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Introduction to Operations Research: Theory and Applications

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Halidi Lyeme

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PREFACE

In a competitive business environment, it has become essential for the prosperity

and growth in the field of Operation Research. The growing importance of new

techniques has been emphasized the need for the developing operation to research

models to provide practical utility. These Operation Research models which con-

stitute the subject matters to develop in readers an understanding of problem

solving methods.

Each chapter begins with introduction, interesting examples and activities at the

end of each chapters. This book will be of immense use for all those who want to

learn how to analyse operation research situation to arrive at optimum decision.

This should be of equal interest to students, professionals and the interested

readers.

Every effort has been made to present the subject matter in easy, clear and sys-

tematic manner. The book is intended to serve as a textbook for students of B.Sc

(Education, Computer Science and Statistics) , BBA who need to understand the

basic concepts of Operations Research and apply them. It also suits the require-

ment for MBA, MSc (Mathematics, Information Technology and Statistics) who

need both theoretical and practical knowledge of Operations Research.

We would like to thank the publisher for the efficient and throughly professional

way in which the whole task was completed. We also thank our family members

for their constant encouragement for writing this book.

Any suggestions to improve in contents or in style are always welcome and will

be appreciated and acknoledged.

Lyeme & Selemani

ii

DEDICATION

To my late son Haris.

iii

TABLE OF CONTENTS

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i

Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii

Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii

List of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii

CHAPTER ONE: General Concept of Operations Research 1

1.1 Background (History of Operation Research) . . . . . . . . . . . . . 1

1.2 Nature and Definition of Operations Research . . . . . . . . . . . . 3

1.3 Characteristics of Operation Research . . . . . . . . . . . . . . . . . 5

1.3.1 System Orientation of Operation Research . . . . . . . . . . . . . . 5

1.3.2 The Use of Interdisciplinary Team. . . . . . . . . . . . . . . . . . . 5

1.3.3 Application of Scientific Method . . . . . . . . . . . . . . . . . . . 5

1.3.4 Quantitative Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3.5 Human Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 Phases of Operation Research . . . . . . . . . . . . . . . . . . . . . 5

1.4.1 The Formulation of the Problem . . . . . . . . . . . . . . . . . . . 6

1.4.2 Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4.3 Driving the Solution from the Model . . . . . . . . . . . . . . . . . 6

1.4.4 Testing the Model and Its Solution . . . . . . . . . . . . . . . . . . 7

1.4.5 Controlling the Solution . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4.6 Implementation of Model . . . . . . . . . . . . . . . . . . . . . . . 7

1.5 Quantitative Techniques of Operation Research . . . . . . . . . . . . 8

1.6 Scope of Operation Research . . . . . . . . . . . . . . . . . . . . . . 9

iv

1.6.1 In Industry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.6.2 In Defence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.6.3 In Planning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.6.4 In Agriculture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.6.5 In Public Utilities . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.7 Model in Operation Research . . . . . . . . . . . . . . . . . . . . . . 10

1.7.1 Physical Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.7.2 Symbolic Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.7.3 Heuristic Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.8 Advantages and Limitations of Operation Research . . . . . . . . . 11

1.9 Applications of Operation Research . . . . . . . . . . . . . . . . . . 12

CHAPTER TWO: LINEAR PROGRAMMING PROBLEM/MODEL 13

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 Linear Programming Problem . . . . . . . . . . . . . . . . . . . . . 13

2.3 Mathematical Formulation of a LPP . . . . . . . . . . . . . . . . . . 14

2.4 Matrix Form of LPP . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.5 Procedure for Formulation of LP Problems . . . . . . . . . . . . . . 15

2.6 Some Important Definitions in LPP . . . . . . . . . . . . . . . . . . 18

2.7 Solution of a LPP . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.8 Advantages of Linear Programming Techniques . . . . . . . . . . . . 19

2.9 Activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

CHAPTER THREE: GRAPHICAL METHOD 21

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2 Procedures for Solving LPP by Graphical Method . . . . . . . . . . 21

v

3.3 Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

CHAPTER FOUR: THE SIMPLEX METHOD 35

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.2 Standard Form of an LPP . . . . . . . . . . . . . . . . . . . . . . . 36

4.3 The Simplex Method . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.3.1 Maximization Case . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.3.2 Minimization Case . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.3.3 The Two-Phase Method . . . . . . . . . . . . . . . . . . . . . . . . 48

4.3.4 The Big - M Method . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.4 Degeneracy in Simplex Method . . . . . . . . . . . . . . . . . . . . . 61

4.5 Types of Linear Programming Solution . . . . . . . . . . . . . . . . 64

4.5.1 Alternative (Multiple) Optimal Solution . . . . . . . . . . . . . . . 64

4.5.2 Unbounded Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 66

4.5.3 Infeasible Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

CHAPTER FIVE: DUALITY IN LINEAR PROGRAMMING 70

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

5.2 Formulation of Dual Linear Programming Problem . . . . . . . . . . 70

5.2.1 Rules for Constructing the Duality from Primal . . . . . . . . . . . 71

5.2.2 Primal - Dual Relationship . . . . . . . . . . . . . . . . . . . . . . 73

5.3 Standard Results on Duality . . . . . . . . . . . . . . . . . . . . . . 73

5.4 Significant of Duality . . . . . . . . . . . . . . . . . . . . . . . . . . 74

5.5 Advantages of Duality . . . . . . . . . . . . . . . . . . . . . . . . . . 74

CHAPTER SIX: SENSITIVITY ANALYSIS IN LINEAR PRO-

GRAMMING 76

vi

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

6.2 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

6.2.1 Change in Objective Function Coefficient (Cj) . . . . . . . . . . . . 76

6.2.2 Change in the Availability of Resources (bi) . . . . . . . . . . . . . 79

6.2.3 Change in the Input-Output Coefficient (a′ijs) . . . . . . . . . . . . 81

6.2.4 Addition of a New Variable (Column) . . . . . . . . . . . . . . . . 82

6.3 Solving LPP using LINDO . . . . . . . . . . . . . . . . . . . . . . . 83

6.4 Using LINDO to Solve LPP . . . . . . . . . . . . . . . . . . . . . . 84

6.5 Interpretation of LINDO Output . . . . . . . . . . . . . . . . . . . . 84

6.6 Tests and Final Examination Questions . . . . . . . . . . . . . . . . 85

REFERENCES 89

vii

LIST OF FIGURES

1.1 Phases of Operation Research . . . . . . . . . . . . . . . . . . . . . 8

4.1 Structure of an algorithms . . . . . . . . . . . . . . . . . . . . . . . 35

viii

LIST OF TABLES

4.1 Initial Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.2 Improved Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.3 Improved Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.4 Optimal Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.5 Initial Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.6 Improved Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.7 Optimal Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.8 Initial Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.9 Improved Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.10 Improved Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.11Modified Simplex Table . . . . . . . . . . . . . . . . . . . . . . . . 52

4.12 Initial Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.13 Optimal but not Feasible Solution . . . . . . . . . . . . . . . . . . 54

4.14 Initial Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.15 Improved Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.16 Improved Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.17 Optimal Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.18 Initial Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.19 Improved Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.20 Optimal Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.21 Initial Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.22 Optimal Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

4.23 Optimal Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.24 Alternative Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 66

ix

4.25 Initiall Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.26 Improved Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.27 Initial Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

4.28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

5.1 Primal-Dual Relationship . . . . . . . . . . . . . . . . . . . . . . . 73

6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

6.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

6.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

1

CHAPTER ONE

General Concept of Operations Research

1.1 Background (History of Operation Research)

I. Pre-World war II:

The roots of OR are as old as science and society. Though the roots of OR

extend to even early 1800s, it was in 1885 when Ferderick W. Taylor emphasized

the application of scientific analysis to methods of production, that the real start

took place. Another man of early scientific management era was Henry L. Gantt.

Most job scheduling methods at that time were rather haphazard. A job, for

instance, may be processed on a machine without trouble but then wait for days

for acceptance by the next machine. Gantt mapped each job from machine to

machine, minimizing every delay. Now with the Gantt procedure it is possible to

plan machine loadings months in advance and still quote delivery dates accurately.

In 1917, A.K.Erlang, a Danish mathematician, published his work on the problem

of congestion of telephone traffic. The difficulty was that during busy periods,

telephone operators were many, resulting in delayed calls. A few years after its

appearance, his work was accepted by the British Post Office as the basis for

calculating circuit facilities. The well known economic order quantity model is

attributed to F.W. Harris, who published his work on the area of inventory control

in 1915.

During the 1930s, H.C. Levinson, an American astronomer, applied scientific anal-

ysis to the problems of merchandising. His work included scientific study of cus-

tomers’ buying habits, response to advertising and relation of environment to the

type of article sold.

However, it was the First Industrial Revolution which contributed mainly to-

wards the development of OR. Before this revolution, most of the industries were

small scale, employing only a handful of men. The advent of machine tools-

the replacement of man by machine as a source of power and improved means of

transportation and communication resulted in fast flourishing industry. It became

increasingly difficult for a single man to perform all the managerial functions ( of

planning, sale, purchase, production, etc.). Consequently, a division of manage-

ment function took place. Managers of production, marketing, finance, personnel,

research and development etc., began to appear. With further industrial growth,

further subdivisions of management functions took place. For example ,produc-

tion department was sub-divided into sections like maintenance, quality control,

2

procurement, production planning, etc.

II. World War II:

During World War II, the military management in England called on a team of

scientists to study the strategic and tactical problems of air and land defence.

This team was under the direction of Professor P.M.S. Blackett of University of

Manchester and a former naval officer. ‘Blackett circus’, as the group was called,

included three physiologist, two mathematical physicists, one astrophysicist, one

army officer, one surveyor, one general physicist and two mathematicians. Many

of these problems were of the executive type. The objective was to find out the

most effective allocation of limited military resources to the military operations

and to the activities within each operation.

The application included the effective use of newly invented radar, allocation of

British Air Force Planes to missions and the determination of best patterns for

searching submarines. This group of scientists formed the first OR team.

The name operations research ( or operational research) was apparently coined

because the team was carrying out research on (military) operation, the encour-

aging results of these effort led to the information of more such teams in British

armed services and the use of scientific teams soon spread to western allies-the

united states, Canada and France. Thus through this scince of operation research

originated in England, the united states soon took the lead, in united state these

OR teams helped in developing strategies from mining operations, inventing new

flight patterns and planning of sea mines.

III. Post-world war II :

Immediately after the war, the success of military teams attracted the attention

of industrial managers who were seeking solutions to their problems. Industrial

operation research in U.K. and U.S.A. developed along different lines. In U.K.,

the critical economic situation required drastic increase in production efficiency

and creation of new markets. Nationalization of a few key industries further

increased the potential field for OR. Consequently OR soon spread from military

to government, industrial, social and economic planning.

In U.S.A. the situation was different. Impressed by its dramatic success in U.K.,

defense operations research in U.S.A was increased. Most of the war experienced

OR workers remained in military service. Industrial executives did not call for

much help because they were returning to the peace-time situation and many of

them believed that it was merely a new application of an old technique. Opera-

tion research by a variety of names in that country such as operational analysis,

operation evaluation, systems analysis, system evaluation, system research and

3

management science.

The progress of industrial operational research in U.S.A. was due to advent of sec-

ond industrial revolution which resulted in automation-the replacement of man by

machine as a source of control, the new revolution began around 1940s when elec-

tronic computers became commercially available. The electronic brains processed

tremendous computational speed and information storage. But for these digital

computers, operation research with its complex computational problems could

not have achieved its promising place in all kinds of operational environments.

In 1950, OR was introduced as a subject for academic study in American univer-

sities since then this subject has been gaining ever increasing importance for the

students of Mathematics, Statistics, Commerce, Economics, Management and En-

gineering. To increase the impact of operation research, the Operations Research

Society of America was formed in 1950. In 1953, the Institute of Management

Sciences (IMS) was established. Other countries followed suit and in 1959 Inter-

national Federation of OR began to appear. Some of them ( in English ) are:

� Operations Research

� Opsearch

� Operational Research Quarterly

� Management Science

� Transportation Science

� Mathematics of Operations Research

� International Journal of Game Theory, etc

Today, the impact of operations research can be felt in many areas. This is shown

by the ever increasing number of educational institutions offering this subject at

degree level. Of late, OR activities have spread to diverse fields such as hospitals,

libraries, Planning, transportation systems, management, defense, etc

1.2 Nature and Definition of Operations Research

Defining Operations Research itself is very difficult. Like many other subjects

that developed pragmatically and shade imperceptibly into adjoining subjects, it

is more easily recognized than defined.

4

Generally speaking, operations research is an approach to the analysis of opera-

tions that to a greater or lesser extent adopts:

(i) Scientific method (observation, hypothesis, deduction and experimentation

as far as possible).

(ii) The explicit formulation of complex relationships.

(iii) An inter-disciplinary nature.

(iv) A non-partisan attitude.

Operational Research can also be regarded as a scientific approach to the analysis

and solution of management problem.

The council of the United Kingdom Operational Research Society defines Oper-

ational Research as “the attack of modern science on complex problems, arising

in the direction and management of large systems of men, machines, materials

and money in industry, business, government and defence. It goes on to state the

distinctive approach as to develop a scientific model of the system; incorporating

measurement of factors such as chance and risk, in order to predict and compare

the outcomes of alternative decisions, strategies and controls. The purpose is to

help management to determine its policy and action scientifically”.

Daellenbach and George, (1978) defined Operation Research as the systematic

application quantitative methods, techniques and tools to the analysis of problems

involving the operation of systems.

Other definitions imply view of Operation Research as being the collection of

models and methods which have developed largely independent of one another.

Thierauf and Klekamp, (1975) defined Operations Research utilizes the planned

approach (updated scientific method) and an interdisciplinary team in order to

represent complex functional relationships as mathematical models for the pur-

pose of providing a quantitative basis for decision-making and uncovering new

problems for quantitative analysis.

It is also worth pointing out that an Operations Research project is often a team

effort that involved people drawn from many different backgrounds including:

Accountants, Engineers, Mathematicians, Statisticians and Scientist as well as

the operations research experts themselves.

5

1.3 Characteristics of Operation Research

1.3.1 System Orientation of Operation Research

One of the most important characteristics of Operations Research study is its

concerned with problem as a whole or its system orientation. This means that an

activity by any part of an organization has some effect on the activity of every

part. Therefore, to evaluate any decision one must identify all possible interactions

and determine their impact on the organization as a whole.

1.3.2 The Use of Interdisciplinary Team.

Operations Research study is performed by a team of scientists whose individuals

members have been drawn from different scientific and engineering disciplines.

For example, one may find a mathematician, statistician, physicist, psychologist,

economist and engineers working together on an Operations Research problem.

1.3.3 Application of Scientific Method

Sometimes, we have to use the scientific method for solving the problem of Op-

erations Research. It is not related to laboratories experiment like physics or

biology or chemistry but it related by to the real life experiment. For example, no

company can risk its failure in order to conduct a successful experiment. Though,

experimentations on subsystem is some time resorted to, by and large, a research

approach that does not involve experimentation on the total system is preferred.

1.3.4 Quantitative Solutions

It provides the management with a quantitative basis for decision making.

1.3.5 Human Factor

Human factor is an important component of the Operations Research study.

Without human factor Operations Research study is incomplete.

1.4 Phases of Operation Research

Operations Research study generally involves the following phases;

6

1.4.1 The Formulation of the Problem

To find the solution of the Operations Research problem, you must have to formu-

late the problem in the form of an appropriate model. The following information

will be required for this;

a) Decision Maker

b) Objective

c) Controllable Factors (Variables)

d) Uncontrollable Factors (Variable)

e) Restrictions or Constraints

It might be of a functional nature as in linear programming or have a logical

structure as in simulation and algorithms. E.g.

