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M.Sc. MathematicsPaper-IVMAT-104

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CONTENTSUnits Page No.

1. Introduction of Operations Research And Convex Sets 1 2. Linear Programming : The Simplex Method 31 3. Duality in Linear Programming 92 4. Dynamic Programming and Theory of Games 160 5. Non-Linear Programming Methods 224

Self-Instructional Material 1

NOTES

Introduction of Operations Research

and Convex Sets

STRUCTURE

1.1 Introduction1.2 Origin of Operations Research1.3 Development of Operations Research1.4 DefinitionsofOperationsResearch1.5 Nature and Features of Operations Research Approach1.6 Models and Modelling in Operations Research1.7 Advantages of Model Building1.8 Methods for Solving Operations Research Models1.9 Methodology of Operations Research1.10 Advantages of Operations Research Study1.11 Opportunities and Shortcomings of the Operations Research Approach1.12 Limitations of Operations Research1.13 Features of Operations Research Solution1.14 Applications of Operations Research1.15 Operations Research Models in Practice1.16 Convex Sets1.17 Properties of Convex Sets1.18 Convex Functions1.19 Convex Combinations of Vectors1.20 Hyperplanes1.21 Separating and Supporting Hyperplane Theorems1.22 Hypersphere

U N I T

1INTRODUCTION OF OPERATIONS

RESEARCH AND CONVEX SETS

A. INTRODUCTION OF OR (Operations Research)

1.1 INTRODUCTIONKnowledge, innovations and technology are changing and hence decision-making

in today’s social and business environment has become a complex task due to little or no precedents. High cost of technology, materials, labour, competitive pressures and

Quantitative analysis is the scientificapproachto decision-making.

2 Self-Instructional Material

NOTES

Operations Research so many economic, social, political factors and viewpoints, have greatly increased the complexity of managerial decision-making. To effectively address the broader tactical and strategic issues, also to provide leadership in the global business environment, decision-makers cannot afford to make decisions based on their personal experiences, guesswork or intuition, because the consequences of wrong decisions can prove to be serious and costly. For example, entering the wrong markets, producing the wrong products,providinginappropriateservices,etc.,maycauseseriousfinancialproblemsfororganizations. Hence, an understanding of the use of quantitative methods to decision-making is desirable to decision-makers.

Theevaluationofeachalternativecanbeextremelydifficultortimeconsumingfortwo reasons: First, the amount and complexity of information that must be processed, and second the availability of large number of alternative solutions. For these reasons, decision-makers increasingly turn to quantitative factors and use computers to arrive at the optimal solution for problems.

There is a need for structural analysis using operations research/quantitative techniques to arrive at a holistic solution to any managerial problem. This can be done by critically examining the levels of interaction between the application process of operations research, and various systems of an organization. Figure 1.1 summarizes the conceptual framework of organizational/management structures and operations research application process.

This book introduces a set of operations research techniques that should help decision-makers in making rational and effective decisions. It also gives a basic knowledge of the use of computer software needed for computational purposes.

1.2 ORIGIN OF OPERATIONS RESEARCHIt is generally agreed that operations research came into existence as a discipline

during World War II when there was a critical need to manage scarce resources. However, a particular model and technique of OR can be traced back as early as in World War I, whenThomasEdison(1914–15)madeanefforttouseatacticalgameboardforfindinga solution to minimize shipping losses from enemy submarines, instead of risking ships in actual war conditions. About the same time AK Erlang, a Danish engineer, carried out experiments to study thefluctuations in demand for telephone facilitiesusingautomatic dialling equipment. Such experiments were later on used as the basis for the development of the waiting-line theory.

Since World War II involved strategic and tactical problems that were highly complicated, to expect adequate solutions from individuals or specialists in a single discipline was unrealistic. Thus, groups of individuals who were collectively considered specialists in mathematics, economics, statistics and probability theory, engineering, behavioural, and physical science, were formed as special units within the armed forces, in order to deal with strategic and tactical problems of various military operations.

SuchgroupswerefirstformedbytheBritishAirForceandlatertheAmericanarmed forces formed similar groups. One of the groups in Britain came to be known as Blackett’s Circus. This group, under the leadership of Prof. P M S Blackett was attached to the Radar Operational Research unit and was assigned the problem of analyzing the coordination of radar equipment at gun sites. Following the success of this group similar mixed-team approach was also adopted in other allied nations.

Decision-maker should consider both qualitative and quantitative factors while solving a problem.

Operations research approach considers environmental influencesalong with both organization structures, and managerial behaviour, for decision-making.


NOTES


and Convex Sets

Fig. 1.1. Conceptual Framework of an Organization and OR Application Process

After World War II, scientists who had been active in the military OR groups made efforts to apply the operations research approach to civilian problems related to business, industry, research, etc. The following three factors are behind the appreciation for the use of operations research approach: (i) The economic and industrial boom resulted in mechanization, automation and

decentralization of operations and division of management functions. This industrialization resulted in complex managerial problems, and therefore the application of operations research to managerial decision making became popular.

(ii) Continued research after war resulted in advancements in various operations research techniques. In 1947, G B Dantzig, developed the concept of linear programming, the solution of which is found by a method known as simplex method. Besides linear programming, many other techniques of OR, such as statistical quality control, dynamic programming, queuing theory and inventory theory were well-developed before 1950’s.

(iii) The use of high speed computers made it possible to apply OR techniques for solving real-life decision problems.

1.3 DEVELOPMENT OF OPERATIONS RESEARCHDuring the 1950s, there was substantial progress in the application of OR

techniques for civilian problems along with the professional development. Many colleges/schools of engineering, public administration, business management, applied

The term ‘operations research’ was coined as a result of conducting research on military operations during World War II


NOTES

Operations Research mathematics, computer science, etc. Today, however, service organizations such as banks, hospitals, libraries, airlines, railways, etc., all recognize the usefulness of OR in improvingefficiency.In1948,anORclubwasformedinEnglandwhichlaterchangedits name to the Operational Research Society of UK. Its journal, OR Quarterly firstappeared in 1950. The Operations Research Society of America (ORSA) was founded in 1952 and its journal, Operations Research wasfirstpublishedin1953.Inthesameyear, The Institute of Management Sciences (TIMS) was founded as an international societytoidentify,extendandunifyscientificknowledgepertainingtomanagement.Its journal, Management Science,firstappearedin1954.

1.4 DEFINITIONS OF OPERATIONS RESEARCHThe wide scope of applications of operations research encouraged various

organizationsandindividualstodefineitasfollows: • Operations research is the application of the methods of science to complex problems

in the direction and management of large systems of men, machines, materials and money in industry, business, government and defence. The distinctive approach is to develop a scientific model of the system incorporating measurements of factors such as chance and risk, with which to predict and compare the outcomes of alternative decisions, strategies or controls. The purpose is to help management in determining its policy and actions scientifically.

– Operational Research Society, UK • The application of the scientific method to the study of operations of large complex

organizations or activities. It provides top level administrators with a quantitative basis for decisions that will increase the effectiveness of such organizations in carrying out their basic purposes.

– Committee on OR National Research Council, USAThedefinitiongivenbyOperationalResearchSocietyofUKhasbeencriticized

because of the emphasis it places on complex problems and large systems, leaving the reader with the impression that it is a highly technical approach suitable only to large organizations. • Operations research is the systematic application of quantitative methods, techniques

and tools to the analysis of problems involving the operation of systems.– Daellenbach and George, 1978

• Operations research is essentially a collection of mathematical techniques and tools which in conjunction with a system’s approach, are applied to solve practical decision problems of an economic or engineering nature.

– Daellenbach and George, 1978ThesetwodefinitionsimplyanotherviewofOR–beingthecollectionofmodels

and methods that have been developed largely independent of one another. • Operations research utilizes the planned approach (updated scientific method) and

an interdisciplinary team in order to represent complex functional relationships as mathematical models for the purpose of providing a quantitative basis for decision-making and uncovering new problems for quantitative analysis.

– Thierauf and Klekamp, 1975

A key person in the post-war development of OR was George B Dantzig.

Operations research is concerned with scientificallydeciding how to best design and operate man-machine systems that usually require the allocation of scarce resources.


NOTES


and Convex Sets • This new decision-making field has been characterized by the use of scientific

knowledge through interdisciplinary team effort for the purpose of determining the best utilization of limited resources. – H A Taha, 1976ThesetwodefinitionsrefertotheinterdisciplinarynatureofOR.However,there

is nothing that can stop one person from considering several aspects of the problem under consideration. • Operations research, in the most general sense, can be characterized as the application

of scientific methods, techniques and tools, to problems involving the operations of a system so as to provide those in control of the operations with optimum solutions to the problems. – Churchman, Ackoff and Arnoff, 1957Thisdefinitionrefersoperationsresearchasa technique forselectingthebest

course of action out of the several courses of action available, in order to reach the desirable solution of the problem. • Operations research has been described as a method, an approach, a set of techniques,

a team activity, a combination of many disciplines, an extension of particular disciplines (mathematics, engineering, economics), a new discipline, a vocation, even a religion. It is perhaps some of all these things.

– S L Cook, 1977 • Operations research may be described as a scientific approach to decision-making

that involves the operations of organizational system.– F S Hiller and G J Lieberman, 1980

• Operations research is a scientific method of providing executive departments with a quantitative basis for decisions regarding the operations under their control.

– P M Morse and G E Kimball, 1951 • Operations research is applied decision theory. It uses any scientific, mathematical,

or logical means to attempt to cope with the problems that confront the executive, when he tries to achieve a thorough-going rationality in dealing with his decision problems. – D W Miller and M K Star, 1969

• Operations research is a scientific approach to problem-solving for executive management. – H M WagnerAs the discipline of operations research grew, numerous names such as Operations

Analysis, Systems Analysis, Decision Analysis, Management Science, Quantitative Analysis, Decision Science were given to it. This is because of the fact that the types of problems encountered are always concerned with ‘effective decision’.

1.5 NATURE AND FEATURES OF OPERATIONS RESEARCH APPROACH

Operations Research (OR) utilizes a planned approach following a scientific method and an interdisciplinary team, in order to represent complex functional relationship as mathematical models, for the purpose of providing a quantitative basis for decision-making and uncovering new problems for quantitative analysis.Thisdefinitionimpliesadditional features of OR approach. The broad features of OR approach in solving any decision problem are summarized as follows:

Operations research is the art of winning wars without actuallyfightingthem.

Operations research is the artoffindingbad answers to problems which otherwise have worse answers.


NOTES

Operations Research Interdisciplinary approach: For solving any managerial decision problem often an interdisciplinary teamwork is essential. This is because while attempting to solve a complex management problem, one person may not have the complete knowledge of all its aspects such as economic, social, political, psychological, engineering, etc. Hence, a team of individuals specializing in various functional areas of management should be organized so that each aspect of the problem can be analysed to arrive at a solution acceptable to all sections of the organization.

Scientific approach: Operations research is the application of scientific methods, techniques and tools to problems involving the operations of systems so as to provide those in control of operations with optimum solutions to the problems (Churchman et al.). Thescientificmethodconsistsofobservinganddefiningtheproblem;formulatingandtestingthehypothesis;andanalysingtheresultsofthetest.Thedatasoobtainedisthen used to decide whether the hypothesis should be accepted or not. If the hypothesis is accepted, the results should be implemented, otherwise not.

Holistic approach: While arriving at a decision, an operations research team examines the relative importance of all conflicting andmultiple objectives. It alsoexamines the validity of claims of various departments of the organization from the perspective of its implications to the whole organization.

Objective-oriented approach: An operations research approach seeks to obtain an optimal solution to the problem under analysis. For this, a measure of desirability (oreffectiveness)isdefined,basedontheobjective(s)oftheorganization.Ameasureofdesirabilitysodefinedisthenusedtocomparealternativecoursesofactionwithrespectto their possible outcomes.

Illustration: TheORapproachattemptstofindasolutionacceptabletoallsectionsof the organizations. One such situation is described below.

A large organization that has a number of management specialists is faced with thebasicproblemofmaintainingstocksoffinishedgoods.Tothemarketingmanager,stocks of a large variety of products are purely a means of supplying the company’s customers with what they want and when they want it. Clearly, according to a marketing manager, a fully stocked warehouse is of prime importance to the company. But the production manager argues for long production runs, preferably on a smaller product range,particularlyifasignificantamountoftimeislostwhenproductionisswitchedfrom one variety to another. The result would again be a tendency to increase the amount of stock carried but it is, of course, vital that the plant should be kept running. On the otherhand,thefinancemanagerseesstocksintermsofcapitalthatisunproductivelytied up and argues strongly for its reduction. Finally, there appears the personnel manager for whom a steady level of production is advantageous for having better labour relations. Thus, all these people would claim to uphold the interests of the organization, but they do so only from their own specialized points of view. They may come up with contradictory solutions and obviously, all of them cannot be right.

In view of such a problem that involves every section of an organization, the decision-maker, irrespective of his/her specialization, may require to seek assistance from OR professionals.

Remark: A system isdefinedasanarrangementof componentsdesigned toachieveaparticular objective or objectives according to plan. The components may either be physical or conceptual or both, but they all share a unique relationship with each other and with the overall objective of the system.

Operations research uses: (i) Interdisci-

plinary,(ii)Scientific,(iii) Holistic, and(iv) Objective-

oriented approaches to decision-making.

Operations research attempts to resolve the conflictsofinterestamong various sections of the organization and seekstofindtheoptimal solution that is in the interest of the organization as a whole.


NOTES


and Convex SetsTEST YOUR KNOWLEDGE – A

1. Brieflytracethehistoryofoperationsresearch.HowdidoperationsresearchdevelopafterWorld War II?

2. Is operations research (i) a discipline (ii) a profession (iii) a set of techniques (iv) a philosophy or a new name for old things? Discuss.

