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OPERATIONS RESEARCHVENKATESHWARA

OPEN UNIVERSITYwww.vou.ac.in

OPERATIONS RESEARCHOPERATIONS RESEARCH

10 MM

VENKATESHWARAOPEN UNIVERSITY

www.vou.ac.in

MA[MEC-203]

OPERATIONS RESEARCH

MA[MEC-203]

S. Kalavathy, Professor, Department of Mathematics, RMD Engineering College, Kaveripettai, Thiruvallur

(Units: 1.3, 2.1-2.2, 2.3.4-2.4, 2.6-2.10, unit-3, 5.1-5.6, 5.8-5.12) © Reserved, 2019

C.R. Kothari, Ex-Associate Prof, Dept. of Economic Administration and Fin. Mgt, University of Rajasthan

(Units: 1.1-1.2, 1.4-1.12, 2.3-2.3.3, 2.5, unit-4, 5.7) © C.R. Kothari, 2019

BOARD OF STUDIES

Prof Lalit Kumar SagarVice Chancellor

Dr. S. Raman IyerDirectorDirectorate of Distance Education

SUBJECT EXPERT

Bhaskar Jyoti NeogDr. Kiran KumariMs. Lige SoraMs. Hage Pinky

Assistant ProfessorAssistant ProfessorAssistant ProfessorAssistant Professor

CO-ORDINATOR

Mr. Tauha KhanRegistrar

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COURSE INTRODUCTION

Operations research approach has become a new tool for managing conventional management problems. In fact,operational research techniques do constitute a scientific methodology of analysing the problems of the businessworld. They provide an improved basis for taking management decisions. The practice of OR helps in tacklingintricate and complex problems such as that of resource allocation, product-mix, inventory management, sequencingand scheduling, replacement, and a host of similar problems of modern business and industry.

With IT facilities becoming widely available, the significance and scope of OR has grown, and is stillgrowing. Hence, OR is now an integral part of courses of computer science, economics, business management,public administration and several other disciplines.

Keeping all these in view, the book, Operations Research, introduces the basic elements of OR in thesimplest possible way and explains how the various approaches in OR are applied by managers.

Designed in keeping with the self-instruction mode format, each unit of the book begins with an outline ofthe Learning Objectives followed by an Introduction to the topic. The content is then presented in a simple andeasy-to-understand manner, and is interspersed with Check Your Progress questions to test the reader’sunderstanding of the topic. A list of Possible Questions is also provided at the end of each unit.

The book is divided into five units.

Unit 1: Introduction to Operations Research

Unit 2: Linear Programming Problem

Unit 3: Transportation and Assignment Models

Unit 4: Game Theory

Unit 5: Network Models

DETAILED SYLLABUS

Unit 1: Nature and Scope of Operations Research, Significance of OR, Problem Formulation, Methodologyof OR, Applications and Scope of OR, Basic OR Models.

Unit 2: Formulation of LP, Solving LP Problems, Special Cases in LP, Concept of Duality and SensitivityAnalysis.

Unit 3: Transportation Problem, Transportation Models, Optimum Solution using MODI Method, SpecialCases, Transshipment Model, Assignment Problem.

Unit 4: Game Theory: Basic Concepts, Pure and Mixed Strategies with Saddle Point, Limitations ofGame Theory.

Unit 5: Construction of Networks, Determination of Floats and Slack Times, Determination of CriticalPaths, Determination of PERT Times, Crashing, Resource Allocation and Levelling.

CONTENTS

UNIT 1 INTRODUCTION TO OPERATIONS RESEARCH 1-321.1 Learning Objectives1.2 Introduction1.3 Nature and Scope of Operations Research1.4 Significance of OR1.5 Problem Formulation1.6 Methodology of OR1.7 Applications and Scope of OR1.8 Basic OR Models

1.8.1 OR Techniques1.9 Let Us Sum Up

1.10 Further Reading1.11 Answers to ‘Check Your Progress’1.12 Possible Questions

UNIT 2 LINEAR PROGRAMMING PROBLEM 33-832.1 Learning Objectives2.2 Introduction2.3 Formulation of LP

2.3.1 Meaning of Linear Programming2.3.2 Fields where Linear Programming can be Used2.3.3 Components of a Linear Programming Problem2.3.4 Formulation of a Linear Programming Problem

2.4 Solving LP Problems2.4.1 Procedure for Solving LPP by Graphical Method2.4.2 Some Important Definitions2.4.3 Canonical and Standard Forms of LPP2.4.4 Simplex Method

2.5 Special Cases in LP2.5.1 Degeneracy2.5.2 Non-Feasible Solution2.5.3 Unbounded Solution2.5.4 Multiple Solution or Alternate Optima

2.6 Concept of Duality and Sensitivity Analysis2.6.1 Formulation of a Dual Problem2.6.2 Definition of a Dual Problem2.6.3 Important Results in Duality2.6.4 Sensitivity Analysis

2.7 Let Us Sum Up2.8 Further Reading2.9 Answers to ‘Check Your Progress’

2.10 Possible Questions

UNIT 3 TRANSPORTATION AND ASSIGNMENT MODELS 85-1423.1 Learning Objectives3.2 Introduction3.3 Transportation Problem3.4 Transportation Models

3.4.1 Finding the Initial Basic Feasible Solution3.4.2 North West Corner Rule3.4.3 Least Cost or Matrix Minima Method3.4.4 Vogel’s Approximation Method (VAM)

3.5 Optimum Solution using MODI Method3.6 Special Cases

3.6.1 Maximization Case in a Transportation Problem3.6.2 Degeneracy in Transportation Problems3.6.3 Unbalanced Transportation3.6.4 Alternate Solution

3.7 Transshipment Model3.8 Assignment Problem

3.8.1 Mathematical Formulation of the Assignment Problem3.8.2 Hungarian Method Procedure3.8.3 Maximization in Assignment Problem

3.9 Let Us Sum Up3.10 Further Reading3.11 Answers to ‘Check Your Progress’3.12 Possible Questions

UNIT 4 GAME THEORY 143-1704.1 Learning Objectives4.2 Introduction4.3 Game Theory: Basic Concepts

4.3.1 Types of Games4.3.2 Value of a Game

4.4 Pure and Mixed Strategies with Saddle Point4.4.1 Method of Dominance4.4.2 Linear Programming Solution to Two-Person Zero-Sum Games

4.5 Limitations of Game Theory4.6 Let Us Sum Up4.7 Further Reading4.8 Answers to ‘Check Your Progress’4.9 Possible Questions

UNIT 5 NETWORK MODELS 171-2165.1 Learning Objectives5.2 Introduction5.3 Construction of Networks

5.3.1 Basic Terms5.3.2 Common Errors5.3.3 Rules of Network Construction5.3.4 Time Analysis

5.4 Determination of Floats and Slack Times5.5 Determination of Critical Paths5.6 Determination of PERT Times

5.6.1 PERT Procedure5.7 Crashing5.8 Resource Allocation and Levelling5.9 Let Us Sum Up


Introduction to Operations Research

1Operations Research

Unit-1

UNIT 1 INTRODUCTION TOOPERATIONS RESEARCH

UNIT STRUCTURE1.1 Learning Objectives1.2 Introduction1.3 Nature and Scope of Operations Research1.4 Significance of OR1.5 Problem Formulation1.6 Methodology of OR1.7 Applications and Scope of OR1.8 Basic OR Models

1.8.1 OR Techniques1.9 Let Us Sum Up


1.1 LEARNING OBJECTIVES

After going through this unit, you will be able to:

Explain the nature and scope of operations research

Explain the methodology of OR

Explain the role of OR in managerial decision-making

Analyse the significance of OR

Explain the basic OR models

1.2 INTRODUCTION

This unit introduces you to operations research, where you will first trace theevolution of OR. You will then learn the various techniques in OR which are appliedin taking decisions by management. You will understand the significance ofoperations research in today’s business environment. You will also learn about someof the basic OR models.

1.3 NATURE AND SCOPE OF OPERATIONS RESEARCH

The term, ‘Operations Research’ was first coined by J.F. McCloskey and F.N.Trefethen in 1940. This new science came into existence in a military context.During World War II, military management called on scientists from variousdisciplines and organized them into teams to assist in solving strategic and tacticalproblems, relating to air and land defence of the country. Their mission was to formulate

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specific proposals and plans for aiding the military commands to arrive at decisions onoptimal utilization of scarce military resources and efforts and also to implement thedecisions effectively. This new approach to the systematic and scientific study of theoperations of the system was called Operations Research (OR) or operational research.Hence, OR can be referred to as ‘an art of winning the war without actually fighting it.’

Definitions

Operations research (OR) has been defined so far in various ways and it is perhaps stilltoo young to be defined in some authoritative way. It is not possible to give uniformly-acceptable definitions of OR. The following are definitions proposed by various specialistsin the field of OR. These have been changed according to the development of thesubject.

OR is a scientific method of providing executive departments with a quantitativebasis for decisions regarding the operations under their control.

Morse and Kimbal (1946)

OR is the scientific method of providing the executive with an analytical andobjective basis for decisions. P.M.S. Blackett (1948)

OR is the art of giving bad answers to problems to which otherwise worseanswers are given. T.L. Saaty (1958)

OR is a systematic method oriented study of the basic structures,characteristics, functions and relationships of an organization to provide theexecutive with a sound, scientific and quantitative basis for decision-making.

E.L. Arnoff and M.J Netzorg

OR is a scientific approach to problem solving for executive management.

H.M. Wagner

OR is an aid for the executive in making his decisions by providing him withthe quantitative information based on the scientific method of analysis.

C.Kittee

OR is the scientific knowledge through interdisciplinary team effort for thepurpose of determining the best utilization of limited resources.

H.A. Taha (2006)

The various definitions given here, bring out the following essentialcharacteristics of operations research:

(i) Systems orientation(ii) Use of interdisciplinary terms

(iii) Application of scientific methods(iv) Uncovering new problems(v) Quantitative solutions

(vi) Human factors

Scope of Operations Research

There is great scope for economists, statisticians, administrators and techniciansworking as a team to solve problems of defence by using the OR approach. Besidesthis, OR is useful in various other important fields such as the following:



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(i) Agriculture

(ii) Finance

(iii) Industry

(iv) Marketing

(v) Personnel management

(vi) Production management

(vii) Research and development

Phases of Operations Research

The procedure to be followed in the study of OR, generally involves the following majorphases:

(i) Formulating the problem

(ii) Constructing a mathematical model

(iii) Deriving a solution from the model

(iv) Testing the model and its solution (updating the model)

(v) Controlling the solution

(vi) Implementation

1.4 SIGNIFICANCE OF OR

OR has gained increasing importance since World War II in the technology of businessand industry administration. It greatly helps in tackling the intricate and complexproblems of modern business and industry. OR techniques are, in fact, examples ofthe use of scientific method of management. The significance of OR can be wellunderstood under the following heads:

(i) OR provides a tool for scientific analysis: OR provides the executives witha more precise description of the cause and effect relationship and risksunderlying the business operations in measurable terms and this eliminatesthe conventional intuitive and subjective basis on which managements usedto formulate their decisions decades ago. In fact, OR replaces the intuitiveand subjective approach of decision-making by an analytical and objectiveapproach. The use of OR has transformed the conventional techniques ofoperational and investment problems in business and industry. As such, ORencourages and enforces disciplined thinking about organizational problems.

(ii) OR provides solution for various business problems: The OR techniquesare being used in the field of production, procurement, marketing, financeand other allied fields. Problems like: ‘How best can the managers andexecutives allocate the available resources to various products so that in agiven time the profits are maximum or the cost is minimum? Is it possiblefor an industrial enterprise to arrange the time and quantity of orders of itsstocks such that the overall profit with given resources is maximum? How far is

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it within the competence of a business manager to determine the number of menand machines to be employed and used in such a manner that neither remains idleand at the same time the customer or the public has not to wait unduly long forservice?’, and similar other problems can be solved with the help of OR techniques.Similarly, we might have a complex of industries—steel, machine tools and others—all employed in the production of one item, say steel. At any particular time wehave a number of choices of allocating resources, such as money, steel and toolsfor producing autos, building steel factories or tool factories. What should be thepolicy which optimizes the total number of autos produced over a given period?OR techniques are capable of providing an answer in such a situation.

Planning decisions in business and industry are largely governed by the picture ofanticipated demands. The potential long range profits of the business may vary inaccordance with different possible demand patterns. The OR techniques serve todevelop a scientific basis for coping with the uncertainties of future demands.Thus, in dealing with the problem of uncertainty over future sales and demands,OR can be used to generate ‘a least risk’ plan.At times, there may be a problem in finding an acceptable definition of long rangecompany objectives. Management may be confronted with different viewpoints—some may stress the desirability of maximizing net profit whereas others mayfocus attention primarily on the minimization of costs. OR techniques (especiallythat of mathematical programming, such as linear programming) can help resolvesuch dilemmas by permitting systematic evaluation of the best strategies forattaining different objectives. These techniques can also be used for estimatingthe worth of technical innovations as also of potential profits associated with thepossible changes in rules and policies.

How much changes can be there in the data on which a planning formulation isbased without undermining the soundness of the plan itself? How accuratelymust managements know the cost coefficients, production performance figuresand other factors before it can make planning decisions with confidence? Manyof the basic data required for the development of long-range plans are uncertain.Such uncertainties though cannot be avoided, but through various ORtechniques, the management can know how critical such uncertainties are andthis in itself is a great help to business planners.

(iii) OR enables proper deployment of resources: OR renders valuable help inproper deployment of resources. For instance, Programme Evaluation andReview Technique (PERT) enables us to determine the earliest and the latesttimes for each of the events and activities and thereby helps in the identificationof the critical path. All these help in the deployment of resources from oneactivity to another to enable the project completion on time. This techniquethus provides for determining the probability of completing an event or projectitself by a specified date.

(iv) OR helps in minimizing waiting and servicing costs: The waiting line orqueuing theory helps the management in minimizing the total waiting and servicing

PERT: Enables us todetermine the earliestand the latest times foreach of the events andactivities and therebyhelps in theidentification of thecritical path.



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costs. This technique also analyses the feasibility of adding facilities and therebyhelps the business people in taking a correct and profitable decision.

(v) OR enables the management to decide when to buy and how much tobuy: The main objective of inventory planning is to achieve balance betweenthe cost of holding stocks and the benefits from stock holding. Hence, thetechnique of inventory planning enables the management to decide when tobuy and how much to buy.

(vi) OR assists in choosing an optimum strategy: Game theory is specially usedto determine the optimum strategy in a competitive situation and enables thebusinessmen to maximize profits or minimize losses by adopting the optimumstrategy.

(vii) OR renders great help in optimum resource allocation: The linearprogramming technique is used to allocate scarce resources in an optimummanner in problems of scheduling, product mix, and so on. This technique ispopularly used by modern managements in resource allocation and in affectingoptimal assignments.

(viii) OR facilitates the process of decision-making: Decision theory enablesthe businessmen to select the best course of action when information is given inthe probabilistic form. Through decision tree (a network showing the logicalrelationship between the different parts of a complex decision and thealternative courses of action in any phase of a decision situation) technique,executive’s judgement can systematically be brought into the analysis of theproblems. Simulation is another important technique used to imitate anoperation or process prior to actual performance. The significance ofsimulation lies in the fact that it enables in finding out the effect of alternativecourses of action in a situation involving uncertainty where mathematicalformulation is not possible. Even complex groups of variables can be handledthrough this technique.

(ix) Through OR management you can know the reactions of the integratedbusiness systems: The Integrated Production Models technique is used tominimize cost with respect to workforce, production and inventory. Thistechnique is quite complex and is usually used by companies having detailedinformation concerning their sales and costs statistics over a long period.Besides, various other OR techniques also help management people in takingdecisions concerning various problems of business and industry. Thetechniques are designed to investigate how the integrated business systemwould react to variations in its component elements and/or external factors.

(x) OR techniques help a lot in the preparation of future (or would be)managers: OR techniques substitute a means for improving the knowledgeand skill of youngsters in the field of management.

Decision theory:Enables thebusinessmen to selectthe best course ofaction wheninformation is given inprobabilistic form.

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CHECK YOUR PROGRESS-1

1. List some characteristic features of OR.

2. Why is OR considered to be a science as well as an art?

3. What happens in cases where analytical as well as numerical methods cannotbe used to derive the solution?

4. What are the various classifications of problems in OR?

5. When is an allocation problem referred to as a linear programming problem?

6. What is the objective in search problems?

1.5 PROBLEM FORMULATION

Nature of the Problem

To formulate a problem correctly one must know what a problem is and mustunderstand the nature of the problem. A problem is said to exist when we have thefollowing conditions satisfied1:

(i) There must be an individual or a group, which has some difficulty or problem.(ii) There must be some objectives to be attained. If one wants nothing, one

cannot have a problem.(iii) There must be alternative means (or the courses of action) for attaining the

objective(s) one wishes to attain. This in other words means that there mustbe atleast two means available. If there is no choice of means, one cannothave a problem.

(iv) There must remain doubt with regard to selection of alternatives, i.e., therelative efficiency of the alternative means must differ.

(v) There must be some environment(s) to which the difficulty pertains.

The nature of the problem must be understood in context of the above pointsbefore its formulation takes place. It is only after this that the problem must bestated along with the objective(s) and the bounds within which it is to be studied. Infact, this involves the task of laying down boundaries within which we shall studythe problem with a predetermined objective in view. In brief, the task of formulatingthe problem consists in making various components of the problem explicit.

Formulating the Objectives

Generally, the objectives of an organization may either be retentive or acquisitive.Retentive objectives are concerned with preserving either resources of value, such as

1 All these conditions have been narrated as components of a research problem by Ackoff Russel, L. in1961. The Design of Social Research. Chicago: Chicago University Press.



Unit-1

money, time, energy, etc., or states such as comfort, safety, stability, and the like. Inother words, retentive objectives are concerned with what is consumed by courses ofaction and as such can easily be translated as minimizing objectives. On the other hand,we may have acquisitive objectives which are concerned with acquiring resources orattaining states and can easily be translated as maximizing objectives. Essentially thenwe may have either minimizing objectives or maximizing objectives. Accordingly, profits,sales, revenues, output and the like are always to be maximized and cost, time and thelike are to be minimized.

Formulation of the problem and the objective often requires a detailed knowledgeof how the concerning system operates, which can be learnt through performing asystem analysis, which is a process analogous to the physical examination that adoctor performs on his patient after initial discussion of symptoms. Such an analysiscan provide the background information required in formulating the problem and themodel required to solve it.2

Measure of Performance

After understanding the nature of the problem and after identifying the courses ofaction, the uncontrolled variables and the objectives, it is necessary to construct ameasure of performance that can be used to determine which alternative is best andwhat function of this measure (usually described as the objective function) shouldbe used as the criterion of best solution of the concerning problem. Selecting acriterion of the best solution often requires a good knowledge of decision theory.The type of appropriate criterion depends on the state of knowledge of outcomesthat we assume to be available. Usually, we may have one of the three situationswith regard to the state of outcomes, viz., deterministic situation or state of certainty,stochastic situation or risk situation, and finally the situation of uncertainty. Theyare discussed as follows.

(i) Deterministic situations: These are situations in which each course of actionresults in only one outcome. In such situations, the probability of each outcomeis known or believed to be either equal to unity or zero. Because of this, thecriterion or the measure of performance used in deterministic situations is thatof the maximization of utility.

(ii) Stochastic situations: These are situations in which each course of action canresult in alternative outcomes; the probabilities of which are known or can beestimated. The criterion that is generally accepted as the most appropriate insuch situations is that of maximizing the expected utility and is often termedas either the Expected Monetary Value (EMV) criterion or the ExpectedOpportunity Loss (EOL) criterion. To select the best alternative, we selectthat alternative of which EMV is highest possible (or EOL is the lowestpossible).

(iii) Uncertainty situations: These are situations in which each course of actioncan result in alternative outcomes, the probabilities of which cannot be assigned,as the decision-maker does not know which outcomes can or will occur. This

Deterministicsituations: Situationsin which each course ofaction results in onlyone outcome.

2 Ackoff, Russell L. and Maurice W. Sasieni. 1968. Fundamental of Operations Research. New York:John Wiley and Sons. pp. 25–29.

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means that the decision-maker has to act with imperfect information in such asituation. Consequently, there is no single best criterion for selecting a strategy todeal with such a situation. Accordingly, various criteria have been developed forselecting a course of action in uncertainty situations. The following criteria deservesmention in this context:

(i) Maximin decision rule(ii) Maximax decision rule(iii) Minimax decision rule (or regret rule)(iv) Hurwicz decision rule(v) Laplace decision rule

Nothing can be said as to which one of these criteria is the best criterion in any givensituation involving uncertainty. The choice for the selection must be left to the decision-maker who must resort to his own judgement, skill and experience.

Thus, problem formulation, which is considered as the first phase of an ORstudy, requires a statement of the problem’s components that include the controllableor the decision variables, the uncontrollable parameters or the environment, theconstraints on the variables and also the objective along with an appropriate measureof performance.

Model Construction

After problem formulation, modelling plays an important part in OR. In fact, it isconsidered as the essence of operations research approach. Modelling permits theoperational researcher to test rigorously the implications of new plans, new schemesand new ideas. The term modelling generally refers to the idea of constructing amodel of the given problem and then deriving the solution from it for the problemunder consideration. Modelling is a device used to arrive at a well-structured viewof reality. By analysing or experimenting on models, we can usually determine howchanges in the relevant system will affect its performance.

What is a model?

A model is a physical or symbolic representation of the relevant aspects of thereality or system with which one is concerned. In other words, a model is a meansof portraying the system or reality of concern to the decision-maker. As such, theconcept of a model generally implies a series of connected and identifiablerelationships that essentially demonstrate the proposition: if this action, then thisresult. Thus, ‘the model, being an abstraction of the assumed real system, identifiesthe pertinent relationships of the system in the form of an objective and a set ofconstraints.’3

Types of models

Models can be of several types, but models commonly referred are as follows:

Iconic models: Iconic model is a pictorial or visual representation of certainaspects of a system. In iconic models, the relevant properties of the real thingare represented by the properties themselves, usually with a change of scale.

3 Taha, Hamdy A. 2003. Operations Research, Fourth edition. New Jersey: Prentice Hall. p. 6.

Model: A physical orsymbolicrepresentation of therelevant aspects of thereality or system withwhich one isconcerned.



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Such models generally look like what they represent, but differ in size. An iconicmodel is said to be ‘scaled down’ or ‘scaled up’ as the dimensions of the modelare smaller or bigger than those of the real thing. For instance, iconic models ofthe sun housed in planetoriums, the model of a huge building or of an airplane orof an automobile for exhibition purposes are scaled down; whereas models of theatom or of a cell used for, say teaching purposes are scaled up. Similarly, maps,photographs, drawings also constitute examples of iconic models. Iconic models,such as models of aircraft, machine, engine or of any tangible thing may also betermed as physical models. Iconic models are generally specific, concrete andeasy to observe, but are difficult to manipulate for experimental purposes and arenot very useful for the purpose of prediction. Moreover, such models represent astatic event only.

Analogue models: Analogue models use one set of properties to represent anotherset of properties. They are more abstract than iconic models. However, suchmodels are easier to manipulate and can represent dynamic situations. Graphsrepresenting time series, stock market changes or other phenomenons, flow charts,demand curves, frequency graphs are examples of analogue models for they usegeometrical magnitudes and location to represent a wide variety of variables andthe relationships between them. Similarly, a hydraulic system can be used as ananalogue model of electrical, traffic and economic systems. Maps in differentcolours, each colour in a map representing a particular characteristic, such asblue colour representing water, brown representing land, and so on, are also termedas analogue models. Such models are generally more useful than iconic modelsbecause of their relatively greater capacity to represent the characteristics of thereal system.

Symbolic or mathematical models: In symbolic models, the components ofwhat is represented and their interrelationships are given by symbols. Thesemodels use letters, numbers and other types of symbols to represent variablesand the relationships between them. Such models generally assume the formof equations or inequalities depicting the relationships amongst the variablesof the system which they represent. For instance, a mathematical equationrepresenting a relationship between constants and variables may be considereda symbolic model. Thus, a symbolic model means a mathematical descriptionof an activity, which expresses the relationships among the various elementswith sufficient accuracy that it can be used to predict the actual outcomeunder any expected set of circumstances. Symbolic models are most generaland precise and also are most abstract in nature. They are usually the easiestto manipulate experimentally.

A great advantage of such models is that they lend themselves easily formanipulation on computers. In most OR applications, it is assumed that theobjective and constraints can be expressed mathematically as functions ofdecision variables. In such a case we say that we are dealing with a symbolicor a mathematical model. Important symbolic or mathematical models oftenused in operations research are as follows:

(i) Allocation models: They are used in finding a solution of optimizing agiven objective such as profit maximization under certain restrictions.

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(ii) Scheduling models: They are used in determining an optimum sequenceof performing certain operations of jobs with a view to minimize overalltime and/or cost. PERT and other network techniques fall under thiscategory.

(iii) Queuing models: They represent the random arrival of customers at aservice station where the facility is limited and the object of such modelshappens to be to minimize the costs of both servicing and waiting.

(iv) Inventory models: These models relate mainly to optimizing inventorylevels. The main emphasis in such models is on minimizing the total costsassociated with inventories.

(v) Replacement models: These models are concerned with replacementstrategies.

(vi) Competitive models: These models are concerned with the determinationof optimum strategies under competitive conditions. Game theory modelsare examples of competitive theory.

These wide variety of models used in OR give rise to a corresponding numberof solution techniques, such as linear, integer, dynamic, non linearprogramming that represent algorithms for solving the models.

Simulation models: At times, however, it may not be possible to depict a realsystem in mathematical formulation, i.e., it may just happen that amathematical model of a given problem cannot be constructed and as suchwe may adopt simulation models.

Simulation models differ from mathematical models in that the relationshipsbetween input and output are not explicitly stated. A simultation model breaks downthe modelled system into basic modules that are then linked to one another by a well-defined logical relationship (in the form of IF/THEN).

Some other models

(i) Functional, operational, and accounting models: A functional model is a modelwithout values specified for the constants and as such the variables can only beidentified in abstract mathematical notation. A functional model specifies theform of the relationship, but not the exact value of the relationship. However,when the values which must be associated with some measurable quantity inthe environment being represented have been identified, we have an operatingmodel. For instance, Y

i= a

i+ b

iX

iis an example of a functional model, but

when we identify the values of the two constants such as a1 = 4 and b

1 = 3 or

a2

= 5 and b2= 4 then we can write the said model either as Y

1 = 4 + 3X

1 or Y

2

= 5 + 4X2 and in this form it is known as an operating model. Certain equations

are essentially tautologies because they merely represent addition of values togive a total value; thus, in a sense they are related to accounts. Suppose, if weare given Y

1and Y

2. The total production, Y

iis equal to Y

1 + Y

2. This is nothing

but an accounting statement. Does it constitute a model? In business world weassume it to be a model (and consider it an accounting model) because it doesappear to represent symbolically relevant aspects of the reality with which weare often concerned.



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(ii) Prediction, improvement and optimization models: Mostly this classification isstated in context of managerial analysis. Optimization models are the result ofquantitative analysis and allow the measurable criterion to assume the highest orthe lowest possible value, given the model structure and constraints. Improvementmodels are the result of system analysis and consequently the highest value of thecriterion is chosen from among the alternatives investigated. Either of these modelsdepends upon inputs which predict outcomes of relationship; thus, they dependupon a prediction model.4

(iii) Static, dynamic, linear and non-linear models: According to Joseph WrightForrester, all models possess differing degrees of various characteristics, viz.,staticness, linearity, stability and openness. In view of this, models can aswell be classified as static model or dynamic model, linear model or non-linear model. A static model is one in which time plays no role, i.e, therelationships do not vary with time. As against this, a dynamic model containstime-varying interactions. A linear model is one in which all of the equationsor inequalities are linear, but a model which contains one or more non-linearequations is known as a non-linear model. In case of non-linear models, thedegree of non-linearity may vary greatly from model to model.

(iv) Stable and unstable models: In the words of Forrester, ‘A stable system isone that tends to return to its initial condition after being disturbed. It mayovershoot and oscillate like a single pendulum that is set in motion, but thedisturbances decline and die out. In an unstable system that starts at rest, aninitial disturbance is amplified, leading to growth or oscillations whoseamplitude increases.’5 Another sub-element of stability is concerned with thepresence of steady state or transient models. Using the words of Forresteragain, we can say that a steady state pattern is one that is repetitive with timein which the behaviour in one period is of the same nature as any other period;whereas the transient behaviour refers to those changes where the characterof the system changes with time.6

(v) Open and close models: On the basis of the characteristic of openness, amodel can be classified either as an open model or as a closed model. Aclosed model is one that internally generates the values of variables throughtime by the interaction of variables one on another; whereas the open modelis one in which some variables are supplied from the external environment.

Properties of a good model

We can list down the properties of a good model as follows:

(i) A good model must represent the reality or the system with which it isconcerned as accurately as possible and must not compromise with any ofthe critical elements of the system. However, this does not mean that a modelshould be as close as possible to reality in every respect for then it wouldrequire an infinite length of time to construct such a model, which at times may

4 Schellenberger. Robert E. 1969. Managerial Analysis. Richard D. Irwin Inc. pp. 88–89.5 Joseph Wright Forrester, Industrial Dynamics. 1961. Massachusetts: Pegasus Communications, p.51.6 Ibid.

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even be beyond human comprehension. Thus, a good model must be economicalof time and also be readily workable. Usable approximate models are muchbetter than more exact models that cannot be used.

(ii) Since a model happens to be a simplified representation of an empirical situation;it must possess the characteristic of predicting outcomes reasonably well andmust be consistent with effective action.

(iii) A good model must be simple enough so as to be readily understood and, ifnecessary, must be capable of being modified quickly and effectively.

(iv) A good model must identify the factors that can influence the course of aparticular decision and must also determine how these factors interact toproduce departures from the expected or desired outcome.7

(v) A model must make certain assumptions about the structure of the problemunder study for no system of equations can produce meaningful results if itconsists entirely of unknown, but the number of assumptions should be assmall as possible in a good model.

(vi) A good model is one that is capable of answering in case when smallperturbations take place in the data. This sensitivity of the model, i.e., itscapacity to answer in context of small changes constitutes an important virtueof a good model.

Advantages of modelling

There are several advantages that result from modelling. Important ones are asfollows:

(i) The main advantage of modelling is that the model formulated under it canbe manipulated in a variety of ways until the best solution of the problemunder consideration is found. The actual system or the organization can thenbe operated in an optimum fashion with a minimum of costly experimentsand adjustments. For instance, models of aircraft are always tested in windtunnels before production is started.

(ii) Models can be used on a ‘what if’ basis to explore the possible futureconsequences of decisions that might be made today and this enables thecreative manager to test the implications of new plans, new schemes and newideas that are most critical to the success or failure of a particular project.Modelling, thus helps in finding avenues for research and improvements.

(iii) Modelling provides an easy and systematic approach for studying the problemconcerned. Through this technique, the problem under study becomescontrollable. Had the models been as complex and difficult to control asreality, then there would have been no advantage in their use. In fact, modellingstrips a natural phenomenon of its bewildering complexity and duplicates theessential behaviour of the natural phenomenon with a few variables, simplyrelated.

Modelling: Providesan easy and systematicapproach for studyingthe problem concerned.

7 Johnson, Rodney D. and Bernard R. Siskin. 1977. Quantitative Techniques for Business Decisons.New Jersey: Prentice Hall p. 3.



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(iv) Modelling provides an opportunity to explore a broad range of solutions and toestimate the sensitivity of the solution to changes in individual elements or incombinations of these elements. This in turn will enable the decision-maker tomove away safely from the precise optimum point by a given margin in orderto accommodate other factors. Thus, modelling indicates the limitations andscope of an activity.

(v) Model building and analysis does not replace intuition and judgement; it rathersupports them with tools for handling complexity and uncertainty with whichthe human intuition cannot cope unaided. In a sense, the use of a model freesthe intuition and permits it to concentrate on those problems to which it isparticularly suited.8

(vi) Modelling enables quick and economical experimentation for finding anoptimum solution for a given problem. The results of such experiments canbe evaluated to determine which strategy is most consistent with a desiredobjective. Besides, the validity of the model (i.e., whether the model resultsin a reliable prediction of the system’s performance?) can be checked and incase of even little doubt, the model can accordingly be modified quickly andeffectively.

Deriving Solutions from Models

(a) Abstraction

Abstraction happens to be the first step in modelling and consists in selecting thecritical factors or variables from the empirical situation. There are usually anuncountable number of ‘facts’ in any empirical situation and the decision-makermust intelligently abstract those factors which he considers to be most relevant tothe problem he faces. Relevance has two charactreristics: (i) There may be variableswhich affect the environment, but whose impact is negligible and as such they arenot considered relevant; (ii) There are some variables which do not affect theenvironment and thus need not be even considered tentatively as a part of the model.Essentially, the model must deal with the relevant aspects of the reality with whichwe are concerned. For instance, one can classify automobiles in several ways: onthe basis of car’s origin one can differentiate between domestic and foreign cars;they can as well be classified on the basis of age, cost or manufacturer. One can aswell classify them on the basis of body style, colour, size of engine or on the basisof the other factor. The point is that all these factors are valid in a particular situation.However, if someone is interested in conducting the study of the effect of the weatheron automobiles, the paint colour would be very useful in such a study. If one is,however, concerned with automobile accidents, then the classification on the basisof colour would be useless and the classification on the basis of engine type andsize would be considered as appropriate and valuable.

(b) Model building

The next step in modelling is that of model building. The relevant factors or variablesselected as stated in (a) are put in some logical manner, so that they form a model of

Abstraction: The firststep in modelling andconsists in selectingthe critical factors orvariables from theempirical situation.

8 Lindsay, Franklin A. 1958. New Techniques for Management Decision Making. New York: McGraw-Hill consultant Reports on Current Business Problems. p. 9.

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the given problem. In model building, it is generally considered desirable to simplifyreality but only to the point where there is no significant loss of accuracy. In fact, thedecision-maker tries his level best to construct the simplest possible model that predictsoutcomes reasonably well and is amenable to quick modification in case there is needfor the same models to be represented in a variety of ways, but in the relatively recentpast, emphasis is laid on the development of symbolic models particularly with regardto business problems. As such, one often talks about symbolic models in context ofmodel building. The question then is: ‘How to construct a mathematical model for agiven problem?’ The answer depends on that the variables and parameters which arerelevant and are also capable of quantitative measurement of some form must be putin logical manner indicating the necessary relationships along with constraints andthe objective function. Such an arrangement is often described as a model of thegiven problem. It is not possible to prepare a manual of constructions for modelbuilding as much depends upon the imagination, insight and experience of the decision-maker himself. However, past experience of model building proves a good guide indeveloping an appropriate model. Accordingly, we give the following types ofillustrative symbolic models:

Case 1 (Total Expense model): Symbolic model can just be presented in theequation form. A very simple model of total expense can be as follows:

T = a + bX

where, T = total costs; a = fixed costs; b = variable costs; X = number ofunits.

This model can be used as a predictor of the total costs associated with agiven level of output, of course with certain assumptions. This model impliesthat total costs are dependent upon the number of units produced. Thus, T isthe dependent variable and X is the independent variable in the given context.On the other hand, a and b are constants (or the parameters) of the givenmodel. The model thus represents the relationship between the total cost onone hand and number of units produced along with given constants on theother. If the object happens to have the total costs at its minimum level, thenwe have to find that value of X corresponding to which T happens to beminimum possible.

Case 2 (Milkman’s model): Consider the situation of a milkman who mustdecide how many cases of milk to order to maximize his expected profits. Hebuys a certain number of milk cases each day and sells some or all of them.He makes a profit on each one he sells. However, the unsold case brings hima loss. The demand for milk cases in the market varies from day to day, butthe probability that any specified number of milk cases will be sold on aparticular day can be determined on the basis of past data. We can developthe model for this type of problem by identifying the relevant variables andassigning symbols to them as follows:

n = The number of milk cases ordered per day

p = The profit made on each milk case sold

l = The loss incurred on each milk case unsold



Unit-1

d = The demand i.e., the number of milk cases that could be sold per dayif n d

pr (d) = The probability that the demand will equal supply on a randomly selectedday

P = Net profit per day (negative P represents a loss)

If the demand on a particular day exceeds the number ordered, i.e., If d > n, hisprofit would be P(d > n) – n.p

On the other hand, if demand does not exceed the number ordered, his profitwould be as follows:

P(n d) = d.p – (n – d).1

Then the exacted net profit per day ( P ) can be expressed as follows:

1 1

– – .n

d d n

P P d dp n d l P d n p

This is a decision model under risk situation, wherein P is the measure ofperformance, n is the controlled variable, d is an uncontrolled variable, and pand I are uncontrolled constants. To solve the problem represented throughthis model, we should find the value of n that maximizes P .

Case 3 (Linear Programming model): A simple linear programming modelcan be stated as follows:

Maximize: Z =1

n

j jj

c X ( j = 1,2,...... n) (Objective Function)

Subject to the constraints:

1

n

j

aij Xj bi

(i = 1,2,...... m) (Constraints concerning resources)

and Xj 0 (Non-negativity condition)

The linear programming model pertains to maximization case. It assumes priorknowledge of all the coefficients, i.e., it operates only under conditions ofcertainty. It has three components, viz., the objective function, constraints andthe non-linearity condition. The various assumptions are the linearity of theobjective function and that of the constraints, additivity of resources anddivisibility of decision variables. Non-negativity condition implies that no Xcan be negative. The model requires the determination ofn production activitiespursued at level X subject to a limited amount of m resources being available.Each unit of the jth activity yields a return of c, and uses an amount aij of the ithresource. Z denotes the optimal value of the objective function for a givensystem.

Case 4 (Simple inventory model concerning economic lot size): After themodel has been built, certain conclusions may be derived about its behaviourby means of logical analysis. Some mathematical calculations are often

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required for obtaining the solution to the model. The solution to the model meansthose values of the controllable variables that optimize the given objective. If thelogic adopted in deriving the conclusions from the model is correct, then the solutionto the model will also serve as the effective solution for the empirical problemrepresented by the said model.

(c) The question of error

It should always be remembered that the solution derived from the model always containssome degree of error because of the abstraction process under which the decision-maker may select the wrong variables, or not enough variables for constructing hismodel. If the error involved is not big enough to disturb the validity of the concernedmodel, then we generally do not mind the error and accordingly accept the solutionprovided by the model. However, when the error becomes significant, then we haveto rethink about the model. The question of when the error becomes so large that theconclusions must be modified before it can be adopted as a solution, is one of judgementof the decision-maker depending upon his experience and knowledge of the givensituation.

(d) Updating the model

A good model is one which gives valid results and permits reliable prediction of thesystem’s performance. As such the model once built up to represent a system must becontinuously updated to take account of the past, present and future specifics of theproblem. This, in other words, means that the model should be modified as soon asone or more of the controlled variables change significantly increasing the size of theerror involved.

A word of caution in modelling

The usefulness of modelling has already been stated, but one must note that the samefollows only when the model happens to be a reasonable approximation of reality. Ifthe model is a good representation of the problem, the solution of the model will be agood solution of the problem. However, the solution of a poor model, even though itmay be an exact one, will not be a good solution of the problem. Hence, the constructionof the model and the translation of the results to the actual problem must be done withutmost care. The key to successful modelling is to represent the decision situation asaccurately as possible and in particular, not to compromise the critical elements of theproblem. For this purpose the problem must be well defined before constructing themodel along with a clear and precise specification of all relevant factors and theinterrelationships among them.

It should also be noted that though many problems can be easily represented inthe form of models, but there may be problems in respect of which we may notconstruct reasonably good models. Accordingly, we must construct an appropriatemodel of a problem if we can, otherwise we must use an intuitive or heuristic approachto solve the given problem. ‘Finally, if a problem is so complex that it cannot bemodelled adequately for solution by one of the available methods or it cannot besolved adequately with a heuristic method, then the problem solver usually resorts tosimulation. Of course, simulation is not the answer to all problems; nevertheless, itdoes have a great deal of merit in studying large, complex systems where thecomponents are highly interrelated.’9

9 Gillett, Billy E. 2008. Introduction to Operations Research, Thirty- second reprint. New York: McGraw-Hill Publishing Company Ltd. p. 6.



