... ) added; t is omitted here for simplicity ... These incomes are listed in the income statement ... ) Other, nonoperating revenues such as sale of assets,license fee income, and royaltiesOE[17 ... A corporation isallowed to remove depreciation, depletion and amortization, and some other ؊ � GI � EBIT are considered separately for tax purposes ... These expenses are tax-deductible for corporations ... Depreciation is not included here since it is not an operating expense ... NOIdeductiblesfrom net operating income in determining the taxable income for a year ... 2]Though there may be subtleties and varying interpretations over time, in essence, the differences between NOI and TI are tax-law-allowed deductibles, such as depreciation ... )Tax rate T is a percentage, or decimalT)[17 ... It is also called net profit after ؊ � TI(1(T)(TI ؊ ) � TI taxes ؊ � TI taxable income� ) T)(TI)[17 ... NOPAT � � applicable tax rate equivalent, of TI that is owed in taxes ... The marginal tax rate is the percentagepaid on the last dollar of income ... The general tax computation relation isIncome taxestaxes (NPAT) ... These are rates for the entire corporation, not for an individualproject, though they are often applied in the after-tax analysis of a single project ... To illustrate the use of the graduated tax rate, assume a company is expected to generate a taxable income of $500,000 in 1 year ... In thiscase, for TI ϭ $500,000,Taxes ϭ 113,900 ϩ 0 ... Taxes ϭ 0 ... 25(75,000 Ϫ 50,000) ϩ 0 ... 39(335,000 Ϫ 100,000) ϩ 0 ... 1Income Tax Terminology and Basic RelationsTABLE 17–1U ... Corporate Income Tax Rate Schedule (2010)If Taxable Income ($) Is:OverBut NotoverTax IsOf theAmountover050,00075,000100,000335,00010,000,00015,000,00018,333,33350,00075,000100,000335,00010,000,00015,000,00018,333,333—15%7,500 ϩ 25%13,750 ϩ 34%22,250 ϩ 39%113,900 ϩ 34%3,400,000 ϩ 35%5,150,000 ϩ 38%35%050,00075,000100,000335,00010,000,00015,000,0000Smallerbusinesses (with TI Ͻ $335,000) receive a slight tax advantage compared to large corporations ... 33 million, there is a flat tax rate of 35% ... The corporate tax rates apply to a corporation as a whole, not to a specific project, unless theproject is the company ... Therefore, thelast dollar of TI is taxed at amarginal rate ... Because the marginal tax rates change with TI, it is not possible to quote directly the percent of TI paid in income taxes ... 5]Average tax rate ϭ ——————— ϭ ———TItaxable incomeReferring to Table 17–1, for a small business with TI ϭ $100,000, the federal income taxburdenaverages $22,250͞100,000 ϭ 22 ... If TI ϭ $15 million, the average tax rate is$5 ... 33% ... For the sake of simplicity, the tax rate used in an economy study is often a single-figure effective tax rate Te, whichaccounts for all taxes ... One reason to use theeffective tax rate is that state taxes are deductible forfederal tax computation ... 6][17 ... 1REI (Recreational Equipment Incorporated) sells outdoor equipment and sporting goods throughretail outlets, the Internet, and catalogs ... Total revenueOperating expensesDepreciation and other allowed deductions$19 ... 6 million$1 ... (b) Find the average federal taxrate paid for the year ... (d) Estimate federal and state taxes using the single-value rate, and compare their total with thetotal in part (a) ... 2] and use Table 17–1 rates for federal taxes due ... 9 million Ϫ 8 ... 8 millionϭ $9 ... 06(TI) ϭ 0 ... 34(8,930,000 Ϫ 335,000) ϭ $3,036,200Total federal and state taxes ϭ3,036,200 ϩ 570,000 ϭ $3,606,200[17 ... 5], the average tax rate paid is approximately 32% of TI ... 3196(c) By Equation [17 ... Te ϭ 0 ... 06)(0 ... 3604(36 ... 5 million from part (a) in Equation [17 ... Taxes ϭ 0 ... 8], this approximation is $182,400 low, a 5 ... It is interesting to understand how corporate taxand individual tax computations differ ... However, for an individual’s taxable income, most of the expenses for living and working are nottax deductible to the same degree as operating expenses are for corporations ... Exemptions are yourself, your spouse, your children, and your other dependents ... Inthe United States, the tax rates for individuals, like those for corporations, are graduated bylevel of TI ... These rates and TI levels are the subject ofongoing political debates at the U ... national level depending upon the balance of power in thecongressional bodies and the economic condition of thecountry ... This process iscalled indexing ... Current information is available on the IRS website www ... gov throughPublication 17, Your Federal Income Tax ... 2 Calculation of Cash Flow after TaxesEarly in the text, the term net cash flow (NCF) was identified as the best estimate of actual cashflow eachyear ... Since then, the annual NCF amounts have been used many times to perform alternative evaluations via the PW,AW, ROR, and B/C methods ... 2Calculation of Cash Flow after Taxeswill be considered, it is time to expand our terminology ... CFBT and CFAT are actual cash flows; that is, theyrepresent the estimated actual flow ofmoney into and out of the corporation that will result from the alternative ... Once theCFAT estimates are developed, the economic evaluation is performed using the same methodsand selection guidelines applied previously ... We learned that net operating incomeS[17 ... Therefore, only in ؉ P ؊ OE ؊ salvage value� GI ؉ initial investment ؊ operating expenses ؊ � gross income (NOI) does not include the purchase or sale of capitalassets ... Incorporating the definitions of gross income andoperating expenses from NOI, CFBT for any year is defined asCFBTyear 0 will the CFBT include P, and only in year n will an S valuebe present ... 10]where taxes are estimated using the relation (T)(TI) or (Te)(TI) ... 2] that depreciation D is considered when calculating TI ... Depreciation is not an operating expense and is a noncash flow ... Therefore, the after-taxengineering economy study must be based on actual cash flow estimates,that is, annual CFAT estimates that do not include depreciation as an expense (negative cash flow) ... Equations [17 ... 10] are now combined ... 11]Suggested table column headings for CFBT and CFAT calculations by hand or byspreadsheet areshown in Table 17–2 ... Expenses OE and initial investment P carry negative signs in alltables and spreadsheets ... It is possible to account for this in a detailed after-tax analysis using carry-forwardand carry-back rules for operating losses ... Rather, the associated negative income tax isconsidered as a taxsavings for the year ... TABLE 17–2YearSuggested Column Headings for Calculation of CFATGrossIncomeGIOperatingExpensesOEInvestmentandSalvageP and S(1)(2)(3)CFBT(4) ϭ(1) ϩ (2) ϩ (3)DepreciationD(5)TaxableIncomeTITaxesCFAT(6) ϭ(7) ϭ (8) ϭ(1) ϩ (2) Ϫ (5) Te(6) (4) Ϫ(7)449450Chapter 17After-Tax Economic AnalysisEXAMPLE 17 ... Wilson plans to purchase listening and detectionequipment for use in the 6-year contract ... Based on the incentive clause in the contract, Wilson estimates that the equipment will increase contract revenue by $200,000 per year andrequire an additional M&O expense of $90,000 per year ... Tabulate and plot the CFBT and CFAT series ... The functions for year 6 are detailed in row 11 ... CFBT: The operating expenses OE and initial investment P are shown as negative cash flows ... CFBT is calculated by Equation[17 ... In year 6, forexample, when the equipment is sold, the function in row 11 indicates thatCFBT6 ϭ 200,000 Ϫ 90,000 ϩ 150,000 ϭ $260,000CFAT: Column F for MACRS depreciation, which is determined using the VDB functionover the 6-year period, writes off the entire $550,000 investment ... TI4 ϭ GI Ϫ OE Ϫ D ϭ200,000 Ϫ 90,000 Ϫ 63,360 ϭ $46,640Taxes4 ϭ (0 ... 35) (46,640) ϭ $16,324CFAT4 ϭ GI Ϫ OE Ϫ taxes ϭ 200,000 − 90,000 Ϫ 16,324 ϭ $93,676In year 2, MACRS depreciation is large enough to cause TI to be negative ($−66,000) ... CommentMACRS depreciates to a salvage value of S ϭ 0 ... Figure 17–taxes in year t)(P͞F, i, t)[17 ... Compare any two depreciation ͚ ) � 1Computation of CFBT and CFAT using MACRS depreciation and Te ϭ 35%, Example 17 ... 17 ... Larger rates in earlieryears of the recovery period require less taxes due to the larger reductions in taxable income ... For the17 ... t� nPWtaxmethods ... On the basis of theseassumptions, the following statements are correct:The total taxes paid are equal for all depreciation methods ... As we learned in Chapter 16, MACRS is the prescribed tax depreciation method in the UnitedStates, and the only alternative is MACRS straight linedepreciation with an extended recoveryperiod ... If the DDB method were still allowed directly, rather than embedded inMACRS, DDB would not fare as well as MACRS ... This is illustrated in Example 17 ... EXAMPLE 17 ... The CFBT for the machine is estimated at $20,000 ... Use a 6-year period for thecomparison to accommodate the half-yearconvention imposed by MACRS ... For classical straight line depreciation with n ϭ 5, Dt ϭ $10,000 for 5 years and D6 ϭ0 (column 3) ... The classical DDB percentage of d ϭ 2͞n ϭ 0 ... The implied salvagevalue is $50,000 − 46,112 ϭ $3888, so not all $50,000 is tax-� deductible ... 35) ϭ $1361 larger than for the classical SL method ... 0Taxes, $(6)D t, $(7)TI, $(8) Total taxes are $24,500,the same as for classical SL depreciation ... Notethe pattern of the curves, especially the lower total taxes relative to the SL model after year 1for MACRS ... 3 ... 53)01( ( 11 ... ) 53)7(and in years 1 through 4 for DDB ... The PWtax values at the bottom of Table 17–3 are calculated usingEquation [17 ... The MACRS PWtax value is the smallest at $18,162 ... It can be shown that a shorterrecovery period will offer a tax advantage over a longer period using the criterion to minimizePWtax... The present worth of taxes is less for smaller n values ... Example 17 ... EXAMPLE 17 ... This iscommon for multinational corporations ... The second set is for foreign government purposes, such asdepreciation and taxes ... Demonstrate thetax advantage for the smaller n if net operating income (NOI)is $30,000 per year, an effectivetax rate of 35% applies, invested money is returning 5% per year after taxes, and classical SLdepreciation is allowed ... SolutionDetermine the annual TI and taxes by Equations [17 ... 3] and the present worth oftaxes using Equation [17 ... 17 ... 35)(20,000) ϭ $7000 peryearPWtax ϭ 7000(P͞A,5%,9) ϭ $49,755Total taxes ϭ (7000)(9) ϭ $63,000Recovery period n ϭ 5 years:Use the same comparison period of 9 years, but depreciation occurs only during the first 5 years ... 35)(30,000 Ϫ 18,000) ϭ $4200(0 ... However, the more rapid write-off forn ϭ 5 results in a present worthtax savings of nearly $2400 (49,755 Ϫ 47,356) ... 4 Depreciation Recapture and CapitalGains (Losses)All the tax implications discussed here are the result of disposing of a depreciable asset before, at,or after its recovery period ... The key is the size of the selling price (or salvage or market value)relativeto the current book value at disposal time and relative to the first cost, which is called theunadjusted basis B in depreciation terminology ... Depreciation recapture DR, also called ordinary gain, occurs when a depreciable asset issold for more than the current book value BVt ... 13]For an after-taxstudy, thetaxeffect is:CG: Taxed at Te(after CL offset)plusDRDR: Taxed at TeSP2DRBookvalue BVtCLSP3CL: Can only offsetCGZero, $0Figure 17–3Summary of calculations and tax treatment for depreciation recapture and capital gains (losses) ... In the United States, an amountequal to the estimated salvagevalue can always be anticipated as DR when the asset is disposed of after the MACRS recovery period ... The amount of DR is treated as ordinary taxable incomein the year of asset disposal ... See Figure 17–3 ... 14]Since future capital gains are difficult to predict, they are usually not detailed in an after-tax economy study ... When the selling price exceeds B, the TI due to the sale is the capital gain plus the depreciation recapture, as shown in Figure 17–3 ... Capital loss CL occurs when a depreciable asset is disposed of for less than its current bookvalue ... 15]An economic analysis does not commonlyaccount for capital loss, simply because it is notestimable for a specific alternative ... For the purposes of theeconomic study, this provides a tax savings in the year of replacement ... These savings are assumed to be offset elsewhere in the corporation by other income-producing assets that generatetaxes ... • U ... tax law defines capital gains as long-term (items retained for more than 1 year) or short-term ... The termcapital gain correctly applies at sale time to property such as investments (stocks and bonds),art, jewelry, land, and the like ... Corporate tax treatment at thistime is the same for both:taxed as ordinary income ... • Capital gains are taxed as ordinary taxable income at the corporation’s regular tax rates ... The terms usedthen are net capital gains (losses) ... • When an asset is disposed of, the tax treatment is referred to as a Section 1231 transaction,which is the IRS rule section of thesame number ... • All these rules apply to corporations ... If the three additional income and tax elements covered here are incorporated into Equation [17 ... 16]17 ... e ... Only when a capital gain or loss must be included due to the nature of the problem will the calculations involve it ... 5Biotech, a medicalimaging and modeling company, must purchase a bone cell analysis system foruse by a team of bioengineers and mechanical engineers studying bone density in athletes ... The effective tax rate is 35% ... Analyzer 1Analyzer 2150,00030,0005225,00010,0005Basis B, $Operating expenses, $ peryearMACRS recovery, yearsAnswer the following questions, solving by hand and spreadsheet ... Which analyzer should be purchased?(b) Assume that 3 years have now passed, and the company is about to sell the analyzer ... Solution by Hand(a) Table 17–4 details the tax computations ... Equation [17... Taxes for the 3-year period are summed, with no consideration ofthe time value of money ... (b) When the analyzer is sold after 3 years of service, there is a depreciation recapture (DR)that is taxed at the 35% rate ... For each analyzer, account for the DR by Equation [17 ... 16], TI ϭ GI − OE − D ϩ DR... TABLE 17–4YearGrossIncomeGI, $Comparison of Total Taxes for Two Alternatives, Example 17 ... 35TI, $150,000120,00072,00043,20040,00022,00041,20014,0007,70014,420Analyzer 10123150,000100,000100,000100,00030,00030,00030,00030,00048,00028,80036,120Analyzer20123225,000100,000100,000100,00010,00010,00010,00045,00072,00043,200225,000180,000108,00064,80045,00018,00046,80015,7506,30016,38038,430455456Chapter 17After-Tax Economic AnalysisDR ϭ 130,000 Ϫ 43,200 ϭ $86,800Year 3 TI ϭ 100,000 Ϫ 30,000 Ϫ 28,800 ϩ 86,800 ϭ$128,000Year 3 taxes ϭ (0 ... 35)(207,000) ϭ $72,450Total taxes ϭ 15,750 ϩ 6300 ϩ 72,450 ϭ $94,500Analyzer 2:Now, analyzer 1 has a considerable advantage in total taxes ($94,500 versus $66,500) ... Similar analysis in rows 14 to 18 results in totaltaxes of $38,430 for analyzer 2, indicating that thecompany should select analyzer 1, basedon taxes only ... The new TI in year 3 has the depreciation recapture incorporated as DR ϭ selling price Ϫ book value ϭ SP Ϫ BV3, whichis shown in the cell tag as the last term (D10 Ϫ F10) ... CommentNote that no time value of money is considered in theseanalyses, as we have used in previousalternative evaluations ... 5 below we will rely upon PW, AW, and ROR analysesat an established MARR to make an after-tax decision based upon CFAT values ... 5 ... 5 After-Tax EvaluationThe required after-tax MARR is established using the market interest rate,the corporation’s effective tax rate, and its weighted average cost of capital ... When positive and negative CFAT values are present, aPW or AW Ͻ 0 indicates the MARR is not met ... The guidelines are:17 ... PW or AW Ն 0, the project is financially viable because the after-tax MARR ismet or exceeded ...Select the alternative with the best (numerically largest) PW or AWvalue ... Assign a plus sign to each saving and apply the guidelines above ... This requirement must be met for everyanalysis—before or after taxes ... For AW analysis: Use the PMT function with an embedded NPV function over one lifecycle ... ϭ ϪPMT(MARR,n,NPV(MARR,year_1: year_n) ϩ year_0)[17 ... (There is an LCM function in Excel ... The general format isϭ ϪPV(MARR,LCM_years,PMT_result_cell )[17 ... 6Paul is designing the interior walls of an industrial building ... Two construction options—stucco on metallath (S) and bricks(B)—each have about the same transmission loss, approximately33 decibels ... Paul has estimated the first costs and after-tax savings each year for both designs ... (b) Use a spreadsheet to select the alternative and determine the required first cost for theplans to break even ... Develop the AW relationsusingthe CFAT values over each plan’s life ... AWS ϭ [−28,800 ϩ 5400(P͞A,7%,6) ϩ 2040(P͞A,7%,4)(P͞F,7%,6)ϩ 2792(P͞F,7%,10)](A͞P,7%,10)ϭ $422AWB ϭ [−50,000 ϩ 14,200(P͞F,7%,1) ϩ · · · ϩ 10,600(P͞F,7%,5)](A͞P,7%,5)ϭ $327Both plans are financially viable; select plan S because AWS is larger ... Usethe AW values and the P͞A factor for theLCM of 10 years to select stucco on metal lath, plan S ... 0236) ϭ $2964PWB ϭ AWB(P͞A,7%,10) ϭ 327(7 ... 6 ... 17], and row 15 shows the 10-year PW that results from the PV function inEquation [17 ... Plan S is chosen by a relatively small margin ... This is asmallreduction from the $−50,000 first cost initially estimated ... If the minus is omitted, the AW and PW valueshave the wrong sign and it appears that the plans are not financially viable in that they do notreturn at least the after-tax MARR ... To utilize the ROR method, apply exactly the same proceduresas in Chapter 7 (single project) and Chapter 8 (two or more alternatives) to the CFAT series ... Multiple roots may exist in the CFAT series, as they can for any cash flow series ... tϭnPresent worth:0ϭ͚ CFAT (P͞F,i*,t)t[17 ... [17 ... 5After-Tax EvaluationSpreadsheet solution for i* is faster for most CFATTe[17 ... To approximate ؊ 1� ——————— series ... 21]If the after-tax ROR is important to the analysis, but the details of an after-tax study are not ofinterest, the before-tax ROR (or MARR) can be adjusted with the effective tax rate Te by usingthe approximating relationafter-tax RORBefore-tax RORtheeffect of taxes without performing the details of an after-tax study, the before-tax MARR can beestimated as0 ... 40If the decision concerns the economic viability of a project and the resulting PW or AW value isclose to zero, the details of an after-tax analysis should be developed ... 7A fiber opticsmanufacturing company operating in Hong Kong has spent $50,000 for a5-year-life machine that has a projected $20,000 annual NOI and annual depreciation of$10,000 for years 1 through 5 ... (a) Determine the after-tax rateof return ... Solution(a) The CFAT in year 0 is $Ϫ50,000 ... (See Equations [17... 9] ... TI ϭ NOI Ϫ D ϭ 20,000 Ϫ 10,000 ϭ $10,000Taxes ϭ Te (TI) ϭ 0 ... 19] ... 125Solution gives i* ϭ 18 ... (b) Use Equation [17 ... 0 ... 30051 Ϫ 0 ... 05%)The actual before-tax i* using CFBT ϭ $20,000 for 5 years is 28 ... 05% is used in a before-taxanalysis ... Solution by spreadsheet is accomplishedusing the incremental CFAT valuesand the IRR function ... 4 through 8 ... You should review and understand these sections before proceeding with this section ... Incremental ROR: Incremental analysis must be performed ... Equal-service requirement: Incremental ROR analysis requires that thealternatives be evaluated over equal time periods ... (The only exception, mentioned in Section 8 ... )Revenue and cost alternatives: Revenue alternatives (positive and negative cash flows) maybe treated differently from cost alternatives (cost-only cash flow estimates) ... Alternatives withi* Ͻ MARR canbe removed from further evaluation ... Breakeven ROROnce the CFAT series are developed, the breakeven ROR can be obtained using a plot of PWversus i* by solving the PW relation for each alternative over the LCM at several interest rates ... The next examples solve CFAT problems usingincremental ROR analysis and the breakevenROR plot of PW versus i ... 8In Example 17 ... Figure 17–5a presented both a PW analysis over 10 years and an AW analysis over the respective lives ... After reviewing this earlier solution, (a) perform an RORevaluation at the after-tax MARR of 7% per yearand (b) plot the PW versus ⌬i graph to determine the breakeven ROR ... Apply the procedure in Section 8 ... Figure 17–6 shows the estimated CFAT for each alternative and the incrementalCFAT series ... Row 14 indicates they do ... 35% ... Plan S isselected, the same as with the PW and AW methods... The graph indicates that the breakeven ⌬i* occurs at 6 ... Whenever the after-tax MARR is above 6 ... CommentNote that the incremental CFAT series has three sign changes ... Accordingly, there may be multiple ⌬i* values ... 17 ... 35%Figure 17–6Incremental evaluation of CFAT and determination ofbreakeven ROR, Example 17 ... EXAMPLE 17 ... 5 an after-tax analysis of two bone cell analyzers was initiated due to a new3-year NBA contract ... The complete solution is in Table 17–4 (hand) and Figure 17–4 (spreadsheet) ... 5: $130,000 for analyzer 1 and $225,000 for analyzer 2 ... SolutionAspreadsheet solution is presented here, but a hand solution is equivalent, just slower ... The CFAT series (column I) are determined by the relation CFAT ϭ CFBT Ϫtaxes, with the taxable income determined using Equation [17 ... Forexample, in year 3 when analyzer 2 is sold for S ϭ $225,000, the CFATcalculation isCFAT3 ϭ CFBT Ϫ (TI)(Te) ϭ GI Ϫ OE Ϫ P ϩ S Ϫ (GI Ϫ OE Ϫ D ϩ DR)(Te)The depreciation recapture DR is the amount above the year 3 book value received at sale time ... CFAT3 ϭ 100,000 Ϫ 10,000 ϩ 0 ϩ 225,000Ϫ(100,000 Ϫ 10,000 Ϫ 43,200 ϩ 160,200)(0 ... 35) ϭ $242,550The cell tags inrow 14 of Figure 17–7 follow this same progression ... These are revenue alternatives, so the overall i* values indicate that both CFAT series areacceptable ... 6% (cell J17) also exceeds MARR ϭ 10%, so analyzer 2 isselected ... 461462Chapter 17After-Tax Economic AnalysisOverall i*ϭB14ϩC14ϪE14ϩ(D14ϪF14)This termcalculatesDR ϭ $160,200CFAT calculationϭ B14ϩC14ϩD14ϪH14Incremental i*ϭ IRR(J11:J14)Figure 17–7Incremental ROR analysis of CFAT with depreciation recapture, Example 17 ... CommentIn Section 8 ... The incrementalROR must be used ... If the larger i*alternativeis chosen, analyzer 1 is incorrectly selected ... For verification, the PW at 10% is calculatedfor each analyzer (column I) ... 17 ... The final decisionmay not be reversed by taxes, but the difference between before-tax AW values may be significantly different from the after-tax difference ...Additionally, the after-tax replacement study considerstax-deductible depreciation and operating expenses not accounted for in a before-tax analysis ... The same procedure as the before-tax replacement study in Chapter 11 is applied here, but forCFAT estimates ... Special attention to Sections 11 ... 5 is

recommended ... 10 presents a solution by hand of an after-tax replacement study using a simplifying assumption of classical SL (straight line) depreciation ... 11 solves the same problem by spreadsheet, but includes the detail of MACRS depreciation ... EXAMPLE 17 ... Management has discovered thatit is technologically and legally outdated now ... If a market value of $400,000 is offered as the trade-in for the currentequipment, perform a replacement study using (a) a before-tax MARR of 10% per year and17 ... Assume an effective tax rate of 34% ... DefenderMarket value, $First cost, $Annual cost,$͞yearRecovery period, yearsChallenger400,000Ϫ100,0008 (originally)Ϫ1,000,000Ϫ15,0005SolutionAssume that an ESL (economic service life) analysis has determined the best life values to be5 more years for the defender and 5 years total for the challenger ... The defender AW uses the market valueas the first cost, PD ϭ $Ϫ400,000 ... 3), we select the better AWvalue ... Thedefender has a $73,280 lower equivalent annual cost compared to the challenger ... (b) For the after-tax replacement study, there are no tax effects other than income tax for thedefender ... Dt ϭ 600,000͞8 ϭ $75,000t ϭ 1 to 8yearsTable 17–5 shows the TI and taxes at 34% ... (Remember that for tax savings in an economicTA BLE 17–5Before-Tax and After-Tax Replacement Analyses, Example 17 ... 34TI, $75,00075,00075,00075,00075,000AW at7%Ϫ175,000Ϫ175,000Ϫ175,000Ϫ175,000Ϫ175,000Ϫ59,500Ϫ59,500Ϫ59,500Ϫ59,500Ϫ59,500ϩ25,000†Ϫ215,000Ϫ215,000Ϫ215,000−215,000Ϫ215,000‡8,500Ϫ73,100Ϫ73,100Ϫ73,100Ϫ73,100Ϫ73,100CFAT, $Defender345678AW at10%012345Ϫ400,000Ϫ100,000Ϫ100,000Ϫ100,000Ϫ100,000Ϫ100,0000Ϫ400,000Ϫ100,000Ϫ100,000Ϫ100,000Ϫ100,000Ϫ100,000Ϫ205,520Ϫ400,000Ϫ40,500Ϫ40,500Ϫ40,500Ϫ40,500Ϫ40,500Ϫ138,056Challenger012345AW at10%*Ϫ1,000,000Ϫ15,000Ϫ15,000Ϫ15,000Ϫ15,000Ϫ15,0000Ϫ1,000,000Ϫ15,000Ϫ15,000Ϫ15,000Ϫ15,000Ϫ15,000Ϫ278,800Minus sign indicates a tax savings for the year ... ‡Assumes challenger’s salvage actually realized is S ϭ 0; no tax ... ) Since only costs are estimated, the annual CFAT is negative,but the$59,500 tax savings has reduced it ... In year0 for the challenger, Table 17–5 includes the following computations to arrive at a taxof $8500 ... 34(25,000) ϭ $8500The SL depreciation is $1,000,000͞5 ϭ $200,000 per year ... 34) ϭ $Ϫ73,100CFAT ϭ CFBT Ϫ taxes ϭ Ϫ15,000 Ϫ (Ϫ73,100) ϭ $ϩ58,100Inyear 5, it is assumed the challenger is sold for $0; there is no depreciation recapture ... Conclusion: By either analysis, retain the defender now and plan to keep it for 5 moreyears ... If andwhen cash flow estimates change significantly, perform another replacement analysis ... The resulting taxsavingswould decrease the CFAT (which is to reduce costs if CFAT is negative) of the challenger ... The CFAT is then $Ϫ1,000,000 Ϫ (Ϫ8500) ϭ$Ϫ991,500 ... 11Repeat the after-tax replacement study of Example 17 ... Assume eitherasset is sold after 5 years for exactly its book value ... SolutionFigure17–8 shows the complete analysis ... Again the defender is selected for retention, but now by an advantage of $44,142 annually ... Therefore, taxes and MACRS have reduced the defender’s economic advantage, butnot enough to reverse the decision to retain it ... There is depreciation recapture in year 0of the challenger due to trade-in of the defender at $400,000, a value larger than the book value of the 3-year-old defender ... 7465After-Tax Value-Added AnalysisDepreciation recaptureϭϪC5 Ϫ F11Challenger sales priceϭ B15 Ϫ F24Figure 17–8After-tax replacement study with MACRS depreciation anddepreciation recapture, Example 17 ... $137,620 (cell G18), is treated as ordinary taxable income ... 34)(137,620) ϭ $46,791(cell H18)See the cell tags and table notes that duplicate this logic ... The entry $57,600 (C23) reflects this assumption, since the forgoneMACRS depreciation in year 6 would be1,000,000(0 ... The spreadsheetrelation ϭ B15 Ϫ F24 determines this value using the accumulated depreciation in F24 ... This implies an additional tax saving of 57,600(0 ... Conversely, if the salvage value exceeds the book value, a depreciation recapture and associatedtax should be estimated ... 7 After-Tax Value-Added AnalysisWhen a person or company is willing to pay more for an item, it is likely that some processinghas been performed on an earlier version of the item to make it more valuable now to the purchaser ... Value added is a term used to indicate that a product or service has added worthfrom theperspective of a consumer, owner, investor, or purchaser ... Value added466Chapter 17After-Tax Economic AnalysisFor an example of highly leveraged value-added activities, consider onions that are grown andsold at the farm level for cents per pound ... 25 per pound ... Thus, from theperspective of the consumer, there has been a large amount of value added by the processing from rawonions in the ground into onion rings sold at a restaurant or fast-food shop ... When value-added analysis is performed after taxes, the approach is somewhat differentfrom that of CFAT analysisdeveloped previously in this chapter ... Value-added analysis starts with Equation [17 ... Depreciation D is included in that TI ϭ GI ϪOE − D ... The term economic value added (EVA) indicates the monetary worth added by an alternativeto the corporation’s bottom line ... ) The technique discussed belowwas first publicized in several articles1 in the mid-1990s, and it has sincebecome very popular as a means to evaluate the ability of a corporation to increase its economicworth, especially from the shareholders’ viewpoint ... That is, EVA indicates the project’s contribution to the netprofit of the corporationThis financial worth is the amount used in public ... 17 (] 1؊i)(BVt ( ؊ ) Te ؊ � TI(1(1 ؊ after-tax interest rate)(book value in year t ؊ ) cost of invested capital� NOPAT ؊ � NOPAT after taxes ... This is the interest incurred by the currentlevel of capital invested in the asset ... ) Computationally,EVAdocuments of the corporation(balance sheet, income statement, stock reports, etc ... The result of an EVA analysis is a series of annual EVA estimates ... If only one project is evaluated, AW Ͼ 0 means the after-tax MARR is exceeded, thus making the project value-adding ... Thus, either method can beused to make a decision ... This comparison is made in Example 17 ... 1A ... 1997, pp ... Freedman, “How Do You Add Up?”Chemical Week, Oct ... 31–34 ... G ... L ... ” The Engineering Economist, vol ... 2 (2000), pp ... 17 ... 12Biotechnics Engineering has developed two mutually exclusive plans forinvesting in newcapital equipment with the expectation of increased revenue from its medical diagnostic services to cancer patients ... (a) Use classical straight linedepreciation, an after-tax MARR of 12%, and an effective tax rate of 40% to perform two annual worth after-tax analyses: EVA and CFAT ...Plan APlan BInitial investment, $Gross income Ϫ expenses, $Ϫ500,000170,000 per yearEstimated life, yearsSalvage valueNoneϪ1,200,000600,000 in year 1, decreasing by 100,000per year thereafter4None4Solution by Spreadsheet(a) Refer to the spreadsheet and function cells (row 22) in Figure 17–9... The net operating profit after taxes (NOPAT) in column H is calculated by Equation [17 ... The book values (column E) are used to determine thecost of invested capital in column I, using the second term in Equation [17 ... This represents the amount of interest at12% per year, after taxes, for thecurrently invested capital as reflected by the book valueat the beginning of the year ... Notice there is no EVA estimate for year 0, since NOPAT and the cost of investedcapital are estimated for years 1 through n ... CFAT evaluation: As shown in function row 22 (plan B for year 3), CFAT estimates(columnK) are calculated as (GI Ϫ OE) Ϫ P Ϫ taxes ... (b) What is the fundamental difference between the EVA and CFAT series in columns J and K?They are clearly equivalent from the time value of money perspective since the AW valuesare numerically the same ... To obtain the AW of EVA estimate of$Ϫ12,617 forFigure 17–9Comparison of two plans using EVA and CFAT analyses, Example 17 ... 467468Chapter 17After-Tax Economic Analysisyears 1 through 4, the initial investment of $500,000 is distributed over the 4-year life usingthe A͞P factor at 12% ... In effect, the yearlyCFAT is reduced by thischarge ... However, the EVA method indicates an alternative’s yearly estimated contribution to the value of the corporation, whereas the CFAT method estimates the actual cashflows to the corporation ... CommentThe calculation P(A͞P,i,n) ϭ $500,000(A͞P,12%,4) is exactly the same as the capitalrecoveryin Equation [6 ... Thus, the cost of investedcapital for EVA is the same as the capital recovery discussed in Chapter 6 ... 17 ... 7], taxes ϭ(Te)(TI) ... Expense deductions vary widely from country tocountry ... CanadaDepreciation: This is deductible and is normally based on DB calculations,although SL may beused ... Theannual tax-deductible allowance is termed capital cost allowance (CCA) ... S ... Class and CCA rate: Asset classes are defined and annual depreciation rates are specified byclass ... There are some 44 classes, and CCA rates vary from 4% per year (the equivalent of a25-year-life asset) for buildings (class 1) to 100% (1-year life) for applications software, chinaware, dies, etc ... Most rates are in the range of 10% to 30% per year ... Expenses related to capital investments are not deductible, since they are accommodated through the CCA ... cra ... ca in theForms andPublications section ... The selected industries and assets can change over time; currently favored industries serve areas such as technology and oil exploration, and equipment subjected to large vibrations during normal usage isallowed accelerated depreciation ... 8After-Tax Analysis for InternationalProjectsRecovery period: Standardized recovery periods are published that vary from 3 years (electronicequipment) to 10 years (aircraft, machinery, and other production equipment) to 20 years (buildings) ... Expenses: Business expenses are deductible with some limitations and some special incentives... Incentives are generous in some cases; for example, 150% of actual expenses is deductible for new technology and new product R&D activities ... worldwide-tax ... MexicoDepreciation: This is a fully deductible allowance for calculating TI ... For some asset types, an immediate deduction of apercentage of the first cost is allowed ... )Class and rates: Asset types are identified, though not as specifically defined as in some countries ... Most rates range from10% to 30% per year ... Most business expenses are deductible ... Tax on Net Assets (TNA): Under some conditions, a tax of 1 ... Internet:The best information is via websites for companies that assist international corporations located in Mexico ... pwcglobal ... JapanDepreciation: This is fully deductible and based on classical SL and DB methods ... Switching to classical SL depreciation must takeplace in the year that the accelerated rateamount falls below the corresponding SL amount ... Expenses: Business expenses are deductible in calculating TI ... mof ... jp ... Some countries levy taxes onlyat the federal level, while others impose taxes at several levels of government (federal, state orprovincial, prefecture, county, and city) ... Theseinclude incometaxes at all reported levels of government within each country; however, other types of taxesmay be imposed by a particular government ... A close examination of international rates shows that they have decreased significantly over the last decade ... 7% (1999) to 25 ... This hasencouraged corporate investment andbusiness expansion within country borders and helped soften the massive economic downturnexperienced in recent years ... kpmg ... aspx) and from country websites on corporate taxation ... These taxes are usually in the form of a value-added tax (VAT), goods andservices tax (GST), and taxes on products imported from outside its borders ... This has been especially true during the first decade of the 21stcentury ... The VAT system is explained now ... 