3. Discuss the following: (a) OR as an interdisciplinary approach. (b)ScientificmethodinOR. (c) OR as more than a quantitative analysis of the problem. 4. What are the essential characteristics of operations research? Mention different phases in

an operations research study. Point out its limitations, if any. 5. Defineoperationsresearchasadecision-makingscience. (a) Give the main characteristics of OR. (b) Discuss the scope of OR. 6. (a) Outline the broad features of the judgement phase and the research phase of the

scientificmethodinOR.Discussindetailanyoneofthesephases. (b)WhatarevariousphasesoftheORproblem?Explainthembriefly. (c) State the phases of OR study and their importance in solving problems. 7. What is the role of operations research in decision-making? 8. Defineoperationsresearch.Explaincriticallythelimitationsofvariousdefinitionsasyou

understand them. 9. Giveanythreedefinitionsofoperationsresearchandexplainthem. 10. (a) Discuss various phases of solving an OR problem. (b) State phases of an OR study and their importance in solving problems. 11. Discusstheroleandscopeofquantitativemethodsforscientificdecision-makinginabusiness

environment. 12. Discuss the advantage and limitations of operations research. 13. Writeacriticalessayonthedefinitionandscopeofoperationsresearch. 14. What post-World War II factors were important in the development of operations research? 15. Does the fact that OR takes the organizational point of view instead of the individual

problem-centered point of view generate constraints on its increased usage? 16. WhatwerethesignificantcharacteristicsofORapplicationsduringWorldWarII?What

caused the discipline of OR to take on these characteristics during that period? 17. Comment on the following statements: (a)ORistheartofwinningwarwithoutactuallyfightingit. (b)ORistheartoffindingbadanswerswhereworseexist. (c) OR replaces management by personality. 18. ExplaincriticallythelimitationsofanythreedefinitionsofORasyouunderstandthem. 19. Discussthesignificanceandscopeofoperationsresearchinmodernmanagement. 20. Quantitative techniques complement the experience and judgement of an executive in

decision-making. They do not and cannot replace it. Discuss. 21. Explainthedifferencebetweenscientificmanagementandoperationsresearch. 22. Decision-makers are quick to claim that quantitative analysis talks to them in a jargon that

does not sound like English. List four terms that might not be understood by a manager. Then explain, in non-technical terms, what each term means.

23. It is said that operations research increases the creative capabilities of a decision-maker. Do you agree with this view? Defend your point of view with examples.


NOTES

Operations Research 24. (a) Why is the study of operations research important to the decision-maker? (b) Operations research increases creative and judicious capabilities of a decision-maker.

Comment. 25. (a)‘Operationsresearchistheapplicationofscientificmethods,techniquesandtoolsto

problems involving the operations of a system so as to provide those in control of the system with optimum solutions to the problem.’ Discuss.

(b) ‘Operations research is an aid for the executive in making his decisions by providing himwiththeneededquantitativeinformation,basedonthescientificmethodanalysis.’Discuss this statement giving examples of OR methods that you know.

26. Comment on the following statements. (a) OR is a bunch of mathematical techniques. (b) OR advocates a system’s approach and is concerned with optimization. It provides a

quantitative analysis for decision making. (c)ORhasbeendefinedsemi-facetiouslyastheapplicationofbigmindstosmallproblems. 27. Discuss the points to justify the fact that the primary purpose of operations research is

toresolvetheconflictsresultingfromthevarioussubdivisionsofthefunctionalareaslikeproduction,marketing,financeandpersonnelinanoptimal,nearoptimalorsatisfyingway.

28. How far can quantitative techniques be applied in management decision-making? Discuss, in detail, with special reference to any functional area of management pointing out their limitations, if any.

1.6 MODELS AND MODELLING IN OPERATIONS RESEARCHModels do not, and cannot, represent every aspect of a real-life problem/system

because of its large and changing characteristics. However, a model can be used to analyze, understand and describe certain aspects (key features) of a system for the purpose of improving its performance as well as to examine changes (if any) without disturbing the ongoing operations.For example, to study theflowofmaterial in amanufacturingfirm,ascaleddiagram(descriptivemodel)onpapershowingthefactoryfloor,positionofequipment, tools,andworkerscanbeconstructed. Itwouldnotbenecessary to give details such as the colour of machines, the heights of the workers, or the temperature of the building.

The key to model-building lies in abstracting only the relevant variables that affect the criteria of the measures-of-performance of the given system and in expressing the relationship in a suitable form. However, a model should be as simple as possible so as to give the desired result. On the other hand, oversimplifying the problem can also lead to a poor decision. Model enrichment is done by changing value of variables, and relaxing assumptions. The essential three qualities of any model are: • Validity of the model – model should represent the critical aspects of the system/

problem under study, • Usability ofthemodel–amodelcanbeusedforthespecificpurposes,and • Value of the model to the user.

Besides these three qualities, other consideration of interest are, (i) cost of the model and its sophistication, (ii) time involved in formulating the model, etc.

Aninformaldefinitionofmodelthatappliestoallofusisatoolforthinkingandunderstanding features of any problem/system before taking action. For example, a model tends to be formulated when (a) we think about what someone will say in response to

A model is an approximation or abstraction of reality which considers only the essential variables (or factors) and parameters in the system along with their relationships.


NOTES


and Convex Sets

our act(s), (b) we try to decide how to spend our money, or (c) we attempt to predict the consequences of some activity (either ours, someone else’s or even a natural event). In other words, we would not be able to derive or take any purposeful action if we do not formamodeloftheactivityfirst.ORapproachusesthisnaturaltendencytocreatemodels. This tendency forces to think more rigorously and carefully about the models we intend to use.

IngeneralmodelsareclassifiedineightcategoriesasshowninTable1.1.Suchaclassificationprovidesausefulframeofreferenceforpractioners/researchers.

Table 1.1.GeneralClassificationofModels

1. Function 4. Degree of certainty 7. Degree of closure

• Descriptive • Predictive • Normative

• Certainty • Conflict • Risk • Uncertainty

• Closed

• Open

2. Structure 5. Time reference 8. Degreeofquantification

• Iconic • Analog • Symbolic

• Static • Dynamic

• Qualitative ♦ Mental ♦ Verbal

• Quantitative ♦ Statistical ♦ Heuristic ♦ Simulation

3. Dimensionality 6. Degree of generality

• Two-dimensional • Multidimensional

• Specialized • General

Asummary classification ofORmodels based ondifferent criteria is given inFig.1.2andfewoftheseclassificationsarediscussedbelow:

1.6.1 ClassificationBasedonStructurePhysical models: These models are used to represent the physical appearance

of the real object under study, either reduced in size or scaled up. Physical models are useful only in design problems because they are easy to observe, build and describe. For example, in the aircraft industry, scale models of a proposed new aircraft are built and tested in wind tunnels to record the stresses experienced by the air frame.

Physical models cannot be manipulated and are not very useful for prediction. Problems such as portfolio selection, media selection, production scheduling, etc., cannot be analysed with the help of these models.

Physicalmodelsareclassifiedintotwocategories. (i) Iconic Models An iconic model is a scaled (small or big in size) version of the

system. Such models retain some of the physical characteristics of the system they represent.

Examples of iconic model are, blueprints of a home, maps, globes, photographs, drawings, air planes, trains, etc.

Physical model represents the physical appearance of the real object under study, either reduced in size or scaled up.


NOTES

Operations Research

Fig. 1.2. ClassificationofModels

An iconic model is used to describe the characteristics of the system rather than explaining the system. This means that such models are used to represent system’s characteristics that are not used in determining or predicting effects that take place due to certain changes in the actual system. For example, (i) colour of an atom does not playanyvitalroleinthescientificstudyofitsstructure,(ii) type of engine in a car has no role to play in the study of the problem of parking, etc. (ii) Analogue Models An analogue model does not resemble physically the system

they represent, but retain a set of characteristics of the system. Such models are more general than iconic models and can also be manipulated.

For example, (i)oildipstickinacarrepresentstheamountofoilintheoiltank;(ii) organizational chart represents the structure, authority, responsibilities and relationship,withboxesandarrows;(iii) maps in different colours represent water, desert and other geographical features, (iv) Graphs of time series, stock-market changes, frequency curves, etc., represent quantitative relationships between any two variables and predict how a change in one variable effects the other, and so on.

Symbolic models: These models use algebraic symbols (letters, numbers) and functions to represent variables and their relationships for describing the properties of the system. Such relationships can also be represented in a physical form. Symbolic models are precise and abstract and can be analysed by using laws of mathematics.

Mathematical model uses mathematical equations and statements to represent the relationships within the model.


NOTES


and Convex Sets

Symbolicmodelsareclassifiedintotwocategories. (i) Verbal Models These models describe properties of a system in written or

spoken language. Written sentences, books, etc., are examples of a verbal model. (ii) Mathematical Models These models use mathematical symbols, letters, num-

bers and mathematical operators (+, –, ÷, ×) to represent relationships among variables of the system to describe its properties or behaviour. The solution to such models is obtained by applying suitable mathematical technique.

Few examples of mathematical model are (i) the relationship among velocity, distance and acceleration. (ii)therelationshipamongcost-volume-profit,etc.

1.6.2 ClassificationBasedonFunction(orPurpose)Descriptive models: These models are used to investigate the outcomes or

consequences of various alternative courses of action (strategies, or actions). Since these models evaluate the consequence based on a given condition (or alternative) rather than on all other conditions, there is no guarantee that an alternative selected is optimal. These models are usually applied (i) in decision-making where optimization models are not applicable, and (ii)whenobjectiveistodefinetheproblemortoassessits seriousness rather than to select the best alternative. These models are especially used for predicting the behaviour of a particular system under various conditions.

Simulation is an example of a descriptive model for conducting experiments with the systems based on given alternatives.

Predictive models: These models represent a relationship between dependent and independent variables and hence measure ‘cause and effect’ due to changes in independent variables.

These models do not have an objective function as a part of the model of evaluating decision alternatives based on outcomes or pay off values. Also, through such models decision-maker does not attempt to choose the best decision alternative, but can only have an idea about the possible alternatives available to him.

For example, the equation S = a + bA + cI relates dependent variable (S) with other independent variables on the right hand side. This can be used to describe how the sale (S) of a product changes with a change in advertising expenditure (A) and disposable personal income (I). Here, a, b and c are parameters whose values must be estimated. Thus, having estimated the values of a, b and c, the value of advertising expenditure (A) can be adjusted for a given value of I, to study the impact of advertising on sales. Also, through such models decision-maker does not attempt to choose the best decision alternative, but can only have an idea about the possible alternative available to him.

Normative (or Optimization) models: These models provide the ‘best’ or ‘optimal’ solution to problems using an appropriate course of action (strategy) subject to certain limitations on the use of resources. For example, in mathematical programming, models are formulated for optimizing the given objective function, subject to restrictions on resources in the context of the problem under consideration and non-negativity of variables. These models are also called prescriptive models because they prescribe what the decision-maker ought to do.


NOTES

Operations Research 1.6.3 ClassificationBasedonTimeReferenceStatic models: Static models represent a system at a particular point of time

and do not take into account changes over time. For example, an inventory model can be developed and solved to determine an economic order quantity assuming that the demand and lead time would remain same throughout the planning period.

Dynamic models: Dynamic models take into account changes over time, i.e., time is considered as one of the variables while deriving an optimal solution. Thus, a sequence of interrelated decisions over a period of time are made to select the optimal course of action in order to achieve the given objective. Dynamic programming is an example of a dynamic model.

1.6.4 ClassificationBasedonDegreeofCertaintyDeterministic models: If all the parameters, constants and functional relationships

are assumed to be known with certainty when the decision is made, the model is said to be deterministic. Thus, the outcome associated with a particular course of action is known, i.e.,foraspecificsetofinputvalues,thereisonlyoneoutputvaluewhichisalsothe solution of the model. Linear programming models are example of deterministic models.

Probabilistic (Stochastic) models: If at least one parameter or decision variable is random (probabilistic or stochastic) variable, then the model is said to be probabilistic. Since at least one decision variable is random, the independent variable, which is the function of dependent variable(s), will also be random. This means consequences (or payoff) due to certain changes in the independent variable(s) cannot be predicted with certainty. However, it is possible to predict a pattern of values of both the variables by their probability distribution.

Insuranceagainstriskoffire,accidents,sickness,etc.,areexampleswherethepattern of events is studied in the form of a probability distribution.

1.6.5 ClassificationBasedonMethodofSolutionorQuantificationHeuristic models: If certain sets of rules (may not be optimal) are applied in

a consistent manner to facilitate solution to a problem, then the model is said to be Heuristic.

Analytical models: Thesemodelshaveaspecificmathematicalstructureandthuscan be solved by the known analytical or mathematical techniques. Any optimization model (which requires maximization or minimization of an objective function) is an analytical model.

Simulation models: These models have a mathematical structure but cannot be solved by the known mathematical techniques. A simulation model is essentially a computer-assisted experimentation on a mathematical structure of a problem in order to describe and evaluate its behaviour under certain assumptions over a period of time.

Simulationmodelsaremoreflexiblethanmathematicalmodelsandcan,therefore,be used to represent a complex system that cannot be represented mathematically. These models do not provide general solution like those of mathematical models.

In a deterministic model all values used are known with certainty.

In a Probabilistic model all values used are not known with certainty;andareoften measured as probability values.


NOTES


and Convex Sets

1.7 ADVANTAGES OF MODEL BUILDINGModels, in general, are used as an aid for analysing complex problems. However,

general advantages of model building are as follows: 1. A model describes relationships between various variables (factors) present in

a system more easily than what is done by a verbal description. That is, models help the decision-maker to understand the system’s structure or operation in a better way. For example, it is easier to represent a factory layout on paper than to constructit.Itischeapertotryoutmodificationsofsuchsystemsbyrearrangementon paper.

2. The problem can be viewed in its entirety, with all the components being considered simultaneously.

3. Models serve as aids to transmit ideas among people in the organization. For example, a process chart can help the management to communicate better work methods to workers.

4. A model allows to analyze and experiment on a complex system which would otherwisebeimpossibleontheactualsystem.Forexample,theexperimentalfiringof INSAT satellite may cost millions of rupees and require years of preparation.