Minimize C = 4x+ 5y (1.1)

Subject to:

x+ 3y ≥ 6 (1.2)

x+ y ≥ 3 (1.3)

x, y ≥ 0, (1.4)

which is a linear programming model.

1.4.2 Data Collection

It involves obtaining quantitative data either from existing records or a new survey

that fits well into the constructed model of the problem.

1.4.3 Driving the Solution from the Model

This involves the manipulation of the model to arrive at the best (optimal) so-

lution to the problem. It may require solving some mathematical equations for

optimal decisions as in calculus or linear programming models. It may also be

a logical approach or a functional approach which does not require solving a

mathematical equation, such as in queuing theory. The optimal solution is then

determined by some criteria.

7

1.4.4 Testing the Model and Its Solution

After getting solution, it is necessary to test the solution for errors if any. This

may be done by re-examining the formulation of the problem and comparing it

with the model that may help to reveal any mistakes.

1.4.5 Controlling the Solution

This phase establishes controls over the solution with any degree of satisfaction.

The model requires immediate modification as soon as the controlled variables

(one or more) change significantly, otherwise the model goes out of control. As

the conditions are constantly changing in the world, the model and the solution

may not remain valid for a long time.

1.4.6 Implementation of Model

The final phase of an Operations Research is to implement the optimum solu-

tion derived by the Operations Research team. As the conditions are constantly

changing in the world, the model and the solution may not remain valid for a long

time. Therefore, as the change occurs, it has to be detected as soon as possible

so that the model, its solution and the resulting course of action can be modified

accordingly. See the figure below

8

Figure 1.1: Phases of Operation Research

1.5 Quantitative Techniques of Operation Research

Operation Research as its name suggests, gives stress on analysis of operations

as a whole. For this purpose, it uses any suitable techniques or tools available

from the fields of mathematics, statistics, cost analysis or numerical calculations.

Some of these techniques are listed bellow;

i. Linear Programming

ii. Non-linear Programming

9

iii. Integer Programming

iv. Dynamic Programming

v. Goal Programming

vi. Game Theory

vii. Inventory Control

viii. Simulation

ix. Queuing Theory

1.6 Scope of Operation Research

1.6.1 In Industry

Operation Research has been successfully applied in industry in the fields of pro-

duction, blending product mix, inventory control, demand forecast, sale and pur-

chase, transportation, repair and maintenance, scheduling and sequencing, plan-

ning and control of projects etc.

1.6.2 In Defence

Operation Research has a wide scope for application in defence operations. All

the defence operations are carried out by a different agencies, namely air force,

army and navy. Operation Research helpful for achieving the desired goals of

different agencies.

1.6.3 In Planning

Operation Research is helpful for planning of various activities of the organization.

Planning is the important function of management, without effective planning,

we cannot achieve the desired goals.

1.6.4 In Agriculture

Operation Research needs to be equally developed in agriculture sector on national

or international basis. Every country is facing the problem of optimum allocation

of land to various crops in accordance with the climatic conditions and optimum

distributions of water from various resources like canal for irrigation purposes.

Thus, there is a need of determining best policies under the prescribed restrictions.

10

1.6.5 In Public Utilities

Operation Research is directly applicable to business and society. It is also

equally applicable for big and small organization. It has been extensively used in

petroleum, paper chemical, metal processing, aircraft, transport and distribution,

mining and textile industries.

1.7 Model in Operation Research

When we present a real life situation in some abstract form whether physical or

mathematical, bringing out the relationships of its important ingredients, we call

it as model. Thus, model need not described all the aspects of this situation,

but it should signify and identify important factors and their interrelationships

to describe the total situation.

There are number of models used in Operation Research. Some of the basic types

are described below.

1.7.1 Physical Models

These models provide a physical appearance of the real object under the study

of either reduced in size or scaled up. Physical models are useful only in design

problems because they are easy to observe, build and describe. Physical models

are classified into the following two categories.

i. Iconic Models: These models represents the system as it is but in different

size. Thus, Iconic Models are obtained by enlarging or reducing the size

of the system. In other words, they are images, examples of iconic models

are blueprints of a home, maps, globes, photographs, drawings, air planes,

trains, etc.

ii. Analogy Models: These models do not look like the real situation but rep-

resent and behave like a system under study. For example, the organiza-

tion chart represents the structure, authority and responsibilities relationship

with boxes and arrows and maps in different colors represent water, desert

and other geographical features.

1.7.2 Symbolic Models

These models use symbols (i.e letters, numbers) and functions to represent vari-

ables and their relationship to describe the properties of the system. These mod-

11

els are also used to represent relationships which can be represented in a physical

form. Symbolic models can be classified into two categories.

i. Verbal Models: These models describe a situation in written or spoken lan-

guage. Written sentences, books, etc are examples of verbal models.

ii. Mathematical Models: These models involve the use of mathematical sym-

bols, letters, numbers and mathematical operators (+, -, x,) to represent

relationships among various variables of the system to describe its properties

or behaviour.

1.7.3 Heuristic Models

These models use intuitive rules or guidelines to solve a particular problem. These

models are not based on any definite mathematical expression or relationships,

but problem solving based on past experience or approach formulated on the basis

of definite stepped procedure. These models need an ample amount of creativity

and experience by the decision maker.

1.8 Advantages and Limitations of Operation Research

Operation Research is useful for improving quality of managerial decision making.

By using various tools and techniques of Operation Research we can get optimal

solution of the problem. However, besides certain advantages, Operation Research

has some limitations.

Advantages

a) It compels the decision maker to be quite explicit about his objective, as-

sumptions and his perspective to constraints.

b) It makes the decision maker to very carefully about what variables influence

the decisions.

c) Quickly points out gaps in the data required to support workable solutions

to a problem.

d) Its models can be solved by a computer, thus the management can get enough

time for decisions that require quantitative approach.

12

Limitations

a) Often solution to a problem is derived either by making it simplified or

simplifying assumptions and thus, such solutions have limitations.

b) Sometimes models do not represent the realistic situations in which decisions

must be made.

c) Often decision maker is not fully aware of the limitations of the models that

he is using.

d) Many real world problems just cannot have an OR solution.

1.9 Applications of Operation Research

Some of the industrial/government/business problems which can be analysed by

OR approach have been arranged by functional areas as follows;

1) Finance and Accounting

2) Marketing

3) Production Management

4) Personnel Management

5) Techniques and General Management

6) Stock re-ordering policies

7) Transport schedules

8) Product mix and Production flows

9) Allocation problems i.e. which jobs should be allocated to which machines

10) Time wasted queuing at issuing, counters

11) Scheduling of activities in a complex project

12) General congestion problem.

13

CHAPTER TWO

LINEAR PROGRAMMING PROBLEM/MODEL

2.1 Introduction

Linear programming comes under the allocation problem, is a problem which

involves the allocation of given number of resources to the job. The objective

of these problems is to optimize the total effectiveness i.e to minimize the total

cost or maximize the total return. Generally, there are three types of allocation

problem;

i. Linear programming problem

ii. Transportation problem

iii. Assignment problem

In this chapter we shall discuss the linear programming problem only.

2.2 Linear Programming Problem

Before formally defining a linear programming problem, we define the concepts of

linear function and linear inequality. A function f(x1, x2, ..., xn) of x1, x2, ..., xn is a

linear function if and only if for some set of constants c1, c2, ..., cn, f(x1, x2, ..., xn) =

c1x1+c2x2+, ...+cnxn. For any linear function f(x1, x2, ..., xn) and any number b,

the inequalities f(x1, x2, ..., xn) ≤ b and f(x1, x2, ..., xn) ≥ b are linear inequalities.

Thus, 2x1 + 3x2 ≤ 3 and 2x1 + x2 ≥ 3 are linear inequalities, but x21x2 ≥ 3 is not

a linear inequality.

Now, the term Linear Programming is the combination of the two term ‘Linear’

and ‘Programming’. The term linear means that all the relations in the particular

problem are linear and the term programming refers to the process determining

particular programme or plan of action.

Therefore, linear programming method is a technique of choosing the best alter-

native from the set of feasible alternatives, in the situations in which the objective

functions as well as constraints can be expressed as linear mathematical function.

The linear function which is to be optimized is called the objective function and

14

the conditions of the problem expressed as simultaneous linear equations (or in-

equalities) are referred as constraints.

Thus, generally we define Linear Programming as the process of transforming

a real life problem into a mathematical model which contains variables repre-

senting decisions that can be examined and solved for an optimal solution using

algorithms.

2.3 Mathematical Formulation of a LPP

A general linear programming problem can be stated as follows; Find x1, x2, x3, ..., xnwhich optimize the linear function.

Z = c1x1 + c2x2 + ...+ cnxn

subjected to the constraintsa11x1 +a12x2 +... +a1jxj + . . . a1nxn(≤=≥)b1a21x1 +a22x2 +... +a2jxj + . . . a2nxn(≤=≥)b2

......

......

...

ai1x1 +ai2x2 +... +aijxj + . . . ainxn(≤=≥)bi...

......

......

am1x1 +am2x2 +... +amixj + . . . amnxn(≤=≥)bn

and non negativity constraints

xj ≥ 0, 1, 2, 3, ..., n

Where all aij, bi and Cj are constants and xi are variables.

2.4 Matrix Form of LPP

The LPP can be expressed in the form of matrix as follows;

Maximize or Minimize Z = CX is the objective function.

Subject to

AX(≤=≥)b constraints equation, b > 0, X ≥ 0 non negativity restrictions.

Where X = (x1, x2, ..., xn) and C = (c1, c2, ..., cn)

b =

⎡⎢⎢⎢⎣

b1b2...

bn

⎤⎥⎥⎥⎦ A =

⎡⎢⎢⎢⎣

a11 a12 . . . a1na21 a22 . . . a2n...

......

am1 am2 amn

⎤⎥⎥⎥⎦

15

2.5 Procedure for Formulation of LP Problems

� Step 1. To write down the decision variables of the problem.

� Step 2. To formulate the objective function to be optimized (Maximized os

Minimized) as a linear function of the decision variable.

� Step 3. To formulate the other conditions of the problem such as resource

limitation, market constraints, interrelations between variables etc, as linear

equations in terms of decision variables.

� Step 4. To add non negativity constraints from the considerations so that

the negative values of the decision variables do not have any valid physical

interpretation.

The objective function, the set of constraints and the non negative constraints

together form a linear programming problem.

Examples 1

A resourceful home decorator manufactures two types of lamps says A and B.

Both lamps go through two technician’s first a cutter, second a finisher. Lamp A

requires 2 hrs of the cutter’s time and 1 hr of the finisher’s time. The cutter has

104 hrs and finisher has 76 hrs of available time each month. Profit per lamp A

is Tsh 600 and per B lamp is Tsh 1100. Assuming that he can sale all that he

produces, how many of each type of lamps should be manufactured to obtain the

best return.

Solution: Formulation of the Mathematical Model of the Problem.

For the clear understanding of the problem, first we have to construct a table;Lamps Cutter Finisher Profit

A 2 hrs 1 hr Tsh 600

B 1 hr 2 hrs Tsh 1100

Available time 104 hrs 76 hrs

Decision Variables:

Let the decorator manufacture x1 and x2 lamps of type A and B respectively.

Objective Functions:

Therefore, the total profit (in Tsh) has to be maximized.

Max(z) = 6x1 + 11x2

Constraints

The manufacturer has limited time for manufacturing the lamp. There are 104

16

hrs available for cutting and 76 hrs available for finishing. Thus, total processing

time is restricted.

2x1 + x2 ≤ 104

x1 + x2 ≤ 76

Finally the complete LPP is;

Max(z) = 6x1 + 11x2

Subject to

2x1 + x2 ≤ 104

x1 + x2 ≤ 76

and

x1, x2 ≥ 0

Example 2: Abdallah, a retired government officer, has recently received his

retirement benefits, viz., provident fund, gratuity, etc. He is contemplating how

much money he should invest in various alternatives open to him so as to maxi-

mize return on investment. The investment alternatives are Government securi-

ties, fixed deposits of a public limited company, equity shares, time deposits in

a bank, and house construction. He has made a subjective estimate of the risk

involved on a five-point scale. The data on the return on investment, the number

of years for which the funds will be blocked to earn this return on investment and

the subjective risk involved are as follows;

Return(%) No.of Years Risk

Government securities 6 15 1

Company deposits 13 3 3

Time deposits 10 5 2

Equity shares 20 6 5

House construction 25 10 1

17

He was wondering as to what percentage of funds he should invest in each al-

ternative so as to maximize the return on investment. He decided that the risk

should not be more than 4, and funds should not be locked up for more than 15

years. He would necessarily invest at least 25% in house construction. Formulate

this problem as an LP model.

Solution: Formulation of an LP model

Decision Variables:

Let x1, x2, x3, x4 and x5 be percentage of the total fund that should be invested

in all given five schemes, respectively.

Objective Functions:

Therefore, the objective function is to maximize the return on investment.

Max Z = 6x1 + 13x2 + 10x3 + 20x4 + 25x5

Constraints:

15x1 + 3x2 + 5x3 + 6x4 + 10x5 ≤ 15

x1 + 3x2 + 2x3 + 5x4 + x5 ≤ 4

x5 ≥ 0.25

x1 + x2 + x3 + x4 + x5 = 1

Finally the complete LP Model is;

Max Z = 6x1 + 13x2 + 10x3 + 20x4 + 25x5

Subject to

15x1 + 3x2 + 5x3 + 6x4 + 10x5 ≤ 15

x1 + 3x2 + 2x3 + 5x4 + x5 ≤ 4

x5 ≥ 0.25

x1 + x2 + x3 + x4 + x5 = 1

and

xj ≥ 0 for all j

18

2.6 Some Important Definitions in LPP

Consider the following LPP

Optimizez = CX

subject to

AX(≤=≥)b

and

X ≥ 0

(i) Objective Function;

Is the function

z = CX = c1x1 + c2x2 + ...+ cnxn

which is to be optimized (maximized or minimized).

(ii) Decision Variables;

The variables x1, x2, ..., xn whose values are to be determined are called de-

cision variables.

(iii) Cost (profit) Coefficients;

The coefficients c1, c2, ..., cn are called cost (profit) coefficients.

(iv) Requirements;

The constraints b1, b2, ..., bn are called requirements.

(v) Solution;

A set of real values of X = (x1, x2, ..., xn) which satisfies the constraint

AX(≤=≥)b is called solution.

(vi) Feasible Solution

A set of real values of X = (x1, x2, ..., xn) which

� Satisfies the constraints AX(≤=≥)b and

� Satisfies the non-negativity restriction X ≥ 0 is called feasible solution.

19

(vii) Optimal Solution;

A set of real values of X = (x1, x2, ..., xn) which

� Satisfies the constraints AX(≤=≥)b

� Satisfies the non-negativity restriction X ≥ 0 and

� Optimizes the objective function Z = CX is called optimal solution.

(viii) Results;

� If an LPP has many optimal solutions, it is said to have multiple solutions.

� If an LPP has only one optimal solution, it is said to have unique solution.

� There may be a case where the LPP may not have any feasible solution at

all (no solution).

� For some LPP the optimum value of Z may be infinity. In this case the LPP

is said to have unbounded solution.

Generally

Objective function is a Mathematical expression that describes the project

objectives.

Constraints are mathematical expressions that describe constraints eg. ca-

pacity constraints.

2.7 Solution of a LPP

In general, we use the following methods for the solution of a LPP;

1. Graphical Method

2. Simplex Method

(i) Big-M Method

(ii) Two-Phase Method

2.8 Advantages of Linear Programming Techniques

The main advantages of linear programming are given below;

1. It indicates how the available resources can be used in the best way.

2. It helps in attaining the optimum use of the productive resources and man-

power.