Unit-1


1. List the three types of states under which decisions are made.

2. What are the two methods of simulation?

3. How is OR useful to personnel management?

4. What are some of the problems concerning defence management?

1.6 METHODOLOGY OF OR

The methodology of OR generally involves the following steps:

(i) Formulating the problem: The first step in an OR study is to formulate theproblem in an appropriate form. Formulating a problem consists of identifying,defining and specifying the measures of the components of a decision model.This means that all quantifiable factors, which are pertinent to the functioningof the system under consideration are defined in mathematical language:variables (factors which are not controllable), and parameters or coefficient,along with the constraints on the variables and the determination of suitablemeasures of effectiveness.

(ii) Constructing the model: The second step consists in constructing the model,by which we mean that appropriate mathematical expressions are formulatedwhich describe interrelations of all variables and parameters. In addition,one or more equations or inequalities are required to express the fact thatsome or all of the controlled variables can only be manipulated within limits.Such equations or inequalities are termed as constraints or the restrictions.The model must also include an objective function, which defines the measureof effectiveness of the system. The objective function and the constraints,together constitute a model of the problem that we want to solve. This modeldescribes the technology and the economics of the system under considerationthrough a set of simultaneous equations and inequalities.

(iii) Deriving the solution: Once the model is constructed, the next step in anOR study is that of obtaining the solution to the model, i.e., finding theoptimal values of the controlled variables—values that produce the bestperformance of the system for specified values of the uncontrolled variables.In other words, an optimum solution is determined on the basis of the variousequations of the model satisfying the given constraints and interrelations ofthe system, and at the same time maximizing profit or minimizing cost orcoming as close as possible to some other goal or criterion. How the solutioncan be derived depending on the nature of the model. In general, there arethree methods available for the purpose, viz., the analytical methods, thenumerical methods and the simulation methods. Analytical methods involve

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expressions of the model by mathematical computations and the kind ofmathematics required depends upon the nature of the model under consideration.This sort of mathematical analysis can be conducted only in some cases withoutany knowledge of the values of the variables, but in others the values of thevariables must be known concretely or numerically. In latter cases, we use thenumerical methods, which are concerned with iterative procedures through theuse of numerical computations at each step. The algorithm (or the set ofcomputational rules) is started with a trial or initial solution and continued witha set of rules for improving it towards optimality. The initial solution is thenreplaced by the improved one and the process is repeated until no furtherimprovement is possible. However, in those cases where the analytical as wellas the numerical methods cannot be used for deriving the solution then simulationmethods are used, i.e., experiments are conducted on the model in whichvalues of the uncontrolled variables are selected with the relative frequenciesdictated by their probability distributions. The simulation methods involve theuse of probability and sampling concepts, and are generally used with the helpof computers. Whichever method is used, our objective is to find an optimal ornearoptimal solution, i.e., a solution which optimizes the measure of effectivenessin a model.

(iv) Testing the validity: The solution values of the model, obtained as stated inthe previous step, are then tested against actual observations. In other words,effort is made to test the validity of the model used. A model is supposed tobe valid if it can give a reliable prediction of the performance of the systemrepresented through the model. If necessary, the model may be modified in thelight of actual observations and the whole process is repeated till a satisfactorymodel is attained. The operational researcher quite often realizes that his modelmust be a good representation of the system and must correspond to reality,which in turn requires this step of testing the validity of the model in an ORstudy. In effect, performance of the model must be compared with the policyor procedure that it is meant to replace.

(v) Controlling the solution: This step of an OR study establishes control over thesolution by proper feedback of information on variables, which might have deviatedsignificantly. As such, the significant changes in the system and its environmentmust be detected and the solution must accordingly be adjusted. This is particularlytrue when solutions are rules for repetitive decisions or decisions thatextend overtime.

(vi) Implementing the results: Implementing the results constitutes the last step ofan OR study. Because the objective of OR is not merely to produce reports but toimprove the performance of systems, the results of the research must beimplemented, if they are accepted, by the decision-makers. It is through this stepthat the ultimate test and evaluation of the research is made and it is in this phaseof the study the researcher has the greatest opportunity for learning.

Thus the procedure for an OR study generally involves some major steps, viz.,formulating the problem, constructing the mathematical model to represent the systemunder study, deriving a solution from, the model, testing the model and the solution soderived, establishing controls over the solution and lastly putting the solution to work-



Unit-1

implementation. (Although the said phases and the steps are usually initiated in the orderlisted in an OR study, it should always be kept in mind that they are likely to overlap intime and interact, i.e., each phase usually continues until the study is completed.)

Flow chart showing OR approach

OR approach can as well be illustrated by the following flow chart shown is Figure 1.1.

Facts, opinionsand symptomsof the problem

Define theproblem

Technical,financial

and economicinformation

from all sources

Determinefactors affecting

the problem(variables,

constraints andassumptions)

Develop objectiveand alternative

solutions

Detailedinformationon factors

Analysealternativesand deriving

optimum solution

Determinationof validity

recommend actioncontrol and

implementation

Computerservice

(may be used)

Modeldevelopment

Toolsof trade

ORTech-niques

Maximizeor

Minimizefunction

Fig. 1.1 OR Approach Flow Chart

1.7 APPLICATIONS AND SCOPE OF OR

OR techniques, having their origin with war operations analysis during the World War IIhave since then assisted defence management to a larger extent. Defence managementbeing vitally concerned with problems of logistics, such as provisioning, distributing, searchand forecasting, OR has given valuable help in decision formation. Although OR originatedin context of military operations, its impact nowadays can be observed in many otherareas. It has successfully entered into different research areas relating to modern businessand industry. In fact, it is being reckoned as an effective scientific technique to solveseveral managerial decision-making problems, such as inventory management problems,resource allocation problems, problems relating to replacement of both men as well asmachines, sequencing and scheduling problems, queuing problems, competitive problems,search problems, and the like.10

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Big and Small Organizations

OR approach, through being practised by big organized business and industrial units, isequally applicable to small organizations. For instance, whenever a departmental storefaces the problem, such as employing additional sales girls, purchasing an additional,van, one can apply some techniques of the operation research to minimize cost andmaximize benefit for each such decision. As such it is often said that wherever there isa problem there is scope for applying operations research techniques.

Military, business and research

In recent years of organized development, operations research has successfully entered,apart from its military and business applications, into several other areas of research.For instance, the basic problem in most of the developing countries in Asia and Africais to remove poverty as quickly as possible. There is a considerable scope to solvethis problem by applying OR techniques. Similarly, with the explosion of populationand consequent shortage of food, every country is facing the problem of optimumallocation of land for various crops in accordance with the climatic conditions andavailable facilities. The problem of optimal distribution of water from a source, like acanal for irrigation purposes, is faced by each developing country. Hence, a goodamount of scientific work related to the OR can be done in this direction. Thus, thereis considerable scope of applying OR techniques for solving problems relating tonational planning. National planning can be largely improved if optimum coordinationcan be arrived at through central planning agency by resorting to operations researchmodels and techniques. In India, there is a promising field for application of operationsresearch techniques in this direction.

Finance

In the field of finance, the progress of operations research has been rather slow.Financial institutions have many situations similar to those prevailing in othercommercial organizations. Investment policy must maximize the return on it, keepingthe factor of risk below a specified level. Long range corporate objectives of aninstitution are always studied along with functional objectives of individualdepartments, consistency between the two being essential for corporate planning.In particular, banking institutions require precise and accurate forecasting on cashmanagement and capital budgeting. Linear programming models for many problemsand quadratic programming formulation for some complex problems may provethe usefulness of operations research techniques to credit institutions.

Hospital management, health planning, transportation system, etc.

The application of operations research techniques can as well be noticed in contextof hospital management, health planning programmes, transportation system andin several other sectors. For instance, hospital management quite often faces theproblem of allotting its limited resources, of its multiphased activities. Optimumallocation of resources, so as to ensure a certain desirable level of service to patients,can be reached through OR. Similarly, for operating a transportation service profitably,

10 All such problems are described later in this unit.



Unit-1

an optimum schedule for vehicles and crews becomes necessary. Punctuality, waitingtime, total travel time, speed of the vehicles and agreement with trade unions of crewsare some of the variables to be taken into account while preparing optimum schedules,which can be worked out utilizing OR techniques.

As stated earlier, OR is generally concerned with problems that are tactical ratherthan strategic in nature, i.e., its use to long range organizational planning problems isvery much limited, but one can hope that with further developments taking place in thefield, ‘Operations Research will be able to deal with organizations in their entirety, ratherthan with slices through them. Managers who currently have to treat the whole withonly intuition and experience as their guides, will then have the methods of science attheir disposal as well11’.

This description brings home the point that during the last four decades, the scopeof OR has extensively been widened and hopefully shall attain new strides by the end ofthe present century. The art of systems analysis, so well developed in the military context,will spread to other contexts—notably such civil government branches as criminal justice,urban problems, housing, health care, education and social services. There is growingawareness amongst the people that unless they make themselves familiar with ORtechniques, they would not be able to understand and appreciate the problems of modernbusiness units. With computer facilities becoming widespread the significance and scopeof OR is likely to grow in the coming years. In fact, the future holds great promise forthe growth of the OR in power, scope and practical importance and utility. OR willcontinue its currently vigorous efforts to reach out to new arenas of exploration andapplication.

OR and Modern Business Management

From what has just been stated, we can say that OR renders valuable service in the fieldof business management. It ensures improvement in the quality of managerial decisionsin all functional areas of management. The role of OR in business management can besummed up as follows:

Helps the directing authority: OR techniques help the directing authority inoptimum allocation of various limited resources, viz., men, machines, money,material, time, etc., to different competing opportunities on an objective basis forachieving effectively the goal of a business unit. They help the chief of executivein broadening management vision and perspectives in the choice of alternativestrategies to the decision problems, such as forecasting manpower, productioncapacities and capital requirements and plans for their acquisition.

Useful to the production management: OR is useful to the productionmanagement in (i) selecting the building site for a plant, scheduling andcontrolling its development and designing its layout; (ii) locating within theplant and controlling the movements of required production materials andfinished goods inventories; (iii) scheduling and sequencing production by adequatepreventive maintenance with optimum number of operatives by proper allocationof machines; and (iv) calculating the optimum product-mix.

11 Ackoff Russell L., and Maurice W. Sasieni. 1968. Fundamentals of Operations Research, New York:John Wiley and Sons. pp. 447.

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Useful in personal management: OR is useful to the personnel managementto find out (i) optimum manpower planning (ii) the number of persons to bemaintained on the permanent or full- time roll; (iii) the number of persons to bekept in a work pool intended for meeting the absenteeism; (iv) the optimum mannerof sequencing and routing of personnel to a variety of jobs; and (v) in studyingpersonnel recruiting procedures, accident rates and labour turnover.

Help in marketing management: OR techniques equally help the marketingmanagement to determine (i) where distribution points and warehousingshould be located, their size, quantity to be stocked and the choice ofcustomers; (ii) the optimum allocation of sales budget to direct selling andpromotional expenses; (iii) the choice of different media of advertising andbidding strategies; and (iv) the consumer preferences relating to size, colour,packaging, etc., for various products as well as to outbid and outwitcompetitors.

Useful in financial management: OR is also very useful to the financialmanagement in (i) finding long-range capital requirements as well as how togenerate these requirements; (ii) determining optimum replacement policies;(iii) working out a plan for the firm; (iv) developing capital investments plans;and (v) estimating credit and investment risks.

In addition to all this, OR provides the business executives an understandingof the business operations which gives them new insights and capability to determinebetter solutions for several decision-making problems with great speed, competenceand confidence. When applied on the level of management where policies areformulated, OR assists the executives in an advisory capacity but on the operationallevel where production, personnel, purchasing, inventory and administrativedecisions are made, it provides management with a means for handling andprocessing information. Thus, in brief, OR can be considered as scientific methodof providing executive departments with a quantitative basis for taking decisionsregarding operations under their control.

Limitations of OR

OR though is a great aid to management, as outlined, but still it cannot be a substitutefor decision-making. The choice of a criterion as to what is actually best for abusiness enterprise is still that of an executive who has to fall back upon hisexperience and judgement. This is so because of the several limitations of OR.Important limitations are given as follows:

(i) The inherent limitations concerning mathematical expressions: ORinvolves the use of mathematical models, equations and similar othermathematical expressions. Assumptions are always incorporated in thederivation of an equation or model and such an equation or model may becorrectly used for the solution of the business problems. This happens whenthe underlying assumptions and variables in the model are present in theconcerning problem. If this caution is not given due care, then there always remainsthe possibility of wrong application of OR techniques. Quite often the operationsresearchers have been accused of having many solutions without being able tofind problems that fit.



Unit-1

(ii) High costs are involved in the use of OR techniques: OR techniques usuallyprove very expensive. Services of specialized persons are invariably calledfor (and along with it the use of computer) while using OR techniques. Assuch only big concerns can think of using such techniques. Even in big businessorganizations we can expect that OR techniques will continue to be of limiteduse simply because they are not in many cases worth their cost. As opposedto this, a typical manager, exercising intuition and judgement, may be able tomake a decision very inexpensively. Thus, the use of OR is a costlier affairand this constitutes an important limitation of OR.

(iii) OR does not take into consideration the intangible factors, i.e., non-measurable human factors: OR makes no allowance for intangible factors,such as skill, attitude, vigour of the management people in taking decisions,but in many instances success or failure hinges upon the consideration ofsuch non-measurable intangible factors. There cannot be any magic formulafor getting an answer to management problems; much depends upon propermanagerial attitudes and policies.

(iv) OR is only a tool of analysis and not the complete decision-making process:It should always be kept in mind that OR alone cannot make the final decision.It is just a tool and simply suggests best alternatives, but in the final analysismany business decisions will involve human element. Thus, OR is at best asupplement rather than a substitute for management; subjective judgement islikely to remain a principal approach to decision making.

(v) Other limitations: Among other limitations of OR, the following deservemention:

a. Bias: The operational researchers must be unbiased. An attempt toshoehorn results into a confirmation of management’s prior preferencescan greatly increase the likelihood of failure.

b. Inadequate objective functions: The use of a single objective functionis often an insufficient basis for decisions. Laws, regulations, publicrelations, market strategies, etc., may all serve to overrule a choice arrivedat in this way.

c. Internal resistance: The implementation of an optimal decision mayalso confront internal obstacles, such as trade unions or individualmanagers with strong preferences for other ways of doing the job.

d. Competence: Competent OR analysis calls for the careful specificationof alternatives, a full comprehension of the underlying mathematicalrelationships and a huge mass of data. Formulation of an industrialproblem to an OR set programme is quite often a difficult task.

e. Reliability of the prepared solution: At times a non-linear relationship ischanged to linear for fitting the problem to linear programming pattern.This may disturb the solution.

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1.8 BASIC OR MODELS

Broadly speaking, problems in OR can be put into one of the following categories:

(i) Allocation

(ii) Replacement

(iii) Sequencing

(iv) Routing

(v) Inventory

(vi) Queuing

(vii) Competitive

(viii) Search

(i) Allocation problems: They involve the allocation of resources to activities insuch a manner that some measure of effectiveness is optimized. If the resourceis people who can each perform any one of the given jobs in different amountsof time and the measure of effectiveness is the total time to perform all of thejobs when one and only one person is allocated to each job, then the allocationproblem is known as an assignment problem. We may as well talk of atransportation problem when a manufacturer produces a certain commodityat a number of factory sites and ships the commodity to different stores locatedat different points, and wants to determine the number of units to ship fromeach factory site to each store to minimize the overall shipping cost whilesatisfying the given constraints, regarding the factory capacity to supply andthe store requirement. In case the measure of effectiveness can be described asa linear function of different variables subject to several linear constraintsinvolving the variables, then the allocation problem is usually given the nameof a linear programming problem. Quite often the problem relating to product-mix, optimizing a given profit or cost function with linearity assumption,constitutes a linear programming problem.

(ii) Replacement problems: They are concerned with situations that arise whensome items (such as machines, electric light bulbs, etc.) need replacementbecause the same may deteriorate with time or may break down completelyor may get outdated due to new developments. Replacement problems thusoccur when one must decide the optimal time to replace equipment for onereason or the other.

(iii) Sequencing problems: They are the problems concerned with placing itemsin a certain sequence or order for service. For instance, N-jobs requiringdifferent amounts of time on different machines must each be processed onM-machines in the same order with no passing between machines, then thequestion: ‘How should the jobs be ordered for processing to minimize thetotal time to process all of the jobs on all of the machines’ constitutes an exampleof a sequencing problem.

(iv) Routing problems: They are problems related to finding the optimal routefrom an origin to a destination when a number of alternative routes are

Allocation problems:Involve the allocationof resources toactivities in such amanner that somemeasure ofeffectiveness isoptimized.



Unit-1

available. For instance, a salesman may wish to visit each of N-cities once andonly once before returning to his headquarter, then his problem is: ‘In what ordershould he visit the cities so that the overall distance travelled is minimized?’ Sucha problem is referred to as a routing problem.

(v) Inventory problems They are problems with regard to holding or storingresources. The decisions required generally entail the determination of howmuch of a resource to acquire or when to acquire it. The problem of decidinghow much of a certain commodity to hold in an inventory is one of realconcern to business and industrial houses. If a customer requires a certainquantity of the commodity but it is not available, then this could mean a lostsale. On the other hand, if the excess of the commodity is carried into aninventory, then the inventory carrying costs may be uneconomical. Hence,the inventory problem is the problem of determining the level of inventorythat will optimize the measure of effectiveness.

(vi) Queuing problems: They are what are known as waiting-line problems andinvolve waiting for service. Queuing problems encircle us from the time werise in the morning until we retire at night. In business, several types ofinterruptions occur: facilities break down and require repair, power failures,workers or the needed material do not show up where and when expected.Allocation of facilities considering such interruptions be done and to do someans solving a queuing problem.

(vii) Competitive problems: They arise when two or more people are competingfor a certain resource which may range from an opponent’s king in a game ofchess to a larger share of the market in business world. Many times acompetitive problem involves bidding for a contract to perform a service.

(viii) Search problems: They are problems concerned with searching forinformation that is required to make a certain decision. The problemconcerning exploring for valuable natural resources like oil or some othermineral is an example of a search problem. Similarly, searching the ocean forenemy ships is another example of a search problem. In case of searchproblems, the objective is to minimize the costs associated with collectingand analysing data, to reduce decision errors and also to minimize the costsassociated with the decision errors themselves.

Since its inception, OR has been applied to a wide variety of problems(as stated already), most of which have been tactical problems rather than strategicones. Characteristics which differentiate the tactical problems from strategic onesare as follows:

(i) One problem is more tactical than another if the effect of its solutionhas a shorter duration. Thus, a problem involving what to producetomorrow is more tactical than that of where to build an additional plant.

(ii) One problem is the more strategic larger portion of the organization isdirectly affected by its solution.

(iii) One problem involves more the determination of ends, goals or objectives.

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Thus, one cannot say that a particular problem is either tactical or strategic in absolutesense, but we can simply say that a particular problem is more or less tactical orstrategic than another with respect to the said characteristics. ORis generally concerned with problems that are tactical, rather than strategic in nature.

1.8.1 OR Techniques

Mathematical models have been constructed for the categorized OR problems andmethods for solving the models are available in many cases. Such methods areusually termed as OR techniques. Some of the important OR techniques often usedby decision-makers in modern times in business and industry are as follows:

(i) Linear programming: This technique is used in finding a solution foroptimizing a given objective, such as profit maximization or cost minimizationunder certain constraints. This technique is primarily concerned with theoptimal allocation of limited resources for optimizing a given function. Thename Linear programming is because of the fact that the model in such casesconsists of linear equations indicating linear relationship between the differentvariables of the system. Linear programming technique solves product-mixand distribution problems of business and industry. It is a technique used toallocate scarce resources in an optimum manner in problems of scheduling,product-mix and so on. Key factors under this technique include an objectivefunction, choice among several alternatives, limits or constraints stated insymbols and variables assumed to be linear.

(ii) Waiting line or queuing theory: Waiting line or queuing theory deals withmathematical study of queues. The queues are formed whenever the currentdemand for service exceeds the current capacity to provide that service.Waiting line technique concerns itself with the random arrival of customersat a service station where the facility is limited. Providing too much of capacitywill mean idle time for servers and will lead to waste of money. On the otherhand, if the queue becomes long there will be a cost due to waiting of units inthe queue. Waiting line theory, therefore, aims at minimizing the costs ofboth servicing and waiting. In other words, this technique is used to analysethe feasibility of adding facilities and to assess the amount and cost of waitingtime. With its help we can find the optimal capacity to be installed, whichwill lead to a sort of an economic balance between cost of service and cost ofwaiting.

(iii) Inventory control/planning: Inventory planning aims at optimizing inventorylevels. Inventory may be defined as a useful idle resource which has economicvalue, e.g., raw materials, spare parts, finished products, etc. Inventoryplanning, in fact, answers the two questions, viz., how much to buy and whento buy? Under this technique the main emphasis is on minimizing costsassociated with holding of inventories, procurement of inventories and theshortage of inventories.

(iv) Game theory: Game theory is used to determine the optimum strategy in acompetitive situation. The simplest possible competitive situation is that oftwo persons playing zero-sum game, i.e., a situation in which two persons areinvolved and one person wins exactly what the other loses. More complex



Unit-1

competitive situations of real life can as well be imagined where game theory canbe used to determine the optimum strategy.

(v) Decision theory: Decision theory concerns with making sound decisions underconditions of certainty, risk and uncertainty. As a matter of fact, there are threedifferent kinds of under which decisions are made, viz., deterministic, stochasticand uncertainty; the decision theory explains how to select a suitable strategyto achieve some object or goal under each of these three states.

(vi) Network analysis: Network analysis involes the determination of an optimumsequence of performing certain operations concerning some jobs in order tominimise overall time and/or cost. Programme Evaluation and ReviewTechnique (PERT), Critical Path Method (CPM) and other network techniques,such as Gantt Chart comes under Network Analysis. Key concepts under thistechnique are network of events and activities, resource allocation, time andcost considerations, network paths and critical paths.

(vii) Simulation: Simulation is a technique of testing a model that resembles a reallife situation. This technique is used to imitate an operation prior to actualperformance. Two methods of simulation are there—Monte Carlo method ofsimulation and System simulation method. The former one, using randomnumbers, is used to solve problems which involve conditions of uncertaintyand the mathematical formulation is impossible. However, in case of Systemsimulation, there is a reproduction of the operating environment and the systemallows for analysing the response from the environment to alternativemanagement actions. This method draws samples from a real population insteadof drawing samples from a table of random numbers.

(viii) Integrated production models: This tecnnique aims at minimizing cost withrespect to workforce, production and inventory. This technique is a highlycomplex one and is used only by big business and industrial units. This techniquecan be used only when sales and costs statistics for a considerable long periodare available.

(ix) Some other OR techniques: In addition, there are several other techniques,such as non-linear programming, dynamic programming, search theory, thetheory of replacement, and so on. A brief mention of some of these is as follows.

a. Non-linear programming: It is that form of programming in which someor all of the variables are curvilinear. In other words, this means thateither the objective function or constraints or both are not in linear form.In most of the practical situations, we encounter non-linear programmingproblems, but for computation purpose we approximate them as linearprogramming problems. Even then there may remain some non-linearprogramming problems which may not be fully solved by presently knownmethods.

b. Dynamic programming: It refers to a systematic search for optimalsolutions to problems that involve many highly complex interrelationsthat are, moreover, sensitive to multistage effects such as successive timephases.

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c. Heuristic programming: It is also known as discovery method and refersto step-by-step search toward an optimum when a problem cannot beexpressed in mathematical programming form. The search procedureexamines successively a series of combinations that lead to stepwiseimprovements in the solution and the search stops when a near optimumhas been found.

d. Integer programming: It is a special form of linear programming inwhich the solution is required in terms of integral numbers (i.e., wholenumbers) only.

e. Algorithmic programming: It is just the opposite of Heuristicprogramming. It may also be termed as near mathematical programming.This programming refers to a thorough and exhaustive mathematicalapproach to investigate all aspects of the given variables in order to obtainoptimal solution.

f . Quadratic programming: It refers to a modification of linearprogramming in which the objective equations appear in quadratic form,i.e., they contain squared terms.

g. Parametric programming: It is the name given to linear programmingwhen the latter is modified for the purpose of inclusion of several objectiveequations with varying degrees of priority. The sensitivity of the solutionto these variations is then studied.

h. Probabilistic programming: It also known as stochastic programmingand refers to linear programming that include an evaluation of relativerisks and uncertainties in various alternatives of choice for managementdecisions.

i. Search theory: It concerns itself with search problems. A search problemis characterized by the need for designing a procedure to collectinformation on the basis of which one or more decisions are made. Thistheory is useful in places in which some events are known to occur, butthe exact location is not known. The first search model was developedduring World War II to solve decision problems connected with air patrolsand their search for submarines. Advertising agencies’ search forcustomers, personnel departments’ search for good executives, etc., aresome of the examples of search theory’s application in business.

j. The theory of replacement: It is concerned with the prediction ofreplacement costs and determination of the most economic replacementpolicy. There are two types of replacement models—one type of modelsdeals in replacing equipment that deteriorate with time and the other typeof models helps in establishing replacement policy for those equipmentwhich fail completely and instantaneously.

All these techniques are not simple, but involve higher mathematics. The tendencytoday is to combine several of these techniques and form than into more sophisticatedand advanced programming models.



Unit-1


1. How do you formulate a problem correctly?

2. What is the requirement for selecting a criterion of best solution?

3. List the requirements of problem formulation.

4. What is the meaning of symbolic model?

ACTIVITY

1. ‘OR is a very powerful tool and analytical process that offers the presentationof an optimum solution in spite of its limitations.’ Discuss.

2. Model building is the essence of the OR approach. Discuss.

1.9 LET US SUM UP

There have been various definitions for Operations Research like applieddeclsion-making, quantitative common sense and making of economic decision.

An OR study generally involves three phases, viz., the judjement phase, theresearch phase and the action phase. Of these three, the research phase is thelongest and the largest, but the remaining two phases are very important sincethey provide, respectively, the basis for and implementation of the researchphase.

Mathematical models have been constructed for the categorized OR problemsand methods for solving the models are available in many cases. Such methodsare usually termed as OR techniques.

OR has gained increasing importance since World War II in the technology ofbusiness and industry administration.

OR, though a great aid to management, cannot be a substitute for decision-making. The choice of a criterion as to what is actually best for a businessenterprise is still that of an executive who has to fall back upon his experienceand judgement.

OR techniques, having their origin with war operations analysis during WorldWar II have since then assisted defence management to a larger extent.

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To formulate a problem correctly, one must know what a problem is and mustunderstand the nature of the problem.

Generally, the objectives of an organization may either be retentive or acquisitive.Retentive objectives are concerned with preserving either resources of valuesuch as money, time and energy, or states such as comfort, safety, stability andthe like.

Modelling plays an important part in operational research. In fact, it is consideredas the essence of operations research approach.

1.10 FURTHER READING

Jensen, Paul A., and Jonathan F. Bard. 2008. Operations Research Models andMethods. New York: John Wiley and Sons.

Taha, Hamdy A. 2010. Operations Research: An Introduction. New Delhi. PrenticeHall.

1.11 ANSWERS TO ‘CHECK YOUR PROGRESS’

Check Your Progress–11. Some of the characteristics of OR are as follows:

(i) Interdisciplinary team approach

(ii) Systems approach

(iii) Scientific method

(iv) Use of models

(v) Reduces complexity

2. As science, OR provides mathematical techniques and algorithms for solvingappropriate decision problems. OR is an art because success in all the phasesthat precede and succeed the solution of a problem largely depends on thecreativity and personal ability of the decision-making analysts.

3. In those cases where analytical as well as numerical methods cannot be usedfor deriving the solution, then simulation methods are used, i.e., experimentsare conducted on the model in which values of uncontrolled variables are selectedwith the relative frequencies dictated by their probability distributions.

4. Broadly speaking, problems in OR can be put into one of the followingcategories:

(i) Allocation

(ii) Replacement

(iii) Sequencing

(iv) Routing

(v) Inventory

(vi) Queuing



Unit-1

(vii) Competitive

(viii) Search

5. In case the measure of effectiveness can be described as a linear function ofdifferent variables subject to several linear constraints involving the variables,then the allocation problem is usually referred to as a linear programming problem.

6. In case of search problems, the objective is to minimize the costs associated withcollecting and analysing data, to reduce decision errors and also to minimize thecosts associated with the decision errors themselves.

Check Your Progress–2

1. There are three different types of states under which decisions are made, viz.,deterministic, stochastic and uncertainty, and the decision theory explains howto select a suitable strategy to achieve some object or goal under each of thesethree states.

2. Two methods of simulation are Monte Carlo method of simulation and Systemsimulation method.

3. OR is useful to personnel management in the following ways:

(i) It finds out the optimum manpower planning.

(ii) It finds out the number of persons to be maintained on a permanent orfull-time roll.

(iii) It finds out the number of workers to be kept in a work pool intended formeeting the absenteeism.

4. Defence management is concerned with problems of logistic, like provisioning,distribution, search and forecasting.

Check Your Progress–3

1. To formulate a problem correctly, one must know what a problem is and mustunderstand the nature of the problem.

2. Selecting a criterion of the best solution often requires a good knowledge ofdecision theory: The type of appropriate criterion depends on the state ofknowledge of outcomes that we assume to be available.

3. Problem formulation, which is considered as the first phase of an OR study,requires a statement of the problem’s components that include the controllableor the decision variables, the uncontrollable parameters or the environment,the constraints on the variables and also the objective along with an appropriatemeasure of performance.

4. A symbolic model is a mathematical description of an activity, which expressesthe relationships among the various elements with sufficient accuracy that itcan be used to predict the actual outcome under any expected set ofcircumstances.

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1.12 POSSIBLE QUESTIONS

Short-Answer Questions

1. Write a short note on the important OR techniques used in modern business andindustrial units.

2. What is the role of OR techniques in business and industry?

3. Write a short note on OR describing some of its important techniques.

4. Write a note on the various phases in solving an OR problem.

5. Write a note on problem formulation in operational research.

6. What are the properties of a good model?

7. Outline the important types of OR models.

8. What are the important aspects pertaining to modelling?

9. ‘Modelling enables quick and economical experimentation for finding an optimumsolution to a given problem.’ Do you agree? Give reasons for your answers.

Long-Answer Questions

1. Define OR. Explain some of its features.

2. Discuss the limitations of OR techniques.

3. Is it possible to make all business decisions with the assistance of OR? Givedetailed reasons to support your answer.

4. Explain the different possible solutions with regard to the state of outcomesand the appropriate criterion to be used for selecting the best course of actionfor each such situation.

5. Enumerate the advantages of modelling in OR.

Linear Programming Problem


Unit-2

UNIT 2 LINEAR PROGRAMMINGPROBLEM

UNIT STRUCTURE2.1 Learning Objectives2.2 Introduction2.3 Formulation of LP

2.3.1 Meaning of Linear Programming2.3.2 Fields where Linear Programming can be Used2.3.3 Components of a Linear Programming Problem2.3.4 Formulation of a Linear Programming Problem

2.4 Solving LP Problems2.4.1 Procedure for Solving LPP by Graphical Method2.4.2 Some Important Definitions2.4.3 Canonical and Standard Forms of LPP2.4.4 Simplex Method

2.5 Special Cases in LP2.5.1 Degeneracy2.5.2 Non-Feasible Solution2.5.3 Unbounded Solution2.5.4 Multiple Solution or Alternate Optima

2.6 Concept of Duality and Sensitivity Analysis2.6.1 Formulation of a Dual Problem2.6.2 Definition of a Dual Problem2.6.3 Important Results in Duality2.6.4 Sensitivity Analysis

2.7 Let Us Sum Up2.8 Further Reading2.9 Answers to ‘Check Your Progress’

2.10 Possible Questions



Explain the meaning of linear programming

Analyse the various fields where LP can be applied

Formulate an LP problem

Write the procedure for solving an LP problem

Solve an LP problem using the simplex method

Explain the concept of duality and sensitivity analysis

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Unit-2

2.2 INTRODUCTION

This unit introduces you to linear programming, where you will first learn the variouscomponents of an LP problem. You will also learn how to formulate an LP problem.The unit describes the steps involved in solving LP problems by the graphical method,which uses two decision variables. You will also learn about the standard andcanonical forms of LPP. There are certain special cases in linear programming,about which you will learn, including degeneracy, non-feasible solutions and alternateoptima. You will also learn about duality and sensitivity analysis and learn toformulate a dual problem.

2.3 FORMULATION OF LP

Decision-making has always been very important in the business and industrialworld, particularly with regard to the problems concerning production ofcommodities. Which commodity/commodities should be produced?; in whatquantities and by which process or processes should they be produced? are themain questions before a production manager. Alfred Marshall, in this connectionpoints out that the businessman always studies his production function and inputprices and substitutes one input for another till his costs become the minimumpossible. This sort of substitution, in the opinion of Marshall, is done by thebusinessman’s trained instinct rather than with formal calculations. But now theredoes exist a method of formal calculations often termed as Linear Programming.This method was first formulated by a Russian mathematician L.V. Kantorovich,but it was developed later in 1947 by George B. Dantzig ‘for the purpose of schedulingthe complicated procurement activities of the United States Air Force’. Today, thismethod is being used in solving a wide range of practical business problems. Theadvent of electronic computers has further increased its applications to solve manyother problems in industry. It is being considered as one of the most versatilemanagement techniques.

2.3.1 Meaning of Linear Programming

Linear Programming (LP) is a major innovation since World War II in the field ofbusiness decision-making, particularly under conditions of certainty. The word‘Linear’ means that the relationships are represented by straight lines, i.e., therelationships are of the form y = a + bx and the word ‘Programming’ means takingdecisions systematically. Thus, LP is a decision-making technique under givenconstraints on the assumption that the relationships amongst the variablesrepresenting different phenomena happen to be linear. In fact, Dantzig originallycalled it ‘programming of interdependent activities in a linear structure’, but lateron shortened it to ‘Linear Programming’. LP is generally used in solvingmaximization (sales or profit maximization) or minimization (cost minimization)problems subject to certain assumptions. Putting in a formal way, ‘LinearProgramming is the maximization (or minimization) of a linear function of variablessubject to a constraint of linear inequalities.’ Hence, LP is a mathematical technique

LinearProgramming:Generally used insolving maximization(sales or profitmaximization) orminimization (costminimization)problems subject tocertain assumptions.



Unit-2

designed to assist the organization in optimally allocating its available resources underconditions of certainty in problems of scheduling, product-mix, and so on.

2.3.2 Fields where Linear Programming can be Used

The problem for which LP provides a solution may be stated as, maximize or minimizefor some dependent variable which is a function of several independent variables whenthe independent variables are subject to various restrictions. The dependent variable isusually some economic objective such as profits, production, costs, workweeks, tonnageto be shipped, and so on. More profits are generally preferred to less profits and lowercosts are preferred to higher costs. Hence, it is appropriate to represent either maximizationor minimization of the dependent variable as one of the firm’s objective. LP is usuallyconcerned with such objectives under given constraints with linearity assumptions. Infact, it is powerful to take in its stride a wide range of business applications. The applicationsof LP are numerous and are increasing everyday. LP is extensively used in solvingresource allocation problems. Production planning and scheduling, transportation, salesand advertising, financial planning, portfolio analysis, corporate planning, etc., are someof its most fertile application areas. More specifically, LP has been successfully appliedin the following fields:

(i) Agriculture: LP can be applied in farm management problems as it relatesto the allocation of resources such as acreage, labour, water supply or workingcapital in such a way that it maximizes net revenue.

(ii) Contract awards: Evaluation of tenders by recourse to LP guarantees thatthe awards are made in the cheapest way.

(iii) Industries: Applications of LP in business and industry are of the mostdiverse type. Transportation problems concerning cost minimization canbe solved by this technique. The technique can also be adopted in solvingproblems of production (product-mix) and inventory control.

Thus, LP is the most widely used technique of decision-making in business andindustry in modern times in various fields.

2.3.3 Components of a Linear Programming Problem

There are certain basic concepts and notations to be first understood for easy adoptionof the LP technique. A brief mention of such concepts is as follows:

(i) Linearity. The term linearity implies straight line or proportionalrelationships among the relevant variables. Linearity in economic theoryis known as constant returns, which means that if the amount of theinput doubles, the corresponding outputs and profits are also doubled.Linearity assumption, thus, implies that two machines and two workerscan produce twice as much as one machine and one worker; fourmachines and four workers twice as much as two machines and twoworkers, and so on.

(ii) Process and its level. Process means the combination of particular inputsto produce a particular output. In a process, factors of production areused in fixed ratios, of course, depending upon technology and as suchno substitution is possible with a process. There may be many processesopen to a firm for producing a commodity and one process can be substituted

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for another. There is, thus, no interference of one process with anotherwhen two or more processes are used simultaneously. If a product can beproduced in two different ways, then there are two different processes (oractivities or decision variables) for the purpose of a linear programme.

(iii) Criterion function. This is also known as the objective function whichstates whether the determinants of the quantity should be maximized orminimized. For example, revenue or profit is such a function when it isto be maximized or cost is such a function when the problem is tominimize it. An objective function should include all the possibleactivities with the revenue (profit) or cost coefficients per unit ofproduction or acquisition. The goal may be either to maximize thisfunction or to minimize this function. In symbolic form, let ZX denotethe value of the objective function at the X level of the activities includedin it. This is the total sum of individual activities produced at a specifiedlevel. The activities are denoted as j =1, 2,..., n. The revenue or costcoefficient of the jth activity is represented by C

j. Thus, 2X

1, implies

that X units of activity j = 1 yields a profit (or loss) of C1= 2.

(iv) Constraints or inequalities. These are the limitations under which onehas to plan and decide, i.e., restrictions imposed upon decision variables.For example, a certain machine requires one worker to operate; anothermachine requires at least four workers (i.e., > 4); there are at most 20machine hours (i.e., < 20) available; the weight of the product shouldbe, say 10 lbs. and so on, are all examples of constraints or what areknown as inequalities. Inequalities like X > C (reads X is greater than Cor X < C (reads X is less than C) are termed as strict inequalities. Theconstraints may be in the form of weak inequalities like X C (reads Xis less than or equal to C) or X C (reads C is greater than or equal toC). Constraints may be in the form of strict equalities like X = C (readsX is equal to C).

Let bidenote the quantity b of resource i available for use in various

production processes. The coefficient attached to resource i is the quantityof resource i required for the production of one unit of product j.

(v) Feasible solutions. These are all those possible solutions which can beworked upon under given constraints. The region comprising all feasiblesolutions is referred to as Feasible Region.

(vi) Optimum solution. Optimum solution is the best of all the feasiblesolutions.

General form of the linear programming model

Linear programming problem mathematically can be stated as follows:

Choose the quantities,

Xj> 0 (j = 1,..., n) ...(2.1)

This is also known as the non-negativity condition and in simple terms means thatno X can be negative.



Unit-2

To maximize,

1

n

j jj

Z C X

...(2.2)

Subject to the constraints,

1

n

ij j ij

a X b

(i = 1,...,m) ...(2.3)

This is the usual structure of a linear programming model in the simplest possible form.This model can be interpreted as a profit maximization situation, where n productionactivities are pursued at level X

jwhich have to be decided upon, subject to a limited

amount of m resources being available. Each unit of the jth activity yields a return C anduses an amount a

ijof the ith resource. Z denotes the optimal value of the objective

function for a given system.

Assumptions or conditions to be fulfilled

LP model is based on the assumptions of proportionality, certainty, additivity, continuityand finite choices.

Proportionality is assumed in the objective function and the constraint inequalities.In economic terminology, this means that there are constant returns to scale, i.e., if oneunit of a product contributes` 5 toward profit, then 2 units will contribute` 10, 4 units`20, and so on.