5. Modelsconsiderablysimplifytheinvestigationandprovideapowerfulandflexibletool for predicting the future state of the system (or process).

1.8 METHODS FOR SOLVING OPERATIONS RESEARCH MODELS

In general, the following three methods are used for solving OR models, where values of decision variables are obtained that optimize the given objective function (a measure of effectiveness).

Analytical (or deductive) method: In this method, classical optimization techniquessuchascalculus,finitedifferenceandgraphsareusedforsolvinganORmodel. The analytical methods are non-iterative methods to obtain an optimal solution of a problem. For example, to calculate economic order quantity (optimal order size), theanalyticalmethodrequiresthatthefirstderivativeofthemathematicalexpression

TC = (D/Q) C0 + (Q/2) Ch

where TC=totalvariableinventorycost; D=annualdemand; C0=orderingcostperorder; Ch=carryingcostpertimeperiod; Q = size of an order

betakenandequatedtozeroasthefirststeptowardscalculating,EOQ(Q) = 2 0DC Ch/ .This is because of the concept of maxima and minima used for optimality.

Numerical (or iterative) method: When analytical methods fail to obtain the solution of a particular problem due to its complexity in terms of constraints or number ofvariables,anumerical(oriterative)methodisusedtofindthesolution.Inthismethod,instead of solving the problem directly, a general algorithm is applied for obtaining a specificnumericalsolution.

The numerical method starts with a solution obtained by trial and error, and a set of rules for improving it towards optimality. The solution so obtained is then replaced by

In general, operations research models are solved using either (i) Analytical,(ii) Numerical, or(iii) Monte Carlo

method


NOTES

Operations Research the improved solution and the process of getting an improved solution is repeated until suchimprovementisnotpossibleorthecostoffurthercalculationcannotbejustified.

Monte Carlo method: This method is based upon the idea of experimenting on amathematicalmodelbyinsertingintothemodelspecificvaluesofdecisionvariablesfor a selected period of time under different conditions and then observing the effect on thecriterionchosen.Inthismethod,randomsamplesofspecifiedrandomvariablesaredrawn to know how the system is behaving for a selected period of time under different conditions. The random samples form a probability distribution that represents the real-life system and from this probability distribution, the value of the desired random variable can then be estimated.

1.9 METHODOLOGY OF OPERATIONS RESEARCHEvery OR specialist may have his/her own way of solving problems. However, the

effective use of OR techniques requires to follow a sequence of steps. Each of these steps is described in detail in Fig. 1.3.

Fig. 1.3. Methodology of Operations Research

1.10 ADVANTAGES OF OPERATIONS RESEARCH STUDYIn addition to learning about various OR models that are useful to arrive at an

optimal decision, the knowledge of operations research techniques can also help the decision-maker to attain several other advantages as listed below. 1. Structured approach to problems: A substantial amount of time and effort can

be saved in developing and solving OR models if a logical and consistent approach isfollowed.Thisimpliesthatthedecision-makerhastobecarefulwhiledefiningvariables, availability of resources, and functional relationships among variables in the objective function and constraints. This will also reduce the chance of conceptual and computational errors. Any such error can also be detected easily and corrected at an early stage.


NOTES


and Convex Sets

2. Critical approach to problem solving: The decision-maker will come to understand various components of the problem and accordingly select a mathematical model for solving the given problem. He will become aware of the explicit and implicit assumptions and limitations of such models. Problem solutions are examined critically and the effect of any change and error in the problem data can be studied through sensitivity analysis techniques.

1.11 OPPORTUNITIES AND SHORTCOMINGS OF THE OPERATIONS RESEARCH APPROACH

The use of quantitative methods is appreciated to improve managerial decision-making. However, besides certain opportunities OR approach has not been without its shortcomings. The main reasons for its failure are due to unawareness on the part of decision-makers about their own role, as well as the avoidance of behavioural/organizational issues while constructing a decision model. A few opportunities and shortcomings of the OR approach are listed below.

OpportunitiesThe OR approach of problem solving facilitates.

•decision-makertodefinedexplicitlyhis/herobjectives,assumptionsandcon-straints.

•decision-makertoidentifyvariablesthatmightinfluencedecisions. • in identifying gaps in the data required to support solution to a problem. • the use of computer software to solve a problem.

Shortcomings • The solution to a problem is often derived by simplifying assumptions. Conse-

quently, such solutions are not optimal. • Models may not represent the realistic situations in which decisions are made. • Often the decision-maker is not fully aware of the limitations of the models.

1.12 LIMITATIONS OF OPERATIONS RESEARCHOperation Research is an extremely powerful tool in the hands of a decision-maker

and to that extent the advantage of OR techniques are immense. Some of them are: (a) It helps in optimum use of resources. LP techniques suggest many methods of

mosteffectiveandefficientwaysofoptimallyusingtheproductionfactors. (b) Quality of decision can be improved by suitable use of OR techniques. If a

mathematical model representing the real life situation is well-formulated representing the real life situation, the computation tables give a clear picture of the happenings (changes in the various elements i.e., variables) in the model. The decision-maker can use it to his advantage, specially if computerised soft-ware can be used to make changes in variables as per requirement.


NOTES

Operations Research The limitations of OR emerge only out of the time and cost involved as also the problem of formulating a suitable mathematical model, otherwise, as suggested above, it is very powerful medium of getting the best out of limited resources. So, the problem is its application rather than its utility , which is beyond doubt. Some of the limitations are: (a) Large number of cumbersome computations. Formulation of mathematical

modelswhichtakesintoaccountallpossiblefactorswhichdefineareallifeproblemisdifficult.Becauseofthis,thecomputationsinvolvedindevelopingrelationships in very large variables need the help of computers. This discour-ages small companies and other organisations from getting the best out of OR techniques.

(b) Quantification of problems. Alltheproblemscannotbequalifiedproperlyasthere are a large number of intangible factors, such a human emotions, human relationship and so on. If these intangible elements / variables are excluded from the problem even though they may be more important than the tangible ones, the best solution cannot be determined.

(c) Difficult to conceptualize and use by the managers. OR applications is a special-ist’s job, these persons may be mathematicians or statisticians who understand theformulationofmodels,findingsolutionsandrecommendingtheimplemen-tation. The managers really do not have the hang of it. Those who recommend a particular OR technique may not understand the problem well enough and those who have to use may not understand the ‘why’ of that recommendation. This creates a ‘gap’ between the two and the results may not be optimal.

1.13 FEATURES OF OPERATIONS RESEARCH SOLUTIONA solution that is expensive as compared to the potential savings from its

implementation, should not be considered. Even, if a solution is well within the budget butdoesnot fulfill theobjective isalsonotacceptable.Fewrequirementsofagoodsolution are: 1. Technically Appropriate: The solution should work technically, meet the

constraints and operate in the problem environment. 2. Reliable: The solution must be useful for a reasonable period of time under the

conditions for which it was designed. 3. Economically Viable: Its economic value should be more than what it costs to

develop it and it should be seen as a wise investment in hiring OR talent. 4. Behaviourally Appropriate: The solution should be behaviourally appropriate

and must remain valid for a reasonable period of time within the organization.

1.14 APPLICATIONS OF OPERATIONS RESEARCHSome of the industrial/government/business problems that can be analysed by the

OR approach have been arranged by functional areas as follows:

FinanceandAccounting • Dividend policies, investment and portfolio management, auditing, balance sheet

andcashflowanalysis

A solution needs to be: (i) technically

appropriate (ii) reliable, and(iii) economically

viable


NOTES


and Convex Sets

• Claim and complaint procedure, and public accounting • Break-evenanalysis,capitalbudgeting,costallocationandcontrol,andfinancial

planning • Establishing costs for by-products and developing standard costs

Marketing • Selection of product-mix, marketing and export planning • Advertising, media planning, selection and effective packing alternatives • Sales effort allocation and assignment • Launching a new product at the best possible time • Predicting customer loyalty

Purchasing,ProcurementandExploration • Optimal buying and reordering with or without price quantity discount • Transportation planning • Replacement policies • Bidding policies • Vendor analysis

ProductionManagementFacilities planning • Location and size of warehouse or new plant, distribution centres and retail outlets • Logistics, layout and engineering design • Transportation, planning and scheduling

Manufacturing • Aggregate production planning, assembly line, blending, purchasing and inventory

control • Employment, training, layoffs and quality control • Allocating R and D budgets most effectively

Maintenance and project scheduling • Maintenance policies and preventive maintenance • Maintenance crew size and scheduling • Project scheduling and allocation of resources

Personnel Management • Manpower planning, wage/salary administration • Designing organization structures more effectively • Negotiation in a bargaining situation • Skills and wages balancing • Scheduling of training programs to maximize skill development and retention


NOTES

Operations Research Techniques and General Management • DecisionsupportsystemandMIS;forecasting • Making quality control more effective • Project management and strategic planning

Government • Economic planning, natural resources, social planning and energy • Urban and housing problems • Military, police, pollution control, etc.

1.15 OPERATIONS RESEARCH MODELS IN PRACTICEThere is no unique set of problems that can be solved by using OR techniques.

Several OR techniques can be grouped into some basic categories as given below. In this book, a large number of OR techniques have been discussed in detail. • Allocation models: Allocation models are used to allocate resources to activities

so as to optimize measure of effectiveness (objective function). Mathematical programming is the broad term for the OR techniques used to solve allocation problems.

Ifthemeasureofeffectivenesssuchasprofit,cost,etc.,isrepresentedasalinearfunction of several variables and limitations on resources (constraints) are expressed asasystemoflinearequalitiesorinequalities,theallocationproblemisclassifiedas a linear programming problem. But, if the objective function or all constraints cannot be expressed as a system of linear equalities or inequalities, the allocation problemisclassifiedasanon-linearprogrammingproblem.

When the value of decision variables in a problem is restricted to integer values or justzero-onevalues,theproblemisclassifiedasanintegerprogrammingproblemor a zero-one programming problem, respectively.

Aproblemhavingmultiple,conflictingandincommensurableobjectivefunctions(goals) subject to linear constraints is called a goal programming problem. If values of decision variables in the linear programming problem are not deterministic, then such a problem is called a stochastic programming problem.

If resources (such as workers, machines or salesmen) have to be assigned to perform a certain number of activities (such as jobs or territories) on a one-to-one basis so as to minimize total time, cost or distance involved in performing a given activity, suchproblemsareclassifiedasassignmentproblems.Butiftheactivitiesrequiremore than one resource and conversely if the resources can be used for more than oneactivity,theallocationproblemisclassifiedasatransportationproblem.

• Inventory models: These models deal with the problem of determination of how much to order at a point in time and when to place an order. The main objective istominimizethesumofthreeconflictinginventorycosts:thecostofholdingorcarrying extra inventory, the cost of shortage or delay in the delivery of items when it is needed and the cost of ordering or set-up. These are also useful in dealing with quantity discounts and selective inventory control.

• Waiting line (or Queuing) models: These models establish a trade-off between costs of providing service and the waiting time of a customer in the queuing system.


NOTES


and Convex Sets

A queuing model describes: Arrival process, queue structure and service process and solution for the measure of performance – average length of waiting time, averagetimespentbythecustomerintheline,trafficintensity,etc.,ofthewaitingsystem.

• Competitive (Game Theory) models: These models are used to characterize the behaviour of two or more competitors (called players) competing to achieve their conflictinggoals.Thesemodelsareclassifiedaccordingtoseveralfactorssuchasnumber of competitors, sum of loss and gain, and the type of strategy which would yield the best or the worst outcomes.

• Network models: These models are applied to the management (planning, controlling and scheduling) of large-scale projects. PERT/CPM techniques help in identifying delay and project critical path. These techniques improve project coordinationandenabletheefficientuseofresources.Networkmethodsarealsoused to determine time-cost trade-off, resource allocation and help in updating activity time.

• Sequencing models: These models are used to determine the sequence (order) in which a number of tasks can be performed by a number of service facilities such as hospital, plant, etc., in such a way that some measure of performance (such as total time to process all the jobs on all the machines) is optimized.

• Replacement models: These models are used to calculate optimal time to replace anequipmentwheneitheritsefficiencydeteriorateswithtimeorfailsimmediatelyand completely.

• Dynamic programming models: These models are used where a problem requires optimization of multistage (sequence of interrelated decisions) decision processes. The method starts by dividing a given problem into stages or sub-problems and then solves those sub-problems sequentially until the solution to the original problem is obtained.

• Markov-chain models: These models are used for analyzing a system which changes over a period of time among various possible outcomes or states. That is, these models describe transitions in terms of transition probabilities of various states.

• Simulation models: These models are used to evaluate alternative courses of action by experimenting with a mathematical model of the problems with random variables. Thus, repetition of the process by using the simulation model provides an indication of the merit of alternative course of action with respect to the decision variables.

• Decision analysis models: These models deal with the selection of an optimal course of action given the possible payoffs and their associated probabilities of occurrence. These models are broadly applied to problems involving decision making under risk and uncertainty.

B.CONVEXSETS

1.16 CONVEX SETS In this lecture we look in more detail at convexity.Wecoverthedefinitionand

properties of convex sets and functions, and provide a toolkit of techniques to prove convexity. We will later use these techniques to show that particular optimization


NOTES

Operations Research problemsareconvex,andcanthusbeefficientlysolvedusingstandardmethodssuchas gradient and subgradient descent.

Definition. A set C is convex if the line segment between any two points in C lies in C, i.e., ∀x1, x2 ∈ C, ∀θ ∈ [0, 1]

θx1 + (1 – θ)x2 ∈ C.

Fig. 1.4. Example of convex set (left) and a non-convex set (right).

Simple examples of convex sets are: • The empty set f, the singleton set {x0}, and the complete space n; • Lines {aT x = b}, line segments, hyperplanes {AT x = b}, and halfspaces {AT x ≤ b}; • Euclidian balls B(x0, e) = {x|||x – x0 ||2 ≤e}.