20

3. It improves the quality of decisions.

4. It reflects the drawbacks of the production process.

5. The necessary modifications of the mathematical solutions is also possible

by using Linear Programming.

6. It helps in re-evaluation of a basic plan with changing conditions.

2.9 Activities

1. A dealer used scooters wishes to stock up his profit. He can select scooter

A, B and C which are valued on wholesale at Tsh 500,000, Tsh 700,000 and

Tsh 800,500 respectively. These can sold at Tsh 600,000, Tsh 800,500 and

Tsh 1,000,500 respectively. For each type of scooter, the probabilities of sale

are as follows;

For every scooters of B-type he should buy one scooter of type A or C. If he

Type of scooter A B C

Prob. of sale in 90 days 0.7 0.8 0.6

has Tsh 10,000,000 to invest, what should he buy to maximize his expected

gain. Formulate this problem as an LP model.

2. A transport company is considering the purchase of new vehicles for the

transportation between Dar es Salaam airport and hotels in the city. There

are three vehicles under consideration: station wagons, mini buses and large

buses. The purchase price would be Tsh 8,000,000 for each station wagon,

Tsh 10,500,000 for a min bus and Tsh 20,000,000 for large bus. The board

of directors has authorized a maximum amount of Tsh 500,000,000 for these

purchases. Because of the heavy air travel, the new vehicles would be utilized

at maximum capacity regardless of the type of vehicles purchased. The

expected annual profit would be Tsh 1,500,000 for the station wagon, Tsh

3,500,000 for the min bus and 4,500,000 for the large bus. The company has

hired 30 new drivers for the new vehicles. They are qualified drivers for all

the three types of vehicles. The maintenance department has the capacity to

handle an additional 80 station wagons. A min bus is equivalent to5

3station

wagons and each large bus is equivalent to two station wagons in terms of

their use of the maintenance department. Formulate this problem as an LP

model to determine optimal number of each type of vehicle to be purchased

to maximize profit.

21

CHAPTER THREE

GRAPHICAL METHOD

3.1 Introduction

If the objective function Z is a function of two variables only the problem can be

solved by graphical method. A problem of three variables can be also solved by

this method but it is complicated.

3.2 Procedures for Solving LPP by Graphical Method

Various procedures of solving LPP by graphical method are as follows;

1. Formulation of the problem into LPP model

The problem is expressed in the form of a mathematical model. Here the

objective function and the constraints are written down.

2. Consider each inequality constraints as equation.

3. Plot each equation on the graph as each equation will geometrically represent

a straight line.

4. Shade the feasible region and identify the feasible solutions

Every point on the line will satisfy the equation of line. If the inequality

constraints corresponding to that line is ≤ then the region below the line

lying in the first quadrant (due to non-negativity of variables) is shaded.

For the inequality constraints with ≥ sign the region above the line in the

first quadrant is shaded. The point lying in common region will satisfy all

the constraints simultaneously. Thus, the common region obtained is called

feasible region. This region is the region of feasible solution. The corner

points of this region are identified.

5. Finding the optimal solutions

The value of Z at various corners points of the region of feasible solution are

calculated. The optimum (maximum or minimum) Z among these values is

noted. Corresponding solution is the optimal solution.

Note: While finding the corner points, greater accuracy is needed, the ordinates

may be obtained by algebraically solving the corresponding equations.

22

Example 1. Solve the following LPP graphically

Max(Z) = 3x1 + 5x2

Subject to constraints

x1 + 2x2 ≤ 2000

x1 + x2 ≤ 1500

x2 ≤ 600

and

x1, x2 ≥ 0

Solution:

� To represent the constraints graphically the inequalities are written as equal-

ities.

� Every equation is represented by a straight line.

� To draw the lines, two points on each of the lines are found as indicated in

the following table (intercepts);

Equation x2 intercept

when x1 = 0

x1 intercept

when x2 = 0

Point (x, y) on the

line

x1 + 2x2 = 2000 x2 = 1000 x1 = 2000 (0,1000)(2000,0)

x1 + x2 = 1500 x2 = 1500 x1 = 1500 (0,1500)(1500,0)

x2 = 600 and x1 = 0, x1 axis x2 = 0, x2 axis. Plot each equation on the

graph.

23

B and C are the point of intersection of lines x1+2x2 = 2000, x1+x2 = 1500 and

x1 + 2x2 = 2000, x2 = 600 on solving we get B = (1000, 500), C = (800, 600)

Corner Points Value of Z = 3x1 + 5x2A(1500, 0) 4500

B(1000, 500) 5500(Max. Value)

C(800, 600) 5400

D(0, 600) 3000

Therefore, the Maximum value of Z occurs at B(1000, 500), hence the optimal

solution is x1 = 1000 and x2 = 500.

Example 2: A and B are two product to be manufactured, unit profits are Tsh

24

40 and Tsh 35 respectively. Maximum materials available are 60 kgs and 96 hrs.

Each units of A needs 2 kg of materials and 3 man-hours, whereas each units of

B needs 4 kg of materials and 3 man-hours. Find optimal level of A and B to be

manufactured.

Solution:First of all we have to formulate the model as follows;

Decision Variables:

Let x1 and x2 be the number of product A and B to be produced by the manu-

facturer respectively.

Objective Functions:

Therefore, the total profit (in Tsh) has to be maximized.

Max(Z) = 40x1 + 35x2

Constraints

The manufacturer has limited materials and labour time for manufacturing prod-

uct A and B. There are 60 kgs available and 96 hrs available for labour. Thus,

total processing is restricted.

2x1 + 3x2 ≤ 60,Material Constraint

4x1 + 3x2 ≤ 96, Labour time Constraint

Finally the complete LPP is;

Max(Z) = 40x1 + 35x2

Subject to

2x1 + 3x2 ≤ 60

4x1 + 3x2 ≤ 96

and

x1, x2 ≥ 0

� To represent the constraints graphically the inequalities are written as equal-

ities.

� Every equation is represented by a straight line.

25

� To draw the lines, two points on each of the lines are found as indicated in

the following table (intercepts);

Equation x2 intercept

when x1 = 0

x1 intercept

when x2 = 0

Point (x, y) on the

line

2x1 + 3x2 = 60 x2 = 20 x1 = 30 (0,20)(30,0)

4x1 + 3x2 = 96 x2 = 32 x1 = 24 (0,32)(24,0)

and x1 = 0, x1 axis x2 = 0, x2 axis. Plot each equation on the graph.

26

B is the point of intersection of lines 2x1+3x2 = 60 and 4x1+3x2 = 96 on solving

we get B = (18, 8)

Corner Points Value of Z = 40x1 + 35x2A(0, 20) 700

B(18, 8) 1000(Max. Value)

C(24, 0) 960

Therefore, the Maximum value of Z occurs at B(18, 8), hence the optimal solution

is x1 = 18 and x2 = 8.

27

3.3 Cases

(i) Multiple Optimal Solution;

Example: Solve the following LPP by graphical method

Max(Z) = 100x1 + 40x2

Subject to

5x1 + 2x2 ≤ 1000

3x1 + 2x2 ≤ 900

x1 + 2x2 ≤ 500

and

x1, x2 ≥ 0

Solution:

– To represent the constraints graphically the inequalities are written as

equalities.

– Every equation is represented by a straight line.

– To draw the lines, two points on each of the lines are found as indicated

in the following table (intercepts);

Equation x2 intercept

when x1 = 0

x1 intercept

when x2 = 0

Point (x, y) on the

line

5x1 + 2x2 = 1000 x2 = 500 x1 = 200 (0,500)(200,0)

3x1 + 2x2 = 900 x2 = 450 x1 = 300 (0,450)(300,0)

x1 + 2x2 = 500 x2 = 250 x1 = 500 (0,250)(500,0)

and x1 = 0, x1 axis x2 = 0, x2 axis. Plot each equation on the graph.

28

B is the point of intersection of lines x1 + 2x2 = 500, 5x1 + 2x2 = 1000 on

solving we get B = (125, 187.5)

Corner Points Value of Z = 100x1 + 40x2A(0, 250) 10,000

B(125, 187.5) 20,000(Max. Value)

C(200, 0) 20,000(Max. Value)

Therefore, the Maximum value of Z occurs at two vertices B and C gives

the maximum value of Z. Thus, there are multiple optimum solution for the

LPP.

(ii) Ubounded Solutions:

Example: Use graphical method to solve the following LPP.

29

Max(Z) = 3x1 + 2x2

Subject to

5x1 + x2 ≥ 10

x1 + x2 ≥ 6

x1 + 4x2 ≥ 12

and

x1, x2 ≥ 0

Solution:

– To represent the constraints graphically the inequalities are written as

equalities.

– Every equation is represented by a straight line.

– To draw the lines, two points on each of the lines are found as indicated

in the following table (intercepts);

Equation x2 intercept

when x1 = 0

x1 intercept

when x2 = 0

Point (x, y) on the

line

5x1 + x2 = 10 x2 = 10 x1 = 2 (0,10)(2,0)

x1 + x2 = 6 x2 = 6 x1 = 6 (0,6)(6,0)

x1 + 4x2 = 12 x2 = 3 x1 = 12 (0,3)(12,0)

and x1 = 0, x1 axis x2 = 0, x2 axis. Plot each equation on the graph.

30

The feasible region is unbounded. Thus, the maximum value of Z occurs at

infinity, hence, the problem has an unbounded solution.

(iii) No Feasible Solution:

Example: Use graphical method to solve the following LPP.

Max(Z) = x1 + x2

Subject to

31

x1 + x2 ≤ 1

−3x1 + x2 ≤ 3

and

x1, x2 ≥ 0

Solution:

– To represent the constraints graphically the inequalities are written as

equalities.

– Every equation is represented by a straight line.

– To draw the lines, two points on each of the lines are found as indicated

in the following table (intercepts);

Equation x2 intercept

when x1 = 0

x1 intercept

when x2 = 0

Point (x, y) on the

line

x1 + x2 = 1 x2 = 1 x1 = 1 (0,1)(1,0)

−3x1 + x2 = 3 x2 = 3 x1 = −1 (0,3)(-1,0)

and x1 = 0, x1 axis x2 = 0, x2 axis. Plot each equation on the graph.

32

In the above graph, there being no point (x1, x2) common to both the shaded

regions. We cannot find a feasible region for this problem. So the problem

can not be solved, hence, the problem has no solution.

Exercise:

1. A furniture manufacturer makes two type of products, chairs and tables.

Processing of these products is done on two machines A and B. A chair

requires 2 hrs on machine A and 6 hrs on machine B. A table requires 5 hrs

on machine A and no time on machine B. There are 16 hrs per day available

on machine A and 30 hrs on machine B. Profit gained by manufacturer from

33

a chair and a table is USD 2 and USD 10 respectively. Solve this problem to

find the daily production of each of the two products.

2. A company manufactures two products Q and W on three machines A, B

and C. Q requires 1 hrs on machine A, 1 hr on machine B and C and yields

a revenue of USD 3. Product W requires 2 hrs on machine A and 1 hr on

machine B and C and yields revenue of USD 5. In the comming period the

available time of three machines A, B and C are 2000 hrs, 1500 hrs and 600

hrs respectively. Find the optimal product mix.

3. A manufacturing company produces two models of TV sets, namely model

A and model B. Model A fetches a profit of Tsh 20,000 whereas madel B

fetches a profit of Tsh 10,000 per set. Both the models use the same type of

picture tubes. Every month there is a supply of 400 picture tubes. Model

A requires 8 hrs labour and model B requires 5 hrs labour. Total labour

available per month is 2600 hrs. Find the optimal production of model A

and B.

4. Use graphical method to solve the following LPP.

Maximize (Z) = 7x1 + 3x2

Subject to the constraints

x1 + 2x2 ≥ 3

x1 + x2 ≤ 4

0 ≤ x1 ≤5

2

0 ≤ x2 ≤3

2and x1, x2 ≥ 0

5. Use graphical method to solve the following LPP.

Maximize (Z) = x1 +x22

Subject to the constraints

3x1 + 2x2 ≤ 12

5x1 = 10

x1 + x2 ≥ 8

−x1 + x2 ≥ 4

and x1, x2 ≥ 0

34

6. A firm makes two type of furniture: chairs and tables. The contribution to

profit by each product as calculated by the accounting department is Tsh

2000 per chair and Tsh 3000 per table. Both products are to be processed on

three machines M1,M2 and M3. The time required in hrs by each product

and total time available in hrs per week on each machine are as follows;

Machine Chair Table Available Time(hrs)

M1 3 3 36

M2 5 2 50

M3 2 6 60

How should the manufacturer schedule his production in order to maximize

profit?

35

CHAPTER FOUR

THE SIMPLEX METHOD

4.1 Introduction

The two variable problem of the LPP can be solved by the graphical method,

but it is very complicated to solve the three or more variable problem by using

the graphical method. In such cases, a simplex and most widely used simplex

method is adopted, which was developed by G. B, Dantzig in 1947. The simplex

method provides an algorithm which is based on the fundamental theorem of

linear programming. See the figure below;

Figure 4.1: Structure of an algorithms

36

4.2 Standard Form of an LPP

We have to convert the LPP into the standard form of LPP before the use of

simplex method. The standard form of the LPP should have the following char-

acteristics;

i) All the constraints should be expressed as equations by adding slack or sur-

plus and / or artificial variables.

ii) The right hand side of each constraints should be made non negative if it is

not, this should be done by multiplying both sides of the resulting constraints

by -1.

iii) The objective function should be of the maximization type

The general standard form of the LPP is expressed as follows;

Optimize Z = c1x1 + c2x2 + ...+ cnxn + 0S1 + 0S2 + . . .+ 0Sm

subjected to the constraintsa11x1 +a12x2 +... +a1jxj + . . . a1nxn + S1(≤=≥)b1a21x1 +a22x2 +... +a2jxj + . . . a2nxn + S2(≤=≥)b2

......

......

...

ai1x1 +ai2x2 +... +aijxj + . . . ainxn + Sn(≤=≥)bi...

......

......

am1x1 +am2x2 +... +amixj + . . . amnxn + Sm(≤=≥)bn

and non negativity constraints

x1, x2, . . . , xn, S1, S2, . . . , Sm ≥ 0

Note:

i) A slack variable represents unused resource, either in the form of time on

a machine, labour hours, money, warehouse space or any number of such

resources in various business problems. Since these variables yield no profit,

therefore such variables are added to the original objective function with zero

coefficients. Slack variables are also defined as the non-negative variables

37

which are added in the LHS of the constraints to convert the inequality ′ ≤′

into an equation.

ii) A surplus variable represents amount by which solution values exceed a re-

source. These variables are also called negative slack variables. Surplus

variables, like slack variable carry a zero coefficient in the objective func-

tion. Surplus variables which are removed from the LHS of the constraints

to convert the inequality ′ ≥′ into an equation.

iii) Artificial variables are also defined as the non-negative variables which are

added in the LHS of the constraints to convert equality into the standard

form of simplex.

4.3 The Simplex Method

4.3.1 Maximization Case

The steps of the simplex algorithm to obtain an optimal solution(if it exists) to the

LPP are as follows. But before you start step 1, first formulate the mathematical

model of the given LPP.

Step 1: Express the Problem in Standard Form

– Check whether the objective function of the formulated LPP is of max-

imization or minimization. If it is of minimization, then convert it into

one of maximization by using the following relationship.

Minimize Z = −Maximize Z∗ where Z∗ = −Z

– Check whether all the bi(i = 1, 2, . . . ,m) values are positive. If any one

of them is negative, then multiply the corresponding constraint by -1

in order to make bi ≥ 0. In doing so, remember to change a ≤ type

constraint to a ≥ type constraint, and vice-versa.