9 Value-Added TaxA value-added tax (VAT) has facetiously been called a sales tax on steroids, because theVATrates on some items in some countries that impose a VAT can be as high as 90% ... A value-added tax is an indirect tax; that is, it is a tax on goods and services rather than onpeople or corporations ... A specific percentage, say 10%, is a charge added to the price of theitem and paid by the buyer ...This process of 10% VAT continues every time the item is resold—as purchased or in a modified form—thus the term value-added ... In some countries, VAT is used inlieu of business or individual income taxes ... Infact, the United States is the only major industrialized country in the world that does nothave aVAT system, though other forms of indirect taxation are used liberally ... A sales tax is used by the U ... government, by nearly all of the states, and by many local entities ... That is, businesses do not pay a sales tax on raw material, unfinished goods,or items they purchase that will ultimately besold to an end user; only the end user pays the salestax ... Total sales tax percentages imposed by multiple government levels can range from 5% to 11%, sometimes largeron specific items ... , Home Depot does not pay a sales tax on the microwave ovens because they will be soldto HD customers whowill pay the sales tax ... Thus, a sales tax is paid only one time, and that17 ... The sales tax is the responsibility of the merchant to collect and remand to the taxing entity ... The seller sends the collected VAT to the taxing entity ... Now, this second seller will send to the taxing entity an amount that isequalto the total tax collected minus the amount of VAT already paid ... S ... Here is how the VATmight work ... As part of the price, Northshore collects $110,000, that is, $100,000 ϩ 0 ... Northshore remits the $10,000 VAT to the U ... Treasury ... Westfall collects $330,000 from GE and then sends $20,000 tothe U ... Treasury, that is, $30,000 it collected in VAT from GE minus $10,000 it paid in taxes to NorthshoreMining ... GE collects $770,000 and then remits $40,000 to the U ... Treasury, that is, $70,000 it collected in taxes from retailers minus $30,000 it paid in VAT toWestfall Steel ... S ... For example, ifGEpaid $5000 in taxes on motors it purchased for the refrigerators, the amount GE would remitto the U ... Treasury would be $35,000 (that is, $70,000 it collected from retailers minus$30,000 it paid in taxes to Westfall Steel minus $5000 it paid in taxes on the motors) ... The retailers remit $25,000 to theU ... Treasury (that is, $95,000 they collectedminus $70,000 they paid previously) ... S Treasury has received $10,000 from Northshore, $20,000from Westfall Steel, $35,000 from GE, $5000 from the supplier of the motors, and $25,000from the retailers, for a total of $95,000 ... TheVAT money wasdeposited into the Treasury at several different times from several differentcompanies ... The taxes that a company collects are called output taxes, and these are forwarded to the taxingentity, less the amount of input taxes the company paid ... Several dimensions of a VAT distinguish it from a sales tax orcorporate income taxes ... • The end user pays all of the value-added taxes, but VATs are not as obvious as a sales tax thatis added to the price of the item at the time of purchase (and displayed on the receipt) ... • Value-added taxes are generally considerably higher than sales taxes, with theaverageEuropean VAT rate at 20% and the worldwide average at 15 ... • The VAT is essentially a “sales tax,” but it is charged at each stage of the product developmentprocess instead of when the product is sold ... • VAT rates vary from country to country and from category to category ... 471472Chapter17After-Tax Economic AnalysisEXAMPLE 17 ... It has three differentmanufacturing units that specialize in manufacturing different transportation-related products, such as trucks, engines and axles, commercial vehicles, utility vehicles, and passenger cars ... In oneparticular accounting period, Tata hadinvoices from four different suppliers (vendors A,B, C, and D) in the respective amounts of $1 ... 8 million, $1 ... The products Tata purchased were subject to VAT rates of 4%, 4%, 12 ... (a) How much total VAT did Tata pay to its vendors?(b) Assume that Tata’s products have a VAT rate of 12 ... If Tata’ssales during the periodwere $9 ... Solve for X and then subtract itfrom the purchase amount to determine the VAT charged by each vendor ... An example computation for vendor A isas follows:X ϩ 0 ... 04X ϭ 1,500,000X ϭ $1,442,308VATA ϭ 1,500,000 Ϫ 1,442,308ϭ $57,692Total VAT paid ϭ 57,692 ϩ146,154 ϩ 122,222 ϩ 162,295ϭ $488,363(b) Total from Tata ϭ total VAT Ϫ VAT paid by vendorsϭ 9,200,000(0 ... 13VendorPurchases, $VAT Rate, %Price before VAT, X, $VAT, $ABCD1,500,0003,800,0001,100,000900,0004 ... 012 ...01,442,3083,653,846977,778737,70557,692146,154122,222162,295Total488,363CHAPTER SUMMARYAfter-tax analysis does not usually change the decision to select one alternative over another;however, it does offer a much clearer estimate of the monetary impact of taxes ... Income tax rates for U... corporations and individual taxpayers are graduated or progressive—higher taxable incomes pay higher income taxes ... Taxes are reduced because of tax deductible items,such as depreciation and operating expenses ... Accordingly, key general cash flow after-tax relations for each year areasfollows:NOI ϭ gross income Ϫ expensesTI ϭ gross income Ϫ operating expenses Ϫ depreciation ϩ depreciation recaptureCFBT ϭ gross income Ϫ operating expenses Ϫ initial investment ϩ salvage valueCFAT ϭ CFBT Ϫ taxes ϭ CFBT Ϫ (Te)(TI)If an alternative’s estimated contribution to corporatefinancial worth is the economic measure,the economic value added (EVA) should be determined ... The equivalent annual worths of CFAT and EVA estimates are the samenumerically, because they interpret the annual cost of the capital investment in different, butequivalent manners when the time valueof money is taken into account ... The replacementstudy procedure of Chapter 11 is applied ... International corporate tax rates have steadily decreased, but indirect taxes, such as valueadded tax (VAT), have increased ... The United States currently has no VAT ... 1 (a)Define the following tax terms:graduated taxrates, marginal tax rate, and indexing ... 17 ... 17 ... 17 ... (a) A new machine had a first-year write-off of$10,500 ... (c) An asset with a book value of $8000 was retired and sold for $8450 ... (e) An asset with a MACRS recovery period of 7years has been owned for 10 years ... (f )(g)(h)(i)Thecost of goods sold in the past year was$3,680,200 ... Based on winners holding these tickets, a rebate of $350was sent to the manager ... The cost to maintain equipment during thepast year was $641,000 ... 5 Two companies have the following values on theirannual tax returns ... Determine the percentof sales revenue eachcompany will pay in federal income tax ... Determine the percentage error made relative to the exact taxes inpart (a) ... 6 Last year, one division of Hagauer ... This year, TI isestimated to be $600,000 ... (a) What was the average federal tax rate paidlast year?(b) What is the marginalfederal tax rate on theadditional TI this year?(c) What will be the average federal tax rate thisyear?(d) What will be the NOPAT on just the additional $350,000 in taxable income?17 ... 5 million for the year ... 3 million ... 2% ... 17 ... If the state income tax rate is 7%, determine the(a) average federal taxrate, (b) overall effectivetax rate, (c) total taxes to be paid based on the effective tax rate, and (d) total taxes paid to the stateand paid to the federal government ... 9 The taxable income for a motorcycle sales and repair business is estimated to be $150,000 this year ... A new engine diagnostics systemwill cost $40,000 andhave an average annual depreciation of $8000 ... Compute the expected change in income taxes forthe year if the new system is purchased ... 10 C ... Jordon Construction Services has operated forthe last 26 years in a northern U ... state where thestate income tax on corporaterevenue is 6% peryear ... F ... Because of pressing labor cost increases, the president wants to move to another state to reduce thetotal tax burden ... You are an engineer with the company and are asked to do the following ... F ... (b) Estimate the state tax rate that would be necessary to reduce theoverall effective tax rateby 10% per year ... F ... After-Tax Economic AnalysisCFBT and CFAT17 ... 17 ... 13 For a year in which there is no initial investment Por salvage value S, derive an equation for CFBTthat contains only the following terms: CFAT,CFBT, D, and Te ... 14 Determine the cash flowbefore taxes for AndersonConsultants when the cash flow after taxes was$600,000, asset depreciation was $350,000 and thecompany’s effective tax rate was 36% ... 15 Four years ago Sierra Instruments of Monterey,California spent $200,000 for equipment to manufacture standard gas flow calibrators... The gross income for year 4 was$100,000, with operating expenses of $50,000 ... TheMACRS depreciation rate for year 4 is 7 ... 17 ... The totalrevenue for year 2 was $48 million, depreciationwas $8 ... Use a federal tax rate of 35% and astate tax rate of 6 ... 17 ... , researchers in medical science, iscontemplating a commercial venture concentrating on proteins based on the newX-ray technology of free-electron lasers ... 5 million CFAT is needed ... Overa 3-year period, the deductible expenses and depreciation are estimated to total $1 ... Of this, 50% is expenses and 50% is depreciation ...18through 17 ... (Show hand and spreadsheet solutions, asinstructed)After 4 years of use, Procter and Gamble has decided toreplace capital equipment used on its Zest bath soap line ... After-tax MARR is 10% per year, and Teis 35% in the United States ... Year01234Purchase, $Ϫ1900Gross income,$800950600300Operating expenses, $Ϫ100 Ϫ150 Ϫ200 Ϫ250Salvage, $70017 ... year 2 if the depreciation method had been straightline instead of MACRS ... 24 Cheryl, an electrical engineering student who isworking on a business minor, is studying depreciation and finance in her engineeringmanagementcourse ... Help her, using assetestimates made for a 6-year study period: P ϭ$65,000, S ϭ $5000, GI ϭ $32,000 per year, AOCis $10,000 per year, SL depreciation, i ϭ 12% peryear, Te ϭ 31% ... 17 ... All cash flows are in $1000 units ... 19 Calculate MACRS depreciation and estimatetheCFAT series over 4 years ... YearGI17 ... 01234—815121017 ... 17 ... Use a spreadsheet to tabulate CFBT, CFAT, NOPAT,and i* before and after taxes for 6 years of ownership ... Salvage is expected to be zero ... 23 An asset purchased by Stratasys, Inc ... In year 2, therevenue was $490,000 withoperating expenses of$140,000 ... 26 Use an effective tax rate of 32% to determine theCFAT and NOPAT associated with the asset shownbelow under two different scenarios: (a) with depreciation at $6000 per year and (b) with depreciation at $6000, $9600, $5760, and $3456 in years 1through 4,respectively ... Estimates, $Year GI OE01234—8151210PDTI— Ϫ30 — —Ϫ2Ϫ4Ϫ3Ϫ5Taxes CFAT NOPAT—Ϫ30—17 ... B ... , an overland freight company, haspurchased new trailers for $150,000 and expects torealize a net $80,000 in gross income over operating expenses for each of the next 3 years ...Assume aneffective tax rate of 35% and an interest rate of15% per year ... Since MACRS takes an additional year to fully depreciate the basis, assume no CFBT beyond year 3, but includeany negative tax as a tax savings ... 17 ... The machinerycosts $200,000, has no salvage value, and theCFBTestimate is $75,000 per year for up to 10years ... Thetwo depreciation methods to consider are:MACRS with n ϭ 5 years and SL with n ϭ8 years (neglect the half-year convention effect) ... 17 ... TheCFBT is estimated at $10,000 for the first 4 yearsand $5000 thereafter as long as the asset is retained ... Inpresent worth dollars, howmuch of the cash flow generated by the asset overits recovery period is lost to taxes?Depreciation Recapture and Capital Gains (Losses)17 ... According to tax law,the simulation is MACRS-depreciated using a3-year recovery period ... (b) Determine by how much the sale willcauseTI and taxes to change in year 5 if Te ϭ 35% ... 31through 17 ... Open Access, Inc ... Different depreciation, recovery period, and tax law practices in the threecountries where depreciable assets are located are summarized in the table ... After-tax MARR ϭ 9% per year and Te ϭ 30% canbe used forall countries ... 31 For country 1, SL depreciation is $20,000 per year ... 17 ... Determine the (a) CFAT series and (b) PWof depreciation, taxes, and CFAT series ... 33 For country 3, DDB depreciation for 5 years is$40,000, $24,000, $14,400, $1600, and 0, respectively ... 17 ... 31 through 17 ... For eachcriterion, select the country that provides the best PWvalue ... (Hint: The PW shouldbe minimized for some criteria and maximized forother criteria ... )17 ... The asset was depreciated by the MACRS method and has a book valueof $100,800 at the time of sale ... 17 ... In order to have fresh water at thesite, thecompany purchased a turnkey reverse osmosis (RO)system for $355,000 ... Four years after the systemwas purchased, water lines from a local water system were extended to the truck stop, so Sun-Texsold the RO system for $190,000 ... TheMACRS depreciation rates are 10%, 18%, 14 ... 52%for years 1 through 4, respectively ... 37 Freeman Engineering paid $28,500 for specializedequipment for use with its new GPS͞GIS system ... Thecompany sold the equipment after 2 years for$5000 when it purchased an upgraded system ... (b) What tax effect will this amount have?17 ... Use them todetermine the amount of income tax effect, if the effective tax rate is 35% ... (b) A hi-tech machine was sold internationallyfor $10,000 more than its purchase price justafter it was in service 1 year ... (c) Land purchased 4 years ago for $1 ... (d) A 21-year-old asset was removed fromservice and sold for$500 ... Classical straight line depreciationwas used for the entire recovery period ... It was sold inthe fourth year of use for $2000 ... 39 Sunnen Products Co ... Louis, Missouri, makesactuator hones for gas meter tubes where light-dutymetal removal is needed ... Use the informationshown to determinethe presence and amount of anydepreciation recapture, capital gain, or capital loss ... 533300,00015,5005,000295,000275,00019,50012,500After-Tax Economic Analysis17 ... 2% ... 17 ... Assume that the company has an effectivetax rate of 35% and it uses MACRS depreciationfor an asset that has a$27,000 salvage value ... 42 Estimate the approximate after-tax rate of returnfor a project that has a first cost of $500,000, asalvage value of 20% of the first cost after 3 years,and annual gross income less operating expensesof (GIϪOE) ϭ $230,000 ... 17 ... No taxes are paid on retirement earnings untilthey are withdrawn; however,she was told by her brother, an accountant, that thisis the equivalent of an 8% per year after-tax return ... 44 Bart is an economic consultant to the textile industry ... If thestated MARR in both companies is 12% per yearafter taxes, determine if management at both companiesshould accept the projects ... Effective tax rates are 34% for the larger corporation and 28% for the small company ... 45 Elias wants to perform an after-tax evaluation ofequivalent methods A and B to electrostatically remove airborne particulate matter from clean roomsused to package liquidpharmaceutical products ... Any tax effects when the equipmentis salvaged were neglected ... Now, useclassical SL depreciation with n ϭ 5 years to evaluate the alternatives ... 46 A corporation uses the following: before-taxMARR of 14% per year, after-tax MARR of7% per year, and Te of 50% ... MachineAFirst cost, $Salvage value, $AOC, $ per yearLife, yearsMachine BϪ15,0003,000Ϫ3,00010Ϫ22,0005,000Ϫ1,50010The machine is retained in use for a total of 10 years,then sold for the estimated salvage value ... (b) After-tax PW analysis, using classical SL depreciation over the 10-year life ... 17 ... TheGIϪOE estimateis made for only 3 years; it is zero when each assetis sold in year 4 ... 48 Offshore platform safety equipment, designed forspecial jobs, will cost $2,500,000, will have nosalvage value, and will be kept in service for exactly 5 years, according to company policy ... The effective tax rate forthe multinational oil company is 30% ... 17 ... This represents areturn of 20% ... If thecorporation wants to realize an after-tax return of10% per year, for how many more years must theequipment remain in service?After-Tax Replacement17 ... (a) How is thegain or loss calculated, and (b) how does itaffectthe AW values in the study?17 ... 52 An asset that was purchased 2 years ago was expectedto be kept in service for its projected life of 5 years, buta new version (the challenger) of this asset promisesto be more efficient and have lower operating costs ... Thedefender had a first cost of $300,000,but its marketvalue now is only $150,000 ... To simplify calculations for thisproblem only, assume that SL depreciation wascharged at $60,000 per year and that it will continueat that rate for the next 3 years ... Assumethe company’s effective tax rate is 35%, and itsafter-tax MARR is 15% per year ... (b)Determine the CFAT in years 1 through 3 forthe challenger and defender ... 17 ... The depreciation charges for the next 3 years will be $69,960,$49,960, and $35,720 ... Assume the company’s aftertax MARR is 12% ... 54 Perform a PW replacement study (hand andspreadsheet solutions, if instructed)from the information shown using an after-tax MARR ϭ 12%per year, Te ϭ 35%, and a study period of 4 years ... Since no revenues areestimated, all taxes are negative and considered“savings” to the alternative ... The effective tax rate is 35%, and the after-taxMARR is 10% per year ... DefenderChallenger17 ... The devicescan be sold on the used equipment market for anestimated $15,000 ... The upgradeinvestment will be depreciated over 3 years withno salvage value ... The new unitswill have operating expenses of $7000 per year ... (b) If thechallenger is known to be salable after 5 yearsforan amount between $2000 and $4000, willthe challenger AW value become more or lesscostly? Why?First cost, $1000Estimated S at purchase, $1000Market value now, $1000AOC, $1000 per yearDepreciation methodRecovery period, yearsUseful life, yearsYears ownedϪ45535Ϫ7SL883Ϫ240—Ϫ8MACRS35—17 ... Pete’s accountant used anafter-tax MARR of 8% per year, Te ϭ 30%, and acurrent market value of $25,000 to determineAW ϭ $2100 ... Estimated CFBT is $15,000 per year ... From the accountant, Ramon learned the currentequipment cost $20,000 when purchased andreached azero book value several years ago ... 17 ... The contract is up for renewalnow for a period of 1 year or 2 years only ... The finance vice president wants to renew thecontract for 2 years without further analysis, butthe vice president for engineering believes it ismore economical to perform themaintenancein-house ... Theestimates for the in-house (challenger) alternative are as follows:First cost, $AOC, $ per yearLife, yearsEstimated salvageMACRS depreciation−800,000−120,0004Loses 25% of P annually:End year 1, S ϭ $600,000End year 2, S ϭ $400,000End year 3, S ϭ $200,000End year4, S ϭ $03-year recovery period17 ... Thesalvage value after 10 years at that time was estimated to be $50,000 ... The new president has recommendedearly replacement of the system with one thatcosts $400,000 and has a 12-year life, a $35,000salvage value, and an estimated AOC of $50,000per year... The president wishes to know thereplacement value that will make the recommendation to replace now economically advantageous ... For solution purposes, use classical SL depreciation for both systems ... Economic Value Added17 ... Why are these preferences predictable?17 ... S ... Onecomponent is a 4-yearproject for a special-purpose transport ship-cranefor use in building permanent storm surge protection against hurricanes on the New Orleans coastline ... MACRS depreciation with a 3-yearrecovery is indicated ... TheCFAT is shown below ... They should havethe same value ...75%and Te ϭ 35% ... 61 Triple Play Innovators Corporation (TPIC) plansto offer IPTV (Internet Protocol TV) service toNorth American customers starting soon ... Let Te ϭ 30% and after-tax MARR ϭ8%; use SL depreciation (neglect half-year convention and MACRS, for simplicity) and a studyperiod of 8years ... 2 million3 ... 62 Sun Microsystems has developed partnershipswith several large manufacturing corporations touse Java software in their consumer and industrialproducts ... One major project involves using Java in commercial and industrialappliances that store and cook food ... For t ϭ 1 to 6years,Annual gross income, GI ϭ 2,800,000 Ϫ 100,000tAnnual operating expenses, OE ϭ 950,000 ϩ 50,000tThe effective tax rate is 30%, the interest rate is12% per year, and the depreciation method chosenfor the $3,000,000 capital investment is a 5-yearMACRS ADS alternative that allows straightlinewrite-off with the half-year convention in years 1and 6 ... Value-Added Tax17 ... 64 In Denmark, VAT is applied at a rate of 25%, withfew exceptions ... Determinethe following ... (b) The amount of tax vendor B sends toDenmark’s Treasury ... The following information is used in Problems 17 ...70Ajinkya Electronic Systems, a company in India thatmanufactures many different electronic products, has topurchase goods and services from a variety of suppliers(wire, diodes, LED displays, plastic components, etc ... It also shows the purchases in$1000 units that Ajinkya made (before taxes) fromeachsupplier in the previous accounting period ... 2 million, and Ajinkya’s products carry a VAT of 12 ... SupplierVAT Rate, %Purchases byAjinkya, $1000ABCDE4 ... 512 ... 332 ... 65 How much VAT did supplier C collect?17 ... 67 What was the total amount of taxes paid by Ajinkyato the suppliers?17 ...68 What was the average VAT rate paid by Ajinkya inpurchasing goods and services?17 ... 71 A graduated income tax system means:(a) Only taxable incomes above a certain levelpay any taxes ... (c) Higher tax rates go with higher taxableincomes ... 17 ... (b) The end user pays value-added taxes ... (d)Value-added taxes are charged only on theraw materials for product development ... 73 A small company has a taxable income that placesit in the 35% tax bracket ... 74 A company that has a 50% effective tax rate hadincome of $200 million in each of the last 2 years ... In the other year, the companyhaddeductions of only $80 million ... 75 Taxable income (TI) is defined as:(a) TI ϭ revenue ϩ operating expensesϪ depreciation(b) TI ϭ revenue Ϫ operating expensesϩ depreciation(c)(d)TI ϭ revenue Ϫ operating expensesϪ depreciation ϩ amortizationTI ϭ revenue Ϫ operating expensesϪ depreciation17 ... 77A subcontractor with an effective tax rate of 25%has gross income of $55,000, other income of$4000, operating expenses of $13,000, and otherdeductions and exemptions of $11,000 ... 78 When a depreciable asset is disposed of for less thanits current book value, the transaction is known as:(a) Anafter-tax expense(b) Capital loss(c) Capital gain(d) Depreciation recapture17 ... If the effective tax rate is 40%, the values for depreciation (D), taxable income (TI), and taxes foryear 1 are closest to:Year012Investment, GI − OE,$$D, $TI, $Taxes,$Ϫ60,00030,00035 ... 80 If the after-tax rate of return for acash flow seriesis 11 ... 8%(b) 5 ... 0%(d) 28 ... 81 An asset purchased for $100,000 with S ϭ $20,000after 5 years was depreciated using the 5-yearMACRS rates ... The assetwas actually sold after 5 years of service for$22,000 ... 53%and 5 ... The after-tax cash flowfrom the sale is closest to:(a)(b)(c)(d)$27,760$17,130$26,870$20,58517 ... Of the following, the statements that are commonly incorrect are:1 ... 2 ... 3 ... 4 ... (a) 1, 2, and 3(b) 1 and 4(c) 2(d) 4CASE STUDYAFTER-TAX ANALYSIS FOR BUSINESS EXPANSIONBackgroundCharles was always a hands-on type of person ... After some 20years, it has grown significantly ... in the Metroplex, specializing in custom-made metal and stone fencing for commercial and residential sites ... Pro-Fence is privately owned by Charles; therefore, the question of how to finance such an expansion has been, and still is,the major challenge ... Taking capitalfrom the retained earnings of Pro-Fence is asecond possibility, but taking too much will jeopardize the current business, especially if the expansion were not an economicsuccess and Pro-Fence were stuck with a large loan to repay ... He knows you are quite economically oriented andthat you understandthe rudiments of debt and equity financing and economic analysis ... You have agreed to help him, as much as you can ... Between his accountant and a small market survey of thebusiness opportunities in Victoria, the following generalizedestimates seem reasonable ... 5 millionAnnual gross income ϭ$700,000Annual operating expenses ϭ $100,000Effective income tax rate for Pro-Fence ϭ 35%Five-year MACRS depreciation for all $1 ... Repayment would be in 5 equal payments of interest and principal ... A range of D-E mixes should be analyzed ... DebtLoanPercentage Amount,$0507090750,0001,050,0001,350,000EquityPercentage100503010InvestmentAmount, $1,500,000750,000450,000150,000Case StudyCase Study Exercises1 ... An after-tax returnof 10% is expected ... )2 ... If the time value of money483is neglected, what is the constant amount by whichthis sumchanges for every 10% increase in equityfunding?3 ... He wants to know why this phenomenon occurs ... After deciding on the 50-50 split of debt and equity financing, Charles wants to know what additional bottomline contributions to the economic worth of the companymay be added by the new Victoriasite ... SECTIONTOPICLEARNING OUTCOME18 ... 18 ... 18 ... 18 ... 18 ... 18 ... This chapter includes several related topics about alternative evaluation ... Then the determination and useof the expected value of a cash flow series are treated ... Economic decisions that involve staged funding are verycommon in professional andeveryday life ... 18 ... Example parameters are first cost, salvage value, AOC, estimated life, production rate, and materials costs ... Economic analysis uses estimates of a parameter’s future value to assist decision makers ... The effect of variation may be determined by usingsensitivity analysis ... Usuallyone parameter at a time is varied, and independence with other parameters is assumed ... Sensitivity analysisIn reality, we have applied this approach (informally) throughout previous chapters to determinethe response to variation in a variety of parameters ... However,variation in the n or AOC value may indicate that the alternative’s measure of worth is very sensitiveto the estimated life or annual operating costs ... For example, if an operating level of 90%of airline seating capacity for a domestic route is compared with 70% for a proposed internationalroute, theoperating cost and revenue per passenger-mile will increase, but anticipated aircraftlife will probably decrease only slightly ... Sensitivity analysis routinely concentrates on the variation expected in estimates of P, AOC,S, n, unit costs, unit revenues, and similar parameters ... Parameters that are interestrate–based are not treated in the same manner ... If performed, sensitivity analysis on them is for specific values or over anarrow range of values ... Plotting the sensitivity of PW, AW, or ROR versus the parameter(s) studied is very helpful ... This is the value at which the two alternatives are economicallyequivalent ... Thus, several charts are constructed, and independence of each parameter is assumed ... However, if the results are sensitive to the parameter value, severalintermediate points should be used to better evaluate the sensitivity, especially if the relationships are not linear ... It maybeperformed one parameter at a time using a spreadsheet or computations by hand or calculator ... Here is a general procedure to follow when conducting a thorough sensitivity analysis ... 2 ... 4 ... Determine which parameter(s) of interest might vary from the most likely estimated value ... Select themeasure of worth ... To better interpret the sensitivity, graphically display the parameter versus the measure ofworth ... When there are two or more alternatives, it is better to usethe PW or AW measure of worth in step 3 ... Example 18 ... EXAMPLE 18 ... expects to purchase a new asset for automatedrice handling ... The MARR for the company varies over awide range from 10% to 25% per year for different types of investments ... Evaluate the sensitivity of PW by varying(a) MARR, while assuming a constant n value of 10 years, and (b) n, while MARR is constantat 15% per year ... Solution byHand(a) Follow the procedure above to understand the sensitivity of PW to MARR variation ... MARR is the parameter of interest ... Select 5% increments to evaluate sensitivity to MARR; the range is 10% to 25% ... The measure of worth is PW ... Set up the PW relation for 10 years ... A plot of MARRversus PW is shown in Figure 18–1 ... If the MARR is established at the upper end of the range, theinvestment is not attractive ... Asset life n is the parameter ... Select 2-year increments to evaluate PW sensitivity over the range 8 to 12 years ... The measure of worth is PW ... Set up the same PW relationas in part (a) at i ϭ 15% ... 1487Determining Sensitivity to Parameter VariationFigure 18–130,000MAPlot of PW versus MARRand n for sensitivityanalysis, Example 18 ... RR20,000Present worth, $n10,0001015682025MARR %12Life n010– 10,000– 20,0005 ... Since the PW measure is positive forallvalues of n, the decision to invest is not materially affected by the estimated life ... This insensitivity to changes in cash flow in the distantfuture is a predictable observation, because the P͞F factor gets smaller as n increases ... The NPV function calculates PW for i values from 10% to25% and n valuesfrom 8 to 12 years ... Figure 18–2Sensitivity analysis ofPW to variation in(a) MARR values and(b) life estimates,Example 18 ... PW computationϭ NPV(C6,B$4:B$13)ϩB$3(a)PW computationϭ NPV(15%,B$4:B$15)ϩB$3(b)488Chapter 18Figure 18–350%Sensitivity analysis graphof percent variationfromthe most likely estimate ... This is sometimes called a spider graph ... The variation in each parameter is indicated as a percentage deviation from the most likely estimate on the horizontal axis ... This is the conclusion for indirectcost in Figure 18–3 ... A reduction of 30%from the expected sales pricereduces the ROR from approximately 20% to −10%, whereas a10% increase in price raises the ROR to about 30% ... Observe the general shape of the sample sensitivity graphs inFigure 18–4 ... Thegraph indicates that the PW of each plan is a nonlinear function of hours of operation ... PW, $ ϫ1000Plan B100PlanA50010002000Hours of operation per year300018 ... Plan B is more attractive due to its relative insensitivity ... It may be necessary to plot the measure of worth at intermediatepoints to better understand the nature of the sensitivity ... 2Columbus, Ohio needs to resurface a 3-kilometerstretch of highway ... The first method is a concrete surface for a cost of$1 ... The second method is an asphaltcovering with a first cost of $1 million and a yearly maintenance of $50,000 ... The city uses the interest rate on bonds, 6% on its last bond issue, as the discount rate ... If the city expects aninterstate to replace this stretch of highway in 10 years, which method should be selected?(b) If the touch-up cost increases by $5000 per kilometer every 3 years, is the decision sensitive to this increase?Solution(a) Use PW analysis to determine the breakeven n value ... , n ... Ϫ500,000 ϩ40,000(P͞A,6%,n) ϩ 75,000(P͞F,6%, j ) ϭ 0j[18 ... 1] switches from negative to positive PW values ... The NPV functionsin column C are the same each year, except that the cash flows are extended 1 year for eachFigure 18–5Sensitivity of the breakeven life between two alternatives, Example 18 ... PW for11 yearsϭ NPV(6%,$B$5:$B15)ϩ$B$4PW for 12 yearsϭ NPV(6%,$B$5:$B16)ϩ$B$4Year counter advances by 1490Chapter 18Sensitivity Analysis and Staged Decisionspresent worth calculation ... 4 years, concrete and asphalt resurfacing break even economically ... (b) The total touch-up cost willincrease by $15,000 every 3 years ... 1] is now[(jϪ3Ϫ500,000 ϩ 40,000(P͞A,6%,n) ϩ 75,000 ϩ15,000 ———3) ] [ ͚(P͞F,6%, j ) ] ϭ 0jNow the breakeven n value is between 10 and 11 years—10 ... The decision has become marginal for asphalt, since theinterstate is planned for 10 years hence ... Oneconclusion is that the asphalt decision becomes more questionable as theasphalt alternative maintenance costs increase; that is, the PW value is sensitive to increasing touch-up costs ... 2 Sensitivity Analysis Using Three EstimatesWe can thoroughly examine the economic advantages anddisadvantages among two or morealternatives by borrowing from the field of project scheduling the concept of making three estimates for each parameter: a pessimistic, a most likely, and an optimistic estimate ... This approach allows us to study measure of worth and alternative selection sensitivitywithin apredicted range of variation for each parameter ... EXAMPLE 18 ... Shehas made three estimates for the salvage value, annual operating cost, and life ... For example, alternative Bhas pessimistic estimates of S ϭ $500, AOC ϭ $Ϫ4000, and n ϭ 2 years ... Perform a sensitivity analysis anddetermine the mosteconomical alternative, using AW analysis at a MARR of 12% per year ... 18 ... 3Alternative AW Values, $EstimatesABCPMLOϪ19,327Ϫ14,548Ϫ9,026Ϫ12,640Ϫ8,229Ϫ5,089Ϫ19,601Ϫ13,276Ϫ8,9272018AW of costs, Ϫ$100016Alternative C14Alternative A12108AlternativeB6420123456789Life n, yearsFigure 18–6Plot of AW of costs for different-life estimates, Example 18 ... SolutionFor each alternative in Table 18–1, calculate the AW value of costs ... Figure 18–6 is a plot of AW versus the three estimates oflife for each alternative ... CommentWhile the alternative thatshould be selected here is quite obvious, this is not normally the case ... Inthis case, it would be necessary to select one set of estimates (P, ML, or O) upon which to basethe decision ... 18 ... This means that probability andsamples are used ... The reason is not that the computations are difficult toperform or understand, but thatrealistic probabilities associated with cash flow estimates are difficult to assign ... The expected value can be interpreted as a long-run average observable if the project is repeatedmany times ... However, even for a single occurrence, the expected value is ameaningfulnumber ... 2]i�1Xi ϭ value of the variable X for i from 1 to m different valuesP(Xi) ϭ probability that a specific value of X will occurwhereProbabilities are always correctly stated in decimal form, but they are routinely spoken of inpercentages and often referred to as chance, such as the chances are about10% ... 2] or any other relation, use the decimal equivalent of 10%,that is, 0 ... In all probability statements the P(Xi) values for a variable X must total to 1 ... iϭm͚ P(X ) ϭ 1 ... If X represents the estimated cash flows, some will be positive and others will be negative ... Ifthe expected value is negative, the