Certainty assumption means the prior knowledge of all the coefficients in theobjective function, the coefficients of the constraints and the resource values. LPmodel operates only under conditions of certainty.

Additivity assumption means that the total of all the activities is given by thesum of each activity conducted separately. For example, the total profit in theobjective function is equal to the sum of the profit contributed by each of the productsseparately.

Continuity assumption means that the decision variables are contiunous.Accordingly, the combinations of output with fractional values, in case of product-mix problems, are possible and obtained frequently.

Finite choices assumption implies that finite number of choices are availableto a decision-maker and the decision variables do not assume negative values.

2.3.4 Formulation of a Linear Programming Problem

The procedure for mathematical formulation of an LPP consists of the followingsteps:

Step 1: The decision variables of the problem are noted.

Step 2: The objective function to be optimized (maximized or minimized) as alinear function of the decision variables is formulated.

Step 3:The other conditions of the problem such as resource limitation, market constraints,interrelations between variables, and so on, are formulated as linear inequations orequations in terms of the decision variables.

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Step 4: The non-negativity constraint from the considerations is added so that the negativevalues of the decision variables do not have any valid physical interpretation.

The objective function, the set of constraints and the non-negative constrainttogether form a linear programming problem.

General formulation of a linear programming problem

The general formulation of the LPP can be stated as follows:

In order to find the values of n decision variables x1

x2

... xn

to maximize orminimize the objective function,

1 1 2 2 n nZ C x C x C x ... (2.4)

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

1 1 2 2

( , , )

( , , )

:

( , , )

:

( )

n n

n n

i i in n i

m m mn n m

a x a x a x b

a x a x a x b

a x a x a x b

a x a x a x b

... (2.5)

Here, the constraints can be inequality or or even in the form of an equation(=) and finally satisfy the non-negative restrictions:

1 20, 0 0nx x x ... (2.6)

Matrix form of a linear programming problem

The LPP can be expressed in the matrix form as follows:

Maximize or minimize Z = Cx Objective function

Subject to Ax () b Constant equation

b > 0, x 0 Non-negativity restrictions

Where, x = 1 2( , )nx x x

C = 1 2( , )nC C C

b =

11 12 11

21 22 22

1 2

:

n

n

mm m mn

a a ab

a a ab A

ba a a

Example 2.1: A manufacturer produces two types of models, M1

and M2. Each

model of the type M1requires 4 hours of grinding and 2 hours of polishing, whereas

each model of the type M2requires 2 hours of grinding and 5 hours of polishing. The

manufacturers have 2 grinders and 3 polishers. Each grinder works 40 hours a weekand each polisher works for 60 hours a week. The profit on M

1model is ` 3.00 and on



Unit-2

model M2is ` 4.00. Whatever is produced in a week is sold in the market. How should

the manufacturer allocate his production capacity to the two types of models, so that hemay make the maximum profit in a week?

Solution:

Decision variables: Let X1

and X2be the number of units of M

1and M

2.

Objective function: Since the profit on both the models are given, we have tomaximize the profit, viz

Max Z = 3X1

+ 4X2

Constraints: There are two constraints: one for grinding and the other forpolishing.

The number of hours available on each grinder for one week is 40 hours.There are 2 grinders. Hence, the manufacturer does not have more than 2 × 40 = 80hours for grinding. M

1requires 4 hours of grinding and M

2requires 2 hours of

grinding.

The grinding constraint is given by,

1 24 2 80X X

Since there are 3 polishers, the available time for polishing in a week is givenby 3 × 60 = 180. M

1requires 2 hours of polishing and M

2requires 5 hours of polishing.

Hence, we have 2X1 + 5X

2 180

Thus, we have,

Max Z = 3X1 + 4X

2

Subject to 1 24 2 80X X

1 22 5 180X X

1 2, 0X X

Example 2.2: A firm produces two different types of products, Product M and ProductN. The firm uses the same machinery for manufacturing both the products. One unitof Product M requires 10 minutes while one unit of Product N requires 2 minutes.The maximum hours the machine can function optimally for a week is 35 hours.The raw material requirement for Product M is 1 kg while that of Product N is 0.5kg. Also, the market constraint on product M is 600 kg, while that of Product N is800 units per week. The cost of manufacturing Product M is ` 5 per unit and it is soldat ` 10; while the cost of Product N is ` 6 per unit and sold at ` 8 per unit. Calculatethe total number of units of Product M and Product N that should be produced perweek, so as to derive maximum profit.

Solution:


and X2 be the number of products of A and B.

Objective function: Cost of product A per unit is ` 5 and is sold at ` 10 per unit.

Profit on one unit of product A = 10 – 5 = 5.

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X1 units of product A contribute a profit of ` 5X

1from one unit of product.

Similarly, profit on one unit of B = 8 – 6 = 2.:. X

2units of product B contribute a profit of ` 2X

2.

The objective function is given by,

Max 1 25 2Z X X

Constraints: Time requirement constraint is given by,

1 2

1 2

10 2 (35 60)

10 2 2100

X X

X X

Raw material constraint is given by,

1 20.5 600X X

Market demand on product B is 800 units every week.

X2 800

The complete LPP is,

Max 1 25 2Z X X

Subject to, 1 2

1 2

2

1 2

10 2 2100

0.5 600

800

, 0

X X

X X

X

X X

Example 2.3: A person requires 10, 12 and 12 units of chemicalsA, B and C respectivelyfor his garden. A liquid product contains 5, 2 and 1 units of A, B and C respectively perjar. A dry product contains 1, 2 and 4 units of A, B, C per carton. If the liquid productsells for ` 3 per jar and the dry product sells for ` 2 per carton, what should be thenumber of jars that needs to be purchased, in order to bring down the cost and meet therequirements?

Solution:


and X2

be the number of units of liquid and dryproducts.

Objective function: Since the cost for the products is given, we have tominimize the cost.

Min Z = 3X1 + 2X

2

Constraints: As there are three chemicals and their requirements are given,we have three constraints for these three chemicals.

1 2

1 2

1 2

5 10

2 2 12

4 12

X X

X X

X X



Unit-2

Hence, the complete LPP is,

Min Z = 3X1 + 2X

2

Subject to,

1 2

1 2

1 2

1 2

5 10

2 2 12

4 12

, 0

X X

X X

X X

X X

Example 2.4: A paper mill produces two grades of paper, X and Y. Because of rawmaterial restrictions, it cannot produce more than 400 tonnes of grade X and 300tonnes of grade Y in a week. There are 160 production hours in a week. It requires0.2 and 0.4 hours to produce a tonne of products X and Y respectively withcorresponding profits of ` 200 and ` 500 per tonne. Formulate this as an LPP tomaximize profit and find the optimum product mix.

Solution:


and X2 be the number of units of the two grades of

paper, X and Y.

Objective function: Since the profit for the two grades of paper X and Y aregiven, the objective function is to maximize the profit.

Max Z = 200X1 + 500X

2

Constraints: There are two constraints, one with reference to raw material,and the other with reference to production hours.

Max Z = 200X1 + 500X

2

Subject to,

1

2

1 2

400

300

0.2 0.4 160

X

X

X X

Non-negative restriction X1, X

2 0

Example 2.5: A company manufactures two products, A and B. Each unit of B takestwice as long to produce as one unit of A and if the company were to produce only A,it would have time to produce 2000 units per day. The availability of the raw materialis enough to produce 1500 units per day of both A and B together. Product B requiringa special ingredient, only 600 units of it can be made per day. If A fetches a profit of` 2 per unit and B a profit of` 4 per unit, find the optimum product mix bythe graphicalmethod.

Solution: Let X1 and X

2 be the number of units of products A and B respectively.

The profit after selling these two products is given by the objective function,

Max Z = 2X1 + 4X

2

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Since the company can produce at the most 2000 units of the product in a day andProduct B requires twice as much time as that of Product A, production restriction isgiven by,

1 22 2000X X

Since the raw material is sufficient to produce 1500 units per day of both

A and B, we have 1 2 1500.X X

There are special ingredients for Product B; hence, we have X2 600.

Also, since the company cannot produce negative quantities X1 0 and

X2 0.

Hence, the problem can be finally put in the form:

Find X1 and X

2 such that the profits, Z = 2X

1 + 4X

2 is maximum.

Subject to,1 2

1 2

2

1 2

2 2000

1500

600

, 0

X X

X X

X

X X

Example 2.6: A firm manufacturers 3 products A, B and C. The profits are ` 3,` 2 and ` 4 respectively. The firm has 2 machines and the following is the requiredprocessing time in minutes for each machine on each product.

4 3 5

3 2 4

Product

A B C

Machines C

D

Machines C and D have 2000 and 2500 machine minutes respectively. Thefirm must manufacture 100 units of A, 200 units of B and 50 units of C, but not morethan 150 units of A. Set up an LP problem to maximize the profit.

Solution: Let X1, X

2, X

3 be the number of units of the product A, B, C respectively.

Since the profits are ` 3, ` 2 and ` 4 respectively, the total profit gained by thefirm after selling these three products is given by,

1 2 33 2 4Z X X X

The total number of minutes required in producing these three products atMachine C is given by 4X

1 + 3X

2 + 5X

3 and at Machine D is given by 3X

1 + 2X

2 +

4X3.

The restrictions on machinesCandDare given by 2000 minutes and 2500 minutes.

1 2 3

1 2 3

4 3 5 2000

3 2 4 2500

X X X

X X X



Unit-2

Also, since the firm manufactures 100 units ofA, 200 units of B and 50 units of C,but not more than 150 units of A, the further restriction becomes,

1

2

3

100 150

200 0

50 0

X

X

X

Hence, the allocation problem of the firm can be finally put in the form:

Find the value of X1, X

2, X

3 so as to maximize,

Z = 3X1 + 2X

2 + 4X

3


1 2 3

1 2 3

4 3 5 2000

3 2 4 2500

X X X

X X X

1 2 3100 150, 200 0,50 0X X X

Example 2.7: A peasant has a 100-acres farm. He can sell all potatoes, cabbage orbrinjals and can increase the cost to get ` 1.00 per kg for potatoes, ` 0.75 a head forcabbage and ` 2.00 per kg for brinjals. The average yield per acre is 2000 kg ofpotatoes, 3000 heads of cabbage and 1000 kg of brinjals. Fertilizers can be boughtat ` 0.50 per kg and the amount needed per acre is 100 kg each for potatoes andcabbage and 50 kg for brinjals. The manpower required for sowing, cultivating andharvesting per acre is 5 man-days for potatoes and brinjals and 6 man-days forcabbage. A total of 400 man-days of labour is available at ` 20 per man-day. Solvethis example as a linear programming model to increase the peasant’s profit.Solution: Let X

1, X

2, X

3 be the area of his farm to grow potatoes, cabbage and

brinjals respectively. The peasant produces 2000X1 kg of potatoes, 3000X

2 heads of

cabbage and 1000X3 kg of brinjals.

The total sales of the peasant will be,

= ` (2000X1 + 0.75 × 3000X

2 + 2 × 1000X

3)

Fertilizer expenditure will be,

= ` 20 (5X1

+ 6X2 + 5X

3)

:. Peasant’s profit will be,Z = Sale (in `) – Total expenditure (in `)= (2000X

1 + 0.75 × 3000X

2 + 2 × 1000X

3) – 0.5 × [100(X

1 + X

2) + 50X

2]

–20 × (5X1 + 6X

2+ 5X

3)

Z = 1850X1 + 2080X

2 + 1875X

3

Since the total area of the farm is restricted to 100 acres,

X1 + X

2 + X

3 100

Also, the total man-days manpower is restricted to 400 man-days.

1 2 35 6 5 400X X X

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Hence, the peasant’s allocation problem can be finally put in the form:Find the value of X

1, X

2 and X


Z = 1850X1 + 2080X

2 + 1875X

3

Subject to,

1 2 3

1 2 3

1 2 3

100

5 6 5 400

, , 0

X X X

X X X

X X X

Example 2.8: ABC Company produces two products: juicers and washing machines.Production happens in two different departments, I and II. Juicers are made inDepartment I and washing machines in Department II. These two items are soldweekly. The weekly production should not cross 25 juicers and 35 washing machines.The organization always employs a total of 60 employees in the two departments. Ajuicer requires two man-weeks’ labour, while a washing machine needs one man-week’s labour. A juicer makes a profit of ` 60 and a washing machine contributes aprofit of` 40. How many units of juicers and washing machines should the organizationmake to achieve the maximum profit? Formulate this as an LPP.


2 be the number of units of juicers and washing machines to be

produced.

Each juicer and washing machine contributes a profit of ` 60 and ` 40. Hence,the objective function is to maximize Z = 60X

1 + 40X

2.

There are two constraints which are imposed: weekly production and labour.

Since the weekly production cannot exceed 25 juicers and 35 washing machines,

1

2

25

35

X

X

A juicer needs two man-weeks of hard work and a washing machine needsone man-week of hard work and the total number of workers is 60.

2 X1 + X

2 60

Non-negativity restrictions: Since the number of juicers and washing machinesproduced cannot be negative, we have X

1 0 and X

2 0.

Hence, the production of juicers and washing machines problem can be finallyput in the form of a LP model given as follows:

Find the value of X1 and X


Z = 60X1 + 40X

2

Subject to,



Unit-2

1

2

1 2

1 2

25

35

2 60

and , 0

X

X

X X

X X


1. List the various fields where LP problems are frequently applied.

2. Define ‘linearity’.3. What does the assumption ‘finite choices’ in an LP problem imply?

2.4 SOLVING LP PROBLEMS

Simple linear programming problems with two decision variables can be easilysolved by the graphical method.

2.4.1 Procedure for Solving LPP by Graphical Method

The steps involved in the graphical method are as follows:

Step 1: Consider each inequality constraint as an equation.

Step 2: Plot each equation on the graph as each will geometrically represent astraight line.

Step 3: Mark the region. If the inequality constraint corresponding to that lineis , then the region below the line lying in the first quadrant (due to non-negativity ofvariables) is shaded. For the inequality constraint sign, the region above the line inthe first quadrant is shaded. The points lying in the common region will satisfy all theconstraints simultaneously. The common region, thus obtained, is called the feasibleregion.

Step 4: Allocate an arbitrary value, say zero, for the objective function.

Step 5: Draw a straight line to represent the objective function with the arbitraryvalue (i.e., a straight line through the origin).

Step 6: Stretch the objective function line till the extreme points of the feasibleregion. In the maximization case, this line will stop farthest from the origin andpasses through at least one corner of the feasible region. In the minimization case,this line will stop nearest to the origin and passes through at least one corner of thefeasible region.

Step 7: Find the coordinates of the extreme points selected in Step 6 and findthe maximum or minimum value of Z.

Note: As the optimal values occur at the corner points of the feasible region,it is enough to calculate the value of the objective function of the corner points ofthe feasible region and select the one which gives the optimal solution, i.e., in the case of

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a maximization problem, the optimal point corresponds to the corner point at which theobjective function has the maximum value and in the case of minimization, the cornerpoint which gives the objective function the minimum value is the optimal solution.

Example 2.9: Solve the following LPP by the graphical method.

Minimize Z = 20X1 + 10X

2

Subject to, X1 + 2X

2 40

3X1

+ X2

30

4X1

+ 3X2

60

X1, X

2 0

Solution: Replace all the inequalities of the constraints by equation,

X1 + 2X

2 = 40 If X

1 = 0 X

2 = 20

If X2 = 0 X

1 = 40

:. X1 + 2X

2= 40 passes through (0, 20) (40, 0).

3X1+ X

2= 30 passes through (0, 30) (10, 0).

4X1+ 3X

2= 60 passes through (0, 20) (15, 0).

Plot each equation on the graph.

3+

=31

2

XX

04

+=

1

2

XX3

60

X1

X1

2

+=40

2X

(4,18)

The feasible region is ABCD.

C and D are points of intersection of lines.

X1+ 2X

2= 40, 3X

1+ X

2= 30

And, 4X1+ 3X

2= 60

On solving, we get C = (4, 18) and D=(6, 12)

Corner Points Value of Z = 20X1

+ 10X2

A (15, 0) 300

B (40, 0) 800

C (4, 18) 260

D (6, 12) 240 (Minimum value)

The minimum value of Z occurs at D (6, 12). Hence, the optimal solution is X1

= 6, X2= 12.



Unit-2

Example 2.10: Find the maximum value of Z = 5X1 + 7X

2.


X1 + X

2 4

3X1 + 8X

2 24

10X1 + 7X

2 35

X1, X

2 > 0

Solution: Replace all the inequalities of the constraints by forming equations.

X1 + X

2= 4 passes through (0, 4) (4, 0)

3X1+ 8X

2= 24 passes through (0, 3) (8, 0)

10X1 + 7X

2= 35 passes through (0, 5) (3.5, 0)

Plot these lines in the graph and mark the region below the line, as the inequalityof the constraint is and is also lying in the first quadrant.

XX1

2+

=4

10+7

=35

1

2

XX

3 +8 =24X

1

2X

X1

X2

(1.6, 2.4)

O

B

C

(1.6, 2.3)

(8, 0)

The feasible region is OABCD.

B and C are points of intersection of lines,

X1 + X

2 = 4, 10X

1 + 7X

2 = 35

And, 3X1 + 8X

2 = 24

On solving we get,

B = (1.6, 2.3)

C = (1.6, 2.4)

Corner Points Value of Z = 5X1+ 7X

2

O (0, 0) 0

A (3.5, 0) 17.5

B (1.6, 2.3) 25.1

C (1.6, 2.4) 24.8 (Maximum value)

D (0, 3) 21

The maximum value of Z occurs at C (1.6, 2.4) and the optimal solution isX

1 = 1.6, X

2= 2.4.

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Example 2.11: A company makes 2 types of hats. Each Hat A needs twice as muchlabour time as Hat B. If the company is able to produce only Hat B, then it can makeabout 500 hats per day. The market limits daily sales of Hat A and Hat B to 150 and 250hats. The profits on Hat A and Hat B are ` 8 and ` 5, respectively. Solve graphically toget the optimal solution.


2 be the number of units of type A and type B hats respectively.

Maximize Z = 8X1 + 5X

2

Subject to, 2X1+ 2X

2 500

X1 150

X2 250

X1, X

2 0

First rewrite the inequality of the constraint into an equation and plot the lines inthe graph.

2X1

+ X2

= 500 passes through (0, 500) (250, 0).

X1

= 150 passes through (150, 0).

X2

= 250 passes through (0, 250).

We mark the region below the lines lying in the first quadrant, as the inequality ofthe constraints are . The feasible region is OABCD. B and C are points of intersectionof lines:

2X1 + X

2 = 500, where X

1 = 150 and X

2 = 250

On solving, we get B = (150, 200)

C = (125, 250)

400 500

X2

X2 = 250

B (150, 200)

X1

=15

0

O X12+

=500

X1

X2

C (125, 250)D (0, 250)


+ 5X2

O (0, 0) 0

A (150, 0) 1200

B (150, 200) 2200

C (125, 250) 2250 (Maximum Z = 2250)

D (0, 250) 1250



Unit-2

The maximum value of Z is attained at C (125, 250).

The optimal solution is X1= 125, X

2= 250

i.e., The company should produce 125 hats of type A and 250 hats of type B inorder to get the maximum profit of ` 2250.

Example 2.12: By graphical method, solve the following LPP.

Maximize Z = 3X1

+ 4X2

Subject to, 5X1 + 4X

2 200

3X1

+ 5X2

150

5X1

+ 4X2

100

8X1 + 4X

2 80

and X1, X

2 0

Solution:X2

(0, 25) D

C , 30(0 )

B 30.8, 11.5(

)3

+ 5= 150

X1

X2

5+

4=

100

X1

X2

8+

4=

80

X1

X2

5+

4=

200

X1

X2

(0 0, )X1

The feasible region is given by OABCD.


+ 4X2

O (20, 0) 60

A (40, 0) 120

B (30.8, 11.5) 138.4 (Maximum value)

C (0, 30) 120

D (0, 25) 100

The maximum value of Z is attained at B (30.8, 11.5).

The optimal solution is X1 = 30.8, X

2 = 11.5.

Example 2.13: Use graphical method to solve the following LPP.

Maximize Z = 6X1+ 4X

2

Subject to, –2X1 + X

2 2

X1

– X2

2

3X1

+ 2X2

9

X1, X

2 0

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Operations Research

Unit-2

Solution:X2

–2+

=2

X 1X 2

3+

2=

9

X1

X2

X1

X 1–

=2

X 2

(13/5

, 3/5)

– –

–

–

–

The feasible region is given by ABC.


+ 4X2

A (2, 0) 12

B (3,0) 18

C (13/5, 3/5)90

185

(Maximum value)

The maximum value of Z is attained at C (13/5, 3/5).

The optimal solution is X1= 13/5, X

2= 3/5.

Example 2.14: Use graphical method to solve the following LPP.

Minimize Z = 3X1 + 2X

2

Subject to, 5X1

+ X2

10

X1

+ X2

6

X1

+ 4X2

12

X1, X

2 0

Solution: Corner Points Value of Z = 3X1+ 2X

2

A (0, 10) 20

B (1, 5) 13 (Minimum value)

C (4, 2) 16

D (12, 0) 36

(1, 5)

5+

=10

X1

X2

(4, 2)

X1 + 4 = 12X2

X1 +

=6

X2

( 00, 1 )



Unit-2

Since the minimum value is attained at B (1,5), the optimum solution is,

X1 = 1, X

2= 5

Note: In this problem, if the objective function is maximization then the solution isunbounded, as maximum value of Z occurs at infinity.

Some more cases

There are some linear programming problems which may have:(i) A unique optimal solution(ii) An infinite number of optimal solutions(iii) An unbounded solution(iv) No solution

The following examples will illustrate these cases.

Example 2.15: Solve the following LPP by the graphical method.


2

Subject to, 5X1+ 2X

2 1000

3X1+ 2X

2 900

X1

+ 2X2

500and X

1 + X

2 0

Solution:

(200

,0)

(0, 250)(125, 187.5)

X1 + 2

= 500X

2

5+

2=

1000

X1

X2

3+

2=

900

X1

X2

The solution space is given by the feasible region OABC.

Corner Points Value of Z = 100X1+ 40X

2

O (0, 0) 0

A (200, 0) 20,000

B (125, 187.5) 20,000 (Maximum value of Z)

C (0, 250) 10,000

The maximum value of Z occurs at two vertices A and B.

Since there are infinite number of points on the line, joining A and B gives thesame maximum value of Z.

Thus, there are infinite number of optimal solutions for the LPP.

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Example 2.16: Solve the following LPP.


2

Subject to, X1 + X

2 1

X1 + X

2 3

X1, X

2 0

Solution: The solution space is unbounded. The value of the objective function at thevertices A and B are Z (A) = 6, Z (B) = 6. But, there exist points in the convex regionfor which the value of the objective function is more than 8. In fact, the maximumvalue of Z occurs at infinity. Hence, the problem has an unbounded solution.

Hence, no feasible solution.

(0, 3)

X2

X1

(2, 1)

X1

X2

–

–

X

X1

2–

=1

X1 +

=3

X2

When there is no feasible region formed by the constraints in conjunction withnon-negativity conditions, then no solution to the LPP exists.

Example 2.17: Solve the following LPP.

Maximize Z = X1+ X

2


X1+ X

2 1

–3X1 + X

2 3

X1, X

2 0

X2

X1–



Unit-2

Solution:

There being no point (X1, X

2) common to both the shaded regions, it is not possible to find

a feasible region for this problem. So, the problem cannot be solved. Hence, the problemhas no solution.

2.4.2 Some Important Definitions

1. A set of values X1, X

2 ... X

nwhich satisfies the constraints of the LPP is called its

solution.

2. Any solution to a LPP which satisfies the non-negativity restrictions of the LPP iscalled its feasible solution.

3. Any feasible solution which optimizes (minimizes or maximizes) the objectivefunction of the LPP is called its optimum solution.

4. Given a system of m linear equations with n variables (m < n), any solutionwhich is obtained by solving m variables keeping the remaining n – m variableszero is called a basic solution. Such m variables are called basic variablesand the remaining variables are called non-basic variables.

5. A basic feasible solution is a basic solution which also satisfies the conditionin which all basic variables are non-negative.

Basic feasible solutions are of two types:(i) Non-degenerate: A non-degenerate basic feasible solution is the basic

feasible solution which has exactly m positive Xi(i = 1, 2, ..., m), i.e.,

none of the basic variables is zero.(ii) Degenerate: A basic feasible solution is said to be degenerate if one or

more basic variables are zero.

6. If the value of the objective function Z can be increased or decreasedindefinitely, such solutions are called unbounded solutions.

2.4.3 Canonical and Standard Forms of LPP

The general LPP can be expressed in the following forms, namely canonical orstandard forms.

In the standard form, irrespective of the objective function, namely maximizeor minimize, all the constraints are expressed as equations. Moreover, RHS of eachconstraint and all variables are non-negative.

Characteristics of the standard form

(i) The objective function is of maximization type.

(ii) All constraints are expressed as equations.

(iii) The right hand side of each constraint is non-negative.

(iv) All variables are non-negative.

Characteristics of the canonical form

(i) The objective function is of maximization type.

(ii) All constraints are of ‘’ type.(iii) All variables X

iare non-negative.

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In the canonical form, if the objective function is of maximization, all the constraintsother than non-negativity conditions are ‘’ type. If the objective function is of minimization,all the constraints other than non-negative conditions are ‘’ type.Notes:

1. Minimization of a function Z is equivalent to maximization of the negativeexpression of this function, i.e., Min Z = –Max (–Z).

2. An inequality in one direction can be converted into an inequality in theopposite direction by multiplying both sides by (–1).

3. Suppose you have the constraint equation,

a11

x1+a

12x

2+...... +a

1nx

n= b

1

This equation can be replaced by two weak inequalities in opposite directions.

a11

x1+ a

12x

2+...... + a

mx

n b

1

a11

x1+ a

12x

2+...... + a

1nx

n b

1

4. If a variable is unrestricted in sign, then it can be expressed as the differenceof two non-negative variables, i.e., if X

1is unrestricted in sign, then X

i= X

i –

Xi, where X

i, X

i, X

i are 0.

5. In the standard form, all constraints are expressed as equations, which ispossible by introducing some additional variables called slack variables andsurplus variables so that a system of simultaneous linear equations is obtained.The necessary transformation will be made to ensure that b

i 0.

Definition

(i) If the constraints of a general LPP be,

1

( 1, 2, ..., ),n

ij i ij

a x b i m

then the non-negative variables S

i, which are introduced to convert the inequalities

() to the equalities1

( 1, 2, ..., )n

ij i i ij

a x S b i m

, are called slack variables.

Slack variables are also defined as the non-negative variables which are addedin the LHS of the constraint to convert the inequality ‘’ into an equation.

(ii) If the constraints of a general LPP be,

1

( 1, 2, ..., )n

ij j ij

a x b i m

,

then the non-negative variables Si

which are introduced to convert the

inequalities ‘’ to the equalities1

– ( 1, 2, ..., )n

ij j i ij

a x S b i m

are called surplus

variables.

Surplus variables are defined as the non-negative variables which are removedfrom the LHS of the constraint to convert the inequality ‘’ into an equation.



Unit-2

2.4.4 Simplex Method

The simplex method is an iterative procedure for solving an LPP in a finite number ofsteps. This method provides an algorithm which consists of moving from one vertex ofthe region of feasible solution to another in such a manner that the value of the objectivefunction at the succeeding vertex is less or more as the case may be at the previousvertex. This procedure is repeated and since the number of vertices is finite, the methodleads to an optimal vertex in a finite number of steps or indicates the existence of anunbounded solution.

Definition

(i) Let XB be a basic feasible solution to the LPP.

Max Z = CX

Subject to AX = b and X 0, such that it satisfies X

B = B–1b,

Here, B is the basic matrix formed by the column of basic variables.

The vector CB = (C

B1, C

B2 … C

Bm), where C

Bj are components of C associated

with the basic variables is called the cost vector associated with the basic feasiblesolution X

B.

(ii) Let XB be a basic feasible solution to the LPP.

Max Z = CX, where A

X = b and X 0.

Let CB be the cost vector corresponding to X

B. For each column vector a

j in

A1, which is not a column vector of B, let

1

m

j ij ji

a a b

Then the number1

m

j Bi iji

Z C a

is called the evaluation corresponding to aj

and the number (Zj – C

j) is called the net evaluation corresponding to j.

Simplex algorithm

While solving an LPP by simplex algorithm, an important assumption is the existenceof an initial basic feasible solution. The following steps are involved in finding anoptimum solution.

Step 1: It is first imperative to find out whether minimization or maximizationis the objective function of the given LPP. If minimization is the objective function,then the LPP should be converted into a maximization problem,

Min Z = –Max (–Z)

Step 2: All values of bi (i = 1, 2, …, m) should be positive. Any negative b

i

value present should be converted into positive value by multiplying with –1.Step 3: The problem should be then expressed in the standard form. This can

be done by introducing slack/surplus variables to convert the inequality constraintsinto equations.

Simplex method: Aniterative procedure forsolving an LPP in afinite number of steps.

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Step 4: Get an initial basic feasible solution to the problem in the formX

B= B–1b and put it in the first column of the simplex table. Form the initial simplex table

shown as follows:

n

n

Step 5: Compute the net evaluations Zj – C

j by using the relation:

Zj – C

j = C

B (a

j – C

j)

Examine the sign of Zj – C

j:

(i) If all Zj – C

j 0, then the initial basic feasible solution X

B is an optimum

basic feasible solution.(ii) If at least one Z

j – C

j> 0, then go to the next step as the solution is not

optimal.

Step 6: To find the entering variable, i.e., key column.

If there are more than one negative Zj – C

j, the most negative of them should

be chosen. Let it be Zr – C

r for some j = r. This gives the entering variable X

r and is

indicated by an arrow at the bottom of the rth column. If there are more than onevariable having the same most negative Z

j – C

j, then any one of the variable can be

selected arbitrarily as the entering variable.(i) If all X

ir 0 (i = 1, 2, …, m) then there is an unbounded solution to the

given problem.(ii) If at least one X

ir> 0 (i = 1, 2, …, m), then the corresponding vector X

r

enters the basis.

Step 7: To find the leaving variable or key row:

Compute the ratio (XBi

/Xkr, X

ir>0)

If the minimum of these ratios be XBi

/Xkr, then choose the variable X

k to leave

the basis called the key row and the element at the intersection of the key row andthe key column is called the key element.

Step 8: Form a new basis by dropping the leaving variable and introducingthe entering variable along with the associated value under C

B column. The leaving

element is converted to unity by dividing the key equation by the key element andall other elements in its column to zero by using the formula:

New element = Old element

Product of elements in the key row and key column–Key element

Step 9: Repeat the procedure of Step (5) until either an optimum solution isobtained or there is an indication of unbounded solution.



Unit-2

Example 2.18: Use the simplex method to solve the following LPP.


2

Subject to, 1 2

1 2

1 2

4

2

, 0

X X

X X

X X

Solution: By introducing the slack variablesS1, S

2, convert the problem into its standard

form.

Max Z = 3X1 + 2X

2 + 0S

1 + 0S

2

Subject to, 1 2 1

1 2 2

1 2 1 2

4

2

, , , 0

X X S

X X S

X X S S

11 2 1 2

2

1

2

41 1 1 0

21 1 0 1

XX X S S

X

S

S

An initial basic feasible solution is given by,

XB

= B–1b,

Here, B = I2, X

B= (S

1, S

2)

i.e., (S1, S

2) = I

2 (4, 2) = (4, 2)

Initial Simplex Table

Zj = C

B a

j

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

0 13 3

0 1

0(1 1) 2 2

0

0(1 0) 0 0

0

0(01) 0 0

0

B

B

B

B

Z C C a C

Z C C a C

Z C C a C

Z C C a C

Cj 3 2 0 0

CB B XB X1 X2 S1 S2 Min1

BX

X

0 S1 4 1 1 1 0 4/1 = 4

0 S2 2 1 –1 0 1 2/1 = 2

Zj 0 0 0 0 0

Zj – Cj –3 –2 0 0

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Since there are some Zj – C

j = 0, the current basic feasible solution is not optimum.

Since Z1 – C

1= –3 is the most negative, the corresponding non-basic variable X

1

enters the basis.

The column corresponding to this X1 is called the key column.

Ratio = Min , 0Biir

ir

XX

X

= Min4 2

,1 1

, which corresponds to S2

The leaving variable is the basic variable S2. This row is called the key row.

Convert the leading element X21

to units and all other elements in its column n, i.e., (X1)

to 0 by using the formula:

New element = Old element –

Product of elements in the key row and key column

Key element

To apply this formula, first we find the ratio, namely

The element to be zero 11

Key element 1

Apply this ratio, for the number of elements that are converted in the keyrow. Multiply this ratio by the key row element shown as follows:

1 × 2

1 × 1

1 × –11 × 0

1 × 1

Now, subtract this element from the old element. The element to be convertedinto 0 is called the old element row. Finally, you have

4 – 1 × 2 = 2 1 – 1 × 1 = 01 – 1 × –1 = 2 1 – 1 × 0 = 1 0 – 1 × 1 = –1 The improved basic feasible solution is given in the following simplex table.



Unit-2

First iteration

Cj 3 2 0 0

CB B XB X1 X2 S1 S2 Min2

BX

X

0 S1 2 0 2 1 –1 2/2 = 1

3 X1 2 1 –1 0 1 –

Zj 6 3 –3 0 0

Zj – Cj 0 –5 0 0

Since, Z2 – C

2 is the most negative, X

2 enters the basis.

To find Min 22

, 0Bi

i

XX

X

Min2

2

This gives the outgoing variables. Convert the leaving element into 1. This isdone by dividing all the elements in the key row by 2. The remaining elements areconverted to 0 by using the following formula.

Here, – 12 is the common ratio. Put this ratio 5 times and multiply each ratio

by the key row element.

12

21

021

22

–1/2 × 1–1/2 × –1Subtract this from the old element. All the row elements which are converted

into 0 are called the old elements.

12 2 3

2

1 – (–1/2 × 0) = 1–1 – (–1/2 × 2) = 00 – (–1/2 × 1) = 1/21 – (–1/2 × 1) = 1/2

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Second iteration

1/2–

S1 S2X2X1X0BCa

Cj

X2

X1

Zj

Zj–Cj 1/2

1/25/2

5/2

1/2 1/2

1/2

3 2 0 0

1

0

2

0

0

1

3

0

11

3

1

3

2

Since all Zj – C

j 0, the solution is optimum. The optimal solution is Max

Z = 11, X1 = 3, and X

2 = 1.

Example 2.19: Solve the LPP when,Maximize Z = 3X

1 + 2X

2

Subject to, 1 2

1 2

1 2

1 2

4 3 12

4 8

4 8

, 0

X X

X X

X X

X X

Solution: Convert the inequality of the constraint into an equation by adding slackvariables S

1, S

2, S

3.

Max Z = 3X1 + 2X

2 + 0S

1 + 0S

2 + 0S

3

Subject to, 1 2 1

1 2 2

1 2 3

1 2 1 2 3

4 3 12

4 8

4 8

, , , , 0

X X S

X X S

X X S

X X S S S

11 2 1 2 3

2

1

2

3

124 3 1 0 0

84 1 0 1 0

84 1 0 0 1

XX X S S S

X

S

S

S

Initial table

Cj 3 2 0 0 0

CB B XB X1 X2 S1 S2 S3 Min1

BX

X

0 S1 12 4 3 1 0 0 12/4 = 3

0 S2 8 4 1 0 1 0 8/4 = 2

0 S3 8 4 –1 0 0 1 8/4 = 2

Zj 0 0 0 0 0 0

Zj – Cj –3 –2 0 0 0



Unit-2

Z1 – C

1 is most negative, X

1 enters the basis and the Min 1

1

, 0Bi

i

XX

X

= Min (3, 2, 2) gives S

3 as the leaving variable.

Convert the leaving element into 1 by dividing the key row elements by 4 and theremaining elements into 0.

First iteration

Cj 3 2 0 0 0

CB B XB X1 X2 S1 S2 S3 Min2

BX

X

0 S1 4 0 4 1 0 –1 4/4 = 1

0 S2 0 0 2 0 1 –1 0/2 = 0

3 X1 2 1 –1/4 0 0 1/4 –

Zj 6 3 –3/4 0 0 3/4

Zj – Cj 0 –11/4 0 0 3/4

4 48 8 0 12 8 4

4 4

4 44 4 0 4 4 0

4 4

4 41 1 2 3 1 4

4 4

4 40 0 0 1 0 1

4 4

4 41 0 1 0 0 0

4 4

4 40 1 1 0 1 1

4 4

Since, 2 23

4Z C is the most negative, X2 enters the basis.

To find the outgoing variable, find Min 22, 0B

ii

X XX

.

Min4 0

, , 14 2

First iteration

Therefore, S2 leaves the basis. Convert the leaving element into 1 by dividing the key

row elements by 2 and the remaining elements in that column into 0 using the formula,

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–Product of elements in the key row and key column

Key element

Cj

3 2 0 0 0

CB

B XB

X1

X2

S1

S2

S3

Min3

BX

S

0 S1

4 0 0 1 –2 1 4/1 = 4

2 X2

0 0 1 0 12 – 1

2 –

3 X1

2 1 0 0 1/8 1/8 2/1/8 = 16

Zj

6 3 2 0 11/8 –5/8Z

j – C

j0 0 0 11/8 –5/8

Second iteration

Since Z5 – C

5 = –5/8 is the most negative, S

3 enters the basis, and

Min 33

4 2, Min ,

1 1/18B

ii

XS

S

Therefore,S1 leaves the basis. Convert the leaving element into 1 and the remaining

elements into 0.

Third iteration

Cj 3 2 0 0 0

CB B XB X1 X2 S1 S2 S3

0 S3 4 0 0 1 –2 1

2 X2 2 0 1 1/2 –1/2 0

3 X1 3/2 1 0 –1/8 3/8 0

Zj 17/2 3 2 5/8 1/8 0

Zj – Cj 0 0 5/8 1/8 0

Since all Zj – C

j 0, the solution is optimum and it is given byX

1 = 3/2, X

2 = 2 and

Max Z = 17/2.

Example 2.20: Using simplex method solve the following LPP.

Maximize Z = X1 + X

2 + 3X

3

Subject to, 1 2 3

1 2 3

1 2 3

3 2 3

2 2 2

, , 0

X X X

X X X

X X X



Unit-2

Solution: Rewrite the inequality of the constraints into an equation by adding slackvariables.

Max Z = X1 + X

2 + 3X

3 + 0S

1 +0S

2

Subject to, 1 2 3 1

1 2 3 2

3 2 3

2 2 2

X X X S

X X X S

The initial basic feasible solution is,

1 2 3

1 2

0

3, 2 and 0

X X X

S S Z

1 2 3 1 2

3 2 1 1 0

2 1 2 0 1

1 1 3 0 0

X X X S S

Cj 3 2 0 0 0

CB B XB X1 X2 X3 S1 S2 Min3

BX

X

0 S1 3 3 2 1 1 0 3/1 = 3

0 S2 2 2 1 2 0 1 2/2 = 1

Zj 0 0 0 0 0 0

Zj – Cj –1 –1 –3 0 0

Since Z3 – C

3 = –3 is the most negative, the variable X

3 enters the basis. The

column corresponding to X3 is called the key column.

To determine the key row or leaving variable, find Min 33

, 0BXX

X

Min3 2

3, 11 2

Therefore, the leaving variable is the basic variable S

2, the row is called the

key row and the intersection element 2 is called the key element.

Convert this element into 1 by dividing each element in the key row by 2 andthe remaining elements in that key column into 0 using the formula,



Key element

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First iteration

Since all Zj – C

j 0, the solution is optimum and it is given by X

1 = 0, X

2 = 0,

X3 = 1, Max Z = 3.

Example 2.21: Use the simplex method to solve the following LPP.

Minimize Z = X2 – 3X

3 + 2X

5

Subject to, 2 3 5

2 3

2 3 5

2 3 5

3 2 7

2 4 12

4 3 8 10

, , 0

X X X

X X

X X X

X X X

Solution: Since the given objective function is of minimization you should convertit into maximization using Min Z = –Max(–Z).