– Replace each unrestricted variable with the difference of two non-negative

variables; replace each non-positive variable with a new non-negative

variable whose value is the negative of the original variable.

– After that express the problem in standard form by introducing slack,

surplus and/or artificial variables, to convert the inequalities into equa-

tions.

38

Step 2:Find the Initial Basic Solution

– In the simplex method, a start is made with a basic feasible solution,

which we shall get by assuming that the objective function value Z=0.

This will be so when decision variables x1, x2, . . . , xn each equal to zero.

These variables are called non-basic variables.

– Substituting x1 = x2 = . . . = xn = 0 in constraint equations we get

S1 = b1, S2 = b2 . . . Sm = bm which is called initial basic feasible solution.

Not that Z = 0 for this solution.

– Variables S1, S2, . . . , Sm are called basic variable (BV).

– The problem in the standard form and the solution obtained above are

now expressed in the form of table, called simplex tableau.

Cj −→ C1 C2 . . . Cn 0 . . . 0

CB B b(=

xB)

x1 x2 . . . xn S1 . . .Sn Min.Ratio

CB1 S1 xB1 =

b1

a11 a12 . . . a1n 1 . . . 0

CB2 S2 xB2 =

b2

a21 a22 . . . a2n 0 . . . 0

......

......

......

......

...

CBm Sm xBm =

bm

am1 am2 . . . amn 0 . . . 1

Z =

ΣCBmxBm

Zj =

ΣCBmxj

0 0 . . . 0 0 . . . 0

. . . . . . Cj−Zj C1 − Z1 C2 − Z2 . . . Cn − Zn 0 . . . 0

Where;

– Cj: Objective row (Coefficient of variable in objective function) it remain

unchanged during succeeding table.

– CBm: Objective column (Coefficient of current basic variable in objective

function)

– Sm: Basic variable in basic. Initially basic variables are slack variables.

– xBm: Values of basic variables column when x1 = x2 = . . . = xn = 0.

– Body Matrix: Coefficient of decision (non-basic) variables in constraints

set (aij).

– Identity Matrix: Coefficient of slack variables in the table.

39

– Z: It presents the profit or loss Z =∑

(CBmxBm) .

– Cj − Zj: It presents the index row.

Step 3: Perform Optimality Test

– Calculate the elements of index row (Cj−Zj), if all the elements in index

row are negative then, current solution is optimum basic solution, if not

then go for next step.

Step 4: Iterate Towards an Optimal Solution

– If step 3 does not holds, then select a variable that has the largest Cj−Zj

value to enter into the new solution. That is Ck − Zk = Max [(Cj −

Zj);Cj − Zj ≥ 0]. The column to be entered is called the key or pivot

column. Such variable indicates the largest per unit improvement in the

current solution.

– Identify key or pivot row, corresponding to smallest non-negative ratio

found by dividing the values. That isxBr

arj= Min

xBi

arj; arj > 0. It should

be noted that division by negative or zero element is not permitted.

– Identify key element, the non-zero positive element at the intersection

of key column and key row, circle the key element.

– Construct new simplex table by calculating the new values for the key

row by dividing every element of the key row by the key element, if the

key element is not 1, otherwise the key row remain unchanged.

– The new values of the elements in the remaining rows for the new simplex

table can be obtained by performing elementary row operations on all

rows so that all elements except the key element in the key column are

zero. We use the following formula for the new row other than key row;

NewRowNo. = (No.inOldRow)− (AssociateNo.inKeyRow)×(CorrespondingNo.inKeyColumn

KeyElement

)

Step 5: Repeat the Procedure

– Go to step 3 and repeat the procedure until either an optimal solution

is reached or there is an indication of unbounded solution. We will see

later on, how you can determine the unbounded solution for the given

LPP.

40

Example 1: Solve the following LPP by using simplex method;

Max Z = 6x1 + 4x2

Subject to

x1 + 2x2 ≤ 720

2x1 + x2 ≤ 780

x1 ≤ 320

Solution:

Step 1: Convert the Following LPP into Standard Form

Max Z = 6x1 + 4x2 + 0S1 + 0S2 + 0S3

Subject to

x1 + 2x2 + S1 = 720

2x1 + x2 + S2 = 780

x1 + S3 = 320

Step 2: Initial Basic Feasible Solution

x1 = 0 and x2 = 0 in the above equation then we have S1 = 720, S2 =

780 and S3 = 320

Step 3: Perform the Optimality Test

Since all Cj − Zj ≥ 0(j = 1, 2), the current solution is not optimal. Variable x1is chosen to enter into the basis as C1 − Z1 = 6 is the largest positive number in

the x1 column, where all elements are positive. This means that for every unit of

variable x1, the objective function will increase in value by 6. The x1 column is

the key column.

41

Table 4.1: Initial Solution

Cj −→ 6 4 0 0 0

CB B b(= xB) x1 x2 S1 S2 S3 Min.Ratio

0 S1 720 1 2 1 0 0720

1= 720

0 S2 780 2 1 0 1 0780

2= 390

0 S3 320 1 0 0 0 1320

1= 320 →

Z = 0 Zj = 0 0 0 0 0

Cj − Zj 6 4 0 0 0

↑

Step 4: Determine the Variable to Leave the Basis

The variable to leave the basis is determined by dividing the value in the xB-

(constant) column by their corresponding elements in the key column as shown in

Table 4.1. Since the exchange ratio, 320 is minimum in row 3, the basic variable

S3 is chosen to leave the solution basis.

Iteration 1:Since the key element enclosed in the circle in Table 4.1 is 1, this row

remain unchanged. The new values of the elements in the remaining rows for the

new Table is obtained by performing the following elementary row operations on

all rows so that all elements except the key element 1 in the key column are zero.

R3(new) →R3(old)

1(keyelement)= (320, 1, 0, 0, 0, 1)

R2(new) → R2(old)− 2R3(new)

R2(new) → (780, 2, 1, 0, 1, 0)− 2(320, 1, 0, 0, 0, 1) = (140, 0, 1, 0, 1,−2)

R1(new) → R1(old)− 1R3(new)

R1(new) → (720, 1, 2, 1, 0, 0)− 1(320, 1, 0, 0, 0, 1) = (400, 0, 2, 1, 0,−1)

Then, the new improved solution is given in 4.2 below;

An improved basic feasible solution can be read from Table 4.2 as: x1 = 320, S2 =

140, S3 = 400 and x2 = 0. The improved value of objective function is Z=1920.

Once again, calculate values of Cj − Zj in the same manner as we have done to

get the improved solution in Table 4.2 to see whether the solution is optimal or

not. Since C2 − Z2 > 0, the current solution is not optimal.

42

Table 4.2: Improved Solution

Cj −→ 6 4 0 0 0

CB B b(= xB) x1 x2 S1 S2 S3 Min.Ratio

0 S1 400 0 2 1 0 -1400

2= 200

0 S2 140 0 1 0 1 -2140

1= 140 →

6 x1 320 1 0 0 0 1

Z = 1920 Zj = 6 0 0 0 6

Cj − Zj 0 4 0 0 -6

↑

Iteration 2:Repeats steps 3 to 4. Table 4.3 is obtained by performing following

row operations to enter x2 into the basis and to drive out S2 from the basis.

R2(new) →R2(old)

1(key element)= (140, 0, 1, 0, 1,−2)

R1(new) → R1(old)− 2R2(new)

R1(new) → (400, 0, 2, 1, 0,−1)− 2(140, 0, 1, 0, 1,−2) = (120, 0, 0, 1,−2, 3)

R3(new) → R3(old)− 0R2(new)

R3(new) → (320, 1, 0, 0, 0, 1)− 0(140, 0, 1, 0, 1,−2) = (320, 1, 0, 0, 0, 1)

Then, the improved solution for iteration 2 is given in Table 4.3 below;

Table 4.3: Improved Solution

Cj −→ 6 4 0 0 0

CB B b(= xB) x1 x2 S1 S2 S3 Min.Ratio

0 S1 120 0 0 1 -2 3120

3= 40 →

4 x2 140 0 1 0 1 -2

6 x1 320 1 0 0 0 1320

1= 320

Z = 2480 Zj = 6 4 0 4 -2

Cj − Zj 0 0 0 -4 2

↑

Iteration 3:Repeats steps 3 to 4. Table 4.4 is obtained by performing following

43

row operations to enter S3 into the basis and to drive out S1 from the basis.

R1(new) →R1(old)

3(key element)= (40, 0, 0, 1/3,−2/3, 1)

R2(new) → R2(old) + 2R1(new)

R2(new) → (140, 0, 1, 0, 1,−2) + 2(40, 0, 0, 1/3,−2/3, 1) = (220, 0, 1, 2/3,−1/3, 0)

R3(new) → R3(old)− 1R1(new)

R3(new) → (320, 1, 0, 0, 0, 1)− 1(40, 0, 0, 1/3,−2/3, 1) = (280, 1, 0,−1/3, 2/3, 0)

Then, the improved solution for iteration 2 is given in Table 4.4 below;

Table 4.4: Optimal Solution

Cj −→ 6 4 0 0 0

CB B b(= xB) x1 x2 S1 S2 S3 Min.Ratio

0 S3 40 0 0 1/3 -2/3 1

4 x2 220 0 1 2/3 -1/3 0

6 x1 280 1 0 -1/3 2/3 0

Z = 2560 Zj = 6 4 2/3 8/3 0

Cj − Zj 0 0 -2/3 -8/3 0

Since all Cj −Zj ≤ 0 corresponding to non - basic variables columns, the current

solution cannot be improved further. This means that the current basic feasible

solution is also the optimal solution. Thus, x1 = 280, x2 = 220 and the value

of objective function is Z=2560. Example 2: Use the simplex method to solve

following LP problem.

Max Z = 6x1 + 17x2 + 10x3

Subject to

x1 + x2 + 4x3 ≤ 2000

2x1 + x2 + x3 ≤ 3600

x1 + 2x2 + 2x3 ≤ 2400

x1 ≤ 30

44

and

x1, x2, x3 ≥ 0

Solution:

� Convert the Following LPP into Standard Form

Max Z = 6x1 + 17x2 + 10x3 + 0S1 + 0S2 + 0S3 + 0S4

Subject to

x1 + x2 + 4x3 + S1 = 2000

2x1 + x2 + x3 + S2 = 3600

x1 + 2x2 + 2x3 + S3 = 2400

x1 + S4 = 30

and

x1, x2, x3, S1, S2, S3, S4 ≥ 0

� Initial Basic Feasible Solution

An initial basic feasible solution is obtained by setting x1 = x2 = x3 = 0.

Thus, the initial solution is: S1 = 2000, S2 = 3600, S3 = 2400, S4 = 30 and

Max Z = 0. The solution can also be read from the initial simplex Table 4.5.

Table 4.5: Initial Solution

Cj −→ 6 17 10 0 0 0 0

CB B b(= xB) x1 x2 x3 S1 S2 S3 S4 Min.Ratio

0 S1 2000 1 1 4 1 0 0 02000

1= 2000

0 S2 3600 2 1 1 0 1 0 03600

1= 3600

0 S3 2400 1 2 2 0 0 1 02400

2= 1200 →

0 S4 30 1 0 0 0 0 0 1 −

Z = 0 Zj = 0 0 0 0 0 0 0

Cj − Zj 16 17 10 0 0 0 0

↑

45

� Perform the Optimality Test

Since all Cj − Zj ≥ 0, the current solution is not optimal. Variable x2 is

chosen to enter into the basis as C2 −Z2 = 17 is the largest positive number

in the x2 column. We apply the following row operations to get a new

improved solution and removing S3 from the basis.

R3(new) −→R3(old)

2(key element)= (1200, 1/2, 0, 0, 0, 1,−1/2, 0)

R1(new) −→ R1(old)−R3(new) = (800, 1/2, 0, 3, 1, 0,−1/2, 0)

R2(new) −→ R2(old)−R3(new) = (2400, 3/2, 0, 0, 0, 1,−1/2, 0)

R4(new) −→ R4(old) = (30, 1, 0, 0, 0, 0, 0, 1)

The new solution is shown in Table 4.6

Table 4.6: Improved Solution

Cj −→ 6 17 10 0 0 0 0

CB B b(= xB) x1 x2 x3 S1 S2 S3 S4 Min.Ratio

0 S1 800 1/2 0 3 1 0 -1/2 0800

1/2= 1600

0 S2 2400 3/2 0 0 0 1 -1/2 02400

3/2= 1600

17 x2 1200 1/2 1 1 0 0 1/2 01200

1/2= 2400

0 S4 30 1 0 0 0 0 0 130

1= 30 →

Z = 20, 000 Zj = 17/2 17 17 0 0 17/2 0

Cj − Zj 15/2 0 -7 0 0 -17/2 0

↑

The solution shown in Table 4.6 is not optimal because C1−Z1 = 15/2 which

is positive in x1 column. Thus, applying the following row operations to get

new improved solution by entering variable x1 into the basis and removing

the variable S4 from the basis.

R4(new) →R4(old)

1(key element)= (30, 1, 0, 0, 0, 0, 0, 1)

R1(new) → R1(old)− (1/2)R4(new) = (785, 0, 0, 3, 1, 0,−1/2,−1/2)

R2(new) → R2(old)− (3/2)R4(new) = (2355, 0, 0, 0, 0, 1,−1/2,−3/2)

R3(new) → R3(old)− (1/2)R4(new) = (1185, 0, 1, 1, 0, 0, 1/2,−1/2)

46

Table 4.7: Optimal Solution

Cj −→ 6 17 10 0 0 0 0

CB B b(= xB) x1 x2 x3 S1 S2 S3 S4

0 S1 785 0 0 3 1 0 -1/2 -1/2

0 S2 2355 0 0 0 0 1 -1/2 -3/2

17 x2 1185 0 1 1 0 0 1/2 -1/2

16 x1 30 1 0 0 0 0 0 1

Z = 20, 625 Zj = 16 17 17 0 0 17/2 15/2

Cj − Zj 0 0 -7 0 0 -17/2 -15/2

Then, the improved solution for this iteration is given in Table 4.7 below;

Since all Cj − Zj ≤ 0 corresponding to non - basic variables columns, the

current solution cannot be improved further. This means that the current

basic feasible solution is also the optimal solution. Thus, x1 = 30, x2 = 1, 185

and x3 = 0 to obtain the maximum value of Z=20,625.

Activities

1. A manufacturer of leather belts makes three types of belts A,B and C which

are processed on three machines M1,M2 and M3. Belts A requires 2 hours

on machines M1 and 3 hours on machine M2 and 2 hours on machine M3.

Belts B requires 3 hours on machine M1, 2 hours on machine M2 and 2 hours

on machine M3 and Belt C requires 5 hours on machine M2 and 4 hours on

machine M3. There are 8 hours of time per day available on machine M1, 10

hours of time per day available on machine M2 and 15 hours of time per day

available on machine M3. The profit gained from belt A is 3 USD per unit,

from belt B is 5 USD per unit, from belt C is 4 USD per unit. What should

be the daily production of each type of belt so that the profit is maximum?

2. A farmers has 1,000 acres of land on which he can grow corn, wheat or

soyabean. Each acre of corn costs 100 USD for preparation, requires 7 men-

days of work and yields a profit of 30 USD. An acre of wheat costs USD 120

to prepare, requires 10 men-days of work and yields a profit of 40 USD. An

acre of soyabean costs 70 USD to prepare, requires 8 men-days of work and

yields a profit of 20 USD. If the farmer has 1,000,000 for preparation and

can count on 8,000 men-days of work, determine how many acres should be

allocated to each crop to maximize profits?