overall outcome is expected to be a cash outflow ... EXAMPLE 18 ... The marketing director estimates that for a typical 24-hour period there is a 50% chance of having a net cash flow of $5000 and a 35% chance of $10,000 ... Determinethe expected net cash flow ... Using Equation [18 ... 5) ϩ 10,000(0 ...05) Ϫ 1000(0 ... 0 and it makes the computation complete ... 4 Expected Value Computations for AlternativesThe expected value computation E(X) is utilized in a variety of ways ... • Evaluate the expected viability of a fully formulated alternative ... 5 illustrates the first situation, and Example 18 ... 18 ...5There are many government incentives to become more energy-efficient ... Theowner pays a portion of the total installation costs, and the government agency pays therest ... She has exceeded the annual budgeted amount of $50 million peryear in each of the previous 2 years ... Over the last 36months, the amount of average monthly payoutand number of months are shown in Table 18–3 ... Provided the same pattern continues, what is the expected value of the dollar increase in annual budget that is needed tomeet the requests?TA BLE 18–3Solar Panel Incentive Payouts, Example 18 ... 54 ...22 ... ,very high) to estimate the probability P(POj)for each level, and make sure the total is 1 ... Level, jVery highHighModerateLowProbability of Payout Level, P(POj)P(PO1) ϭ 15͞36 ϭ 0 ... 278P(PO3) ϭ 7͞36 ϭ 0 ... 1111 ... 2] ... 5(0 ... 7(0 ... 2(0 ... 9(0 ... 711 ϩ 1 ... 621 ϩ 0 ... 961 ($4,961,000)The annualexpected budget need is 12 ϫ 4 ... 532 million ... 532 million per year ... 6Lite-Weight Wheelchair Company has a substantial investment in tubular steel bending equipment ... Estimated cash flows(Table 18–4) depend on economic conditions classified as receding, stable, or expanding ... Apply expectedvalue and PW analysis to determine if the equipment should be purchased ... 493494Chapter 18Sensitivity Analysis and Staged DecisionsTA BLE 18–4Equipment Cash Flow and Probabilities, Example 18 ... ϭ 0 ... ϭ 0 ... ϭ 0 ... 2] ... The PW values for the three scenarios arePWR ϭ Ϫ5000 ϩ2500(P͞F,15%,1) ϩ 2000(P͞F,15%,2) ϩ 1000(P͞F,15%,3)ϭ Ϫ5000 ϩ 4344 ϭ $Ϫ656PWS ϭ Ϫ5000 ϩ 5708 ϭ $ϩ708PWE ϭ Ϫ5000 ϩ 6309 ϭ $ϩ1309Only in a receding economy will the cash flows not return the 15% to justify the investment ... 4) ϩ 708(0 ... 2)ϭ $283At 15%, E(PW) Ͼ 0; the equipment is justified,using an expected value analysis ... ComputingE(cash flow) first may be easier in that it reduces the number of PW computations ... E(CF0) ϭ $Ϫ5000E(CF1) ϭ 2500(0 ... 4) ϩ 2000(0 ... 5 Staged Evaluation of AlternativesUsing a Decision TreeAlternative evaluation may require a series of decisions inwhich the outcome from one stage is important to the next stage of decision making ... A decision tree includes:••••••More than one stage of alternative selection ... Expected results from a decision at each stage ... Estimates of economic value (cost or revenue) for each outcome ... 18 ... 5Probabilitynode0... 3(b) Probability node with outcomesDDDFinal outcomes(c) Tree structureFigure 18–7Decision and probability nodes used to construct a decision tree ... • A square represents a decision node with the possible alternatives indicated on the branchesfrom the decision node (Figure 18–7a) ... • Thetreelike structure in Figure 18–7c results, with outcomes following a decision ... These cash flows are expressed in terms of PW,AW, or FW values and are shown to the right of each final outcome branch ... This process, called solving the tree or rollback, is explainedafter Example 18 ... EXAMPLE 18 ... S... He was recently approached by an international supermarket chain that wants tomarket in-country its own brand of frozen microwaveable dinners ... The current decision involves two alternatives: (1) Lease a facility in the United ArabEmirates (UAE) from the supermarket chain, which has agreed toconvert a current processingfacility for immediate use by Jerry’s company; or (2) build and own a processing and packaging facility in the UAE ... The decision choices 2 years hence are dependent upon the lease-or-own decision madenow ... This will be a mutual decision between the supermarket chainand Jerry’s company ... Outcomes for thefuture decisions are, again, good and poor market responses ... If market response is good, the decision alternatives are four or two times original levels ... Construct the tree of decisions and outcomes for Hill Products and Services ... Identify the decision nodesand branches, and then develop the tree using the branches and the outcomes ofgood and poor market for each decision ... Stage 1 (decision now):Label it D1 ... Outcomes: good and poor markets ... 5ϫPoorGoodGood0 ... 18 ... Outcomes: good market, poor market, and out of business ... 5ϫ); stopproduction (0ϫ)The alternatives for future production levels (D2 through D5) are added to the tree and followed by the market responses of good and poor ... To utilize the decision tree for alternative evaluation and selection, the following additionalinformation is necessary for each branch:• The estimated� )͚ outcome ( probability that each outcome may occur ... 0for each set of outcomes (branches) that result from a decision ... Decisions are made using the probability estimate and economic value estimate for each outcome branch ... 2] ... Start at the top right of the tree ... 2 ... E(decisionestimate)P(outcome)[18 ... 3 ... 4 ... 5 ... EXAMPLE 18 ... If the product is marketed, thenext decision is to take it international or national ... The probabilities for each outcome and PW of CFBT(cash flow before taxes) are indicated ... Determine thebest decision at the decision node D1 ... 1 ... 2 ... 3] ... 2in ovalsare determined asE(international decision) ϭ 12(0 ... 5) ϭ 14E(national decision) ϭ 4(0 ... 4) Ϫ 1(0 ... 2The expected PW values of 4 ... 497498Chapter 18Sensitivity Analysis and Staged Decisions1414InternationalD2PW ofCFBT($ million)0 ... 50 ... 16LowMarket4 ... 229NationalD10 ... 20 ... 40 ...24 ... 40 ... 212164–3–16–36–229Sell1 ... 8 ... Select the larger expected value at each decision node ... 2 (international) at D3 ... Calculate the expected PW for the two D1 branches ... 2) ϩ 4 ... 8) ϭ 6 ... 0) ϭ 9The expected value for the sell decision is simple since the one outcome has a payoff of 9 ... 5... 18 ... When the decision to invest more or less can be delayed into the future, theproblem is called staged funding ... Decision makers may opt to (1) build the capacity tosupply100,000 per month to market immediately or (2) build capacity to supply 25,000 permonth now and test the market’s receptivity... Of course, if an aggressive competitorenters the scene, or the economy falters, the staged funding decision will change as warranted ... Before we go further, some definitions are needed ... 18 ... The options usually involve physical (real) assets, buildings, equipment, materials, andthe like, thus, theword real ... The investment alternatives present varying amounts of risk, which is estimated by probabilities of occurrence for predictable future events ... The estimated cash flows and otherconsequences of these delays are analyzed with risk taken into account to the degree possible ... g ... Adecisionmay be to expand, continue as is, contract, abandon, or replicate the alternative at thetime the option must be exercised ... After some illustrations of real options, we will discuss the probabilisticdimensions ... Industrial SettingNew markets—Purchase equipment and staff to enter an expandinginternational market overthe next 5 years ... Removing car models—Ford Motor Company can decide to maintain production on an established car model with dwindling sales for the next 3 years or can opt to discontinue the modelin stages over a 1- or 2-year period ... The drilling may not be justified atthis time, but the contract offersthe option to drill were it to become economically advantageous based on events such as increased oil prices or improved recovery technology ... The price of the option isthe cost of the extended warranty ... House insurance—When a homeowner has no mortgage to pay,maintaining house insuranceis an option ... g ... Self-insurance, wheremoney is set aside for potential damages while accepting the risk that a major event will takeplace, is an option for the homeowner ... • Anticipated future options and cash flow estimates (double production with annual net cashflowsestimated) ... • Market and risk-free interest rates (expected market MARR of 12% per year and inflationrate estimate of 4% per year) ... • Economic criterion used to make a decision (PW, ROR, or other measure of worth) ... Example 18 ... EXAMPLE 18 ... , hasdeveloped and field-tested a modularized,scalar solar thermal electric (STE) generationsystem that is relatively inexpensive to purchase and has an efficiency considerably betterthan traditional photovoltaic (PV) panels ... Additionally, a contractwith a consortium of sunbelt states has been offered, but not accepted thus far, for a total of$1 ... Bycontract, the units will be marketed throughthe state energy departments with all revenue going to the state treasuries ... ConditionOptionCIF FundingConsortium Contract2 ϫ productionlevelAdditional $10 million inyear 2Additional 8 years;$4 million in years 3–10Sales are excellent1 ϫ production(3000–5000 units͞year)levelNothing; no salvage after10 yearsAdditional 8 years;$1 ... 5 million after 5 yearsAdditional 3 years;$1 ... (b) The base case is the 1 ϫ production level with the 8-year follow-on contract from the consortium ... (c) Determine the PW values for each possible final outcome at 10% per year,and identify thebest economic option when the stage 2 funding decision must be made ... There are 2 outcome branches initially (accept option; decline option) and four final branches for theaccept decision at D1, based upon sales level ... (SolarScale has other ways to pursue revenue that are notrepresented in this abbreviated example ... The resulting PW1ϫ Ͻ 0 as shownbelow indicates the contract is not justified economically ... 5(P͞A,10%,10)ϭ $Ϫ0 ... The i* values are also shown using the IRR function ... 5 million (½ ϫ level) or after 2 years for $5 million(stop) is included ... 18 ... 9 ... 9 ... IfSolarScale and CIF, the financial backers, are not convinced that the sales level will exceed5000 units per year, the contract option should be declined ... Of the characteristics listed above for real option situations, the primary one absent inExample 18 ... Decision making underrisk is covered moreextensively in the next chapter; however, we can use the following definition for this discussion of real options analysis ... Risk502Chapter 18Sensitivity Analysis and Staged DecisionsA coin has two sides ... If the coin is intentionally biased in weight, such that 58% ofthe time it lands heads up, then thelong-run probability of heads is P(heads) ϭ 0 ... Since the sumof probabilities across all possible values must add to 1, the biased coin has P(tails) ϭ 0 ... There are a couple of points worth mentioning about risk and the calculated PW values for realoptions analysis ... (We shall see this in Example 18 ... )•Since one of the objectives of real options analysis is to evaluate the economic consequencesof delaying a decision, a PW value of the base case that is moderately positive means that theproject is justified and should be accepted immediately, without the decision delay ... Thus, the project should berejected now, notdelayed for a future decision ... EXAMPLE 18 ... 9 about SolarScale’s state-consortium contract offer, some expected sales information was collected ... The results are as follows:Probability of OutcomeExcellentSolarScaleCIFConsortiumPoor0 ... 80 ... 50 ... 4Use these probabilityestimates to determine the expected PW value, provided equal weightingis given to each representative’s input ... Excellent: From 2 ϫ level and 1 ϫ level, select 2 ϫ with PW ϭ $1 ... Poor: From ½ ϫ level and stop now, select ½ ϫ with PW ϭ $Ϫ2 ... In $ million, E(PW) for each organization isE(PW forSolarScale) ϭ 1 ... 5) Ϫ 2 ... 5) ϭ $Ϫ0 ... 97(0 ... 76(0 ... 02E(PW for consortium) ϭ 1 ... 6) Ϫ 2 ... 4) ϭ $0 ... 33(Ϫ0 ... 02 ϩ 0 ... 23 ($230,000)The base case of 1 ϫ level production in part (b) of Example 18 ... When compared with the positive E(PW) result here, we see that with consideration of the differentoptions of production level and probabilities for sales level, the expectedPW has increased to a positive value ... Problems503There are many other examples and dimensions of real options analysis in engineering economics and in the area of financial analysis, where options analysis got its start someyears ago ... CHAPTER SUMMARYIn this chapter the emphasis is on sensitivity to variation in one or more parameters using aspecific measure of worth ... When several parameters are expected to vary over a predictable range, the measure of worthis plotted and calculated using three estimates for aparameter—most likely, pessimistic, andoptimistic ... Independence between parameters is assumed in all these analyses ... Decision trees are used to make a series of alternative selections ... It is necessary to make several types of estimates for a decision tree: outcomes for each possible decision,cash flows, and probabilities ... Staged funding over time can be approached using the evolving area of real options ... PROBLEMSSensitivity to Parameter Variation18 ... The company willmake the investment only if it will result in a rateof return of 15% per year or higher ... 18 ... For the past 5 years theyhave beenable to invest one of their salaries ($50,000 peryear, which includes employer contributions)while living off the other one ... If theyhave gotten a rate of return of 10% per year ontheir investments and expect to continue at thisROR, is reaching their goal of $2 ... e ... 18 ... The company can invest$80,000 now, 1 year from now,or 2 years from now ... That is,the savings will be $25,000, $26,000, or $29,000per year if the investment is made now (year 0),in 1 year, or in 2 years, respectively ... 18 ... Option 1 is a low-pressure seawater reverse osmosis (SWRO) system that will operate at500 psi witha fixed cost of $465 per day and anoperating cost of $0 ... A secondoption is a higher-pressure SWRO system offeredby vendor X that operates at 800 psi and will havea lower fixed cost of $328 per day (because offewer membranes); however, its operating costwill be $1 ... A third option isalso a high-pressure SWRO system from vendor Y,who claims that its system will have a lower operating cost of $1 ... Determine if theselection of a low- or high-pressure system is dependent on the lower operating cost offered byvendor Y ... 5 A machine that is currently used in manufacturingcircuit board cardlocks has AW ϭ $–63,000 peryear ... Threedifferent engineers have given their opinion, aboutwhat the salvage value of the new machine will be3 years from now: $10,000, $13,000, and $18,000 ... 6 An equipment alternative is being economicallyevaluated separately by three engineers atRaytheon ...The engineers disagree, however, on theestimated revenue the equipment will generate ... Janestates that this is too low and estimates $14,000,while Carlos estimates $18,000 per year ... 18 ... The firstcost of the equipment is $250,000, and it willlikely have a salvage value of $90,000 in 3 years,at whichtime he will not need the equipmentanymore ... Alternatively, the owner cansubcontract the work for $175,000 per year ... He thinks itmight be worth as little as $10,000 in 3 years(a scrap value) ... Sensitivity Analysis and Staged Decisions18 ... 5 million for aplant expansion is not sure what the interestratewill be when it applies for the loan ... The company will onlymove forward with the project if the annual worthof the expansion is below $5 ... The M&Ocost is fixed at $3 ... The salvagecould be $2 million if the interest rate is 10% or$2 ... Is the decision tomove forward with the project sensitive to theinterest rate and salvage value estimates?18 ... Estimated cash flows are asfollows:BatchFirst cost, $Annual cost, $ per yearSalvage value for any year, $Life, yearsContinuousϪ80,000Ϫ55,00010,0003–10Ϫ130,000Ϫ30,00040,0005The chief operating officer (COO) has asked youto determine if the batchoption would ever have alower annual worth than the continuous flow system, using interest rates over a range of 5% to 15%for the batch option but only 15% for the continuous flow system ... )18 ... Fixed cost ϭ $300,000 per yearCost per unit ϭ $40Revenue per unit ϭ $70(a)(b)What is the range inbreakeven quantity ifthere is possible variation in the fixed costfrom $200,000 to $400,000 per year? (Use$50,000 increments ... 11through 18 ... A new online patient diagnostics system for surgeons willcost $200,000 to install, cost $5000 annually to maintainand will have an expected life of 5 years ...Examine the sensitivity of present worthto variation in selected parameter estimates, while othersremain constant ... 11 Sensitivity to first cost variation: $150,000 to$250,000 (−25% to ϩ25%) ... 12 Sensitivity to revenue variation: $45,000 to $75,000(−25% to ϩ25%) ... 13 Sensitivity to life variation: 4years to 7 years (−20%to ϩ40%) ... 18 (a)18 ... 18 ... She expects to retire and start to drawa total of R ϭ $60,000 per year 1 year after the20th deposit ... Determine and comment onthe sensitivity of the size of the annual withdrawal R for variations in A and i ... (a) Variation of Ϯ5% in the annual deposit A... 18 ... Testson metal composites that rely upon scanning electron microscope results can be subcontracted, orthe labs can purchase new equipment ... Use theAW method and plot the results on a sensitivitygraph (like Figure 18–3) ... 17 Titan manufactures and sells gas-powered electricity generators... Cost andsavings estimates are made, but the savings estimate is unreliable at this time ... (b)Ϫ50,000Ϫ7,50015,0005,0005Ϫ37,500Ϫ8,00013,0003,7005Graph the sensitivity of what a personshould be willing to pay now for a 9%,$10,000 bond due in 10 years if there is a30% change in (1) face value, (2)dividendrate, or (3) required nominal rate of return,which is expected to be 8% per year, compounded semiannually ... If the investor did purchase the $10,000 facevalue bond at a premium of 5% (i ... , 5%above face value) and all your other estimates were correct, that is, 0% change, didhe pay too muchor too little? How much?Three Estimates18 ... The company is considering making the partsin-house through the purchase of equipment thatwill have a first cost of $240,000 with an estimated salvage value of $30,000 after 5 years ... Determine ifthe company should purchase the equipmentunder any ofthe operating cost scenarios ... 18 ... The $30,000 per yearlease agreement will be a net, net, net lease,which means that the lessee (Astor) will pay thereal estate taxes on the leased space, the buildinginsurance on the leased space, and the commonarea maintenance ... A newbuilding will cost $880,000to purchase, but thereis considerable uncertainty about what it will beworth in 20 years, which is the planning periodselected ... Determine if Astor shouldpurchase the building under any of the estimatedresale values at i ϭ 10% per year ... 21 Holly Farms is considering two environmentalchambers toaccomplish detailed laboratory confirmations of online bacteria tests in chickenmeat for the presence of E ... There is some uncertaintyabout how long the D103 chamber will be useful ... The estimatedsalvage value will remain the same ... Chamber D103 Chamber 490GInstalled cost, $AOC, $ peryearSalvage value at 10% of P, $Life, yearsϪ400,000Ϫ4,00040,0002, 3, or 6Ϫ250,000Ϫ3,00025,000218 ... However, ina receding economy the expected return is 8% ... An expandingeconomy causes the estimates of asset life to godown about 20%, and a receding economy makesthe n values increaseabout 10% ... (c) Considering all theanalyses, under which scenario, if any, should planM or Q be rejected?Plan MInitial investment, $Cash flow, $ per yearLife, yearsPlan QϪ100,000ϩ15,00020Ϫ110,000ϩ19,00020Expected Value18 ... The probabilities are 20%, 50%, and 30% for revenues of $800,000,$1,000,000, and $1,100,000per year, respectively, and operating expenses areconstant at $200,000 per year ... 24 A company that manufactures amplified pressuretransducers is trying to decide between a dualspeed and a variable-speed machine ... The results can be summarized as follows: there is a32% chance of getting $20,000, a45% chance of getting $28,000, and a 13%chance of getting $34,000 ... Calculate theexpected salvage value ... 25 The average success probability for a wildcat oilwell drilled in the Wind River basin 7 miles fromthe nearest existing production well is estimatedto be 13%... 5 million, $1 ... 4 million, what is the expected income fromthe well?18 ... Determine the expected value of themonthly income, if economic conditions remainthe same ... 27 Determine the expected maximum rainfall intensity in El Paso, Texas for the month of July usingthe estimated probabilities shown... 40 ... 20 ... 28 There are four estimates made for the anticipatedcycle time to produce a subcomponent ... (a) Ifequal weight is placed on each estimate, what isthe expected cycle time? (b) If the largest time isdisregarded, what is the percent reduction in theexpected time?18 ... Your office partner toldyou that the low bid is$3200 per year ... 30 A total of 40 different proposals were evaluated bythe IRAD (Industrial Research and Development)committee during the past year ... Their rate of return estimates are summarized with the i* values rounded to the nearest507Problemsinteger ... ProposalROR,i*%Number ofProposalsϪ8Ϫ5058101511552332018 ... The amount of rainfall experienced in a short time may cause damage in varying amounts, and the wall increases in cost inorder to protect against larger and faster rainfalls ... 31 Beckman Electronics has performed an economicanalysis of proposedservice in a new region of thecountry ... The optimistic and pessimistic values each have an estimated 20% chanceof occurring ... OptimisticFW value, $$35,000 per year ... Most Likely50,000Ϫ25,000Probability ofGreater RainfallFirst Costof Wall, $2 ... 252 ... 03 ... 30 ... 050 ...005200,000225,000300,000400,000450,000Pessimistic300,000Rainfall, Inchesper 30 Minutes18 ... Because of its location, there is a 30% chance ofa 120-day season of good outdoor weather, a50% chance of a 150-day season, and a 20%chance of a 165-day season ... Thefeature will cost $375,000 toconstruct and require a $25,000 rework each 4 years; and the annual maintenance and insurance costs will be$56,000 ... If a life of 10 years is anticipated and a 12% peryear return is expected, determine if the additionis economically justified ... 33 The owner of Ace Roofing may invest $200,000 innewequipment ... The annual extra revenue will depend upon the state ofthe housing and construction industry ... Realestate economists estimate a 50% chance of theslump lasting 3 years and they give it a 20% chanceof continuing for 3 additional years ... The principal and interest will be repaid overa 10-year period ... A discount rate of6% per year is applicable ... Decision Trees18 ... (This decision branch is part of alarger tree ... 40 ... 3Value, $55– 3010D30 ... 4– 17018 ... If decisions D1, D2, and D3 areall options in a 1-year period, find the decision508Chapter 18Sensitivity Analysis and StagedDecisionspath that maximizes the outcome value ... annual equivalent cost for each alternative is dependent upon specific circumstances of the plant, producer, or contractor ... Construct and solve a decision tree to determine the least-cost alternative toprovide the subassemblies ... 9D30 ... 60 ...Make500$80902 ... 30 ... 40 ... 5$303 ... 37 Decision D4, which has three possible alternatives—x, y, or z—must be made in year 3 of a 6-yearstudy period in order to maximize the expectedvalue of present worth ... Investment Cash flow, $1000Required, $OutcomeYears6 probability3455050500 ...33040500 ... 55– 350,000 1901701500 ... 3– 200,000LowxHighD4yLowzHighLow– 75,000Plant:ABCQuantity:Ͻ5000, pay premium5000 availableϾ5000, forced to buyDelivery:Timely deliveryLate delivery, thenbuy some off shelfProbability0 ... 50 ... 20 ... 1Ϫ550,000Ϫ250,000Ϫ290,0000 ...5Ϫ175,000Ϫ450,000200– 100$20HighOutcomesAnnual Cost for5000 Units, $per Year18 ... The subassemblies can be obtained in one of three ways:(1) Make them in one of three plants owned by thecompany; (2) buy them off the shelf from the oneand only manufacturer; or (3) contract to havethemmade to specifications by a vendor ... 39 The president of ChemTech is trying to decidewhether to start a new product line or purchase asmall company ... To make the product for a 3-year period willrequire an initial investment of $250,000 ... 5), $90,000 (0 ... 1) ... Market surveys indicate a55%chance of increased sales for the company anda 45% chance of severe decreases with an annualcash flow of $25,000 ... Increased sales could be $100,000 the first 2 years ... This expansion could generate cash flowswith indicated probabilities as follows: $120,000(0 ... 3), and $175,000 (0 ... Ifexpansion is not chosen, the current size will be maintained with anticipated sales to continue ... Usethe description given and a 15% per year return todo the following ... (b) Determine the expected PW values at the“expansion/no expansion” decision node after2 years, provided sales are up ...509Additional Problems and FE Exam Review Questions(d)Explain in words what would happen to theexpected values at each decision node if theplanning horizon were extended beyond3 years and all cash flow values continued asforecasted in the description ... 40 A privately held company that makeschips that areessential for high-volume data storage is valued at$3 billion ... 1 billion ... 41 A company that is considering adding a new product line has determined that the first cost would be$80 million ... Instead of expanding now, the company could implement a test program for 1 year in a limitedareathat will cost $4 million ... ) This will provide the companywith the option to move forward or cancel the project ... In this case, the pessimistic estimate will be eliminated, and equal probability willbe placed on the remaining revenue projections ... 42 Dow Chemical is considering licensing a low liquiddischarge (LLD) water treatment system from asmall company that developed the process ... 8 millionplus 25% of sales ... 9 million plus 30% of sales ... 18 ... There is a 1-year sales warranty with the purchase; however, an extended warranty is availablefor $2500 that will cover the same repairs andcomponent failures as the 1-year warranty for 3 additional years ... To help withher decision, the salesman provided three typicalsets of historical data on estimated repair costs forused cars ... Year1Repair cost, $ per year:ABC234000Ϫ500Ϫ10000Ϫ1200Ϫ1400Ϫ500Ϫ850Ϫ400Ϫ2000The salesman saidcase C is the base case, since itshows that the extended warranty is not needed because the cost of repairs equals the warranty cost ... (a) If Abby assumes that each repair cost scenariohas equal probability of occurring with hercar, and money is worth 5% per year to her,how much should she be willingto pay for theextended warranty that is offered at $2500?(b) If the base case actually occurs for her carand she does not purchase the warranty, whatis the PW value of the expected future costsat i ϭ 5% per year?ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS18 ... 45 When themeasure of worth is plotted versus percentchange for several parameters, the parameter that isthe most sensitive in the economic analysis is the one:(a) That has the steepest curve(b) That has the flattest curve(c) With the largest present worth(d) With the shortest life510Chapter 18Sensitivity Analysisand Staged Decisions18 ... 47 For annual worth values of $30,000, $40,000, and$50,000 with chances of 20%, 20%, and 60%, respectively, the expected AW is closest to:(a) $34,000(b) $40,000(c) $44,000(d) $48,00018 ... A challenger will have afirst cost of $90,000, an operating cost of $29,000,and asalvage value after 5 years that may vary considerably ... 49 A decision tree includes all of the following except:(a) Probability estimates for each outcome(b) Measure of worth as the selection criterion(c) Expected results from a decision at eachstage(d) The MARR18 ... 51 A small manufacturing companyneeds to purchase a machine that will have a first cost of$70,000 ... If thecompany’s MARR is 10% per year, the maximumamount the company should pay for the option isclosest to:(a) $5850(b) $6365(c) $6845(d) $7295CASE STUDYSENSITIVITY TO THE ECONOMIC ENVIRONMENTBackground andInformationCase Study QuestionsBerkshire Controllers usually finances its engineering projects with a combination of debt and equity capital ... Normally, a 7% per year return is expected ... The following estimates are the most likely values for two expansionplans currently being evaluated ... Allmonetaryestimates are in $1000 units ... 1 ... Are the PW values sensitive to varying life estimates?3 ... Ascities grow and extend their boundaries to outlying areas,they often inherit water systems that were not constructedaccording to city codes ... To avoid these problems, city officials sometimes installwater systems beyond the existing city limitsin anticipation of future growth ... From about a dozen suggested plans, five methods weredeveloped by an executive committee as alternative waysof providing water to the study area ... Six attributes or factors wereused in the initial rating: ability to serve thearea, relativecost, engineering feasibility, institutional issues, environmental considerations, and lead time requirement ... After the top three alternatives were identified, each was subjected to a detailed economic evaluation for selection of the best alternative ... The annual cost (an AW value) wasthendivided by the population served to arrive at a monthlycost per household ... Alternatives 1A, 3, and 4were determined to be the three best and were chosen for further evaluation ... Alternative 1ACapital costLand with water rights: 1720 hectares@ $5000 per hectarePrimary treatment plantBooster stationat plantReservoir at booster stationSite costTransmission line from riverTransmission line right-of-wayPercolation bedsPercolation bed pipingProduction wellsWell field gathering systemDistribution systemAdditional distribution systemReservoirsReservoir site, land, and developmentSubtotalEngineeringand contingencies$8,600,0002,560,000221,42550,32540,2603,020,00023,3502,093,50060,400510,00077,0001,450,0003,784,800250,00017,00022,758,0605,641,940Total capital investment$28,400,000Results of Rating Six Factors for Each Alternative, CaseStudyFactorsAlternativeDescription1AReceive citywater andrecharge wellsJoint city andcounty plantCounty treatmentplantDesaltgroundwaterDevelop militarywater34812AbilitytoSupply Relative Engineering Institutional EnvironmentalLead TimeAreaCostFeasibilityIssuesConsiderations RequirementTotal5434532454434323443343211211341255413119512Chapter 18Sensitivity Analysis and Staged DecisionsAlternative 4Maintenance and operation costs (annual)Pumping 9,812,610 kWh per year@ $0 ... 76 per monthHousehold costϭ 28,400,000(A͞P,8%,20)ϩ 1,060,419On the basis of the lowestmonthly household cost, alternative 3 (joint city and county plant) is the most economicallyattractive ... 95ϭ $69 ... 38 per monthTA BLE 18–61 ... If the ability to supply area and relative cost factors wereeach weighted 20% and the other four factors 15% each,which alternatives would be ranked in the topthree?3 ... If alternative 1A served 100% of the households insteadof 95%, by how much would the monthly householdcost decrease?5 ... Three estimates are made for eachparameter in Table 18–6 ... The estimated number of households (4980) is determined to be the pessimistic estimate ... (b)Consider the monthly cost per household for alternative 4, the optimistic estimate ... What isthe number of households that would have to beavailable in order for this option to have exactlythe same monthly household cost as that for alternative 3 at the optimistic estimate of 5230 households?CHAPTER19More onVariation andDecisionMaking underRiskL E A R N I N GO U T C O M E SPurpose: Incorporate decision making under risk into an engineering economy evaluation using probability, sampling,and simulation ... 1Risk versus certainty• Understand the approaches to decision makingunder riskand certainty ... 2Probability and distributions• Construct a probability distribution andcumulative distribution for one variable ... 3Random sample• Obtain a random sample from a cumulativedistribution using a random number table ... 4�, �, and �2• Estimate the population expected value,standarddeviation, and variance from a random sample ... 5Simulation• Use Monte Carlo sampling and spreadsheetbased simulation for alternative evaluation ... Fundamentals discussed include variables; probability distributions, especially their graphs and propertiesof expected value and dispersion;random sampling; and the use of simulation toaccount for estimate variation in engineering economy studies ... These techniques are more time-consuming than using estimates made with certainty, so they should be used primarily for critical parameters ... 1 Interpretation of Certainty,Risk, andUncertaintyAll things in the world vary—one from another, over time, and with different environments ... Except for the use of breakeven analysis, sensitivity analysis, and a verybrief introduction to expected values, virtually all our estimates have been certain; that is, novariation in the amount has enteredinto the computations of PW, AW, ROR, or any relationsused ... Decision making under certainty is, of course, not present in the real world now and surely not in thefuture ... To allow a parameter of an engineering economy study to vary implies that risk, and possiblyuncertainty, is introduced ... Virtuallyall decision making isperformed under risk ... Decision making under uncertainty means there are two or more values observable, but thechances of their occurring cannot be estimated or no one is willing to assign the chances ... For example, consider the states of nature to be the rate of national inflationin a particularcountry during the next 2 to 4 years: remain low, increase 2% to 6% annually, or increase 6%to 8% annually ... Example 19 ... EXAMPLE 19 ... One of each piece of equipment must be purchased ... The total for each piece of equipment is 100% ... (a) Consider the ratings as the chance outof 100 that the bid will be chosen, and plot costversus chance for each vendor ... Plot thisrange with an equal chance of 1 in 14 that any amount in between these limits is possible ... Risk516Chapter 19Figure 19–1Equipment AEquipment BBid, $1000Rating, %Bid, $1000Rating,%3,0005,00010,0006525108,00010,00015,00033 ... 333 ... 333 ... 3789Cost, $ million10111225272920012345613141516(a) Specific valuesTotal cost1—14ChancePlot of cost estimatesversus chance for (a)each piece of equipment and (b) total costrange, Example 19 ... More on Variation and DecisionMaking under Risk079111315171921Cost, $ million23(b) Continuous rangeSolution(a) Figure 19–1a plots the specific bids for equipment A and B ... No values between the specific bids have any chance of occurring, according to the single-estimate bids from the three vendors ... Tom decided to make hisestimate of total cost continuous between these two extremes ... Rather the entire range from $11 million to $25 million with a chance forevery total cost in between is included ... Now, the sum is a continuous value ... In the graph for the sum of the cost for equipment A and B (Figure 19–1b),the y axisvalues are continuous over a specific range ... 19 ... A summary of themeaning and use for each type of analysis follows ... Deterministic estimates are made and entered into measure of worth relations—PW, AW, FW,ROR, B͞C—and decision making is based on the results ... A typical example isanasset’s first cost estimate made with certainty, say, P ϭ $50,000 ... The term deterministic, in lieu of certainty, is often used when single-value or single-pointestimates are used exclusively ... The resulting measure of worth values are calculated and graphicallyportrayed to determine the decision’ssensitivity to different estimates for one or moreparameters ... However, it is more difficult to make a clear decision because the analysis attempts to accommodate variation ... The estimates will be expressed as in Example 19 ... Fundamentally,there are two ways to consider risk in an analysis:Expectedvalue analysis ... 