Wecangeneralize thedefinitionofa convexsetabove fromtwopoints toanynumber of points n. A convex combination of points x1, x2, ..., xk ∈ C is any point of form

θ1x1 + θ2x2 + ... + θkxk, where θi ≥ 0, i = 1, ... k, i

k

i=∑ =1

1θ . Then, a set C is convex iff any convex combination of points in C is in C.

(a)(b)

Fig. 1.5. (a) Representation of a convex set as the convex hull of a set of points. (b) Represen-tationofaconvexsetastheintersectionofa(possiblyinfinite)numberofhalfspaces.

We can take this even further to infinite countable sums: C convex iff ∀xi ∈ C,

θi ≥ 0, i = 1, 2, ..., i

i=

∞

∑ =1

1θ :

ii ix C

=

∞

∑ ∈1

θ

if the series converges.Most generally, C is convex iff for any random variable X over C, (X ∈ C) = 1, its

expectation is also in C:(X) ∈ C.


NOTES


and Convex Sets

1.17 PROPERTIES OF CONVEX SETSIn different contexts, different representations of a convex set may be natural

or useful. In the following sections we introduce the convex hull and intersection of halfspaces representations, which can be used to show that a set is convex, or prove general properties about convex sets.

1.17.1ConvexHullDefinition: The convex hull of a set C is the set of all convex combinations of points

in C:conv(C) = {θ1x1 + ... + θkxk |xi ∈ C, θi ≥ 0, i = 1, ...k,

i

k

i=∑ =

11θ }

The convex hull of a set C is the smallest convex set which includes C: • conv(C) is convex • C ⊆ conv(C) • ∀C ′, C ′ convex, C ⊆ C′ ⇒ conv(C) ⊆ C′

Lemma: Any closed convex set C can be written as the convex hull of a possibly infinite set of points X:

C = hull(X )Indeed, any closed convex set is the convex hull of itself. However, we may by

abletofindasetXofmuchsmallerdimensionalitythanC,suchthatwestillhave C = hull(X ).

1.17.2 IntersectionofHalfspacesLemma: Any closed set C can be written as the possibly infinite intersection of a

set of halfplanes: C = ∩i {x|aix + bi ≤ 0 |}Indeed, any closed convex set is the intersection of all halfspaces that contain it: C = ∩{H|H halfspaces, C ⊆H }.However,wemaybeabletofindamuchsmallersetofhalfspacessuchthatthe

representation still holds.

1.17.3 ConvexityPreservingOperationsA standard way to prove that a set (or later, a function) is convex is to build it up

from simple sets for which convexity is known, by using convexity preserving operations. We present some of the basic operations below: • Intersection If C, D are convex sets, then C ∩ D is also convex. • Affine transform If C is a convex set, C ⊆ n, A ∈ mxn, b ∈ m, then

AC + b = {Ax + b|x ∈ C } ⊆ m

is also convex. In particular, the following operations also preserve convexity: – Translation C + b – Scaling αC


NOTES

Operations Research – Orthogonal projection T = {x1 ⊆ n1 |(x1, x2) ∈ C for x2 ∈ n2} – Set sum C1 + C2 = {c1 + c2 |c1 ∈ C1, c2 ∈ C2} Indeed, the set sum C1 + C2 is the image of the cartesian product C1 × C2 through

theaffinetransformf (x) = f (x1, x2) = x1 + x2 = (1, 1) . x. • Perspective transform If C ∈ n × ++ is a convex set, then the perspective

transform P(C) with P (x) = P (x1, x2, ..., xn, t) = (x1/t, x2/t, ..., xn/t) ∈Rn

is also convex. We canprove convexitypreservationunder intersectionandaffine transforms

triviallyfromthedefinitionofaconvexset.Forperspectivetransformsweshowthat segments between any two points x, y ∈ C are mapped to segments between the transforms of the points P(x), P(y) ∈ P(C), and thus any line segment between two points in P(C) is also in P(C).

An example of operation that does not preserve convexity is set union.

1.17.4 ProvingaSetisConvexTo recap, there are multiple different ways to prove that a set C is convex. Some

of the most common ones we’ve seen are: • Usingthedefinitionofaconvexset • Writing C as the convex hull of a set of point X, or the intersection of a set of

halfspaces • Building it up form convex sets using convexity preserving operations

1.18 CONVEX FUNCTIONSA convex function isa functiondefinedonaconvexdomainsuchthat, forany

two points in the domain, the segment between the two points lies above the function curvebetweenthem.Wewillshowbelowthatthisdefinitioniscloselyconnectedtotheconcept of a convex set: a function f is convex if and only if its epigraph, the set of all points above the function graph, is a convex set. We restate these results more precisely:

Definition: A function f : n → is convex if dom (f) is a convex set and if ∀x, y ∈dom(f), ∀θ∈ [0, 1], we have

– f(θx + (1 – θ)y) ≤ θ f (x) + (1 – θ) f(y)The epigraph of a function f : n → is the set of point epi(f) = {(x, t)|x ∈ dom (f),

t ≥ f (x)}.Lemma: The function f is convex iff the set epi (f) is convex.

1.18.1 CriteriaforConvexityAs with sets, there are multiple ways to characterize a convex function, each of

which may be convenient or insightful in different contexts. Below we explain the most commonly used three criteria.


NOTES


and Convex Sets

1.18.1.1 Zeroth OrderTheorem: A function is convex iff it is convex when restricted to any line that

intersects its domain. In other words, f is convex iff ∀x ∈ dom (f ), ∀v, the function g(t) = f (x + tv) is convex, when restricted to its domain {t |x + tv ∈ dom (f ) |}.

This property is useful because it allows us to reduce the problem of checking the convexity of a multivariate function to checking the convexity of a uni-variate function, for which we can use much simpler criteria.

We can intuitively visualize why the property holds by imagining e.g., a 3D convex cup-shaped function, choosing any point (x0, y0) on the function graph, and taking a vertical slice through (x0, y0). The resulting plane intersects the domain of f on a line, x0 + tv, and generates a new 2D function g(t). Intuitively, we expect this slice function to be convex (in this case, a 2D cup), and indeed this is what the zeroth-order property states.

f(y)

(x, f(x))

f(x) + f(x) (y – x)� T

Fig. 1.6. If f is convex and differentiable, then f(x) + ∇ f(x)T (y – s) ≤ f (y) for all x, y ∈ dom(f).

1.18.1.2 First OrderTheorem: Let f be a differentiable function. Then f is convex iff dom (f) is a convex

set and: f (y) ≥ f (x) + ∇ f (x)T (y – x).The function g(y) = f(x) + ∇ f(x)T (y – s)isthefirst-orderTaylorapproximationof

f near x.Theinequalityabovestatesthatforaconvexfunction,thefirst-orderTaylorapproximation is in fact a global underestimator of the function (See Figure 1.6). Conversely, if thefirst-orderTaylor approximation of a function is alwaysa globalunderestimator, then the function is convex.

The inequality above shows that from local information about a convex function (i.e. its value and derivative at a point) we can derive global information (i.e. a global underestimator of it). This is perhaps the most important property of convex function and convex optimization problems, and begins to explain why convex optimization problems are so much more easy to solve than non-convex optimization problems. As one simple example, the inequality above shows that if ∆ f (x) = 0 then for all y ∈dom(f), f(y) ≥ f(x), so x is a global minimizer of f.

Itgivesaneatproofofthefirst-ordercriterionbyfirstshowingitholdsforthe n = 1 case, and then using the zeroth order criterion for the case n > 1 to reduce the problem back to n = 1. An additional intuition for why the above Theorem holds is that the term g(x) = f (x) + ∇ f (x)T (y – x) represents a hyperplane through x which has ∇ f (x) slopes with respect to each axis of the input, and which supports the epigraph of f. We expect such a hyperplane exists because, since f is convex, so is epi(f).Thus,thefirst-order criterion is just an analytical expression of the supporting hyperplane theorem for convex sets!


NOTES

Operations Research 1.18.1.3 Second OrderIf f is twice differentiable, we have a simple second-order criterion, which is the

most common way to show well behaved functions are convex:Theorem: Let f be a twice differentiable function on an open domain dom(f). The

f is convex iff dom (f) is convex and its Hessian is positive semidefinite: H(x) = ∇2 f(x) ≥ 0.[Remember that a matrix A ∈ n×n ispositivesemidefinite,A ≥ 0, iff ∀x ∈ n,

xT AX ≥ 0.]For a function on , this reduces to the simple condition f ″(x) ≥ 0, which means

thatthefirstderivativef ′(x) is non-decreasing.

1.18.2 ConvexityPreservingOperationsJust as for convex sets, we consider standard operations which preserve function

convexity. These are useful tools for proving a complicated function is convex by showing it is constructed from simpler convex functions using legal operations. If f, fi are convex functions, then the following functions are also convex:

• Non-negative weighted sum ∀ w1, ..., wk ≥ 0, i

k

i if=∑

1ω is convex

• Composition with affine mapping f(Ax + b) is convex • Composition with monotone convex g(f(x)) is convex for f convex, g convex and

non-decreasing. We can see this easily in the case f : → :h″(x) = [g(f (x))]″ = [g′(f (x))f ′(x)]′ = g″(f (x)) [f ′ (x)]2 + g′(f (x)) f ″ (x) ≥ 0

If f, g are convex than g″ (f (x)) and f ″ (x) are both ≥ 0. Further, if g is non-decreasing, g′ (f (x)) is also ≥ 0, making all the terms in h″ (x) ≥0.

In this form we can also easily see other combinations of properties of f and g which lead to known properties of h. For example, if f is concave, g convex and non-increasing, then h is again convex. • Pointwise maximum f(x) = maxi fi(x) is convex This follows directly from the fact that the epigraph of a max/sup of functions is the

intersection of the epigraphs of the functions, and that set convexity is preserved under intersection.

• Minimum over a convex set g(x) = in fy∈C f(x, y) is convex. We can also get an intuition for this result by looking at it in terms of epigraphs.

Indeed,assumingthatallinfimumsareactuallyachieved,foranyx, g(x) = f (x, y) for some y ∈ C. Thus, (x, t) is in the epigraph of g iff (x, y, t) is in the epigraph of f for some y ∈ C. In other words: epi g = {(x, t) |(x, y, t) ∈ epi f, y ∈ C }

and the epigraph of g is the projection of the epigraph of f on one of its components. Since we’ve shown that projection on components preserves set convexity, we must have epi g convex, and thus the function g is also convex.

Fig. 1.7. The point-wise maximum of convex functions is convex.


NOTES


and Convex Sets

1.18.3 ProvingaFunctionisConvexAlready in the previous section, we’ve used three different techniques (!) for proving

functionconvexity:thedefinitionofaconvexfunction,fornon-negativeweightedsumsand compositionwith affinemapping; the 2nd order criteria, for compositionwithmonotoneconvex;andtheconnectionwithconvexsets,formaximumandminimum,Let’s review all the different ways one can prove a function is convex: • Usingthedefinitionofaconvexfunction • Showing that the function’s epigraph is a convex set • Using one of the 0th, 1st or 2nd order criteria:

– 0th order: g(t) = f(x + tv) is a convex function of t for all x, v– 1st order: f(y) ≥ f(x) + ∇ f(x)T (y – x) for all x, y– 2nd order: ∇2 f(y) ≥ 0 for all x

• Building it up from convex functions using convexity preserving operations.

1.19 CONVEX COMBINATIONS OF VECTORSGiven a set of points P0, P1, ..., Pn,wecanformaffinecombinationsofthesepoints

by selecting α0, α1, ..., αn, with α0 + α1 + ... + αn = 1 and form the point P = α0P0 + α1P1 + ... + αnPn

If each αi is such that 0 ≤ αi ≤ 1, then the points P is called a convex combination of the points P0, P1, ..., Pn.

To give a simple example of this, consider two points P0 and P1. Any point P on the line passing through these two points can be written as P = α0P0 + α1P1 which is anaffinecombinationofthetwopoints.ThepointsQ and RinthefollowingfigureareaffinecombinationsofP0 and P1.

P0 Q

P1

R

Fig. 1.8

However, the point Q is a convex combination, as 0 ≤ α0, α1 ≤ 1, and any point on the line segment joining P0 and P1 can be written in this way.

Given any set of points, we say that the set is a convex set, if given any two points of the set, any convex combination of these two points is also in the set. The following figureillustratesbothaconvexset(ontheleft)andanon-convexset(ontheright).

This concept is actually quite intuitive, in that if one can draw a straight line between two points of the set that is not completely contained within the set, the set is non-convex.

Fig. 1.9


NOTES

Operations Research The set of all points P that can be written as convex combinations of P0, P1, ..., Pn is called the convex hull of the points P0, P1, ..., Pn. This convex hull is the smallest convex set that contains the set of points P0, P1, ..., Pn.Thefollowingfigureillustratesthe convex hull of a set of six points:

P3

P5

P0 P1

P4P2

Fig. 1.10

One of the six points does not contribute to the boundary of the convex hull. If one lookedcloselyatthecoordinatesofthepoint,onewouldfindthatthispointcouldbewrittenasaconvexcombinationoftheotherfive.

1.20 HYPERPLANESHyperplane: Choose a vector a = (a1, a2, a3, ..., an) ≠ 0 in En and a scalar b. The

set of the points u = (x1, x2, x3, ..., xn) satisfying a1x1 + a2x2 + ... + anxn = bi.e. II = {u ∈ En: a . u = b}is called hyperplane in En.

a

a · u b�a · u b�

u

Fig. 1.11

In E2, hyperplanes are straight lines. In E3, hyperplanes are planes.For any n, a hyperplane in En divides En into two half-spaces, as pictured in

Figure 1.11Note: If b = 0, then the hyperplane is said to pass through the origin and its equation is

then written as a1x1 + a2x2 + ... + anxn = 0.Closed (Open) half-spaces: The sets H = {u ∈ En: a . u ≤ b} and H+ = {u ∈ En:

a . u ≥ b} are called closed half-spaces. If we replace the weak inequalitites with strict inequalities, we have open half-spaces.Thatis,thesetsdefinedby{u ∈ En: a . u < b} and {u ∈ En: a . u > b} are called open half-spaces.