47

4.3.2 Minimization Case

In certain cases it is difficult to obtain an initial basic feasible solution, such case

arise;

� When the constraints are of the ≤ type

n∑j=1

aijxj ≤ bi, xj ≥ 0

but some right-hand side constants are negative (bi < 0). In this case, after

adding the non-negative slack variable Si, the initial solution so obtained will

be Si = −bi for some i. It is not the feasible solution because it violates the

non-negativity condition of slack variables.

� When the constraints are of ≥ type

n∑j=1

aijxj ≥ bi, xj ≥ 0

� In this case to convert the inequalities into equation form, add surplus (neg-

ative slack) variables

n∑j=1

aijxj − Si = bi, xj, Si ≥ 0

� Letting xj = 0, we get an initial solution −Si = bi or Si = −bi. It is also

not a feasible solution as it violates the non-negativity condition of surplus

variables.

� In this case we add artificial variables Ai to get an initial basic feasible

solution. The resulting system of equations then becomes;

n∑j=1

aijxj − Si + Ai = bi, xj, Si, Ai ≥ 0, i = 1, 2, 3, ...,m

and has m equations and (n +m +m) variables (i.e n-decision variables, m

artificial variables and m surplus variables).

� To get back to the original problem, artificial variables must be dropped

out of the optimal solution. There are two methods for eliminating these

variables from the solution

1. Two - Phase Method

2. Big-M Method or Method of Penalties.

48

4.3.3 The Two-Phase Method

� In the first phase of this method the sum of all artificial variables is minimized

subject to the given constraints to get a basic feasible solution of the LPP.

� The second phase minimizes the original objective function starting with the

basic feasible solutio obtained at the end of the first phase. The steps of the

algorithm is given bellow;

Phase I:

1. (a) If all the constraints in the given LPP are ≤ type then go to Phase

II. Otherwise, add some surplus and artificial variables to get equality con-

straints.

(b) If the given LPP is of minimization then convert to maximization.

2. Assign zero coefficients to each of the decision variables xj and to the surplus

variables and assign -1 coefficient to each of the artificial variables. This

yields the following auxiliary LPP;

Max Z∗ =m∑i=1

(−1)Ai

subject to

n∑j=1

aijxj + Ai = bi, xj, Ai ≥ 0, i = 1, 2, 3, ...,m

3. Apply the simplex algorithm to solve this auxiliary LPP. The following three

cases may arise at optimality;

� Max Z∗ = 0 and atleast one artificial variable is present in the basis with

positive value. Then no feasible solution exists for the original LPP.

� Max Z∗ = 0 and no artificial variable is present in the basis. Then the

basis consists of only decision variables x′js and hence we may move to

Phase II to obtain an optimal basic feasible solution on the original LPP.

� Max Z∗ = 0 and atleast one artificial variable is present in the basis at

zero value. Then a feasible solution to the above LPP is also a feasible

solution to the original LPP. Now we may proceed direct to Phase II.

Phase II:

49

� Assign actual coefficients to the variables in the objective function and zero

to the artificial variables which appear at zero value in the basis at the end

of Phase I. Then apply the usual simplex algorithm to the modified simplex

table to get optimal solution to the original problem. Artificial variables

which do not appear in the basis may be removed.

Example 1:Solve the following LP model using Two-Phase Method;

Max Z = 5x1 − 4x2 + 3x3

subject to

2x1 + x2 − 6x3 = 20

6x1 + 5x2 + 10x3 ≤ 76

8x1 − 3x2 + 6x3 ≤ 50

and

x1, x2, x3 ≥ 0

Solution:

� After adding surplus variables S1 and S2 and artificial variable A1 the prob-

lem becomes;

Max Z = 5x1 − 4x2 + 3x3

subject to

2x1 + x2 − 6x3 + A1 = 20

6x1 + 5x2 + 10x3 + S1 = 76

8x1 − 3x2 + 6x3 + S2 = 50

and

x1, x2, x3, S1, S2, A1 ≥ 0

Phase I:

� Construction of Auxiliary LP model

Max Z∗ = −A1

subject to

2x1 + x2 − 6x3 + A1 = 20

6x1 + 5x2 + 10x3 + S1 = 76

8x1 − 3x2 + 6x3 + S2 = 50

50

and

x1, x2, x3, S1, S2, A1 ≥ 0

� Solution of an Auxiliary LP model

Table 4.8: Initial Solution

Cj −→ 0 0 0 -1 0 0

CB B b(= xB) x1 x2 x3 A1 S1 S2 Min.Ratio

-1 A1 20 2 1 -6 1 0 020

2= 10

0 S1 76 6 5 10 0 1 076

6= 12.66

0 S2 50 8 -3 6 0 0 150

8= 6.25 →

Z = −20 Zj = -2 -1 6 -1 0 0

Cj − Zj 2 1 -6 0 0 0

↑

� Slack variable S2 is removed from the basis since it has minimum ratio and

variable x1 is entering the basis since it has highest positive value into Cj−Zj

row.

Iteration 1: The improved solution is obtained by performing the following

elementary row operations.

R3(new) →R3(old)

8(key element)= (25/4, 1,−3/8, 3/4, 0, 0, 1/8)

R1(new) → R1(old)− (2)R3(new) = (15/2, 0, 7/4,−15/2, 1, 0,−1/4)

R2(new) → R2(old)− (6)R3(new) = (77/2, 0, 29/4, 11/2, 0, 1,−3/4)

� The improved solution is given in Table 4.9

Iteration 2: To remove A1 from the solution shown in Table 4.9 above,

enter x2 in the basis by applying the following elementary row operations.

R1(new) →R1(old)

7/4(key element)= (30/7, 0, 1,−30/7, 4/7, 0,−1/7)

R2(new) → R2(old)− (29/4)R1(new) = (52/7, 0, 1, 256/7, 1, 2/7)

R3(new) → R3(old)− (−3/8)R1(new) = (55/7, 1, 0,−6/7, 3/4, 0, 1/14)

� The improved solution is given in Table 4.10

51

Table 4.9: Improved Solution

Cj −→ 0 0 0 -1 0 0

CB B b(= xB) x1 x2 x3 A1 S1 S2 Min.Ratio

-1 A1 15/2 0 7/4 -15/2 1 0 -1/415/2

7/4= 30/7 →

0 S1 77/2 0 29/4 11/2 0 1 -3/477/2

29/4= 154/29

0 x1 25/4 1 -3/8 3/4 0 0 1/8 -

Z = −15/2 Zj = 0 -7/4 15/2 -1 0 1/4

Cj − Zj 0 7/4 -15/2 0 0 -1/4

↑

Table 4.10: Improved Solution

Cj −→ 0 0 0 -1 0 0

CB B b(= xB) x1 x2 x3 A1 S1 S2

0 x2 30/7 0 1 -30/7 4/7 0 -1/7

0 S1 52/7 0 1 256/7 -29/7 1 2/7

0 x1 55/7 1 0 -6/7 3/4 0 1/14

Z = 0 Zj = 0 0 0 0 0 0

Cj − Zj 0 0 0 -1 0 0

� Since all Cj−Zj ≤ 0 an optimal solution to the auxiliary LP model has been

obtained and Max Z=0 with no artificial variable in the basis.

item However, this solution may or may not be the basic feasible solution

to the original LPP. Thus, go to Phase II to get an optimal solution to our

original LPP.

Phase II

� The modified simplex table from Table 4.10 is as follows;

� Since all Cj − Zj ≤ 0 for all non-basic variables, the current basic feasible

solution is also optimal. Hence, an optimum feasible solution to the given

LPP is x1 = 55/7, x2 = 30/7, x3 = 0, S1 = 52/7, S2 = 0, S3 = 0 and Max.

Z = 155/7.

Example 2: Solve the following LPP by using two-phase method;

Min Z = x1 − 2x2 − 3x3

52

Table 4.11: Modified Simplex Table

Cj −→ 5 -4 0 0 0

CB B b(= xB) x1 x2 x3 S1 S2

-4 x2 30/7 0 1 -30/7 0 -1/7

0 S1 52/7 0 1 256/7 1 2/7

5 x1 55/7 1 0 -6/7 0 1/14

Z = 155/7 Zj = 5 -4 90/7 0 13/14

Cj − Zj 0 0 -69/7 0 -13/14

subject to

−2x1 + 3x2 + 3x3 = 2

2x1 + 3x2 + 4x3 = 1

and

x1, x2, x3 ≥ 0

Solution:

� After converting the objective function into maximization and adding arti-

ficial variables A1 and A2 in the constraints of the given LPP, the problem

becomes;

Max Z∗ = −x1 + 2x2 + 3x3

subject to

−2x1 + 3x2 + 3x3 + A1 = 2

2x1 + 3x2 + 4x3 + A2 = 1

and

x1, x2, x3, A1, A2 ≥ 0 where Z∗ = −Z

Phase I:

� Construction of Auxiliary LP model

Max Z∗ = −A1 − A2

subject to

−2x1 + 3x2 + 3x3 + A1 = 2

2x1 + 3x2 + 4x3 + A2 = 1

53

and

x1, x2, x3, A1, A2 ≥ 0

� The initial solution of an Auxiliary LPP is given bellow;

Table 4.12: Initial Solution

Cj −→ 0 0 0 -1 -1

CB B b(= xB) x1 x2 x3 A1 A1 Min.Ratio

-1 A1 2 -2 1 3 1 02

3= 0.67

-1 A2 1 2 3 4 0 11

4= 0.25 →

Z∗ = −3 Zj = 0 -4 -7 -1 -1

Cj − Zj 0 4 7 0 0

↑

� Artificial variable A2 is removed from the basis since it has minimum ratio

and variable x3 is entering the basis since it has highest positive value into

Cj − Zj row.

Iteration 1: The improved solution is obtained by performing the following

elementary row operations.

R2(new) →R2(old)

4(key element)= (1/4, 1/2, 3/4, 1, 0)

R1(new) → R1(old)− (3)R2(new) = (5/4,−7/2,−5/4, 0, 1)

� The improved solution so obtained is given in Table 4.13. Since in Table

4.13, Cj −Zj ≤ 0 corresponds to non-basic variables, the optimal solution is

x1 = 0, x2 = 0, x3 = 1/4, A1 = 5/4 and A2 = 0 with Max Z∗ = −5/4. But

at the same time, the value of Z∗ < 0 and the artificial variable A1 appears

in the basis with positive value 5/4. Hence the given original LPP does not

possess any feasible solution.

Activity

1. Use two phase method to solve the following LP problems;

(a) Min Z = x1 − 2x2 − 3x3

54

Table 4.13: Optimal but not Feasible Solution

Cj −→ 0 0 0 -1 -1

CB B b(= xB) x1 x2 x3 A1 A1

-1 A1 5/4 -7/2 -5/4 0 1 0

0 x3 1/4 1/2 3/4 1 0 1

Z∗ = −3 Zj = 7/2 5/4 0 -1 -1

Cj − Zj -7/2 -5/4 0 0 0

subject to

−2x1 + x2 + 2x3 = 2

2x1 + 3x2 + 2x3 = 1

and

x1, x2, x3 ≥ 0

(b) Min Z = 2x1 + x2 + x3

subject to

4x1 + 6x2 + 3x3 = 8

3x1 − 6x2 − 4x3 = 1

2x1 + 3x2 − 5x3 = 4

and

x1, x2, x3 ≥ 0

4.3.4 The Big - M Method

The Big - M method is another method of removing artificial variables from the

basis. In this method we assign coefficients to artificial variables, undesirable

from the objective function. If objective function Z is to be minimized, then a

very large positive price (called penalty) is assigned to each artificial variable.

Similarly, if Z is to be maximized, then a very large negative price (also called

penalty) is assigned to each of these variables. The penalty will designated by

−M for a maximization problem and +M for a minimization problem, where

M > 0. The following are steps of the Algorithm for solving LPP by the Big - M

method;

55

(i) Express the LPP in the standard form by adding slack variables, surplus

variables and artificial variables. Assign a zero coefficient to both slack and

surplus variables and a very large positive coefficient +M (for min. case)

and −M (for max. case) to artificial variable in the objective function.

(ii) The initial basic feasible solution is obtained by assigning zero value to orig-

inal variables.

(iii) Calculate the value of Cj−Zj in last row of simplex table and examine these

values.

– If all Cj − Zj ≥ 0 then the current basic feasible solution is optimal.

– If for a column k, Ck−Zk is most negative and all entries in this column

are negative, then the problem has unbounded optimal solution.

– If one or more Cj −Zj < 0 (minimization case), then select the variable

to enter into the basis with the largest negative Cj − Zj value. That is

Ck − Zk = Min{Cj − Zj} : Cj − Zj < 0.

(iv) Determine the key row and key element in the same manner as discussed in

the simplex algorithm of the maximization case.

Remarks

At any iteration of the simplex algorithm any one of the following cases may

arise;

1. If at least one artificial variable is present in the basis with zero coefficient

of M in each case Cj−Zj ≥ 0, then the given LPP has no solution. That

is, the current basic feasible solution is degenerate.

2. If at least one artificial variable is present in the basis with positive value

and the coefficient of M in each Cj − Zj ≥ 0, then given LPP has no

optimum basic feasible solution. In this case the given LPP has a pseudo

optimum basic feasible solution.

Example 1:Solve the following LPP using penalty (Big - M) method;

Max Z = x1 + 2x2 + 3x3 − x4

subject to

x1 + 2x2 + 3x3 = 15

2x1 + x2 + 5x3 = 20

x1 + 2x2 + x3 + x4 = 10

56

and

x1, x2, x3 ≥ 0

Solution:

� Since all constraints of the given LPP are equation, therefore adding only

artificial variables A1 and A2 in the constraints. The standard form of the

problem becomes;

Max Z = x1 + 2x2 + 3x3 − x4 −MA1 −MA2

subject to

x1 + 2x2 + 3x3 + A1 = 15

2x1 + x2 + 5x3 + A2 = 20

x1 + 2x2 + x3 + x4 = 10

and

x1, x2, x3, A1, A2 ≥ 0

� The initial basic feasible solution is given in Table4.14 below;

Table 4.14: Initial Solution

Cj −→ 1 2 3 -1 -M -M

CB B b(= xB) x1 x2 x3 x4 A1 A2 Min.Ratio

-M A1 15 1 2 3 0 1 015

3= 5

-M A2 20 2 1 5 0 0 120

5= 4 →

-1 x4 10 1 2 1 1 0 010

1= 10

Z = −35M − 10 Zj = -3M-1 -3M-2 -8M-1 -1 -M -M

Cj − Zj 3M+2 3M+4 8M+4 0 0 0

↑

� Since the value of C3−Z3 in Table 4.14 has largest positive value the variable

x3 is chosen to enter into the basis. To get an improved basic feasible solution,

apply the following row operations and removing A2 from the basis.

R2(new) →R2(old)

5(key element)= (4, 2/5, 1/5, 1, 0, 0)

R1(new) → R1(old)− (3)R2(new) = (3,−1/5, 7/5, 0, 0, 1)

R3(new) → R3(old)− (1)R1(new) = (6, 3/5, 9/5, 0, 1, 0)

57

� The improved solution is shown in Table 4.15

Table 4.15: Improved Solution

Cj −→ 1 2 3 -1 -M

CB B b(= xB) x1 x2 x3 x4 A1 Min.Ratio

-M A1 3 -1/5 7/5 0 0 13

7/5= 15/7 →

3 x3 4 2/5 1/5 1 0 04

1/5= 20

-1 x4 6 3/5 9/5 0 1 06

9/5= 30/9

Z = −3M + 6 Zj = (M/5)-3/5 -(7M/5)-6/5 3 -1 -M

Cj − Zj -(M/5)-2/5 (7M/5)+16/5 0 0 0

↑

� The solution shown in Table 4.15 is not optimal because C2 −Z2 is positive.