2] ... To select the alternative, choose the most favorableexpected value of the measure of worth ... The computations may become more elaborate, but the principle is fundamentally the same ... Use the chance and parameter estimates to generate repeated computations of the measureof worth relation by randomly sampling from a plot for each varyingparameter similar to those in Figure 19–1 ... Usually, graphics are animportant part of decision making via simulation analysis ... Decision Making under Uncertainty When chances are not known for the identifiedstates of nature (or values)of the uncertain parameters, the use of expected value–based decision making under risk as outlined above is not an option ... If it is possible to agree that each state isequally likely, then all states have the same chance, and the situation reduces to one of decision making under risk, because expectedvalues can be determined ... In an engineering economy study, observed parameter values will vary from the value estimated at the time of the study ... Those that are estimable with a relatively high degreeof certainty should be fixed for the study ... Parameters such as P, AOC, material and unit costs,sales price, revenues, etc ... Anticipated variation in interest rates is more commonly addressed by sensitivity analysis ... Sections 19 ... 4 provide foundation material necessary to design and correctly conduct a simulation analysis (Section 19 ... 517(a) Discrete and continuous variable scales and(b)scales for a variableversus its probability ... 00 ... 6Estimated life,probability vs ... 40 ... 00 ... 6Rate of return,probability vs ... 40 ... 2 Elements Important to DecisionMaking under RiskSome basics of probability and statistics are essential to correctly perform decision making underrisk via expected valueor simulation analysis ... (If you are already familiar withthem, this section will provide a review ... Variables are classified as discrete or continuous ... The estimated life of an asset is a discrete variable ... The rate of return is an example of acontinuous variable; i can vary from −100% to ϱ, that is, −100%Յ i Ͻ ϱ ... (Inprobability texts, capital letters symbolize a variable, say X, and small letters x identify aspecific value of the variable ... )Probability is a number between 0 and 1 ... Probability is simply the amount of chance, divided by 100 ... (Actually, for a continuous variable, the probability at asingle valueis zero, as shown in a later example ... 0,19 ... The probability scale, like the percentage scale for chance inFigure 19–1, is indicated on the ordinate (y axis) of a graph ... 0range of probability for the variables n and i ... Discrete variable distributions look significantly different from continuous� sum of all probabilities through ( variabledistributions, as indicated by the inset at the right ... 1]The distribution may be developed in one of two ways: by listing each probability value for eachpossible variable value (see Example 19 ... 3) ... Identified by F(Xi), each cumulative value is calculated asF(Xithe value Xi� P(X Յ Xi)[19 ... Examples 19 ... 3 illustrate cumulativedistributions that correspond to specific probability distributions ... EXAMPLE 19 ... He is planning to start prescribing an antibiotic that may reduce infection in patients with flesh wounds ... If no drug is used, there is always a positiveprobability that the infectionwill be reduced by a person’s own immune system ... e ... Use the probabilities listed below to construct a probability distribution and a cumulative distribution forthe total number of treatments per day ... 070 ... 100 ... 130 ... 25SolutionDefine the random variable T as thenumber of added treatments per day ... The probability of infection reduction islisted for each value in column 2 of Table 19–1 ... 2] by adding all P(Ti) values through Ti, as indicated in column 3 ... The summing of probabilities to obtain F(Ti) gives the cumulative distribution thestair-stepped appearance,and in all cases the final F(Ti) ϭ 1 ... 0 ... 2(1)Number per DayTi(a) Probability distribution P(Ti) and(b) cumulative distribution F(Ti) forExample 19 ... ProbabilityP(Ti)(3)CumulativeProbabilityF(Ti)0123456Figure 19–3(2)0 ... 080 ... 120 ... 250 ... 070 ... 250 ... 500 ... 00F(Ti)P(Ti)0 ... 001 ... 25 0 ... 80 ... 20 ...500 ... 130 ... 370 ... 07 0 ... 30 ... 150 ... {0 ... 080 ... 120 ... 250 ... 070 ... 25F(Ti) ϭ 0 ... 500 ... 00T1 ϭ 0T2 ϭ 1T3 ϭ 2T4 ϭ 3T5 ϭ 4T6 ϭ 5T7 ϭ 6In basic engineering economy situations, the probability distribution for a continuous variableis commonly expressed as a mathematical function, such as auniform distribution, a triangulardistribution (both discussed in Example 19 ... For continuous variable distributions, the symbol f(X) isroutinely used instead of P(Xi), and F(X) is used instead of F(Xi), simply because the point probability for a continuous variable is zero ... EXAMPLE 19 ... Sallie hasconcluded the following about the distribution of these monthly cash flows:19 ... Sallie assumes cash flow to bea continuous variable referred to as C ... SolutionAll cash flow values are expressed in $1000 units ... Probability and cumulative probability take the following general forms ... 3]low value Յ C1 Յhigh valueL Յ C1 Յ H[19 ... Figure 19–4 is a plot of f(C1) and F(C1) from Equations [19 ... 4] ... 25C1 Ϫ 10F(C1) ϭ ————5$10 Յ C1 Յ $15$10 Յ C1 Յ $15(b) The probability that client 1 has a monthly cash flow of no more than $12 is easily determined from the F(C1) plot as 0 ... If the F(C1) relation isused directly, thecomputation is12 Ϫ 10F($12) ϭ P(C1 Յ $12) ϭ ———— ϭ 0 ... 3 ... 00 ... 60 ... 20 ... This probability distribution has the shape of an upward-pointing triangle with the peak at the mode M, and downward-sloping lines joining the x axis on eitherside at the low (L) and high (H) values ...2f(mode) ϭ f(M) ϭ ———HϪL[19 ... 6]For C2, the low value is L ϭ $20, the high is H ϭ $30, and the most likely cash flow is themode M ϭ $28 ... 5] is22f(28) ϭ ———— ϭ —— ϭ 0 ... Using Equation [19 ... 830 Ϫ 20Figure 19–5 presents the plots for f(C2) and F(C2) ... (b) From the cumulative distribution inFigure 19–5, there is an estimated 31 ... Therefore,F($30) Ϫ F($25) ϭ P(C2 Ն $25) ϭ 1 Ϫ 0 ... 6875CommentThe general relations f(C2) and F(C2) are not developed here ... Therefore, it requires the use of an integral to find cumulative probability values from the probability distribution f(C2) ... 00 ... 60 ...3125Modef (C2)0 ... 20020283020C2Figure 19–5Triangular distribution for client 2 monthly cash flow, Example 19 ... 25C2283019 ... 3 Random SamplesEstimating a parameter with a single value in previous chapters is the equivalent of taking a random sample of size 1 from an entire population ofpossible values ... Each estimate is a sample of size 1from an entire population of possible values for each parameter ... If all values in the population were known, the probability distribution and cumulative distribution would be known ... When we perform an engineering economy study and utilizedecision making under certainty,we use one estimate for each parameter to calculate a measure of worth (i ... , a sample of size 1for each parameter) ... We know that all parameters will vary somewhat; yet some are important enough, or willvary enough, that a probability distribution should be determinedor assumed for it and the parameter treated as a random variable ... This approach complicates the analysis somewhat; however, it alsoprovides a sense of confidence (or possibly a lack of confidence in some cases) about the decisionmade concerning the economic viability of the alternative based onthe varying parameter ... )A random sample of size n is the selection in a random fashion of n values from a populationwith an assumed or known probability distribution, such that the values of the variable have thesame chance of occurring in the sample as they are expected to occur in the population ...For a two-crew aircraft, there are three parachutes on board ... ”Yvon is relatively sure that nationwide the probability distribution of N, the specific number ofchutes fully ready, may be described by the probability distribution{0 ... 015P(N ϭ Ni) ϭ0 ... 920N ϭ 0 chutes readyN ϭ 1 chute readyN ϭ 2 chutesreadyN ϭ 3 chutes readyThis means that the safety standard is clearly not met nationwide ... If the sample is truly random and Yvon’sprobability distribution is a correct representation of actual parachute readiness, the observed N values in the 200 aircraft will approximate the same proportions as thepopulationprobabilities, that is, 1 aircraft with 0 chutes ready, etc ... However, if the results are relatively close, thestudy indicates that the sample results may be useful in predicting parachute safety acrossthe nation ... 1for Xi ϭ 0, 1, 2, … , 9In tabular form, the random digits so generated are commonlyclustered in groups of two digits,three digits, or more ... This format is very useful because the numbers 00 to 99 conveniently relate to the cumulative distribution values 0 ... 00 ... To apply this logic manually and develop a random sample of size n from a known523524Chapter 19TA BLE 19–2More onVariation and Decision Making under RiskRandom Digits Clustered into Two-Digit Numbers51 82 88 18 19 81 03 88 91 46 39 19 28 94 70 76 33 15 64 20 14 5273 48 28 59 78 38 54 54 93 32 70 60 78 64 92 40 72 71 77 56 39 2710 42 18 31 23 80 80 26 74 71 03 90 55 61 61 28 41 49 00 79 96 7845 4479 29 81 58 66 70 24 82 91 94 42 10 61 60 79 30 01 26 31 4268 65 26 71 44 37 93 94 93 72 84 39 77 01 97 74 17 19 46 61 49 6775 52 14 99 67 74 06 50 97 46 27 88 10 10 70 66 22 56 18 32 06 24discrete probability distribution P(X ) or a continuous variable distribution f(X ), the followingproceduremay be used ... Develop the cumulative distribution F(X) from the probability distribution ... 2 ... For the parachute safety example, the probabilities from 0 ... 15 arerepresented by the random numbers 00 to 14 ... 3 ... Any direction and pattern is acceptable, but thescheme should be used consistently forone entire sample ... Select the first number from the RN table, enter the F(X ) scale, and observe and record thecorresponding variable value ... 5 ... These may include•••••Plotting the sample probability distribution ... Comparing sample results with the assumed population distribution ... 4) ... 5) ...4Develop a random sample of size 10 for the variable N, number of months, as described by theprobability distribution{0 ... 500 ... 7]SolutionApply the procedure above, using the P(N ϭ Ni) values in Equation [19 ... 1 ... 2 ... 2; 50 numbers to N2 ϭ 30; and 30 numbers to N3 ϭ 36 ... Initially select anyposition in Table 19–2, and go across the row to the right and onto therow below toward the left ... )4 ... 5 ... RN45447929815866702482N3030363036303036303619 ... 0Cumulative distributionwith random number values assigned in proportion to probabilities,Example 19 ... 70–99F(Ni )0 ... 200–19243036Ni, monthsNow, using the 10 values, develop the sample probabilities ... 7]Probability2430360640 ... 600 ... 20 ... 3With only 10 values, we can expect the sample probability estimates to be different from thevalues in Equation [19 ... Only the value N ϭ 24 months is significantly different, sinceno RNof 19 or less occurred ... To take a random sample of size n for a continuous variable, the procedure above is applied,except the random number values are assigned to the cumulative distribution on a continuousscale of 00 to 99 corresponding to the F(X ) values ... 3 ... 2 for all values between L

and H (all values are divided by $1000) ... If the two-digit RN of 45 is chosen, the corresponding C1 is graphically estimated to be$12 ... It can also be linearly interpolated as $12 ... 1 ... 6590 ... 2Random numbersassigned to thecontinuous variable ofclient 1 cash flows inExample 19 ... 990 ...25131415C1, $1000526Chapter 19More on Variation and Decision Making under RiskFor greater accuracy when developing a random sample, especially for a continuous variable,it is possible to use 3-, 4-, or 5-digit RNs ... In computer-based sampling, most simulation software packages have anRNgenerator built in that will generate values in the range of 0 to 1 from a continuous variable uniform distribution, usually identified by the symbol U(0, 1) ... 00000 and 0 ... The Excel functions RAND and RANDBETWEEN are described in Appendix A, Section A ... An initial question in random samplingusually concerns the minimum size of n required toensure confidence in the results ... However, since reality doesnot follow theory exactly, and since engineering economy often deals with sketchy estimates,samples in the range of 100 to 200 are the common practice ... 19 ... If the entire population for avariable were known, these properties would becalculated directly ... The following is a brief introduction to the interpretation and calculation of these properties using a randomsample of size n from the population ... True Population MeasureSymbolExpected value� or E(X)______Standard deviation � or√Var(X)___or √�2NameSample EstimateSymbolName__Mu or true meanSigma or truestandard deviationSample meanX__s or √s2Sample standarddeviationThe expected value E(X) is the long-run expected average if the variable is sampled many times ... Equation [18 ... 8], is used to compute the E(X)� —————————— sample size⌺Xi ⌺fi Xi� —— � ——— nn[19 ... 9]The fi in the second form of Equation [19 ... The resulting X is not necessarily an observed value � ⌺Xi P(Xi)__sum of sample valuesX ( of a probability distribution, and Equation [19 ... Population:Probability distribution:Sample:�E(Xof the variable; it is the long-run average value and can take on any value within the range of the variable ... )19 ... 5Kayeu, an engineer with Pacific NW Utilities, is planning to test several hypotheses aboutresidential electricity bills in North American and Asian countries ... S ... Two small samples havebeen collected from different countries of North America and Asia ... Do the samples (from a nonstatistical viewpoint) appear to bedrawn from one population of electricity bills or from two different populations?North American, Sample 1, $40667592107Asian, Sample 2,$8490104187190159275SolutionUse Equation [19 ... nϭ7nϭ5Sample 1:Sample 2:⌺Xi ϭ 814⌺Xi ϭ 655__X ϭ $116 ... 00Based solely on the small sample averages, the approximate $15 difference, which isonly11% of the larger average bill, does not seem sufficiently large to conclude that thetwopopulations are different ... (Check a basic statistics text to learnabout them ... The sample averageis the most popular, but the mode and the median are also good measures ... 3 for a triangular distribution ... Themedian is the middle value of the sample ... The two medians in the samples are $92 and$104 ... The standard deviation __ or s(X ) is the dispersion or spread of values about the expected valuesE(X ) or sample average X ... A probability distribution for data with strongcentral tendency is more closely clustered about the center of the data, and has a smaller s, thana wider, more disperseddistribution ... f (X)P(X)s2s3s4s1X1s1 Ͼ s2X2s4 Ͼ s3Figure 19–8Sketches of distributions with different separate lines standard deviation values ... The standard deviationis simply the square root of the variance, so either measure can be used ... Mathematically, the formulasand symbols for variance and� —————————————————— sample size � ⌺[Xi Ϫ E(X)]2P(Xi)sum of (sample value Ϫ sample average)2Sample:s2 ( � √ 2� � √ Var(X)Probability distribution:Var(X � Var(X)and_______� standard deviation of a discrete variable and a random sample ofsize n are as follows:Population:___�2To accurately__measure the dispersion in both directions from the average, the quantity (X Ϫ X ) is squared ... 11] is extracted ... The fi in the second form of Equation [19 ... 19]1؊n 1؊1n؊n 1؊� ——— ؊ ——— X 2n � ——— ؊ ——— X 2 � √ s2Equation [19 ... ⌺Xi2⌺fiXi2n __n __s2 Ϫ 1[19 ... 11]__s... One simple way to combine the average and standard deviation is to determine the percentageor fraction of the sample that is within Ϯ1, Ϯ2, or Ϯ3 standard deviations of the average, that is,__X Ϯ tsfor t ϭ 1, 2, or 3[19 ... 14]__Virtually all the sample values will always be within the Ϯ3s range of X, but thepercent within__Ϯ1s will vary depending on how the data points are distributed about X ... 6 illustrates__the calculation of s to estimate � and incorporates s with the sample average using X Ϯ ts ... 6(a) Use the two samples of Example 19 ... (b) Determine the percentages of each sample that are insidethe rangesof 1 and 2 standard deviations from the mean ... For sample 1 (North American) with n ϭ 7, use X to identify the values ... 11], with X ϭ $116 ... The resulting s2 and s values are37,743 ... 576s ϭ $79 ... 4Expected Value and Standard DeviationTABLE 19–3 Computation of StandardDeviation__Using Equation [19 ... 29, Example 19 ... 29Ϫ50 ... 29Ϫ24 ... 29ϩ42 ... 715,820 ... 081,704 ... 0086 ... 1425,188 ... 40TABLE 19–4 Computation of Standard Deviation__Using Equation [19 ... 6Y, $Y284901041871907,0568,10010,81634,96936,10065597,041__For sample 2 (Asian), use Y toidentify the values ... 12] ... 25 Ϫ 1 ... 31) ... 13] determines the ranges of X Ϯ 1s and X Ϯ 2s ... See Figure 19–9for a plot of the data and the standard deviation ranges ... 29– 80– 4004080120–X ± 1s–X ± 2s160200240280X240Y(a)–Y = 13104080120–Y ± 1s–Y ± 2s160200(b)Figure 19–9Values, averages,and standard deviation ranges for (a) North American and(b) Asian samples, Example 19 ... 529530Chapter 19More on Variation and Decision Making under RiskNorth American sample__X Ϯ 1s ϭ 116 ... 31for a range of $36 ... 60Six out of seven values are within this range, so the percentage is 85 ...__X Ϯ 2s ϭ 116 ... 62for a range of $Ϫ42 ... 91__There are still six of the seven values within the X Ϯ 2s range ... 33 is meaningful only from the probabilistic perspective; from the practical viewpoint, use zero, thatis, no amount billed ... __Y Ϯ 2s ϭ 131 Ϯ 106for a range of $25 to $237__All five of the valuesare within the Y Ϯ 2s range ... In the two samples here, the range estimates are $235 and $106 ... Calculators andspreadsheets all have functions to determine these values by simply entering the data ... 8] through [19 ... The primary differences are that thesummation symbol is replaced by the integralover the defined range of the variable, which weidentify as R, and that P(X) is replaced by the differential element f (X) dX ... 15]Var(X) ϭ ͐R X2f(X) dX Ϫ [E(X)]2Variance:[19 ... 3 (Figure 19–4) overthe range R from $10 to $15 ... 1$10 Յ X Յ $15f(X) ϭ — ϭ 0 ... 2) dX ϭ 0 ... 1(225 Ϫ 100) ϭ $12 ... 2 15Var(X)ϭ ͐ X2(0 ... 5)2 ϭ ——X3 10 Ϫ (12 ... 06667(3375 Ϫ 1000) Ϫ 156 ... 08|____� ϭ √ 2 ... 44Therefore, the uniform distribution between L ϭ $10 and H ϭ $15 has an expected value of$12 ... 44 ... 7Christy is the regional safety engineer for a chain of franchise-based gasoline and food stores ... ) on concretesurfaces ... 4Expected Value and Standard DeviationCorporate management has authorized each regional engineer to contract locally to apply to allexterior concrete surfaces a newly marketed product that absorbs up to 100 times its ownweight in liquid and to charge a home office account for theinstallation ... You have been asked to write a brief but thorough summary about the normal distribution,explain the $8000 to $12,000 range statement, and explain the phrase “random samples thatassume a normal population ... Normal distribution, probabilities, and random samplesThe normaldistribution is also referred to as the bell-shaped curve, the Gaussian distribution, or the error distribution ... It places exactly one-half of the probability on either side of the mean orexpected value ... Thenormal distribution is found to accurately predict many types of outcomes, such as IQ values;manufacturing errors about a specified size, volume, weight, etc ... The normal distribution, identified by the symbol N(�, �2), where � is the expected value ormean and �2 is the variance, or measure of spread, can be described as follows:• The mean � locates the probability distribution (Figure 19–10a), and the spread of the distribution varies with variance (Figure 19–10b), growing wider and flatter for larger variancevalues ... • The normal probability distribution f(X) for a variable X is quite complicated, because itsformula is{[(X Ϫ �)21___f(X) ϭ ——— exp Ϫ ————2�2�√2�]}where exprepresents the number e ϭ 2 ... Since f(X) is so unwieldy, random samples and probability statements are developed using atransformation, called the standard normal distribution (SND), which uses � and � (popula__tion) or X and s (sample) to compute values of the variable Z ... 17]__Sample:XϪXZ ϭ———s[19 ... Therefore, the probabilityvalues under the SND curve can be stated exactly ... 17] for X:X ϭ Z� ϩ �[19 ... �1�2�2�1• Equal dispersion(�1 = �2 = �3)• Increasing means�3�3X(a)f (X)• Increasingdispersion(�1 < �2 < �3)• Two differentmeans�1�2�3�1 = �3�2X(b)f (X)Normaldistribution� – 3�� + 3��X0 ... 9546f (Z)0 ... 3413 0 ... 0013 0 ... 1360–20 ... 0214 0 ... Variable X RangeProbabilityVariable Z Range� ϩ 1�� Ϯ 1�� ϩ 2�� Ϯ 2�� ϩ 3�� Ϯ 3�0 ... 68260 ... 95460 ... 99740 to ϩ1Ϫ1 to ϩ10 to ϩ2Ϫ2 to ϩ20 to ϩ3Ϫ3 to ϩ3As an illustration, probability statements from thistabulation and Figure 19–10c for X and Zare as follows:The probability that X is within 2� of its mean is 0 ... The probability that Z is within 2� of its mean, which is the same as between the valuesϪ2 and ϩ2, is also 0 ... 19 ... (Tables of SND values are available in many statistics books ... 10,ϩ1 ...Translation from the Z value back to the sample values for X is via Equation [19 ... Interpretation of the home office memoThe statement that virtually all the local contract amounts should be between $8000 and$12,000 may be interpreted as follows: A normal distribution is assumed with a mean of µ�Ϫ2 ... 12Ϫ0 ... 24Ϫ2 ... 99X ϭ (Ϫ2 ... 12)(667) ϩ 10,000 ϭ $12,081X ϭ (Ϫ0 ... 24)(667) ϩ 10,000 ϭ ؉ � Z� ϭ$10,000 and a standard deviation for � ϭ $667, or a variance of �2 ϭ ($667)2; that is, anN[$10,000, ($667)2] distribution is assumed ... 74%) is within 3� of the mean, as stated above ... 19]X$10,827X ϭ (Ϫ2 ... 99)(667) ϩ 10,000 ϭ $9340In this sample of six typical concrete surfacing contract amounts for sites in our region, theaverage is $9825 and five of six values are within the range of $8000 to $12,000, with the sixthbeing only $81 above the upper limit ... 5 Monte Carlo Sampling andSimulation AnalysisUp to this point, all alternative selections have been made using estimates with certainty, possibly followed by some testing of the decision via sensitivity analysis or expected values ... The random sampling technique discussed in Section 19 ... The general procedure outlined belowuses Monte Carlo sampling to obtain samples of size n forselected parameters of formulated alternatives ... All other parameters in an alternative are considered certain; that is, they are known, or they can be estimated with enough precision to consider them certain ... All parameters are independent; thatis, one variable’s distribution does not affect the value ofany other variable of the alternative ... The simulation approach to engineering economy analysis is summarized in the followingbasic steps ... Formulate alternative(s) ... Determine the form of the relation(s) to calculate the measureof worth ...Parameters with variation ... Estimate values for all other (certain) parameters for the analysis ... Step 4 ... Step 6 ... More on Variation and Decision Making under RiskDetermine probability distributions ... Use standard distributions, where possible, to simplify the sampling process and to prepare forcomputer-based simulation ... Incorporate the random sampling procedure of Section 19 ... This results in the cumulative distribution,assignment of RNs, selection of the RNs, and a sample of size n for each variable ... Compute n values of the selected measure ofworth from the relation(s) determined instep 1 ... (This is when the propertyof independent random variables is actually applied ... Construct the probability distribution of the measure of worth, using between 10 and 20 cells of data, and calculate measures such as____X, s, X Ϯ ts, and relevant probabilities ... Draw conclusions about eachalternative, and decide which is to beselected ... Example 19 ... 9 utilizes spreadsheet simulation for the same estimates ... 8Yvonne Ramos is the CEO of a chain of 50 fitness centers in the United States and Canada ... As an enticement, the offer includes a guarantee of annualrevenue for one of thesystems for the first 5 years ... Details for the two systems follow:System 1 ... No guarantee for annual net revenue is offered ... First cost is P ϭ $8000, there is no salvage value, and there is a guaranteedannual net revenue of $1000 for each of the first 5 years, but after this period, there isno guarantee... Cancellation anytime after the initial 5 years is allowed, with nopenalty ... If theMARR is 15% per year, use PW analysis to determine if neither, one, or both of the systemsshould be installed ... Step 1 ... Using PW analysis, the relations for system 1 and system 2 are developed ... PW1 ϭ ϪP1 ϩNCF1(P͞A,15%,n1)[19 ... 21]Step 2 ... Yvonne summarizes the parameters estimated withcertainty and makes distribution assumptions about three parameters treated as random variables ... 5Monte Carlo Sampling and Simulation AnalysisSystem 1Certainty ... Variable ... System 2Certainty ... Variable ...Variable ... Now, rewrite Equations [19 ... 21] to reflect the estimates made with certainty ... 1604)[19 ... 4972)[19 ... Figure 19–11 (left side) shows the assumedprobability distributions for NCF1, NCF2, and n2 ... Random sampling ... 3 ... The RNsfor NCF2 identify the x axis values so that all net cashflows will be in even $1000amounts ... However, we round the numberto the next higher value of n2 because it is likely the contract may be canceled on ananniversary date ... Once the first RN is selected randomly from Table 19–2, the sequence (step 3)used will be to proceed down the RN table columnand then up the column to the left ... Step 5 ... With the five sample values in Table 19–5, calculate the PW values using Equations [19 ... 23] ... 1 ... 3 ... 5 ... 2 ... 4 ... PW1 ϭ Ϫ12,000 ϩ (Ϫ2200)(4 ... 1604)PW1 ϭ Ϫ12,000 ϩ (Ϫ1100)(4 ... 1604)PW1 ϭ Ϫ12,000 ϩ 3100(4 ... 4972)PW2 ϭ Ϫ4648 ϩ1000(P͞A,15%,5)(0 ... 4972)PW2 ϭ Ϫ4648 ϩ 3000(P͞A,15%,10)(0 ... 4972)ϭ $Ϫ21,153ϭ $Ϫ3679ϭ $Ϫ16,576ϭ $Ϫ15,744ϭ $ϩ897ϭ $Ϫ2579ϭ $Ϫ2981ϭ $ϩ6507ϭ $ϩ2838ϭ $Ϫ107Now, 25 more RNs are selected for each variable from Table 19–2, and the PW values are calculated ... Measure of worthdescription ... 535536Chapter 19More on Variation and Decision Making under RiskFigure 19-11F(NCF1 )Distributions used forrandom samples,Example 19 ... RN1 ... 860–790 ... 420–39f (NCF1)Continuous variable1100 ... 0083–990 ... 6750–660 ... 33P(NCF2)17–32Discrete variable160 ... 0888–9990... 6444–6650 ... 2000–2210610n2, years15681012n2, years14 15TA BLE 19–5 Random Numbers and Variable Values for NCF1, NCF2, andn2, Example 19 ... 39 ... 714 ... 3121013158*Randomly start with row 1, column 4 in Table 19–2 ... ‡Start with row 4, column 6 ... 19 ... 8 ... Sample values rangefrom $−24,481 to $ϩ12,962 ... Sample values range from $−3031 to $ϩ10,324 ... Conclusions ... Of course, many conclusions are possible once the PW distributions areknown, but the following seem clear at this point ... Based on this small sample of 30 observations, do not accept thisalternative ... 27 (8out of 30 values) that the PW will be__positive, and X1 is a large negative ... A largersample may alter this analysis somewhat ... If Yvonne is willing to accept the longer-term commitment that mayincrease the NCF some years out, the sample of 30 observations indicates to accept this alternative ...__However, the probability of observing PW within the X Ϯ 1s limits ($Ϫ1612 and$7060) is 0 ... Conclusion at this point ... 537538Chapter 19More on Variation and Decision Making under RiskCommentThe estimates in Example 13 ... The alternatives wereevaluated by the payback period method atMARR ϭ 15%, and the first alternative was selected ... 5 selected alternative 2 based, inpart, upon the anticipated larger cash flow in the later years ... 9Help Yvonne Ramos set up a spreadsheet simulation for the three random variables and PWanalysis in Example 19 ... Does the PW distribution varyappreciably from that developedusing manual simulation? Do the decisions to reject the system 1 proposal and accept the system 2 proposal still seem reasonable?Solution by SpreadsheetFigures 19–13 and 19–14 are spreadsheet screen shots that accomplish the simulation portionof the analysisdescribed above in steps 3 (determine probability distribution) through 6 (measure of worth description) ... Figure 19–13 shows the results of a small sample of 30 values from the three distributionsusing the RAND and IF functions ... 3 in Appendix A ... The spreadsheet relation in column Btranslates RN1values (column A) into NCF1 amounts ... Column D cells displayNCF2 in the $1000 increments using the logical IF function to translate from the RN2 values ... The results in column F are integer values obtained using the INT function operating on the RN3 values ... 9 ... 5Monte Carlo Sampling andSimulation AnalysisϭSUM(G13:G42)ϭAVERAGE(F13:F42)ϭSTDEV(F13:F42)Figure 19–14Simulation results for 30 PW values, Example 19 ... Figure 19–14 presents the two alternatives’ estimates in the top section ... 22] and [19 ... The tabular approach used here tallies the number of PW valuesbelowzero ($0) and equal to or exceeding zero using the IF operator ... 22] ... Sample averages and standard deviations are also indicated ... System 1 PWSystem 2 PWX, $s, $No ... ofPW Ն 0Ϫ7,729Ϫ7,10510,19013,1998102,7241,6494,3363,8712019__HandSpreadsheet__For the spreadsheet simulation, 10(33%) of the PW1 values exceed zero, while the manual simulation included 8 (27%) positive values ... (It is possible to define RAND to keep the same RN values ... )The conclusion to reject the system 1 proposal and accept system 2 is still appropriate forthe spreadsheet simulation as it was for the handsolution, since there are comparable chancesthat PW Ն 0 ... Assumptions about the shape of the variable’s probabilitydistribution are used to explain how the estimates of parameter values may vary ... In this chapter, we learned several of the simple, but useful, discrete and continuous populationdistributions used in engineering economy—uniform and triangular—as wellas specifying our own distribution or assuming the normal distribution ... The results are used to make probability statements about the parameter, which help make thefinal decision with risk considered ... The results ofsuch ananalysis can then be compared with decisions when parameter estimates are made withcertainty ... 1 Identify the following variables as either discreteor continuous ... He wants to use E(output) in the decisionmaking process ... Probability and Distributions19 ... Data collectedfrom stripper wells in anestablished oil field wereused to develop the probability-royalty relationship shown below ... 2 For each situation below, determine (1) if the variable is discrete or continuous and (2) if the information involves certainty, risk, and͞or uncertainty ... (b) The raises for engineers and technical staffemployees willbe 3%, or 5%, with one-halfgetting 3% and one-half getting 5% ... (d ) The salvage value for an old machine will be$500 (i ... , its asking price) or $0 (it will bethrown away) ... 19 ... 10 0 ... 320 ... 090 ... Determine the probability that the royaltyincome will be at least $12,600 per year ... 5 Daily revenuefrom vending machines placed invarious buildings of a major university is as follows:20, 75, 43, 62, 51, 52, 78, 33, 28, 39, 61, 56, 43, 49, 48, 49,71, 53, 57, 46, 42, 41, 63, 36, 51, 59, 40, 32, 37, 29, 26(a)(b)(c)(d)Construct a frequency distribution table witha cell size of 12 starting with 19 ... e ... 5–31 ... 5–43 ... Determine the probability distribution ... 6 A survey of households included a question aboutthe number of operating automobiles N currentlyowned by people living at the residence and theinterest rate i on the lowest-rate loan for the cars ... Number of Cars NHouseholds0123Ն412562633PurchaseAlternative13141938124Estimated Value0 ... 01Ϫ44 ... 01Ϫ88 ... 01Ϫ12(b)(c)(d)19 ... Use the parameter estimates and assumed distribution data shown to plot the probability distributionson one graph for each corresponding parameter ... HouseholdsLoan Rate i, %(a)ParameterProbability(a)(b)(c)025101000 ... 0450 ... 0130 ... Calculate the expected value of the distribution of dollars per ticket ... 8 Bob is working on two separate probability-relatedprojects ... Thevariable N is described by the formula (0 ... The second involves a batterylife L that varies between 2 and 5 months ... 7 An officer ofthe state lottery commission sampledlottery ticket purchasers over a 1-week period atone location ... Plot the probability distributions and cumulative distributions for N and i ... Distribution, $5 months, which is the design life ... (a) Write out and plot the probabilitydistributions and cumulative distributions forBob ... Ϫ3,0002,000849,0005,000Life, yearsAOC, $ per yearAssumedDistributionUniform;continuousTriangular;mode at $2500Triangular;mode at 6Uniform;continuousLease AlternativeEstimated ValueParameterHighLowLease first cost, $20001800AOC, $ per year90005000Lease term,years33AssumedDistributionUniform;continuousTriangular;mode at $7000Certainty19 ... She has collected debt-to-equity mix data on mature (M) andyoung (Y) companies ... Carla has definedDM as a variable for the mature companies from0 to 1, with DM ϭ 0 interpreted as the low of 20%debt and DM ϭ1 ... Thevariable for young corporation debt percentagesDY is similarly defined ... (b) What canyou comment about the probability that a maturecompany or a young company will have a low debtpercentage? A high debt percentage?542Chapter 19More on Variation and Decision Making under Risk19 ...A sample of size 50 results in the followingprobability estimates:Xi1236910P(Xi)0 ... 20 ... 10 ... 1(a)(b)(c)Write out and graph the cumulativedistribution ... Use the cumulative distribution to show thatP(X ϭ 7 or 8) ϭ 0 ... Even though this probability is zero, the statement about X is that itcan take on integervalues of 1 to 10 ... 14 Develop a discrete probability distribution of yourown for the variable G, the expected grade in thiscourse, where G ϭ A, B, C, D, F, or I (incomplete) ... Now plot the probability values fromthe sample for each G value ... 15 Use the RAND function in Excel to generate100 values froma U(0,1) distribution ... 5,the expected value for a random sample between 0 and 1 ... 1 width, that is 0 ... 1,0 ... 2, etc ... Determine the probability for each grouping from the results ... 12 A discrete variable X can take on integer values of1 to 5 ... Xi(b)2345P(Xi)(a)10 ... 30 ... 16 An engineer was asked todetermine whether theaverage air quality in a vehicle assembly plant waswithin OSHA guidelines ... 1Use the following random numbers to estimate the probabilities for each value of X ... Compare the sample results withthe probabilities in the problem statement ... 13 The percent price increase p on avariety of retailfood prices over a 1-year period varied from 5% to10% in all cases ... 