Max Z = –X2 + 3X

3 – 2X

5

Subject to,2 3 5

2 3

2 3 5

3 2 7

2 4 12

4 3 8 10

X X X

X X

X X X

You have to rewrite the inequality of the constraints into an equation by addingslack variables S

1, S

2, S

3 and the standard form of LPP becomes,

Max Z = –X2 + 3X

3 – 2X

5 + 0S

1 + 0S

2 + 0S

3

Subject to, 2 3 5 1

2 3 2

2 3 5 3

2 3 5 1 2 3

3 2 7

2 4 12

4 3 8 10

, , , , , 0

X X X S

X X S

X X X S

X X X S S S

The initial basic feasible solution is given by,

S1=7, S

2=12, S

3=10. (X

2=X

3=X

5=0)



Unit-2

Initial table

Cj –1 3 –2 0 0 0

CB B XB X2 X3 X5 S1 S2 S3 Min3

BX

X

0 S1 7 3 –1 2 1 0 0 –

0 S2 12 –2 4 0 0 1 0 12/4=3

S3 10 –4 3 8 0 0 1 10/3=3.33

Zj 0 0 0 0 0 0 0

Zj – Cj 1 –3 2 0 0 0

Since, Z2 – C

2 = – 3 < 0, the solution is not optimum.

The incoming variable is X3 (key column) and the outgoing variable (key row) is

given by,

33

12 10Min , 0 Min ,

4 3B

ii

X XX

Hence, S2 leaves the basis.

First iterationCj –1 3 –2 0 0 0

CB B XB X2 X3 X5 S1 S2 S3 Min2

BX

X

0 S1 10 5/2 0 2 1 1/4 0 10/5/2=4

3 X3 3 –1/2 1 0 0 1/4 0 –

S3 1 –5/2 0 8 0 –3/4 1 –

Zj 9 –3/2 3 0 0 3/4 0

Zj – Cj –1/2 0 2 0 3/4 0

Since Z1 – C

1 < 0, the solution is not optimum. Improve the solution by allowing

the variable X2 to enter into the basis and the variable S

1 to leave the basis.

Second iteration

1/2–

Since, Zj – C

j 0, the solution is optimum.

The optimal solution is given by Max Z = 11

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X2 = 4, X

3 = 5, X

5 = 0

Min Z = –Max (–Z) = –11 Min Z = –11, X

2 = 4, X

3 = 5, X

5 = 0

Example 2.22: Solve the following LPP using the simplex method.


2+ 9X

3+ 2X

4

Subject to, 2X1+ X

2+ 5X

3+ 6X

4 20

3X1+ X

2+ 3X

3+ 25X

4 24

7X1 + X

4 70

X1, X

2, X

3, X

40

Solution: Rewrite the inequality of the constraint into an equation by adding slackvariables S

1, S

2and S

3. The standard form of LPP becomes,

Max Z = 15X1+ 6X

2 + 9X

3 + 2X

4 + 0S

1 + 0S

2 + 0S

3

Subject to,2X1 + X

2 + 5X

3 + 6X

4 + S

1= 20

3X1 +X

2+ 3X

3 + 25X

4 + S

2= 24

7X1+ X

4+ S

3= 70

X1, X

2, X

3, X

4, S

1, S

2, S

3 0

The initial basic feasible solution is S1 = 20, S

2 = 24, S

3 = 70

(X = X2 = X

3 = X

4 = 0 non-basic)

The intial simplex table is given by,Cj 15 6 9 2 0 0 0

CB B XB X1 X2 X3 X4 S1 S2 S3 Min1

BX

X

0 S1 20 2 1 5 6 1 0 0 20/2=10

0 S2 24 3 1 3 25 0 1 0 24/3=8

S3 70 7 0 0 1 0 0 1 70/7=10

Zj 0 0 0 0 0 0 0 0

Zj – Cj –15 –6 –9 –2 0 0 0

As some of Zj – C

j 0, the current basic feasible solution is not optimum.

Z1 – C

1 = –15 is the most negative value, and hence, X

1 enters the basis and the

variable S2 leaves the basis.

First iterationCj 15 6 9 2 0 0 0


BX

X

0 S1 4 0 1/3 3 –32/3 1 –2/3 0 4/1/3=12

15 X1 8 1 1/3 1 25/3 0 1/3 0 8/1/3=24

S3 14 0 –7/3 –7 –172/3 0 –7/3 1 –

Zj 120 15 5 15 125 0 5 0

Zj – Cj 0 –1 6 123 0 5 0



Unit-2

Since Z2 – C

2 = –1 < 0; the solution is not optimal, and therefore, X

2enters the

basis and the basic variable S1leaves the basis.

Second iterationCj 15 6 9 2 0 0 0


BX

X

0 S1 4 0 1/3 3 –32/3 1 –2/3 0 4/1/3=12

15 X1 8 1 1/3 1 25/3 0 1/3 0 8/1/3=24

S3 14 0 –7/3 –7 –172/3 0 –7/3 1 –

Zj 120 15 5 15 125 0 5 0

Zj – Cj 0 –1 6 123 0 5 0

Since all Zj – C

j 0, the solution is optimal and is given by,

Max Z = 132, X1 = 4, X

2 = 12, X

3 = 0, X

4 = 0

Example 2.23: Solve the following LPP using the simplex method.Maximize Z=3X

1 + 2X

2 + 5X

3

Subject to, X1 + 2X

2 + X

3 430

3X1 + 2X

3 460

X1+ 4X

2 420

X1, X

2, X

3 0

Solution: Rewrite the constraint into an equation by adding slack variables S1, S

2,

S3. The standard form of LPP becomes,


2 + 5X

3 + 0S

1 + 0S

2 + 0S

3

Subject to,X1 + 2X

2 + X

3 + S

1= 430

3X1 + 2X

3 + S

2= 460

X1 + 4X

2 + S

3= 420

X1, X

2, X

3, S

1, S

2, S

3 0

The initial basic feasible solution is,

S1 = 430, S

2 = 460, S

3 = 420 (X

1 = X

2 = X

3 = 0)

Initial tableC

j3 2 5 0 0 0

CB

B XB

X1

X2

X3

S1

S2

S3

3

Min BX

X

0 S1

430 1 2 1 1 0 0 430/1=430

0 S2

460 3 0 2 0 1 0 460/2=230

0 S3

420 1 4 0 0 0 1

Zj

0 0 0 0 0 0 0

Zj–C

j–3 –2 –5 0 0 0

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Since some of Zj – C

j 0, the current basic feasible solution is not optimum. Since

Z3 – C

3 = –5 is the most negative, the variable X

3 enters the basis. To find the variable

leaving the basis, find

Min 33

, 0Bi

i

XX

X

= Min

430 460430, 230

1 2

The variable S2 leaves the basis.

First iterationC

j3 2 5 0 0 0

CB

B XB

X1

X2

X3

S1

S2

S3

2

Min BX

X

0 S1

200 –1/2 2 0 1 1/2 0 200/2=100

5 X3

230 3/2 0 1 0 1/2 0

0 S3

420 1 4 0 0 0 1 420/4=105

Zj

1150 15/2 0 0 0 5/2 0

Zj–C

j9/2 –2 0 0 5/2 0

Since Z2 – C

2 = –2 is negative, the current basic feasible solution is not

optimum. Therefore, the variable X2 enters the basis and the variable S

1 leaves the

basis.

2.5 SPECIAL CASES IN LP

2.5.1 Degeneracy

Sometimes, a linear programming problem may have a degenerate situation.Degeneracy is revealed when a basic variable acquires a zero value (rather than anegative or positive value) or in the final solution, either the number of basic variablesis not equal to the number of constraints, or the number of zero variables does notequal the number of decision variables. The instance of degeneracy is usuallypreceded by a tie for an existing variable and an arbitrary selection for it. If this isresolved by a proper selection of pivot element degeneracy can be avoided.

2.5.2 Non-Feasible Solution

A linear programming problem may be unsolved mathematically due to thecontradictory nature of the constraints. Such an instance is referred to as a non-feasible solution. A solution is also non-feasible if an artificial variable appears inthe basis of the solution which was considered to be optimal.

2.5.3 Unbounded Solution

If the coefficient of the entering variable is either negative or zero, implying thatthis variable can be increased indefinitely without ever violating the feasibility, themaximization problem has an unbounded solution.



Unit-2

2.5.4 Multiple Solution or Alternate Optima

The optimal solution may not be unique if one of the non-basic variables has a zerocoefficient. The non-basic variable in the basis will neither increase nor decreasethe value of the objective function. Thus, the problem has an alternate solutionwhich is also optimal.

2.6 CONCEPT OF DUALITY AND SENSITIVITYANALYSIS

Every LPP (called the primal) is associated with another LPP (called its dual). Eitherof the problem can be considered as primal with the other as dual.

Two reasons are attributed to the importance of the concept of duality:

(i) If a large number of constraints and a lesser number of variables constitutethe primal, then the procedure for computation can be minimized by convertingthe primal into its dual and then finding its solution.

(ii) While taking future decisions regarding the activities being programmed, theinterpretation of the dual variables from the cost or economic point of viewproves extremely useful.

2.6.1 Formulation of a Dual Problem

In the formulation of a dual problem, it should be first converted into its canonicalform. Formulation of a dual problem involves the following modifications:

(i) The objective function of minimization in the primal should be convertedinto that of maximization in the dual and vice versa.

(ii) The number of constraints in the dual will be equal to the number of variablesin its primal and vice versa.

(iii) The right-hand side constraints in the dual will be constituted by the costcoefficients C1, C2 ... Cn in the objective function of the primal and viceversa.

(iv) While forming the constraints for the dual, the transpose of the body matrixof the primal problem should be considered.

(v) The variables in both problems should be positive, that is, there should be nonegative values.

(vi) If the variable in the primal is unrestricted in sign, then the correspondingconstraint in the dual will be an equation and vice versa.

2.6.2 Definition of a Dual Problem

Let the primal problem be

Max Z = C1x

1 + C

2x

2 + ... + C

nx

n

Subject to 11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

1 2, 0

n n

n n

m m mn n m

n

a x a x a x b

a x a x a x b

a x a x a x b

x x x

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Dual: The dual problem is defined as

Min Z1 = b1w

1 + b

2w

2 + ... + b

mw

m

Subject to 11 1 21 2 1 1

12 1 22 2 2 2

1 1 2 2

1 2, 0

m m

m m

n n mn n n

m

a w a w a w C

a w a w a w C

a w a w a w C

w w w

Here, w1, w

2, w

3 ... w

mare called dual variables.

Example 2.24: Write the dual of the following primal LP problem.

Max Z = x1+ 2x

2 + x

3

Subject to 1 2 3

1 2 3

1 2 3

1 2 3

2 2

2 5 6

4 6

, , 0

x x x

x x x

x x x

x x x

Solution: Since the problem is not in the canonical form, we interchange theinequality of the second constraint.

Max Z = x1+ 2x

2 + x

3

Subject to 1 2 3

1 2 3

1 2 3

1 2 3

2 2

2 5 6

4 6

and , , 0

x x x

x x x

x x x

x x x

Dual: Let w1, w

2, w

3be the dual variables.

Min Z1 = 2w1+ 6w

2 + 6w

3

Subject to 1 2 3

1 2 3

1 2 3

1 2 3

2 2 4 1

2

5 1

, , 0.

w w w

w w w

w w w

w w w

Example 2.25: Find the dual of the following LPP.

Max Z = 3x1– x

2 + x

3

Subject to 1 2

1 2 3

1 3

1 2 3

4 8

8 3 12

5 6 13

, , 0

x x

x x x

x x

x x x



Unit-2

Solution: Since the problem is not in the canonical form, we interchange the inequalityof the second constraint.

M a x Z = 3x1– x

2 + x

3

Subject to 1 2 3

1 2 3

1 2 3

1 2 3

4 0 8

8 3 12

5 0 6 13

, , 0

x x x

x x x

x x x

x x x

Max Z = Cx

Subject to

0

Ax B

x

1

2

3

8

(3 11) 12

13

x

C x b

x

4 1 0

8 1 3

5 0 6

A

Dual: Let w1, w

2, w

3be the dual variables. The dual problem is

Min Z1 = bTW

Subject to ATW CT and W 0

i.e., Min Z1 =1

2

3

(8 12 13)

w

w

w

Subject to1

2

3

4 8 5 3

1 1 0 1

0 3 6 1

w

w

w

Min Z1 = 8w

1 – 12w

2 + 13w

3

Subject to 1 2 3

1 2 3

1 2 3

1 2 3

4 8 5 3

0 1

0 3 6 1

, , 0

w w w

w w w

w w w

w w w

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Unit-2

Example 2.26: Write the dual of the following LPP.

Min Z = 2x2 + 5x

3

Subject to 1 2

1 2 3

1 2 3

1 2 3

2

2 6 6

3 4

, , 0

x x

x x x

x x x

x x x

Solution: Since the given primal problem is not in the canonical form, we interchangethe inequality of the constraint. Also, the third constraint is an equation. This equationcan be converted into two inequations.

Min Z = 0x1 + 2x

2 + 5x

3

Subject to 1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

0 2

2 6 6

3 4

3 4

, , 0

x x x

x x x

x x x

x x x

x x x

Again on rearranging the constraint, we have

Min Z = 0x1 + 2x

2 + 5x

3

Subject to 1 2 3

1 2 3

1 2 3

1 2 3

1 2 3

0 2

2 6 6

3 4

3 4

, , 0

x x x

x x x

x x x

x x x

x x x

Dual: Since there are four constraints in the primal, we have four dual variables,n a m e l y w

1, w

2, w

3, w

3

Max Z = 2w1 – 6w

2 + 4w

3 – 4w

3

Subject to 1 2 3 3

1 2 3 3

1 2 3 3

1 2 3 3

2 0

2

0 6 3 3 5

, , , 0

w w w w

w w w w

w w w w

w w w w

Let 3 3 3w w w

Max 1 2 3 32 6 4( )Z w w w w

Subject to 1 2 3 3

1 2 3 3

2 ( ) 0

( ) 2

w w w w

w w w w



Unit-2

Finally, we have, 1 2 3 30 6 3( ) 5w w w w

Max 1 2 32 6 4Z w w w

Subject to 1 2 3

1 2 3

1 2 3

1 2 3

2 0

2

0 6 3 5

, 0, is unrestricted.

w w w

w w w

w w w

w w w

Example 2.27: Give the dual of the following problem.

Max Z = x + 2y

Subject to 2 3 4

3 4 5

0 and unrestricted.

x y

x y

x y

Solution: Since the variable y is unrestricted, it can be expressed as ,y y y

, 0y y . On reformulating the given problem, we have

Max 2( )Z x y y

Subject to 2 3 ( ) 4

3 4 ( ) 5

3 4 ( ) 5

, , 0

x y y

x y y

x y y

x y y

Since the problem is not in the canonical form we rearrange the constraints.

Max Z = x + 2y – 2y

Subject to 2 3 3 4

3 4 4 5

3 4 4 5.

x y y

x y y

x y y

Dual: Since there are three variables and three constraints, in dual we havethree variables, namely w

1, w

2, w

2

Min 1 2 24 5 5Z w w w

Subject to 1 2 2

1 2 2

1 2 2

1 2 2

2 3 3 1

3 4 4 2

3 4 4 2

, , 0

w w w

w w w

w w w

w w w

Let w2 = w

2 – w

2, so that the dual variable w

2 is unrestricted in sign. Finally, the

dual is

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Unit-2

Min 11 2 24 5 ( 5 )Z w w w

Subject to1 2 2

1 2

1 2

2 3 ( ) 1

3 4 ( ) 2

3 4 ( ) 2

w w w

w w w

w w w

i.e., Min Z1 = –4w1 + 5w

2

Subject to 1 2

1 2

1 2

2 3 1

3 4 2

3 4 2

w w

w w

w w

w1 0 and w

2 is unrestricted.

i.e., Min Z1 = –4w1 + 5w

2

Subject to 1 2

1 2

1 2

2 3 1

3 4 2

3 4 2

w w

w w

w w

i.e., Min Z1 = –4w1 + 5w

2

Subject to –2w1 + 3w

2 1

–3w1 + 4w

2 2, w

1 0 and w

2 is unrestricted.

Example 2.28: Write the dual of the following primal LPP.

Min Z = 4x1 + 5x

2 – 3x

3

Subject to 1 2 3

1 2 3

1 2 3

22

3 5 2 65

7 4 120

x x x

x x x

x x x

x1 + x

2 0 and x

3 is unrestricted.

Solution: Since the variable x3 is unrestricted, 3 3 3x x x . Also, bring the problem

into the canonical form by rearranging the constraints.

Min Z = 4x1 + 5x

2 – 3 3 3( )x x

Subject to 1 2 3 3

1 2 3 3

1 2 3 3

1 2 3 3

1 2 3 3

( ) 22

22

3 5 2 ( ) 65

7 4 ( ) 120

, , 0

x x x x

x x x x

x x x x

x x x x

x x x x

Min Z = 4x1 + 5x

2 – 3 33 3x x



Unit-2

Subject to 1 2 3 3

1 2 3 3

1 2 3 3

1 2 3 3

1 2 3 3

22

22

3 5 2 2 65

7 4 4 120

, , 0

x x x x

x x x x

x x x x

x x x x

x x x x

Dual: Since there are four constraints in the primal problem, in dual there are four

variables, namely 1 2 2 3, , ,w w w w so that the dual is given by

Max 1 1 2 322( ) 65 120Z w w w w

Subject to 1 1 2 3

1 1 2 3

1 1 2 3

1 1 2 3

1 1 2 3

1 1 2 3

3 4

5 7 5

2 4 3

2 4 3

2 4 3

, , , 0

w w w w

w w w w

w w w w

w w w w

w w w w

w w w w

Let 1 1 11w w w , i.e., the variable w1 is unrestricted.

i.e. Max 1 1 2 322( ) 65 120Z w w w w

Subject to 1 1 2 3

1 1 2 3

1 1 2 3

1 1 2 3

3 4

5 7 5

( ) 2 4 3

( ) 2 4 3

w w w w

w w w w

w w w w

w w w w

i.e., Max 1 2 322 65 120Z w w w

Subject to 1 2 3

1 2 3

1 2 3

1 2 3

3 4

5 7 5

2 4 3

2 4 3

w w w

w w w

w w w

w w w

Finally, we have

Min 1 2 322 65 120Z w w w

Subject to 1 2 3

1 2 2

1 2 3

3 4 4

5 7 5

2 4 3

w w w

w w w

w w w

w2, w

3 0 and w

1 is unrestricted.

2.6.3 Important Results in Duality

1. A primal will always constitute the dual of a dual.

2. In a dual, if one is a minimization problem, then the other will be a maximizationproblem.

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3. Both the LPP and its dual should have feasible solution, so as to arrive at anoptimal solution.

4. The fundamental duality theorem states that in an LPP if the primal problemhas a finite optimal solution, then its dual also has a finite optimal solution.Moreover, both the problems have the same optimal values of their objectivefunction, that is, Max Z = Min Z. The solution of the other problem can be readfrom the Zj – Cj row below the columns of slack and surplus variables.

5. The existence Theorem states that there will be no feasible solution for a problemif either problem in an LPP has an unbounded solution.

6. The complementary slackness theorem states that:(i) If a primal variable is positive, then its dual constraint will be an equation

at the optimum and vice versa.(ii) If a primal constraint is an inequality, then the corresponding dual

variable is zero at the optimum and vice versa.

2.6.4 Sensitivity Analysis

The optimal solution of a linear programming problem is formulated using variousmethods. You have learned the use and importance of dual variables to solve anLPP. Here, you will learn how sensitivity analysis helps to solve repeatedly the realproblem in a little different form. Generally, these scenarios crop up as an end resultof parameter changes due to the involvement of new advanced technologies and theaccessibility of well-organized latest information for key (input) parameters or the‘what-if’ questions. Thus, sensitivity analysis helps to produce optimal solution ofsimple perturbations for the key parameters. For optimal solutions, consider the simplexalgorithm as a ‘black box’ which accepts the input key parameters to solve LPP asshown below:

SimplexAlgorithm

OptimalSolution

LinearProgram

Example 2.29: Illustrate sensitivity analysis using simplex method to solve thefollowing LPP:

Maximize Z = 20x1 + 10x

2

Subject to, 1 2

1 2

3

3 7

x x

x x

And x1, x

2 0

Solution: Sensitivity analysis is done after making the initial and final tableau usingthe simplex method. Add slack variables to convert it into equation form.

Sensitivity analysis:Helps to produceoptimal solution ofsimple perturbationsfor the key parameters.



Unit-2

Maximize Z = 20x1 + 10x

2 + 0S

1 + 0S

2

Subject to, 1 2 1 2

1 2 1 2

0 3

3 0 7

x x S S

x x S S

Where x1, x

2 0

To find basic feasible solution we put x1 = 0 and x

2 = 0. This gives Z = 0, S

1 = 3 and

S2 = 7. The initial table will be as follows:

Initial table

Cj 20 10 0 0

CB B XB X1 X2 S1 S2 Min B

i

X

X

0 S1 3 1 1 1 0 3/1 = 3

0 S2 7 3 1 0 1 7/3 = 2.33

Zj 0 0 0 0 0

Zj – Cj –20 –10 0 0

FindB

i

X

X for each row and also find minimum for the second row. Here,

Zj – C

j is maximum negative (–20). Hence, x

1 enters the basis and S

2 leaves the

basis. It is shown with the help of arrows.

Key element is 3, key row is second row and key column is x1. Now convert

the key element into entering key by dividing each element of the key row by keyelement using the following formula:



Key element

The following is the first iteration table:

Cj 20 10 0 0

CB B XB X1 X2 S1 S2

0 S1 2/3 0 2/3 1 –1/3 2

3/

2

3=1

20 X1 7/3 1 1/3 0 1 7

3/

3

1=7

Zj 140/3 20 20/3 0 20

Zj – Cj – 0 –10/3 0 20

Since Zj – C

j has one value less than zero, i.e., negative value hence this is not

yet the optima solution. Value –10/3 is negative hence x2 enters the basis and S

1

leaves the basis. Key row is upper row.

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Cj 20 10 0 0

CB B XB X1 X2 S1 S2

X2 1 0 1 3/2 –1/2

20 X1 4/3 1 0 0 4/3

Zj 110/3 20 10 0 25

Zj – Cj 0 0 0 25

Zj – C

j 0 for all, hence optimal solution is reached, where 1

4

3x , x

2 = 1,

80

3Z + 10 =

110

3.


1. What are the two types of basic feasible solutions?

2. What are unbounded solutions?

3. What is the role of a slack variable?

4. List the various characteristics of the canonical form.

5. What are surplus variables?

6. What is a non-feasible solution?

ACTIVITY

1. Egg contains 6 units of vitamin A per gram and 7 units of vitamin B pergram and 12 units of vitamin B per gram and costs 20 paise per gram. Thedaily minimum requirement of vitamin A and vitamin B are 100 units and120 units respectively. Find the optimal product mix.

2. A firm has available 240, 370 and 180 kg of wood, plastic and steelrespectively. The firm produces two products A and B. Each unit of Arequires 1, 3 and 2 kg of wood, plastic and steel respectively. Thecorresponding requirement for each unit of B are 3, 4 and 1 respectively.If A sells for ` 4, and B for ` 6, determine how many units of A and Bshould be produced in order to obtain the maximum gross income. Usethe simplex method.



Unit-2

2.7 LET US SUM UP

Linear programming is a decision-making technique under given constraintson the assumption that the relationships among the variables representingdifferent phenomena happen to be linear.

LP is the most widely used technique of decision-making in business andindustry. Thus, it finds application in production planning and scheduling,transportation, sales and advertising, financial planning, portfolio analysis,corporate planning, and many other fields.

Criterion function, which is also known as objective function, is a componentof LP problems. It states whether the determinants of quantity should bemaximized or minimized.

Feasible solutions are all those possible solutions which can be worked uponunder given constraints.

Simple linear LP problems with two decision variables can be easily solvedusing the graphical method.

A set of values X1, X

2, …X

n, which satisfies the constraints of the LPP is

called its solution.

An LP problem can have either a unique optimal solution, or an infinite numberof optimal solutions or an unbounded solution or it can have no solution.

There are two types of basic feasible solutions, namely degenerate and non-degenerate.

In the standard form of LPP, all constraints are expressed as equations; whilein the standard form all constraints are expressed as equations by introducingadditional variables called slack variables and surplus variables.

Simplex method is an iterative procedure used for solving an LPP in a finitenumber of steps.

2.8 FURTHER READING

Gupta, P.K. and D.S. Hira. 2002. Introduction to Operations Research. New Delhi:S. Chand & Co.

Jensen, Paul A. and Jonathan F. Bard. 2003. Operations Research Models andMethods. New York: John Wiley & Sons.

Kalavathy, S. 2002. Operations Research. New Delhi: Vikas Publishing House Pvt.Ltd.

Kothari, C.R. 1992. An Introduction to Operational Research. New Delhi: VikasPublishing House Pvt. Ltd.

Taha, H.A. 2006. Operations Research: An Introduction. New Delhi: Prentice-Hallof India.

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Check Your Progress–11. LP has been successfully applied in the following fields:

(i) Agriculture: LP can be applied in farm management problems as itrelates to the allocation of resources such as acreage, labour, water supplyor working capital in such a way that it maximizes net revenue.

(ii) Contract awards: Evaluation of tenders by recourse to LP guaranteesthat the awards are made in the cheapest way.

(iii) Industries: Applications of LP in business and industry are of the mostdiverse type. Transportation problems concerning cost minimization canbe solved by this technique. The technique can also be adopted in solvingproblems of production (product-mix) and inventory control.

2. The term linearity implies straight line or proportional relationships amongthe relevant variables.

3. Finite choices assumption implies that finite number of choices are availableto a decision-maker and the decision variables do not assume negative values.

Check Your Progress–21. Basic feasible solutions are of two types:

(i) Non-degenerate: A non-degenerate basic feasible solution is the basicfeasible solution which has exactly m positive X

i(i = 1, 2, ..., m), i.e.,

none of the basic variables is zero.(ii) Degenerate: A basic feasible solution is said to be degenerate if one or

more basic variables are zero.

2. If the value of the objective function Z can be increased or decreasedindefinitely, such solutions are called unbounded solutions.

3. Slack variables are used to convert a problem into the standard form.

4. The following are the various characteristics of the canonical form:(i) The objective function is of maximization type.

(ii) All constraints are of ‘’ type.(iii) All variables X

iare non-negative.

5. Surplus variables are defined as the non-negative variables which are removedfrom the LHS of the constraint to convert the inequality ‘’ into an equation.

6. A linear programming problem may be unsolved mathematically due to thecontradictory nature of the constraints. Such an instance is referred to as anon-feasible solution.



1. What is meant by proportionality in linear programming?2. What do you understand by certainty in linear programming?

3. What are the basic constituents of an LP model?



Unit-2

4. What are slack variables? Where are they used? Explain in brief.

5. What is the simplex method?

6. Does every LPP solution have an optimal solution? Explain.


1. A company manufactures 3 products A, B and C. The profits are: ` 3, ` 2 and `4 respectively. The company has two machines and given below is the requiredprocessing time in minutes for each machine on each product.

ProductsMachines A B C

I 4 3 5II 2 2 4

Machines I and II have 2000 and 2500 minutes respectively. The company mustmanufacturers 100 A’s 200 B’s and 50 C’s but no more than 150 A’s. Find thenumber of units of each product to be manufactured by the company to maximizethe profit. Formulate the above as a LP Model.

2. A company produces two types of leather belts A and B. A is of superior qualityand B is of inferior quality. The respective profits are ` 10 and ` 5 per belt. Thesupply of raw material is sufficient for making 850 belts per day. For belt A, aspecial type of buckle is required and 500 are available per day. There are 700buckles available for belt B per day. Belt A needs twice as much time as thatrequired for belt B and the company can produce 500 belts if all of them were ofthe type A. Formulate a LP Model for the above problem.

3. The standard weight of a special purpose brick is 5 kg and it contains two ingredientsB

1and B

2,where B

1costs` 5 per kg and B

2costs` 8 per kg. Strength considerations

dictate that the brick contains not more than 4 kg ofB1and a minimum of 2 kg ofB

2

since the demand for the product is likely to be related to the price of the brick.Formulate the above problem as a LP Model.

4. In a chemical industry two products A and B are made involving two operations.The production of B also results in a by product C. The product A can be soldat ` 3 profit per unit and B at ` 8 profit per unit. The by product C has a profitof ` 2 per unit but it cannot be sold as the destruction cost is ` 1 per unit.Forecasts show that upto 5 units of C can be sold. The company gets 3 units ofC for each units of A and B produced. Forecasts show that they can sell all theunits of A and B produced. The manufacturing times are 3 hours per unit for Aon operation one and two respectively and 4 hours and 5 hours per unit for B onoperation one and two respectively. Because the product C results fromproducing B, no time is used in producing C. The available times are 18, and21 hours of operation one and two respectively. How much of A and B need tobe produced keeping C in mind, to make the highest profit. Formulate theabove problem as LP Model.

5. A company produces two types of hats. Each hat of the first type requires asmuch labour time as the second type. If all hats are of the second type only, thecompany can produce a total of 500 hats a day. The market limits daily sales ofthe first and second type to 150 and 250 hats. Assuming that the profits per hat

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are ` 8 for type B, formulate the problem as a linear programming model in orderto determine the number of hats to be produced of each type so as to maximizethe profit.

6. Solve the following by graphical method:

(i) Max Z = X1 – 3X

2

Subject to, X1 + X

2 300

X1 – 2X

2 200

2X1 + X

2 100

X2 200

X1, X

2 0

(ii) Max Z = 5X + 8YSubject to, 3X + 2Y 36

X + 2Y 203X + 4Y 42 X, Y 0

(iii) Max Z = X – 3YSubject to, X + Y 300

X – 2Y 200 X + Y 100 Y 200and X, Y 0

7. Solve graphically the following LPP:

Max Z = 20X1 + 10X

2

Subject to, X1 + 2X

2 40

3X1 + X

2 30

4X1 + 3X

2 60

and X1, X

2 0

8. A company produces two different products A and B. The company makes aprofit of ` 40 and ` 30 per unit on A and B respectively. The productionprocess has a capacity of 30,000 man hours. It takes 3 hours to produce oneunit of A and one hour to produce one unit of B. The market survey indicatesthat the maximum number of units of product A that can be sold is 8000 andthose of B is 12000 units. Formulate the problem and solve it by graphicalmethod to get maximum profit.

9. Using simplex method, find non-negative values of X1, X

2 and X

3 when

(i) Max Z = X1 + 4X

2 + 5X

3


3X1 + 6X

2 + 3X

3 22

X1 + 2X

2 + 3X

3 14 and

3X1 + 2X

2 14

X1, X

2, X

3 0



Unit-2

(ii) Max Z = X1 + X

2 + 3X

3

Subject to, 3X1 + 2X

2 + X

3 2

2X1 + X

2 + 2X

3 2

X1, X

2, X

3 0

(iii) Max Z = 10X1 + 6X

2

Subject to, X1 + X

2 2

2X1 + X

2 4

3X1 + 8X

2 12

X1, X

2 0

(iv) Max Z = 30X1 + 23X

2 + 29X

3


6X1 + 5X

2 + 3X

3 52

6X1 + 2X

2 + 5X

3 14

X1, X

2, X

3 0

(v) Max Z = X1 + 2X

2 + X

3

Subject to, 2X1 + X

2 – X

3 –2

–2X1 + X

2 – 5X

3 6

4X1 + X

2 + X

3 6

X1, X

2, X

3 0

10. A manufacturer is engaged in producing 2 products X and Y, the contributionmargin being ` 15 and ` 45 respectively. A unit of product X requires 1 unit offacility A and 0.5 unit of facility B. A unit of product Y requires 1.6 units offacility A, 2.0 units of facility B and 1 unit of raw material C. The availabilityof total facility A, B and raw material C during a particular time period are240, 162 and 50 units respectively.

Find out the product-mix which will maximize the contribution margin bysimplex method.

Transportation and Assignment Models


Unit-3

UNIT 3 TRANSPORTATION ANDASSIGNMENT MODELS

UNIT STRUCTURE3.1 Learning Objectives3.2 Introduction3.3 Transportation Problem3.4 Transportation Models

3.4.1 Finding the Initial Basic Feasible Solution3.4.2 North West Corner Rule3.4.3 Least Cost or Matrix Minima Method3.4.4 Vogel’s Approximation Method (VAM)

3.5 Optimum Solution using MODI Method3.6 Special Cases

3.6.1 Maximization Case in a Transportation Problem3.6.2 Degeneracy in Transportation Problems3.6.3 Unbalanced Transportation3.6.4 Alternate Solution

3.7 Transshipment Model3.8 Assignment Problem

3.8.1 Mathematical Formulation of the Assignment Problem3.8.2 Hungarian Method Procedure3.8.3 Maximization in Assignment Problem

3.9 Let Us Sum Up3.10 Further Reading3.11 Answers to ‘Check Your Progress’3.12 Possible Questions



Find an initial basic feasible solution

Explain the various methods of finding an initial solution

Use the MODI method for finding an optimal solution

Explain the various special cases of a transportation problem

Solve an assignment problem using the Hungarian method

Explain how to maximize an assignment problem

3.2 INTRODUCTION

After reading this unit, you will be able to explain the transportation problem, whichis a subclass of an LPP (linear programming problem). Transportation problems dealwith the objective of transporting various quantities of a single homogeneouscommodity initially stored at various origins, to different destinations, in a way thatkeeps transportation cost at a minimum.

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You will learn about the applications of the transportation problem andthe solutionsand rules to solve such problems. The solution of any transportation problem is obtainedin two stages, initial solution and optimal solution. There are three methods of obtainingan initial solution. These are: North West Corner Rule, Least Cost Method and Vogel’sApproximation Method (VAM). VAM is preferred since the solution obtained thisway is very close to the optimal solution. The optimal solution of any transportationproblem is a feasible solution that minimizes the total cost. An optimal solution is thesecond stage of a solution obtained by improving the initial solution. MODI method isused to obtain optimal solutions and optimality tests.

The unit also discusses the basic concepts of an assignment problem. It can bepresented in the form of an n×n cost matrix of real numbers. You will also learn tosolve an assignment problem using the Hungarian method. The unit also explainsmaximization in assignment problems.

3.3 TRANSPORTATION PROBLEM

The transportation problem is one of the subclasses of LPP (Linear ProgrammingProblem) in which the objective is to transport various quantities of a singlehomogeneous commodity that are initially stored at various origins to differentdestinations in such a way that the transportation cost is minimum. To achieve thisobjective we must know the amount and location of available supplies and thequantities demanded. In addition we must know the costs that result from transportingone unit of commodity from various origins to various destinations.

Elementary Transportation Problem

Consider a transportation problem with m origins (rows) and n destinations(columns). Let C

ij be the cost of transporting one unit of the product from the ith

origin to jth destination, ai the quantity of commodity available at origin i, b

j the

quantity of commodity needed at destination j. Xij is the quantity transported from

ith origin to jth destination. This transportation problem can be stated in the followingtabular form.

X X X X

X X X X

X X X

X X X X

i = 1 j = 1

Transportationproblem: Theobjective is to transportvarious quantities of asingle homogeneouscommodity that areinitially stored atvarious origins todifferent destinations insuch a way that thetransportation cost isminimum.



Unit-3

The linear programming model representing the transportation problem is given by,

1 1

Minimizem n

ij iji j

Z C X


1

1, 2, ...,n

ij ij

X a i n

(Row Sum)

1

1, 2, ...,m

ij ji

X b j n

(Column Sum)

0ijX For all i and j

The given transportation problem is said to be balanced if,

i.e., the total supply is equal to the total demand.

Definitions

Feasible solution: Any set of non-negative allocations (Xij>0) which satisfies

the row and column sum (rim requirement) is called a feasible solution.

Basic feasible solution: A feasible solution is called a basic feasible solutionif the number of non-negative allocations is equal to m + n – 1, where m is thenumber of rows and n the number of columns in a transportation table.

Non-degenerate basic feasible solution: Any feasible solution to atransportation problem containing m origins and n destinations is said to be non-degenerate, if it contains m + n – 1 occupied cells and each allocation is inindependent positions.

The allocations are said to be in independent positions if it is impossible toform a closed path. Closed path means by allowing horizontal and vertical lines andwhen all the corner cells are occupied.

The allocations in the following tables are not in independent positions.

The allocations in the following tables are in independent positions.

Feasible solution: Anyset of non-negativeallocations (Xij>0)which satisfies the rowand column sum (rimrequirement).

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Degenerate basic feasible solution: If a basic feasible solution contains lesst h a n m + n – 1 non-negative allocations it is said to be degenerate.

3.4 TRANSPORTATION MODELS

3.4.1 Finding the Initial Basic Feasible Solution

Optimal solution is a feasible solution (not necessarily basic) which minimizes thetotal cost.

The solution of a transportation problem (TP) can be obtained in two stages,namely initial solution and optimum solution.

Initial solution can be obtained by using any one of the three methods, viz.(i) North West Corner Rule (NWCR)

(ii) Least Cost Method or Matrix Minima Method(iii) Vogel’s Approximation Method (VAM)

VAM is preferred over the other two methods, since the initial basic feasiblesolution obtained by this method is either optimal or very close to the optimalsolution.

The cells in the transportation table can be classified into occupied cells andunoccupied cells. The allocated cells in the transportation table are called occupiedcells and the empty cells in the transportation table are called unoccupied cells.

The improved solution of the initial basic feasible solution is called optimalsolution which is the second stage of solution, that can be obtained by MODI(Modified Distribution Method).

3.4.2 North West Corner Rule

Step 1: Starting with the cell at the upper left corner (North West) of the transportationmatrix we allocate as much as possible so that either the capacity of the first rowis exhausted or the destination requirement of the first column is satisfied, i.e.,X

11 = Min (a

1,b

1).

Step 2: If b1>a

1, we move down vertically to the second row and make the second

allocation of magnitude X22

= Min(a2

b1– X

11) in the cell (2, 1).

If b1<a

1, move right horizontally to the second column and make the second

allocation of magnitude X12

= Min (a1, X

11– b

1) in the cell (1, 2).

If b1 = a

1, there is a tie for the second allocation. We make the second allocations

of magnitude,

Optimal solution: Afeasible solution (notnecessarily basic)which minimizes thetotal cost.



Unit-3

or,12 1 1 1

21 2 1 1

Min ( , ) 0 in the cell (1,2)

Min ( , ) 0 in the cell (2,1)

X a a b

X a b b

Step 3: Repeat steps 1 and 2 moving down towards the lower right corner of thetransportation table until all the rim requirements are satisfied.

Example 3.1: Obtain the initial basic feasible solution of a transportation problemwhose cost and rim requirement table is as follows:

Solution: Since ai = 34 = b

j, there exists a feasible solution to the transportation

problem. We obtain the initial feasible solution as follows.

The first allocation is made in the cell (1, 1), the magnitude being

X11

= Min (5, 7) = 5.

The second allocation is made in the cell (2, 1) and the magnitude of theallocation is given by X

21 = Min (8, 7 – 5) = 2.

The third allocation is made in the cell (2, 2) the magnitude X22

= Min (8 – 2, 9)= 6.

The magnitude of the fourth allocation is made in the cell (3, 2) given byX

32 = Min (7, 9 – 6) = 3.

The fifth allocation is made in the cell (3, 3) with magnitude X33

= Min (7 – 3, 14)= 4.

The final allocation is made in the cell (4, 3) with magnitude X43

= Min(14, 18 – 4) = 14.

Hence, we get the initial basic feasible solution to the given TP and is given by,

X11

= 5; X21

= 2; X22

= 6; X32

= 3; X33

= 4; X43

= 14

Total Cost = 2 × 5 + 3 × 2 + 3 × 6 + 3 × 4 + 4 × 7 + 2 × 14

=10 + 6 + 18 + 12 + 28 + 28 = ` 102

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Example 3.2: Determine an initial basic feasible solution to the following transportationproblem using North West Corner Rule.