Thus, applying the following row operations for entering variable x2 into the

basis and removing variable A1 from the basis.

R1(new) →R1(old)

7/5(key element)= (15/7,−1/7, 1, 0, 0)

R2(new) → R2(old)− (1/5)R1(new) = (25/7, 3/7, 0, 1, 0)

R3(new) → R3(old)− (9/5)R1(new) = (15/7, 6/5, 0, 0, 1)

� The new solution is shown in Table 4.16

Table 4.16: Improved Solution

Cj −→ 1 2 3 -1

CB B b(= xB) x1 x2 x3 x4 Min.Ratio

2 x2 15/7 -1/7 1 0 0 -

3 x3 25/7 3/7 0 1 025/7

3/7= 25/3

-1 x4 15/7 6/7 0 0 115/7

6/7= 5/2 →

Z = 90/7 Zj 1/7 2 3 -1

Cj − Zj 6/7 0 0 0

↑

58

� Again, the solution shown in Table 4.16 is not optimal. Thus, applying the

following row operations by entering x1 into the basis and removing variable

x4 from the basis.

R3(new) →R3(old)

6/7(key element)= (15/6, 1, 0, 0, 7/6)

R2(new) → R2(old)− (3/7)R3(new) = (15/6, 0, 0, 1,−1/2)

R1(new) → R1(old)− (−1/7)R3(new) = (15/6, 0, 1, 0, 1/6)

� The new solution is shown in Table 4.17

Table 4.17: Optimal Solution

Cj −→ 1 2 3 -1

CB B b(= xB) x1 x2 x3 x4

2 x2 15/6 0 1 0 1/6

3 x3 15/6 0 0 1 -1/2

1 x1 15/6 1 0 0 7/6

Z = 15 Zj 1 2 3 0

Cj − Zj 0 0 0 -1

Since all Cj − Zj ≤ 0 in Table 4.17. Thus, an optimal solution has been

arrived with values of variables as x1 = 15/6, x2 = 15/6, x3 = 15/6, x4 = 0

and Max Z = 15.

Example 2:Solve the following LPP using penalty (Big - M) method;

Main Z = 600x1 + 500x2

subject to

2x1 + x2 ≥ 80

x1 + 2x2 ≥ 60

and

x1, x2 ≥ 0

Solution:

� By introducing surplus variables S1 and S2 and artificial variables A1 and A2

in the constraints. The standard form of the problem becomes;

Main Z = 600x1 + 500x2 + 0S1 + 0S2 +MA1 +MA2

59

subject to

2x1 + x2 − S1 + A1 = 80

x1 + 2x2 − S2 + A2 = 60

and

x1, x2, S1, S2, A1, A2 ≥ 0

� The initial basic feasible solution is obtained by setting x1 = x2 = S1 = S2 =

0 as shown in Table4.18;

Table 4.18: Initial Solution

Cj −→ 600 500 0 0 M M

CB B b(= xB) x1 x2 S1 S2 A1 A2 Min.Ratio

M A1 80 2 1 -1 0 1 080

1= 80

M A2 60 1 2 0 -1 0 160

2= 30 →

Z = 140M Zj 3M 3M -M -M M M

Cj − Zj 600-3M 500-3M M M 0 0

↑

� Since the value of C2−Z2 in Table 4.18 has largest negative value, therefore

enter variable x2 to replace basic variable A2 into the basis. To get an

improved basic feasible solution, apply the following row operations.

R2(new) →R2(old)

2(key element)= (30, 1/2, 1, 0,−1/2, 0)

R1(new) → R1(old)− (1)R2(new) = (50, 3/2, 0,−1, 1/2, 1)

� The improved solution is shown in Table 4.19

� The solution shown in Table 4.19 is not optimal because C1 − Z1 is largest

negative. Thus, applying the following row operations by entering variable

x1 into the basis and removing variable A1 from the basis.

R1(new) →R1(old)

3/2(key element)= (100/3, 1, 0,−2/3, 1/3)

R2(new) → R2(old)− (1/2)R1(new) = (40/3, 0, 1, 1/3,−2/3)

60

Table 4.19: Improved Solution

Cj −→ 600 500 0 0 M

CB B b(= xB) x1 x2 S1 S2 A1 Min.Ratio

M A1 50 3/2 0 -1 1/2 150

3/2= 33.33 →

500 x2 30 1/2 1 0 -1/2 030

1/2= 60

Z = 15000 + 50M Zj (3M/2)+250 500 -M (M/2)-250 M

Cj − Zj 350-3M 0 M 250-M/2 0

↑

Table 4.20: Optimal Solution

Cj −→ 600 500 0 0

CB B b(= xB) x1 x2 S1 S2

600 x1 100/3 1 0 -2/3 1/3

500 x2 40/3 0 1 1/3 -2/3

Z = 80, 000/3 Zj 600 500 -700/3 -400/3

Cj − Zj 0 0 700/3 400/3

� The new solution is shown in Table 4.20

� In Table 4.20, all the numbers in the Cj − Zj row are either zero or positive

and also both artificial variables have been reduced to zero, an optimum

solution has been arrived at with x1 = 100/3, x2 = 40/3 and total minimum

cost, Z = 80, 000/3.

Activity

1. Use Big - M method to solve the following LPP.

Max Z = 8x1 + 15x2 + 25x3 + x4

subject to

x1 + 2x2 + 3x3 = 15

2x1 + x2 + 5x3 = 20

x1 + 2x2 + x3 + x4 = 10

61

and

x1, x2, x3, x4 ≥ 0

2. Use Big - M method to solve the following LPP.

Max Z = 12x1 + 20x2 + 18x3 + 40x4

subject to

4x1 + 9x2 + 7x3 + 10x4 ≤ 6000

x1 + x2 + 3x3 + 40x4 ≤ 4000

and

x1, x2, x3, x4 ≥ 0

4.4 Degeneracy in Simplex Method

A basic feasible solution of a simplex method is said to be degenerate basic feasible

solution if at least one of the basic variable is zero and at any iteration of the

simplex method more than one variable is eligible to leave the basis and hence

the next simplex iteration produces a degenerate solution in which at least one

basic variable is zero. This concept is known as tie.

A situation may arise at any iteration when two or more columns may have

exactly the same Cj − Zj value (+ve or -ve depending on the type of LPP).

In order to break this tie, the selection for key column (entering variable) can

be made arbitrary,. However, the number of iterations required to arrive at the

optimal solution can be minimized by adopting the following rules;

� If there is a tie between two decision variables, then the selection can be

made arbitrarily.

� If there is a tie between decision variable and slack (or surplus) variable, then

select the decision variable to enter into the basis first.

� If there is a tie between two slack (or surplus) variables, then selection can

be made arbitrarily.

Again, while solving LPP the situation may arise in which there is a tie between

two or more basic variables for leaving the basis i.e minimum ratio to identify

the basic variable to leave the basis is not unique or values of one or more basic

62

variables in the solution values column (xB) become equal to zero. This causes the

problem of degeneracy. However, if minimum ration is zero, then the iterations of

simplex method are repeated (cycle) indefinitely without arriving at the optimal

solution.

In most of the cases when there is a tie in the minimum ratios, the selection

is made arbitrarily. However, the number of iterations required to arrive at the

optimal solution can be minimized by applying the following rules;

� Divide the coefficients of slack variables in the simplex table where degener-

acy is detected by the corresponding positive numbers of the key column in

the row, starting from left to right.

� The row which contains smallest ratio comparing from left to right column-

wise becomes the key row.

Remark:When there is a tie between a slack and artificial variables to leave the

basis, the preference shall be given to artificial variable to leave the basis and there

is no need to apply the procedure for resolving degeneracy under such cases.

Example: Solve the following LPP

Max Z = 3x1 + 9x2

subject to the constraints

x1 + 4x2 ≤ 8

x1 + 2x2 ≤ 4

and

x1, x2 ≥ 0

Solution:

� Adding slack variables S1 and S2 to the constraints, the problem can be

expressed as;

Max Z = 3x1 + 9x2 + 0S1 + 0S2

subject to the constraints

x1 + 4x2 + S1 = 8

x1 + 2x2 + S2 = 4

and

x1, x2, S1, S2 ≥ 0

63

� The initial basic feasible solution is given in Table 4.21

Table 4.21: Initial Solution

Cj −→ 3 9 0 0

CB B b(= xB) x1 x2 S1 S2 Min.Ratio

0 S1 8 1 4 1 08

4= 2

0 S2 4 1 2 0 14

2= 2

Z = 0 Zj 0 0 0 0

Cj − Zj 3 9 0 0

↑

� from the Table 4.21, C2 − Z2 is the largest positive value, therefore variable

x2 is selected to enter into the basis. However, both variables S1 and S2.

This is an indication of the existence of degeneracy. To obtain the unique

key row, apply the following procedure for resolving degeneracy.

� Write coefficients of the slack variables as follows;

Column

Row S1 S2

S1 1 0

S2 0 1

� Dividing the coefficients by the corresponding element of the key column, we

obtain the following ratios;

Column

Row S1 S2

S1 1/4=1/4 0/4=0

S2 0/2=0 1/2=1/2

� Comparing the ratios of the previous step from left to right column-wise,

the minimum ratio occurs for the second row. Therefore, the variable S2 is

selected to leave the basis. The new solution is obtain by performing the

following row operations and shown in Table 4.22

R2(new) −→R2(old)

2(keyelement)= (2, 1/2, 1, 0, 1/2)

R1(new) −→ R1(old)− 4R2(new) = (0,−1, 0, 1,−2)

64

Table 4.22: Optimal Solution

Cj −→ 3 9 0 0

CB B b(= xB) x1 x2 S1 S2

0 S1 0 -1 0 1 -2

9 x2 2 1/2 1 0 1/2

Z = 18 Zj 9/2 9 0 9/2

Cj − Zj -3/2 0 0 -9/2

� Since all Cj −Zj ≤ 0 in Table 4.22. Therefore, an optimal solution is arrived

at x1 = 0, x2 = 2 and Max Z = 18.

4.5 Types of Linear Programming Solution

4.5.1 Alternative (Multiple) Optimal Solution

The alternative optimal solution can be obtained by considering the Cj−Zj row of

the simplex table. We know that an optimal solution to a maximization problem

is reached if all Cj −Zj ≤ 0. What will happen if Cj −Zj = 0 for some non-basic

variable columns in the optimal simplex table? Each entry in the Cj−Zj indicates

the contribution per unit of a particular variable in the objective function value

if is entered into the basis. Thus, if a non-basic variable corresponding to which

Cj − Zj = 0 is entered into the basis, a new solution will be arrived at but the

value of the objective function will not change.

Example: Solve the following LPP;

Max Z = 6x1 + 4x2

subject to the constraints

2x1 + 3x2 ≤ 30

3x1 + 2x2 ≤ 24

x1 + x2 ≥ 3

and

x1, x2 ≥ 0

Solution:

� Adding slack variables S1, S2, surplus variable S3 and artificial variable A1

65

in the constraint set the LPP becomes;

Max Z = 6x1 + 4x2 + 0S1 + 0S2 + S3 −MA1

subject to the constraints

2x1 + 3x2 + S1 = 30

3x1 + 2x2 + S2 = 24

x1 + x2 − S3 + A1 = 3

and

x1, x2, S1, S2, S3, A1 ≥ 0

� The optimal solution for this LPP is presented in Table 4.23

Table 4.23: Optimal Solution

Cj −→ 6 4 0 0 0

CB B b(= xB) x1 x2 S1 S2 S3 Min.Ratio

0 S1 14 0 5/3 1 -2/3 014

15/3= 42/5 →

0 S3 5 0 -1/3 0 1/3 1 -

6 x1 8 1 2/3 0 1/3 08

2/3= 12

Z = 48 Zj 6 4 0 2 0

Cj − Zj 0 0 0 -2 0

↑

� The optimal solution shown in Table 4.23 is x1 = 8, x2 = 0 and Max Z=48.

� From the Table 4.23, C2 − Z2 = 0 corresponding to a non-basic variable,

x2 = 0. Thus, an alternative optimal solution can also be obtained by

entering variable x2 into the basis and removing S1 from the basis. The new

solution is shown in Table 4.24

� The optimal solution shown in Table 4.24 is x1 = 12/5, x2 = 42/5 and Max

Z=48.

� Further observe that in Table 4.24, C3 − Z3 = 0 and variable S1 is not in

the basis. This again indicates that an alternative optimal solution exists,

thus for each alternative solution (infinite number of solutions) the value of

objective function will remain the same.

66

Table 4.24: Alternative Solution

Cj −→ 6 4 0 0 0

CB B b(= xB) x1 x2 S1 S2 S3

4 x2 42/5 0 1 3/5 -2/5 0

0 S3 39/5 0 0 1/5 1/5 1

6 x1 12/5 1 0 -2/5 3/5 0

Z = 48 Zj 6 4 0 2 0

Cj − Zj 0 0 0 -2 0

4.5.2 Unbounded Solution

In maximization LPP, if Cj − Zj > 0(Cj − Zj < 0 for a maximization case) for

a column not in the basis and all entries in this column are negative, then for

determining key row, we have to calculate minimum ratio corresponding to each

basic variable having negative or zero value in the denominator. Negative value in

the denominator can not be considered, as it would indicate the entry of non-basic

variable in the basis with a negative value (an infeasible solution will occur). A

zero value in the denominator would result in ratio having a +∞. This implies

that the entering variable could be increased indefinitely with any of the current

basic variables being removed from the basis. In general, an unbounded solution

occurs due to wrong formulation of the problem within the constraint set, and

thus needs reformulation.

Example: Solve the following LPP;

Max Z = 3x1 + 5x2

subject to the constraints

x1 − 2x2 ≤ 6

x1 ≤ 10

x2 ≥ 1

and

x1, x2 ≥ 0

Solution:

� Adding slack variables S1, S2, surplus variable S3 and artificial variable A1

in the constraint set the LPP becomes;

Max Z = 3x1 + 5x2 + 0S1 + 0S2 + 0S3 −MA1

67

subject to the constraints

x1 − 2x2 + S1 = 6

x1 + S2 = 10

x2 − S3 + A11

and

x1, x2, S1, S2, S3, A1 ≥ 0

� The initial solution to this LPP is shown in Table 4.25

Table 4.25: Initiall Solution

Cj −→ 3 5 0 0 0 -M

CB B b(= xB) x1 x2 S1 S2 S3 A1 Min.Ratio

0 S1 6 1 -2 1 0 0 0 -

0 S2 10 1 0 0 1 0 0 -

-M A1 1 0 1 0 0 -1 11

1= 1 →

Z = −M Zj 0 -M 0 0 M -M

Cj − Zj 3 5+M 0 0 -M 0

↑

� From Table 4.25, C2 − Z2 has largest positive value, thus variable x2 enters

the basis and A1 leaves the basis. The new solution is shown in Table 4.26

Table 4.26: Improved Solution

Cj −→ 3 5 0 0 0 -M

CB B b(= xB) x1 x2 S1 S2 S3 A1

0 S1 8 1 0 1 0 -2 2

0 S2 10 1 0 0 1 0 0

5 x2 1 0 1 0 0 -1 1

Z = 5 Zj 0 5 0 0 -5 5

Cj − Zj 3 0 0 0 5 -M-5

� From the Table 4.26, C1 − Z1 = 3 and C5 − Z5 = 5 entries are positive and

C5−Z5 ≥ C1−Z1. Therefore, variable S3 should enter into the basis. Here it

may be noted that coefficients in the ’S3’ column are all negative or zero. This

68

indicates that S3 cannot be entered into the basis. However, the value of S3

can be increased infinitely without removing any one of the basic variables.

