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Introduction to Operations Research: Theory and Applications
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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
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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