Solution: The problem is a balanced TP, as the total supply is equal to the total demand.Using the steps we find the initial basic feasible solution as given in the following table.

The solution is given by,

X11

= 6; X12

= 8; X22

= 2; X23

= 14; X33

= 1; X34

= 4

Total Cost = 6 × 6 + 4 × 8 + 2 × 9 + 2 × 14 +6 × 1 + 2 × 4

= 36 + 32 + 18 + 28 + 6 + 8 = ` 128

3.4.3 Least Cost or Matrix Minima Method

Step 1 : Determine the smallest cost in the cost matrix of the transportation table. Let itbe C

ij. Allocate X

ij = Min (a

i, b

j) in the cell (i, j).

Step 2: If Xij = a

i cross off the ith row of the transportation table and decrease

bjby a

i. Then go to Step 3.

If Xij = b

j cross off the jth column of the transportation table and decrease a

iby b

j. Go to Step 3.

If Xij = a

i = b

j cross off either the ith row or the jth column, but not both.

Step 3: Repeat steps 1 and 2 for the resulting reduced transportation table until all therim requirements are satisfied. Whenever the minimum cost is not unique, make anarbitrary choice among the minima.

Example 3.3: Obtain an initial feasible solution to the following TP using MatrixMinima Method.



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Solution: Since ai = b

j = 24, there exists a feasible solution to the TP using the steps

in the least cost method, the first allocation is made in the cell (3, 1) the magnitude beingX

31 = 4. This satisfies the demand at the destination D

1 and we delete this column from

the table as it is exhausted.

The second allocation is made in the cell (2, 4) with magnitude X24

= Min (6, 8) =6. Since it satisfies the demand at the destination D

4, it is deleted from the table. From

the reduced table, the third allocation is made in the cell (3, 3) with magnitudeX33

= Min(8, 6) = 6. The next allocation is made in the cell (2, 3) with magnitudeX

23 of Min (2, 2)

= 2. Finally, the allocation is made in the cell (1, 2) with magnitude X12

= Min (6, 6) = 6.Now, all the requirements have been satisfied and hence, the initial feasible solution isobtained.


X12

= 6; X23

= 2; X24

= 6; X31

= 4; X33

= 6

Since the total number of occupied cells = 5 < m + n – 1We get a degenerate solution.

Total cost×××××` 28

Example 3.4: Determine an initial basic feasible solution for the following TP, using theLeast Cost Method.


j, there exists a basic feasible solution. Using the steps in

least cost method we make the first allocation to the cell (1, 3) with magnitude X13

= Min (14, 15) = 14 (as it is the cell having the least cost).

This allocation exhausts the first row supply. Hence, the first row is deleted.From the reduced table, the next allocation is made in the next least cost cell (2, 3)which is chosen arbitrarily with magnitude X

23 = Min (1, 16) = 1. This exhausts the

third column destination.

From the reduced table, the next least cost cell is (3, 4) for which allocationis made with magnitude Min (4, 5) = 4. This exhausts the destination D

4 requirement.

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Delete this fourth column from the table. The next allocation is made in the cell(3, 2) with magnitude X

32 = Min (1, 10) = 1 which exhausts the third origin capacity.

Hence, the third row is exhausted. From the reduced table, the next allocation is given tothe cell (2,1) with magnitude X

21 = Min (6, 15) = 6. This exhausts the first column

requirement. Hence, it is deleted from the table.

Finally, the allocation is made to the cell (2, 2) with magnitude X22

= Min(9, 9) = 9 which satisfies the rim requirement. These allocations are shown in thefollowing transportation table:

(I Allocation) (II Allocation)

(III Allocation) (IV Allocation)

(V, VI Allocation)

The following table gives the initial basic feasible solution.


X13

=14; X21

=6; X22

= 9; X23

= 1; X32

= 1; X34

= 4

Transportation cost,

= 14 × 1 + 6 × 8 + 9 × 9 + 1 × 2 + 3 × 1 + 4 × 2

= 14 + 48 + 81 + 2 + 3 + 8 = ` 156



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3.4.4 Vogel’s Approximation Method (VAM)

The steps involved in this method for finding the initial solution are as follows.

Step 1: Find the penalty cost, namely the difference between the smallest and the nextsmallest costs in each row and column.

Step 2: Among the penalties as found in Step (1) choose the maximum penalty. If thismaximum penalty is more than one (i.e., if there is a tie) choose any one arbitrarily.

Step 3: In the selected row or column as by Step (2) find out the cell having the leastcost. Allocate to this cell as much as possible depending on the capacity and requirements.

Step 4: Delete the row or column which is fully exhausted. Again, computethe column and row penalties for the reduced transportation table and then go toStep (2). Repeat the procedure until all the rim requirements are satisfied.

Note: If the column is exhausted, then there is a change in row penalty and viceversa.

Example 3.5: Find the initial basic feasible solution for the following transportationproblem using VAM.


j = 950, the problem is balanced and there exists a feasible

solution to the problem.

First, we find the row and column penalty PI as the difference between the

least and the next least cost. The maximum penalty is 5. Choose the first columnarbitrarily. In this column, choose the cell having the least cost name (1, 1). Allocateto this cell with minimum magnitude (i.e., Min (250, 200) = 200.) This exhausts thefirst column. Delete this column. Since a column is deleted, there is a change inrow penalty P

II while column penalty P

II remains the same. Continuing in this manner,

we get the remaining allocations as given in the following table.

II AllocationI Allocation

11

1

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III Allocation IV Allocation

V Allocation VI Allocation

Finally, we arrive at the initial basic feasible solution which is shown in the followingtable.

There are six positive independent allocations which equals to m + n –1 =3 + 4 – 1. This ensures that the solution is a non-degenerate basic feasible solution.

Transportation cost

11 × 200 + 13 × 50 + 18 × 175 + 10 × 125 + 13 × 275 + 10 × 125

= ` 12,075

Example 3.6: Find the initial solution to the following TP using VAM.


j, the problem is a balance TP. Hence, there exists a feasible

solution.



Unit-3

Finally, we have the initial basic feasible solution as given in the followingtable.

There are six independent non-negative allocations equal to m + n – 1 = 3 +4 – 1 = 6. This ensures that the solution is non-degenerate basic feasible.

Transportation cost

= 3 × 45 + 4 × 30 + 1× 25 + 2 × 80 + 4 × 45 + 1 + 75

= 135 + 120 + 25 + 160 + 180 + 75

= ` 695


1. What is a transportation problem?

2. What must one know to achieve the objectives of a transportation problem?

3. How are transportation problems presented?

4. Give the mathematical presentation of a transportation problem.

5. Define feasible, basic, non-degenerate solutions of a transportation problem.

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3.5 OPTIMUM SOLUTION USING MODI METHOD

Optimality Test

Once the initial basic feasible solution has been computed, the next step in the problemis to determine whether the solution obtained is optimum or not.

Optimality test can be conducted to any initial basic feasible solution of a TP,provided such allocations has exactly m + n – 1 non-negative allocations, where m isthe number of origins and n is the number of destinations. Also, these allocationsmust be in independent positions.

To perform this optimality test, we shall discuss the modified distribution method(MODI). The various steps involved in the MODI method for performing optimalitytest are as follows.

MODI Method

Step 1: Find the initial basic feasible solution of a TP by using any one of the threemethods.

Step 2: Find out a set of numbers ui and v

j for each row and column satisfying

ui + v

j= C

ij for each occupied cell. To start with, we assign a number ‘0’ to any row or

column having the maximum number of allocations. If this maximum number ofallocations is more than 1, choose any one arbitrarily.

Step 3: For each empty (unoccupied) cell, we find the sum ui and v

j written in the

bottom left corner of that cell.

Step 4: Find out for each empty cell the net evaluation value ij = C

ij–(u

i+v

j) and

which is written at the bottom right corner of that cell. This step gives the optimalityconclusion,

(i) If all ij > 0 (i.e., all the net evaluation values) the solution is optimum

and a unique solution exists.

(ii) If ij 0 then the solution is optimum, but an alternate solution exists.

(iii) If at least one ij < 0 , the solution is not optimum. In this case, we go to

the next step, to improve the total transportation cost.

Step 5: Select the empty cell having the most negative value of ij. From this cell we

draw a closed path by drawing horizontal and vertical lines with the corner cellsoccupied. Assign signs ‘+’ and ‘–’ alternately and find the minimum allocation fromthe cell having the negative sign. This allocation should be added to the allocationhaving the positive sign and subtracted from the allocation having the negative sign.

Step 6: The previous step yields a better solution by making one (or more) occupiedcell as empty and one empty cell as occupied. For this new set of basic feasibleallocations, repeat from Step (2) till an optimum basic feasible solution is obtained.



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Example 3.7: Solve the following transportation problem.

Solution: We first find the initial basic feasible solution by using VAM. Sincea

i= b

j the given TP is a balanced one. Therefore, there exists a feasible solution.

Finally, we have the initial basic feasible solution as given in the followingtable.

From this table, we see that the number of non-negative independent allocationsis 6 = m+ n – 1 = 3 + 4 –1.

Hence, the solution is non-degenerate basic feasible.

The initial transportation cost

= 11 × 13 + 3 × 14 + 4 × 23 + 6 × 17 + 17 × 10 + 18 × 9 = ` 711

To find the optimal solution: We apply the MODI method in order to determine theoptimum solution. We determine a set of numbers u

i and v

j for each row and column,

with ui+v

j= C

ij for each occupied cell. To start with we give u

2 = 0 as the second

row has the maximum number of allocation.

Now, we find the sum ui and v

j for each empty cell and enter at the bottom left

corner of that cell.

C21

= u2 + v

1 = 17 = 0 + v

1 = 17 v

1 = 17

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C23

= u2 + v

3 = 14 = 0 + v

3 = 14 v

3 = 14

C24

= u2 + v

4 = 23 = 0 + v

4 = 23 v

4 = 23

C14

= u1 + v

4 = 13 = u

1 + 23 = 13 u

1 = 10

C33

= u3 + v

3 = 18 = u

3 + 14 = 18 u

3 = 4

Next, we find the net evaluations ji = C

ij – (u

i+ v

j) for each unoccupied cell and

enter at the bottom right corner of that cell.

Initial table

Since all ij > 0, the solution is optimal and unique. The optimum solution is given

by,

X14

= 11; X21

= 6; X23

= 3; X24

= 4; X32

= 10; X33

= 9

The minimum transportation cost

= 11 × 13 + 17 × 6 + 3 × 14 + 4 × 23 + 10 × 17 + 9 × 18 = ` 711

Example 3.8:Solve the following transportation problem starting with the initial solutionobtained by VAM.


j the problem is a balanced TP. Therefore, there exists a

feasible solution.



Unit-3

Finally, the initial basic feasible solution is given as follows:

Since the number of occupied cells = 6 = m + n – 1 and are also independent,there exists a non-degenerate basic feasible solution.


= 3 × 2 + 3 × 5 + 4 × 4 + 7 × 1 + 6 × 3 + 6 × 1 = ` 68

To find the optimal solution, applying the MODI method, we determine a setof numbers u

i and v

j for each row and column, such that u

i + v

j = C

ij for each

occupied cell. Since the third now has the maximum number of allocations we givenumber u

3 = 0. The remaining numbers can be obtained as follows.

C31

= u3 + v

1 = 7 = 0 + v

1 =7 v

1 = 7

C32

= u3 + v

2 = 6 = 0 + v

2 = 6 v

2 = 6

C33

= u3 + v

3 = 6 = 0 + v

3 = 6 v

3 = 6

C23

= u2 + v

3 = 5 = u

2 + 6 = 5 u

2 = –1

C24

= u2 + v

4 = 4 = –1 + v

4 = 4 v

4 = 5

C11

= u1 + v

1 = 2 = u

1 + 7 = 2 u

1 = –5

We find the sum ui and v

j for each empty cell and enter at the bottom left corner

of the cell. Next, we find the net evaluation ij given by

ij = C

ij – (u

i+ v

j) for each

empty cell and enter at the bottom right corner of the cell.

Initial table

v1 v2 v3 v4

Since all ij > 0, the solution is optimum and unique. The solution is given by,

X11

= 3; X23

= 3; X24

= 4, X31

= 1; X32

= 3; X33

= 1

The total transportation cost

= 2 × 3 + 3 × 5 + 4 × 4 + 7 × 1 + 6 × 3 + 6 × 1 = ` 68

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1. List the approaches used with transportation problems for determining thestarting solution.

2. Define the optimal solution to a transportation problem.

3. State the necessary and sufficient conditions for the existence of a feasiblesolution to a transportation problem.

4. What is the purpose of the MODI method?

3.6 SPECIAL CASES

3.6.1 Maximization Case in a Transportation Problem

In this, the objective is to maximize the total profit for which the profit matrix isgiven. For this, first we have to convert the maximization problem into minimizationby subtracting all the elements from the highest element in the given transportationtable. This modified minimization problem can be solved in the usual manner.

Example 3.9: There are three factories A, B and C which supply goods to fourdealers D

1, D

2, D

3 and D

4. The production capacities of these factories are 1000,

700 and 900 units per month respectively. The requirements from the dealers are900, 800, 500 and 400 units per month respectively. The per unit return (excludingtransportation cost) are ` 8, ` 7 and ` 9 at the three factories. The following tablegives the unit transportation costs from the factories to the dealers.

Determine the optimum solution to maximize the total returns.

Solution: Profit/Return–Transportation cost. With this we form a transportationtable with profit.



Unit-3

Profit Matrix

This profit matrix is converted into its equivalent loss matrix by subtracting all theelements from the highest element, namely 8. Hence, we have the following loss matrix.

We use VAM to get the initial basic feasible solution.

The initial solution is given in the following table:

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Since the number of allocated cells = 5 < m + n – 1 = 6, the solution is degenerate.To resolve this degeneracy, we add an empty cell (1, 3) by allocating a non-negativequantity.

This cell is the least cost cell and is of independent position. The initial basic feasiblesolution is given as follows:

Initial table

The number of allocations = 6 = m + n – 1 and the 6 allocations are in independentposition. Hence, we can perform the optimality test using MODI method.

A

B

C

uiD1 D2 D3 D4

22 2 4

64 2 3

23 1 0

200 800

700

500 400

ui

1 2 1 1

4 2 4 2 03

310

2

–1

2 2 2 1

Since all the net evaluation ji > 0 are non-negative, the initial solution is optimum.

The optimum distribution is

A D1 = 200 units

A D2 = 800 units

A D3 = units

B D1 = 700 units

C D3 = 500 units

A D4 = 400 units

The total profit or max. return

= 200 6 + 6 800 + 4 700 + 7 500 + 8 400

= ` 15,500



Unit-3

Example 3.10: Solve the following transportation problem to maximize the profit.

5

Solution: Since the given problem is to maximize the profit, we convert this into lossmatrix and minimize it. For converting it into minimization type, we subtract all theelements from the highest element 90. Hence, we have the following loss matrix.

Since ai = b

j there exists a feasible solution which can be obtained by VAM.

The initial basic feasible solution is given in the following table.

As the number of independent allocated cells = 6 = m + n – 1, the solution is non-degenerate.

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Optimality test using MODI method

Since all the net evaluationsij > 0 the solution is optimum and unique. The optimum

solution is given by,

X12

= 23; X21

= 6; X24

= 30; X31

= 17; X33

= 16

The optimum profit = 23 × 51 + 6 × 80 + 42 × 8 + 81 × 30 + 90 × 17 + 66 × 16

= ` 7005

3.6.2 Degeneracy in Transportation Problems

In a TP, if the number of non-negative independent allocations is less than m + n – 1,where m is the number of origins (rows) and n is the number of destinations, (columns)then it results in degeneracy. This may occur either at the initial stage or at thesubsequent iteration.

To resolve this degeneracy, we adopt the following steps.

Step 1: Among the empty cells, we choose an empty cell having the least cost whichis of an independent position. If there are more than one cell, choose any onearbitrarily.

Step 2: To the cell as chosen in Step (1) we allocate a small positive quantity > 0.

The cells containing ‘’ are treated like other occupied cells and degeneracyis removed by adding one (more) accordingly. For this modified solution, we adoptthe steps involved in MODI method till an optimum solution is obtained.

Example 3.11: Solve the transportation problem for minimization.

Solution: Since ai=b

j the problem is a balanced TP. Hence, there exists

a feasible solution. We find the initial solution by North West Corner rule asgiven here.



Unit-3

Since, the number of occupied cells = 5 = m + n – 1, and all the allocations areindependent, we get an initial basic feasible solution.


= 10 × 2 + 4 × 10 + 5 × 1 + 10 × 3 + 1 × 30 = 20 + 40 + 5 + 30 + 30 = ` 125

To find the optimal solution (MODI method): We use the previous table to applythe MODI method. We find out a set of numbers u

i and v

j for which u

i+ v

j = C

ij, only

for occupied cells. To start with, as the maximum number of allocations is 2 inmore than one row and column, we choose arbitrarily column 1, and assign a number0 to this column, i.e., v

1 = 0. The remaining numbers can be obtained as follows.

C11

= u1 + v

1 = 2 = u

1 + 0 = 2 u

1 = 2

C21

= u2 + v

1 = 4 u

2 = 4 – 0 = 4

C22

= u2 + v

2 = 1 v

2 = 1 – u

2 = 1 – 4 = –3

C32

= u3 + v

2 = 3 = u

3 = 3 – v

2 = 3 – (–3) = 6

C33

= u3 + v

3 = 1 = v

3 = 1 – u

3 = 1 – 6 = –5

Initial table

v1 v2 v3

We find the sum of ui and v

j for each empty cell and write at the bottom left

corner of that cell. Find the net evaluations ij= C

ij– (u

i+ v

j) for each empty cell and

enter at the bottom right corner of the cell.

The solution is not optimum as the cell (3, 1) has a negative ij value. We

improve the allocation by making this cell (3, 1) as an allocated cell. We draw aclosed path from this cell and assign the signs ‘+’ and ‘’ alternately. From the cellhaving the negative sign we find the minimum allocation given by Min (10, 10) =10. Hence, we get two occupied cells.

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(2, 1) (3, 2) becomes empty and the cell (3, 1) is occupied and results in a degeneratesolution. (Degeneracy in subsequent iteration).

Number of allocated cells = 4 < m n 1 5

We get degeneracy. To resolve the degeneracy we add the empty cell (1, 2) andallocate >0. This cell namely (1, 2) is added as it satisfies the two steps for resolvingthe degeneracy. We assign a number 0 to the first row, namely u

1= 0; we get the

remaining numbers as follows.

C11

C12

C31

C33

C22

Next, we find the sum of uj and v

j for the empty cell which is entered at the

bottom left corner of that cell and also the net evaluation ij

Cij– (u

i+ v

j) for each

empty cell which is entered at the bottom right corner of the cell.

Iteration table

The modified solution is given in the following table. This solution is also optimaland unique as it satisfies the optimality condition that all

ij > 0.



Unit-3

X11

= 10: X22

= 15; X33

= 30; X12

= ; X31

= 10

Total cost = 10 × 2 + 2 × + 15 × 1 + 10 × 1 + 30 × 1

= 75 + 2 = ` 75

Example 3.12: Solve the following transportation problem whose cost matrix is givenbelow.


j, the problem is a balanced transportation problem. Hence,

there exists a feasible solution. We find the initial solution by North West CornerRule.

We get the total number of allocated cells = 7 = 4 + 4 – 1. As all the allocationsare independent, the solution is non-degenerate.


= 1 × 21 + 5 × 13 + 3 × 12 + 3 × 1 + 12 × 2 + 2 × 2 + 17 × 4

= ` 221

To find the optimal solution (MODI Method): We determine a set of numbers ui

and vj for each row and each column with u

j + v

j = C

ij for each occupied cell. To start

with we give 0 to the third column as it has the maximum number of allocations.

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Initial table

A B C D

C23

= u2 + v

3 = 1 u

2 = 1 – 0

C33

= u3 + v

3 = 2 u

3 = 2 – v

3 = 2 – 0 = 2

C43

= u4 + v

3 = 2 u

4 = 2 – 0 = 2

C44

= u4 + v

4 = 4 v

4 = 4 – 2 = 2

C22

= u2 + v

2 = 3 v

2 = 3 – u

2 = 2

C12

= u1 + v

2 = 5 u

1 = 5 – v

2 = 3

C11

= u1 + v

1 = 1 v

1 = 1 – u

1 = 1 – 3 = –2

We find the sum ui and v

j for each empty cell and enter at the bottom left corner

of that cell. We find the net evaluation ij = C

ij – (u

i + v

j) for each empty cell and enter

at the bottom right corner of that cell. The solution is not optimum as some of ij<0.

We choose the most negative ij, i.e., –2. There is a tie between the cell (1, 4) and (3,

2), but we choose the cell (3, 2) as it has the least cost. From this cell, we draw aclosed path and assign ‘+’ and ‘–’ signs alternately and find the minimum allocationfrom the cell having ‘–’ sign.

Thus, we get, Min (12, 12) = 12. Hence, one empty cell (3, 2) becomes occupiedand two occupied cells (2, 2) (3, 3) becomes empty resulting in degeneracy(Degeneracy in subsequent iteration). By adding and subtracting this minimumallocation, we get the modified allocation as given in the following table. For thismodified allocations, we repeat the steps in the MODI method.

First iteration table



Unit-3

The number of allocations 6 < m n 1 7. We add the cell (3,3) as it is theleast cost empty cell which is of independent position. Give a small quantity > 0.This removes the degeneracy.

The modified allocation is given in the following table.

The solution is not optimum. The next negative value of ij = –4 (the cell

(1, 4)).

The minimum allocation is Min (13, , 17) = . Proceeding in the samemanner, we have the second iteration table given as follows.

Second iteration table

As the solution is not optimum, we improve the solution by using the stepsinvolved in the MODI method. The most negative value of

ij = – 2. The minimum

allocation is Min (13 – , 15, 17 – ) = 13 – .

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Third iteration table

The solution is improved by adding and subtracting the new allocation givenby Min (2 + , 4) = (2 +).

Fourth iteration table

Since all ij 0, the solution is optimum (alternate solution exists). The solution

is given by

The total transportation cost21 × 1 + 3 × 13 + 3 (13 – ) + 2 (2 + ) + 2 (12 + ) + 2 (17 + ) + 4 (2 – )= 169 – = ` 169

Example 3.13: A company has three plants A, B, C and three warehouses X, Y, Z.The number of units available at the plants is 60, 70, 80 and the demand at X, Y, Zare 50, 80, 80 respectively. The unit cost of the transportation is given in the followingtable.



Unit-3

Find the allocation so that the total transportation cost is minimum.

Solution:

Since ai = b

j = 210, the problem is a balanced one. Hence, there exists a

feasible solution. Let us find the initial solution by the least cost method.

Here, the least cost cell is not unique, i.e., the cells (2, 1) (1, 3) and (3, 2) havethe least value 3. So, choose the cell arbitrarily. Let us choose the cell (2, 1) andallocate with magnitude Min (70, 50) = 50. This exhausts the first column. Now,delete this column. The reduced transportation table is given by,

Continuing in this manner, we finally arrive at the initial solution which isshown in the following table.

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The number of allocated cells is 4 < m + n – 1 = 5. This results in degeneracy(Initial stage). To remove this degeneracy, we add an empty cell (3, 3) whose cost isminimum and is of independent position. Allocate to this cell a small quantity >0.Hence, we have the initial solution given in the following table:

The total number of occupied cells = 5 = m + n – 1 = 5. These are also ofindependent position. This solution is non-degenerate.

The solution is given by

X13

= 60; X21

= 50; X23

= 20; X32

= 80; X33

=


= 3 × 60 + 3 × 50 + 9 × 20 + 3 × 80 + 5 ×

= ` 750 + 5 = ` 750

To find the optimal solution: We apply the steps involved in the MODI method tothe previous table. We find a set of numbers u

i and v

j for which u

i + v

j = C

ij is

satisfied for each of the occupied cell. To start with, we assign a number 0 to thethird column (i.e., v

3 = 0) as it has the maximum number of allocations. The remaining

numbers are obtained as follows.

Since, all ij > 0 we obtain an optimum solution.


X13

= 60; X21

= 50; X23

= 20; X32

= 80; X33

=

Total transportation cost = 3 × 60 + 3 × 50 + 9 × 20 + 3 × 80 + 5 ×

= ` 750 + 5 = ` 750



Unit-3

3.6.3 Unbalanced Transportation

The given TP is said to be unbalanced ifaib

j, i.e., if the total supply is not equal to

the total demand.

There are two possible cases.

Case (i):i=1 1

m n

i jj

a b

If the total supply is less than the total demand, a dummy source (row) is includedin the cost matrix with zero cost; the excess demand is entered as a rim requirement forthis dummy source (origin). Hence, the unbalanced transportation problem can beconverted into a balanced TP.

Case (ii):i=1 1

m n

i jj

a b

i.e., the total supply is greater than the total demand. In this case, the unbalancedTP can be converted into a balanced TP by adding a dummy destination (column) withzero cost. The excess supply is entered as a rim requirement for the dummy destination.

Example 3.14: Solve the transportation problem when the unit transportation costs,demands and supplies are as given below:

Solution: Since the total demand bj215 is greater than the total supply a

i195,

the problem is an unbalanced TP.

W e c o n v e r t t h i s i n t o a b a l a n c e d T P b y i n t r o d u c i n g a d u m m y o r i g i n O4with cost

zero and giving supply equal to 215 – 195 = 20 units. Hence, we have the convertedproblem as follows:

UnbalancedTransportation: Thegiven TP is said to beunbalanced if the totalsupply is not equal tothe total demand.

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As this problem is balanced, there exists a feasible solution to this problem. UsingVAM, we get the initial solution.

The initial solution to the problem is given by,

There are seven independent non-negative allocations equal tom + n – 1. Hence,the solution is non-degenerate.


= 6 × 65 + 5 × 1 + 5 × 30 + 2 × 25 + 4 × 25 + 7 × 45 + 20 × 0

= ` 1010

To find the optimal solution: We apply the steps in the MODI method to the previoustable.



Unit-3

11

10

5

25 45

0

Since all ij 0, the solution is not optimum; hence, we introduce the cell (3,1) as

this cell has the most negative value of ij. We modify the solution by adding and

subtracting the minimum allocation given by Min (65, 30, 25). While doing this, theoccupied cell (3, 3) becomes empty.


As the number of independent allocations are equal to m+n–1, we check theoptimality.

Since all ij0, the solution is optimal and an alternate solution exists as

14

= 0. Therefore, the optimum allocation is given by,

X11

= 40; X12

= 30; X22

= 5; X23

= 50; X31

= 25; X34

= 45; X41

= 20

The optimum transportation cost,

= 6 × 40 + 1 × 30 + 5 × 5 + 2 × 50 + 10 × 25 + 7 × 45 + 0 × 20

= ` 960

Example 3.15: A product is produced by 4 factories F1, F

2, F

3 and F

4. Their unit

production costs are ` 2, 3, 1 and 5 respectively. The production capacity of thefactories are 50, 70, 40 and 50 units respectively. The product is supplied to 4 storesS

1, S

2, S

3 and S

4, the requirements of which are 25, 35, 105 and 20 respectively.

Find the transportation plan such that the total production and transportationcost is minimum. The unit costs of transportation are given as follows:

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Solution: We form the transportation table which consists of both production andtransportation costs.

Total capacity = 200 units

Total demand = 185 units

Therefore, ai > b

j. Hence, the problem is unbalanced. We convert it into a

balanced one by adding a dummy store S5 with cost 0 and the excess supply is given

as the rim requirement to this store, namely (200–185) units.

The initial basic feasible solution is obtained by the least cost method. We getthe solution containing eight non-negative independent allocations equal to m +n –1. So the solution is a non-degenerate solution.


= 4 × 25 + 6 × 5 + 8 × 20 + 10 × 50 + 8 × 20 + 4 × 30 + 13 × 55 + 0 × 15

= ` 1785

To find the optimal solution: We apply MODI method to the previous table as it hasm + n – 1 independent non-negative allocations.



Unit-3

Initial table

The solution is not optimum as the cell (4, 5) has a negative net evaluation value,i.e.,

45= –3. We draw a closed path from this cell and have a modified allocation by

adding and subtracting the allocation Min (35, 20)= 20. This modified allocation is givenin the following table.


Since all the values of ij 0 the solution is optimum, but an alternate solution exists.

The optimum solution or the transportation plan is given by,

X11

= 25 units; X32

= 30 units

X12

= 5 units; X43

= 15 units

X13

= 20 units; X44

= 20 units

X23

= 70 units; X45

= 15 units

This is the surplus capacity that is not transported but is manufactured in factoryF

4. The optimum production with transportation cost

= 4 × 25 + 6 × 5 + 8 × 20 + 10 × 70 + 4 × 30 + 8 × 20 + 13 × 15 + 0 × 15

= ` 1465

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3.6.4 Alternate Solution

You know that the optimal solution of any LPP occurs at the extreme point of thefeasible solution when the solution is unique. No other solution will give the similarvalue of the objective function. However, sometimes a given LPP may have morethan one solution and the result is similar for the optimal objective function valueand is termed as alternate solution. For an alternate solution, the following twoconditions are necessary:

1. The slope of the objective function should be similar as constraint whichforms the edge of the feasible region.

2. The constraint must be an active constraint, i.e., in the path of optimal progressof the objective function.

Basically, an active constraint is the constraint which passes through one of theextreme points of the feasible region.

3.7 TRANSSHIPMENT MODEL

In a transshipment model, the objects are supplied from various specific sources tovarious specific destinations. It is also economic if the shipment passes via thetransient nodes which are in between the sources and the destinations. It is differentfrom transportation problem where the shipments are directly sent from a specificsource to a specific destination, whereas in the transshipment problem the maingoal is to reduce the total cost of shipments. Hence, the shipment passes via one ormore intermediary nodes before it reaches its desired specific destination. Basically,there are two methods of evaluating transshipment problems as discussed below.

The following is the schematic illustration of the sources and destinationsacting as transient nodes of a simple transshipment problem.

S1 D1

S2 D2

S3 D3

Sj

Sm

Di

Dn

Fig. 3.1 Schematic Diagram of Simple Transshipment Model

Alternate solution: Agiven LPP may havemore than one solutionand the result is similarfor the optimalobjective functionvalue.



Unit-3

Figure 3.1 shows the shipment of objects from source S1 to destination D

2.

Shipment from source S1 can pass via S

2 and D

1 before it reaches the desired destination

D2. Because the shipment passes via the particular transient nodes, this arrangement

is named as transshipment model. The goal of the transshipment problem is to discoverthe optimal shipping model so that the total transportation cost is reduced.

Figure 3.2 shows a different approach where the number of first starting nodesand also the number of last ending nodes is the sum of the total number of sourcesand destinations of the original problem. Let B be the buffer which should bemaintained at every transient source and transient destination. Considering it as abalanced problem, buffer B at the least may be equal to the sum of total supplies orthe sum of total demands. Therefore, a constant B is further added to all the startingnodes and the ending nodes as shown below:

S1 S1

D1 D1

S2 S2

D2 D2

S3 S3

D3 D3

Sj Sj

Sm Sm

Di Di

Dn Dn

Sources Destinations

a B1 +

a B2 +

a B3 +

a Bj +

a Bm +

B

B

B

B

B

B

B

B

B

B

b B1 +

b2 + B

b3 + B

bi + B

bn + B

Fig. 3.2 Modified Version of Simple Transshipment Problem

Here in the modified version of simple transshipment model, the destinationsD

1, D

2, D

3, ..., D

i,..., D

nare incorporated as added starting nodes which basically

acts as the transient nodes. Hence, these do not have the original supplies and atleast the supply of every transient node must be equal to B. Therefore, every transientnode is assigned B units of supply value. Also, the sources S

1, S

2, S

3,..., S

j..., S

mare

incorporated as added ending nodes which basically act as the transient nodes. Thesenodes too do not have the original demands but every transient node is assigned Bunits of demand value. To make it a balanced problem, B is further added to everystarting node and to the ending nodes. Hence, the problem resembles a usualtransportation problem and can be solved to obtain the optimum shipping plan.

Example 3.16: The following is the transshipment problem with 4 sources and 2destinations. The supply values of the sources S

1, S

2, S

3 and S

4are 100, 200, 150 and

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350 units respectively. The demand values of destinations D1 and D

2are 350 and 450

units respectively. Transportation cost per unit between various defined sources anddestinations are given in the following table. Solve the transshipment problem.

Destination

S1 S2 S3 S4 D1 D2

S1 0 4 20 5 25 12

S2 10 0 6 10 5 20

S3 15 20 0 8 45 7

S4 20 25 10 0 30 6

D1 20 18 60 15 0 10

Source

D2 10 25 30 23 4 0

Solution: In the above table the number of sources is 4 and the number of destinationsis 2. Therefore, the total number of starting nodes and the ending nodes of thetransshipment problem will be 4 + 2 = 6.

The following is the detailed format of the transshipment problem after includingtransient nodes for the sources and destinations. Here, the value of B is added to allthe rows and columns.

Destination Supply

S1 S2 S3 S4 D1 D2

S1 0 4 20 5 25 12 100+800=900

S2 10 0 6 10 5 20 200+800=1000

S3 15 20 0 8 45 7 150+800=950

S4 20 25 10 0 30 6 350+800=1150

D1 20 18 60 15 0 10 800

D2 10 25 30 23 4 0 800

Source

800 800 800 800 350+800=1150 450+800=1250

The optimal solution and the corresponding total cost of transportation is` 5,600. The allocations defined in the main diagonal cells are ignored. Thediagrammatic representation of the optimal shipping pattern of the shipments relatedto the off-diagonal cells is shown below:



Unit-3

S1

D1

S2

S4

D2

S3

100300

50150

350

Optimal Shipping Pattern

3.8 ASSIGNMENT PROBLEM

The assignment problem is one of the fundamental combinatorial optimization problems.It helps to find a maximum weight identical in nature in a weighted bipartite graph. Theassignment problem is also termed as a special case of transportation problem.

Suppose there are n jobs to be performed and n persons are available for doingthese jobs. Assume that each person can do each job at a time, though with varyingdegrees of efficiency. Let C

ij be the cost if the ith person is assigned to the jth job. The

solution to the problem is to find an assignment (which job should be assigned to whichperson on one-one basis) so that the total cost of performing all jobs is minimum. Problemsof this kind are known as assignment problems.

The assignment problem can be stated in the form of n × n cost matrix [Cij] of

real numbers as given in the following table:

Jobs

Persons

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

1 2 3

1 2 3

1 2 3... ...

... ...1

... ...2

... ...3

... ...

... ...

j n

j n

j n

i i i ij in

n n n nj nn

j n

C C C C C

C C C C C

C C C C C

C C C C Ci

C C C C Cn

3.8.1 Mathematical Formulation of the Assignment Problem

Mathematically, the assignment problem can be stated as,

Minimize Z =1 1

1,2, ,

1,2, ,

n n

ij iji j

C x i n

j n

Subject to the restrictions:

Assignment problem:Helps to find amaximum weightidentical in nature in aweighted bipartitegraph.

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1 if the th person is assigned th job

0 if notij

i jx

1

1n

ijj

x

(one job is done by the ith person)

and1

1n

iji

x

(only one person should be assigned the jth job)

Where xij denotes that the jth job is to be assigned to the ith person.

3.8.2 Hungarian Method Procedure

The solution of an assignment problem can be arrived at by using the Hungarian method.The steps involved in this method are as follows:

Step 1: Prepare a cost matrix. If the cost matrix is not a square matrix then adda dummy row (column) with zero cost element.

Step 2: Subtract the minimum element in each row from all the elements of therespective rows.

Step 3: Further modify the resulting matrix by subtracting the minimum elementof each column from all the elements of the respective columns. Thus, obtain the modifiedmatrix.

Step 4: Then, draw the minimum number of horizontal and vertical lines to coverall zeroes in the resulting matrix. Let the minimum number of lines beN. Now, there aretwo possible cases.

Case (i) If N = n, where n is the order of matrix, then an optimal assignmentcan be made. So make the assignment to get the required solution.

Case (ii) If N < n, then proceed to Step 5.

Step 5: Determine the smallest uncovered element in the matrix (elementnot covered by N lines). Subtract this minimum element from all uncovered elementsand add the same element at the intersection of horizontal and vertical lines. Thus,the second modified matrix is obtained.

Step 6: Repeat steps (3) and (4) until we get the Case (i) of Step 4.

Step 7: (To make zero assignment) Examine the rows successively until arow with exactly single zero is found. Circle (O) this zero to make the assignment.Then mark a cross (×) over all zeroes if lying in the column of the circled zeroshowing that they cannot be considered for future assignment. Continue in thismanner until all the zeroes have been examined. Repeat the same procedure for thecolumns also.

Step 8: Repeat Step 6 successively until one of the following situation arises: Case (i) If no unmarked zero is left, then the process ends.Case (ii) If there are more than one unmarked zero in any column or row, then

circle one of the unmarked zeroes arbitrarily and mark a cross in the cellsof remaining zeroes in its row or column. Repeat the process until nounmarked zero is left in the matrix.



Unit-3

Step 9 Thus, exactly one marked circled zero in each row and each column ofthe matrix is obtained. The assignment corresponding to these marked circled zeroeswill give the optimal assignment.

Example 3.17: Using the following cost matrix determine (i) Optimal job assignment(ii) The cost of assignments.

Job

Mechanic

Job

1 2 3 4 5

10 3 3 2 8

9 7 8 2 7

7 5 6 2 4

3 5 8 2 4

9 10 9 6 10

A

B

C

D

E

Solution: Select the smallest element in each row and subtract this smallest elementfrom all the other elements in its row.

Subtract the minimum element from each column and subtract this elementfrom all the elements in its column. With this we get the first modified matrix.

In this modified matrix, we draw the minimum number of lines to cover allzeroes (horizontal and vertical).

Number of lines drawn to cover all zeroes is N = 4.

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The order of matrix is n = 5.

Hence, N < n, less than the order of matrix.

Now, we get the second modified matrix by subtracting the smallest uncoveredelement from the remaining uncovered elements and adding it to the element at thepoint of intersection of lines.

Number of lines drawn to cover all zeroes is N = 5.

The order of matrix is n = 5.

Hence, N = n, which is order of matrix. Now, we determine the optimumassignment.

Assignment

The first row contains more than one zero. So proceed to the second row. Ithas exactly one zero. The corresponding cell is (B, 4). Circle this zero, thus makingan assignment. Mark (×) for all other zeroes in its column to show that they cannotbe used for making other assignments. Now, Row 5 has a single zero in the cell(E, 3). Make an assignment in this cell and cross the second zero in the third column.

Now, Row 1 has a single zero in Column 2, i.e., in the cell (A, 2).

Make an assignment in this cell and cross the other zeroes in the secondcolumn. This leads to a single zero in Column 1 of the cell (D, 1). Make an assignmentin this cell and cross the other zeroes in the fourth row. Finally, we have a singlezero left in the third row making an assignment in the cell (C, 5). Thus, we have thefollowing assignment.

Job Mechanic Cost

1 D 3

2 A 3



Unit-3

3 E 9

4 B 2

5 C 4

` 21

Therefore, 1D, 2A, 3E, 4B, 5C with minimum cost ` 21.

Example 3.18: A company has 5 jobs to be done on five machines. Any job can bedone on any machine. The cost of doing the jobs in different machines are as follows.Assign the jobs for different machines so as to minimize the total cost.

Solution: We form the first modified matrix by subtracting the minimum elementfrom all the elements in the respective rows and also in the respective columns.

Since each column has the minimum element 0, we have the first modifiedmatrix. Now we draw the minimum number of lines to cover all zeroes.

The number of lines drawn to cover zeroes is N = 4, which is less than theorder of matrix which is n = 5.

We find the second modified matrix by subtracting the smallest uncoveredelement from all the uncovered elements and adding to the element which is at thepoint of intersection of lines.

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The number of lines drawn to cover all zeroes is N = 5 = n = 5, which is the orderof matrix. Hence, we can form an assignment.

Assignment

All the five jobs have been assigned to five different machines.