Further, since S3 is associated with x1 in the third constraint, x1 will also be

increased infinitely because it can be expressed as x1 = 1 + S3 −A1. Hence,

the solution to the given LPP is unbounded.

4.5.3 Infeasible Solution

In the final simplex table, if atleast one of the artificial variable appears with

a positive value, no feasible solution exists, because it is not possible to remove

such an artificial variable from the basis using the simplex algorithm. When an

infeasible solution exists, the LP Model should be reformulated. This may be

because of the fact that the model is either improperly formulated or two or more

of the constraints are incompatible.

Example:

Max Z = 6x1 + 4x2

subject to the constraints

x1 + x2 ≤ 5

x2 ≥ 8

and

x1, x2 ≥ 0

Solution:

� By adding slack, surplus and artificial variables, the LPP becomes;

Max Z = 6x1 + 4x2 + 0S1 + 0S2 −MA1

subject to the constraints

x1 + x2 + S1 = 5

x2 − S2 + A1 = 8

and

x1, x2, S1, S2, A1 ≥ 0

� The initial solution to this LPP is shown in Table 4.27

69

Table 4.27: Initial Solution

Cj −→ 6 4 0 0 -M

CB B b(= xB) x1 x2 S1 S2 A1 Min.Ratio

0 S1 5 1 1 1 0 0 5

1= 5 →

-M A1 8 0 1 0 -1 1 8

1= 8

Z = −8M Zj 0 -M 0 M -M

Cj − Zj 6 4+M 0 -M 0

↑

Table 4.28:

Cj −→ 6 4 0 0 -M

CB B b(= xB) x1 x2 S1 S2 A1

4 x2 5 1 1 1 0 0

-M A1 3 -1 0 -1 -1 1

Z = 20− 3M Zj 4+M 4 4+M M -M

Cj − Zj 2-M 0 -4-M -M 0

� Variable x2 enters the basis and S1 leaves the basis. The new solution is

shown in Table 4.28

� Since all Cj − Zj ≤ 0, the solution shown in Table 4.28 ia optimal. But this

solution is not feasible for the given problem since it has x1 = 0 and x2 = 5

(recall that in the second constraint x2 ≥ 8). The fact that artificial variable

A1 = 3 is in the solution also indicates that the final solution violates the

second constraint.

70

CHAPTER FIVE

DUALITY IN LINEAR PROGRAMMING

5.1 Introduction

The term dual in general sense implies two or double. In the context of linear

programming duality implies that each LPP can be analysed in two different

ways but having equivalent solution. Moreover, whenever the LPP contains a

large number of constraints and a smaller number of variables then the labour

of computational can be considerably reduced by converting it into the dual and

then solve it. Every LPP is associated with another LPP called the dual based

on the same data. The original problem is called the primal.

5.2 Formulation of Dual Linear Programming Problem

Let the primal LPP be;

Max Zx = c1x1 + c2x2 + ...+ cnxn

subject to the constraints

a11x1 +a12x2 + . . .+ a1nxn ≤ b1a21x1 +a22x2 + . . .+ a2nxn ≤ b2

......

...

am1x1 +am2x2 + . . .+ amnxn ≤ bm

and

x1, x2, . . . , xn ≥ 0

Then the corresponding dual is defined as:

Min Zy = b1y1 + b2y2 + ...+ bmym

subject to the constraints

a11y1 +a12y2 + . . .+ a1nym ≤ c1a21y1 +a22y2 + . . .+ a2nym ≤ c2

......

...

an1y1 +an2y2 + . . .+ amnym ≤ cn

71

and

y1, y2, . . . , ym ≥ 0

5.2.1 Rules for Constructing the Duality from Primal

1. Change the objective of maximization in the primal into minimization in the

dual and vice-versa.

2. For a maximization primal with all ≤ type constraints, there exists a min-

imization dual problem with all ≥ type constraints and vice-versa. The

inequality sign for non-negativity constraint is unreversed.

3. The number of variables in the primal will be the number of constraints in

the dual and vice-versa.

4. The cost of coefficients c1, c2, . . . , cn in the objective function of the primal

will be the RHS constant of the constraints in the dual and vice-versa.

5. For the constraints of dual, transpose the body matrix of the primal problem.

6. If the ith primal variable is unrestricted in sign, then the jth dual constraint

is = type and vice-versa.

Example 1: Write the dual of the following LPP;

Max Zx = 3x1 + x2 + x3

Subject to

4x1 − x2 ≤ 8

8x1 + x2 + 3x3 ≥ 12

5x1 − 6x3 ≤ 13

and

x1, x2, x3 ≥ 0

Solution: Let y1, y2 and y3 be the dual variables, then the corresponding dual is;

Min Zy = 8y1 + 12y2 + 13y3

Subject to

4y1 − 8y2 + 5y3 ≥ 13

−y1 − y2 ≥ −1

−3y1 − 6y3 ≥ 1

72

and

y1, y2, y3 ≥ 0

Example 2: Write the dual of the following LPP;

Min Zx = 3x1 − 2x2 + 4x3

Subject to

3x1 + 5x2 + 4x3 ≥ 7

6x1 + x2 + 3x3 ≥ 4

7x1 − 2x2 − x3 ≤ 10

x1 − 2x2 + 5x3 ≥ 3

4x1 + 7x2 − 2x3 ≥ 2

and

x1, x2, x3 ≥ 0

Solution:Since the objective function is of minimization type all inequalities have

to be changed to ≥ type. Constraint No:3 will change to;

−7x1 + 2x2 + x3 ≥ −10

Let y1, y2, y3, y4 and y5 are dual variable corresponding to five primal constraints,thus

the dual to this LPP is;

Max Zy = 7y1 + 4y2 − 10y3 + 3y4 + 2y5

Subject to

3y1 + 6y2 − 7y3 + y4 + 4y5 ≤ 3

5y1 + y2 + 2y3 − 2y4 + 7y5 ≤ −2

4y1 + 3y2 + y3 + 5y4 − 2y5 ≤ 4

and

y1, y2, y3, y4, y5 ≥ 0

Example 3:Obtain the dual of the following LPP;

Max Zx = x1 − 2x2 + 3x3

Subject to

−2x1 + x2 + 3x3 = 2

2x1 + 3x2 + 4x3 = 1

73

and

x1, x2, x3 ≥ 0

Solution: Since both the primal constraints are = type, the corresponding dual

variables y1, y2 will be unrestricted in sign, the dual to this LPP is;

Min Zy = 2y1 + y2

Subject to

−2y1 + 2y2 ≥ 1

y1 + 3y2 ≥ −2

3y1 + 4y2 ≥ 3

and

y1, y2 unrestricted in sign

5.2.2 Primal - Dual Relationship

A summary of the general relationship between primal and dual LPP is given in

Table 5.1

Table 5.1: Primal-Dual Relationship

If Primal Then Dual

i)Objective is to maximise i)Objective is to minimise

ii)Variable xj ii)Constraint j

iii)Constraint i iii)Variable yi

iv)Variables xj unrestricted insign iv)Constraint j is = type

v)Constraint i is = type v)Variable yi is unrestricted in sign

vi)≤ type constraints vi)≥ type constraints

vii)xj unrestricted in sign vii)jth constraint is an equatio

5.3 Standard Results on Duality

You can make a proof of the following standard results;

1. The dual of the dual LPP is again the primal problem.

74

2. If either the primal or the dual has an unbounded objective function value,

the other problem has no feasible solution.

3. If either the primal or dual problem has a finite optimal solution, the other

one also possesses the same, and the optimal value of the objective function

of the two problems are equal i.e. Max Zx = Min Zy. This analytical result

is known as the fundamental primal-dual relationship.

4. Complementary slackness property of primal-dual relationship states that for

a positive basic variable in the primal, the corresponding dual variable will

be equal to zero. Alternatively, for a non-basic variable in the primal (which

is zero), the corresponding dual variable will be basic and positive.

5.4 Significant of Duality

The importance of dual LPP is in terms of the information which it provides

about the value of the resources. The economic analysis is concerned in deciding

whether or not to secure more resources and how much to pay for these additional

resources. The significance of the study of dual is as follows;

1. The dual variables provide the decision-maker a basis for deciding how much

to pay for additional units of resources.

2. The maximum amount that should be paid for one additional unit of a re-

source is called its shadow price (also called simplex multiplier).

3. The total marginal value of the resources equals the optimal objective func-

tion value. The dual variables equal the marginal value of resources (shadow

prices).

5.5 Advantages of Duality

1. It is advantageous to solve the dual of primal having less number of con-

straints, because the number of constraints usually equals the number of

iterations required to solve the problem.

2. It avoids the necessity for adding surplus or artificial variables and solves

the problem quickly (the technique is known as the primal-dual method).

In economics, duality is useful in the formulation of the input and output

systems. It is also in physics, engineering, mathematics, etc.

3. The dual variables provide an important economic interpretation of the final

solution of an LPP.

75

4. It is quite useful when investigating changes in the parameters of an LPP

(the technique is known as the sensitivity analysis).

5. Duality is used to solve an LPP by the simplex method in which the initial

solution is infeasible (the technique is known as the dual simplex method)

Problem:

Write the dual of the following primal LPP;

1.

Max Zx = 2x1 + 5x2 + 6x3

Subject to

5x1 + 6x2 − x3 ≤ 3

−2x1 + x2 + 4x3 ≤ 4

x1 − 5x2 + 3x3 ≤ 1

−3x1 − 3x2 + 7x3 ≤ 6

and

x1, x2, x3 ≥ 0

2.

Max Zx = 2x1 + 3x2 + x3

Subject to

4x1 + 3x2 + x3 = 6

x1 + 2x2 + 5x3 = 4

and

x1, x2, x3 ≥ 0

3.

Min Zx = 2x1 + 3x2 + x3

Subject to

2x1 + 3x254x3 ≥ 2

3x1 + x2 + 5x3 = 3

x1 + 4x2 + 6x3 ≤ 5

and

x1, x2 ≥ 0, x3 is unrestricted.

76

CHAPTER SIX

SENSITIVITY ANALYSIS IN LINEAR PROGRAMMING

6.1 Introduction

In LP model, the input data (also known as parameters) are assumed constant

and known with certainty during a planning period. These parameters are such

as;

(i) Profit (cost) contribution (Cj) per unit of decision variable.

(ii) Availability of resources (bi).

(iii) Consumption of resources per unit of decision variable (aij).

However, in real-world situations some data may change over time because of

the dynamic nature of the business such changes may raise doubt on the validity

of the optimal solution of the given LP model. Thus, a decision-maker in such

situation would like to know, how sensitive the optimal solution is to the changes

in the original input data values.

6.2 Sensitivity Analysis

Sensitivity analysis is the study of sensitivity of the optimal solution of an LPP

due to discrete variation (changes) in its parameters.

The degree of sensitivity of the solution due to these variations can range from

no change at all to a substantial change in the optimal solution of the given LPP.

Thus, in sensitivity analysis we determine the range over which the LP model

parameters can change without affecting the current optimal solution.

The process of studying the sensitivity of the optimal solution of an LPP is also

called post-optimality analysis because it is done after an optimal solution, as-

suming a given set of parameters, has been obtained for the model. Different

categories of parameter changes in the original LP model includes the following;

6.2.1 Change in Objective Function Coefficient (Cj)

The question is; what happens to the optimal solution and the objective function

when this coefficient is changed? There are three cases in this change;

77

Case I: Change in the coefficient of a non-basic variable

– The current optimal solution for a maximization LPP will remain opti-

mal as long as all Cj − Zj ≤ 0 for all j.

– Let Ck be the coefficient of non-basic variable xk in the objective func-

tion.

– Ck does not affect any of the Cj values listed in the CB column. Cal-

culation of Zj = CBB−1aj values do not involve Cj, therefore change in

Cj does not alter Zj values and hence Cj −Zj values.remain unchanged

except Ck − Zk do to change in Ck.

– To retain optimality of the current optimal solution for a change �Ck

in Ck we must have;

(Ck +�Ck)− Zk ≤ 0 or

(Ck −�Ck)− Zk ≤ 0

– Hence for a maximization LPP, the value of Ck may be increased up

to the value of Zk and decrease to −∞ without affecting the optimal

solution.

Case II: Change in the coefficient of a basic variable

– In the maximization LPP the change in the coefficient say Ck of a basic

variable xk affects the Cj − Zj values corresponding to all non-basic

variables in the simplex table. It is because the Ck is listed in the CB

column of the simplex table and affects the calculation of Zj values.

– The sensitivity limits for the Ck of a basic variable are calculated as

under;

Lower limit = (Original V alue Ck)− (Lower absolute

value of improvement ratio −∞)

Upper limit = (Original V alue Ck) + (Lower positive value of

improvement ratio ∞)

where

Improvement ratio =Per unit Impr.Ratio

Coefficient in the V ariable row=

Cj − Zj

akj

Note: While performing sensitivity analysis, the artificial variable columns

in the simplex table are ignored because artificial variable has no eco-

nomic interpretation.

78

Case III: Change in the coefficient of non-basic variable in cost

min. problem

– The procedure for calculating sensitivity limits to a cost minimization

LPP when the objective function coefficients are unit costs is identical

to the Case I above.

– In this case the unit cost coefficient can be increased to any arbitrary

level but it cannot be decreased by more than per unit improvement

value without making it eligible so that a non-basic variable can be

entered into the new solution mix.

– The sensitivity limits can be calculated as;

Lower limit = (Original V alue)− (Unit improvement value)

Upper limit = Infinity (∞)

Example: Given the optimal solution Table 6.1

Table 6.1:

Cj −→ 4 6 2 0 0

CB B b(= xB) x1 x2 x3 S1 S2

4 x1 1 1 0 -1 4/3 -1/3

6 x2 2 0 1 2 -1/5 1/3

Z = 16 Zj 4 6 8 10/3 2/3

Cj − Zj 0 0 -6 -10/3 -2/3

(a) Effect of change in C3 of non-basic variable x3 is shown in Table 6.2

Table 6.2:

Cj −→ 4 6 2 +�C3 0 0

CB B b(= xB) x1 x2 x3 S1 S2

4 x1 1 1 0 -1 4/3 -1/3

6 x2 2 0 1 2 -1/5 1/3

Z = 16 Zj 4 6 8 10/3 2/3

Cj − Zj 0 0 �C3 − 6 -10/3 -2/3

� Then

�C3 − 6 ≤ 0

�C3 ≤ 6

79

� Recall that

C3 − 6 ≤ 2 +�C3

C3 ≤ 2 + 6

C3 ≤ 8

or −∞ ≤ C3 ≤ 8

(b) Effect of change in the coefficient of C1 of basic variable x1

Table 6.3:

Cj −→ 4 6 2 0 0

CB B b(= xB) x1 x2 x3 S1 S2

4 +�C1 x1 1 1 0 -1 4/3 -1/3

6 x2 2 0 1 2 -1/5 1/3

Z = 12 + (4 +�C1) Zj 4 +�C1 6 8−�C1 (10/3)−

4�C1/3

(2/3)−�C1/3

Cj − Zj 0 0 �C1 − 6 (−10/3)−

4�C1/3

(�C1/3)− 2/3

� For the solution to remain optimal we must have all Cj − Zj ≤ 0. That is

�C1 − 6 ≤ 0 i.e �C1 ≤ 6

(−10/3)− 4�C1/3 ≤ 0 i.e �C1 ≥ −5/2

(�C1/3)− 2/3 ≤ 0 i.e �C1 ≤ 2

� Thus, the range of values within which C1 may change without affecting the

current optimal solution is;

−5/2 ≤ �C1 ≤ 2 or

4− (5/2) ≤ C1 ≤ 4 + 2

i.e 3/2 ≤ C1 ≤ 6

6.2.2 Change in the Availability of Resources (bi)

Case I: While slack variable is not in the solution mix (i.e not in the basis)

– The procedure for finding the range for resource value within which

current optimal remain unchanged is;

80

1. Calculate exchange ratio (minimum ratio) for every row for slack

variables.