It is easy to verify that all hyperplanes, open and closed half-spaces, are convex.Polyhedral: A convex set is polyhedral ifitisasolutionsetoffinitelymanylinear

inequalities and equalities.


NOTES


and Convex Sets

Let a1, a2, ..., am denote m vectors in En, and let b1, b2, ..., bm denote the corresponding scalars.

Let Hj = {uj ∈ En: aj . uj ≤ bj}. Each Hj is half-space. The set S = j

m

jH= 1

is a

polyhedral convex set. The set S is a set of solution to the system of inequalities: a11x1 + a12x2 + ... + a1nxn ≤ b1 a21x1 + a22x2 + ... + a2nxn ≤ b2 ............................................... am1x1 + am2x2 + ... + amnxn ≤ bmBounded convex polyhedra are called convex

polytopes. The direction vectors of an unbounded convex polyhedron are vectors in the direction of its bounding rays. Note that a convex polytope has no direction vectors. The hyperplanes producing the halfspaces are called the generating hyperplanes of the polytope. A polyhedral convex set is drawn in Figure 1.12.

Extreme point of a convex set: An extreme point or vertex of a convex set is a corner point of polyhedron. More formally, an extreme point is a point which cannot be expressed as a convex combination of other points in the polyhedron.

Thus, a point u of a convex set S is an extreme point of the set, if there does not exist any pair of points u1, u2 in S, such that

u = θu1 + (1 – θ)u2 0 < θ < 1Note:Anextremepointisaboundarypointoftheset;however,

not all boundary points of a convex set are necessarily extreme points.For example, a and b are extreme points in Figure 1.13.

1.21 SEPARATING AND SUPPORTING HYPERPLANE THEOREMS

An important idea that we will use later in the analysis of convex optimization problemsistheuseofthehyperplanesoraffinefunctionstoseparateconvexsetsthatdo not intersect. The two main results are the separating hyperplane theorem and the supporting hyperplane theorem.

1.21.1SeparatingHyperplaneTheoremThe separating hyperplane theorem states that, for any two convex sets C and D

which do not intersect, C ∩ D = f, there exists a hyperplane such that C and D are on opposite sides of the hyperplane. More rigurously:

Theorem: Let C, D be convex sets such that C ∩ D = f. Then there exists a ≠ 0, b such that:

∀x ∈ C, aTx ≤ b and ∀y ∈ D, aTy ≥ b.This separating hyperplane theorem is geometrically intuitive.

a2

a1

a5

a4

a3

Fig. 1.12. A polyhedra convex

a b

Fig. 1.13


NOTES

Operations Research Now the proof for the special case of two sets C and D whichareafinitedistance dist (C, D) = minc∈C,d∈D ||c – d||2 apart, and for which there exist two points c ∈ C, d ∈ D which actually achieve this minimum. In this case it can be shown that any hyperplane perpendicular on the segment [c, d], which goes through the segment [c, d], is a separating

hyperplane. In particular, the hyperplane through c d+2 is a separating hyperplane.

The proof is a proof by contradiction, which essentially shows that, if one of the sets (sayD)wouldintersectourproposedhyperplane,thanwecouldfindanotherd′ closer to c than d.

D

Ca

a x bT �

a x bT �

Fig. 1.14. The hyperplane x |aTx = b separates the disjoint convex sets C and D.Theaffine function aTx – b is non-positive on C and non-negative on D.

Cx0

a

Fig. 1.15. The hyperplane x |aTx = atx0 supports C at x0.

A related concept is strict separation, where the separating hyperplane does not intersect either C or D:

∀x ∈ C, aTx < b and ∀y ∈ D, aTy > b.

1.22 HYPERSPHEREA hypersphere in En with centre at a and radius e > 0 is defined to be the set of points X = {x : | x – a | = e}

i.e., the equation of a hypersphere in En is (x1 – a1)2 + (x2 – a2)2 + ... + (xn – an)2 = e2

where a = (a1, a2, ..., an), x = (x1, x2, ..., xn)which represent a circle in E2 and sphere in E3.


NOTES


and Convex Sets

1.22.1 An e-neighbourhoodAn e-neighbourhood about the point a is defined as the set of points lying inside the

hypersphere with centre at a and radius e > 0.i.e., the e-neighbourhood about the point is the set of points,

X = {x : |x – a | < e}

1.22.2 AnInteriorPointA point a is an interior point of the set S if there exists an e-neighbourhood about

a which contains only points of the set S.An interior point of S must be an element of S.

TESTYOURKNOWLEDGE–B 1. Model building is the essence of the operations research approach. Discuss. 2. What is meant by a mathematical model of a real situation? Discuss the importance of

models in the solution of OR problems. 3. What is the purpose of a mathematical model? How does a model achieve this purpose? In

your answer consider the concept of ‘a model an abstraction of reality.’ 4. Whatisamodel?Discussvariousclassificationschemesofmodels. 5. (a) Discuss, in brief, the role of OR models in decision-making. What areas of OR have

madeasignificantimpactonthedecision-makingprocess? (b) Discuss the importance of operations research in the decision-making process. (c) Give some examples of various types of models. What is a mathematical model? Develop

two examples of mathematical models. 6. Explain how and why operations research methods have been valuable in aiding executive

decisions. 7. StatethedifferenttypesofmodelsusedinOR.Explainbrieflythegeneralmethodsfor

solving these OR models. 8. Explain the scope and methodology of OR, the main phases of OR and techniques used in

solving an OR problem. 9. HowcanORmodelsbeclassified?Whatisthebestclassificationintermsoflearningand

understanding the fundamentals of OR? 10. Explain the essential characteristics of the following types of models: (a) Allocation models (b) Competitive games (c) Inventory (d) Waiting line. 11. Define an OR model and give four examples. State their properties, advantages and

limitations. 12. What are the advantages and disadvantages of operations research models? Why is it

necessary to test models and how would you go about testing a model? 13. Operations research is an ongoing process. Explain this statement with examples. 14. Distinguish between model results that recommend a decision and model result that are

descriptive. 15. ‘Mathematics … tends to lull the unsuspecting into believing that he who thinks elaborately

thinks well’. Do you think that the best OR models are the ones that are most elaborate and mathematically complex? Why?

16. ‘The hard problems are those for which models do not exist’ interpret this statement. Give some examples.


NOTES

Operations Research 17. Definehypersphere. 18. Defineconvexhull. 19. Write a short note on intersection of half spaces. 20. Defineconvexfunction. 21. What is convex sets? Write down the properties of convex sets. 22. Write the criteria for convexity? 23. What do you mean by convexity preserving operations? 24. What is convex combinations of vector? Explain with the help of an example. 25. What is hyperplane? Discuss about extreme point of a convex set. 26. Explain how convex sets are related to separating and supporting hyperplane theorem.

SUMMARY

• Operationsresearchisthescientificapproachtodecisionmaking.Thisapproachincludesdefining theproblem,developingamodel, solving themodel, testingthe solution, analyzing the results, and implementing the results. In using operations research approach, however, there can be potential problems, including conflictingviewpoints, the impactofquantitativeanalysisonvarioussectionsof the organization, beginning assumptions, outdated solutions, understanding the model, obtaining relevant input data, obtaining optimal solution, testing the solution, and analyzing the results. In using the operations research approach, implementationisnotthefinalstepbecausetherecanbealackofcommitmentto the approach and resistance to change.

• A set C is convex if the line segment between any two points in C lies in C, i.e., ∀x1, x2 ∈ C, ∀θ ∈ [0, 1]

• The convex hull of a set C is the set of all convex combinations of points in C:

Conv (C) = {θ1x1 + … + θkxk | xi ∈ C, θi ≥ 0, i = 1, …, k, i

k

i=∑ =

11θ }

• The function f is convex iff the set epi (f) is convex. • A vector a = (a1, a2, a3, ..., an) ≠ 0 in En and a scalar b. The set of the points

u = (x1, x2, x3, ..., xn) satisfying a1x1 + a2x2 + ... + anxn = bi.e. II = {u ∈ En: a . u = b}is called hyperplane in En.

• Anextremepointisaboundarypointoftheset;however,notallboundarypointsof a convex set are necessarily extreme points.

NOTES

Linear Programming: The Simplex Method


STRUCTURE

2.1 Introduction2.2 Standard Form of an LP Problem2.3 Simplex Algorithm (Maximization Case)2.4 Simplex Algorithm (Minimization Case)2.5 Types of Linear Programming Solutions2.6 Slack and Surplus Variables2.7 Basic Feasible Solution (Reduction From Feasible Solution)2.8 Improving a Basic Feasible Solution2.9 The Fundamental Theorem of Linear Programming2.10 Alternative Optimal Solution2.11 Unbounded Solution2.12 Condition of Optimality

U N I T

2LINEAR PROGRAMMING: THE SIMPLEX METHOD

2.1 INTRODUCTIONMost real-life problems when formulated as an LP model have more than two

variables and therefore need a more efficient method to suggest an optimal solution for such problems. In this chapter, we shall discuss a procedure called the simplex method for solving an LP model of such problems. This method was developed by G B Dantzig in 1947.

The simplex, an important term in mathematics, represents an object in an n-dimensional space, connecting n + 1 points. In one dimension, the simplex represents a line segment connecting two points; in two dimensions, it represents a triangle formed by joining three points; in three dimensions, it represents a four-sided pyramid.

The concept of simplex method is similar to the graphical method. In the graphical method, extreme points of the feasible solution space are examined in order to search for the optimal solution that lies at one of these points. For LP problems with several variables, we may not be able to graph the feasible region, but the optimal solution will still lie at an extreme point of the many-sided, multidimensional figure (called an n-dimensional polyhedron) that represents the feasible solution space. The simplex

NOTES

Operations Research


method examines these extreme points in a systematic manner, repeating the same set of steps of the algorithm until an optimal solution is found. It is for this reason that it is also called the iterative method.

Since the number of extreme points of the feasible solution space are finite, the method assures an improvement in the value of objective function as we move from one iteration (extreme point) to another and achieve the optimal solution in a finite number of steps. The method also indicates when an unbounded solution is reached.

2.2 STANDARD FORM OF AN LP PROBLEMThe use of the simplex method to solve an LP problem requires that the problem

be converted into its standard form. The standard form of the LP problem should have the following characteristics: (i) All the constraints should be expressed as equations by adding slack or surplus

and/or artificial variables. (ii) The right-hand side of each constraint should be made non-negative (if not).

This is done by multiplying both sides of the resulting constraint by –1. (iii) The objective function should be of the maximization type.

The standard form of the LP problem is expressed as:Optimize (Max or Min) Z = c1 x1 + c2 x2 + … + cn xn + 0s1 + 0s2 + … + 0sm

subject to the linear constraints a11 x1 + a12 x2 + … + a1n xn + s1 = b1

a21 x1 + a22 x2 + … + a2n xn + s2 = b2

am1 x1 + am2 x2+ … + amn xn + sm = bm

and x1, x2, … , xn, s1, s2, …, sm ≥ 0This standard form of the LP problem can also be expressed in the compact form

as follows:

Optimize (Max or Min) Z = j

n

j ji

m

ic x s= =∑ ∑+1 1

0 (Objective function)

subject to the linear constraints

j

n

ij ja x=∑

1 + si = bi; i = 1, 2, ..., m (Constraints)

and xj, si ≥ 0, for all i and j (Non-negativity conditions)In matrix notations the standard form is expressed as:Optimize (Max or Min) Z = cx + 0s

subject to the linear constraints Ax + s = b, and x, s ≥ 0

where, c = (c1, c2, …, cn) is the row vector; x = (x1, x2, ..., xn), b = (b1, b2, ..., bm)T and s = (s1, s2, . . ., sm) are column vectors, and A is the m × n matrix of coefficients, aij of variables x1, x2, …, xn in the constraints.

Simplex method examines corner points of the feasible region, using matrix row operations, until an optimal solution is found.

NOTES



Remarks: Any LP problem (maximization or minimization) may have 1. (a) no feasible solution, i.e. value of decision variables, xj( j = 1, 2, …, n) may not satisfy

every constraint. (b) a unique optimum feasible solution. (c) more than one optimum feasible solution, i.e. alternative optimum feasible solutions. (d) a feasible solution for which the objective function is unbounded (value of the objective

function can be made as large as possible in a maximization problem or as small as possible in a minimization problem).

2. Any minimization LP problem can be converted into an equivalent maximization problem by changing the sign of cj’s in the objective function. That is,

Minimize c xj jj

n

=∑

1 = Maximize ( )−

=∑ c xj jj

n

1

3. Any constraint expressed by equality (=) sign may be replaced by two weak inequalities. For example, a11 x1 + a12 x2 + + a1n xn = b1 is equivalent to following two simultaneous constraints,

a11 x1 + a12 x2 + … + a1n xn ≤ b1 and a11 x1 + a12 x2 + + a1n xn ≥ b1 4. Three types of additional variables, namely (i) slack variables (s) (ii) surplus variables

(– s), and (iii) artificial variables (A) are added in the given LP problem to convert it into the standard form for the following reasons:

(a) These variables allow us to convert inequalities into equalities, thereby converting the given LP problem into its standard form. Such form will help in getting solution of the LP problem.

(b) These variables help decision-makers to draw economic interpretation from the final solution.

(c) These variables are used to get an initial feasible solution represented by the columns of the identity matrix.

The summary of the additional variables to be added in the given LP problem in order to convert it into a standard form is given in Table 2.1.