17 Carol sampled the monthly maintenance costs for automated soldering machines a total of 100 times during 1 year ... She indicated the number of times(frequency) each cell value was observed ... CellMidpoint,$when p ϭ 5%1when p ϭ 10%For a continuous variable the cumulative distribution F(X) is the integral of f (X) over the samerange of the variable ... Transform the X values into interest rates ... (a)(b)Frequency6008001000120014001600180020000ՅXՅ10Determine the sample mean ... Determine thenumber of values and percentof values that fall within Ϯ1 standard deviation of the mean ... What is the best estimate of the percentage ofcosts that will fall within 2 standard deviations of the mean?543Additional Problems and FE Exam Review Questions(c)(d)Develop a probability distribution ofthemonthly maintenance costs from Carol’ssample, and indicate the answers to the previous two questions on it ... Simulation19 ... The additional CFAT of $2800 in year 10 isthe salvage value of capital assets ... 18 (a) Determine the values of sample average andstandard deviation of the data in Problem19 ... (b) Determine the values 1 and 2 standard deviations from the mean ... 19 (a) Use the relations in Section 19 ... 10 ... 19 ... 4 for continuous variables to determine the expected valueand variance for the distribution of DM inProblem 19 ... f (DM) ϭ 3(1 Ϫ DM)2(b)0 Յ DM Յ 1Determine the probabilitythat DM is within 2standard deviations of the expected value ... 19 ... 21 Calculate the expected value for the variable N inProblem 19 ... 19 ... Data collected over a 30-weekperiod are summarized by the following probability distribution ... Y copiesP(Y)37101͞41͞3CFAT,$01Ϫ67Ϫ1010Ϫ28,8005,4002,0402,800The PW value at the current MARR of 7% peryear isPW ϭ Ϫ28,800 ϩ 5400(P͞A,7%,6)ϩ 2040(P͞A,7%,4)(P͞F,7%,6)ϩ 2800(P͞F,7%,10)ϭ $2966Carl believes the MARR will vary over a relativelynarrow range, as will the CFAT, especially duringthe out years of 7 through10 ... Use the following probability distribution assumptions for MARRand CFAT to perform a simulation—hand- orspreadsheet-based ... Uniform distribution over the range6% to 10% ... Uniform distribution over the range $1600 to $2400 for eachyear ... Should the planbe accepted using decision makingunder certainty? Under risk?19 ... 23, except use the normal distribution for the CFAT in years 7 through 10 withan expected value of $2000 and a standard deviation of $500 ... 25 When there are at least two values for a parameterand it is possible to estimate the chance that eachmay occur, this situationis known as:(a) Uncertainty(b) Risk(c) Standard deviation(d) Cost estimating19 ... 27 Decision making under risk includes all of the following except:(a) Expected value analysis(b) Simulation(c) Using only single-value estimates(d) Probability19 ... 31 The revenue from an oil dispersant producthasaveraged $15,000 per month for the past 12 months ... 32 A survey of the types of cars parked at an NFLfootball stadium revealed that there were equalprobabilities of finding cars identified as type A, B,C, and D ... 19 ... 15(b) 0 ... 32(d) 0 ... 150 ... 320 ... 090 ... 30 The symbol that represents the truepopulationmean is:(a) �(b) s(c) �__(d) XCar TypeAssigned RNABCD0 through 2425 through 4950 through 7475 through 99The sample probability of a type B car from the12 random numbers shown is closest to:RN: 75, 52, 14, 99, 67, 74, 06, 50, 97, 46, 27, 88(a)(b)(c)(d )0 ... 250 ... 42CASESTUDYUSING SIMULATION AND THREE-ESTIMATE SENSITIVITY ANALYSISBackgroundThe Knox Brewing company makes specialty-named sodasand flavored drinks for retail grocery chains throughoutNorth and Central America ... Dr ... You just handedhim the first cost bids from three vendors forthe machine ... You were quite surprised, as this was the first time you hadbeen in his office, and most other engineers at Knox have agreat fear of “the Old Man ... After a few minutes, he told you to take these numbersand use some of that “new engineering knowledge” youacquired in college todetermine which, if any, of these threebids made the best economic sense ... M ... In addition, youdeveloped some possible distributions for the parameters thatDr ... These are summarized in Table 19–7 ... Case Study ExercisesFirst, learn to use the RNG (random number generator) inExcel, if you havenot already done so ... dev ... dev ... dev ... RNG is part of the Analysis Tool-Pakaccessed through the Office Button, Excel Options, AddIns path ... Prepare the simulation using a spreadsheet; determinewhich of the vendors offers the best machine from aneconomic perspective, and take into account theestimates made by Dr ... Use a sample size of at least50, and base your conclusions on the AW measure ofworth ... Prepare a short presentation for Dr ... APPENDIX AUSING SPREADSHEETSAND MICROSOFT EXCEL©This appendix explains the layout of a spreadsheet and the use of Microsoft� sign is required to perform any formula or function Excel (hereaftercalled Excel) functions in engineering economy ... Some specific commands and entries refer to Excel 2007 andmay differ slightly from your version ... 1 Introduction to Using ExcelEnter a Formula or Use an Excel FunctionThecomputation in a cell ... The symbol ` is usually in the upper left of the keyboard with the ~ (tilde) symbol ... 1 ... 2 ... (Move the pointer to C3 and left-click ... Type ϭ PV(5%,12,10) and ϽEnterϾ ... Another example: To calculate the future value of 12 payments of $10 at 6% per year interest, dothefollowing:1 ... 3 ... 5 ... 7 ... Move to cell B3, and type INTEREST ... Move to cell B4, and type PAYMENT ... Move to cell B5, and type NUMBER OF PAYMENTS ... Move to cell B7, and type FUTURE VALUE ... The answer will appearin cell C7 ... Move to cell C3 and type ϭ5͞100 (the previous value will bereplaced) ... The value in cell C7 will update ... This approach defines the referenced cell as a global variable for the worksheet ... Relative References If a cell reference is entered, for example, A1, into a formula or functionthat is copied or dragged into another cell, the reference is changed relative to themovement ofthe original cell ... This feature is used when dragging a function through several cells, and the source entries must change with the column or row ... For example,548Appendix AUsing Spreadsheets and Microsoft Excel©ϭ $A$1 will retain the formula when it is moved anywhere on theworksheet ... Absolute references are used in engineering economy for sensitivity analysis of parameterssuch as MARR, first cost, and annual cash flows ... Print the SpreadsheetFirst define the portion (or all) of the spreadsheet to be printed ... Move the pointer to the top left cell of your spreadsheet ...Hold down the left-click button ... )3 ... 4 ... (It is ready to print ... Left-click the Office button (see Figure A–1) ... Move the pointer down to select Print and left-click ... In the dialog box, left-click the Print option (or similar command) ... Save the SpreadsheetYou can save your spreadsheet at any time duringor after completing your work ... 1 ... 2 ... option ... Type the file name, e ... , Prob 7 ... To save the spreadsheet after it has been saved the first time, i ... , a file name has been assignedto it, left-click the Office button, move the pointer down, and left-click on Save ... It plots pairs of data and can placemultiple series of entries on the Y axis ... 1 ... 2 ... Column A, cell A1 through A6: Rate i%, 4, 6, 8, 9, 10Column B, cell B1 through B6: $ for A, 40, 55, 60, 45, 10Column C, cell C1 through C6: $ for B, 100, 70, 65, 50, 30 ... Move the mouse to A1, left-click, and hold while dragging to cell C6 ... 4 ... Afterdragging over one column ofdata, momentarily release the left click, then move to the top of the next (nonadjacent) column of the chart ... 5 ... 6 ... The graph appears with alegend (Figure A–1) ... Note that only the bottom row of the title can be highlighted ... on the legend ... 2Office buttonOrganization(Layout) of the SpreadsheetInsert forscatter chartFigure A–1Scatter chart for data entries and location of commonly used buttons ... To get general help information, left-click on the “?” (upper right) ... Enter the topic or phrase ... 3 ... You can browse through the options by left-clickingon any item ... 2Organization (Layout) of the SpreadsheetA spreadsheet can be used in several ways to obtain answers to numerical questions ... For example, to find the future worth in a single-cell operation, move the pointer to any cell andenter ϭ FV(8%,5,-2500) ... 50 is the 8% future worth at the end of year 5of fiveequal payments of $2500 each ... Some fundamental guidelines in spreadsheetorganization are presented here ... As solutions becomemore complex, organization of the spreadsheet becomes increasingly important, especially forpresentation to an audience via PowerPoint or similar software ... It isadvisable to organize the given or estimated data in thetop left of the spreadsheet ... Then B1 can be the referenced cell for allentries requiring the MARR ... Often, the answers are best placed at the bottom or top of the column of entriesused in the formula or predefined function ... Each column or rowshould be labeled so its entries areclear to the reader ... Enter income and cost cash flows separately ... and first cost, salvage value, and annual costs (usually negative, with salvage a positive number)be entered into two adjacent columns ... There are two immediate advantages to this practice: fewererrors are madewhen performing the summation and subtraction mentally, and changes for sensitivity analysisare more easily made ... The use of absolute and relative cell references is a must when any changesin entries are expected ... The absolute cell referenceentry $B$1 in the three functionsallows the MARR to be changed one time, not three ... When the formulas and functionsare kept relatively simple, the final answer can be obtained using the SUM function ... This practice is especially useful when the cashflow series are complex ... If a chart (graph) will be developed, plan ahead byleaving sufficientroom on the right of the data and answers ... Placement on the same worksheet is recommended, especially when theresults of sensitivity analysis are plotted ... 3 Excel Functions Important to EngineeringEconomy (alphabetical order)DB (Declining Balance)Calculates the depreciationamount for an asset for a specified period n using the declining balance method ... This is Equation [16 ... Threedecimal-place accuracy is used for d ... Salvage value ... The period, year, for which the depreciation is to be calculated ... A ... At the end of10 years, the salvage value of the machine is$50,000 ... Simply use the optional factor entry for rates other than d ϭ 2͞n ... A factor can also be entered for some other declining balance depreciation method byspecifying a factor in the function ... Salvage value of the asset ... The period, a year, for which the depreciation is to be calculated ... If, forexample,the entry is 1 ... Example A new machine costs $200,000 and is expected to last 10 years ... Calculate the depreciation of the machine for the first and the eighth years ... Depreciation for the first year: ϭ DDB(200000,10000,10,1)Depreciation for the eighth year: ϭ� EFFECT(nominal, npery)nominalnperyNominal interest rate for the year ... Example Claude has applied for a $10,000 loan ... What effective annual rate will Claude pay? ... DDB(200000,10000,10,8)Depreciation for the fifth year using 175% DB: ϭ DDB(200000,10000,10,5,1 ... Excel uses Equation [4Effective annual rate: ϭ EFFECT(8%,12)EFFECT can also be used to find effective rates other than annually ... Example Interest is stated as 3 ... Find the effectivesemiannual rate ... Effective semiannual rate: ϭ EFFECT(7%,2)551552Using Spreadsheets and Microsoft Excel©Appendix AFV (FutureValue)Calculates the future value (worth) based on periodic payments at a specific interest rate ... Number of compounding periods ... The present value amount ... (optional entry) Either 0 or 1 ... If omitted, 0 is assumed ... He will deposit$12,000 to start the account and plans to add $500 to the accountat the beginning of each monthfor the next 24 months ... 25% per month ... 25%,24,500,12000,1)IF (IF Logical Function)Determines which of two entries is entered into a cell based on the outcome of a logical check onthe outcome of another cell ... If the response is a text string, place it between quotemarks(“ ”) ... Up to seven IF functions can be nested forvery complex logical tests ... Result if the logical_test argument is true ... Example The entry in cell B4 should be “selected” if the PW value in cell B3 is greater thanor equal to zero and “rejected” if PW Ͻ 0 ... Entry in cell C5:ϭIF(C4Ͻ0,“rejected”,IF(C4Ͼϭ200,“fantastic”,“selected”))IPMT (Interest Payment)Calculates the interest accrued for a given period n based on constant periodic payments andinterest rate ... Period for which interest is to be calculated ... Present value ... Future value ... The fvcan also be considered a cash� IRR(values, guess)valuesguessA set of numbers in a spreadsheet column (or row) ... balance after the last payment is made ... A 0 represents payments made at theend of the period, and 1 represents payments made at the beginning ofthe period ... A ... The interest rate is 0 ... Interest due: ϭ IPMT(0for which the rate ofreturn will be calculated ... Negative numbers denote a paymentmade or cash outflow, and positive numbers denote income or cashinflow ... In most cases, a guess is not required, and a10% rate of return is initially assumed ... Inputting different guess values makesit possible todetermine the multiple roots for the rate of return equationof a nonconventional cash flow series ... He will need $25,000 in capital and anticipates that the business will generate the following incomes during the first 5 years ... Year 1Year 2Year 3Year 4Year 5$5,000$7,500$8,000$10,000$15,000Set up anarray in the spreadsheet ... In cell A2, type 5000 (positive for income) ... In cell A4, type 8000 ... In cell A6, type 15000 ... Note that any years with a zero cash flow must have a zero entered to ensure thatthe year value is correctly maintained for computation purposes ... To calculate the internal rate of� MIRR(values, finance_rate, reinvest_rate)valuesRefers to an array of cells in the spreadsheet ... The series of payments and income must occur atregular periods and must contain at least one positive ( ... return after 5 years and specify a guess value of 5%, move tocell A8, and type ϭ IRR(A1:A6,5%numberand one negative number ... 9]) ... 9]) ... See Section 7 ... )Example Jane opened a hobby store 4 years ago ... Since then, the business has yielded $10,000 thefirst year, $15,000 the second year, $18,000 the third year, and $21,000 the fourth year ... What is the modified rate of return after 3years andafter 4 years?In cell A1, type Ϫ50000 ... In cell A3, type 15000 ... In cell A5, type 21000 ... To calculate the modified rate of return after 4 years, move to cell A7, and typeϭ MIRR(A1:A5,12%,8%) ... This function is designed to display only nominal annual rates ... Number of times that interest iscompounded per year ... 55% per year ... Nominal annual rate, quarterly compounding: ϭ NOMINAL(12 ... 55%,100000)NPER (Number of Periods)Calculates the number of periods for the present worth of an investment to equal the future valuespecified, based on uniform regular payments and a statedinterest rate ... Amount paid during each compounding period ... (optional entry) Future value or cash balance after the last payment ... (optional entry) Enter 0 if payments are due at the end of thecompounding period and 1 if payments are due at the beginning of theperiod ... A ... 25% per month ... How� NPV(rate, series)rateseriesInterest rate per compounding period ... Example Mark is considering buying a sports store for $100,000 and expects to receive thefollowing income during ... many paymentsdoes she have to make to accumulate $25,000 to buy a new car?Number of payments: ϭ NPER(0� PMT(rate, nper, pv, fv, type)ratenperpvfvtypeInterest ... the next 6 years of business: $25,000, $40,000, $42,000, $44,000,$48,000, $50,000 ... In cells A1 through A7, enter Ϫ100,000, followed by the six annual incomes ... Any year with a zero cash flow must have a 0 enteredto ensure a correct resultrate per compounding period ... Present value ... (optional entry) Enter 0 for payments due at the end of thecompounding period and 1 if payment is due at the start of thecompounding period ... Example Jim plans to take a $15,000 loan to buy a new car ... He wants to pay the loan off in 5 years (60� PV(rate, nper, pmt, fv, type)ratenperpmtfvtypeInterest rate per compounding period ... Cash flow at ... � PPMT(rate, per, nper, pv, fv, type)ratepernperpvfvtypeInterest rate per compounding period ... Total number of periods ... Future value ... If omitted, 0 is assumed ... The interest rate is 5% ( ... monthsregular intervals ... Future value or cash balance at the end of the last period ... If omitted, 0 is assumed ... Example Jose is considering leasing a car for $300 a month for 3 years (36 months) ... Using an interest rate of 8% per year, findthe present value of this option ... RAND (Random Number)Returnsan evenly distributed number that is (1) Ն 0 and Ͻ 1; (2) Ն 0 and Ͻ 100; or (3) betweentwo specified numbers ... Example Grace needs random numbers between 5 and 10 with 3 digits after the decimal ... Random number:ϭ RAND()*5 ϩ 5Example Randi wants to generate random numbers between the� RATE(nper, pmt, pv, fv, type, guess)nperpmtpvfvtypeguessTotal number of periods ... Present value ... (optional entry) Enter 0 for payments due at the end of thecompounding period and 1 if payments are due at the start of ... limits of Ϫ10 and 25 ... Random number:ϭ RAND()*35 Ϫ 10Aeachcompounding period ... (optional entry) To minimize computing time, include a guessedinterest rate ... This function usually converges to a solution ifthe rate is between 0% and 100% ... She will make an initial deposit of$1000 to open the account and plans to deposit $100 at the beginning of each� SYD(cost, salvage, life, ... � SLN(cost, salvage, life)costsalvagelifeFirst cost or basis of the asset ... Depreciation life ... The machine has an allowed depreciation life of 8 years and an estimated salvage value of $15,000 ... month ... At the end of 3 years, she wants to have at least $5000period)costsalvagelifeperiodFirst cost or basis of the asset ... Depreciation life ... Example Jack bought equipment for $100,000 that has a depreciation life of 10 years ... What is the depreciation for year 1 and year 9?Depreciation for year 1: ϭ SYD(100000,10000,10,1)Depreciation for year 9: ϭ� VDB (cost, salvage, life, start_period, end_period, factor, ... SYD(100000,10000,10,9)VDB (Variable Declining Balance)Calculates the depreciation using the declining balance method with a switch to straight linedepreciation in the year in which straight line has a larger depreciation amountno_switch)costsalvagelifestart_periodend_periodfactorno_switchFirst cost of the asset ... Depreciation life ... Last period for depreciation to be calculated ... Other entriesdefine the declining balance method, for example, 1 ... (optional entry) If omitted or entered as FALSE, the function willswitch fromdeclining balance to straight line depreciation whenthe latter is greater than DB depreciation ... Example Newly purchased equipment with a first cost of $300,000 has a depreciable life of10 years with no salvage value ... Depreciation for first year, with switching: ϭ VDB(300000,0,10,0,1,1 ...75)Depreciation for first year, no switching: ϭ VDB(300000,0,10,0,1,1 ... 75,TRUE)VDB (for MACRS Depreciation)The VDB function can be adapted to generate the MACRS annual depreciation amount, when thestart_period and end_period are replaced with the MAX and MIN functions, respectively ...factor)Example Determine the MACRS depreciation for year 4 for a $350,000 asset that has a 20%salvage value and a MACRS recovery period of 3 years ... Depreciation for year 4: ϭ VDB(350000,0,3,MAX(0,4Ϫ1 ... 5),2)Example Find the MACRS (, 5 ... 1؊The VDB format is� VDB(cost,0,life,MAX(0,tdepreciation in year 16 for a $350,000-asset with a recoveryperiod of n ϭ 15 years ... 5 is required here, since MACRS starts with150% DB for n ϭ 15-year and 20-year recovery periods ... Depreciation for year 16: ϭ VDB(350000,0,15,MAX(0,16Ϫ1 ... 5),1 ... Thesecan be viewed by clicking the Formulastab on the Excel toolbar ... 4 Goal Seek—A Tool for Breakeven andSensitivity AnalysisGoal Seek is found on the Excel toolbar labeled Data, followed by What-if Analysis ... It is a good tool for sensitivity analysis, breakeven analysis, and “what if?” questionsA ... ϭ B$5ϭ B$5ϩ500Figure A–4Use of GoalSeek to determine an annual cash flow to increase the rate of return ... The initial Goal Seek template is picturedin Figure A–3 ... Only a single cell can be identifiedas the changing cell; however, this limitation can be avoided by using equations rather than specific numerical inputs in any additional cellsalso to be changed ... Example A new asset will cost $25,000, generate an annual cash flow of $6000 over its 5-yearlife, and have an estimated $500 salvage value ... 94% ... Figure A–4 (top left) shows the cash flows and return displayed using the functionϭ IRR(B4:B9) prior to the use of Goal Seek ...The $500 salvage is added for thelast year ... The tool finds the required cash flow of $6506 to approximate the 10% peryear return ... Clicking OK saves allchanged cells; clicking Cancel returns to the original values ... 5 Solver—An Optimizing Tool for Capital Budgeting,Breakeven, and SensitivityAnalysisSolver is a powerful spreadsheet tool to change the value in multiple (one or more) cells based onthe value in a specific (target) cell ... (Section 12 ... ) The initial Solver template is shown in Figure A–5 ... Set Target Cell box ... The target cell itself must contain a formula or function ... By ChangingCells box ... Each cell must be directly or indirectly related to the target cell ... TheGuess button will list all possible changing cells related to the target cell ... Enter any constraints that may apply, for example,$C$1 Ͻ $50,000 ... Options box ... Also, linear andnonlinear model assumptions can be set here... 0001 ... If tolerance remains atthe default value of 5%, a project may be incorrectly included in the solution set at a very lowlevel ... This appears after Solve is clicked and a solution appears ... It is possible to update thespreadsheet by clicking Keep Solver Solution, or return to the original entries usingRestoreOriginal Values ... 6 Error MessagesIf Excel is unable to complete a formula or function computation, an error message is displayed ... #N͞ARefers to a value that is not available ... #NULL!Specifies an invalid intersection of two areas ... #REF!Refers to a cell that is not valid ... #####Produces aresult, or includes a constant numeric value,that is too long to fit in the cell ... )APPENDIX BBASICS OF ACCOUNTINGREPORTS AND BUSINESS RATIOSThis appendix provides a fundamental description of financial statements ... B ... The fiscal year (FY) is commonly not the calendar year (CY) for acorporation ... S ... For example, October2011 through September 2012 is FY2012 ... At the end of each fiscal year, a company publishes a balance sheet ... This is a yearly presentation of the state of thefirm at a particular time, for example, May 31, 2012; however, a balance sheet is alsousuallyprepared quarterly and monthly ... Assets ... Thereare two main classes of assets ... ), which is more easily converted to cash, usually within 1 year ... Conversion of these holdingsto cash in a short time would require a major corporate reorientation ... This section is a summary of all financialobligations (debts, mortgages, loans,etc ... Bond indebtedness is included here ... Also called owner’s equity, this section provides a summary of the financialvalue of ownership, including stocks issued and earnings retained by the corporation ... For example,current assets is comprised of cash, accounts

receivable, etc ... B ... The income statement summarizes the profits or losses of the corporation for a stated period of time ... The major categories of an income statement areRevenues ... Expenses ... Some expense amounts are itemized in other statements, for example, cost ofgoods sold ... 1 and 17 ...� profit (or loss)The cost of goods sold is an important accounting term ... Cost of goods sold may also be called factory cost ... The total ofthe cost of goods sold statement is entered expenses ؊ The income statement, published at the same time as the balance sheet, uses the basic equationRevenuesas an expense item on the income statement ... 1]B ... Indirect cost allocation methods are discussed in Chapter 15 ... 3 Business RatiosAccountants, financial analysts, and engineering economists frequently utilize business ratioanalysis to evaluate the financial health (status) of a company over time andin relation to industry norms ... For comparison purposes, it is necessaryto compute the ratios for several companies in the same industry ... The ratios are classified according to their role in measuring the corporation ... Assess ability to meet short-term and long-term financial obligations ... Measuremanagement’s ability to use and control assets ... Evaluate the ability to earn a return for the owners of the corporation ... Current Ratio This ratio is utilized to analyze the company’s working capital condition ... Note thatonly balance sheet data are utilized in the current ratio; that is, no association withrevenues orexpenses is made ... 0626,700Since current liabilities are those debts payable in the next year, the current ratio value of 3 ... Current ratiovalues of 2 to 3 are common ... Often, however, a better idea of a company’s immediate financial position canbe obtained by using the acid test ratio ... ForJAGBA Corporation,81,700 Ϫ 52,000Acid test ratio ϭ ——————— ϭ 1 ... However, an acid test ratio of approximately 1 ... Debt Ratio This ratio is a measure of financial strength since it is defined astotal liabilitiesDebt ratio ϭ ——————total assetsFor JAGBA Corporation,62,700Debt ratio ϭ ————ϭ 0 ... 6% creditor-owned and 86 ... A debt ratio in the range of 20%or less usually indicates a sound financial condition, with little fear of forced reorganization becauseof unpaid liabilities ... The debt-equity (D-E) mix is another measure of financial strength ... Itis defined asnet profitReturn on sales ϭ———— (100%)net salesNet profit is the after-tax value from the income statement ... For JAGBA Corporation,78,925Return on sales ϭ ———— (100%) ϭ 15 ... 5% to 4 ... In truth, for a relatively large-volume, high-turnover business, an income ratio of 3% is quite healthy ... Return on Assets Ratio Thisis the key indicator of profitability since it evaluates the ability of the corporation to transfer assets into operating profit ... 1%462,700Efficient use of assets indicates that the company should earn a high return, while low returnsusually accompany lower values of this ratio compared to the industry groupratios ... They both indicate the numberof times the average inventory value passes through the operations of the company ... 3Business Ratioswhere average inventory is the figure recorded in the balance sheet ... 7152,000This means that the average value of the inventory has been sold 9 ... Values ofthis ratio vary greatly from one industry to another ... This ratio is commonly used as a measure of the inventoryturnover rate in manufacturing companies ... For JAGBA, using the values in Table B–3,290,000Cost of goods sold to inventory ϭ ————————— ϭ 5 ... EXAMPLE B ... Compare thecorresponding JAGBA Corporation values with these norms, and comment ondifferences and similarities ... 41 ... 3%40 ... 40 ... 8%8 ... 21 ... 1%8 ... 61 ... 4%5 ... SOURCE: L ... SolutionIt is not correct to compare ratios for one company with indexes in different industries, that is,with indexes for differentNAICS codes ... The corresponding values for JAGBA areCurrent ratio ϭ 3 ... 11Debt ratio ϭ 13 ... 1%JAGBA has a current ratio larger than all four of these industries, since 3 ... 6 and much less in the case of the “average”air transportation corporation ... Return on assets, which is a measure of ability toturn assets into profitability, is not as high at JAGBA as motor vehicles, butJAGBA competes well with the other industry sectors ... Corporate assets are classified in categories by$100,000 units, such as 100 to 250, 1001 to 5000, over 250,000, etc ... nspe ... Appendix CCode of Ethics forEngineers567Code of Ethics for EngineersPreambleEngineering is an important and learned profession ... Engineering has a direct and vital impact on the quality of life forall people ... Engineers must perform under a standard ofprofessional behavior that requires adherence to the highest principles ofethicalconduct ... Fundamental CanonsEngineers, in the fulfillment of their professional duties, shall:1 ... 2 ... 3 ... 4 ... 5 ... 6 ... II ... Engineers shall hold paramount the safety, health, and welfareof the public ... If engineers’ judgment is overruled under circumstances thatendanger life or property, they shallnotify their employer or clientand such other authority as may be appropriate ... Engineers shall approve only those engineering documents that arein conformity with applicable standards ... Engineers shall not reveal facts, data, or information without theprior consent of the client or employer except asauthorized orrequired by law or this Code ... Engineers shall not permit the use of their name or associate inbusiness ventures with any person or firm that they believe isengaged in fraudulent or dishonest enterprise ... Engineers shall not aid or abet the unlawful practice of engineeringby a person or firm... Engineers having knowledge of any alleged violation of this Codeshall report thereon to appropriate professional bodies and, whenrelevant, also to public authorities, and cooperate with the properauthorities in furnishing such information or assistance as may berequired ... Engineers shall performservices only in the areas of theircompetence ... Engineers shall undertake assignments only when qualified byeducation or experience in the specific technical fields involved ... Engineers shall not affix their signatures to any plans or documentsdealing with subject matter in which they lack competence,nor toany plan or document not prepared under their direction andcontrol ... Engineers may accept assignments and assume responsibility forcoordination of an entire project and sign and seal the engineeringdocuments for the entire project, provided that each technicalsegment is signed and sealed onlyby the qualified engineers whoprepared the segment ... Engineers shall issue public statements only in an objective andtruthful manner ... Engineers shall be objective and truthful in professional reports,statements, or testimony ... b ... c ... 4 ... a ... b ... c ... d ... e ... 5 ... a ... Theyshall not misrepresent orexaggerate their responsibility in or for thesubject matter of prior assignments ... b ... They shall not offer any gift or other valuableconsideration in order to secure work ... III ... Engineers shall be guided in all their relations by the highest standardsof honesty and integrity ... Engineers shall acknowledgetheir errors and shall not distort oralter the facts ... Engineers shall advise their clients or employers when they believea project will not be successful ... Engineers shall not accept outside employment to the detriment oftheir regular work or interest ... d ... e ... 2 ... a ... b ... If the client or employer insists onsuch unprofessionalconduct, they shall notify the proper authorities and withdraw fromfurther service on the project ... Engineers are encouraged to extend public knowledge andappreciation of engineering and its achievements ... Engineers are encouraged to adhere to the principles ofsustainabledevelopment1 in order to protect the environment for futuregenerations ... Engineers shall avoid all conduct or practice that deceives the public ... Engineers shall avoid the use of statements containing a materialmisrepresentation of fact or omitting a material fact ... Consistent with theforegoing, engineers may advertise forrecruitment of personnel ... Consistent with the foregoing, engineers may prepare articles forthe lay or technical press, but such articles shall not imply credit tothe author for work performed by others ... Engineers shall not disclose, without consent, confidentialinformationconcerning the business affairs or technical processes of any present orformer client or employer, or public body on which they serve ... Engineers shall not, without the consent of all interested parties,promote or arrange for new employment or practice in connectionwith a specific project forwhich the engineer has gained particularand specialized knowledge ... Engineers shall not, without the consent of all interested parties,participate in or represent an adversary interest in connection with aspecific project or proceeding in which the engineer has gainedparticular specialized knowledge onbehalf of a former client oremployer ... Engineers shall not be influenced in their professional duties byconflicting interests ... Engineers shall not accept financial or other considerations,including free engineering designs, from material or equipmentsuppliers for specifying their product ... Engineers shallnot accept commissions or allowances, directly orindirectly, from contractors or other parties dealing with clients oremployers of the engineer in connection with work for which theengineer is responsible ... Engineers shall not attempt to obtain employment or advancement orprofessional engagements byuntruthfully criticizing other engineers,or by other improper or questionable methods ... Engineers shall not request, propose, or accept a commission on acontingent basis under circumstances in which their judgment maybe compromised ... Engineers in salaried positions shall accept part-timeengineeringwork only to the extent consistent with policies of the employer andin accordance with ethical considerations ... Engineers shall not, without consent, use equipment, supplies,laboratory, or office facilities of an employer to carry on outsideprivate practice ... Engineers shall not attempt to injure,maliciously or falsely, directlyor indirectly, the professional reputation, prospects, practice, oremployment of other engineers ... a ... b ... c ... 8 ... a ... b ... 9 ... a ... b ... c ... d ... The employershould indemnify the engineer for use of the information for anypurpose other than the original purpose ... Engineersshall continue their professional development throughouttheir careers and should keep current in their specialty fields byengaging in professional practice, participating in continuingeducation courses, reading in the technical literature, and attendingprofessional meetings and seminars ... As Revised July2007“By order of the United States District Court for the District of Columbia,former Section 11(c) of the NSPE Code of Ethics prohibiting competitivebidding, and all policy statements, opinions, rulings or other guidelinesinterpreting its scope, have been rescinded as unlawfully interfering with thelegal rightof engineers, protected under the antitrust laws, to provide priceinformation to prospective clients; accordingly, nothing contained in the NSPECode of Ethics, policy statements, opinions, rulings or other guidelines prohibitsthe submission of price quotations or competitive bids for engineering servicesatany time or in any amount ... ”It is further noted that as made clear in the Supreme Court decision:1 ... 