Exchange Ratio =Solution V alue XB

Coefficient in Slack V ariable Column

2. Find both lower and upper sensitivity limits

Lower limit = (Original V alue)− (Smallest positive

ratio or −∞)

Upper limit = (Original V alue) + (Smallest absolute negative

ratio or ∞)

Example: Given an optimal solution in Table 6.4

Table 6.4:

Cj −→ 4 6 2 0 0

CB B b(= xB) x1 x2 x3 S1 S2

4 x1 1 1 0 -1 4/3 -1/3

6 x2 2 0 1 2 -1/5 1/3

Z = 16 Zj 4 6 8 10/3 2/3

Cj − Zj 0 0 -6 -10/3 -2/3

– No slack variable in the solution mix (B).

– Thus ratios are calculated as;

B XB Coef.in S1

Column

Exchange

ratio

Coef.in S2

Column

Exchange

ratio

x1 1 4/3 1/(4/3)=3/4 -1/3 1/(-1/3)=-3

x2 2 -1/3 2/(-1/3)=-6 1/3 2/(1/3)=6

– Lower and upper limits for resources for constraints are calculated as;

Suppose b1 = 3 and b2 = 9 for constraint 1 and 2 respectively;

bi Solution Mix Lower Limit Upper Limit

3 x1(Manpower) 3-(3/4)=9/4 3+3=6

9 x2(Row material) 9-6=3 9+6=15

Case II: When slack variable is in the basis (Column CB)

– The procedure for finding the range of variation for corresponding right

hand of the constraint is as follows;

LL = Original V alue bi − Solution of Slack V ariable (XB)

LU = Infinity (∞)

81

Case III: Change in right hand side when constraints are mixed type

– When surplus is not in the basis (B)

LL = (Original V alue)− (Smallest absolute value of − ve exchange

ratio or −∞)

LU = (Original V alue)− (Smallest + ve minimum ratio or ∞)

– When surplus variable is in the basis

LL = Minus infinity (−∞)

LU = Original value+ Solution value of surplus variable

6.2.3 Change in the Input-Output Coefficient (a′ijs)

Case I: When a non-basic column ak ∈ B Changed to a∗k

– The only effect of such change will be on optimality condition. Thus the

solution will remain optimal if;

Ck − Z∗

k = Ck − CBB−1a∗k ≤ 0

otherwise the simplex method is continued, after column k of the simplex

table is updated, by introducing the non-basic variable xk into the basis.

– The range of discrete change �aij in the coefficient of non-basic variable

xj in the constraint i can be determined as follows;

Max = {Cj − Zj

CBβi > 0} ≤ aij ≤ Min{

Cj − Zj

CBβi < 0}

– βi is the ith column of B−1. If CBβi = 0 then �aij is unrestricted in

sign. B−1 is matrix of coefficients corresponding to slack variables in the

optimal solution table.

Case II: When a basic variable column ak ∈ B is changed to a∗k

– The condition to maintain both feasibility and optimality of the current

optimal solution are;

(a) Maxk=p{−xBk

xBkβpi − xBpβki > 0} ≤ �aij ≤

Mink=p{−xBk

xBkβpi − xBpβki < 0}

(b) Max{Cj − Zj

(Cj − Zj)βpi − ypjCBβi > 0} ≤ �aij ≤

Min{Cj − Zj

(Cj − Zj)βpi − ypjCBβi < 0}

82

6.2.4 Addition of a New Variable (Column)

– Let an extra variable xn+1 with coefficient Cn+1 be added in the system

of original constraint AX = B, X ≥ 0. Thus, it creates an extra column

an+1 in the matrix A of coefficients.

– To see the impact of this addition on the current optimal solution, we

compute;

yn+1 = B−1an+1 and

Cn+1 − Zn+1 = Cn+1 − CByn+1

– Two cases of the maximization LP model may arise;

1. If Cn+1 −Zn+1 ≤ 0, then XB = 0 and hence current solution remain

optimal.

2. If Cn+1 − Zn+1 ≥ 0, then the current optimal solution can be im-

proved by introducing a new column an+1 into the basis to find a

new optimal solution.

Example: Given the optimal solution in Table 6.5, discuss the effect on

optimality by adding a new variable with column coefficients (3,3,3) and

coefficient 5 in the objective function (Minimization)

Table 6.5:

Cj −→ 3 8 0 0 M

CB B b(= xB) x1 x2 S1 S2 A1

0 S2 60 0 0 -1 1 1

3 x1 80 1 0 1 0 0

8 x2 120 0 1 -1 0 1

Z = 1, 200 Zj 3 8 -5 0 8

Cj − Zj 0 0 5 0 M-8

� Solution: To see the changes in the optimal solution, we were given C7 =

5, a7 = (3, 3, 3)T then we calculate;

C7 − Z7 = C7 − CBB−1a7

= 5− (0, 3, 8)

⎡⎣ −1 1 1

1 0 0

−1 0 1

⎤⎦⎡⎣333

⎤⎦ = −4

a∗7 =

⎡⎣ −1 1 1

1 0 0

−1 0 1

⎤⎦⎡⎣333

⎤⎦ =

⎡⎣330

⎤⎦

83

� New column with variable x7 added as shown in Table 6.6

Table 6.6:

Cj −→ 3 8 0 0 M 5

CB B b(= xB) x1 x2 S1 S2 A1 x7 Min.Ratio

0 S2 60 0 0 -1 1 1 3 60/3→

3 x1 80 1 0 1 0 0 3 80/3

8 x2 120 0 1 -1 0 1 0 -

Z = 1, 200 Zj 3 8 -5 0 8 9

Cj − Zj 0 0 5 0 M-8 -4

� From the above table, variable x7 must enter into the solution and S2 should

leave it. The new solution is shown in Table 6.7

Table 6.7:

Cj −→ 3 8 0 0 M 5

CB B b(= xB) x1 x2 S1 S2 A1 x7

5 x7 20 0 0 -1/3 1/3 1/3 1

3 x1 20 1 0 2 -1 -1 0

8 x2 120 0 1 -1 0 1 0

Z = 1, 120 Zj 3 8 -11/3 -4/3 -4/3 5

Cj − Zj 0 0 11/3 4/3 M+4/3 0

� Since all Cj − Zj ≥ 0, optimal solution x1 = 20, x2 = 120, x7 = 20 and Min

Z=1,120

6.3 Solving LPP using LINDO

Introduction

Computer programs designed to solve LP problems are now widely available. Most

large LP problems can be solved with just a few minutes of computer time. Small

LP problems usually require only a few seconds. We will use LINDO to solve our

LP problems.

84

6.4 Using LINDO to Solve LPP

You can Download Free LINDO Software and follow the instructions which ex-

plains how to enter (and solve) a program in LINDO.

6.5 Interpretation of LINDO Output

We will discuss the following output:

� Objective function value

� Values of the decision variables

� Reduced costs

� Slack/surplus

Reduced Cost

The reduced cost for a decision variable whose value is 0 in the optimal solution

is the amount the variable’s objective function coefficient would have to improve

(increase for maximization problems, decrease for minimization problems) before

this variable could assume a positive value. The reduced cost for a decision

variable with a positive value is 0.

Example

Consider the following objective function:

Min 2x1 + 5x2 + 4x3 (6.1)

Suppose the optimal value ofx1 is zero, with a reduced cost of 1.2. Since this is

a minimization problem, this tells us that the current coefficient of x1 , which is

2, must be decreased by 1.2 in order for the optimal value of x1 to be non zero.

Thus if the objective function coefficient of x1 was 0.8 (or less), resolving the LP

would yield a non zero value of x1.

Slack/Surplus

The slack for “less than or equal to” constraints is the difference between the right

hand side of an equation and the value of the left hand side after substituting the

85

optimal values of the decision variables.

The slack represents the amount of unused units of the right hand side resources.

The surplus for “greater than or equal to]] constraints is the difference between the

right hand side of an equation and the value of the left hand side after substituting

the optimal values of the decision variables.

The surplus represents the number of units in which the optimal solution causes

the constraint to exceed the right hand side lower limit.

6.6 Tests and Final Examination Questions

FACULTY OF SCIENCE

DEPARTMENT OF MATHEMATICS

MTH 242: INTRODUCTION TO OPERATIONS RESEARCH

TEST NO. 1

Instructions:

1. Do all questions and answer each question in a separate page .

2. Write your registration number only.

3. Show all your work clearly to merit marks.

Saturday, 28th April, 2012 Time 90 min

1. What is OR? Give a brief historical development of OR.

2. A company makes two types of sofas, regular and long, at two locations, one

in Msamvu and one in Kihonda. The plant in Msamvu has a daily operating

budget of Tsh 45,000,000 and can produce at most 300 sofas daily in any

combination. It costs Tsh 150,000 to make a regular sofa and Tsh 200,000

to make a long sofa at the Msamvu plant. The Kihonda plant has a daily

operating budget of Tsh 36,000,000, can produce at most 250 sofas daily in

any combination and makes a regular sofa for Tsh 135,000 and a long sofa

for Tsh 180,000. The company wants to limit production to a maximum of

250 regular sofas and 350 long sofas each day. If the company makes a profit

86

of Tsh 50,000 on each regular sofa and Tsh 70,000 on each long sofa, how

many of each type should be made at each plant in order to maximize profit?

Formulate this problem as an LP model.

3. A patient in a hospital is required to have at least 84 units of drug A and

120 units of drug B each day. Each gram of substance M contains 10 units

of drug A and 8 units of drug B, and each gram of substance N contains

2 units of drug A and 4 units of drug B. Now suppose that both M and N

contain an undesirable drug C, 3 units per gram in M and 1 unit per gram

in N. How many grams of substances M and N should be mixed to meet the

minimum daily requirements at the same time minimize the intake of drug

C? How many units of the undesirable drug C will be in this mixture?

4. Under what condition an LP problem has

(a) Alternate (or multiple) optimal solution.

(b) Unbounded solution.

(c) Infeasible solution (No solution).

FACULTY OF SCIENCE

DEPARTMENT OF MATHEMATICS

MTH 242: INTRODUCTION TO OPERATIONS RESEARCH

TEST NO. 2

Instructions:

1. Do all questions and answer each question in a separate page .

2. Write your registration number only.

3. Show all your work clearly to merit marks.

Saturday, 19th May, 2012 Time 90 min

1. Define slack and surplus variables in a linear programming problem.

2. A diet is to contain at least 20 ounces of protein and 15 ounces of carbohy-

drate. There are three foods A, B and C available in the market, costing 2

USD, 1 USD and 3 USD per unit, respectively. Each unit of A contains 2

ounces of protein and 4 ounces of carbohydrate, each unit of B contains 3

87

ounces of protein and 2 ounces of carbohydrate and each unit of C contains

4 ounces of protein and 2 ounces of carbohydrate. How many units of each

food should the diet contain so that the cost per unit diet is minimum?

3. A transistor radio company manufactures four models A, B, C and D which

have profit contributions of USD 8, USD 15 and USD 25 on models A, B and

C respectively and a loss of USD 1 on model D. Each type of radio requires a

certain amount of time for the manufacturing of components for assembling

and for packing. A dozen units of model A require an hour of manufacturing,

two hours for assembling and one hour for packing. The corresponding figure

for a dozen units of model B are 2, 1 and 2 and for a dozen units of C are

3, 5 and 1, while a dozen units of model D require 1 hour of packing only.

During the forthcoming week, the company will be able to make available

15 hours of manufacturing, 20 hours of assembling and 10 hours of packing

time. Determine the optimal production schedule for the company.

FACULTY OF SCIENCE

DEPARTMENT OF MATHEMATICS

MTH 242: INTRODUCTION TO OPERATIONS RESEARCH

TEST NO. 3

Instructions:

1. Do all questions and answer each question in a separate page .

2. Write your registration number only.

3. Show all your work clearly to merit marks.

Saturday, 02nd June, 2012 Time 120 min

1. (i) Explain the term Degeneracy in the context of LPP and discuss a method

to resolve the degeneracy in LPP.

(ii) Solve the following with the help of Simplex method

Max Z = 3x1 + 9x2

subject to

x1 + 4x2 ≤ 8

x1 + 2x2 ≤ 4

88

and

x1, x2 ≥ 0

2. (i) Write the dual of the given primal;

Max Zx = x1 − 10x2 + 2x3 − 3x4

subject to

5x1 − x2 + 3x3 + 6x4 ≤ 15

−x1 + 2x2 − x3 + 7x4 = 30

and

x1, x2 ≥ 0 and x3, x4 unrestricted in sign.

(ii) What are advantages of Duality in LPP.

3. A company produce three products: A, B and C. Each product requires two

raw materials: steel and aluminium. The following LP model describes the

company’s product mix problem.

Max Z = 30xA + 10xB + 50xC

subject to

6xA + 3xB + 5xC ≤ 450(Steel)

3xA + 4xB + 5xC ≤ 300(Aluminium)

and

xA, xB, xC ≥ 0

The optimal product plan is the following table where SS and SA are the

Cj −→ 30 10 50 0 0

CB B b(= xB) xA xB xC SS SA

0 SS 150 3 -1 0 1 -1

50 xC 60 3/5 4/5 1 0 1/5

Z = 16 Zj 30 40 50 0 10

Cj − Zj 0 -30 0 0 -10

slack variables for unused steel and aluminium quantity, respectively.

(a) Determine the sensitivity limits for the available steel and aluminium

within which the present product mix will remain optimal.

(b) Find the new optimal solution when available steel is 300 tonnes and

aluminium is 400 tonnes.

89

REFERENCES

Ackoff, R.L and M. W. Sasieni (1968), Fundamentals of Operations Research,

John Wiley & Sons.

Cooper, L and D. Steinberg (1974), Methods and Applications of Linear Program-

ming, W. B. Saunders Company, Philadelphia.

Dantzig, G., and M. Thapa (1997), Linear Programming 1: Introduction, Springer,

New York.

Ecker, J. G and M. Kupferschmid (1988), Introduction to Operations research,

John Wiley & Sons.

Fabryckey, W. J., P. M. Ghare and P. E. Torgersen (1987), Applied Operations

Research and Management Science, Prentice-Hall of India.

Hiller, L. S. and G. J. Lieberman (1985), Introduction to Operations Research

(4thEdn), C. B. S. Publishers and Distributors (India).

Hamdy: T.A. (1987), Operations Research: an Introduction, MacMillan Publish-

ing Company New York.

Schrage, L. (1999), Optimization Modeling with LINGO, LINDO Systems, Inc.,

Chicago.

Sharma, J. K (2001), Quantitative Techniques For Managerial Decisions, Macmil-

lan India Ltd, New Delhi.

Sharma, J. K (2003), Operations Research: Theory and Application (2nd Ed),

Macmillan India Ltd, New Delhi

Shepard, R., D. Hartley, P. Hasman, L. Thorpe, and M. Bathe (1988), Applied

Operations Research, Plenum Press, New York.

Thierauf, R. J and R. L. Klekamp, (1970), Decision Making Through Operations

Research, John Wiley & Sons, New York.

WinstonW.C. (1994), Operations Research: Application and Algorithms, Duxbury

Press California.

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