Here, the optimal assignment is as follows:

Job Machine

1 B

2 E

3 C

4 A

5 D

Minimum (Total cost) = 8+12+4+6+12 =` 42

Example 3.19: Four different jobs can be done on four different machines and downtime costs are prohibitively high for changeovers. The following matrix gives the cost inrupees of producing Job i on Machine j:

MachineJobs M1 M2 M3 M4

J1 5 7 11 6J2 8 5 9 6J3 4 7 10 7J4 10 4 8 3

How should the jobs be assigned to the various machines so that the total cost isminimized?

Solution:First we form a modified matrix by subtracting the least element in the respectiverow and respective column.



Unit-3

0 2 6 1

3 0 4 1

0 3 6 3

7 1 5 0

M M M M1 2 3 4

J

J

J

J

1

2

3

4

Since the third column has no zero element, we subtract the smallest element 4from all the elements.

0 2 2 1

3 0 0 1

0 3 2 3

7 1 1 0

M M M M1 2 3 4

J

J

J

J

1

2

3

4

0 2 2 1

3 0 0 1

0 3 2 3

7 1 1 0

M M M M1 2 3 4

J

J

J

J

1

2

3

4

Now, we draw the minimum number of lines to cover all zeroes.

The number of lines drawn to cover all zeroes is N = 3, which is less than theorder of matrix, which is n = 4.

Hence, we form the second modified matrix, by subtracting the smallestuncovered element from all the uncovered elements and adding to the element whichis at the point of intersection of lines.

0 1 1 1

4 0 0 2

0 2 1 3

7 0 0 0

M M M M1 2 3 4

J

J

J

J

1

2

3

4

N = 3 < n = 4

0 0 0 0

5 0 0 2

0 1 0 2

8 0 0 0

M M M M1 2 3 4

J

J

J

J

1

2

3

4

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N = 4 = n = 4 = Order of matix

0 0 0 0

5 0 0 2

0 1 0 2

8 0 0 0

M M M M1 2 3 4

J

J

J

J

1

2

3

4

× × ×××

× ×

Hence, we can make an assignment.

Since no rows and no columns have single zero, we have different assignments(Multiple solution).

Optimal assignment

Job Machine

J1

M4

J2

M2

J3

M1

J4

M3

Minimum (Total cost)

6+5+4+8 = ` 23

Alternate solution

0 0 0 0

5 0 0 2

0 1 0 2

8 0 0 0

M M M M1 2 3 4

J

J

J

J

1

2

3

4

J1 M

1; J

2 M

2; J

3 M

3; J

4M

4

Minimum (Total cost) 5+5+10+3 = ` 23

Example 3.20: Solve the following assignment problem in order to minimize thetotal cost. The following cost matrix gives the assignment cost when different operatorsare assigned to various machines.

Operators

Machines

30 25 33 35 36

23 29 38 23 26

30 27 22 22 22

25 31 29 27 32

27 29 30 24 32

I II III IV V

A

B

C

D

E



Unit-3

Solution: We form the first modified matrix by subtracting the least element from allthe elements in the respective rows and then in the respective columns.

Since each column has the minimum element 0, the first modified matrix isobtained. We draw the minimum number of lines to cover all zeroes.

The number of lines drawn to cover all zeroes is N = 4, which is less than theorder of matrix n = 5. Hence, we form the second modified matrix by subtractingthe smallest uncovered element from the remaining uncovered elements and add tothe element which is at the point of intersection of lines.

5 0 5 10 8

0 6 12 0 0

11 8 0 3 0

0 6 1 2 4

3 5 3 0 5

I II III IV V

A

B

C

D

E

N = 5, i.e, the number of lines drawn to cover all zeroes is equal to the orderof matrix. Hence, we can make the assignment.

Here, N = 5 = n = 5 = Order of matrix

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The optimum assignment is

Operators Machines

I D

II A

III C

IV E

V B

The optimum cost is given by,

25 + 25 + 22 + 24 + 26 = ` 122


1. What is an assignment problem?

2. How is an assignment problem presented to find the minimum cost?

3. Name the method used for solving assignment problems.

3.8.3 Maximization in Assignment Problem

In this, the objective is to maximize the profit. To solve this, we first convert thegiven profit matrix into the loss matrix by subtracting all the elements from thehighest element of the given profit matrix. For this converted loss matrix we applythe steps in Hungarian method to get the optimum assignment.

Example 3.21: The owner of a small machine shop has four mechanics availableto assign jobs for the day. Five jobs are offered with expected profit for each mechanicon each job which are as follows:

Job

Mechanic

A B C D E

1

2

3

4

62 78 50 101 82

71 84 61 73 59

87 64 111 71 81

48 64 87 77 80

Find using the assignment method, the assignment of mechanics to the job thatwill result in a maximum profit. Which job should be declined ?



Unit-3

Solution: The given profit matrix is not a square matrix as the number of jobs is notequal to the number of mechanics. Hence, we introduce a dummy Mechanic 5 withall the elements 0.

Job

Mechanic

1 62 78 50 101 82

2 71 84 61 73 59

3 87 92 111 71 81

4 48 64 87 77 80

5 0 0 0 0 0

A B C D E

Now, we convert this profit matrix into loss matrix by subtracting all theelements from the highest element 111.

Loss Matrix

1 49 33 61 10 29

2 40 27 50 38 52

3 24 19 0 40 30

4 63 47 24 34 31

5 111 111 111 111 111

A B C D E

We subtract the smallest element from all the elements in the respective row.

1 39 23 51 0 19

2 13 0 23 11 25

3 24 19 0 40 30

4 39 23 0 10 7

5 0 0 0 0 0

A B C D E

Since each column has minimum element as zero, we draw the minimumnumber of lines to cover all zeroes.

39 23 51 0 19

13 0 23 11 25

24 19 0 40 30

39 23 0 10 7

0 0 0 0 0

A B C D E

1

2

3

4

5

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Here, the number of lines drawn to cover all zeroes isN = 4, which is less than theorder of the matrix.

The second modified matrix is formed by subtracting the smallest uncoveredelement from the remaining uncovered elements and adding to the element whichis at the point of intersection of lines.

39 30 58 0 19

6 0 23 4 18

17 19 0 33 23

32 23 0 3 0

0 7 7 0 0

A B C D E

1

2

3

4

5

Here, N = 5 = n = 5. Order of matrix.

We make the assignment.

39 30 58 0 19

6 0 23 4 18

17 19 0 33 23

32 23 0 3 0

0 7 7 0 0

A B C D E

1

2

3

4

5

The optimum assignment is

Job MechanicA 5B 2C 3D 1E 4

Since the fifth mechanic is a dummy, Job A assigned to fifth Mechanic isdeclined.

The maximum profit is given by 84 + 111 + 101 + 80 = ` 376

Example 3.22: A marketing manager has 5 salesmen and there are 5 sales districts.Considering the capabilities of the salesmen and the nature of 12 districts, theestimates made by the marketing manager for the sales per month (in 1000 rupees)for each salesmen in each district would be as follows.



Unit-3

Find the assignment of salesmen to the districts that will result in the maximumsales.

Solution: We are given the profit matrix. To maximize the profit, first we convert itinto loss matrix which can be minimized. To convert it in to loss matrix, we subtractall the elements from the highest element 41. Subtract the smallest element from allthe elements in the respective rows and colums, to get the first modified matrix.

Loss Matrix

1 8 2 0 12 0

2 0 16 12 19 4

3 0 14 8 11 4

4 19 3 0 5 5

5 11 7 0 5 1

A B C D E

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We now draw the minimum number of lines to cover all zeroes.

8 0 0 7 0

0 14 12 14 4

0 12 8 6 4

9 1 0 0 5

11 5 0 0 1

A B C D E

1

2

3

4

5

N = 4 < n = 5

We subtract the smallest uncovered element from the remaining uncoveredelements and add to the elements at the point of intersection of lines, to get thesecond modified matrix.

9 0 1 8 0

0 13 12 14 3

0 11 8 6 3

9 0 0 0 4

11 4 0 0 0

A B C D E

1

2

3

4

5

Again, N = 4 < n = 5. Repeat the above step.

12 0 1 8 0

0 10 9 11 0

0 8 5 3 0

12 0 0 0 4

14 4 0 0 0

A B C D E

1

2

3

4

5

N = 5 = n = 5 = Order of matrix. Hence, we make the assignment.

Assignment

12 0 1 8 0

0 10 9 11 0

0 8 5 3 0

12 0 0 0 4

14 4 0 0 0

A B C D E

1

2

3

4

5



Unit-3

Since no row or column has single zero, we get a multiple solution. (i) The optimum assignment is:

1 B, 2 A, 3 E, 4 C, 5 D.With maximum profit (38 + 40 + 37 + 41 +35) =` 191

(ii) The optimum assignment is:1 B, 2 A, 3 E, 4 D, 5 C.Maximum profit (38 + 40 + 37 + 36 + 40) = ` 191


1. What is an unbalanced assignment? How can it be balanced?

2. How is the presence of an alternate optimal solution established?

3. How do you convert the maximization assignment problem into aminimization one?

4. How will you solve an assignment problem if profit is to be maximized?

ACTIVITY

1. Vogel’s Approximation Method results in the most economical initial basicfeasible solution. How?

2. What kind of solution would you get when the net change in value of allunoccupied cells is non-negative?

3. If 4 jobs are to be assigned to 5 machines, how will you proceed to solvethis problem?

3.9 LET US SUM UP

A transportation problem is a subclass of linear programming problem. It issolved in two stages: i) Finding an initial solution and ii) Finding the optimalsolution.

There are three methods to obtain the initial solution. Of these, Vogel’sApproximation Method is the most preferred since the initial basic feasiblesolution obtained by this method is either optimal or very close to the optimalvalue.

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The optimal solution is found using the MODI method.

A given transportation problem is said to be unbalanced if the total supply is notequal to the total demand.

If the number of non-negative independent allocations in a transportation problemis less than m + n – 1, where m is the number of origins and n is the number ofdestinations, it results in degeneracy.

An assignment problem is one of the fundamental combinatorial optimizationproblems and helps to find a maximum weight identical in nature in a weightedbipartite graph.

The solution of an assignment problem can be arrived at using the Hungarianmethod.

An assignment problem is balanced if the cost matrix is a square matrix;otherwise, it is termed as unbalanced.

To convert an unbalanced assignment problem into a balanced problem,dummy rows or columns are added with all entries as 0s.




Kalavathy, S. Operations Research. 2002. New Delhi: Vikas Publishing House Pvt.Ltd.




Check Your Progress–11. Transportation problem deals with transportation of various quantities of a

single homogeneous commodity initially stored at various origins to differentdestinations at the minimum cost.

2. To achieve the objectives of transportation problem, one must know:(i) Amount and location of available supplies, (ii) Quantity demanded atdestination and (iii) Costs of transporting one unit of commodity from variousorigins to various destinations.

3. Transportation problems are presented in the tabular form with m rows (aso r i g i n s ) a n d n columns (as destinations). Every cell of this table has cost as C

ij



Unit-3

of transporting one unit of product from ith origin to jth destination and Xijwhich

is the quantity transported from ith origin to jth destination.

4. If there are m origins and n destinations and Cij be the cost of transporting one

unit of product from ith origin to jth destination and Xijbe the quantity transported

from ith origin to jth destination, and Z is the cost given as: Z = Cij X

ijwhere

1<= i <= n and 1 <= j <= m.

5. Feasible solution: Any set of non-negative allocations (Xij>0) which satisfies

the row and column sum (rim requirement) is called a feasible solution.

Basic feasible solution: A feasible solution is called a basic feasible solutionif the number of non-negative allocations is equal to m + n – 1, where m is thenumber of rows and n the number of columns in a transportation table.

Non-degenerate basic feasible solution: Any feasible solution to atransportation problem containing m origins and n destinations is said to benon-degenerate, if it contains m + n – 1 occupied cells and each allocation isin independent positions.

Check Your Progress–21. The following approaches are used:

(i) North West Corner Rule(ii) Least Cost Method (Matrix Minima)

(iii) Vogel’s Approximation Method2. The basic feasible solution to a transportation problem is said to be optimal if

it minimizes the total transportation cost.

3. The necessary and sufficient condition for the existence of a feasible solutionis a solution that satisfies all conditions of supply and demand.

4. The purpose of MODI method is to get the optimal solution of a transportationproblem.

Check Your Progress–31. An assignment problem is a problem of finding an assignment of jobs among

persons so that total cost of performing all jobs is at minimum.

2. An assignment problem can be stated in the form of n × n matrix [Cij] with i

rows and j columns of real numbers. Here, Cij

is the cost when ith person isassigned jth job.

3. Hungarian method is used to solve assignment problems.

Check Your Progress–41. In Assignment problem the cost matrix of the assignment problem is not a

square matrix, the problem is said to be unbalanced. To make it balanced,extra row(s) or column(s) are added with zero elements.

2. If the final cost matrix contains more than the required number of zeroes atindependent positions, then it indicates the presence of an alternate optimalsolution.

3. The maximization assignment problem can be converted into a minimizationassignment problem by subtracting all the elements in the given profit matrixfrom the highest element in that matrix.

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4. The given profit matrix can be converted into a loss matrix or minimization typeby subtracting all the elements from the highest element of the given matrix. Forthis minimization problem steps of the Hungarian method are applied to get anoptimal assignment.



1. What is an unbalanced transportation problem?

2. How will you convert an unbalanced transportation problem into a balanced one?

3. List the merits and limitations of the North West Corner rule.

4. How will you identify if a transportation problem has an alternate optimal solutionor not?

5. What is the number of non-basic variables in a balanced transportation problem?

6. What are the numbers of non-basic variables for 4 rows and 5 columns?

7. While dealing with North West Corner rule, when does one move to the next cellin next column?

8. What is the coefficient of Xij

of constraints in a transportation problem?

9. What is an initial basic feasible solution?

10. When does degeneracy occur in an m × n transportation problem?

11. What is a cost matrix?

12. How does a cost matrix help solve assignment problems?

13. Write the first three steps of the Hungarian procedure for solving an assignmentproblem.

14. While solving a problem using the Hungarian method, at what stage are youable to make an assignment? Explain in brief.

15. While solving a problem using the Hungarian method, what is the conditionthat indicates an unique optimal solution?

16. What do you mean by maximization in an assignment problem?

17. State the mathematical form of an assignment problem.


1. What do you understand by transportation model?

2. Define feasible solution, basic solution, non-degenerate solution, optimalsolution in a transportation problem.

3. Explain the following briefly with examples:(i) North West Corner Rule

(ii) Least Cost Method(iii) Vogel’s Approximation Method



Unit-3

4. Explain degeneracy in a transportation problem. Describe a method to resolve it.

5. What do you mean by an unbalanced transportation problem? Explain the processof converting an unbalanced transportation problem into a balanced one.

6. Give the mathematical formulation of a transportation problem.

7. Write an algorithm to solve a transportation problem.

8. Obtain the initial solution for the following transportation problem using(i) North West Corner Rule (ii) Least Cost Method (iii) VAM

DestinationA B C Supply

1 2 7 4 52 3 3 1 83 5 4 7 74 1 6 2 14

Sour

ce

Demand 7 9 18 34

9. Solve the following transportation problem where the cell entries denote theunit transportation costs.

5 4 2 6 20

8 3 5 7 30

5 9 4 6 50

10 40 20 30 100

Destination

A B C D Supply

P

Origin Q

R

Demand

10. Solve the following transportation problem.

2 2 3 10

4 1 2 15

1 3 1 40

20 15 30

1 2 3

1

2

3

Destination

Capacity

Source

Demand

11. Find the minimum transportation cost.

19 30 50 10 7

70 30 40 60 9

40 8 70 30 18

5 8 7 14

1 2 3 4

1

2

3

Warehouse

D D D D Supply

F

Factory F

F

Demand

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12. Solve the following transportation problem using Vogel’s Approximation Method.

9 12 9 6 9 10 5

7 3 7 7 5 5 6

6 5 9 11 3 11 2

6 8 11 2 2 10 9

4 4 6 2 4 2

A B C D E F

1

2

3

4

Warehouse

Available

Factory

Requirement


1 2 3 4 6

4 3 2 0 8

0 2 2 1 10

4 6 8 6

1

2

3

Destination

A B C D Supply

Source

Demand


11 20 7 8 50

21 16 20 12 40

8 12 8 9 70

30 25 35 40

1

2

3

Destination

A B C D Supply

Source

Demand

15. Solve the following transportation problem to maximize the profit.

40 25 22 33 100

44 35 30 30 30

38 38 28 30 70

40 20 60 30

1

2

3

Destination

A B C D Supply

Source

Demand

16. Explain the assignment problem using a suitable example.

17. Explain the significance and applications of an assignment problem.

18. Explain the algorithm for solving an assignment problem.

19. How can you maximize an objective function in an assignment problem?

20. Write the steps for finding a solution using the Hungarian method.

21. Solve the following assignment problems:



Unit-3

(i)

1 4 6 3

9 7 10 9

4 5 11 7

8 7 8 5

A B C D

I

II

III

IV

(ii)

10 25 15 20

15 30 5 15

35 20 12 24

17 25 24 20

A B C D

I

II

III

IV

Men

Job

(iii)

1 3 2 8 8

2 4 3 1 5

5 6 3 4 6

3 1 4 2 2

1 5 6 5 4

A B C D E

I

II

III

IV

V

Men

Tasks

22. There are five jobs to be assigned, one each to 5 machines. The associatedcost matrix is as follows.

Jobs

1 2 3 4 5

11 17 8 16 20

9 7 12 6 15

13 16 15 12 16

21 24 17 28 26

14 10 12 11 15

A

B

C

D

E

Machine

How should the job be assigned to various machines?

23. A company is faced with the problem of assigning 4 machines to 6 differentjobs (one machine to one job only). The profits are estimated as follows.

Job

1 3 6 2 6

2 7 1 4 4

3 3 8 5 8

4 6 4 3 7

5 5 2 4 3

6 5 7 6 4

A B C D

Machine

Solve the problem to maximize the profit.

24. Determine the optimum assignment schedule for the following assignmentproblem. The cost matrix is as follows.

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Machine

1 2 3 4 5 6

11 17 8 16 20 15

9 7 12 6 15 13

13 16 15 12 16 8

21 24 17 28 2 15

14 10 12 11 15 6

A

B

C

D

E

If job C cannot be assigned to machine 6, will the optimum solution change?

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UNIT 4 GAME THEORY

UNIT STRUCTURE4.1 Learning Objectives4.2 Introduction4.3 Game Theory: Basic Concepts

4.3.1 Types of Games4.3.2 Value of a Game

4.4 Pure and Mixed Strategies with Saddle Point4.4.1 Method of Dominance4.4.2 Linear Programming Solution to Two-Person Zero-Sum Games

4.5 Limitations of Game Theory4.6 Let Us Sum Up4.7 Further Reading4.8 Answers to ‘Check Your Progress’4.9 Possible Questions



Explain the concept of game theory

Discuss two-person zero-sum games and the value of a game

Deal with pure and mixed strategies with saddle point

Reduce the size of a game using dominance

Solve a game using the linear programming method

4.2 INTRODUCTION

Competition is a watchword of modern life. We say that a competitive situationexists if two or more individuals make decisions in a situation that involvesconflicting interests and in which the outcome is controlled by the decisions of allthe parties concerned.We assume that in a competitive situation, each participantacts in a rational manner and tries to resolve the conflict of interests in his favour. Itis in this context that the game theory has been developed. Professor John vonNeumann and Oscar Morgenstern, in their book The Theory of Games and EconomicBehaviour, provided a new approach to many problems involving conflictsituations—an approach now widely used in economics, business administration,sociology, psychology and political science, as well as in military training.Fundamentally, the theory of games attempts to provide an answer to the question:what may be considered a rational course of action for an individual confrontedwith a situation whose outcome depends not only upon his own actions, but alsoupon the actions of others who, in turn, are faced with a similar problem of choosinga rational course of action? In fact, game theory is simply the logic of rationaldecisions.

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In this unit, you will look into the basic concepts of game theory. You will alsolearn about pure and mixed strategies, the use of dominance and linear programming.

4.3 GAME THEORY: BASIC CONCEPTS

The term ‘game’ represents a conflict between two or more parties. Game theory isreally the ‘science of conflict’. It is not concerned with finding an optimum orwinning strategy for a particular conflict situation, but it provides general rulesconcerning the logic that underlies strategic behaviour of all types.

Game theory applies to those competitive situations that are technically knownas ‘competitive games’ or simply ‘games’. Situations, in order to be termed games,must possess the following properties:

(i) The number of competitors is finite.

(ii) There is a conflict of interests between the participants.

(iii) Each of the participants has available to him a finite list of possiblecourses of action, i.e., several choices of appropriate actions; this listbeing not necessarily the same for each competitor.

(iv) The rules governing these choices are specified and known to all theplayers; a play of the game results when each of the players chooses asingle course of action from the list of courses available to him.

(v) The outcome of the game is affected by the choices made by all theplayers; the choices are made simultaneously so that no competitorknows his opponent’s choice until he is already committed to his own.

(vi) The outcome for all the specific sets of choices by all the players isknown in advance and numerically defined. The outcome of a playconsists of the particular set of courses of action undertaken by thecompetitors. Each outcome determines a set of payments (positive,negative or zero), one to each competitor.

Illustration of a Game

When a competitive situation meets all the above stated criteria, we can call it agame. This can be made clear by an example of a simple game. Suppose there aretwo opponents X and Y. We can think of them as sitting across a table from eachother and each with two buttons in front of him. We shall denote player X’s buttons, m and n and player Y’s buttons as r and t; thus each player has two choices open tohim. We also presume a partition between them so that neither can see in advancewhich button his opponent is going to press. At a signal from a third party, eachplayer presses one of his buttons. The results of each of the possible four combinationsis known in advance to both of the players; the uncertainty inherent in the gamearises from the fact that neither player knows what button his opponent will pressnext. Every time the third party signals, each player presses one of his buttons andthe game thus continues. At the end of, say, a hundred ‘plays’, the game is over andthe points won by each of the players are totalled and the winner is determined. It isassumed that both players are of equal intelligence and that each actively attempts towin the game.

Game: Represents aconflict between two ormore parties.

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This sort of simple game can be illustrated in tabular form as follows:

A Simple Game in Tabular FormPlayer Y

Button r Button t

Button X wins X winsm 2 points 3 points

Player XButton Y wins Y wins

n 3 points 1 point

In the above table, the plays open to each of the opponents and the resultinggains or losses (called pay-offs) have been shown. This game is biased againstPlayer Y because if Player X presses button m for each play of the game, Player Ycannot win; in fact, Player Y faces a choice between losing 2 points on each play ifhe responds by pressing Button r or losing 3 points on each play if he responds bypressing Button t. Player Y will respond each time by pressing Button r since thisrepresents his least loss alternative. Thus, Player X will win 2 points on each play ofthe game. We can similarly present a game biased against Player X wherein X hasno chances winning and at best an minimize his losses like Player Y in the abovestated case.

But all simple games cannot be said to have such simple solutions. This canbe illustrated by the following example of a game:

A Simple Game in Tabular FormPlayer Y

Button r Button t

Button Y wins X winsm 4 points 3 points

Player XButton X wins Y wins

n 1 point 2 points

In the above case, it is not as simple as it was in the earlier example to determinethe individual player’s strategies. We can see from the game that Player X wouldnot press his button m, hoping to win 3 points on each play simply because Player Ycould counter with button r and win 4 points himself. Similarly, Player Y would notpress button hoping to win 4 points on every play because Player X could counterwith his button n and win 1 point himself. In such a situation, therefore, it isadvantageous for the players to play each of their choices (buttons) a part of thetime only. How to calculate the proportion of time to allot to each choice shall bediscussed a little later.

Standard Conventions in Game Theory

In order to eliminate the necessity for written descriptions of the ‘pay-offs’ (as wehave shown in the above two examples), a standard set of conventions has beenestablished in game theory. It is the usual practice to omit a description like ‘Player Xwins two points’ and replace it with integer 2. The positive algebraic sign which isassumed to accompany this number indicates that it is Player X who benefits from this

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pay-off. Similarly, instead of saying ‘Player Y wins three points’ one simply indicates thiswith the value –3, the minus sign indicating that it is Player Y who benefits from thisparticular pay-off.

Another standard convention that is usually followed is that PlayerX has choicesbetween the rows and Player Y has choices between the columns.

Keeping these two conventions in view we can write the above stated twoillustrations of games as follows:

Illustration 1 Illustration 2

Player Y Player Y

2 3 –4 3Player X Player X

–3 –1 1 –2

A further convention in game theory is known as matrix notation. In thiscase, games are represented in the form of a matrix (a rectangular array of numbersderiving from matrix algebra). When games are expressed in this fashion, the resultingmatrix is commonly known as a pay-off matrix. The above stated two illustrationscan be put in the form of pay-off matrices as follows:

Illustration 1 Illustration 2

Player Y Player Y

Player X2 3

3 1

Player X4 3

1 2

The term strategy is often talked about in game theory. It refers to the totalpattern of choices employed by any player. It may be defined as a complete set ofplans of action specifying precisely what the player will do under every possiblefuture contingency that might occur during the play of the game. For example, ifPlayer X chooses to play his first row half of the time and his second row half of thetime, his strategy for the game is 1/2, 1/2. Thus, the strategy of a player is thedecision rule he uses for making the choice from his list of courses of action. Thestrategy could be a pure or mixed one. When a player plays one row all the time (orone column all the time) he is said to be adopting a pure strategy. In a mixed strategy,Player X will play each of his rows a certain part of the time and Player Y will playeach of his columns a certain part of the time. In business, a close analogy is when,for example, a manager follows a certain course of action A until an alternate courseof action B appears to be more profitable. Later on, should Action A appear moreattractive again, the manager switches back to it.

4.3.1 Types of Games

Games can be of several types. The important ones are as follows:

1. Two-person games and n-person games. In two-person games, the playersmay have many possible choices open to them for each play of the game butthe number of players remains only two. But games can also involve manypeople as active participants, each with his own set of choices for each play

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of the game. A game of three players can be named as three-person game. Thus,in case of more than two persons, the game is generally named an n persongame.

2. Zero-sum and non-zero-sum games. A zero-sum game is one in which the sumof the payments to all the competitors is zero for every possible outcome of thegame. In other words, in such a game the sum of the points won equals the sumof the points lost, i.e., one player wins at the expense of the other (others). A two-person matrix game is always a zero-sum game since one player loses what theother wins. In a non-zero-sum game, the sum of the pay-offs from any play of thegame may be either positive or negative but not zero. A well-known example of anon zero-sum game is the case of two competing firms, each with a choiceregarding its advertising campaign. We may assume that the market is quite smallso that sales remains at nearly the same level regardless of whether the firmsengage in advertising or not. If neither firm advertises, they both save the cost ofadvertising and thus each receives a positive benefit or pay-off (representing anon-zero-sum outcome). In the event both firms spend equal amounts of moneyon advertisement, neither firms’ sales will increase (given the equal quality of theadvertising campaigns) and thus both firms will have incurred the cost of advertisingwithout any benefits. Since both firms will lose money, this is clearly a non-zero-sum outcome.

3. Games of perfect information and games of imperfect information. Whateverstrategy is adopted by either player, if the same can also be discovered by hiscompetitor, then such games are known as games of perfect information. Ingames of imperfect information, neither player knows the entire situationand must be guided in part by guesswork as to what the real situation is.

4. Games with finite, i.e., limited number of moves (or plays) and games withunlimited number of moves. Games with finite number of moves are thosewhere the number of moves is limited to a fixed magnitude before play begins.If the game could continue over an extended period of time and no limit isput on the number of moves, it is referred to as a game with an unlimitednumber of moves.

5. Constant-sum games. In some situations, a zero-sum game is correctly referredto as a constant-sum game. Take for example, a competitive struggle for marketshare by two firms (a duopoly). In such a case, every percentage point gainedby one of the firms is necessarily lost by the other. This situation is ordinarilycalled a two-person, zero-sum game. But strictly speaking, this is a constant-sum rather than a zero-sum game because the sum of the two market shares isthe fixed number 100 (per cent), not zero. It should, however, be rememberedthat ‘there is no significant analytic difference between the constant-sum andthe zero-sum games. This is because there is no change in strategic possibilitiesfrom a given game to another game in which some constant amount that cannotbe changed by players is added to the original pay-offs.’

4. 2 × 2 two-person games and 2 × m and m × 2 games. Two person zero-sumgames with only two choices open to each player are denoted as 2 × 2 twoperson games, but games in which one of the players has more than two

Zero-sum and non-zero-sum games: Thesum of the payments toall the competitors iszero for every possibleoutcome of the game.

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choices of rows or columns and in which the other player has exactly two choicesare referred to as m × 2 or 2 × m games respectively.

7. 3 × 3 and larger games. Two-person zero-sum games can be of size 3 × 3and larger. If each of the two players has three choices then the game is of3 × 3 type and if the choices open to any of the player or to both the playersare more than three, the game is referred to as of a larger size.3 × 3 and larger games quite often present complications and in suchsituations, linear programming as a solution method may allow us to findthe optimum strategies.

8. Negotiable (or cooperative) and non-negotiable (or non-cooperative) games.Negotiation amongst players is possible in n-person and non zero-sum gamesbut the same is not necessarily required. On this basis, we can divide theanalysis of such games into two parts—games in which the participants cannegotiate and games in which negotiation is not permitted. The former typeof games are known as negotiable games and the latter are known as non-negotiable games.

4.3.2 Value of a Game

In game theory, the concept ‘value of a game’ is considered very important. It refersto the average pay-off per play of the game over an extended period of time. Thiscan be explained by an example. Suppose the two games are as follows:

Player Y Player Y

Player X3 4

6 2

Player X7 2

3 1

Game one Game two

(with a positive value) (with a negative value)

In game one, Player X would play his first row on each play of the game andY would respond by playing his first column each time in order to minimize hislosses. Since Player X wins three points on each play of the game, his averagewinnings per play will also be 3 as long as the game is played. The value of the gameis thus 3 with an implicit positive algebraic sign which denotes that Player X wins thegame. But in game two, Player Y plays his first column on each play of the game andPlayer X responds by playing his second row on each play of the game in order tominimize losses. As a result, Y wins 3 points on each play of the game. Since theaverage pay-off per play is –3, the value of this game is –3, the minus sign indicatingthat Y is the winner.

But determining the value of the game is not always as simple as in the case ofthese two examples. In case the players determine that their best alternative is to playeach row or each column a certain part of the time, calculating the value of the gamebecomes a bit more complex.

Now we shall see the process of determination of optimum strategies and thevalue of a game with the help of some illustrations.

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Example 4.1: Determine the optimum strategies for the two players X and Y and findthe value of the game from the following pay-off matrix:

Player Y

Player X

3 1 4 2

1 3 7 0

4 6 2 9

Solution: For determining the optimum strategies, the cautious approach is to assumethe worst and act accordingly. If Player X plays with first row strategy, then PlayerY will play with second column the win one point, otherwise he will lose 3, 4 and 2if he plays with column, 13 and 4 respectively. If Player X plays with second rowstrategy, then the worst would happen to him only when Player Y plays with thethird column because in that case Y would win 7 points. If Player X plays with thirdrow strategy, then the worst he can expect is losing 9 points when Y plays with thefourth column. In this problem then, Player X should adopt first row strategy becauseonly then his loss will be minimum. Thus, Player X can make the best of the situationby aiming at the highest of these minimal pay-offs. This decision rule is known as‘maximin strategy’.

Looking from the perspective of Player Y, we can say that if Y plays withcolumn first strategy the maximum he can lose is 4 points if Player X adopts thestrategy of row three. If Player Y plays with column 2 strategy, there is no question ofany loss whatever may be the strategy of Player X. In such a case, player X will adoptthe strategy of row 1, for only then his loss will be minimum. If Player Y plays withcolumn 3 strategy, the maximum he can lose is 4 points if X adopts the strategy ofrow 1. If Y adopts the strategy of column 4, he can lose at the most 2 points if X adoptsthe strategy of row 1. In this problem then, Y should adopt the column 2 strategy andthereby ensure a victory of 1 point which is the maximum in the given case. Thus,Player Y can make the best of the situation by aiming at the lowest of these maximumpay-offs (viz. 4, –1, 4, 2). Thus, he should seek the minimum among the maximumpay-offs. This decision rule is known as ‘minimax strategy’.

Thus, Player Y will play his second column on each play and Player X willrespond by playing his first row on each play. In this way, Y will win 1 point and Xwill lose 1 point in each play. Hence, this is a two-person zero-sum game with apure strategy. Since Y will win 1 point and X will lose 1 point in each play, the valueof the game is –1. This pay-off –1 is then a saddle point in the given game and canbe marked (encircled) as follows:

Player Y

Player X

3 1 4 2

1 3 7 0

4 6 2 9

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4.4 PURE AND MIXED STRATEGIES WITHSADDLE POINT

The saddle point in a pay-off matrix is the smallest value in its row and the largestvalue in its column. The saddle point is also known as the equilibrium point in the theoryof games. An element of a matrix that is simultaneously the minimum of the row inwhich it occurs and the maximum of the column in which it occurs is a saddle point ofthe matrix game. In a game having a saddle point, the optimum strategy for Player X isalways to play the row containing a saddle point and for PlayerY to play the column thatcontains a saddle point. The saddle point also gives the value of such a game. The saddlepoint in the pay-off matrix of a game may or may not exist. If there is a saddle point, wecan easily find out the optimum strategies and the value of the game by what is knownas solution by saddle point without having to do too many calculations. But when thesaddle point is not there, then we have to use algebraic methods for working out thesolutions of the game problems.

Game Problems of Mixed Strategy

Example 4.2: Find the optimum strategies and the value of the game from the followingpay-off matrix concerning a two-person game:

Player Y

Player X1 4

5 3

Solution: In the given game, there is no saddle point because there is no one valuewhich is smallest in its row and largest in its column. Therefore, the players have toresort to mixed strategy, i.e., Player X will play each of his rows a certain part oftime and Player Y will play each of his columns a certain part of the time. Thequestion then is to determine the proportion of time a player should spend on hisrespective rows and columns. This can be done by the use of the following algebraicmethod:

Let Q equal the proportion of time Player X spends playing the first row, then1 – Q must equal the time he spends playing his second row (because one equals thetime available for play). Similarly, suppose Player Y spends time R in playing thefirst column and 1 – R be the time he spends on playing the second column. All thesecan be stated as follows:

Player Y

Player X1 4

1 5 3

Q

Q

Now, we must find the values of Q and R. Let us analyse the situation from X’sviewpoint. He would like to devise a strategy that would maximize his winning(or minimize his losses) irrespective of what his opponent Y does. For this, X wouldlike to divide his play between his rows in such a manner that his expected winnings

Saddle point: Thesmallest value in itsrow and the largestvalue in its column.

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or losses, when Y plays the first column equal to his expected winnings or losses whenYplays the second column. (Expected winnings indicate the sum, overtime, of the pay-offs that will be obtained multiplied by the probabilities that these pay-offs will obtain). Inour case, we can calculate the same as follows:

X’s Expected WinningsWhen Y plays When Y playscolumn one column two

X plays row one: (1) (Q) (4) (Q)

Q of the time

X plays row two: 5(1– Q) 3(1 – Q)

1 – Q of the time

X’s total expected winnings Q + 5(1 – Q) 4Q + 3 (1 – Q)

Equating the expected winnings of X when Y plays column 1 with when Y playscolumn 2, we can find the value of Q as follows:

Q + 5(1 – Q) = 4Q + 3(1 – Q)

Q + 5 – 5Q = 42 + 3 – 3Q

–5Q = –2or Q = 2/5

1 – Q = 3/5

This means that Player X should play his first row 2/5 of the time and his secondrow 3/5 of the time if he wants to maximize his expected winnings from the game.

Similarly, the losses of Y can be worked out as follows:

Y’s Expected Losses

Equating the expected losses of Y when X plays row 1 with when X playsrow 2, we can find the value of R as follows:

R + 4(l – R) = 5R + 3(1 – R)

R + 4 – 4R = 5R + 3 – 3R

or –5R = –1or R = 1/5

1 – R = 4/5

This means that Player Y should play his first column 1/5 of the time and hissecond column 4/5 of the time if he wants to minimize his expected losses in thegame.

Now we can illustrate the original game with the appropriate strategies foreach of the player as follows:

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Player Y

2 / 5

3/ 5

1 4Player

5 3X

An alternative method (or the short-cut method) for finding the abovestrategies is as follows:

Original game

Y

1 4

5 3X

Step 1: Subtract the smaller pay-off in each row from the larger one and the smallerpay-off in each column from the larger one.

Y

1 4 3

5 3 2X

(i.e., 4 – 1 = 3)

(i.e., 5 – 3 = 2)(i.e., 5 – 1 = 4) (i.e., 4 – 3 = 1)

Step 2: Interchange each of these pairs of subtracted numbers found in Step 1 above.

Y

1 4 2

5 3 3

1 4

X

Step 3: Put each of the interchanged numbers over the sum of the pair of the numbers.

Y

1 4 2 /(2 3)

5 3 3/(2 3)

1/(1 4) 4 /(1 4)

X

Step 4: Simplify the fraction to obtain the proper proportions or the required strategies.

Y

1 4 2 /5

5 3 3/5

1/5 4 /5

X

Now we determine the value of the game. Looking at the game from X’s pointof view we can argue that:

(i) During the 1/5 of the time that Y plays column 1, X wins 1 point 2/5 of thetime (when X plays row 1) and 5 points 3/5 of the time (when X playsrow 2).

(ii) During the 4/5 of the time that Y plays column 2, X wins 4 points 2/5 ofthe time (when X plays row 1) and 3 points 3/5 of the time (when X playsrow 2).

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Thus, the total expected winnings of player X is the sum of the above twostatements, that is,

1 2 3 4 2 3(1) (5) (4) (3)

5 5 5 5 5 5

1 17 4 17

5 5 5 5

17 8

25 2585

2517

5

Thus, the value of the game is 17/5 which means that Player X can expect to winan average pay-off of 17/5 points for each play of the game if he adopts the strategy wehave determined above. If the value of the game determined above had a negative sign,it would simply signify that Y was the winner. The same result can also be achieved bylooking at the game from Y’s point of view and carrying out similar calculations.First Alternative Method (or Short-cut Method) for Determining the Value of aGame

Under this, we calculate:X’s expectations when Y plays column 1X’s expectations when Y plays column 2Y’s expectations when X plays row 1Y’s expectations when X plays row 2

In all these four cases, the answer remains the same. Hence, any one of thesecalculations is sufficient to determine the value of the game. The logic behind thisapproach is the same as that of determining the optimum strategies. (In the case ofPlayer X, we determined a strategy that guaranteed X the same winnings, irrespectiveof his opponent’s choice of columns). This can be illustrated in Figure 4.1 as follows:

(1× ) + (5× ) =25

35

Player X

175

25

3

1

5

15

45 2

5

Player 15

45

175

175

175

Fig. 4.1 Diagrammatic Form of Determining the Value of a Game

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Second Alternative Method for Determining the Value of a Game Using theProbabilities of each Pay-off

In a simple 2 × 2 game without a saddle point, each player’s strategy consists of twoprobabilities denoting the portion of the time he spends on each of his rows orcolumns. Since each player plays a random pattern, we can list the probabilities ofeach pay-off (in our given question) as follows:

Pay-off Strategies which produce Probability of thisthis pay-off pay-off

1 Row 1 column 12 1 2

5 5 25


5 5 25


5 5 25


5 5 25

Sum = 1.0

The value of the game can be found out by taking the sum of the products ofeach of these pay-offs and their respective probabilities.

For the given problem, this can be worked out as follows:

Pay-off Probability of the Product of the first twopay-off columns

12

25

2

25

48

25

32

25

53

25

15

25

312

25

36

25

Total =85

25

=17

5 (Value of the game)

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4.4.1 Method of Dominance

The concept of dominance is very useful for reducing the size of the game. Applyingthis concept we can, at times, convert a bigger game into a smaller game. Hence,we should examine the possibility of reducing the game size by the principle ofdominance.