Table 2.1. Summary of Additional Variables Added in an LP Problem

Types of Constraint Extra Variable Needed

Coefficient of Additional Variables

in the Objective Function

Presence of Additional

Variables in the Initial Solution

Max Z Min Z• Less than or equal to (≤) A slack variable is

added0 0 Yes

• Greater than or A surplus variable 0 0 No equal to (≥) is subtracted, and

an artificial variable is added

– M + M Yes

• Equal to (=) Only an artificial variable is added

– M + M Yes

Remark: A slack variable represents an unused resource, either in the form of time on a machine, labour hours, money, warehouse space, etc. Since these variables don’t yield any profit, therefore such variables are added to the objective function with zero coefficients. A surplus variable represents the amount by which solution values exceed a resource. These variables are also called negative slack variables. Surplus variables, like slack vari-ables carry a zero coefficient in the objective function.

Slack variable represents an unused quantity of resource; it is added to less-than or equal-to type constraints in order to get an equality constraint.

NOTES

Operations Research


DefinitionsBasic solution: Given a system of m simultaneous linear equations with n (> m)

variables: Ax = B, where A is an m × n matrix and rank (A) = m. Let B be any m × m non-singular submatrix of A obtained by reordering m linearly independent columns of A. Then, a solution obtained by setting n – m variables not associated with the columns of B, equal to zero, and solving the resulting system is called a basic solution to the given system of equations.

The m variables (may be all non zero) are called basic variables. The m × m non-singular sub-matrix B is called a basis matrix and the columns of B as basis vectors.

If B is the basis matrix, then the basic solution to the system of equations will be xB = B–1b.

Basic feasible solution: A basic solution to the system of simultaneous equations, Ax = b is called basic feasible if xB ≥ 0.

Degenerate solution A basic solution to the system of simultaneous equations, Ax = b is called degenerate if one or more of the basic variables assume zero value.

Cost vector: Let xB be a basic feasible solution to the LP problem.Maximize Z = cx

subject to the constraints Ax = b, and x ≥ 0.Then the vector cB = (cB1, cB2, …, cBm), is called cost vector that represents the

coefficient of basic variable, xB in the basic feasible solution.

2.3 SIMPLEX ALGORITHM (MAXIMIZATION CASE)The steps of the simplex algorithm for obtaining an optimal solution (if it exists)

to a linear programming problem are as follows:Step 1: Formulation of the mathematical model

(i) Formulate the LP model of the given problem. (ii) If the objective function is of minimization, then convert it into equivalent

maximization, by using the following relationship. Minimize Z = – Maximize Z*, where Z* = – Z.

(iii) Check whether all the bi (i = 1, 2, …, m) values are positive. If any one of them is negative, then multiply the corresponding constraint by – 1 in order to make bi > 0. In doing so, remember to change a ≤ type constraint to a ≥ type constraint, and vice versa.

(iv) Express the given LP problem in the standard form by adding additional variables in constraints (as per requirement) and assign a zero-cost coefficient to these variables in the objective function.

(v) Replace each unrestricted variable (if any) with the difference of the two non-negative variables.

Step 2: Set-up the initial solution: Write down the coefficients of all the variables in the LP problem in a tabular form, as shown in Table 2.2, in order to get an initial basic feasible solution [xB = B–1b].

Surplus variable represents the amount of resource usage above the minimum required and is added to greater-than or equal-to constraints in order to get equality constraint.

NOTES



Table 2.2. Initial Simplex Tablecj → c1 c2 … cn 0 0 … 0

Basic VariablesCoefficient

(cB)

BasicVariables

B

Basic Variables

Valueb(= xB)

x1 x2Variables

…xn s1 s2 … sm

cB1 s1 xB1 = b1 a11 a12 … a1n 1 0 … 0cB2 s2 xB2 = b2 a21 a22 … a2n 0 1 … 0

cBm sm xBm = bm am1 am2 … amn 0 0 … 1

Z = ∑ cBi xBi zj = ∑ cBi xj 0 0 … 0 0 0 … 0cj – zj c1 – z1 c2 – z2 … cn – zn 0 0 … 0

Step 3: Test for optimality: Calculate the cj – zj value for all non-basic variables. To obtain the value of zj multiply each element under ‘Variables’ column (columns, aj of the coefficient matrix) with the corresponding elements in cB column. Examine values of cj – zj. The following three cases may arise: (i) If all cj – zj ≤ 0, then the basic feasible solution is optimal. (ii) If at least one column of the coefficients matrix (i.e. ak ) for which ck – zk > 0 and

all other elements are negative (i.e. aik < 0), then there exists an unbounded solution to the given problem.

(iii) If at least one cj – zj > 0, and each of these columns have at least one positive element (i.e. aij ) for some row, then this indicates that an improvement in the value of objective function Z is possible.

Step 4: Select the variable to enter the basis: If Case (iii) of Step 3 holds, then select a variable that has the largest cj – zj value to enter into the new solution. That is,

ck – zk = Max {(cj – zj); cj – zj > 0}The column to be entered is called the key or pivot column. Obviously, such a

variable indicates the largest per unit improvement in the current solution.Step 5: Test for feasibility (variable to leave the basis): After identifying

the variable to become the basic variable, the variable to be removed from the existing set of basic variables is determined. For this, each number in xB-column (i.e. bi values) is divided by the corresponding (but positive) number in the key column and a row is selected for which this ratio is non-negative and minimum. This ratio is called the replacement (exchange) ratio. That is,

xaBr

rj = Min

xa

aBr

rjrj; >

0

Step 6: Finding the new solution (i) If the key element is 1, then the row remains the same in the new simplex

table. (ii) If the key element is other than 1, then divide each element in the key row

(including the elements in xB-column ) by the key element, to find the new values for that row.

Non-basic variables are those which are not in the ‘basis’ and have zero-value.

cj – zj row values represent net profit or loss resulting from introducing one unit of any variable into the ‘basis’ or solution mix.

NOTES

Operations Research


(iii) The new values of the elements in the remaining rows of the new simplex table can be obtained by performing elementary row operations on all rows so that all elements except the key element in the key column are zero. In other words, for each row other than the key row, we use the formula:

Number innew row = Number inold row

± Number above orbelow key element

Corresponding number inth

× ee new row, that is rowreplaced in Step 6 ( )ii

The new entries in cB and xB columns are updated in the new simplex table of the current solution.

Step 7: Repeat the procedure: Go to Step 3 and repeat the procedure until all entries in the cj – zj row are either negative or zero.

Remark: The flow chart of the simplex algorithm for both the maximization and the minimization LP problem is shown in Fig. 2.1.

Example 1. Use the simplex method to solve the following LP problem.Maximize Z = 3x1 + 5x2 + 4x3

subject to the constraints (i) 2x1 + 3x2 ≤ 8, (ii) 2x2 + 5x3 ≤ 10, (iii) 3x1 + 2x2 + 4x3 ≤ 15 and x1, x2, x3 ≥ 0

Sol. Step 1: Introducing non-negative slack variables s1, s2 and s3 to convert the given LP problem into its standard form:

Maximize Z = 3x1 + 5x2 + 4x3 + 0s1 + 0s2 + 0s3subject to the constraints (i) 2x1 + 3x2 + s1 = 8, (ii) 2x2 + 5x3 + s2 = 10, (iii) 3x1 + 2x2 + 4x3 + s3 = 15 and x1, x2, x3, s1, s2, s3 ≥ 0

Step 2: Since all bi (RHS value) > 0, (i = 1, 2, 3), therefore choose the initial basic feasible solution as:

x1 = x2 = x3 = 0; s1 = 8, s2 = 10, s3 = 15and Max Z = 0

This solution can also be read from the initial simplex Table 2.3 by equating row-wise values in the basis (B) column and solution values (xB) column.

Step 3: To see whether the current solution given in Table 2.3 is optimal or not, calculate

cj – zj = cj – cB B–1 aj = cj – cB yjfor non-basic variables x1, x2 and x3 as follows.

zj = (Basic variable coefficients, cB) × (jth column of data matrix)That is, z1 = 0 (2) + 0 (0) + 0 (3) = 0 for x1-column z2 = 0 (3) + 0 (2) + 0 (2) = 0 for x2-column z3 = 0 (0) + 0 (5) + 0 (4) = 0 for x3-columnThese zj values are now subtracted from cj values in order to calculate the net

profit likely to be gained by introducing (performing) a particular variable (activity) x1, x2 or x3 into the new solution mix. This is done by introducing one unit of each variable x1, x2 and x3 into the new solution mix.

c1 – z1 = 3 – 0 = 3, c2 – z2 = 5 – 0 = 5, c3 – z3 = 4 – 0 = 4

NOTES



The zj and cj – zj rows are added into the Table 2.3.The values of basic variables, s1, s2 and s3 are given in the solution values (xB)

column of Table 2.3. The remaining variables that are non-basic at the current solution have zero value. The value of objective function at the current solution is given by

Z = (Basic variables coefficient, cB) × (Basic variables value, xB) = 0(8) + 0(10) + 0(15) = 0

Table 2.3. Initial Solution

cj → 3 5 4 0 0 0

Basic Variables Coefficient

cB

Basic Variables

B

Basic Variables

Value b(=xB)

x1 x2 x3 s1 s2 s3Min Ratio

xB / x2

0 s1 8 2 3 0 1 0 0 8/3 →

0 s2 10 0 2 5 0 1 0 10/2

0 s3 15 3 2 4 0 0 1 15/2

Z = 0 zj 0 0 0 0 0 0

cj – zj 3 5 4 0 0 0

↑

Since all cj – zj ≥ 0 ( j = 1, 2, 3), the current solution is not optimal. Variable x2 is chosen to enter into the basis because c2 – z2 = 5 is the largest positive number in the x2-column, where all elements are positive. This means that for every unit of variable x2, the objective function will increase in value by 5. The x2-column is the key column.

Step 4: The variable that is to leave the basis is determined by dividing the values in the xB-column by the corresponding elements in the key column as shown in Table 2.3. Since the ratio, 8/3 is minimum in row 1, the basic variable s1 is chosen to leave the solution (basis).

Step 5: (Iteration 1): Since the key element enclosed in the circle in Table 2.3 is not 1, divide all elements of the key row by 3 in order to obtain new values of the elements in this row. The new values of the elements in the remaining rows for the new Table 2.4 can be obtained by performing the following elementary row operations on all rows so that all elements except the key element 1 in the key column are zero.

R1 (new) → R1(old) ÷ 3 (key element) → (8/3, 2/3, 3/3, 0/3, 1/3, 0/3, 0/3) = (8/3, 2/3, 1, 0, 1/3, 0, 0)

R2 (new) → R2 (old) – 2R1 (new) R3 (new) → R3 (old) – 2R1 (new)

10 – 2 × 8/3 = 14/3 15 – 2 × 8/3 = 29/3

0 – 2 × 2/3 = –4/3 3 – 2 × 2/3 = 5/3

2 – 2 × 1 = 0 2 – 2 × 1 = 0

5 – 2 × 0 = 5 4 – 2 × 0 = 4

0 – 2 × 1/3 = – 2/3 0 – 2 × 1/3 = – 2/3

1 – 2 × 0 = 1 0 – 2 × 0 = 0

0 – 2 × 0 = 0 1 – 2 × 0 = 1

NOTES

Operations Research


Table 2.4. Improved Solutioncj → 3 5 4 0 0 0

Basics Variables Coefficient

cB

Basic Variables

B

Basic Variables

Value b(= xB)


xB / x3

5 x2 8/3 2/3 1 0 1/3 0 0 –

0 s2 14/3 – 4/3 0 5 – 2/3 1 0 (14/3)/5 →

0 s3 29/3 5/3 0 4 – 2/3 0 1 (29/3)4

Z = 40/3 zj 10/3 5 0 5/3 0 0

cj – zj – 1/3 0 4 – 5/3 0 0↑

The improved basic feasible solution can be read from Table 2.4 as: x2 = 8/3, s2 = 14/3, s3 = 29/3 and x1 = x3 = s1 = 0. The improved value of the objective function becomes:

Z = (Basic variable coefficients, cB) × (Basic variable values, xB) = 5 (8/3) + 0 (14/3) + 0 (29/3) = 40/3Once again, calculate values of cj – zj to check whether the solution shown in Table

2.4 is optimal or not. Since c3 – z3 > 0, the current solution is not optimal.Iteration 2: Repeat Steps 3 to 5. Table 2.5 is obtained by performing following

row operations to enter variable x3 into the basis and to drive out s2 from the basis. R2(new) = R2(old) ÷ 5(key element) = (14/15, – 4/15, 0, 1, – 2/15, 1/5, 0)

R3(new) → R3(old) – 4R2(new)

29/3 – 4 × 14/15 = 89/15

5/3 – 4 × – 4/15 = 41/15

0 – 4 × 0 = 0

4 – 4 × 1 = 0

– 2/3 – 4 × – 2/15 = – 2/15

0 – 4 × 1/5 = – 4/5

1 – 4 × 0 = 0

Table 2.5 is completed by calculating the new zj and cj – zj values and the new value of objective function:

z1 = 5(2/3) + 4(– 4/15) + 0(41/15)= 34/15 z4 = 5(1/3) + 4(– 2/15) + 0(2/15) = 17/15 z5 = 5(0) + 4(1/5) + 0(– 4/5) = 4/5 c1 – z1 = 3 – 34/15 = 11/15 for x1-column c4 – z4 = 0 – 17/15 = – 17/15 for s1-column c5 – z5 = 0 – 4/5 = – 4/5 for s2-columnThe new objective function value is: Z = (Basic variables coefficient, cB) × (Basic variables value, xB) = 5(8/3) + 4(14/15) + 0(89/15) = 256/15

NOTES



The improved basic feasible solution is shown in Table 2.5.Table 2.5. Improved Solution

cj → 3 5 4 0 0 0


cB

Basic Variables

B

Basic Variables

Value b (= xB)


xB / x1

5 x2 8/3 2/3 1 0 1/3 0 0 (8/3)/(2/3) = 44 x3 14/15 – 4/15 0 1 – 2/15 1/5 0 –0 s3 89/15 41/15 0 0 2/15 – 4/5 1 (89/15)/(41/15)

= 2.17 →Z = 256/15 zj 34/15 5 4 17/15 4/5 0

cj – zj 11/15 0 0 – 17/15 – 4/5 0↑

Iteration 3: In Table 2.5, since c1 – z1 is still a positive value, the current solution is not optimal. Thus, the variable x1 enters the basis and s3 leaves the basis. To get another improved solution as shown in Table 2.5 perform the following row operations in the same manner as discussed earlier.