2 ... 3 ... 4 ... 5 ... State registration boards with authority toadopt rules of professional conduct may adopt rules governing procedures toobtain engineering services ... As noted by the Supreme Court,“nothing in the judgment prevents NSPE andits members from attempting to influence governmental action ... The Code deals with professional services, which services must be performed by realpersons ... The Code is clearly written to apply to the Engineer,and it is incumbent on members of NSPE toendeavor to live up to its provisions ... 1420 King StreetAlexandria, Virginia 22314-2794703/684-2800 • Fax:703/836-4875www ... orgPublication date as revised: July 2007 • Publication #1102APPENDIX DALTERNATE METHODS FOREQUIVALENCE CALCULATIONSThroughout the text, engineeringeconomy factor formulas, tabulated factor values, or built-inspreadsheet functions have been used to obtain a value of P, F, A, i, or n ... An overview of the possibilities is presented here ... From the summation ofthe series, it is possible to perform calculator-based computations to obtain P, F, or A values... 2 ... 1 Using Programmable CalculatorsA basic way to calculate one parameter, given the other four, is to use a calculator that allows apresent worth relation to be encoded ... For example, consider a PW relation in which all five parameters are included ... The interest rate i can be coded for entry as apercentor decimal ... This is the approach taken by relatively simple scientific calculators, such as the HewlettPackard (HP) scientific series, for example, HP 33s ... The initial investment is called B rather than P, and the equal uniform amount is termed P rather than A ... Another example that offersfreedom from the spreadsheet and tables is an engineering calculator that has the same functions built in as those on a spreadsheet to determine P, F, A, i, or n ... The functions are basicallythe same as those on a spreadsheet ... For example, the tvmPV function formatistvmPV(n,i,Pmt,FV,PpY,CpY,PmtAt)wheren ϭ number of periodsi ϭ annual interest rate as a percentPmt ϭ equal uniform periodic amount AFV ϭ future amount FPpY ϭ payments per year (optional; default is 1)CpY ϭ compounding periods per year (optional; default is 1)PmtAt ϭ beginning- or end-of-period payments (optional; default is 0 ϭ end)570Appendix DAlternate Methods for Equivalence CalculationsIt is easy to understand why it is possible to do a lot on a calculator with relatively wellbehaved cash flow series ... However, the use of tables or factor formulas is not necessary ... 2 Using theSummation of a Geometric SeriesA geometric progression is a series of n terms with a common ratio or base r ... 1]Ristroph1 and others have explained how the recognition that equivalence computations aresimple applications of geometric series can be used to determine F, P, and A using Equation[D ...Before explaining how to apply this approach, we define the base r as follows when a futureworth F or present worth P is sought ... This is the same as using a geometricseries of only one term ... 1)10ϭ 100(2 ... 37This is identical to using the tabulated value (or formula) for the F/P factor ... 5937)ϭ $259... To calculate F, we can moveeach A value forward to year 10 ... F ϭ A1(1 ϩ i)9 ϩ A2(1 ϩ i)8 ϩ ⋅ ⋅ ⋅ ϩ A9(1 ϩ i)1 ϩ A10(1 ϩ i)0Figure D–1F=?Future worth of a singleamount in year 0 ... H ... D ... P= ?i = 10%0 12345678910YearA ϭ $1005678910TimesdiscountedFigure D–3Present worth of a shifteduniform series ... With r ϭ 1 ϩ i, the geometric seriesand its sum S (from Equation [D ... Removing the subscript on A,jϭ9F = A[r9 ϩ r8 ϩ ⋅ ⋅ ⋅ ϩ r1 ϩ r0] ϭ A⌺rjjϭ010r10 Ϫ r0 (1 ... 9374rϪ10 ... F ϭ 100(S) ϭ 100(15 ... 74F ϭ 100(F/A,10%,10) ϭ 100(15 ... 74As a final demonstration of the geometric seriesapproach to equivalence computations, consider the shifted cash flow series in Figure D–3 ... If the tables are used, the solution isP ϭ A(P/A,10%,6)(P/F,10%,4) = 100(4 ... 6830)ϭ $297 ... The last A term is discounted 10 years, making n ϭ 10 ... Again using Equation [D ... 1)11 Ϫ (1/1 ... 1) Ϫ 1]](0 ...9091)5ϭ 100 —————————— ϭ 100(2 ... 9091 Ϫ 1ϭ $297 ... Proponents of this approach point out the use of standardmathematical notation; the removal of a need to derive any factors; no need to remember theplacement of P, F, and A values based on factor formula development; and the easyuse of a calculator to determine the equivalence relations ... When the series becomequite involved, or when sensitivity analysis is required to reach an economic decision, it maybe beneficial to use a spreadsheet ... However, once again, the use of tables and factors is not necessary ... 1 ImportantConcepts and GuidelinesThe following elements of engineering economy are identified throughout the text in the marginby this checkmark and a title below it ... Time Value of Money It is a fact that money makes money ... (1)Economic Equivalence A combination of time value of money and interest ratethat makesdifferent sums of money at different times have equal economic value ... Revenues arecash inflows and carry a positive (+) sign; expenses are outflows and carry a negative (−) sign ... g ... (1, 9)End-of-Period Convention To simplify calculations, cash flows (revenues and costs) areassumed tooccur at the end of a time period ... A half-year convention is often used in depreciation calculations ... COC is usuallya weighted average that involves the cost of debt capital (loans, bonds, and mortgages) and equity capital (stocks and retained earnings) ... Also called the hurdle rate, MARR is based oncost ofcapital, market trend, risk, etc ... (1, 10)Opportunity Cost A forgone opportunity caused by the inability to pursue a project ... Stated differently, it is the ROR of the first project rejected because of unavailability offunds ... Effective interestrate is the actual rate over a period of time becausecompounding is imputed; for example, 1% permonth, compounded monthly, is an effective 12 ... Inflation or deflation is not considered ... The A or AW is a series of equal,end-of-period cash flows for n consecutive periods, expressed as money per time (say,$/year; /year) ... (2, 3)Title574AppendixEGlossary of Concepts and TermsPlacement of Gradient Present Worth (PG ; Pg) The (P/G,i%,n) factor for an arithmeticgradient finds the PG of only the gradient series 2 years prior to the first appearance of the constant gradient G ... The (P/A,g,i,n) factor for a geometric gradient determines Pg for thegradient and initial amountA1 two years prior to the appearance of the first gradient amount ... (2, 3)Equal-Service Requirement Identical capacity of all alternatives operating over thesame amount of time is mandated by the equal-service requirement ... PW analysis requires evaluation over thesamenumber of years (periods) using the LCM (least common multiple) of lives; AW analysis isperformed over one life cycle ... (5, 6, 8)LCM or Study Period To select from mutually exclusive alternatives under the equal-servicerequirement for PW computations, use the LCM of lives with repurchase(s) asnecessary ... (5, 6, 11)Salvage/Market Value Expected trade-in, market, or scrap value at the end of the estimatedlife or the study period ... MACRS depreciation always reduces the book value to a salvage of zero ... DN is status quo; it generates no new costs, revenues, or savings ... Incrementalevaluation requires comparison with DN for revenue alternatives ... (5)Rate of Return An interest rate that equates a PW or AW relation to zero ... (7, 8)Project Evaluation For a specified MARR, determine a measure of worth for net cash flowseries over the life or study period ... (5, 6, 7, 9, 17)Presentworth: If PW ≥ 0Future worth: If FW ≥ 0Benefit/cost: If B/C ≥ 1 ... 0ME Alternative Selection For mutually exclusive (select only one) alternatives, comparetwo alternatives at a time by determining a measure of worth for the incremental (∆) cash flowseries over the life or study period, adhering to the equal-service requirement ... Rate of return: Order by initial cost, perform pairwise ∆i* comparison; if ∆i* ≥ MARR,select larger cost alternative; continue until one remains ... 0, select larger cost alternative; continue until one remains ... 1Important Concepts and GuidelinesCost-effectiveness ratio: For servicesector alternatives; order by effectiveness measure;perform pairwise ∆C/E comparison using dominance; select from nondominated alternativeswithout exceeding budget ... Calculate a measure of worth and select using the guidelines below ... Rate of return: No incremental comparison; select all projectswith overall i * ≥ MARR ... 0 ... When a capital budget limit is defined, independent projects are selected using the capital budgeting process based on PW values ... Capital Recovery CR is the equivalent annual amount an asset or system must earn to recoverthe initial investment plus a stated rate ofreturn ... The salvage value is considered in CR calculations ... (11)Sunk Cost Capital (money) that is lost and cannot be recovered ... They should be handled using tax laws and write-offallowances, not the economic study ... Inflationoccurs when the value of a currency decreases ... (1, 14)Breakeven Fora single project, the value of a parameter that makes two elements equal,e ... , sales necessary to equate revenues and costs ... Breakeven analysis is fundamental to make-buy decisions, replacement studies, payback analysis, sensitivity analysis,breakeven ROR analysis, and many others ... (8,13)Payback Period Amount of time n before recovery of the initial capital investment is expected ... (13)Direct / Indirect Costs Direct costs are primarily human labor, machines, and materials associated with a product, process, system, or service ... (15)Value Added Activities have added worth to aproduct or service from the perspective of aconsumer, owner, or investor who is willing to pay more for an enhanced value ... Parameters may be any cost factor, revenue,life, salvage value, inflation rate, etc ... Risk represents an absence of or deviation from certainty ... (10, 18, 19, 20)E ... The numbersin parentheses indicate sections where the term is introduced and used in variousapplications ... 5, 6 ... DescriptionAnnual operating costAOCEstimated annual costs to maintain and support analternative (1 ... Benefit/cost ratioB/CRatio of a project’s benefits to costs expressed in PW, AW,or FW terms (9... Book valueBVRemaining capital investment in an asset after depreciationis accounted for (16 ... Breakeven pointQBEQuantity at which revenues and costs are equal, or twoalternatives are equivalent (13 ... Capital budgetbAmount of money available for capital investment projects(12 ... CapitalrecoveryCR or AEquivalent annual cost of owning an asset plus the requiredreturn on the initial investment (6 ... Capitalized costCC or PPresent worth of an alternative that will last forever (or along time) (5 ... Cash flowCFActual cash amounts that are receipts (inflow) and disbursements (outflow) (1 ...Cash flow before orafter taxesCFBT orCFATCash flow amount before relevant taxes or after taxes areapplied (17 ... CompoundingfrequencymNumber of times interest is compounded per period (year)(4 ... Cost-effectiveness ratioCERRatio of equivalent cost to effectiveness measure to evaluate servicesector projects (9 ... Cost estimating relationshipsC2 or CTRelations that use design variables and changing costs overtime to estimate current and future costs (15 ... Cost of capitalWACCInterest rate paid for the use of capital funds; includes bothdebt and equity funds ... 9, 10 ... Debt-equity mixD-EPercentages of debt and equity investment capital used by acorporation (10 ... DepreciationDReduction in the value of assets using specific models andrules; there are book and tax depreciation methods (16 ... Depreciation ratedtAnnual rate for reducing the value of assets using differentdepreciationmethods (16 ... Economic service lifeESL or nNumber of years at which the AW of costs is a minimum(11 ... Effectiveness measureEA nonmonetary measure used in the cost-effectiveness ratiofor service sector projects (9 ... E ... 3, 19 ... Expenses, operatingOEAll corporate costs incurred in transactingbusiness (17 ... First costPTotal initial cost—purchase, construction, setup, etc ... 3, 16 ... Future amount or worthF or FWAmount at some future date considering time value ofmoney (1 ... 4) ... 5) ... 6) ... 1) ... 1) ... 4) ... 4, 4 ... Interest rate, inflationadjustedifInterest rate adjusted to take inflation intoaccount (14 ... Life (estimated)nNumber of years or periods over which an alternative orasset will be used; the evaluation time (1 ... Life-cycle costLCCEvaluation of costs for a system over all stages: feasibilityto design to phaseout (6 ... Measure of worthVariesValue, such as PW, AW, i*, used to judgeeconomicviability (1 ... Minimum attractiverate of returnMARRMinimum value of the rate of return for an alternative to befinancially viable (1 ... 1) ... 5) ... 6) ... 1) ... 1) ... Payback periodnpNumber of years to recover the initial investment and astated rate of return (13 ... Present amountor worthP orPWAmount of money at the current time or a time denoted aspresent (1 ... 2) ... 2) ... 2, 12 ... Random variableXParameter or characteristic that can take on any one ofseveral values; discrete and continuous (19 ... Rate of returni* or RORCompound interest rate on unpaid or unrecovered balancessuchthat the final amount results in a zero balance (7 ... 577578Appendix EGlossary of Concepts and TermsTermSymbolDescriptionRecovery periodnNumber of years to completely depreciate an asset (16 ... Return on investedcapitali'' or ROICUnique ROR when a reinvestment rate ii is applied tomultiple-rate cash flows (7 ... Salvage/market valueS or MVExpected trade-in or market value when an asset is tradedor disposed of (6 ... 1, 16 ... Standard deviations or σMeasure of dispersion or spread about the expected valueor average (19 ... Study periodnSpecified number of years over which an evaluationtakesplace (5 ... 5) ... 1) ... 1) ... 1) ... 7) ... 1) ... 7) ... 9) ... T ... Tarquin: Basics of Engineering Economy, McGraw-Hill, New York, 2008 ... S ... , PearsonPrentice Hall, Upper Saddle River, NJ, 2003 ... E ... G ... , PearsonPrentice Hall, Upper Saddle River, NJ, 1992 ... R ... G ... J ... A ... , Pearson PrenticeHall, Upper Saddle River, NJ, 2005 ... A ... R ... , Pearson Prentice Hall,Upper Saddle River, NJ, 1999 ... F ... Loulakis: Design-Build Contracting Handbook, 2d ed ... Eschenbach, T ... : Engineering Economy: Applying Theory to Practice, 3d ed ... Fabrycky, W ... , G ... Thuesen, and D ... , PearsonPrentice Hall,Upper Saddle River, NJ, 1998 ... M ... M ... Bernhardt, and M ... Hartman, J ... : Engineering Economy and the Decision Making Process, Pearson Prentice Hall, UpperSaddle River, NJ, 2007 ... M ... Newnan, D ... , J ... Lavelle, and T ... Eschenbach: Engineering Economic Analysis, 10th ed... Ostwald, P ... : Construction Cost Analysis and Estimating, Pearson Prentice Hall, Upper Saddle River,NJ, 2001 ... F ... S ... Park, C ... : Contemporary Engineering Economics, 5th ed ... Park, C ... : Fundamentals of Engineering Economics, 2d ed ... Peurifoy, R ... , and G ... Oberlender: EstimatingConstruction Costs, 5th ed ... Riggs, J ... , D ... Bedworth, and S ... Randhawa: Engineering Economics, 4th ed ... Stewart, R ... , R ... Wyskida, and J ... Johannes: Cost Estimator’s Reference Manual, 2d ed ... Sullivan, W ... , E ... Wicks, and C ... Koelling: Engineering Economy, 15th ed ... Thuesen, G ... ,and W ... Fabrycky: Engineering Economy, 9th ed ... White, J ... , K ... Case, D ... Pratt, and M ... Agee: Principles of Engineering Economic Analysis,5th ed ... Using Excel 2007Gottfried, B ... : Spreadsheet Tools for Engineers Using Excel® 2007, McGraw-Hill, New York, 2010 ... E ... S ... J ... ,WadsworthCengage Learning, Belmont, CA, 2009 ... W ... Schinzinger: Introduction to Engineering Ethics, 2d ed ... 580Reference MaterialsWebsitesConstruction cost estimation index: www ... comThe Economist: www ... comFor this textbook: www ... com/blankPlant cost estimation index: www ... com/pciRevenueCanada: www ... gc ... S ... irs ... online ... comU ... Government Publications (available atwww ... gov)Corporations, Publication 544, Internal Revenue Service, GPO, Washington, DC, annually ... Your Federal Income Tax, Publication 17, Internal Revenue Service, GPO, Washington, DC, annually ...Harvard Business Review, Harvard University Press, Boston, bimonthly ... 581Compound Interest Factor Tables0 ... 25%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmeticGradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniform SeriesA͞G1234567891011121314151617181920212223242526272829303640485052556072758490961001081201321442403604801 ...00501 ... 01001 ... 01511 ... 02021 ... 02531 ... 03041 ... 03561 ... 04081 ... 04601 ... 05121 ... 05651 ... 06181 ... 06711 ... 07241 ... 07781 ... 10501 ... 13301 ... 14721 ... 19691 ... 23341 ... 27091 ... 30951 ... 39041 ... 82082 ... 31510 ... 99500 ... 99010 ... 98510 ... 98020 ... 97530 ... 97050 ... 96560 ...96080 ... 95610 ... 95130 ... 94660 ... 94180 ... 93710 ... 93250 ... 92780 ... 90500 ... 88260 ... 87170 ... 83550 ... 81080 ... 78690 ... 76360 ... 71920 ... 54920 ... 30161 ... 499380 ... 249060 ... 165630 ... 123910 ... 098880 ... 082190 ... 070280 ... 061340 ... 054380 ... 048820 ... 044270 ... 040480 ... 037270... 034520 ... 032140 ... 023800 ... 018800 ... 016980 ... 012690 ... 010710 ... 009230 ... 008080 ... 006400 ... 003050 ... 001081 ... 00253 ... 01505 ... 03767 ... 07049 ... 113311 ... 166413 ... 229815 ... 303517 ... 387619 ... 482221 ... 587223 ... 702825 ... 829027 ... 965830 ... 113337 ... 013250 ... 188755... 881964 ... 779482 ... 3419100 ... 3474113 ... 8093139 ... 1582173 ... 3020582 ... 05951 ... 501880 ... 251560 ... 168130 ... 126410 ... 101380 ... 084690 ... 072780 ... 063840 ... 056880 ... 051320 ... 046770 ... 042980 ... 039770 ... 037020 ... 034640 ... 026300 ... 021300 ... 019480 ... 015190 ... 013210... 011730 ... 010580 ... 008900 ... 005550 ... 003580 ... 99252 ... 97514 ... 94786 ... 91078 ... 863910 ... 807312 ... 741014 ... 665016 ... 579518 ... 484520 ... 380022 ... 266024 ... 142626 ... 009927 ... 867934 ... 019945 ... 946248 ... 326455 ... 816968 ... 681380 ... 254688 ... 5453103 ... 3121120 ...3109237 ... 34180 ... 98015 ... 900714 ... 722327 ... 406144 ... 913364 ... 205388 ... 2441116 ... 9917148 ... 4106183 ... 4634222 ... 1131264 ... 3230310 ... 0566360 ... 2776413 ... 4988728 ... 061125 ... 591353 ... 082265 ... 613029 ... 873886 ... 244829 ... 116950 ... 411939936264538210 ... 99831 ...99502 ... 99003 ... 98344 ... 97505 ... 96506 ... 95347 ... 94018 ... 92519 ... 908510 ... 890111 ... 870212 ... 848513 ... 825214 ... 230619 ... 020923 ... 937726 ... 751434 ... 830540 ... 816245 ... 421651 ... 508461 ... 1949107 ... 8902192 ... 5%TABLE 20 ... 00501 ... 01511 ... 02531 ... 03551 ... 04591 ...05641 ... 06701 ... 07771 ... 08851 ... 09941 ... 11041 ... 12161 ... 13281 ... 14421 ... 15561 ... 19671 ... 27051 ... 29611 ... 34891 ... 45361 ... 56661 ... 64671 ... 81941 ... 05083 ... 022610 ... 99500 ... 98510 ... 97540 ... 96570 ... 95610 ... 94660 ... 93720 ... 92790 ... 91870 ... 90960 ... 90060 ... 89160 ...88280 ... 87400 ... 86530 ... 83560 ... 78710 ... 77160 ... 74140 ... 68790 ... 63830 ... 60730 ... 54960 ... 48760 ... 16600 ... 000000 ... 331670 ... 198010 ... 140730 ... 108910 ... 088660 ... 074640 ... 064360 ... 056510 ... 050300 ... 045280 ... 041130 ... 037650 ... 034690 ... 032130 ... 025420 ... 018490 ...016890 ... 014330 ... 011020 ... 008830 ... 007730 ... 006100 ... 004760 ... 001000 ... 00002 ... 01504 ... 05036 ... 10598 ... 182110 ... 279212 ... 397214 ... 536516 ... 697318 ... 879720 ... 084023 ... 310425 ... 559127 ... 830429 ... 124432 ... 336144 ... 097856 ... 218063 ... 770086 ... 7265104 ... 3109122... 3337142 ... 8793186 ... 1502462 ... 521991 ... 005000 ... 336670 ... 203010 ... 145730 ... 113910 ... 093660 ... 079640 ... 069360 ... 061510 ... 055300 ... 050280 ... 046130 ... 042650 ... 039690 ... 037130 ... 030420 ... 023490 ... 021890 ... 019330 ... 016020 ... 013830 ... 012730 ... 011100 ... 009760 ...006000 ... 99501 ... 97023 ... 92595 ... 86217 ... 77919 ... 677011 ... 556213 ... 416615 ... 258617 ... 082418 ... 888020 ... 675722 ... 445624 ... 198026 ... 933027 ... 871036 ... 580344 ... 689747 ... 725660 ... 413668 ... 331376 ... 542683 ... 073596 ... 4747139 ... 7916181 ... 99012 ... 90119 ... 655220 ...175534 ... 386552 ... 213674 ... 583599 ... 4238128 ... 6634160 ... 2322195 ... 0611233 ... 0820275 ... 2281319 ... 4332367 ... 6324557 ... 3347959 ... 701113 ... 271448 ... 352163 ... 662976 ... 183562 ... 374823 ... 596451 ... 49880 ... 49381 ... 48552 ... 47383 ... 45894 ... 44065 ... 41906 ... 39407 ...36588 ... 33429 ... 299310 ... 261111 ... 219512 ... 174713 ... 126516 ... 835922 ... 462424 ... 744728 ... 350434 ... 576341 ... 684545 ... 675853 ... 310362 ... 1131128 ... 7949583Compound Interest Factor Tables0 ... 75%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform SeriesPaymentsArithmetic GradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniformSeriesA͞G1234567891011121314151617181920212223242526272829303640485052556072758490961001081201321442403604801 ... 01511 ... 03031 ... 04591 ... 06161 ... 07761 ... 09381 ... 11031 ... 12701 ... 14401 ... 16121 ... 17871 ... 19641 ... 21441 ... 23271 ... 25131 ... 34831 ... 45301 ... 50831... 71261 ... 87321 ... 04892 ... 24112 ... 68132 ... 009214 ... 10990 ... 98520 ... 97060 ... 95620 ... 94200 ... 92800 ... 91420 ... 90070 ... 88730 ... 87420 ... 86120 ... 84840 ... 83580 ... 82340 ... 81120 ... 79920 ... 74160 ... 68830 ... 66300 ... 58390 ... 53380 ... 48810 ... 44620 ... 37300 ... 16640 ... 02771 ...498130 ... 247210 ... 163570 ... 121760 ... 096670 ... 079950 ... 068010 ... 059060 ... 052100 ... 046530 ... 041980 ... 038180 ... 034980 ... 032230 ... 029850 ... 021530 ... 016560 ... 014760 ... 010530 ... 008590 ... 007150 ... 006040 ... 004460 ... 001500 ... 000211 ... 00753 ... 04525 ... 11367 ... 21329 ...344311 ... 507613 ... 703415 ... 932318 ... 194720 ... 491222 ... 822325 ... 188527 ... 590329 ... 028232 ... 502941 ... 446557 ... 394363 ... 768875 ... 0070100 ... 4269127 ... 8562148 ... 4832193 ... 1748257 ... 88691830 ... 321 ... 505630 ... 254710 ... 171070 ... 129260 ... 104170 ... 087450 ... 075510 ...066560 ... 059600 ... 054030 ... 049480 ... 045680 ... 042480 ... 039730 ... 037350 ... 029030 ... 024060 ... 022260 ... 018030 ... 016090 ... 014650 ... 013540 ... 011960 ... 009000 ... 007710 ... 97772 ... 92614 ... 84566 ... 73668 ... 599610 ... 434912 ... 243014 ... 024315 ... 779217 ... 508019 ... 211221 ...889122 ... 542224 ... 170725 ... 775131 ... 446940 ... 566442 ... 931648 ... 476857 ... 154065 ... 258470 ... 839478 ... 606487 ... 1450124 ... 64090 ... 94085 ... 705814 ... 180826 ... 254442 ... 817461 ... 763284 ... 9876110 ... 3887139 ... 8671171 ... 3253206 ... 6682243 ... 8029284 ... 6387327 ... 0867373... 9924637 ... 8404953 ... 591128 ... 521791 ... 222308 ... 002853 ... 753419 ... 564583 ... 589494 ... 49810 ... 49071 ... 47822 ... 46083 ... 43844 ... 41105 ... 37866 ... 34137 ... 29898 ... 25169 ... 199410 ... 142211 ... 080012 ... 012813 ... 940716 ... 505822 ... 947623 ... 122327 ... 288233 ... 135739 ...810743 ... 315450 ... 823258 ... 4210107 ... 6620584Compound Interest Factor Tables1%TABLE 41%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmeticGradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniform SeriesA͞G1234567891011121314151617181920212223242526272829303640485052556072758490961001081201321442403604801 ...02011 ... 04061 ... 06151 ... 08291 ... 10461 ... 12681 ... 14951 ... 17261 ... 19611 ... 22021 ... 24471 ... 26971 ... 29531 ... 32131 ... 34781 ... 48891 ... 64461 ... 72851 ... 04712 ... 30672 ... 59932 ... 92893 ... 71904 ... 892635 ... 64770 ... 98030 ... 96100 ... 94200 ... 92350 ... 90530 ... 88740 ... 87000 ...85280 ... 83600 ... 81950 ... 80340 ... 78760 ... 77200 ... 75680 ... 74190 ... 67170 ... 60800 ... 57850 ... 48850 ... 43350 ... 38470 ... 34140 ... 26890 ... 09180 ... 00841 ... 497510 ... 246280 ... 162550 ... 120690 ... 095580 ... 078850 ... 066900 ... 057940 ... 050980 ... 045420 ... 040860 ... 037070 ... 033870... 031120 ... 028750 ... 020460 ... 015510 ... 013730 ... 009550 ... 007650 ... 006250 ... 005180 ... 003680 ... 001010 ... 000081 ... 01003 ... 06045 ... 15207 ... 28579 ... 462211 ... 682513 ... 947416 ... 257918 ... 614720 ... 019023 ... 471625 ... 973528 ... 525630 ... 129133 ... 784943 ... 886461 ... 463267... 852581 ... 7099110 ... 6723144 ... 9273170 ... 8926230 ... 8959319 ... 25543494 ... 010000 ... 340020 ... 206040 ... 148630 ... 116740 ... 096450 ... 082410 ... 072120 ... 064260 ... 058050 ... 053030 ... 048890 ... 045410 ... 042450 ... 039900 ... 033210 ... 026330 ... 024760 ... 022240 ... 019020 ...016900 ... 015870 ... 014350 ... 013130 ... 010290 ... 99011 ... 94103 ... 85345 ... 72827 ... 56609 ... 367611 ... 133713 ... 865114 ... 562316 ... 226018 ... 857019 ... 455821 ... 023222 ... 559624 ... 065825 ... 107532 ... 974039 ... 394242 ... 955051 ... 587156 ... 160961 ... 028965 ... 700573 ... 137290 ...218399 ... 98032 ... 80449 ... 320519 ... 381233 ... 843550 ... 568771 ... 422194 ... 2734120 ... 9957149 ... 4664181 ... 5663216 ... 1800252 ... 1957292 ... 5047333 ... 0021494 ... 8561820 ... 4176939 ... 811192 ... 871702 ... 322240 ... 432605 ... 423334 ... 694177 ... 608720 ... 160 ... 99341 ... 98012 ...96023 ... 93374 ... 90055 ... 86076 ... 81437 ... 76138 ... 70179 ... 635410 ... 562611 ... 483111 ... 397112 ... 304413 ... 428518 ... 597622 ... 268624 ... 533331 ... 379335 ... 872439 ... 342644 ... 834951 ... 867675 ... 699595 ... 25%TABLE 51 ... 01251 ... 03801 ... 06411 ... 09091 ... 11831 ... 14641 ...17531 ... 20481 ... 23511 ... 26621 ... 29811 ... 33071 ... 36421 ... 39851 ... 43371 ... 56391 ... 81541 ... 90781 ... 10722 ... 53882 ... 05883 ... 46343 ... 44025 ... 982519 ... 5410388 ... 98770 ... 96340 ... 93980 ... 91670 ... 89420 ... 87230 ... 85090 ... 83000 ... 80960 ... 78980 ... 77040 ... 75150 ... 73300 ...71500 ... 69750 ... 63940 ... 55090 ... 52420 ... 47460 ... 39390 ... 32690 ... 28870 ... 22520 ... 16720 ... 01140 ... 000000 ... 329200 ... 195060 ... 137590 ... 105670 ... 085370 ... 071320 ... 061030 ... 053160 ... 046960 ... 041940 ... 037800 ... 034320 ... 031370 ... 028820 ... 022170 ... 015330 ... 013770 ...011290 ... 008120 ... 006070 ... 005070 ... 003630 ... 002510 ... 000140 ... 00002 ... 03774 ... 12666 ... 26808 ... 463410 ... 713912 ... 021115 ... 386317 ... 811120 ... 296822 ... 845025 ... 457427 ... 135430 ... 880933 ... 695436 ... 115551 ... 228468 ... 627178 ... 5745115 ... 1035147 ... 7050183 ...0723226 ... 2171332 ... 60211497 ... 28310161 ... 509390 ... 257860 ... 174030 ... 132130 ... 107000 ... 090260 ... 078310 ... 069350 ... 062380 ... 056820 ... 052270 ... 048490 ... 045290 ... 042550 ... 040180 ... 031920 ... 027020 ... 025250 ... 021150 ... 019300 ... 017950 ... 016920 ... 015510 ... 013170... 012530 ... 96312 ... 87814 ... 74606 ... 56818 ... 345510 ... 079311 ... 770613 ... 420315 ... 029516 ... 599318 ... 130619 ... 624221 ... 081322 ... 502524 ... 888928 ... 326935 ... 012938 ... 601742 ... 292548 ... 822253 ... 724656 ... 086561 ... 478166 ... 942379 ... 79420 ... 90235 ... 516014 ... 657125 ...148741 ... 820159 ... 507280 ... 0519104 ... 3021130 ... 1115159 ... 3392191 ... 8499224 ... 5132260 ... 2040298 ... 8019337 ... 2830559 ... 2296811 ... 9409946 ... 841428 ... 791778 ... 832127 ... 242468 ... 573109 ... 615101 ... 906284 ... 49690 ... 48451 ... 46382 ... 43483 ... 39754 ... 35205 ... 29826 ...23627 ... 16598 ... 08749 ... 000610 ... 905611 ... 802412 ... 691113 ... 571516 ... 851521 ... 929522 ... 893625 ... 204731 ... 325836 ... 179339 ... 773745 ... 223451 ... 176475 ... 7619586Compound Interest Factor Tables1 ... 5%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniformSeries PaymentsArithmetic GradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniformSeriesA͞G1234567891011121314151617181920212223242526272829303640485052556072758490961001081201321442403604801 ... 03021 ... 06141 ... 09341 ... 12651 ... 16051 ... 19561 ... 23181 ... 26901 ... 30731 ... 34691 ... 38761 ... 42951 ... 47271 ... 51721 ... 56311 ... 81402 ... 10522 ... 26792... 92123 ... 49263 ... 17584 ... 99275 ... 13708 ... 6328212 ... 700 ... 97070 ... 94220 ... 91450 ... 88770 ... 86170 ... 83640 ... 81180 ... 78800 ... 76490 ... 74250 ... 72070 ... 69950 ... 67900 ... 65910 ... 63980 ... 55130 ... 47500 ... 44090 ... 34230 ... 28630 ... 23950 ... 20030 ... 14010 ... 02810 ... 00081 ...496280 ... 244440 ... 160530 ... 118580 ... 093430 ... 076680 ... 064720 ... 055770 ... 048810 ... 043250 ... 038700 ... 034920 ... 031730 ... 029000 ... 026640 ... 018430 ... 013570 ... 011830 ... 007810 ... 006020 ... 004720 ... 003760 ... 002440 ... 000430 ... 000011 ... 01503 ... 09095 ... 22967 ... 43289 ...702711 ... 041214 ... 450416 ... 932419 ... 489421 ... 123724 ... 837627 ... 633530 ... 514032 ... 481535 ... 538747 ... 267969 ... 682877 ... 529696 ... 0772136 ... 1726187 ... 7202228 ... 1778331 ... 1354502 ... 8514114845801 ... 511280 ... 259440 ... 175530 ... 133580 ... 108430 ... 091680 ... 079720 ...070770 ... 063810 ... 058250 ... 053700 ... 049920 ... 046730 ... 044000 ... 041640 ... 033430 ... 028570 ... 026830 ... 022810 ... 021020 ... 019720 ... 018760 ... 017440 ... 015430 ... 015010 ... 95592 ... 85444 ... 69726 ... 48598 ... 222210 ... 907511 ... 543413 ... 131314 ... 672616 ... 168617 ... 620819 ...030420 ... 398622 ... 726723 ... 015827 ... 915834 ... 999735 ... 271539 ... 844744 ... 578649 ... 701751 ... 313755 ... 325758 ... 795766 ... 61420 ... 88335 ... 422913 ... 401825 ... 612540 ... 856858 ... 945478 ... 6974101 ... 9400126 ... 5084154 ... 2453184 ... 0006216 ... 6310249 ... 0002284 ... 9779321... 8303524 ... 5462749 ... 8774868 ... 16741279 ... 561568 ... 541847 ... 452112 ... 712588 ... 583870 ... 724415 ... 49630 ... 48141 ... 45662 ... 42193 ... 37724 ... 32275 ... 25826 ... 18397 ... 09978 ... 00579 ... 901810 ... 788111 ... 664612 ... 531312 ... 388315 ... 527720 ... 427722 ... 289425 ... 189330 ...966834 ... 438137 ... 617142 ... 157947 ... 736864 ... 2883587Compound Interest Factor Tables2%TABLE 72%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmeticGradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniform SeriesA͞G1234567891011121314151617181920212223242526272829303640485052556072758490961001081201321442403604801 ...04041 ... 08241 ... 12621 ... 17171 ... 21901 ... 26821 ... 31951 ... 37281 ... 42821 ... 48591 ... 54601 ... 60841 ... 67341 ... 74101 ... 81142 ... 20802 ... 69162 ... 97173 ... 16114 ... 27735 ... 69297 ... 488310 ... 652817 ... 88871247 ... 98040 ... 94230 ... 90570 ... 87060 ... 83680 ... 80430 ... 77300 ... 74300... 71420 ... 68640 ... 65980 ... 63420 ... 60950 ... 58590 ... 56310 ... 49020 ... 38650 ... 35710 ... 30480 ... 22650 ... 16830 ... 13800 ... 09290 ... 05780 ... 00080 ... 000000 ... 326750 ... 192160 ... 134510 ... 102520 ... 082180 ... 068120 ... 057830 ... 049970 ... 043780 ... 038780 ... 034670 ... 031220 ...028290 ... 025780 ... 019230 ... 012600 ... 011110 ... 008770 ... 005860 ... 004050 ... 003200 ... 002050 ... 001230 ... 000021 ... 02003 ... 12165 ... 30817 ... 58309 ... 949712 ... 412114 ... 973917 ... 639320 ... 412322 ... 297425 ... 299028 ... 421932 ... 670935 ... 051238 ... 568151 ... 402079 ... 579490 ...5865114 ... 0570170 ... 8666247 ... 6467312 ... 4129488 ... 6415815 ... 44623281 ... 515050 ... 262620 ... 178530 ... 136510 ... 111330 ... 094560 ... 082600 ... 073650 ... 066700 ... 061160 ... 056630 ... 052870 ... 049700 ... 046990 ... 044650 ... 036560 ... 031820 ... 030140 ... 026330 ... 024680 ...023510 ... 022670 ... 021580 ... 020170 ... 020000 ... 94162 ... 80774 ... 60146 ... 32558 ... 98269 ... 575311 ... 106212 ... 577714 ... 992015 ... 351417 ... 658018 ... 913919 ... 121020 ... 281321 ... 396525 ... 355530 ... 423632 ... 174834 ... 984138 ... 525541 ... 529443 ... 109545 ... 337847 ... 568649 ...99630 ... 84585 ... 240313 ... 903524 ... 572038 ... 997755 ... 947574 ... 202196 ... 5554119 ... 8139144 ... 7959171 ... 3309199 ... 2592229 ... 4311259 ... 7064291 ... 0405461 ... 9657642 ... 7849733 ... 69751034 ... 641230 ... 171409 ... 751569 ... 421833 ... 792374 ... 572498 ... 49500 ... 47521 ... 44232... 39613 ... 33674 ... 26425 ... 17866 ... 07997 ... 96818 ... 84339 ... 705510 ... 554710 ... 391011 ... 214512 ... 025115 ... 888519 ... 442021 ... 105723 ... 223428 ... 361631 ... 137033 ... 577437 ... 567641 ... 911049 ... 9643588Compound Interest Factor Tables3%TABLE 83%Discrete Cash Flow:Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmetic GradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniformSeriesA͞G1234567891011121314151617181920212223242526272829303132333435404550556065707580848590961081201 ... 06091 ... 12551 ... 19411 ... 26681 ... 34391 ... 42581 ... 51261 ... 60471 ... 70241 ... 80611 ... 91611 ... 03282 ... 15662 ... 28792 ... 42732 ... 57512 ... 73192 ... 26203 ...38395 ... 89166 ... 91789 ... 640911 ... 335714 ... 075524 ... 71100 ... 94260 ... 88850 ... 83750 ... 78940 ... 74410 ... 70140 ... 66110 ... 62320 ... 58740 ... 55370 ... 52190 ... 49190 ... 46370 ... 43710 ... 41200 ... 38830 ... 36600 ... 30660 ... 22810 ... 16970 ... 12630 ... 09400 ... 08110 ... 05860 ... 02881 ...492610 ... 239030 ... 154600 ... 112460 ... 087230 ... 070460 ... 058530 ... 049610 ... 042710 ... 037220 ... 032750 ... 029050 ... 025940 ... 023290 ... 021020 ... 019050 ... 017320 ... 013260 ... 008870 ... 006130 ... 004340 ... 003110 ... 002650 ... 001870 ... 000891 ... 03003 ... 18365 ... 46847 ... 892310 ...463912 ... 192015 ... 086318 ... 156921 ... 414425 ... 870428 ... 536832 ... 426536 ... 553040 ... 930945 ... 575450 ... 502855 ... 730260 ... 401392 ... 7969136 ... 0534194 ... 5941272 ... 3630365 ... 8570443 ... 8502778 ... 701 ... 522610 ... 269030 ... 184600 ... 142460 ... 117230 ... 100460 ... 088530 ...079610 ... 072710 ... 067220 ... 062750 ... 059050 ... 055940 ... 053290 ... 051020 ... 049050 ... 047320 ... 043260 ... 038870 ... 036130 ... 034340 ... 033110 ... 032650 ... 031870 ... 030890 ... 91352 ... 71714 ... 41726 ... 01977 ... 53029 ... 954010 ... 296111 ... 561113 ... 753514 ... 877515 ... 936916 ...935517 ... 876818 ... 764119 ... 600420 ... 388820 ... 131821 ... 114824 ... 729826 ... 675628 ... 123429 ... 200830 ... 631231 ... 381231 ... 37300 ... 77295 ... 888813 ... 954723 ... 611936 ... 533051 ... 419668 ... 000286 ... 0280106 ... 2788126 ... 5496148 ... 6566170 ... 4336194 ... 7309217 ... 4137241 ...3609265 ... 4642289 ... 6267361 ... 6325477 ... 7411583 ... 2010676 ... 6978756 ... 5434791 ... 6302858 ... 6013963 ... 49260 ... 46311 ... 41382 ... 34503 ... 25654 ... 14855 ... 02106 ... 87427 ... 70818 ... 52298 ... 31869 ... 095410 ... 853511 ... 593011 ... 314112 ... 016913 ... 701814 ... 650217 ... 557519... 067422 ... 214524 ... 035325 ... 834926 ... 361528 ... 7737589Compound Interest Factor Tables4%TABLE 94%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmeticGradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniform SeriesA͞G12345678910111213141516171819202122232425262728293031323334354045505560657075808590961081201441 ...08161 ... 16991 ... 26531 ... 36861 ... 48021 ... 60101 ... 73171 ... 87301 ... 02582 ... 19112 ... 36992 ... 56332 ... 77252 ... 99873 ... 24343 ... 50813 ... 79433 ... 80105 ... 10678 ... 519612 ... 571618 ... 049828 ... 119343 ... 1195110 ... 66180 ... 92460 ... 85480 ... 79030 ... 73070 ... 67560 ... 62460 ...