Rules for Dominance

(a) If all the elements in a column are greater than or equal to the correspondingelements in another column, then that column is dominated.

(b) Similarly, if all the elements in a row are less than or equal to the correspondingelements in another row, then that row is dominated.

Dominated rows or columns may be deleted, which reduces the size of thegame. Always look for dominance when solving a game.

This becomes clearer from the following illustration:

Example 4.3: Determine the optimum strategies and the value of the game fromthe following 2×m pay-off matrix game for X:

Player Y

6 3 1 0 3Player

3 2 4 2 1X

Solution: There is no saddle point in the above game. Hence, mixed strategies willhave to be adopted by the two players. Further, if we look at the given pay-offmatrix from Y’s point of view, we find that he chooses not to play columns 1, 2 or 4,since either column 3 or column 5 (both with two negative pay-offs) offers a betteralternative irrespective of the actions of Player X. We can say then that columns 1,2 and 4 are dominated by the remaining two columns, viz. number 3 and 5, andhence, would never be played by Y. As soon as Y decides never to play his first,second and fourth columns, the game is reduced to size 2 × 2 as stated below:

Player Y

1 3Player

4 1X

Now, the optimum strategies and the value of the game can be easily foundout as per the method stated in Example 4.2. The calculations can be shown asfollows:

Games as stated above.

Y

1 3

4 1X

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Step 1: Substracting the pay-offs

Y

1 3 2 (i.e., 1 ( 3) 2)

4 1 3 (i.e., 1 ( 4) 3)

3 2

X

(i.e., –1 – (–4) = 3 (i.e., –1 – (–3) = 2)

Step 2: Pairs interchanged

Y

1 3 2

4 1 3

2 3

X

Step 3: Pairs over the sum

1 3 3 / (3 2)

4 1 2 / (3 2)X

2 / (2 + 3) 3 / (3 + 2)

Step 4: Fractions simplifed and optimum strategies determined

2/5 3/5

3/5 1 3

2/5 4 1

Y

X

Value of the Game

3 21 4

5 5

3 8 11

5 5 5

Example 4.3 is an example of solving the problem by what is known as the methodof dominance.

Graphical method. The graphical method can only be used in games with nosaddle point and having pay-off m × n matrices where either m or n is two. Thegraphical method enables us to substitute a much simpler 2 × 2 matrix for the originalm × 2 or 2 × n matrix. We will illustrate this in Example 4.4.

We will apply the graphical short-cut by plotting on two different vertical axis,the two pay-offs corresponding to each of the five columns. The pay-off numbers inthe first row are plotted on axis 1 and those in the second row on axis 2, which

Game Theory


Unit-4

should be drawn at some distance away from the first axis but should be parallel to thefirst axis as shown in the following figure:

A —6543210

–1

— — — — — — — —

–2 —–3K —

–4 —–5 —–6 —

————————— –2— –3— –4— –5— –6

6543210

–1

Axis 1 Axis 2

B

T

L

Thus, the two pay-off numbers 6 and 3 in the first column are denoted respectivelyby point A on axis 1 and point B on axis 2. Line AB then denotes Y’s move of the firstcolumn. By plotting the pay-off numbers of each of the remaining four columns on thetwo axes, we obtain five lines like the lineAB, which correspond to the given five movesof Y.

If using a thick line we draw the segments which bound the figure from thebottom, namely the segments AT and LT and mark the highest point (7) on thisboundary, the two lines passing through it identify the two critical moves of Y,which combined with the two of X, yield the following 2×2 matrix:

1 3

4 1X

The optimal strategies now can be determined in the way explained earlier.

Example 4.4: Determine the optimum strategies and the value of the game fromthe following pay-off matrix concerning a 2 person 4×2 game.

6 2

3 4

2 9

7 1

Y

X

Solution: There is no saddle point in the given game. This game also cannot bereduced by dominance (as we did in Example 4.3) because there is no row which isalways preferred by X irrespective of what Y might do.

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Player X can actually think of this 4 × 2 game as six sub-games, each of size 2 × 2(because he can choose not to play any two of his rows if he so desires). These sixsub-games can be listed as follows:

325 5

15

45

6 2- . ( )

3 4

Y

X Sub game No i

Strategies have been noted in this matrix pay-off.

The value of the game is6 12

5 5

=

18

5 = –3.60

7 815 15

1115

415

6 2- . ( )

2 9

Y

X Sub game No ii

Strategies have been noted in this matrix pay-off.


6 215 15

=66 8 58

3.8715 15 15

6 2- . ( )

7 1

Y

X Sub game No iii

There is a saddle point here. Hence, the value of this sub-game is –6.

3 4- . ( )

2 9

Y

Sub game No iv

There is saddle point here. Hence value of this sub-game is –4. Y

3 47 7

67

17

3 4- . ( )

7 1X Sub game No v

Strategies have been noted in the matrix pay-off.


7 7 .

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Unit-4

=25

7 = –3.57

Y

8 97 7

617

1117

2 9- . ( )

7 1X Sub game No vi

Strategies have been noted in the matrix pay-off.

The value of the game is,

12 77 653.82

17 17 17

If we look at the values of all these sub-games, we find that the values beingnegative, Y will win and X will lose in all the cases. Player X, who is making thechoice, will naturally prefer the sub-game with the smallest negative value and thisis sub-game No. (v) in our example. Thus, X will play a two-row mixed strategybetween the second and fourth rows of the original game and he will expect to losean average of 357 points per play of the game, which will be his minimum possibleloss. Player Y will also adopt a strategy consisting of the same number of columns.In brief, sub-game No.(v)’s strategies will be adopted by players X and Y with agame value equal to –3.57 in the case of the given example.

Solution through Graphic Method

The above example can also be easily worked out with the help of the graphicalmethod as follows.

The case of pay-off matrices having only two columns but more than two rowsis similar, except that in the diagram, we thicken the line segments which bound thefigure from the top and take the lowest point on this boundary. The following diagramshows the pay-off numbers from each row represented as points on two vertical axes,1 and 2.

Axis 1 Axis 2

5

K0

–5

–10

–5

–10

0K

L

5

P

MB1

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Thus, line B joins the first pay-off number –6 and –2 of the first row. Similarly,other lines have been drawn representing pay-off numbers in the second, third andfourth rows. The segments KP, PM and ML drawn in thick lines bound the figurefrom the top and their lowest intersection M, through which two lines pass, definingthe following 2 × 2 matrix relevant for our purpose:

3 4

1 1

Y

X

The optimal strategies can now be determined as usual.

4.4.2 Linear Programming Solution to Two-PersonZero-Sum Games

If there is no saddle point in a game and reduction of the game by dominance is alsonot possible, then the linear programming technique offers an efficient solution.Hence, we shall work out the following problem by the linear programming method:

Example 4.5: Determine the optimum strategies and the value of the game for thefollowing pay-off matrix concerning a 3 × 3 game.

Player

1 2 1

Player 2 1 1

2 0 1

Y

X

Solution: Let us designate X1, X

2and X

3 to represent the proportion of the time

Player X should spend playing his respective rows. Similarly, designate Y1,Y

2and Y

3

to represent the proportion of the time Player Y should spend playing his respectivecolumns. We can show all these as follows:

1 2 3

1

2

3

Player

1 2 1

Player 2 1 1

2 0 1

Y

Y Y Y

X

X X

X

Our linear programming technique must give an answer to these six unknowns,viz. X

1, X

2, X

3, Y

1, Y

2 and Y

3.

Starting first with the determination of Player Y’s optimum strategies, hisexpectations can be written as follows:

If X adopts strategy X1, Y’s expectation is:

1Y1+2Y

2–1Y

3 V


–2Y1+1Y

2+1Y

3 V

Game Theory


Unit-4


2Y1+0Y

2+1Y

3 V

where V = the value of the game

The V term on the right-hand side of the above inequalities can be removed bydividing all the terms by V as shown below:

31 2 11 2 YY Y

V V V

31 2 12 1 YY Y

V V V

31 2 12 0 YY Y

V V V

If we take Y =Y

V, then the above stated inequalities can be put as:

1 2 3

1 2 3

1 2 3

2

2

2 0

Y Y Y

Y Y Y

Y Y Y

We know that the value of Y1+ Y

2+ Y

3 must be equal to one (because the total

of one’s strategies must be equal to unity), which is an important constraint underwhich we must solve the game. To put this constraint into the Y form, we firstdivide all its terms by V as follows:

31 2 1YY Y

V V V V

and then, usingY

YV

we get,

1 2 3

1Y Y Y

V . Thus, the various constraints now are,

1 2 3

1Y Y Y

V ...(i)

1 2 32 1Y Y Y ...(ii)

1 2 3 1Y Y Y ...(iii)

1 2 32 0 1Y Y Y ...(iv)

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Y’s objective in any game is to minimize the value of V and this is identical to

maximizing 1/V. Thus, the player’s objective is to maximize the value of 1 2 3Y Y Y ,

subject to three linear constraints [number (ii), (iii) and (iv)] stated above. All thesecan be put in a formal way as follows for a linear programming solution:

Maximize:

1 2 3Y Y Y

Subject to,

1 2 32 1Y Y Y

1 2 3 1Y Y Y

1 2 32 0 1Y Y Y

We can work out this problem by the simplex method. Since the inequalitiesare of less than type, we need to add a slack variable to each of the inequalities forobtaining the equations for the simplex method. Taking S

1, S

2 and S

3 as the slack

variables, we can state the problem as follows:

Maximize:

1 2 3 1 2 30 0 0Y Y Y S S S

Subject to,

1 2 32 1Y Y Y

1 2 3 1Y Y Y

1 2 32 0 1Y Y Y

From the above, the first simplex tableau can be set up as follows:

Cj

1 1 1 0 0 0 Ratio

Basic Value ofVariable the Basic 1Y 2Y 3Y 1S 2S 3S

X1: Variable

0 S1

1 1 2 –1 1 0 1 1

0 S2

1 –2 1 1 0 1 0 –1/2

0 S3

2 2 0 1 0 0 1 1/2 }Key Row

Zj

0 0 0 0 0 0 0Index Row

Cj – Z

j+1 +1 +1 0 0 0

Key Number Key Column

Since there is a tie in Zi. C

i row, let 1Y be the first entering variable.

Game Theory


Unit-4

The second simplex tableau is as follows:

Cj

1 1 1 0 0 0 Ratio


Variable

0 S1

1/2 0 2 –3/2 1 0 -1/2 1

0 S2

2 0 1 2 0 1 0 –1/2

1 1Y 1/2 1 0 1/2 0 0 1/2 }Key Row

Zj

1/2 1 0 1/2 0 0 1/2Index Row

Cj – Z

j0 –1 –1/2 0 0 -1/2

Key Number Key Column (i.e., 2Y enters next)

The third simplex tableau is as follows:

Cj

1 1 1 0 0 0 Ratio


Variable

12Y 1/4 0 1 -3/4 1/2 0 -1/4 -1/3

0 S2

7/4 0 0 11/4 -1/2 1 5/4 7/11

1 1Y 1/2 1 0 1/2 0 0 1/2 }Key Row

Zj

3/4 1 1 -1/4 1/2 0 1/4Index Row

Cj – Z

j0 0 –5/4 1/2 0 -1/4

Key Number Key Column (i.e., 3Y enters next)

The fourth simplex tableau is as follows:

Cj

1 1 1 0 0 0 Ratio


Variable

12Y 8/11 0 1 0 4/11 3/11 1/11

03Y 7/11 0 0 1 -2/11 4/11 5/11

1 1Y 2/11 1 0 0 1/11 –2/11 3/11

Zj

17/11 1 1 1 3/11 5/11 9/11

Cj – Z

j0 0 0 -3/11–5/11 -9/11

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Since there is no positive value in the Cj– Z

j row, this indicates that we have

reached the optimum solution.

The fourth of the final simplex tableau worked out as above shows that thevalue of the objective function is 17/11 and this was taken as 1/V in our problem.Hence, the value of the game V = 11/21.

On the basis of this value, we can now convert our Y terms into Y terms asfollows:

11 1 1 1

2 11 2or . or

11 17 17

YY Y Y V Y

V

22 2 2 2

8 11 8or . or

11 17 17

YY Y Y V Y

V

33 3 3 3

7 11 7or . or

11 17 17

YY Y Y V Y

V

Hence, the optimum strategies for Player Y are to play 2/17, 8/17 and 7/17 thefirst, second and third columns respectively.

The optimum strategies for X can also be found out from the final simplextableau prepared above. They appear as entries in the C

j– Z

j row under the slack

variable columns, viz. –3/11, –5/11 and –9/11. Like Y, these values are also X valuesand must be reduced as follows to obtain the optimum strategies of Player X.

1 1

3 11 3. or

11 17 17X X V

2 2

5 11 5. or

11 17 17X X V

3 3

9 11 9. or

11 17 17X X V

Hence, the optimum strategies for Player X are to play 3/17, 5/17, and 9/17the first, second and third rows respectively.

Short-cut Method

A short-cut method can also give the solution to the above problem which we havejust solved through the linear programming technique. But this short-cut methodapplies only to games having square pay-off matrices, i.e., matrices having as manycolumns as the number of rows and having a saddle point, i.e., pure strategy is notavailable to any player.

We take the same illustration to explain this short-cut method:

Player Y

Player X

1 2 1

2 1 1

2 0 1

Game Theory


Unit-4

Since the above is a 3 × 3 matrix, i.e., a square matrix and there is no saddle point,we can use the short-cut technique.

If the probabilities of three Y-moves which Y must mix for optimal play are p, q,r then the three pay-offs to Y corresponding to each of the three counter moves of hisopponent X must all be equal to the optimal value V of the game. Since Y’s pay-offsagainst the three X-moves are respectively,

1 2 1

29 1 1

2 0 1

p q r

q r

p q r

we obtain three equations by equating each of these three Y- weighted pay-offswith V, that is,

(1p + 2q – 1r) = (–2p + 1q + 1r)

= (2p + 0q + 1r) = V

or 3p + 1q – 2r = 0

and –p + 2q – 2r = 0

Hence,2 8 7

p q r

Consequently, Y’s mixture of his three moves, viz. first, second and third columnmoves for optimal play must be in the proportion 2:8:7 (i.e., 2/17, 8/17,7/17).

Similarly, denoting the probabilities used by X for blending his three moves bythe symbols p, q, r, we find that X’s three average pay-offs against Y’s three movesare as follows:

1 ' 2 ' 2 '

2 ' 1 ' 0 '

1 ' 1 ' 1 '

p q r

p q r

p q r

The equality of the three pay-offs to the optimal value of the game leads to thethree equations:

(1p – 2q + 2r) = (2p + 1q + 0r)

= (–p + 1q + 1r) = V

or –p – 3q + 2r = 0

and 2p – 3q + 1r = 0

Hence,' ' '

3 5 9

p q r

Thus for optimal play, X must blend his three moves viz., first, second andthird row moves in the proportion 3:5:9 (i.e., 3/17, 5/17, 9/17).

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4.5 LIMITATIONS OF GAME THEORY

Game theory has been used in business and industry to develop bidding tactics, pricingpolicies, advertising strategies and timing of the introduction of new models in themarket. The theory provides the basis for rational decisions and thus improves thedecision-making process. In spite of all these, the theory of games suffers from certainlimitations and this questions the relevancy of game theory in practical life. Theimportant limitations are as follows:

(i) Businessmen do not have all the knowledge required for the theory of games.They do not even know all of the alternative strategies available to themselves,let alone the strategies open to their rivals.

(ii) There is a great deal of uncertainty that cannot conveniently be inserted intogame theory models. The outcome of a game may not be known with certaintyin many cases. It may, for example, depend upon the toss of a coin or uponcutting a deck of cards. As such, we usually restrict to games with knownoutcomes. This, therefore, is an important limitation.

(iii) The minimax strategy implies that the businessman minimizes the chance ofthe maximum loss. It is thus a very conservative rule. It is a much less accuratedescription of the dynamic businessman who is constantly in quest of profit.

(iv) The techniques of solving games involving mixed strategies particularly incase of a larger pay-off matrix is very complicated and this lessens thesignificance of this sort of analysis.

(v) Though the game theory points out some of the important aspects ofcompetitive situations but still it is not a model that offers complete solution,particularly when the game does not happen to be a strictly adversary game.For example, in an oligopoly situation a reduction in price of a commoditymay result in the expansion of the entire quantity demanded and not merelyshift demand units from one firm, say A to another firm say B. The gametheory will not be able to provide a complete solution to this sort of oligopolygame problem.


1. What is a pay-off matrix?

2. What are two-person games?

3. When is a player said to have adopted a pure strategy?

4. Define the saddle point of a pay-off matrix.

5. What do you understand by the principle of dominance?

Game Theory


Unit-4

ACTIVITY

1. Make a list of common examples of application of game theory ineveryday life.

2. Prepare a chart showing various types of game theory games.

4.6 LET US SUM UP

The word ‘game’ represents a conflict between two or more parties. Gametheory provides the general rules of the logic that underlies strategic behaviourof all types.

There are several types of games such as two-person games, constant-sumgames, games of perfect information, and so on.

The concept of value of a game refers to the average pay-off per play of thegame over an extended period of time.

The saddle point in a pay-off matrix is the smallest value in its row and thelargest value in its column.

The concept of dominance is used for reducing a bigger game into a smallergame.

If there is no saddle point in a game and reduction of the game by dominanceis also not possible, then linear programming techniques can offer an efficientsolution.

4.7 FURTHER READING

Ackoff, R.L. and M.W. Sasieni. 1968. Fundamentals of Operations Research. NewYork: John Wiley & Sons Inc.

Bierman, Harold, Charles Bonini and W.H. Hausman. 1973. Quantitative Analysisfor Business Decisions. Homewood (Illinois): Richard D. Irwin.

Enrick, Norbert Lloyd. 1965. Management Operations Research. New York: Holt,Rinehart & Winston.

Gillett, Billy E. 2007. Introduction to Operations Research. New Delhi: TataMcGraw-Hill.

Herbert, G. Hicks. 1967. The Management of Organisations. New York: McGraw-Hill.

Kalavathy, S. 2002. Operations Research. New Delhi: Vikas Publishing House.

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Kothari, C.R. 1982. An Introduction to Operational Research. New Delhi: VikasPublishing House.

Lindsay, Franklin A. New Techniques for Management Decision Making. New York:McGraw-Hill.

Massie, Joseph L. 1986. Essentials of Management. New Jersey: Prentice-Hall.

Swarup, Kanti, P.K. Gupta and Manmohan. 2001. Operations Research. New Delhi:Sultan Chand & Sons.


Check Your Progress–11. When games are represented in the form of a matrix (a rectangular array of

numbers deriving from matrix algebra), the resulting matrix is known as apay-off matrix.

2. In a two-person game, the players may have many possible choices open tothem for each play of the game, but the number of players remains only two.

3. When a player plays one row all the time (or one column all the time), he issaid to have adopted a pure strategy.

4. The saddle point in a pay-off matrix is the smallest value in its row and thelargest value in its column.

5. Sometimes, one of the pure strategies of either player is always inferior to atleast one of the remaining ones. Such superior strategies are said to dominatethe inferior ones. In such cases of dominance, we reduce the size of the pay-off matrix by deleting those strategies which are dominated by others.



1. List the general rules of dominance.

2. Write a short note on the value of a game.

3. What are zero-sum and non-zero-sum games?

4. Define maximum and minimax strategies.


1. What are the properties that a situation should possess to be termed a game?

2. Discuss the various types of games with examples.

3. For what type of business can game theory be useful? Explain.

4. The following is a pay-off matrix in rupees for a hypothetical game:

Game Theory


Unit-4

Strategies open to A Strategies open to B

x y z m

a 2 3 4 15

b 9 11 8 13

c 5 13 6 4

Determine the optimum strategies for A and B and also find the value of thegame. Can it be described as a game of pure strategy? If so, why?

5. For each of the following game matrices, find the optimal strategy for eachplayer. What is the saddle point and the value of each game? Are the gamesfair?

1

3

0

1

2

–34

3

3

–1–2

0

A

B

C

Blue

( )a

Red

1

–46

5

2

–3–2–4

3

4

3

–5

A

B

C

White

( )b

Black

6. Use graphical methods in solving the following games:

B

(a)3 3 4

1 1 3A

Y

(b)2 5

4 1X

7. The following matrix represents the pay-off to P1 in a rectangular game

between two persons P1 and P

2.

2

1

8 15 4 –219 15 17 16

0 20 15 5

P

P

By the notion of dominance, reduce the game to a 2 × 4 game and solve itgraphically.

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8. Solve the following game.

Player BPlayer

3 2 4 0

3 4 2 4

4 2 4 0

0 4 0 8

BI II III IV

I

II

III

IV

9. Using dominance, solve the pay-off matrix given by(i) Player B

2 2 4 1

6 1 12 3Player

3 2 0 6

2 3 7 7

A

(ii) Player B

1 7 3 4

Player 5 6 4 5

7 2 0 3

A

Network Models


Unit-5

UNIT 5 NETWORK MODELS

UNIT STRUCTURE5.1 Learning Objectives5.2 Introduction5.3 Construction of Networks

5.3.1 Basic Terms5.3.2 Common Errors5.3.3 Rules of Network Construction5.3.4 Time Analysis

5.4 Determination of Floats and Slack Times5.5 Determination of Critical Paths5.6 Determination of PERT Times

5.6.1 PERT Procedure5.7 Crashing5.8 Resource Allocation and Levelling5.9 Let Us Sum Up




Explain network analysis for large projects

List the applications of PERT and CPM for handling projects

Construct a network of activities and events for analysis

Highlight the importance of CPM and PERT analysis

Explain the PERT procedure

5.2 INTRODUCTION

There are various aspects of network analysis such as network scheduling, planning andcontrol. Network scheduling is a technique used for planning and scheduling large projectsin the field of construction, maintenance, fabrication, etc. It is a tool for minimizingproblems in the execution and controlling of critical factors in a project. Program EvaluationReview Technique (PERT) and Critical Path Method (CPM) are two planning andcontrol techniques for keeping a project schedule on track to complete within thescheduled time.

In this unit, you will learn the basic terms associated with the network analysistechnique. A network is a graphic representation of logically and sequentially connectedarrows and nodes, representing the activities and events, respectively of a project. Anevent is the beginning and end point of an activity and is represented by a node. You willlearn how to construct a network after going through the rules of network construction.

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You will learn how to carry out the time analysis of a network model from itsestimated completion time, earliest start time, earliest finishing time, latest starttime and latest finishing time. You will also learn how to analyse the critical path ofa project to complete it within the scheduled time.

5.3 CONSTRUCTION OF NETWORKS

Network scheduling is a technique used for planning and scheduling large projectsin the field of construction, maintenance, fabrication, purchasing computer system,etc. The technique is a method of minimizing the trouble spots such as production,delays and interruptions, by determining critical factors and coordinating various partsof the overall job.

There are two basic planning and control techniques that utilize a network tocomplete a predetermined project or schedule. These are Programme EvaluationReview Technique (PERT) and Critical Path Method (CPM).

A project is defined as a combination of interrelated activities all of which mustbe executed in a certain order for its completion. The work involved in a project canbe divided into three phases corresponding to the management functions of planning,scheduling and control.

Planning: This phase involves setting the objectives of the project and the assumptionsto be made. Also, it involves the listing of tasks or jobs that must be performed tocomplete a project under consideration. In this phase, men, machines and materialsrequired for the project, in addition to the estimates of costs and duration of the variousactivities of the project, are also determined.

Scheduling: This consists of carrying out the activities according to the precedenceorder and determining the following:

(i) The start and finish times for each activity

(ii) The critical path on which the activities require special attention

(iii) The slack and float for the non-critical paths

Controlling: This phase is exercised after the planning and scheduling, which involvesthe following:

(i) Making periodical progress reports

(ii) Reviewing the progress

(iii) Analysing the status of the project

(iv) Management decisions regarding updating, crashing and resource allocation

5.3.1 Basic Terms

To understand the network techniques, one should be familiar with few basic termsof which both CPM and PERT are special applications.

Network scheduling:A technique used forplanning andscheduling largeprojects in the field ofconstruction,maintenance,fabrication, purchasingcomputer system, etc.

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Network: It is the graphic representation of logically and sequentially-connected arrowsand nodes representing the activities and events of a project. Networks are also calledarrow diagrams.

Activity: An activity represents some action and is a time consuming effort necessaryto complete a particular part of the overall project. Thus, each activity has a point oftime at which it begins and a point at which it ends.

It is represented in the network by an arrow,

i jA

Here, A is called the activity.

Event: The beginning and end points of an activity are called events or nodes.Event is a point in time and does not consume any resource. It is represented by anumbered circle. The head event called the jth event has always a number higherthan the tail event called the ith event.

i jActivity

Tail Head

Merge and burst events: It is not necessary for an event to be the ending event ofonly one activity, but can be the ending event of two or more activities. Such an eventis defined as a merge event.

If the event happens to be the beginning event of two or more activities it isdefined as a burst event.

Preceding, succeeding and concurrent activities: Activities which must beaccomplished before a given event can occur are termed as preceding activities.

Activities which cannot be accomplished until an event has occurred aretermed as succeeding activities.

Activities, which can be accomplished concurrently are known as concurrentactivities.

This classification is relative, which means that one activity can be precedingto a certain event, and the same activity can be succeeding to some other event or itmay be a concurrent activity with one or more activities.

Dummy activity: Certain activities which neither consume time nor resources, butare used simply to represent a connection or a link between events are known as

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dummies. It is shown in the network by a dotted line. The objectives of introducing adummy activity are as follows:

(i) To maintain uniqueness in the numbering system as every activity may have adistinct set of events by which the activity can be identified

(ii) To maintain a proper logic in the network

5.3.2 Common Errors

The following are the three common errors in a network construction:

1. Looping (cycling): In a network diagram, looping error is also known as cyclingerror. Drawing an endless loop in a network is known as error of looping. Aloop can be formed if an activity is represented as going back in time.

2. Dangling: To disconnect an activity before the completion of all the activitiesin a network diagram is known as dangling.

3. Redundancy: If a dummy activity is the only activity emanating from an eventand which can be eliminated it is known as redundancy.

5.3.3 Rules of Network Construction

There are a number of rules in connection with the handling of events and activities ofa project network that should be followed.

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(i) Try to avoid arrows which cross each other.

(ii) Use straight arrows.

(iii) No event can occur until every activity preceding it has been completed.

(iv) An event cannot occur twice, i.e., there must be no loops.

(v) An activity succeeding an event cannot be started until that event has occurred.

(vi) Use arrows from left to right. Avoid mixing two directions. Vertical and standingarrows may be used if necessary.

(vii) Dummies should be introduced if it is extremely necessary.

(viii) The network has only one entry point called the start event and one point ofemergence called the end or terminal event.

Numbering the Events (Fulkerson’s Rule)After the network is drawn in a logical sequence every event is assigned a number.The number sequence must reflect the flow of the network. In numbering the events,the following rules should be observed:

(i) Event numbers should be unique.

(ii) Event numbering should be carried out on a sequential basis from left toright.

(iii) The initial event which has all outgoing arrows with no incoming arrow shouldbe numbered as 1.

(iv) All arrows emerging from all the numbered events should be deleted. Thiswill create at least one new start event out of the preceding events.

(v) All new start events should be numbered 2, 3, and so on. This process shouldbe repeated until the terminal event without any successor activity is reached.The terminal node should be then numbered suitably.

Note: The head of an arrow should always bear a number higher than the one assignedto the tail of the arrow.

Example 5.1: Construct a network for the project whose activities and theirprecedence relationships are given as follows:

– A A – D B,C,E F D G,HActivities A B C D E F G H I

Immediate Predecessor

Solution: From the given constraints, it is clear that A, D are the starting activitiesand I the terminal activity. B, C are starting with the same event and are both thepredecessors of the activity F. Also, E has to be the predecessor of both F and H.Hence, we have to introduce a dummy activity.

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D1 is the dummy activity.

Finally, we have the following network.

Example 5.2: Construct a network for each of the projects whose activities and theirprecedence relationships are given as follows:

– – – A B B C D E H,I F,GActivity A B C D E F G H I J K

Predecessor

Solution: A, B, C are the concurrent activities as they start simultaneously. B becomesthe predecessor of activity E and F. Since the activities J, K have two precedingactivities, a dummy may be introduced (if possible).

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Finally we have,

Example 5.3: Construct a network of the project whose activities are given as follows:

A<C, D, I; B<G, F; D<G, F; F<H, K; G, H<J; I, J, K<E

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Solution: Given A<C, which means that C cannot be started until A is completed, i.e., Ais the preceding activity to C. The above constraints can be given in the following table:

– – A A I,J,K B,D B,D F A G,H FActivity A B C D E F G H I J K

Predecessor

A, B are the starting activities, and E is the terminal activity.

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Finally we have,

Example 5.4: Construct the network for the project whose activities and precedencerelationship is given below. Show also the dummy activity.

– – A,B B B A,B F,D F,D C,GActivities A B C D E F G H I

Immediate Predecessor

Solution: A, B are concurrent activities as they start simultaneously. I is the terminalactivity. Since the activities C and F arise from both the activities A, B we need tointroduce a dummy activity.

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Example 5.5: Make a network of the project having activities and precedencerelationship as given below:

A, B, C can start simultaneously.

A<D, I; B<G, F; D<G, F; C<E; E<H, K; F<H, K; G, H<J

Solution: The given constraints can be formatted into a table.

– – – A C B,D B,D B,F A G E,FActivity A B C D E F G H I J K

Predecessor Activity

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J

5.3.4 Time Analysis

Once the network of a project is constructed, the time analysis of the network becomesessential for planning various activities of the project. Activity time is a forecast ofthe time an activity is expected to take from its starting point to its completion (undernormal conditions).

The following notations can be used for basic scheduling computations.

( i, j) = Activity (i, j) with tail event i and head event j

Tij

= Estimated completion time of activity (i, j)

ESij

= Earliest starting time of activity (i, j)

EFij

= Earliest finishing time of activity (i, j)

(LS)ij

= Latest starting time of activity (i, j)

(LF)ij

= Latest finishing time of activity (i, j)

The basic scheduling computation can be put under the following two groups.

Forward pass computations (for earliest event time)

Before starting computations, the occurrence time of the initial network event is fixed.The forward pass computation yields the earliest start time and the earliest finish timefor each activity (i, j) and indirectly the earliest occurrence time for each event, namelyE

i. This consists of the following three steps:

Step 1: The computations begin from the start node and move towards the end node.Let zero be the starting time for the project.

Step 2: Earliest starting time (ES)ij = E

iis the earliest possible time when an activity

can begin, assuming that all the predecessors are also started at their earliest startingtime. Earliest finish time of activity (i, j) is the earliest starting time together withthe activity time.

(EF )ij

= (ES)ij + t

ij

Step 3: Earliest event time for event j is the maximum of the earliest finish time ofall the activities ending at that event.

Activity time: Aforecast of the time anactivity is expected totake from its startingpoint to its completion(under normalconditions).

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Ej

= j iji

Max E t

The computed ‘E’ values are put over the respective rectangle representingeach event.

Backward pass computations (for latest allowable time)

The latest event time (L) indicates the time by which all activities entering into anevent must be completed without delaying the completion of the project. This can becalculated by reversing the method of calculations used for the earliest event time.This is done in the following steps:

Step 1: For ending an event, assume E = L.

Step 2: Latest finish time for activity (i, j) is the target time for completing the project.

(LFij) = L

j

Step 3: Latest starting time of the activity (i,j) = Latest completion time (i,,j) of the

activity time.LS

ij= LF

ij– t

ij

= Lj – t

ij

Step 4: Latest event time for event i is the minimum of the latest start time of allactivities originating from the event.

–i j ijj

L Min L t

The computed ‘L’ values are put over the respective triangle representingeach event.

5.4 DETERMINATION OF FLOATS AND SLACKTIMES

Float is defined as the difference between the latest and the earliest activity time.

Slack is defined as the difference between the latest and the earliest eventtime.

Hence, the basic difference between the slack and float is that slack is usedfor events, only whereas float is used for activities.

There are mainly three types of floats as given below:

(i) Total float: It refers to the duration of time by which the completion of anactivity could be delayed beyond the earliest expected completion time withoutaffecting the overall project duration time.

Mathematically, the total float of an activity (i, j) is the difference between thelatest start time and the earliest start time of that activity.

Hence, the total float for an activity (i, j) denoted by (TF)ij is calculated by the

formula,

Float: The differencebetween the latest andthe earliest activitytime.

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(TF)ij= (Latest start – Earliest start) for activity (i, j)

i.e.,( ) –

or, – –

ij ij ij

j i ijij

TF LS ES

TF L E t

Here, Ei, L

j are the earliest time and latest time for the tail event i and head event

j and tij is the normal time for the activity (i, j). This is the most important type of float

as it concerns the overall project duration.

(ii) Free float: The time by which the completion of an activity can be delayedbeyond the earliest finish time without affecting the earliest start of a subsequentsucceeding activity.

Mathematically, the free float for activity (i, j) denoted by (FF)ij can be calculated

by the formula,

Total float – Head event slack

– –ij j i ij

ij

FF E E t

FF

Head event slack = Lj– E

j

This float is concerned with the commencement of the subsequent activity.

The free float can take values from zero up to total float, but it cannot exceedthe total float. This float is very useful for rescheduling the activities with minimumdisruption of earlier plans.

(iii) Independent float: The amount of time by which the start of an activity can bedelayed without affecting the earliest start time of any immediately followingactivities assuming that the preceding activity has finished at its latest finishtime.

Mathematically, independent float of an activity (i, j) denoted by IFij can be

calculated by the formula,

– –

or

Free float –Tail event slack

ij j i ij

ji

IF E L t

IF

Where tail event slack is given by,

Tail event slack = Li – E

i

The negative independent float is always taken as zero. This float is concerned withprior and subsequent activities.

IFij

FFij

TFij

Notes: 1. If the total float TFij for any activity (i, j) is zero, then that activity is called

critical activity.

2. The float can be used to reduce the project duration. While doing this, the float ofnot only that activity but that of other activities would also change.

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Slack. The term slack is normally asociated with events. It indicates the amount oflatitude (or the lee-way) that is available for an event to occur. It is worked outas follows:

Slack of an event = (Latest occurance time of the event) – (earliest occurancetime of the event) or simply slack = L

T – E

T

It can be positive or negative depending upon whether the targetted date ofcompletion is later or earlier than the earliest finish time of the task respectively.When used for activities, the term slack should be used for activities, (Activityslack is synonymous to float). Since slack is associated with events, each activitywill have two slacks, viz. the slack of its head event or the head slack and the slackof its tail event or the tail slack.

5.5 DETERMINATION OF CRITICAL PATHS

Critical activity: An activity is said to be critical if a delay in its start will cause afurther delay in the completion of the entire project.

Critical path: The sequence of critical activities in a network is called the criticalpath. It is the longest path in the network from the starting event to the ending eventand defines the minimum time required to complete the project. In the network, it isdenoted by a double line. This path identifies all the critical activities of the project.Hence, for the activity (i, j) to lie on the critical path, the following conditions mustbe satisfied.

(i) ESi= LF

i

(ii) ESj= LF

j

(iii) ESj– ES

i= LF

j– LF

i = t

ij

ESi, ES

j, are the earliest start and finish times of the events i and j.

LFi, LF

j are the latest start, finish times of the events i and j.

The iterative procedure of determining the critical path is as follows.

Step 1: List all the jobs and then draw arrow (network) diagram. Each job is indicatedby an arrow with the direction of the arrow showing the sequence of jobs. Thelength of the arrows has no significance. The arrows are placed based on thepredecessor, successor, and concurrent relation within the job.

Step 2: Indicate the normal time (tij) for each activity (i, j) above the arrow which is

deterministic.

Step 3: Calculate the earliest start time and the earliest finish time for each eventand write the earliest time E

i for each event i in the . Also calculate the latest

finish and latest start time. From this we calculate the latest time Lj for each event j

and put in the .

Step 4: Tabulate the various times, namely normal time, earliest time and latesttime on the arrow diagram.

Step 5: Determine the total float for each activity by taking the difference between theearliest start and the latest start time.

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Step 6: Identify the critical activities and connect them with the beginning event and theending event in the network diagram by double line arrows. This gives the critical path.

Step 7: Calculate the total project duration.

Note: The earliest start, finish time of an activity, and the latest start, finish time of anactivity will be shown in the table. These are calculated by using the following hints.

To find the earliest time we consider the tail event of the activity. Let the startingtime of the project be ES

i= 0. Add the normal time with the starting time to get the

earliest finish time. The earliest starting time for the tail event of the next activity isgiven by the maximum of the earliest finish time for the head event of the previousactivity.

Similarly, to get the latest time, we consider the head event of the activity.

The latest finish time of the head event of the final activity is given by the targettime of the project. The latest start time can be obtained by subtracting the normaltime of that activity. The latest finish time for the head event of the next activity isgiven by the minimum of the latest start time for the tail event of the previous activity.

Example 5.6: A project schedule has the following characteristics.

( ) 4 1 1 1 6 5 4 8

( ) 1 2 5 7

Time days

Time days

Activity 1– 2 1– 3 2 – 4 3 – 4 3 – 5 4 – 9 5 – 6 5 –7

Activity 6 – 8 7 – 8 8 – 10 9 – 10

From the above information, you are required to:

(i) Construct a network diagram.

(ii) Compute the earliest event time and latest event time.

(iii) Determine the critical path and total project duration.

(iv) Compute total, free float for each activity.

Solution: First we construct the network with the given constraints. Here we get thisby just connecting the event numbers.

The following table gives the critical path, total, free floats calculation.

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Time

The earliest and latest calculations are shown below:

Forward pass calculation: In this we estimate the earliest start ESi and finish time

ESj. The earliest time for the event i is given by,

Backward pass calculation: In this, we calculate the latest finish and the latest starttime. The latest time L for an event i is given by L

i= Min

j (LF

j– t

ij).

Here, LFj is the latest finish time for the event j, t

ij is the normal time of the

activity.

10

9 10 9,10

8 10 8,10

7 8 7,8

6 8 6,8

5 6 5,6 7 5,7

4 9 4,9

3 4 3,4 5 3,5

2 4 2,4

22

22 7 15

22 5 17

17 2 15

17 1 16

Min ( , )

Min (16 4,15 8) 7

15 5 10

Min ( , )

Min (10 1, 7 6) 1

10 1

L

L L t

L L t

L L t

L L t

L L t L t

L L t

L L t L t

L L t

1 2 12 3 13

9

Min ( , ) Min (9 4,1 1) 0.L L t L t

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10

9 10 9,10

8 10 8,10

7 8 7,8

6 8 6,8

5 6 5,6 7 5,7

4 9 4,9

3 4 3,4 5 3,5

2 4 2,4

22

22 7 15

22 5 17

17 2 15

17 1 16

Min ( , )

Min (16 4,15 8) 7

15 5 10

Min ( , )

Min (10 1, 7 6) 1

10 1

L

L L t

L L t

L L t

L L t

L L t L t

L L t

L L t L t

L L t

1 2 12 3 13

9

Min ( , ) Min (9 4,1 1) 0.L L t L t

These calculations are shown in the given table.

To find the TF (Total Float): Considering the activity 1–2, TF of (1–2) = Latest start–Earliest start.

So, TF = 5 – 0 = 5Similarly, TF(2–4) = LS – ESSo, TF = 9 – 4 = 5Free float = TF – Head event slackConsider the activity 1 – 2FF of 1 – 2 = TF of 1 – 2 – Slack for the head event 2So, FF = 5 – (9 – 4) (from the figure for event 2) FF = 5 – 5 = 0FF of 2 – 4 = TF of 2 – 4 – Slack for the head event 4So, FF = 5 – (10 – 5) = 5 – 5 = 0Like this we calculate the TF and FF for the remaining activities.

From the above table we observe that the activities 1–3, 3–5, 5–7, 7–8, 8–10 are thecritical activities as their total float is 0.

Hence, we have the following critical path.

1 3 5 7 8 10, with the total project duration of 22 days.