R3(new) → R3(old) × 15/41 (key element) → (89/15 × 15/41, 41/15 × 15/41, 0 × 15/41, 0 × 15/41, – 2/15 × 15/41, – 4/5 × 15/41, 1 × 15/41) → (89/41, 1, 0, 0, – 2/41, – 12/41, 15/41)

R1(new) → R1(old) – (2/3) R3(new) R2 (new) → R2 (old) + (4/15) R3 (new)

8/3 – 2/3 × 89/3 = 50/41 14/15 + 4/15 × 89/41 = 62/412/3 – 2/3 × 1 = 0 – 4/15 + 4/15 × 1 = 0

1 – 2/3 × 0 = 1 0 + 4/15 × 0 = 00 – 2/3 × 0 = 0 1 + 4/15 × 0 = 0

1/3 – 2/3 × – 2/41 = 15/41 – 2/15 + 4/15 × – 2/41 = – 6/410 – 2/3 × – 2/41 = 8/41 1/5 + 4/15 × –12/41 = 5/410 – 2/3 × 15/41 = – 10/41 0 + 4/15 × 15/41 = 4/41

Table 2.6. Optimal Solutioncj → 3 5 4 0 0 0

BasicVariables

Coefficient cB

BasicVariables

B

BasicVariables Value

b (= xB)x1 x2 x3 s1 s2 s3

5 x2 50/41 0 1 0 15/41 8/41 – 10/414 x3 62/41 0 0 1 – 6/41 5/41 4/413 x1 89/41 1 0 0 – 2/41 – 12/41 15/41

Z = 765/41 zj 3 5 4 45/41 24/41 11/41cj – zj 0 0 0 – 45/41 – 24/41 – 11/41

In Table 2.6, all cj – zj < 0 for non-basic variables. Therefore, the optimal solution is reached with, x1 = 89/41, x2 = 50/41, x3 = 62/41 and the optimal value of Z = 765/41.

NOTES

Operations Research


2.4 SIMPLEX ALGORITHM (MINIMIZATION CASE)

In certain cases, it is difficult to obtain an initial basic feasible solution of the given LP problem. Such cases arise

(i) when the constraints are of the ≤ type, aijj

n

=∑

1 xj ≤ bi, xj ≥ 0

and value of few right-hand side constants is negative [i.e. bi < 0]. After adding the non-negative slack variable si (i = 1, 2, …, m), the initial solution so obtained will be si = – bi for a particular resource, i. This solution is not feasible because it does not satisfy non negativity conditions of slack variables (i.e. si ≥ 0).

(ii) when the constraints are of the ≥ type, aijj

n

=∑

1 xj ≥ bi, xj ≥ 0

After adding surplus (negative slack) variable si, the initial solution so obtained will be – si = bi or si = –bi

aijj

n

=∑

1 xj – si = bi, xj ≥ 0, si ≥ 0.

This solution is not feasible because it does not satisfy non-negativity conditions of surplus variables (i.e. si ≥ 0). In such a case, artificial variables, Ai (i = 1, 2, …, m) are added to get an initial basic feasible solution. The resulting system of equations then becomes:

aijj

n

=∑

1 xj – si + Ai = bi

xj, si, Ai ≥ 0, i = 1, 2, …, mThese are m simultaneous equations with (n + m + m) variables (n decision

variables, m artificial variables and m surplus variables). An initial basic feasible solution of LP problem with such constraints can be obtained by equating (n + 2m – m) = (n + m) variables equal to zero. Thus the new solution to the given LP problem is: Ai = bi (i = 1, 2, …, m), which is not the solution to the original system of equations because the two systems of equations are not equivalent. Thus, to get back to the original problem, artificial variables must be removed from the optimal solution. There are two methods for removing artificial variables from the solution. • Two-Phase Method • Big-M Method or Method of Penalties

NOTES



The simplex method, both for the minimization and the maximization LP problem, may be summarized through a flow chart shown in Fig. 2.1.

Fig. 2.1. Flow Chart of Simplex Algorithm

Remark: Artificial variables have no meaning in a physical sense and are only used as a tool for generating an initial solution to an LP problem. Before the optimal solution is reached, all artificial variables must be dropped out from the solution mix. This is done by assigning appropriate coefficients to these variables in the objective function. These variables are added to those constraints with equality (=) and greater than or equal to (≥) sign.

2.4.1 Two-Phase MethodIn the first phase of this method, the sum of the artificial variables is minimized

subject to the given constraints in order to get a basic feasible solution of the LP problem. The second phase minimizes the original objective function starting with the basic feasible solution obtained at the end of the first phase. Since the solution of the LP problem is completed in two phases, this method is called the two-phase method.

Advantages of the method 1. No assumptions on the original system of constraints are made, i.e. the system

may be redundant, inconsistent or not solvable in non-negative numbers.

NOTES

Operations Research


2. It is easy to obtain an initial basic feasible solution for Phase I. 3. The basic feasible solution (if it exists) obtained at the end of phase I is used as

initial solution for Phase II.Example 2. Use two-phase simplex method to solve the following LP problem:Minimize Z = x1 + x2

subject to the constraints (i) 2x1 + x2 ≥ 4, (ii) x1 + 7x2 ≥ 7and x1, x2, ≥ 0

Sol. Converting the given LP problem objective function into the maximization form and then adding surplus variables s1 and s2 and artificial variables A1 and A2 in the constraints, the problem becomes: (i) 2x1 + x2 – s1 + A1 = 4, (ii) x1 + 7x2 – s2 + A2 = 7and x1, x2, s1, s2, A1, A2 ≥ 0where Z* = – Z

Phase I: This phase starts by considering the following auxiliary LP problem:Maximize Z* = – A1 – A2

subject to the constraints (i) 2x1 + x2 – s1 + A1 = 4, (ii) x1 + 7x2 – s2 + A2 = 7and x 1, x2, s1, s2, A1, A2 ≥ 0

The initial solution is presented in Table 2.7Table 2.7. Initial solution

cj → 0 0 0 0 – 1 – 1


cB

Basic Variables

B

Basic VariablesValueb(= xB)

x1 x2 s1 s2 A1 A2

– 1 A1 4 2 1 – 1 0 1 0

– 1 A2 7 1 7 0 – 1 0 1 →

Z* = – 11 zj – 3 – 8 1 1 – 1 – 1

cj – zj 3 8 – 1 – 1 0 0

↑

Artificial variables A1 and A2 are now removed, one after the other, maintaining the feasibility of the solution.

Iteration 1: Applying the following row operations to get an improved solution by entering variable x2 in the basis and first removing variable A2 from the basis. The improved solution is shown in Table 2.8. Note that the variable x1 cannot be entered into the basis as this would lead to an infeasible solution.

R2 (new) → R2 (old) ÷ 7 (key element); R1 (new) → R1 (old) – R2 (new)

NOTES



Table 2.8. Improved Solutioncj → 0 0 0 0 – 1 – 1


cB

Basic Variables B

Basic Variables

Value b(= xB)x1 x2 s1 s2 A1 A2*

– 1 A1 3 13/7 0 – 1 1/7 1 – 1/7 →

0 x2 1 1/7 1 0 – 1/7 0 1/7

Z* = – 3 zj – 13/7 0 1 – 1/7 – 1 1/7

cj – zj 13/7 0 – 1 1/7 0 – 8/7↑

* This column can permanently be removed at this stage.Iteration 2: To remove A1 from the solution shown in Table 2.8, we enter variable

s2 in the basis by applying the following row operations. The new solution is shown in Table 2.9. It may be noted that if instead of s2, variable x1 is chosen to be entered into the basis, it will lead to an infeasible solution

R1 (new) → R1 (old) × 7; R2 (new) → R2 (old) + (1/7) R1 (new)

Table 2.9. Improved Solution

cj → 0 0 0 0 – 1 – 1

Basic Variables

Coefficient cB

Basic Variables

B


x1 x2 s1 s2 A*1 A*

2

0 s2 21 13 0 – 7 1 7 – 1

0 x2 4 2 1 – 1 0 1 0

Z* = 0 zj 0 0 0 0 0 0

cj – zj 0 0 0 0 – 1 – 1

* Remove columns A1 and A2 from Table 2.9.Since all cj – zj ≤ 0 correspond to non-basic variables, the optimal solution: x1 = 0,

x2 = 4, s1 = 0, s2 = 21, A1 = 0, A2 = 0 with Z* = 0 is arrived at. However, this solution may or may not be the basic feasible solution to the original LP problem. Thus, we have to move to Phase II to get an optimal solution to our original LP problem.

Phase II: The modified simplex table obtained from Table 2.9 is represented in Table 2.10.

Table 2.10cj → – 1 – 1 0 0

Basic Variables

Coefficient cB

BasicVariables

B


x1 x2 s1 s2Min Ratio

xB /x1

0 s2 21 13 0 – 7 1 21/13 →– 1 x2 4 2 1 – 1 0 4/2

Z* = – 4 zj – 2 – 1 1 0cj – zj 1 0 – 1 0

↑

NOTES

Operations Research


Iteration 1: Introducing variable x1 into the basis and removing variable s2 from the basis by applying the following row operations:

R1 (new) → R1(old) ÷ 13(key element); R2(new) → R2(old) – 2R1(new)The improved basic feasible solution so obtained is given in Table 2.11 Since in

Table 2.11, cj – zj ≤ 0 for all non-basic variables, the current solution is optimal. Thus, the optimal basic feasible solution to the given LP problem is: x1 = 21/13, x2 = 10/13 and Max Z* = – 31/13 or Min Z = 31/13.

Table 2.11. Optimal Solution

cj → – 1 – 1 0 0


cB

Basic VariablesB


x1 x2 s1 s2

– 1 x1 21/13 1 0 – 7/13 1/13

– 1 x2 10/13 0 1 1/13 – 2/13

Z* = – 31/13 zj – 1 – 1 6/13 1/13

cj – zj 0 0 – 6/13 – 1/13

Example 3. Solve the following LP problem by using the two-phase simplex method.Minimize Z = x1 – 2x2 – 3x3

subject to the constraints (i) – 2x1 + x2 + 3x3 = 2, (ii) 2x1 + 3x2 + 4x3 = 1and x1, x2, x3 ≥ 0.

Sol. After converting the objective function into the maximization form and by adding artificial variables A1 and A2 in the constraints, the given LP problem becomes:

Maximize Z* = – x1 + 2x2 + 3x3

subject to the constraints (i) – 2x1 + x2 + 3x3 + A1 = 2, (ii) 2x1 + 3x2 + 4x3 + A2 = 1and x1, x2, x3, A1, A2 ≥ 0where Z* = – Z

Phase I: This phase starts by considering the following auxiliary LP problem:Maximize Z* = – A1 – A2

subject to the constraints (i) – 2x1 + x2 + 3x3 + A1 = 2 (ii) 2x1 + 3x2 + 4x3 + A2 = 1and x1, x2, x3, A1, A2 – 0

The initial solution is presented in Table 2.12.

NOTES



Table 2.12. Initial Solutioncj → 0 0 0 – 1 – 1

Basic Variables Coefficient cB

Basic Variables B

Basic Variables

Value b(= xB)x1 x2 x3 A1 A2

– 1 A1 2 – 2 1 3 1 0

– 1 A2 1 2 3 4 0 1 →

Z* = – 3 zj 0 – 4 – 7 – 1 – 1

cj – zj 0 4 7 0 0↑

To first remove the artificial variable A2 from the solution shown in Table 2.12, introduce variable x3 into the basis by applying the following row operations:

R1 (new) → R2 (old) ÷ 4 (key element) ; R1 (new) → R1 (old) – 3R2 (new)The improved solution so obtained is given in Table 2.13. Since in Table 2.13,

cj – zj ≤ 0 corresponds to non-basic variables, the optimal solution is: x1 = 0, x2 = 0, x3 = 1/4, A1 = 5/4 and A2 = 0 with Max Z* = – 5/4. But at the same time, the value of Z* < 0 and the artificial variable A1 appears in the basis with positive value 5/4. Hence, feasible solution to the given original LP problem does not exist.

Table 2.13. Optimal but not Feasible Solutioncj → 0 0 0 – 1


cB

Basic Variables

B

Basic Variables Value b(= xB)

x1 x2 x3 A1

– 1 A1 5/4 –7/2 –5/4 0 10 x3 1/4 1/2 3/4 1 0

Z* = – 5/4 zj 7/2 5/4 0 – 1

cj – zj – 7/2 – 5/4 0 0

Example 4 Use two-phase simplex method to solve following LP problem.Maximize Z = 3x1 + 2x2 + 2x3

subject to the constraints (i) 5x1 + 7x2 + 4x3 ≤ 7, (ii) – 4x1 + 7x2 + 5x3 ≥ – 2, (iii) 3x1 + 4x2 – 6x3 ≥ 29/7and x1, x2, x3 ≥ 0.

Sol. Since RHS of constraint 2 is negative, multiplying it by – 1 on both sides and express it as: 4x1 – 7x2 – 5x3 ≤ 2.

Phase I: Introducing slack, surplus and artificial variables in the constraints, the standard form of LP problem becomes:

Minimize Z* = A1subject to the constraints (i) 5x1 + 7x2 + 4x3 + s1 = 7, (ii) 4x1 – 7x2 – 5x3 + s2 = 2, (iii) 3x1 + 4x2 – 6x3 – s3 + A1 = 29/7and x1, x2, x3, s1, s2, s3, A1 ≥ 0.