57750 ... 53390 ... 49360 ... 45640 ... 42200 ... 39010 ... 36070 ... 33350 ... 30830 ... 28510 ... 26360 ... 20830 ... 14070 ... 09510 ... 06420 ... 04340 ... 02930 ... 01450 ... 00351 ... 490200 ... 235490 ... 150760 ... 108530 ... 083290 ... 066550 ... 054670 ... 045820 ... 038990 ... 033580 ... 029200 ... 025590... 022570 ... 020010 ... 017830 ... 015950 ... 014310 ... 010520 ... 006550 ... 004200 ... 002750 ... 001810 ... 001210 ... 000590 ... 000141 ... 04003 ... 24655 ... 63307 ... 214210 ... 006113 ... 025816 ... 291920 ... 824523 ... 645427 ... 778131 ... 248036 ... 082641 ... 311747 ... 967652 ... 084959 ... 701566... 857973 ... 0255121 ... 6671191 ... 9907294 ... 2905448 ... 2450676 ... 98331054 ... 992741 ... 551 ... 530200 ... 275490 ... 190760 ... 148530 ... 123290 ... 106550 ... 094670 ... 085820 ... 078990 ... 073580 ... 069200 ... 065590 ... 062570 ... 060010 ... 057830 ... 055950 ... 054310 ... 050520 ... 046550... 044200 ... 042750 ... 041810 ... 041210 ... 040590 ... 040140 ... 88612 ... 62994 ... 24216 ... 73277 ... 11098 ... 38519 ... 563111 ... 652312 ... 659313 ... 590314 ... 451114 ... 247015 ... 982816 ... 663116 ... 292017 ... 873618 ... 411218 ... 792820 ... 482222 ... 623523 ... 394523 ... 915424 ... 267324 ...638324 ... 91190 ... 70255 ... 554712 ... 065722 ... 801333 ... 377247 ... 454661 ... 735577 ... 958194 ... 8933111 ... 3414129 ... 1284147 ... 1040165 ... 1385183 ... 1206201 ... 9556218 ... 5634236 ... 8768286 ... 4028361 ... 6890422 ... 2014472 ... 0408511 ... 9384540 ... 9312576 ... 2428610 ... 49020 ...45101 ... 38572 ... 29443 ... 17734 ... 03435 ... 86596 ... 67207 ... 45307 ... 20918 ... 94079 ... 64799 ... 331210 ... 990911 ... 627411 ... 241112 ... 832413 ... 476515 ... 812217 ... 697219 ... 196120 ... 371821 ... 282622 ... 414623 ... 4906590Compound Interest Factor Tables5%TABLE 105%Discrete CashFlow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmetic GradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniformSeriesA͞G123456789101112131415161718192021222324252627282930313233343540455055606570758085909596981001 ... 10251 ... 21551 ... 34011 ... 47751 ... 62891 ... 79591 ... 97992 ... 18292 ... 40662 ... 65332 ... 92533 ... 22513 ... 55573 ... 92014 ... 32194 ... 76495 ... 25335 ... 04008 ...467414 ... 679223 ... 426438 ... 561463 ... 7304103 ... 1864119 ... 50130 ... 90700 ... 82270 ... 74620 ... 67680 ... 61390 ... 55680 ... 50510 ... 45810 ... 41550 ... 37690 ... 34180 ... 31010 ... 28120 ... 25510 ... 23140 ... 20990 ... 19040 ... 14200 ... 08720 ... 05350 ... 03290 ... 02020 ... 01240 ... 00920 ...00761 ... 487800 ... 232010 ... 147020 ... 104720 ... 079500 ... 062830 ... 051020 ... 042270 ... 035550 ... 030240 ... 025970 ... 022470 ... 019560 ... 017120 ... 015050 ... 013280 ... 011760 ... 008280 ... 004780 ... 002830 ... 001700 ... 001030 ... 000630 ... 000470 ... 000381 ... 05003 ... 31015 ... 80198 ...549111 ... 577914 ... 917117 ... 598621 ... 657525 ... 132430 ... 066035 ... 505241 ... 502047 ... 113554 ... 402662 ... 438870 ... 298880 ... 067090 ... 7998159 ... 3480272 ... 5837456 ... 5285756 ... 22881245 ... 612040 ... 732365 ... 031 ... 537800 ... 282010 ... 197020 ... 154720 ... 129500 ... 112830 ...101020 ... 092270 ... 085550 ... 080240 ... 075970 ... 072470 ... 069560 ... 067120 ... 065050 ... 063280 ... 061760 ... 058280 ... 054780 ... 052830 ... 051700 ... 051030 ... 050630 ... 050470 ... 050380 ... 85942 ... 54604 ... 07575 ... 46327 ... 72178 ... 86339 ... 898610 ... 837811 ... 689612 ... 462212 ...163013 ... 798614 ... 375214 ... 898115 ... 372515 ... 802716 ... 192916 ... 159117 ... 255918 ... 929319 ... 342719 ... 596519 ... 752319 ... 815119 ... 84790 ... 63475 ... 236911 ... 232120 ... 126831 ... 498843 ... 987956 ... 288070 ... 140584 ... 327598 ... 6673112 ... 0087127 ... 2275141 ... 2226155 ...9126168 ... 2333181 ... 1351194 ... 5807229 ... 3145277 ... 5104314 ... 6910340 ... 0721359 ... 8007372 ... 6774378 ... 2139381 ... 48780 ... 43911 ... 35792 ... 24453 ... 09914 ... 92195 ... 71336 ... 47366 ... 20347 ... 90308 ... 57308 ... 21409 ... 826610 ... 411410 ... 969111 ... 500511 ... 006312 ... 377514... 223315 ... 606217 ... 621218 ... 352618 ... 871219 ... 104419 ... 2337591Compound Interest Factor Tables6%TABLE 116%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmeticGradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniform SeriesA͞G123456789101112131415161718192021222324252627282930313233343540455055606570758085909596981001 ... 12361... 26251 ... 41851 ... 59381 ... 79081 ... 01222 ... 26092 ... 54042 ... 85433 ... 20713 ... 60353 ... 04894 ... 54944 ... 11175 ... 74356 ... 45346 ... 25107 ... 285713 ... 420224 ... 987744 ... 075979 ... 7960141 ... 4645253 ... 7590301 ... 30210 ... 89000 ... 79210 ... 70500 ... 62740 ... 55840 ... 49700 ... 44230... 39360 ... 35030 ... 31180 ... 27750 ... 24700 ... 21980 ... 19560 ... 17410 ... 15500 ... 13790 ... 09720 ... 05430 ... 03030 ... 01690 ... 00950 ... 00530 ... 00370 ... 00291 ... 485440 ... 228590 ... 143360 ... 101040 ... 075870 ... 059280 ... 047580 ... 038950 ... 032360 ... 027180 ... 023050 ... 019680 ...016900 ... 014590 ... 012650 ... 011000 ... 009600 ... 006460 ... 003440 ... 001880 ... 001030 ... 000570 ... 000320 ... 000220 ... 000181 ... 06003 ... 37465 ... 97538 ... 897511 ... 180814 ... 869918 ... 015123 ... 672528 ... 905733 ... 785639 ... 392346 ... 815654 ... 156463 ... 528173 ... 058284 ... 889897 ...1838111 ... 7620212 ... 3359394 ... 1282719 ... 93221300 ... 602342 ... 084209 ... 655016 ... 371 ... 545440 ... 288590 ... 203360 ... 161040 ... 135870 ... 119280 ... 107580 ... 098950 ... 092360 ... 087180 ... 083050 ... 079680 ... 076900 ... 074590 ... 072650 ... 071000 ... 069600 ... 066460 ... 063440 ...061880 ... 061030 ... 060570 ... 060320 ... 060220 ... 060180 ... 83342 ... 46514 ... 91735 ... 20986 ... 36017 ... 38388 ... 29509 ... 105910 ... 827611 ... 469911 ... 041612 ... 550412 ... 003213 ... 406213 ... 764813 ... 084014 ... 368114 ... 046315 ... 761915 ... 161416 ... 384516 ... 509116 ... 578716 ...604716 ... 61750 ... 56924 ... 934511 ... 449719 ... 576829 ... 870240 ... 962951 ... 554663 ... 401175 ... 306287 ... 113698 ... 7007110 ... 9732121 ... 8600132 ... 3096142 ... 2864152 ... 7681161 ... 7427185 ... 1096217 ... 3222239 ... 9450253 ... 4527262 ... 8096268 ... 4375270 ... 4491272 ... 48540 ...42721 ... 33042 ... 19523 ... 02204 ... 81135 ... 56355 ... 27946 ... 95977 ... 60517 ... 21668 ... 79519 ... 34149 ... 856810 ... 342210 ... 798811 ... 227611 ... 359013 ... 796414 ... 790915 ... 461315 ... 903316 ... 189116 ... 308116 ... 3711592Compound Interest Factor Tables7%TABLE 127%Discrete CashFlow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmetic GradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniformSeriesA͞G123456789101112131415161718192021222324252627282930313233343540455055606570758085909596981001 ... 14491 ... 31081 ... 50071 ... 71821 ... 96722 ... 25222 ... 57852 ... 95223 ... 37993 ... 86974 ... 43044 ... 07245 ... 80746 ... 64887 ... 61238 ... 71539 ... 978110 ... 974521 ...457041 ... 946481 ... 9894159 ... 2344314 ... 1030618 ... 9766757 ... 71630 ... 87340 ... 76290 ... 66630 ... 58200 ... 50830 ... 44400 ... 38780 ... 33870 ... 29590 ... 25840 ... 22570 ... 19710 ... 17220 ... 15040 ... 13140 ... 11470 ... 10020 ... 06680 ... 03390 ... 01730 ... 00880 ... 00450 ... 00230 ... 00150 ...00121 ... 483090 ... 225230 ... 139800 ... 097470 ... 072380 ... 055900 ... 044340 ... 035860 ... 029410 ... 024390 ... 020410 ... 017190 ... 014560 ... 012390 ... 010590 ... 009070 ... 007800 ... 005010 ... 002460 ... 001230 ... 000620 ... 000310 ... 000160 ... 000110 ... 000081 ... 07003 ... 43995 ... 15338 ...259811 ... 816415 ... 888520 ... 550525 ... 888130 ... 999037 ... 995544 ... 005753 ... 176763 ... 676574 ... 697787 ... 4608102 ... 2182118 ... 2588138 ... 6351285 ... 5289575 ... 52041146 ... 132269 ... 064478 ... 198823 ... 5210813123821 ... 553090 ... 295230 ... 209800 ... 167470 ... 142380 ... 125900 ...114340 ... 105860 ... 099410 ... 094390 ... 090410 ... 087190 ... 084560 ... 082390 ... 080590 ... 079070 ... 077800 ... 075010 ... 072460 ... 071230 ... 070620 ... 070310 ... 070160 ... 070110 ... 070080 ... 80802 ... 38724 ... 76655 ... 97136 ... 02367 ... 94278 ... 74559 ... 44669 ... 059110 ... 594010 ...061211 ... 469311 ... 825811 ... 137112 ... 409012 ... 646612 ... 854012 ... 331713 ... 800713 ... 039214 ... 160414 ... 222014 ... 253314 ... 264114 ... 26930 ... 50604 ... 646710 ... 714918 ... 140427 ... 466537 ... 330247 ... 446157 ... 592367 ... 599177 ... 339387 ... 720196 ... 6765104 ... 1656113 ...1622120 ... 6550128 ... 6435134 ... 1353152 ... 7559172 ... 1243185 ... 1452193 ... 1035198 ... 5717200 ... 5581201 ... 9651202 ... 48310 ... 41551 ... 30322 ... 14653 ... 94614 ... 70255 ... 41675 ... 08976 ... 72257 ... 31637 ... 87258 ... 39238 ... 87739 ... 32899 ... 74879 ... 138110 ... 498710 ... 423312 ...528712 ... 232113 ... 666213 ... 927314 ... 081214 ... 140514 ... 1703593Compound Interest Factor Tables8%TABLE 138%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmeticGradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniform SeriesA͞G123456789101112131415161718192021222324252627282930313233343540455055606570758085909596981001 ... 16641... 36051 ... 58691 ... 85091 ... 15892 ... 51822 ... 93723 ... 42593 ... 99604 ... 66105 ... 43655 ... 34126 ... 39647 ... 62719 ... 062710 ... 737112 ... 690114 ... 724531 ... 901668 ... 2571148 ... 6064321 ... 9548693 ... 921497 ... 891885 ... 760 ... 85730 ... 73500 ... 63020 ... 54030 ... 46320 ... 39710 ... 34050... 29190 ... 25020 ... 21450 ... 18390 ... 15770 ... 13520 ... 11590 ... 09940 ... 08520 ... 07300 ... 04600 ... 02130 ... 00990 ... 00460 ... 00210 ... 00100 ... 00060 ... 00051 ... 480770 ... 221920 ... 136320 ... 094010 ... 069030 ... 052700 ... 041300 ... 032980 ... 026700 ... 021850 ... 018030 ... 014980 ...012510 ... 010490 ... 008830 ... 007450 ... 006300 ... 003860 ... 001740 ... 000800 ... 000370 ... 000170 ... 000080 ... 000050 ... 000041 ... 08003 ... 50615 ... 33598 ... 636612 ... 486616 ... 977121 ... 214927 ... 324333 ... 450241 ... 762050 ... 456860 ... 764873 ... 954487 ... 3388103 ... 2832123 ...2135145 ... 6267172 ... 0565386 ... 7702848 ... 211847 ... 084002 ... 948655 ... 080000 ... 388030 ... 250460 ... 192070 ... 160080 ... 140080 ... 126520 ... 116830 ... 109630 ... 104130 ... 099830 ... 096420 ... 093680 ... 091450 ... 089620 ... 088110 ... 086850 ... 085800 ... 082590 ... 081180 ... 080540 ...080250 ... 080120 ... 080050 ... 080040 ... 92591 ... 57713 ... 99274 ... 20645 ... 24696 ... 13907 ... 90388 ... 55958 ... 12169 ... 60369 ... 016810 ... 371110 ... 674810 ... 935211 ... 158411 ... 349811 ... 513911 ... 654611 ... 108412 ... 318612 ... 416012 ... 461112 ... 482012 ... 491712 ... 493412 ... 85732... 65017 ... 523314 ... 806121 ... 976830 ... 633939 ... 472347 ... 264056 ... 842665 ... 089873 ... 925780 ... 299787 ... 184294 ... 5687100 ... 4558106 ... 8575111 ... 7924116 ... 0422133 ... 5928144 ... 3000149 ... 5326152 ... 8001154 ... 9925155 ... 4112155 ... 61070 ... 94871 ... 84652 ... 69373 ... 49103... 23954 ... 94025 ... 59455 ... 20376 ... 76977 ... 29407 ... 77868 ... 22548 ... 63638 ... 01339 ... 35849 ... 67379 ... 961110 ... 044711 ... 690211 ... 060212 ... 265812 ... 377212 ... 436512 ... 448012 ... 09001 ... 29501 ... 53861 ... 82801 ... 17192 ... 58042 ... 06583 ... 64253 ... 32764 ... 14175 ... 10886 ...25797 ... 62319 ... 245111 ... 172213 ... 461815 ... 182018 ... 414031 ... 327374 ... 4083176 ... 8460416 ... 1909986 ... 932335 ... 503916 ... 685529 ... 91740 ... 77220 ... 64990 ... 54700 ... 46040 ... 38750 ... 32620 ... 27450 ... 23110 ... 19450 ... 16370 ... 13780 ... 11600 ... 09760 ... 08220 ... 06910 ...05820 ... 04900 ... 02070 ... 00870 ... 00370 ... 00160 ... 00070 ... 00030 ... 00020 ... 000000 ... 305050 ... 167090 ... 108690 ... 076800 ... 056950 ... 043570 ... 034060 ... 027050 ... 021730 ... 017620 ... 014380 ... 011810 ... 009730 ... 008060 ... 006690 ... 005560 ... 004640 ... 001900 ... 000790 ...000330 ... 000140 ... 000060 ... 000030 ... 000020 ... 00002 ... 27814 ... 98477 ... 200411 ... 021015 ... 560320 ... 953426 ... 360933 ... 973741 ... 018551 ... 764562 ... 531976 ... 700993 ... 7231112 ... 1354136 ... 5752164 ... 8003196 ... 7108337 ... 8587815 ... 091944 ... 294619 ...23109511685525939399174351051696614231 ... 568470 ... 308670 ... 222920 ... 180670 ... 155820 ... 139650 ... 128430 ... 120300 ... 114210 ... 109550 ... 105900 ... 103020 ... 100720 ... 098850 ... 097340 ... 096100 ... 095080 ... 092960 ... 091230 ... 090510 ... 090220 ... 090090 ... 090040 ... 090020... 090020 ... 75912 ... 23973 ... 48595 ... 53485 ... 41776 ... 16077 ... 78628 ... 31268 ... 75568 ... 12859 ... 44249 ... 70669 ... 929010 ... 116110 ... 273710 ... 406210 ... 517810 ... 757410 ... 961711 ... 048011 ... 084411 ... 099811 ... 106411 ... 108311 ... 10910 ... 38604 ... 111010 ... 374616 ... 571124 ...248132 ... 073139 ... 806947 ... 282154 ... 386861 ... 050968 ... 235974 ... 926579 ... 124184 ... 842289 ... 102493 ... 931496 ... 3590105 ... 5561114 ... 0362118 ... 3344121 ... 9646122 ... 7533122 ... 1287123 ... 1963123 ... 47850 ... 39251 ... 24982 ... 05123 ... 79784 ... 49104 ... 13265 ... 72456 ... 26876... 76747 ... 22327 ... 63847 ... 01568 ... 35718 ... 66578 ... 94369 ... 19339 ... 795710 ... 429510 ... 768310 ... 942710 ... 029911 ... 072611 ... 086611 ... 0930595Compound Interest Factor Tables10%TABLE 1510%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform SeriesPaymentsArithmetic GradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniformSeriesA͞G123456789101112131415161718192021222324252627282930313233343540455055606570758085909596981001 ... 21001 ... 46411 ... 77161 ... 14362 ... 59372 ... 13843 ... 79754 ... 59505 ... 55996 ... 72757 ... 14038 ... 849710 ... 918213 ... 421015 ... 449419 ... 113823 ... 547728 ... 259372... 3909189 ... 4816490 ... 74701271 ... 403298 ... 028556 ... 3411389137810 ... 82640 ... 68300 ... 56450 ... 46650 ... 38550 ... 31860 ... 26330 ... 21760 ... 17990 ... 14860 ... 12280 ... 10150 ... 08390 ... 06930 ... 05730 ... 04740 ... 03910 ... 02210 ... 00850 ... 00330 ... 00130 ... 00050 ... 00020 ... 00010... 00011 ... 476190 ... 215470 ... 129610 ... 087440 ... 062750 ... 046760 ... 035750 ... 027820 ... 021930 ... 017460 ... 014010 ... 011300 ... 009160 ... 007450 ... 006080 ... 004970 ... 004070 ... 002260 ... 000860 ... 000330 ... 000130 ... 000050 ... 000020 ... 000010 ... 000011 ... 10003 ... 64106 ... 71569... 435913 ... 937418 ... 384324 ... 975031 ... 949740 ... 599251 ... 275064 ... 402779 ... 497398 ... 1818121 ... 2099148 ... 4940181 ... 1378222 ... 4767271 ... 5926718 ... 911880 ... 824893 ... 471270920474329805312085557941131 ... 576190 ... 315470 ... 229610 ... 187440 ... 162750 ... 146760 ...135750 ... 127820 ... 121930 ... 117460 ... 114010 ... 111300 ... 109160 ... 107450 ... 106080 ... 104970 ... 104070 ... 102260 ... 100860 ... 100330 ... 100130 ... 100050 ... 100020 ... 100010 ... 100010 ... 73552 ... 16993 ... 35534 ... 33495 ... 14466 ... 81377 ... 36677 ... 82378 ... 20148 ... 51368 ... 77158... 98479 ... 16099 ... 30669 ... 42699 ... 52649 ... 60869 ... 77919 ... 91489 ... 96729 ... 98739 ... 99519 ... 99819 ... 99899 ... 99930 ... 32914 ... 86189 ... 763116 ... 421522 ... 396329 ... 377236 ... 152043 ... 581949 ... 582755 ... 109560 ... 146265 ... 696469 ... 777373 ... 414677 ... 639580 ... 485682 ...987288 ... 454494 ... 561997 ... 470598 ... 331799 ... 712099 ... 877399 ... 905299 ... 47620 ... 38121 ... 22362 ... 00453 ... 72554 ... 38844 ... 99555 ... 54935 ... 05266 ... 50816 ... 91897 ... 28817 ... 61867 ... 91378 ... 17628 ... 40918 ... 61498 ... 09629 ... 57049 ... 80239 ... 91139 ... 96099 ... 98319 ...98989 ... 9927596Compound Interest Factor Tables11%TABLE 1611%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmeticGradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniform SeriesA͞G1234567891011121314151617181920212223242526272829303132333435404550556065707580851 ... 23211 ... 51811 ...87042 ... 30452 ... 83943 ... 49853 ... 31044 ... 31095 ... 54367 ... 06238 ... 933611 ... 239213 ... 079916 ... 579920 ... 892325 ... 205631 ... 752138 ... 0009109 ... 5648311 ... 0572883 ... 022507 ... 117119 ... 90090 ... 73120 ... 59350 ... 48170 ... 39090 ... 31730 ... 25750 ... 20900 ... 16960 ... 13770 ...11170 ... 09070 ... 07360 ... 05970 ... 04850 ... 03940 ... 03190 ... 02590 ... 00910 ... 00320 ... 00110 ... 00040 ... 00011 ... 473930 ... 212330 ... 126380 ... 084320 ... 059800 ... 044030 ... 033230 ... 025520 ... 019840 ... 015580 ... 012310 ... 009790 ... 007810 ... 006260 ... 005020 ... 004040 ... 003260 ...001720 ... 000600 ... 000210 ... 000070 ... 000030 ... 00002 ... 34214 ... 22787 ... 783311 ... 164016 ... 561422 ... 211630 ... 405439 ... 500850 ... 939564 ... 265181 ... 1479102 ... 4133127 ... 0786159 ... 3972199 ... 9132247 ... 5292306 ... 5896581 ... 63861668 ... 204755 ... 79135182278538401647141... 583930 ... 322330 ... 236380 ... 194320 ... 169800 ... 154030 ... 143230 ... 135520 ... 129840 ... 125580 ... 122310 ... 119790 ... 117810 ... 116260 ... 115020 ... 114040 ... 113260 ... 111720 ... 110600 ... 110210 ... 110070 ... 110030 ... 90091 ... 44373 ... 69594 ... 71225 ... 53705 ... 20656 ... 74996 ...19097 ... 54887 ... 83937 ... 07518 ... 26648 ... 42178 ... 54788 ... 65018 ... 73318 ... 80058 ... 85528 ... 00799 ... 06179 ... 08069 ... 08739 ... 08960 ... 27404 ... 62409 ... 187215 ... 352021 ... 694527 ... 929033 ... 870939 ... 409545 ... 485649 ... 077154 ... 186458 ... 832261 ... 043364 ... 854267 ... 301669... 422871 ... 253875 ... 155179 ... 771281 ... 881982 ... 339782 ... 52450 ... 93061 ... 79232 ... 58632 ... 31443 ... 97884 ... 58224 ... 12755 ... 61805 ... 05746 ... 44916 ... 79696 ... 10457 ... 37547 ... 61317 ... 82107 ... 00218 ... 15948 ... 67638 ... 91358 ... 01729 ... 06109 ... 0790597Compound InterestFactor Tables12%TABLE 1712%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmetic GradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniformSeriesA͞G1234567891011121314151617181920212223242526272829303132333435404550556065707580851 ... 25441 ... 57351 ... 97382 ... 47602 ... 10583 ... 89604 ... 88715 ... 13046 ... 69008 ... 646310 ... 100313 ... 178617 ... 040121 ... 883926 ... 959933 ... 581742 ... 142552 ... 0510163 ...0022509 ... 59691581 ... 804913 ... 48152590 ... 79720 ... 63550 ... 50660 ... 40390 ... 32200 ... 25670 ... 20460 ... 16310 ... 13000 ... 10370 ... 08260 ... 06590 ... 05250 ... 04190 ... 03340 ... 02660 ... 02120 ... 01070 ... 00350 ... 00110 ... 00040 ... 00010 ... 000000 ... 296350 ... 157410 ... 099120 ...067680 ... 048420 ... 035680 ... 026820 ... 020460 ... 015760 ... 012240 ... 009560 ... 007500 ... 005900 ... 004660 ... 003690 ... 002920 ... 002320 ... 00740 ... 000240 ... 000080 ... 000020 ... 000011 ... 12003 ... 77936 ... 115210 ... 299714 ... 548720 ... 133128 ... 392637 ... 753348 ... 749763 ... 052481 ...5026104 ... 1552133 ... 3339169 ... 6989214 ... 3327271 ... 8477342 ... 5210431 ... 09141358 ... 024236 ... 64131742322340934721461 ... 591700 ... 329230 ... 243230 ... 201300 ... 176980 ... 161440 ... 150870 ... 143390 ... 137940 ... 133880 ... 130810 ... 128460 ... 126650 ... 125240 ... 124140 ...123280 ... 122600 ... 121300 ... 120420 ... 120130 ... 120040 ... 120010 ... 89291 ... 40183 ... 60484 ... 56384 ... 32825 ... 93776 ... 42356 ... 81096 ... 11967 ... 36587 ... 56207 ... 71847 ... 84317 ... 94267 ... 02188 ... 08508 ... 13548 ... 17558 ... 28258 ... 31708 ... 32818 ... 33168 ... 33280 ... 22084 ...39708 ... 644314 ... 356320 ... 128825 ... 702431 ... 920236 ... 697340 ... 997944 ... 818848 ... 177651 ... 104654 ... 636956 ... 814158 ... 676160 ... 261261 ... 605265 ... 734267 ... 408268 ... 058169 ... 303169 ... 39350 ... 92461 ... 77462 ... 55122 ... 25743 ... 89534 ... 46834 ... 98035 ... 43535 ... 83756... 19136 ... 50106 ... 77086 ... 00497 ... 20717 ... 38117 ... 53027 ... 65777 ... 05728 ... 22518 ... 29228 ... 31818 ... 3278598Compound Interest Factor Tables14%TABLE 1814%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmeticGradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniform SeriesA͞G1234567891011121314151617181920212223242526272829303132333435404550556065707580851 ... 29961 ... 68901 ...19502 ... 85263 ... 70724 ... 81795 ... 26137 ... 13729 ... 575212 ... 743515 ... 861020 ... 212226 ... 166634 ... 204544 ... 950258 ... 214875 ... 052898 ... 8835363 ... 23301348 ... 924998 ... 641853035677686930 ... 76950 ... 59210 ... 45560 ... 35060 ... 26970 ... 20760 ... 15970 ... 12290 ... 09460 ... 07280... 05600 ... 04310 ... 03310 ... 02550 ... 01960 ... 01510 ... 01160 ... 00530 ... 00140 ... 00040 ... 00010 ... 000000 ... 290730 ... 151280 ... 093190 ... 062170 ... 043390 ... 031160 ... 022810 ... 016920 ... 012660 ... 009540 ... 007230 ... 005500 ... 004190 ... 003200 ... 002450 ... 001880 ... 001440 ...000390 ... 000100 ... 000030 ... 000011 ... 14003 ... 92116 ... 535510 ... 232816 ... 337323 ... 270732 ... 581143 ... 980459 ... 394178 ... 0249104 ... 4360138 ... 6586181 ... 3327238 ... 8892312 ... 7868407 ... 8202532 ... 5199693 ... 032590 ... 529623 ... 140000 ... 430730 ... 291280 ... 233190 ... 202170... 183390 ... 171160 ... 162810 ... 156920 ... 152660 ... 149540 ... 147230 ... 145500 ... 144190 ... 143200 ... 142450 ... 141880 ... 141440 ... 140390 ... 140100 ... 140030 ... 140010 ... 140000 ... 64672 ... 91373 ... 88874 ... 63894 ... 21615 ... 66035 ... 00216 ... 26516 ... 46746 ... 62316 ... 74296 ... 83516... 90616 ... 96076 ... 00277 ... 03507 ... 05997 ... 10507 ... 13277 ... 14017 ... 14217 ... 14277 ... 76952 ... 89575 ... 251110 ... 102815 ... 990620 ... 639924 ... 900928 ... 705732 ... 038035 ... 913538 ... 365840 ... 437142 ... 172843 ... 617645 ... 813246 ... 797947 ... 605347 ... 237649 ... 437550 ... 835750... 963250 ... 003051 ... 46730 ... 33701 ... 12182 ... 82463 ... 44903 ... 99984 ... 48194 ... 90115 ... 26305 ... 57345 ... 83815 ... 06246 ... 25146 ... 41006 ... 54236 ... 65226 ... 74316 ... 93007 ... 07147 ... 11977 ... 13567 ... 14067 ... 15001 ... 52091 ... 01142 ... 66003 ... 51794 ... 65245 ... 15287 ... 13719... 761312 ... 231816 ... 821521 ... 891528 ... 919037 ... 535350 ... 575566 ... 143587 ... 6998115 ... 1755267 ... 76931083 ... 624384 ... 791773635673717510 ... 75610 ... 57180 ... 43230 ... 32690 ... 24720 ... 18690 ... 14130 ... 10690 ... 08080 ... 06110 ... 04620 ... 03490 ... 02640 ... 02000 ... 01510 ...01140 ... 00860 ... 00370 ... 00090 ... 00020 ... 00011 ... 465120 ... 200270 ... 114240 ... 072850 ... 049250 ... 034480 ... 024690 ... 017950 ... 013190 ... 009760 ... 007270 ... 005430 ... 004070 ... 003060 ... 002300 ... 001730 ... 001310 ... 000560 ... 000140 ... 000030 ... 000011 ... 15003 ... 99346 ...753711 ... 726816 ... 303724 ... 001734 ... 504747 ... 717565 ... 836488 ... 4436118 ... 6316159 ... 1678212 ... 7120283 ... 1041377 ... 7451500 ... 1005664 ... 3654881 ... 093585 ... 721452429220587791 ... 615120 ... 350270 ... 264240 ... 222850 ... 199250 ... 184480 ... 174690 ... 167950 ... 163190 ...159760 ... 157270 ... 155430 ... 154070 ... 153060 ... 152300 ... 151730 ... 151310 ... 150560 ... 150140 ... 150030 ... 150010 ... 150000 ... 86961 ... 28322 ... 35223 ... 16044 ... 77165 ... 23375 ... 58315 ... 84745 ... 04726 ... 19826 ... 31256 ... 39886 ... 46416 ... 51356 ... 55096 ... 57916 ... 60056 ... 61666... 65436 ... 66366 ... 66596 ... 66656 ... 66660 ... 07123 ... 77517 ... 192412 ... 754816 ... 128921 ... 135224 ... 693028 ... 782831 ... 421333 ... 644835 ... 498837 ... 031438 ... 289039 ... 314640 ... 146641 ... 818442 ... 358643 ... 805144 ... 255844 ... 390344 ... 429244 ... 44020 ... 90711 ... 72282 ...44982 ... 09223 ... 65493 ... 14384 ... 56504 ... 92515 ... 23075 ... 48835 ... 70405 ... 88345 ... 03196 ... 15416 ... 25416 ... 33576 ... 40196 ... 58306 ... 64146 ... 65936 ... 66466 ... 6661600Compound Interest Factor Tables16%TABLE 2016%Discrete Cash Flow: Compound Interest FactorsSinglePaymentsUniform Series PaymentsArithmetic GradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniform SeriesA͞G12345678910111213141516171819202224262830323435363840455055601 ...34561 ... 81062 ... 43642 ... 27843 ... 41145 ... 93606 ... 98759 ... 748012 ... 462516 ... 460826 ... 236447 ... 800485 ... 5196155 ... 3141209 ... 4515378 ... 44381670 ... 057370 ... 86210 ... 64070 ... 47610 ... 35380 ... 26300 ... 19540 ... 14520 ... 10790 ... 08020 ... 05960 ... 03820 ... 02110 ... 01160 ...00640 ... 00480 ... 00260 ... 00060 ... 00011 ... 462960 ... 197380 ... 111390 ... 070220 ... 046900 ... 032410 ... 022900 ... 016410 ... 011880 ... 008670 ... 004670 ... 002550 ... 001400 ... 000890 ... 000570 ... 000200 ... 000050 ... 00002 ... 50565 ... 87718 ... 413914 ... 518521 ... 732930 ... 786243 ...659560 ... 673084 ... 6032115 ... 4150213 ... 0883392 ... 3117715 ... 26981120 ... 031752 ... 764965 ... 160000 ... 445260 ... 305410 ... 247610 ... 217080 ... 198860 ... 187180 ... 179360 ... 173950 ... 170140 ... 166350 ... 163450 ... 161890 ... 161040 ... 160770 ... 160420 ... 160100 ... 160020 ... 60522 ...79823 ... 68474 ... 34364 ... 83325 ... 19715 ... 46755 ... 66855 ... 81785 ... 92886 ... 07266 ... 15206 ... 19596 ... 21536 ... 22786 ... 24216 ... 24826 ... 74322 ... 68145 ... 63809 ... 896213 ... 039917 ... 847221 ... 217524 ... 124127 ... 582829 ... 632132 ... 697034 ... 707336 ... 993037 ... 632737 ... 079938... 659838 ... 953439 ... 46300 ... 31561 ... 07292 ... 73883 ... 31873 ... 81894 ... 24644 ... 60864 ... 91305 ... 16665 ... 54905 ... 80415 ... 97066 ... 05486 ... 11456 ... 19346 ... 23436 ... 18001 ... 64301 ... 28782 ... 18553 ... 43555 ... 17597 ... 599410 ... 973714 ... 672219 ... 214427 ... 142153 ... 9490102 ...3706199 ... 9638327 ... 0368538 ... 37831716 ... 368984 ... 84750 ... 60860 ... 43710 ... 31390 ... 22550 ... 16190 ... 11630 ... 08350 ... 06000 ... 04310 ... 02620 ... 01350 ... 00700 ... 00360 ... 00260 ... 00130 ... 00030 ... 000000 ... 279920 ... 139780 ... 082360 ... 052390 ... 034780 ... 023690 ... 016400 ...011490 ... 008100 ... 004850 ... 002470 ... 001260 ... 000650 ... 000470 ... 000240 ... 000050 ... 00002 ... 57245 ... 15429 ... 141515 ... 085923 ... 755134 ... 218750 ... 965372 ... 0680103 ... 4135146 ... 3448289 ... 2721566 ... 94801103 ... 691816 ... 652988 ... 219531 ... 180000 ... 459920 ... 319780 ...262360 ... 232390 ... 214780 ... 203690 ... 196400 ... 191490 ... 188100 ... 184850 ... 182470 ... 181260 ... 180650 ... 180470 ... 180240 ... 180050 ... 180010 ... 56562 ... 69013 ... 49763 ... 07764 ... 49414 ... 79324 ... 00815 ... 16245 ... 27325 ... 35275 ... 45095 ... 50165 ... 52775 ... 53865 ... 54525 ...55235 ... 55495 ... 71821 ... 48285 ... 08348 ... 829212 ... 352515 ... 481118 ... 157621 ... 388523 ... 212324 ... 681326 ... 772528 ... 053729 ... 819130 ... 177330 ... 415230 ... 700630 ... 826830 ... 45870 ... 29471 ... 02522 ... 65582 ... 19363 ... 64703 ... 02504 ... 33694 ... 59164 ... 