Example 5.7: A small maintenance project consists of the following jobs whoseprecedence relationships is given below:

15 15 3 5 8 12 1 14 3 14

Job 1 – 2 1 – 3 2 – 3 2 – 5 3 – 4 3 – 6 4 – 5 4 – 5 5 – 6 6 –7Duration(days)

(i) Draw an arrow diagram representing the project.

(ii) Find the total float for each activity.

(iii) Find the critical path and the total project duration.

Solution:

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Forward pass calculation: In this, we estimate the earliest start and the earliest finishtime ES

j given by,

( )j i iji

ES Max ES t , where ESi is the earliest start time and t

ij is the normal time

for the activity (i, j).

1

2 1 15

3 2 23 1 13

4 3 34

5 2 25 4 45

6 3 36 4 46 5 56

7

0

0 15 15

Max ( , )

Max (15 3, 0 15) 18

18 8 26

Max ( , )

Max (15 5, 26 1) 27

Max ( , , )

Max (18 12, 26 14, 27 3)

40

ES

ES ES t

ES ES t ES t

ES ES t

ES ES t ES t

ES ES t ES t ES t

ES ES

6 67 40 14 54t

Backward pass calculation: In this, we calculate the latest finish and latest start timeLF

i, given by LF

i= Min

i(LF

j–t

ij), where LF

j is the latest finish time for the event j.

7

6 7 67

5 6 56

4 5 45 6 46

3 4 34 6 36

2 5 25 3 23

1 3 13

54

54 14 40

40 3 37

Min ( , )

Min (37 1, 40 14) 26

Min ( , )

Min (26 8, 40 12) 18

Min ( , )

Min (37 5, 18 3) 15

Min ( ,

LF

LF LF t

LF LS t

LF LS t LS t

LF LF t LF t

LF LF t LF t

LF LF t

2 12 )

Min (18 15, 15 15) 0

LF t

The following table gives the calculation for critical path and total float.

Time Float

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From the above table we observe that the activities 1 – 2, 2 – 3, 3 – 4, 4 – 6, 6 – 7 are thecritical activities and the critical path is given by, 1 2 3 4 6 7.

The total project completion is given by 54 days.

Example 5.8: The following table shows the jobs of a project with their duration indays. Draw the network and determine the critical path. Also, calculate all the floats.

Jobs 1–2 1–3 1–4 2–5 3–7 4–6 5–7 5–8 6–7 6–9 7–10Duration 10 8 9 8 16 7 7 7 8 5 12Jobs 8–10 9–10 10–11 11–12Duration 10 15 8 5

Solution: First we construct the network as shown below:

Forward pass calculation: In this we calculate the earliest start and the earliestfinish time for the activity and the earliest time for each event.

1

2 1 12

3 1 13

4 1 14

5 2 25

6 4 46

0

0 10 10

0 8 8

0 9 9

10 8 18

9 7 16

ES

ES ES t

ES ES t

E E t

E E t

E E t

1

2 1 12

3 1 13

4 1 14

5 2 25

6 4 46

7 3 37 5 57 6 67

8 5 58

9 6 69

10 7 7,10

0

0 10 10

0 8 8

0 9 9

10 8 18

9 7 16

Max ( , , )

Max (8 16, 16 8, 18 7) 25

18 7 25

16 5 21

Max ( ,

ES

ES ES t

ES ES t

E E t

E E t

E E t

E E t E t E t

E E t

E E t

E E t E

8 8,10 9 9,10

11 10 10,11

12 11 11,12

, )

Max (25 12, 25 10, 21 15) 37

37 8 45

45 5 50

t E t

E E t

E E t

1

2 1 12

3 1 13

4 1 14

5 2 25

6 4 46

7 3 37 5 57 6 67

8 5 58

9 6 69

10 7 7,10

0

0 10 10

0 8 8

0 9 9

10 8 18

9 7 16

Max ( , , )

Max (8 16, 16 8, 18 7) 25

18 7 25

16 5 21

Max ( ,

ES

ES ES t

ES ES t

E E t

E E t

E E t

E E t E t E t

E E t

E E t

E E t E

8 8,10 9 9,10

11 10 10,11

12 11 11,12

, )

Max (25 12, 25 10, 21 15) 37

37 8 45

45 5 50

t E t

E E t

E E t

Backward pass calculation: In this, we calculate the latest finish and the latest startt i m e LF

i given by,

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12 12

11 12 11,12

10 11 10,11

9 10 9,10

8 10 8,10

7 10 7,10

6 9 69 7 67

Min

50 (the target completion time)

50 5 45

45 8 37

37 15 22

37 10 27

37 12 25

Min ( , )

Min (22 5, 25

i j ijj

LF LF t

L E

L L t

L L t

L L t

L L t

L L t

L L t L t

5 8 58 7 57

4 6 46

3 7 37

2 5 25

1 4 14 3 13 2 12

8) 17

Min ( , )

Min (27 7, 25 7) 18

17 7 10

25 16 9

18 10 10

Min ( , , )

Min (10 9, 9 8, 10 10) 0

L L t L t

L L t

L L t

L L t

L L t L t L t

The computation of critical path and all the floats is given in the following table.

Time

TF = Total float = LS – ES or LF – EFFF = Free float = TF – Head event slackIF = Independent float = FF – Tail event slackFrom the above calculation we observe that the activities 1 – 2, 2 – 5, 5 – 7,

7 – 10, 10 – 11, 11 – 12 are the critical activities as their total float is 0. Hence, wehave the critical path 1 2 5 7 10 11 12 with the total projectduration as 50 days.

Example 5.9: A project consists of a series of tasks labelled A, B, ... H, I with thefollowing constraints:

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A<D, E; B, D<F; C<G; B<H; F, G<I; W<X, Y means X, and Y cannot start untilW is completed. You are required to construct a network using this notation. Also, findthe minimum time of completion of the project when the time of completion of each taskis given as follows.

23 8 20 16 24 18 19 4 10Time(days)

Task A B C D E F G H I

Solution: The given constraints can be shown in the following table:

– – – A A B,D C B F,GPreceding Activity

Activity A B C D E F G H I

To determine the minimum time of completion of the project, we computeES

i and LF

i for each of the tasks (i, j) of the project. The critical path calculations

are as follows:

2323

3939

6767

5757

2038

00

(23)A

2

13

F(20)

(19)

(18)

H (4)

E

B (8)(16)

D

C

4G

5

6

(24)

(10)I

Critical path 1 – 2 – 3 – 5 – 6

The above table shows that the critical activities are 1 –2, 2–3, 3–5, 5–7 as theirtotal float is zero. Hence, we have the critical path, 1 2 3 5 7 with thetotal project duration (the least possible time to complete the entire project) as 67days.

Example 5.10: Tasks A, B, ... H, I constitute a project. The notation X<Y means thatthe task X must be completed before Y is started. With the notations,

A<D; A<E; B<F; D<F; C<G; C<H; F<I; G<I

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draw a graph to represent the sequence of tasks and find the minimum time of completion ofthe project, when the time (in days) of completion of each task is as follows:

The above constraints can be given in the following table.

8 10 8 10 16 17 18 14 9Time(days)


Solution: The above constraints can be given in the following table.

– – – A A B,D C B F,GPreceding Activity

Activity A B C D E F G H I

Time calculation: Using forward and backward pass calculation, we first estimatethe earliest and the latest time for each event.

1 1

2 1 12

3 1 13 2 23

4 1 14

5 3 35 4 45

6 2 26 4 46 5 56

0

0 8 8

Max ( , )

Max (0 10, 8 10) 18

0 8 8

Max ( , )

Max (18 17, 8 18) 35

Max ( , , )

Max (8 16, 8 14, 35 9) 44

ES E

E E t

E E E E t

E E t

E E t E t

E E t E t E t

The value of the latest time can now be obtained.

6 6

5 6 56

4 6 46 5 45

3 5 35

2 6 26 3 23

1 4 14 3 13 2

44 (Target completion time for the project)

44 9 35

Min ( , )

Min (44 14, 35 18) 17

35 17 18

Min ( , )

Min (44 16, 18 10) 8

Min ( , ,

L E

L L t

L L t L t

L L t

L L t L t

L L t L t L

12 )

Min (17 8, 18 10, 8 8) 0

t

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To evaluate the critical events, all these calculations are put in the following table.

NormalTime/Days

The above table shows that the critical events are the tasks 1 – 2, 2 – 3, 3 – 5,5 – 6 as their total float is zero.

The critical path is given by 1 2 3 5 6 or A D F I with thetotal project duration as 44 days.

5.6 DETERMINATION OF PERT TIMES

The network methods discussed so far may be termed as deterministic, sinceestimated activity times are assumed to be known with certainty. However, in aresearch project or design of gear box or a new machine, various activities arebased on judgement. It is difficult to obtain a reliable time estimate due to thechanging technology. Time values are subject to chance variations. For such caseswhere the activities are non-deterministic in nature, PERT was developed. Hence,PERT is a probabilistic method where the activity times are represented by aprobability distribution. This probability distribution of activity times is based uponthree different time estimates made for each activity. These are as follows:

(i) Optimistic time estimate

(ii) Most likely time estimate

(iii) Pessimistic time estimate

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(i) Optimistic time estimate: It is the smallest time taken to complete the activity ifeverything goes on well. There is very little chance that the activity can be donein time less than the optimistic time. It is denoted by t

0 or a.

(ii) Most likely time estimate: It refers to the estimate of the normal time the activitywould take. This assumes normal delays. It is the mode of the probabilitydistribution. It is denoted by t

mor (m).

(iii) Pessimistic time estimate: It is the longest time that an activity would take ifeverything goes wrong. It is denoted by t

p or b. These three time values are

shown in the following figure.

Fre

quen

cy0 tpt0 tm

Time Distribution Curve

From these three time estimates, we have to calculate the expected time of anactivity. It is given by the weighted average of the three time estimates,

0 4

6m p

e

t t tt

distribution with weights of 1,4,1, for to, t

mand t

p estimates respectively.

Variance of the activity is given by,

2 =2

0–6

pt t

The expected length (duration), denoted by Tc of the entire project is the

length of the critical path, i.e., the sum of the Tcs of all the activities along the

critical path.

The main objective of the PERT analysis is to find the completion time for aparticular event within the specified date T

S, given by P (Z D) where,

D =Due date – Expected date of completion

Project variance

Here, Z stands for standard normal variable.

5.6.1 PERT Procedure

Step 1: Draw the project network.

Step 2: Compute the expected duration of each activity using the formula.

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0 4

6m p

e

t t tt

Also, calculate the expected variance 2 of each activity.

Step 3: Compute the earliest start, earliest finish, latest start, latest finish and totalfloat of each activity.

Step 4: Find the critical path and identify the critical activities.

Step 5: Compute the project length variance 2 which is the sum of the variance ofall the critical activities and hence, find the standard deviation of the projectlength .

Step 6: Calculate the standard normal variable Z =–s eT T

.

Here,

TS is the scheduled time to complete the project.

Teis normal expected project length duration.

is expected standard deviation of the project length.

Using the normal curve, we can estimate the probability of completing theproject within a specified time.

Example 5.11: The following table shows the jobs of a network along with theirtime estimates.

( ) 1 2 2 2 7 5 5 3 8

( ) 7 5 14 5 10 5 8 3 17

13 14 26 8 19 17 29 9 32( )

a days

m days

b days

Job 1 – 2 1 – 6 2 – 3 2 – 4 3 – 5 4 – 5 6 –7 5 – 8 7 – 8

Draw the project network and find the probability that the project is completedin 40 days.

Solution: First we calculate the expected time and standard deviation for eachactivity.

Activity4

6o m p

e

t + t + tt =

–

22σ =

6p ot t

1–21 4 7 13

76

213 1

46

1–62 4 5 14

66

214 2

46

2–32 4 14 26

146

226 2

166

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2–42 5 4 8

56

28 2

16

3–57 4 10 19

116

219 7

46

4–55 5 4 17

76

217 5

46

6–75 8 4 29

116

229 5

166

5–83 3 4 9

46

29 3

16

7–88 4 17 32

186

232 8

166

Expected project duration = 36 days

Critical path 1 2 3 5 8

Project length variance = 2= 4 +16 + 4 + 1

= 25

= 5

Probability that the project will be completed in 40 days is given by,

P(Z )

=– 40 36 4

0.85 5

s eT T

Area under the normal curve for = 0.8,

P(Z 0.8)

= 0.5 + (0.8) [(8) = 0.2881 (from table)]

= 0.5 + 0.2881 = 0.7881 = 78.81%

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Conclusion: If the project is performed 100 times under the same conditions, there willbe 78.81 occasions for this job to be completed in 40 days.

Example 5.12: A small project is composed of seven activities whose time estimatesare listed in the following table.

Duration (Weeks)Likely

You are required to:

(a) Draw the project network.

(b) Find the expected duration and variance of each activity.

(c) Calculate the early and late occurrence for each event and the expected projectlength.

(d) Calculate the variance and standard deviations of project length.

(e) What is the probability that the project will be completed?(i) 4 weeks earlier than expected.

(ii) Not more than 4 weeks later than expected.(iii) If the project due date is 19 weeks, what is the probability of meeting

the due date.

Solution: The expected time and variance of each activity is computed as shown inthe following table.

Activity a m b 4

6m

e

a + t + bt = 2 =

6

b a 2

σ

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The earliest and the latest occurrence time for each is calculated as follows:

E1

= 0; E2= 0 + 2 = 2

E3

= 0 + 4 = 4

E4

= 0 + 3 = 3

E5

= Max (2 + 1, 4 + 6) = 10

E6

= Max (10 + 7, 3 + 5) = 17

To determine the latest expected time we start from E6 being the last event and

move backwards subtracting te from each activity. Hence, we have

L6

= E6= 17

L5

= L6– 7 = 17 – 7 = 10

L4

= 17 – 5 = 12L

3= 10 – 6 = 4

L2

= 10 – 1 = 9L

1= Min (9 – 2, 4 – 4, 12 – 3) = 0

Using the above information we get the following network, where the criticalpath is shown by the double line arrow.

We observe the critical path of the above network as 1 3 5 6.

The expected project duration is 17 weeks, i.e., Te = 17 weeks.

The variance of the project length is given by,

2 = 1 + 4 + 4 = 9

(i) The probability of completing the project within 4 weeks earlier than expectedis given by,

–( ), where ST T

P Z D Ds

Due date Expected date of completion

Project varianceD

17 4 17 13 17 4

3 3 31.33

D

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( 1.33) 0.5 (1.33)

0.5 0.4082 (from the table)

= 0.0918 = 9.18%

P Z

Conclusion: If the project is performed 100 times under the same conditions, then therewill be 9 occasions for this job to be completed in 4 weeks earlier than expected.

(ii) The probability of completing the project not more than 4 weeks later than expectedis given by,

P (Z D)

Here, s eT TD

where, Ts = 17 + 4 = 21

21 17 41.33

3 3D

P (Z 1.33)

= 0 . 5 + (1.33)

= 0.5 + 0.4082 (from the table)

= 0.9082 = 90.82%

Conclusion: If the project is performed 100 times under the same conditions, thenthere will be 90.82 occasions when this job will be completed not more than 4weeks later than expected.

(iii) The probability of completing the project within 19 weeks is given by,

19 17 2( ) where, 19

3 3 SP Z D D= T

= 0.666P (Z 0.666) = 0.5 + (0.666)= 0.5 + 0.2514 (from the table)= 0.7514 = 75.14%

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Conclusion: If the project is performed 100 times under the same conditions, then therewill be 75.14 occasions for this job to be completed in 19 weeks.

Example 5.13: Consider the following project.

Activity Time Estimate in Weeks Predecessor

to tm tp

A 3 6 9 None

B 2 5 8 None

C 2 4 6 A

D 2 3 10 B

E 1 3 11 B

F 4 6 8 C,D

G 1 5 15 E

Find the path and standard deviation. Also, find the probability of completing theproject in 18 weeks.

Solution: First we calculate the expected time and variance of each activity as in thefollowing table:

Activity ot tm tp

4

6o m p

e

t + t + tt =

–

22σ =

6p ot t

A 3 6 93 4 6 9

66

[(9 – 3)/6]2 = 1

B 2 5 830

56

[(8 – 2)/6]2 = 1

C 2 4 6 24/6 = 4 [(6 – 2)/6]2 = 0.444

D 2 3 10 4 1.777

E 1 3 11 4 2.777

F 4 6 8 6 0.444

G 1 5 15 6 5.444

We construct the network with the help of the predecessor relation given inthe data.

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Critical path is 1 2 4 6 or A C F

The project length = 16 weeks

Project length variance 2= 1 + 0.444 + 0.444 = 1.888

Standard deviation = = 1.374

The probability of completing the project in 18 weeks is given by:

P(Z D)

Here, D =–s eT T

Ts = 18; T

e = 16; = 1.374

D =18 –161.374

= 1.4556

P(Z D) = P(Z 1.4556) = 0.5 + (1.4456)

= 0.5 + 0.4265

= 0.9265 = 92.65%

Conclusion: If the project is performed 100 times under the same conditions, then therewill be 92.65 occasions when this job will be completed by 18 weeks.

Example 5.14: The following table shows the jobs of a network along with their timeestimates. The time estimates are in days:

3 2 6 2 5 3 1 3 4

6 5 12 5 11 6 4 9 19

15 14 30 8 17 15 7 27 28

a

m

b

Job 1 – 2 1 – 6 2 – 3 2 – 4 3 – 5 4 – 5 5 – 8 6 –7 7 – 8

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(i) Draw the project network.

(ii) Find the critical path.

(iii) Find the probability that the project will be completed in 31 days.

Solution: First we calculate the expected time and variance for each activity as shownin the following table.

Activity a m b te = (a + 4m + b)/6–

22 ( )σ =

6b a

1–2 3 6 15 7 41–6 2 5 14 6 42–3 6 12 30 14 162–4 2 5 8 5 13–5 5 11 17 11 44–5 3 6 15 7 45–8 1 4 7 4 16–7 3 9 27 11 167–8 4 19 28 27 16

(i) Project network:

(ii) The critical path is given by 1 6 7 8 and the project length is givenby 44 days.

The project length variance 2 = 4 + 16 + 16 = 36

Standard deviation = = 36 = 6

(iii) The probability of completing the project within 31 days is given by,

P (Z D), where D = s eT T

31 442.1666

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i.e., P(Z –2.1666) = 0 . 5 – (2.166)

= 0.5 – 0.4886 = 0.0114 = 1.14%

Conclusion: If the project is performed 100 times under the same conditions, then therewill be 1.14 occasions when this job will be completed in 31 days.

Example 5.15: Using the data given in the previous example, find the probability ofcompleting the jobs on the critical path in 41 days.

Solution: Refer the previous example to find the critical path, project length and projectlength variance.

The probability of completing the project within 41 days is given by,

P (Z D), where D = s eT T

41 44 30.5

6 6

i.e., P (Z – 0.5) = 0.5 – (0.5)

= 0.5 – 0.1915 (from the table)= 0.3085 = 30.85%

Conclusion: If the project is performed 100 times under the same condition, then therewill be 30.85 occasions, when the project will be completed in 41 days.

Example 5.16: Assuming that the expected times are normally distributed, find theprobability of meeting the schedule date as given for the network.

Activity Days(i–j) Optimistic Most Likely Pessimistic

a m b

1–2 2 5 141–3 9 12 152–4 5 14 173–4 4 4 104–5 8 17 203–5 6 6 12

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Scheduled project completion date is 30 days. Also, find the date on which theproject manager can complete the project with a probability of 0.90.

Solution: The expected time te and variance for each activity is calculated in the

following table:Activity te = (a + 4m + b)/6 2 = ((b – a)/)2

1–2 6 41–3 12 12–4 13 43–4 5 13–5 16 44–5 7 1

To determine the critical path, the earliest expected time and the latestallowable time, first we draw the project network as follows:

The critical path is given by 1 3 5 and the project duration is given by 28 days.Project length variance = 2 = 1 + 4 = 5. Standard deviation = 2 = 2.236.

The probability of completing the project within 30 days is given by,

P (Z D), where D =30 28

0.89442.236

s eT T

P (Z 0.8944) = 0.5 + (0.8944)

= 0.8133

= 81.33%

Conclusion: If the project is performed 100 times under the same conditions, thenthere will be 81.33 occasions when the project will be completed within 30 days.

If the probability for the completion of the project is 0.90 then thecorresponding value of Z = 1.29.

1.29s eT TZ

28

i.e., 1.292.236sT

Ts = (1.29) (2.236) + 28

Ts = 30.88 weeks

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1. What are the management functions for the three phases of workinvolved in a project?

2. What is planning?

3. What is an activity?

4. What is an event?

5. What is a network?

6. What are merge events and burst events?

7. What is a dummy activity?

8. What is a redundant activity or redundancy?

9. What is a critical activity and critical path?

5.7 CRASHING

Crashing of a project means intentionally reducing the duration of a project byallocating more resources to it. A project can be crashed by crashing its criticalactivities (because the duration of a project is dependent upon the duration of itscritical activities). The use of more workers, better equipment, overtime, etc, wouldgenerate higher direct costs for individual activities. However, shortening the overalltime of the project would also reduce certain fixed and overhead expenses ofsupervision, as well as indirect costs that vary with the length of the project.

We know that by adding more resources, the duration of an activity can bereduced. If an activity gets completed in ten days with five men working on it, thesame activity can be finished in say, six days with ten men (exact mathematicalrelationships do not work here) working on it. The initial direct cost was 50 man-days (5 men x 10 days) and now it is 60 man-days (10 men x 6 days). Therefore, thedirect cost has increased by 10 man-days.

At the same time, because of the decrease in duration of the activity by fourdays, the indirect cost (cost of supervision) decreases. Hence, we can conclude thatthe direct and indirect costs are inversely proportional to each other, i.e., when oneincreases, the other decreases.

An activity can be crashed by adding more resources only up to a definitelimit. Beyond this limit, the duration of the activity does not decrease by addingmore resources. This is due to decreasing efficiency of labour, and also increasingconfusion due to a large number of resources. In our example, if we increase thenumber of workers to 15, the same activity can probably be done in four days; butby adding ten more men (so that 20 men work on this activity), the activity timemay not decrease further.

Crashing:Intentionally reducingthe duration of aproject by allocatingmore resources to it.

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The crash time is the shortest time that could be achieved if all efforts(at any reasonable cost) were made to reduce the activity time. The limit beyondwhich the duration of the activity does not decrease by adding any amount of resourcesis called the crash time. It is the shortest possible activity time.

The direct cost associated with each crash time is called the crash cost.

The normal time (10 days in our example) can be defined as the duration of anactivity when the minimum possible resources required for its performance aredeployed. The lowest expected activity costs are called the normal costs.

Project direct cost is the direct cost involved in all the activities of the project.

Project indirect cost is the costs associated with sustaining a project. It includesthe cost of supervision during the implementation of the project, overheads, facilities,penalty costs and lost incentive payments. The salaries paid to the project manager/supervisor, etc., miscellaneous costs due to delays in the project, and rewards to theproject team members for its early completion are indirect costs. Project indirect costis dependent upon other length of duration of the project. A project having a longerduration will have a higher indirect cost (due to supervision required for longerduration).

In any project, there is a direct relationship between the time taken to completean activity and the project cost. On one hand, it costs money to expedite a project.Costs associated with expediting a project are called activity direct costs, and aredifferent from project direct costs. Some examples of activity direct costs are—hiringmore workers, buying or leasing additional equipment, drawing on additional supportfacilities, etc.

If activity direct costs rise, project indirect costs will fall. Therefore, in a realsituation, we need to have a time–cost trade-off; this means the sum of activity directcosts and project indirect costs must be minimum.

During the process of crashing a project, the critical path may get changed. Atsome stage of crashing, there may even be two or more critical paths (having thesame duration) simultaneously. In such situations, one activity is chosen from each ofthe critical paths and these activities are crashed by unit time to reduce the duration ofthe project by unit time.

Time–cost models search for the optimum reductions in time. We seek to shortenthe length of a project to the point where the savings in indirect project costs is offsetby the increased direct expenses incurred in the individual activities.

Example 5.17: A network has four activities with expected times as shown. Theminimum feasible times and cost per day to gain reductions in the activity times arealso shown. If fixed project costs are ` 90 per day, what is the lowest cost timeschedule?

Crash time: Theshortest time that couldbe achieved if allefforts (at anyreasonable cost) weremade to reduce theactivity time.

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(All costs are in ` ‘000)

1

2

3

46

4

4

5

Activity Minimum Time in Days Direct Costs of Time Reduction (`)1 – 21 – 32 – 43 – 4

2243

40 ( each day)35 (first day, 80 (second day)None possible45 (first day), 110 (other days)

Solution:

First we must determine the critical path and critical path time cost.

Path Path Times Total Project Cost1–2, 2–41–3, 3–4

5 + 4 = 94 + 6 = 10 10 days × ` 90/day = $ 900

For ease of reference, let us call the paths A and B respectively. Path B that is, 1–3–4 is the critical path with duration 10 days and cost ` 900.

Next, we must select the activity that can reduce critical path time at the leastcost. Select activity 1–3 at ` 35 per day, which is less than the ` 90 per day fixedcost. Reduce activity 1–3 to 3 days. Revise the critical path time cost.

Revised Path Times Total Fixed Cost Savings over Previous ScheduleA: 5 + 4 = 9B: 3 + 6 = 9

9 × ` 90 = ` 810 ` 900 – ( 810 + 35) = ` 55

Both paths are now critical, so we must select an activity on each path. Selectactivity 1–2 at ` 40 per day and 3–4 at ` 45 per day. Reduce activity 1–2 to 4 daysand 3–4 to 5 days. Revise the critical path time cost.

Revised Path Times Total Fixed Cost Savings over Previous Schedule

A: 4 + 4 = 8B: 3 + 5 = 8

8 × 90 = ` 720 810 – (720 + 40 + 45) = ` 5

Let us see if we can reduce the time of both paths any further. Activity 1–2 isa good candidate on Path A, for it is still at 4 days and can go to 3 for a ` 40 cost. Butwhen this cost is combined with the ` 80 cost for reducing activity 1–3 another day,the sum is greater than ` 90, so further reduction is not economically justified.

The final step in time–cost analysis is to compare the crash times and the costsassociated with them (crash costs). A sufficient number of intermediate schedules arecomputed such that the total of the direct and indirect (fixed) project costs can beplotted.

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Example 5.18: Graph the total relevant costs for the previous example, and indicatethe optimal time–cost trade-off value.

ProjectLength(days)

IndirectCost

ActivityReduced

Relevant DirectCost

Relevant Total Cost

109876

900810720630540

None1 – 31 – 2 and 3 – 41 – 2 and 1 – 31 – 2 and 3 – 4

0 3535 + 85 = 120120 + 120 = 240240 + 150 = 390

900810 + 35 = 845720+40+45+35 = 840870930

cost .920

.880

.840 .

800

6 7 8 9 10 days

This graph is called the crash–time diagram for completing the project. Thelowest total cost is to complete the project in 8 days at a cost of ` 840,000. However,extending it to 9 days adds only ` 5000 to this cost.

5.8 RESOURCE ALLOCATION AND LEVELLING

Resource allocation (also known as resource scheduling) implies the task of allocationof resources to various activities in such a manner that the allocation is consideredas acceptable under the given situation. The task of allocation of resources is importantas the final schedule depends upon the quantity of deployment of resources. Thebasic question then is, How should the resources be allocated?

This depends upon several factors such as availability of resources, requirementsrestrictions in regard to completion date, and so on. Various types of problems maybe encountered in this connection, but we shall consider only two of such problems:(i) Resource levelling (ii) Limited resource allocation.

(i) Resource levelling (also known as local smoothing) means the resourcescheduling exercise in which the resource demand is evened out or levelledas much as possible. In other words, resource levelling refers to the schedulingof activities within the limits of the available floats in such a way that variationsin resources requirements are minimized.

(ii) Limited resource allocation We often find that projects require costly itemsof plant and equipment for the execution of work of which only a limited

Resource levelling:The resourcescheduling exercise inwhich the resourcedemand is evened outor levelled as much aspossible.

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number is available. Such limited resources must be allocated with a lot of care sothat the total requirement should not exceed the ceiling and the utilization factorremains high. This necessitates rescheduling of some or all of the activities andmay even involve delay in the overall completion of the project.

For resource levelling the methodology involves the following steps:

(i) Prepare the list of resources that would be required for the execution of variousactivities.

(ii) Prepare the resource profiles for each resource by resource aggregationexercise.

(iii) Identify the periods of peak and low demands.

(iv) Make an attempt to lower the peak to fill up the troughs, i.e., to make thedemand as uniform as possible. This can be done by altering the times of startand finish of non-critical activities in accordance of their floats withoutaffecting the overall completion date of the project:

Example 5.19: Workout the aggregate resource requirements, day by day for thefollowing network.

1

2

3

4

5

6

7

8

(9)

(3)

(6) (4)(7)

(2)

(1)

(8)4

3

2

3

2

2

4

The figures over the arrows in brackets indicate the requirement of labour and figuresbelow the arrows show the duration of an activity (in days). Presuming no resourceconstraint, smooth out the requirements of the resource (i.e., labour) to the extentpossible and outline the revised schedule.

Solution:

For finding the aggregate resource requirements day by day, it would be helpful topresent the given network as shown below:

1 2 3 4 5 6 7 8 9 10 11

22 22 24 10 2 2 1 1 1 1

DAYS

1

24

2

8764

5

3

(9)

(6)

(4)

(7) (1)

(8)

(2)(3)

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The manpower requirements vary from 1–24 and variations of this size are not acceptable.So it is necessary to smoothout the aggregate requirements. This can be done by modifyingthe input of resources to non-critical activities. To begin with activities 1–4, 4–6, 6–7,and 7–8 are critical and have to be completed without delay. Activities 1–2, 1–3, 3–7 and1–5 are non-critical activities. They can be delayed with in their floats. Accordingly, letus now compute the float of each activity.

Start Finish

Activity ET LT ET LT Float

1 – 2 0 7 4 11 71 – 3 0 2 2 4 21 – 4 0 0 2 2 01 – 5 0 2 2 4 23 – 7 2 4 4 7 24 – 6 2 2 5 4 06 – 7 4 4 7 7 07 – 8 7 7 11 11 0

Activity 1–2 can be delayed for 7 days without affecting the overall projectcompletion time. If we postpone this activity for 7 days we get the following schedule:

1 2 3 4 5 6 7 8 9 10 11

13 1513 15 1010101010 2 2

1

2

3

4

5

6 7 8

(3)

(4)

(2) (1)

(9)

Thus, with above stated modification, the variation in manpower requirements hasbeen smoothened from 1–24 to 2–15. We can make some further modifications aswell. Activity 1–3 can be delayed for 2 days without affecting the overall projectcompletion time and if we do so, we get the following schedule:

1 2 3 4 5 6 7 8 9 10 11

1010 10 10 10 10 10 10 10 10 10

Days

MainpowerRequirements

1

2

3

4

5

6 7 8(6) (7) (2) (1)

(9)

(3)

(4)

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Thus, the pattern of resource demand has further improved. Perfect levelling has beenmade possible in this particular case.

The methodology for limited resource problem involves the following:

(i) Attempt to lower the peak requirements of the resources by staggering the inputson non-critical activities. At times, it may become necessary to tackle some criticalactivities for the purpose because of limited resource constraint and this maymean even delays in completion of the work.

(ii) It is advantageous to re-arrange the activities in the descending order of themagnitude of their floats as resources can be diverted from the activitieshaving large amounts of floats. Critical activities should only be tackled inthe end, if necessary.

(iii) The following four rules (heuristics) for resource allocation in projectscheduling with limited resources facilitate the work:(a) Allocate the resources to activities serially in line (i.e. in the order the

activities become due for start.)(b) When several activities compete for the same resources give preference

to activities having minimum float.(c) If two competing activities happen to have equal floats, allocate the

resources to the activity having lower duration.(d) Reschedule non-critical jobs; it possible to free resources for scheduling

critical jobs.


1. What is PERT? Where is it used?

2. What are optimistic time, most likely time and pessimistic time foractivities in a project?

3. What is expected time for an activity in a project?

ACTIVITY

1. The total float of an activity i – j is 18. The latest and earliest occurrence ofevents i and j are 15, 12 and 22, 10 respectively. Find the free float.

2. What is independent float when the total float of an activity i – j is 18?Latest and earliest occurrence of events i and j are 15, 12 and 22, 10respectively.

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5.9 LET US SUM UP

Network scheduling is a technique used for planning and scheduling large projectsin the field of construction, maintenance, fabrication, purchasing computer system,etc.

PERT and CPM are the two basic planning and control techniques that utilize anetwork to complete a predetermined project or schedule.

Dummies are activities which neither consume time nor resources, but are usedsimply to represent a connection between events.

Activity time is a forecast of the time an activity is expected to take from itsstarting point to its completion.

Latest event time indicates the time by which all activities entering into thatevent must be completed without delaying the completion of the project.

Float is the difference between the latest and the earliest activity times.

An activity is said to be critical if a delay in its start will cause a further delayin the completion of the entire project.

The critical path method involves the preparation of a network in the form ofan arrow diagram and its analysis to indicate the critical path.

PERT is a probabilistic method, where the activity times are represented by aprobability distribution.

Crashing of a project means intentionally reducing the duration of a projectby allocating more resources to it. A project can be crashed by crashing itscritical activities.

Resource allocation implies the task of allocation of resources to variousactivities in such a manner that the allocation is considered as acceptableunder the given situation.




Kalavathy, S. Operations Research. 2002. New Delhi: Vikas Publishing House Pvt.Ltd.



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Check Your Progress–11. Management functions involved in three phases of work involved in a project

are planning, scheduling and controlling.

2. Planning is setting of objectives of the project by listing the tasks to be performedand resources available to complete the project.

3. An activity represents an action. It is an effort that consumes time that is neededto complete a part of the overall project.

4. An event is either start or end of an activity.

5. Network is a graphic representation of logically connected activities and eventswhere activities are presented as arrows and events as nodes.

6. Events that are ending of more than one activity are known as merge eventsand those which are beginning of more than one activity are known as burstevents.

7. An activity that consumes neither any resource nor time, but it is there on thenetwork to show a link between events, is known as a dummy activity.

8. A dummy activity that is the only activity emanating from an event is a redundantactivity which can be eliminated, and this phenomenon is known as redundancy.

9. An activity is critical if delay in its start will cause further delay in completionof the entire project and critical path is the sequence of all such activities inthe network.

Check Your Progress–21. PERT stands for Program Evaluation Review Technique. It is a probabilistic

method where activity times are represented by a probability distribution.PERT is used where activities involved in a project are non-deterministic innature.

2. Smallest time taken to complete an activity assuming that everything goeswell is optimistic time. Most likely time is the normal time taken by an activityassuming normal delay. Pessimistic time is the longest time required for anactivity to complete assuming everything going wrong.

3. Expected time for an activity in a project is denoted in a PERT chart and isgiven by the formula as given below:

Expected time = (Optimistic time + 4 Most likely time + Pessimistic time)/6.



1. What is a project?

2. What is dangling in a network? How can it be avoided?

3. Write basic two differences between PERT and CPM.

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Unit-5

4. How are time estimates used in PERT and CPM?

5. What are the different types of float?

6. Differentiate between float and slack.

7. What is the total float of a critical activity?

8. What do you mean by resource levelling?


1. The following table gives the activities and duration of a constructionproject.

20 25 10 12 6 10Duration(days)

Activity 1 - 2 1 - 3 2 - 3 2 - 4 3 - 4 4 - 5

(i) Draw the network for the project.(ii) Find the critical path.

2. A small project consits of 11 activities A, B, C, ..., K. The precedencerelationship A, B can start simultaneously. Given A<C, D, I; B<G, F; D<G, F;F<H, K; G, H<J; I, J, K<E. The duration of the activities are as follows.

5 3 10 2 8 4 5 6 12 8 9Duration(Days)

Activity A B C D E F G H I J K

Draw the network of the project. Summarize the CPM calculations in a tabularform computing the total, and free floats of activities and hence determine thecritical path.

3. Draw the network and determine the critical path for the given data. Also,calculate all the floats involved in CPM.

6 5 10 3 4 6 10 9Duration

Jobs 1 - 2 1 - 3 2 - 4 3 - 4 3 - 5 4 - 5 4 - 6 5 - 6

4. A small maintenance project consists of the following 12 jobs.

2 7 3 3 5 3 5 8 4 4 1 7Duration (Days)

Jobs 1 - 2 2 - 3 2 - 4 3 - 4 3 - 5 4 - 6 5 - 8 6 -7 6 - 10 7 - 9 8 - 9 9 - 10

Draw the arrow network of the project. Summarize CPM calculations in atabular form calculating the three types of floats and hence determine thecritical path.

5. Consider the following data for activities in a given project.

Predecessor – A – B,C C D,ETime (Days) 5 4 7 3 4 2

Activity A B C D E F

Network Models


Unit-5

Draw the arrow diagram for the project. Compute the earliest and the latestevent times. What is the minimum project completion time? List the activitieson the critical path.

6. For the following project, determine the critical path and its duration.

Predecessors – A A B B D,E D C,F,GTime (Days) 2 4 8 3 2 3 4 8

Activity A B C D E F G H

7. A project has the following time schedule.

Duration2 2 1 4 8 5 3 1 5 4 3

(Month)

Activity 1 - 2 1 - 3 1 - 4 2 - 5 3 - 6 3 -7 4 - 6 5 - 8 6 - 9 7 - 8 8 - 9

Construct the network and compute: (i) Total float for each activity(ii) Critical path and its duration

8. The data for a small PERT project is as given below, where a representsoptimistic time, m the most likely time and b the pessimistic time. Estimates(in days) of the activities A, B, ..., J, K.

3 2 6 2 5 3 3 1 4 1 2

6 5 12 5 11 6 9 4 19 2 4

5 14 30 8 17 15 27 7 28 9 12

a

m

b

Activity A B C D E F G H I J K

A, B, C can start simultaneously; A D, I; B < G, F; D < G, F; C < E; E < H,K; F < H, K; G, H < J. (i) Draw the arrow network of the project. (ii) Calculate the earliest and the latest expected times to each event and

find critical path.(iii) What is the probability that the project will be completed 2 days later

than expected?

9. The three estimates for the activities of a project are given below:

Estimate Duration (Days)Activity a m b

1–2 5 6 7

1–3 1 1 7

1–4 2 4 12

2–5 3 6 15

3–5 1 1 1

4–6 2 2 8

5–6 1 4 7

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Unit-5

Draw the project network. Find out the critical path of the project and projectduration. What is the probability that the project will be completed at least 5days earlier than expected?

What is the probability that the project will be completed by 22 days?

10. Consider the network shown in the figure below. The estimate to, t

m and t

p are

shown in this order for each of the activities on the top of the arcs denotingthe respective activities.

Find the probability of completing the project in 25days.

11. A project is represented by the network shown below and has the followingtable:

Least time 5 18 26 16 15 6 7 7 3

Greatest time 10 22 40 20 25 12 12 9 5

Most likely time 8 20 33 18 20 9 10 8 4


Determine the following:(i) Expected tasks time and their variance.

(ii) The earliest and the latest expected time to reach each mode.(iii) The critical path.(iv) The probability of completing the project within 41.5 weeks.

12. Consider a project having the following activities and their time estimates.

Draw an arrow diagram for the project. Identify the critical path and computethe expected project completion time. What is the probability that the projectwill require atleast 75 days?

Activity Predecessor Days

t0 tm tp

A – 2 4 6

B A 8 12 16

C A 14 16 30

D B 4 10 16

E C,B 6 12 18

F E 6 8 22

G D 18 18 30

H F,G 8 14 32

13. Define ‘crashing’. When is crashing preferred? Explain the procedure ofcrashing a project.

OPERATIONS RESEARCHVENKATESHWARA

OPEN UNIVERSITYwww.vou.ac.in

OPERATIONS RESEARCH

MA [ECONOMICS][MEC-104]

OPERATIONS RESEARCH10 MM

VENKATESHWARAOPEN UNIVERSITY

www.vou.ac.in

venkateshwara open university research_ma... · 2020. 10. 29. · introduction to operations...

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