79784 ... 09505 ...28105 ... 39455 ... 44855 ... 48495 ... 52935 ... 54945 ... 20001 ... 72802 ... 48832 ... 58324 ... 15986 ... 43018 ... 699312 ... 407018 ... 186126 ... 948038 ... 206179 ... 4755164 ... 3763341 ... 2235590 ... 80191020 ... 773657 ... 44226450 ... 69440 ... 48230 ... 33490 ... 23260 ... 16150 ... 11220 ... 07790 ...05410 ... 03760 ... 02610 ... 01260 ... 00610 ... 00290 ... 00170 ... 00100 ... 00030 ... 000000 ... 274730 ... 134380 ... 077420 ... 048080 ... 031100 ... 020620 ... 013880 ... 009440 ... 006460 ... 003690 ... 001760 ... 000850 ... 000410 ... 000280 ... 000140 ... 000020 ... 00002 ... 64005 ... 44169 ... 915916 ...798925 ... 150439 ... 496659 ... 035187 ... 9306128 ... 7400186 ... 0307392 ... 3773819 ... 881704 ... 122948 ... 015098 ... 8618281454971 ... 654550 ... 386290 ... 300710 ... 260610 ... 238520 ... 225260 ... 216890 ... 211440 ... 207810 ... 205360 ... 202550 ... 201220 ... 200590 ... 200340 ... 200200 ...200050 ... 200010 ... 52782 ... 58872 ... 32553 ... 83724 ... 19254 ... 43924 ... 61064 ... 72964 ... 81224 ... 86964 ... 93714 ... 96974 ... 98544 ... 99154 ... 99514 ... 99864 ... 99980 ... 85193 ... 90616 ... 25519 ... 433512 ... 233015 ... 588317 ... 509519 ... 041920 ... 243921 ... 554623 ... 646023 ... 262824 ...603824 ... 710824 ... 846924 ... 969824 ... 45450 ... 27421 ... 97882 ... 57562 ... 07393 ... 48413 ... 81753 ... 08514 ... 29754 ... 46434 ... 69434 ... 82914 ... 90614 ... 94064 ... 96274 ... 98774 ... 9976603Compound Interest Factor Tables22%TABLE 2322%Discrete Cash Flow: Compound InterestFactorsSingle PaymentsUniform Series PaymentsArithmetic GradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniformSeriesA͞G123456789101112131415161718192022242628303234353638404550551 ... 48841 ... 21532 ... 29734 ... 90775 ... 30468 ... 872213 ... 182219 ... 085629 ... 849043 ... 357679 ... 2050175 ... 8637389 ... 1156863 ... 401285 ... 822847 ... 7120797562070 ... 67190 ... 45140 ... 30330 ... 20380 ...13690 ... 09200 ... 06180 ... 04150 ... 02790 ... 01870 ... 00850 ... 00380 ... 00170 ... 00090 ... 00050 ... 00011 ... 450450 ... 181020 ... 095760 ... 056300 ... 034890 ... 022280 ... 014490 ... 009530 ... 006310 ... 004200 ... 001880 ... 000840 ... 000380 ... 000210 ... 000120 ... 000030 ... 00002 ... 70845 ...739610 ... 739617 ... 670028 ... 962044 ... 745969 ... 1922104 ... 0201158 ... 2535237 ... 4432532 ... 16531185 ... 082632 ... 204783 ... 058690 ... 220000 ... 489660 ... 349210 ... 292780 ... 264110 ... 247810 ... 237940 ... 231740 ... 227750 ... 225150 ... 222810 ... 221260 ... 220570 ... 220260 ... 220170... 220080 ... 220010 ... 81971 ... 04222 ... 86363 ... 41553 ... 78633 ... 03544 ... 20284 ... 31524 ... 39084 ... 44154 ... 48824 ... 51964 ... 53384 ... 54024 ... 54194 ... 54394 ... 54524 ... 67191 ... 12754 ... 12397 ... 041710 ... 610012 ... 743814 ... 451916 ... 783817 ... 802518 ... 570219 ... 563519 ... 096220... 374820 ... 490520 ... 560120 ... 631920 ... 65630 ... 86831 ... 60901 ... 22972 ... 74092 ... 15513 ... 48553 ... 74513 ... 94654 ... 10094 ... 26494 ... 39684 ... 46834 ... 50604 ... 51744 ... 53144 ... 54314 ... 24001 ... 90662 ... 93163 ... 50775 ... 93108 ... 657113 ... 386320 ... 195631 ... 740848 ... 567973 ...5735174 ... 5121412 ... 8199976 ... 851861 ... 713548 ... 9115995468900 ... 65040 ... 42300 ... 27510 ... 17890 ... 11640 ... 07570 ... 04920 ... 03200 ... 02080 ... 01350 ... 00570 ... 00240 ... 00100 ... 00050 ... 00030 ... 00011 ... 446430 ... 175930 ... 091070 ... 052290 ... 031600 ... 019650 ... 012420 ...007940 ... 005100 ... 003290 ... 001380 ... 000580 ... 000250 ... 000130 ... 000070 ... 000020 ... 00002 ... 77765 ... 048410 ... 615319 ... 712531 ... 237950 ... 109780 ... 8151126 ... 2534195 ... 0328303 ... 0563723 ... 631716 ... 924062 ... 387750 ... 281478122729666401 ... 686430 ... 415930 ... 331070 ...292290 ... 271600 ... 259650 ... 252420 ... 247940 ... 245100 ... 243290 ... 241380 ... 240580 ... 240250 ... 240130 ... 240070 ... 240020 ... 240000 ... 45681 ... 40432 ... 02053 ... 42123 ... 68193 ... 85143 ... 96164 ... 03334 ... 07994 ... 11034 ... 14284 ... 15664 ... 16244 ... 16644 ... 16554 ... 16644 ...16660 ... 69932 ... 33275 ... 03928 ... 445810 ... 431312 ... 996013 ... 191514 ... 084615 ... 740615 ... 401116 ... 893017 ... 136917 ... 255217 ... 288617 ... 327417 ... 356317 ... 44640 ... 23461 ... 88982 ... 42362 ... 84993 ... 18433 ... 44203 ... 63763 ... 78403 ... 89223 ... 02844 ... 09874 ... 13384 ... 14794... 15604 ... 16394 ... 1663605Compound Interest Factor Tables25%TABLE 25Single Paymentsn1234567891011121314151617181920222426283032343536384045505525%Discrete Cash Flow: Compound Interest FactorsUniform Series PaymentsArithmeticGradientsCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresent WorthP͞GGradientUniform SeriesA͞G1 ... 56251 ... 44143 ... 81474 ... 96057 ... 313211 ... 551918 ... 737428 ... 527144 ... 511269 ... 7362135 ... 7582330 ... 9879807... 181972 ... 193081 ... 827523 ... 80000 ... 51200 ... 32770 ... 20970 ... 13420 ... 08590 ... 05500 ... 03520 ... 02250 ... 01440 ... 00740 ... 00300 ... 00120 ... 00050 ... 00030 ... 00011 ... 444440 ... 173440 ... 088820 ... 050400 ... 030070 ... 018450 ... 011500 ... 007240 ... 004590 ... 002920 ... 001190 ...000480 ... 000200 ... 000100 ... 000050 ... 000011 ... 25003 ... 76568 ... 258815 ... 841925 ... 252942 ... 207768 ... 9495109 ... 1085173 ... 0446273 ... 9447538 ... 03291319 ... 953227 ... 717884 ... 76123221925530089918311 ... 694440 ... 423440 ... 338820 ... 300400 ... 280070 ... 268450 ... 261500 ...257240 ... 254590 ... 252920 ... 251190 ... 250480 ... 250200 ... 0250100 ... 250050 ... 250010 ... 250000 ... 44001 ... 36162 ... 95143 ... 32893 ... 57053 ... 72513 ... 82413 ... 88743 ... 92793 ... 95393 ... 98113 ... 99233 ... 99683 ... 99843 ... 99923 ... 99983 ... 00000 ... 66402 ... 20355 ... 77257 ... 02079 ...846011 ... 261712 ... 326013 ... 108514 ... 674114 ... 232615 ... 637315 ... 831615 ... 922915 ... 948115 ... 976615 ... 996915 ... 44440 ... 22491 ... 86832 ... 38722 ... 79712 ... 11453 ... 35593 ... 53663 ... 66983 ... 76673 ... 88613 ... 94573 ... 97463 ... 98583 ... 99213 ... 99803 ... 9997606CompoundInterest Factor Tables30%TABLE 2630%Discrete Cash Flow: Compound Interest FactorsSingle PaymentsUniform Series PaymentsArithmetic GradientsnCompoundAmountF͞PPresentWorthP͞FSinkingFundA͞FCompoundAmountF͞ACapitalRecoveryA͞PPresentWorthP͞AGradientPresentWorthP͞GGradientUniform SeriesA͞G12345678910111213141516171819202224252628303234351 ... 69002 ... 85613 ... 82686 ... 157310 ... 785817 ... 298130 ... 373851 ... 541786 ... 4554146 ... 0496321 ... 8008705 ... 33331550 ... 004427 ... 979727 ... 76920 ... 45520 ... 26930 ... 15940 ... 09430 ...05580 ... 03300 ... 01950 ... 01160 ... 00680 ... 00310 ... 00140 ... 00060 ... 00020 ... 00011 ... 434780 ... 161630 ... 078390 ... 041920 ... 023460 ... 013450 ... 007820 ... 004580 ... 002690 ... 001590 ... 000550 ... 000330 ... 000110 ... 000040 ... 00002 ... 99006 ... 043112 ... 582823 ... 015042 ... 405374 ...6250127 ... 2863218 ... 0139371 ... 9734630 ... 281806 ... 803054 ... 318729 ... 300000 ... 550630 ... 410580 ... 356870 ... 331240 ... 317730 ... 310240 ... 305980 ... 303510 ... 302070 ... 300940 ... 300430 ... 300190 ... 300070 ... 300030 ... 36091 ... 16622 ... 64272 ... 92473 ... 09153 ... 19033 ... 24873 ...28323 ... 30373 ... 31583 ... 32723 ... 32973 ... 33213 ... 33293 ... 59171 ... 55243 ... 66565 ... 48007 ... 88728 ... 91739 ... 64379 ... 142610 ... 478810 ... 701910 ... 943310 ... 004511 ... 068711 ... 094511 ... 43480 ... 17831 ... 76542 ... 21562 ... 55122 ... 79522 ... 96853 ... 08923 ... 17183 ... 22753 ...28903 ... 30503 ... 32193 ... 32883 ... 35001 ... 46043 ... 48406 ... 172211 ... 893720 ... 143936 ... 469766 ... 1585121 ... 3138221 ... 4619404 ... 78861342 ... 782447 ... 118128 ... 74070 ... 40640 ... 22300 ... 12240 ... 06710 ... 03680 ... 02020 ... 01110 ... 00610 ... 00330 ... 00140 ... 00060 ... 00020 ...00011 ... 425530 ... 150760 ... 069260 ... 034890 ... 018320 ... 009820 ... 005320 ... 002900 ... 001580 ... 000870 ... 000260 ... 000140 ... 000040 ... 000010 ... 00002 ... 17256 ... 954414 ... 491928 ... 696454 ... 6967101 ... 4848187 ... 7385344 ... 6109630 ... 74831152 ... 253833 ... 506989 ... 350000 ...589660 ... 450460 ... 398800 ... 375190 ... 363390 ... 357220 ... 353930 ... 352140 ... 351170 ... 350480 ... 350190 ... 350080 ... 350020 ... 350010 ... 28941 ... 99692 ... 38522 ... 59822 ... 71502 ... 77922 ... 81442 ... 83372 ... 84432 ... 85012 ... 85502 ... 85602 ... 85682 ... 85702 ... 54871 ... 26483 ...98284 ... 35155 ... 33636 ... 00497 ... 44217 ... 72067 ... 89467 ... 00178 ... 10618 ... 12968 ... 15178 ... 15948 ... 42550 ... 13411 ... 66981 ... 05972 ... 33382 ... 52052 ... 64432 ... 72462 ... 77562 ... 80752 ... 83932 ... 84652 ... 85352 ... 85592 ... 40001 ... 74403 ... 37827 ... 541414 ... 661028 ... 495756 ...3715111 ... 5681217 ... 9135426 ... 6304836 ... 903214 ... 886299 ... 71430 ... 36440 ... 18590 ... 09490 ... 04840 ... 02470 ... 01260 ... 00640 ... 00330 ... 00170 ... 00060 ... 00020 ... 00011 ... 416670 ... 140770 ... 061260 ... 029070 ... 014320 ... 007180 ... 003630 ... 001850 ... 000940 ... 000480 ...000120 ... 000060 ... 000020 ... 00002 ... 36007 ... 945616 ... 853434 ... 152669 ... 7391139 ... 9287275 ... 4202541 ... 78371064 ... 582089 ... 248033 ... 400000 ... 629360 ... 491360 ... 441920 ... 420340 ... 410130 ... 405100 ... 402590 ... 401320 ... 400670 ... 400240 ... 400090 ... 400030 ... 400010 ...400000 ... 22451 ... 84922 ... 16802 ... 33062 ... 41362 ... 45592 ... 47752 ... 48852 ... 49412 ... 49702 ... 49922 ... 49962 ... 49992 ... 50002 ... 51021 ... 02002 ... 42783 ... 47134 ... 16965 ... 61065 ... 87885 ... 03766 ... 12996 ... 18286 ... 22946 ... 23876 ... 24666 ... 24906 ... 41670 ... 09231 ... 58111 ...91852 ... 14192 ... 28452 ... 37292 ... 42622 ... 45772 ... 47612 ... 49252 ... 49592 ... 49882 ... 49962 ... 50002 ... 37505 ... 593811 ... 085925 ... 443457 ... 4976129 ... 6195291 ... 8939656 ... 26131477 ... 843325 ... 83168342525137877852230 ... 44440 ... 19750 ... 08780 ... 03900 ... 01730 ... 00770 ...00340 ... 00150 ... 00070 ... 00030 ... 00011 ... 400000 ... 123080 ... 048120 ... 020300 ... 008820 ... 003880 ... 001720 ... 000760 ... 000340 ... 000150 ... 000030 ... 000010 ... 00002 ... 75008 ... 187520 ... 171949 ... 8867113 ... 9951257 ... 2390581 ... 78781311 ... 522953 ... 686648 ... 500000 ... 710530... 575830 ... 531080 ... 513350 ... 505850 ... 502580 ... 501140 ... 500510 ... 500230 ... 500070 ... 500020 ... 500010 ... 500000 ... 500000 ... 11111 ... 60491 ... 82441 ... 92201 ... 96531 ... 98461 ... 99311 ... 99701 ... 99861 ... 99941 ... 99991 ... 99992 ... 00002 ... 00002 ... 44441 ... 62962 ... 59532 ...21963 ... 58383 ... 78423 ... 89043 ... 94523 ... 97293 ... 98683 ... 99693 ... 99853 ... 99973 ... 99993 ... 40000 ... 01541 ... 42261 ... 67521 ... 82351 ... 90681 ... 95191 ... 97561 ... 98781 ... 99401 ... 99861 ... 99931 ... 99981 ... 00002 ... Chapter 2Opener: Royalty-Free/CORBIS ... Chapter 4Opener: ChadBaker/Getty Images ... Chapter 6Opener: Stockdisc ... Chapter 8Opener: PhotoLink/Getty Images ... Chapter 10Opener: © Digital Vision/PunchStock ... Chapter 12Opener: Ariel Skelley/Blend Images/Getty Images ... Chapter 14Opener: Photodisc/Getty Images ... Chapter 16Opener: Arthur S ... Chapter17Opener: Photodisc/Getty Images ... Chapter 19Opener: © Brand X/JupiterImages ... See also Gradient, arithmeticAlternative depreciation system (ADS), 426–27Allocation variance, 399Alternativescost, 131defined, 6do-nothing, 130–31independent, 130–31, 132, 157, 203, 207, 244, 247 (seealsoCapital budgeting)infinite life, 138, 157–60mutually exclusive, 130–31, 132, 155, 208, 238, 248revenue, 131selection, 6service, 131in simulation, 533–40Amortization, 415Annual interest rateeffective, 99–105nominal, 99–105Annual operating costs (AOC), 6, 153, 297–98and estimation, 388AnnualPercentage Rate (APR), 97Annual Percentage Yield (APY), 97Annual worthadvantages, 151after-tax analysis, 456–58of annual operating costs, 297–98and B/C analysis, 235, 238–42and breakeven analysis, 345–48and capital-recovery-plus-interest, 153–54equivalent uniform, 151evaluation by, 155–60and EVA, 153, 465–68and future worth, 151and incremental rate of return, 213–14of infinite-life projects, 157–60and inflation, 151, 377–78and present worth, 151and rate of return, 175, 213–14and replacement analysis, 302–06, 307–10, 462–65spreadsheet solutions, 156, 159, 458when to use,262AOC ... See Gradient, arithmeticAssets ... See Expected valueAverage cost per unit, 345Average tax rate, 447BBalance sheet, 267, 561–63Base amountdefined, 50and shifted gradients, 80–82Basis, unadjusted, 416B/C ... See Normal distributionBenefit and cost difference, 236Benefit/costratiocalculation, 235–36conventional, 235incremental analysis, 238–39modified, 235–36for three or more alternatives, 242–46for two alternatives, 238–42when to use, 262Benefitsdirect versus implied, 242in public projects, 231, 235, 242␤ (beta), 274Bondsand debt financing, 271–72and inflation, 372,385interest computation, 190payment periods, 190present worth, 191–92for public sector projects, 232rate of return, 190–92types, 191Book depreciation, 415, 417, 466Book valuedeclining balance method, 419–22defined, 416double declining balance method, 419–22and EVA, 466612IndexBook value(continued)MACRS method, 423versus market value, 416straight line method, 418sum-of-years-digits method, 430, 557unit-of-production method, 431Borrowed money, 267Borrowing rate, 185–87Bottom-up approach, 388–89Breakeven analysis ... i graphaverage cost per unit, 345fixed costs, 341andGoal Seek, 353–54and make-buy decisions, 341, 347and payback, 351–54and rate of return, 210–12, 460versus sensitivity analysis, 341, 485single project, 341–45spreadsheet application, 212, 352–54three or more alternatives, 348two alternatives, 345–47variable costs, 341Breakeven point, 341,489Budgeting ... See also Cost of capitaldebt, 27, 267, 271–73equity, 27, 267, 273–75mixed (debt and equity), 275–77Capital gainsdefined, 454short-term and long-term, 454taxes for, 454Capital lossesdefined, 454taxes for, 454Capital rationing ... See also A/P factor; Depreciationdefined, 153andeconomic service life, 297–99and EVA, 468and inflation, 377–78and replacement analysis, 306Capital recovery factor, 43–45and equivalent annual worth, 153Capitalized costin alternative evaluation, 138–42and annual worth, 138and public projects, 244–46CAPM ... See Cash flow after taxesCFBT ...See also CompoundingCompoundingannual, 99–105continuous, 114–16and effective interest rate, 96frequency, 97–98interperiod, 113–14period, 97–98and simple interest, 21–23Compounding periodcontinuous, 114–16defined, 97and effective annual rate, 102and payment period, 106–14Concepts,fundamental, summary, 573–76Contingent projects, 323Continuous compounding, 114–16Contracts, types, 234Conventional benefit/cost ratio, 235Conventional cash flow series, 180Conventional gradient, 51–57Corporationsand capital, 4financial worth, 466–68leveraged, 275–77Cost alternative, 131,204, 216, 457, 460Cost-effectivenessanalysis, 246–50ratio, 246Cost, life-cycle, 160–63Cost-capacity equations, 394–95Cost centers, 397, 401Cost components, 387–88Cost depletion, 427–29Cost drivers, 401Cost-estimating relationships, 394–97Cost estimationaccuracy, 389approaches, 388–89cost-capacity method, 394–95and cost indexes, 391–94factor method, 395–97and inflation, 12, 377unit method, 390Cost of capitaland debt-equity mix, 269–71for debt financing, 271–73defined, 26, 267for equity financing, 273–75versus MARR, 26, 267weighted average, 27, 270, 275Cost of goods sold, 397,562–63, 565Cost of invested capital, 466–67Costs ... See also Depreciation recapture; Rate ofdepreciation; Replacement analysisaccelerated, 416, 419ACRS, 422alternative system, 426–27and amortization, 415basis, 416book, 415, 417, 466declining balance, 419–22, 550–51defined, 415doubledeclining balance, 419–22, 551and EVA, 466general depreciation system (GDS), 426half-year convention, 416, 424, 427and income taxes, 415, 417, 445–68MACRS, 417, 422–27present worth of, 432–35property class, 426recovery period for, 416, 418, 426rate of, 416recovery rate,416613614IndexDepreciation (continued)straight line, 418–19, 557straight line alternative, 426–27sum-of-years digits, 430, 557switching methods, 432–38, 557–58tax, 415, 417unit-of-production, 431Depreciation recapturedefinition, 453in replacement studies, 462–65and taxes, 453, 461Descartes’ rule,181–84, 219Design-build contracts, 234Design stages, preliminary and detailed, 161, 163, 389Design-to-cost approach, 388–89Direct benefits, 242, 244Direct costs, 387, 390–97Disbenefits, 231, 235Disbursements, 15, 177Discount rate, 129, 232Discounted cash flow, 129Discounted payback analysis,349Discrete cash flowscompound interest factors (tables), 581–609discrete versus continuous compounding, 114–16and end-of-period convention, 15–16Disposal phase, 161Distribution ... See EquivalenceEconomic service life (ESL), 294, 296–302Economic value added, 153, 465–68EFFECT function,103, 551Effective interest rateannual, 99–100for any time period, 105–06of bonds, 191–92and compounding periods, 100, 105for continuous compounding, 114–16defined, 96and nominal rate, 96–97Effective tax rate, 447, 462Efficiency ratios, 563End-of-period convention, 15–16EngineeringeconomyConcepts, summary, 573–76defined, 3study approach, 4–7terminology and symbols, 13Equal service requirement, 131, 151, 213, 217, 240, 457Equity financing, 26, 267cost of, 273–75Equivalencecalculations without tables, 569–72compounding period greater than payment period, 112–14compounding period less than payment period, 107–12defined, 19–21Equivalent uniform annual cost ... See Annual worthError distribution ... See Annual worthEUAW ... See also Spreadsheet, usage in examplesabsolute cell reference, 9, 547, 550basics, 547–50charts, 548–49displaying functions,30embedding functions, 75, 156–57error messages, 560functions, in engineering economy, 28, 550–58Goal Seek tool, 558–59introduction, 27–30, 547–50and linear programming, 330–31random number generation, 556Solver tool, 559–60spreadsheet layout, 549–50Expected valuecomputation, 492–94, 526–27, 530and decisions under risk, 517and decision trees, 497–98defined, 492, 526and real options, 501–02in simulation, 538–39Expenses, operating, 6, 153, 297–98External rate of return, 185–90 ... See also BondsFixed percentage method ... See also Single payment factorsFuture worthandannual worth, 151and effective interest rate, 100evaluation by, 137and inflation, 374–77and multiple rates of return, 186–87from present worth, 137of shifted series, 73, 77–78and spreadsheet solutions, 42, 48when to use, 262FV function, 28, 552and shifted uniform series, 78–79and single payments,41and uniform series, 46GGains and losses, 454Gaussian distribution ... See Public sector projectsGradient, arithmeticbase amount, 50conventional, 51decreasing, 51, 56defined, 50factors, 52–54increasing, 51–55shifted, 80–82, 83–86spreadsheet use, 57Graduated tax rates, 446–47, 448Grossincome, 445, 448615HHalf-year convention, 416, 424, 427Highly leveraged corporations, 275–77Hurdle rate ... See First costInstallment financing, 175Intangible factors, 6 ... See also Interest rate(s)compound, 22, 24, 29continuous compounding, 114–16defined, 10interperiod, 112–14rate, 10, 12simple,21, 24Interest period, 10, 12Interest rate(s) ... See also Rate of returnand annual worth, 175definition, 173versus external ROR, 185and present worth, 175spreadsheet solution, 177International aspectscontracts, 234corporate taxes, 468–70cost estimation, 388–89deflation, 368–69depreciation, 416–17,

470–72dumping, 368inflation aspects, 368, 377value-added tax, 470–72Interperiod interest, 112–14Interpolation, in interest rate tables, 48–50Inventory turnover ratio, 564–65Invested capital, cost of, 466–67Investment(s) ... S ... See also Annualoperating costMake-or-buy decisions, 341, 347 ... SeeMinimum attractive rate of returnMean ... See Economic service lifeMIRR function, 186, 553–54M&O costs ... See DepletionNet cash flow, 15Net operating income (NOI), 446, 459Net investment procedure, 187–90Net operating profit after taxes (NOPAT), 446, 466Net present value ... i graphs, 178–79sensitivity analysis, 486–87, 489and shifted series, 75, 86NSPE (National Society of Professional Engineers), 7,404, 566–68OObsolescence, 294One-additional-year replacement study, 302–05Operating costs ... See Indirect costsOwner’s equity, 267, 561PP, 13, 40P/A factor,43, 53 ... See PaybackanalysisPercentage depletion ... See also Gradient, arithmeticPhaseout phase, 161Planning horizon ... See Present worthPresent worthafter-tax analysis, 456–58and annual worth, 151assumptions, 134and B/C analysis, 235of bonds, 191–92and breakeven analysis, 345–46and capital budgeting, 325–29of depreciation, 432for equal lives, 132–33evaluation method, 261–62geometric gradient series, 82–86income taxes, 451–53and independent projects, 325–29index, 332and inflation, 369–74and multiple interest rates, 181–84and profitability index, 237and rate of return, 175–76, 182–84, 458–62andsensitivity analysis, 486–90in shifted series, 73, 76, 80–86in simulation, 534–39single-payment factor, 40–41for unequal lives, 133–37Present worth factorsgradient, 50–53, 58–60single payment factor, 40–41uniform series, 43–45Probabilityin decision trees, 495–98defined, 492, 518and expected value,492–94, 526–27and standard deviation, 528Probability distributionof continuous variables, 519–22defined, 519of discrete variables, 519–20properties, 526–28, 530and samples, 523–26in simulation, 533–39Probability node, 495Productive hour rate, 399Profitability index, 237, 332Profitability ratios,563Profit-and-loss statement, 562Project net-investment, 187–90Property class, 426Property of independent random variables, 533Public-private partnerships, 234–35Public sector projects, 230–35and annual worth, 157–58B/C analysis, 235–38capitalized cost, 138–42characteristics, 231–32design-build contracts, 234–35profitability index, 237public-private partnerships, 234–35Purchasing power, 367, 374, 376PV function, 28, 44, 556versus NPV function, 556and present worth, 41, 44and single payment, 41and uniform series present worth, 44PW vs ... See also Incremental rate of returnafter-tax,458–62and annual worth, 175, 213–14, 218of bonds, 190–92, 272breakeven, 210–12, 460in capital budgeting, 332–34cautions, 179–80on debt capital, 272–73, 276defined, 12, 173, 175on equity capital, 273–74, 276evaluation method, 261–62external, 185–90on extra investment, 206incremental, 206,207–10and independent projects, 207and inflation, 12–13, 368, 374–75Indexinstallment financing, 175internal, 173, 175, 185minimum attractive (see Minimum attractive rate of return)modified ROR approach, 185–87multiple, 180–90and mutually exclusive alternatives, 206, 207–18and present worth,175, 177–79, 207–09, 218ranking inconsistency, 213, 216, 462return on invested capital (ROIC) approach, 185, 187–90spreadsheet solution, 177, 179, 211–13, 216–19Ratios, accounting, 563–65Real interest rate, 368, 370, 374–75Real options, 498–503and decision trees, 500definition, 499Realproperty, 416, 423–24, 426Recovery perioddefined, 416effect on taxes, 452–53MACRS, 423–24, 426–27straight line option, 426Recovery rate ... See Investment rateRepayment of loans, 24–25Replacement analysis, 292–313, 462–65after-tax, 462–65annual worth, 295, 302–06and capital losses, 454,462cash flow approach, 306depreciation recapture, 462–64and economic service life, 296–99, 305first costs, 294–96and marginal costs, 300–01market value, 294, 301need for, 294one-additional year, 302–05opportunity cost approach, 306overview, 302and study periods, 307–12sunk costs,295terminology, 294viewpoint, 295Replacement life ... See Economic service lifeReturn on assets ratio, 564Return on invested capital, 187–90Return on investment (ROI), 12, 173 ... See Return on investmentRoot mean square deviation, 528ROR ... See also Market valueand capital recovery, 153,297defined, 6, 153and depreciation, 416, 418, 420, 423, 430and market value, 294, 297–98and public projects, 235in PW analysis, 132, 134–37in replacement analysis, 294, 297, 303–06, 464and trade-in value, 294, 464Sampling, 523–26Savings, tax, 449, 463–65Scatter charts ... See also Breakevenanalysisdescription, 485and Excel cell referencing, 29, 547of one parameter, 485–87spider graph, 488with three estimates, 490–91two alternatives, 488–90Service alternative ... See also Excelannual worth, 155–56, 159, 218, 299, 301, 304,305, 353, 458, 467B/C analysis, 245breakeven analysis, 352,353–54cash flow after tax (CFAT), 450, 456, 458, 461, 462compound interest, 29–30depreciation, 422, 425, 434EVA, 467and factor values, 49independent projects, 328, 331inflation, 372layout, 549–50multiple attributes, 283nominal and effective interest, 103–04619620IndexSpreadsheet(continued)present worth, 136, 209, 212, 218, 311, 354, 458,461, 489, 501, 539rate of return, 63, 182, 184, 189, 192, 209, 212,217, 218, 461, 462, 501replacement analysis, 299, 301, 304, 305, 311, 465replacement value, 312sensitivity analysis, 487, 489simulation, 538–39Staged funding, 494–503Standard deviationfor continuous variable, 530definition, 527–28for discrete variable, 528–29Standard normal distribution, 531–33StocksCAPM model, 274common, 267, 274in equity financing, 273–75, 276preferred, 267, 273Straight line alternative, in MACRS, 426–27Straight line depreciation, 418–19Straight line rate, 418Study periodand AW evaluation, 155and equal service, 133and FW analysis, 137and PW evaluation, 133–34and replacement analysis, 302, 307–11and salvage value, 153spreadsheet example, 136, 310–11Sum-of-years digits depreciation, 430Sunk costs, 295SYD function, 430,557System, phases of, 160–61UTWACC ... See After-tax; Income tax; Taxable incomeTime, 13Time value of moneydefined, 4and equivalence, 19factors to account for, 39–61and no-return payback, 349–50Total cost relation, 342–45 ... See also Market value; Salvage valueTreasury securities, 26,190Triangular distribution, 520, 522Unadjusted basis, 416Uncertainty, 515, 517Uniform distribution, 520–21, 535, 538Uniform gradient ... See Declining balance depreciationUniform seriescompound amount (F/A) factor, 46compounding period greater than payment period, 112–14compounding periodless than payment period, 109–12description, 13present worth (P/A) factor, 43shifted, 73–80Unit method, 390–91Unit-of-production depreciation, 431Unknown interest rate, 61–63Unknown years (life), 61, 63–64Unrecovered balance, 173–74VValue, resale, 6 ... See Economic value addedValue-addedFind more such useful books onwww ... netLearn more and make your parents proud :)Regardswww ... netTitle: Solution manual of Engineering Economics 7th Ed Description: ... 2 ... 2 ... 2 ... 2 ... 2 ... 0؊tax, 470–72Variable ... Place a minus in front of thefunction to retain the same sign ... 5),MIN(n, tSolution manual of Engineering Economics 7th Ed by Leland Blank and Anthony

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