NOTES

Operations Research


Setting variables x1 = x2 = x3 = s3 = 0, the basic feasible solution shown in Table 2.14 to the auxiliary LP problem is: s1 = 7, s2 = 2, A1 = 29/7

Table 2.14. Initial Solutioncj → 0 0 0 0 0 0 1

Basic Variables

Coefficient cB

Basic Variables

B

Basic Variables

Value b(= xB )

x1 x2 x3 s1 s2 s3 A1

Min Ratio xB /x2

0 s1 7 5 7 4 1 0 0 0 1 →0 s2 2 4 – 7 – 5 0 1 0 0 —1 A1 29/7 3 4 – 6 0 0 – 1 1 29/28

Z = 29/7 zj 3 4 – 6 0 0 – 1 1cj – zj – 3 – 4 6 0 0 1 0

↑

Since c2 – z2 = – 4 is the largest negative value under x2-column in Table 2.14, replacing basic variable s1 with the non-basic variable x2 into the basis. For this, apply following row operations:

R1(new) → R1(old) ÷ 7(key element); R2(new) → R2(old) + 7R1(new);R3(new) → R3(old) – 4R1(new)to get an improved solution as shown in Table 2.15.

Table 2.15. An Improved Solution

cj → 0 0 0 0 0 0 1


cB

Basic Variables

B

Basic Variables

Value b(=xB )

x1 x2 x3 s1 s2 s3 A1

Min Ratio xB /x1

0 x2 1 5/7 1 4/7 1/7 0 0 0 7/5

0 s2 9 9 0 – 1 1 1 0 0 1

1 A1 1/7 1/7 0 – 58/7 – 4/7 0 –1 1 1

Z = 1/7 zj 1/7 0 – 58/7 – 4/7 0 – 1 1

Cj – zj –1/7 0 58/7 4/7 0 1 0

↑

Solution shown in Table 2.15 is not optimal and there is a tie between the s2 and A1-rows for the key row. The A1-row is selected as the key row because preference is given to the artificial variable. Thus (1/7) is the key element. Replace basic variable A1 in the basis with the non-basic variable x1. Applying following row operations to get the new solution as shown in Table 2.16.

R3(new) → R3(old) ÷ (1/7) key element; R1(new) → R1(old) – (5/7) R3 (new);R2(new) → R2(old) – 9 R3(new)

Tie

NOTES



Table 2.16. Basic Feasible Solution

cj → 0 0 0 0 0 0


cB

Variables in Basis

B

Basic Variables

Value b(= xB )

x1 x2 x3 s1 s2 s3

0 x2 2/7 0 1 42 3 0 5

0 s2 0 0 0 521 37 1 63

0 x1 1 1 0 – 58 – 4 0 – 7

Z = 0 zj 0 0 0 0 0 0

cj – zj 0 0 0 0 0 0

In Table 2.16, all cj – zj = 0 under non-basic variable columns. Thus, the current solution is optimal. Also Min Z* = 0 and no artificial variable appears in the basis, and hence this solution is the basic feasible solution to the original LP problem.

Phase II: Using the solution shown in Table 2.16 as the initial solution for Phase II and carrying out computations to get optimal solution as shown in Table 2.17.

Table 2.17. Initial Solutioncj → 3 2 2 0 0 0


cB

Basic Variables

B

Basic Variables

Value b(= xB )


xB /x3

2 x2 2/7 0 1 42 3 0 5 1/1470 s2 0 0 0 521 37 1 63 0 →3 x1 1 1 0 – 58 – 6 0 – 7 —

Z = 25/7 zj 3 2 – 90 – 6 0 – 11cj – zj 0 0 92 6 0 11

↑

In Table 2.17, c3 – z3 = 92 is the largest positive value corresponding non basic variable x3, replacing basic variable s2 with non-basic variable x3 into the basis. For this apply necessary row operations as usual. The new solution is shown in Table 2.18.


cj → 3 2 2 0 0 0


cB

Basic Variables

B

Basic Variables

Value b(= xB)x1 x2 x3 s1 s2 s3

2 x2 2/7 0 1 0 9/521 – 42/521 – 41/521

2 x3 0 0 0 1 37/521 1/521 63/521

3 x1 1 1 0 0 62/521 58/521 7/521

Z = 25/7 zj 3 2 2 278/521 92/521 65/521

cj – zj 0 0 0 –278/521 – 92/521 – 65/521

NOTES

Operations Research


Since all cj – zj ≤ 0 in Table 2.18, the current solution is the optimal basic feasible solution: x1 = 1, x2 = 2/7, x3 = 0 and Max Z = 25/7.

2.4.2 Big-M MethodBig-M method is another method of removing artificial variables from the basis. In

this method, large undesirable (unacceptable penalty) coefficients to artificial variables are assigned from the point of view of the objective function. If the objective function Z is to be minimized, then a very large positive price (called penalty) is assigned to each artificial variable. Similarly, if Z is to be maximized, then a very large negative price (also called penalty) is assigned to each of these variables. The penalty is supposed to be designated by – M, for a maximization problem, and + M, for a minimization problem, where M > 0. The Big-M method for solving an LP problem can be summarized in the following steps:

Step 1: Express the LP problem in the standard form by adding slack variables, surplus variables and/or artificial variables. Assign a zero coefficient to both slack and surplus variables. Then assign a very large coefficient + M (minimization case) and – M (maximization case) to artificial variable in the objective function.

Step 2: The initial basic feasible solution is obtained by assigning zero value to decision variables, x1, x2, ..., etc.

Step 3: Calculate the values of cj – zj in last row of the simplex table and examine these values. (i) If all cj – zj ≥ 0, then the current basic feasible solution is optimal. (ii) If for a column, k, ck – zk is most negative and all entries in this column are

negative, then the problem has an unbounded optimal solution. (iii) If one or more cj – zj < 0 (minimization case), then select the variable to enter

into the basis (solution mix) with the largest negative cj – zj value (largest per unit reduction in the objective function value). This value also represents the opportunity cost of not having one unit of the variable in the solution. That is,

ck – zk = Min {cj – zj : cj – zj < 0}Step 4: Determine the key row and key element in the same manner as discussed

in the simplex algorithm for the maximization case.Step 5: Continue with the procedure to update solution at each iteration till

optimal solution is obtained.Remarks: At any iteration of the simplex algorithm one of the following cases may arise:

1. If at least one artificial variable is a basic variable (i.e., variable that is present in the basis) with zero value and the coefficient it M in each cj – zj (j = 1, 2, ..., n) values is non-negative, then the given LP problem has no solution. That is, the current basic feasible solution is degenerate.

2. If at least one artificial variable is present in the basis with a positive value and the coefficients M in each cj – zj (j = 1, 2, …, n) values is non-negative, then the given LP problem has no optimum basic feasible solution. In this case, the given LP problem has a pseudo optimum basic feasible solution.

Example 5. Use penalty (Big-M) method to solve the following LP problem.Minimize Z = 5x1 + 3x2

NOTES



subject to the constraints (i) 2x1 + 4x2 ≤ 12, (ii) 2x1 + 2x2 = 10, (iii) 5x1 + 2x2 ≥ 10and x1, x2 ≥ 0.

Sol. Adding slack variable, s1; surplus variable, s2 and artificial variables, A1 and A2 in the constraints of the given LP problem, the standard form of the LP problem becomes.

Minimize Z = 5x1 + 3x2 + 0s1 + 0s2 + MA1 + MA2

subject to the constraints (i) 2x1 + 4x2 + s1 = 12, (ii) 2x1 + 2x2 + A1 = 10 (iii) 5x1 + 2x2 – s2 + A2 = 10and x1, x2, s1, s2, A1, A2 ≥ 0

An initial basic feasible solution: s1 = 12, A1 = 10, A2 = 10 and Min Z = 10M + 10M = 20M is obtained by putting x1 = x2 = s2 = 0. It may be noted that the columns that correspond to the current basic variables and form the basis (identity matrix) are s1 (slack variable), A1 and A2 (both artificial variables). The initial basic feasible solution is given in Table. 2.19

Since the value c1 – z1= 5 – 7M is the smallest value, therefore variable x1 is chosen to enter into the basis (solution mix). To decide a current basic variable to leave the basis, calculate minimum ratio as shown in Table 2.19.


cj → 5 3 0 0 M M


cB

Basic Variables

B

Basic Variables

Value b (= xB )x1 x2 s1 s2 A1 A2

Min Ratio xB /x1

0 s1 12 2 4 1 0 0 0 12/2 = 6

M A1 10 2 2 0 0 1 0 10/2 = 5

M A2 10 5 2 0 – 1 0 1 10/5 = 2 →

Z = 20M zj 7M 4M 0 – M M M

cj – zj 5 – 7M 3 – 4M 0 M 0 0

↑

Iteration 1: Introduce variable x1 into the basis and remove A2 from the basis by applying the following row operations.

R3(new) → R3(old) ÷ 5(key element); R2(new) → R2(old) – 2R3(new).R1(new) → R1(old) – 2R3(new).The improved basic feasible solution is shown in Table 2.20

NOTES

Operations Research



cj → 5 3 0 0 M


cB

Basic Variables

B

Basic Variables

Value b(= xB )

x1 x2 s1 s2 A1

Min RatioxB /x2

0 s1 8 0 16/5 1 2/5 0 8/(16/5)= 5/2 →

M A1 6 0 6/5 0 2/5 1 6/(6/5) = 5

5 x1 2 1 2/5 0 – 1/5 0 2/(2/5) = 5

Z = 10 + 6M zj 5 (6M/5) + 2 0 (2M/5) – 1 M

cj – zj 0 (– 6M/5) + 1 0 (– 2M/5) + 1 0

↑

Iteration 2: Since the value of c2 – z2 in Table 2.20 is the largest negative value, variable x2 is chosen to replace basic variable s1 in the basis. Thus, to get an improved basic feasible solution, apply, the following row operations:

R1(new) → R1(old) × 5/16 (key element); R2(new) → R2(old) – (6/5) R1(new).R3(new) → R3(old) – (2/5) R1(new).The new solution is shown in Table 2.21.

Table 2.21. Improved Solutioncj → 5 3 0 0 M


cB

Basic Variables

B

Basic Variables

Value b(= xB)

x1 x2 s1 s2 A1

Min Ratio xB /x2

3 x2 5/2 0 1 5/16 1/8 0 (5/2)/(1/8) = 40

M A1 3 0 0 – 3/8 1/4 1 3/(1/4) = 12 →

5 x1 1 1 0 – 1/8 – 1/4 0Z = 25/2 + 3M zj 5 3 – 3M/8 + 5/16 M/4 – 7/8 M

cj – zj 0 0 3M/8 – 5/16 – M/4 + 7/8 0↑

Iteration 3: Since c4 – z4 < 0 (negative) in s2-column, the current solution is not optimal. Thus, non-basic variable s2 is chosen to replace artificial variable A1 in the basis. To get an improved basic feasible solution, apply the following row operations:

R2(new) → R2(old) × 4(key element); R1(new) → R1(old) – (1/8) R2(new)R3(new) → R3(old) + (1/4) R2(new).The improved basic feasible solution is shown in Table 2.22.

NOTES



Table 2.22. Optimal Solutioncj → 5 3 0 0

Basic Variables

Coefficient cB

Basic Variables

B

Basic Variables

Value b(= xB )x1 x2 s1 s2

3 x2 1 0 1 1/2 00 s2 12 0 0 – 3/2 15 x1 4 1 0 – 1/2 0

Z = 23 zj 5 3 – 1 0cj – zj 0 0 1 0

In Table 2.17, all cj – zj ≥ 0. Thus an optimal solution is arrived at with the value of variables as: x1 = 4, x2 = 1, s1 = 0, s2 = 12 and Min Z = 23.

Example 6. Use penalty (Big-M) method to solve the following LP problem.Maximize Z = x1 + 2x2 + 3x3 – x4

subject to the constraints (i) x1 + 2x2 + 3x3 = 15, (ii) 2x1 + x2 + 5x3 = 20, (iii) x1 + 2x2 + x3 + x4 = 10and x1, x2, x3, x4 ≥ 0

Sol. Since all constraints of the given LP problem are equations, therefore only artificial variables A1 and A2 are added in the constraints to convert given LP problem to its standard form. The standard form of the problem is stated as follows:

Maximize Z = x1 + 2x2 + 3x3 – x4 – MA1 – MA2

subject to the constraints (i) x1 + 2x2 + 3x3 + A1 = 15, (ii) 2x1 + x2 + 5x3 + A2 = 20 (iii) x1 + 2x2 + x3 + x4 = 10and x1, x2, x3, x4, A1, A2 ≥ 0

An initial basic feasible solution is given in Table 2.23.

Table 2.23. Initial Solutioncj → 1 2 3 –1 – M – M


cB

Basic Variables

B

Basic Variables

Value b(= xB)

x1 x2 x3 x4 A1 A2Min Ratio

xB /x3

– M A1 15 1 2 3 0 1 0 15/3 = 5– M A2 20 2 1 5 0 0 1 20/5 = 4 →– 1 x4 10 1 2 1 1 0 0 10/1 = 10

Z = – 35M – 10 zj – 3M – 1 – 3M – 2 – 8M – 1 –1 – M – Mcj – zj 3M + 2 3M + 4 8M + 4 0 0 0

↑

Since value of c3 – z3 in Table 2.23 is the largest positive value, the non-basic variable x3 is chosen to replace the artificial variable A2 in the basis. Thus, to get an improved solution, apply the following row operations.

NOTES

Operations Research


R2(new) →R1(old) ÷ 5(key element) ; R1(new) → R1(old) – 3 R2(new).R3(new) → R3(old) – R2(new)The improved basic feasible solution is shown in Table 2.24.


cj → 1 2 3 – 1 – M


cB

Basic Variables

B

Basic Variables

Value b(= xB )

x1 x2 x3 x4 A1Min Ratio

xB /x2

– M A1 3 – 1/5 7/5 0 0 1 37 5

157/

= →

3 x3 4 2/5 1/5 1 0 0 41 5/

= 20

– 1 x4 6 3/5 9/5 0 1 0 69 5

309/

=

Z = – 3M + 6 zj M/5 + 3/5 – 7M/5 – 6/5 3 – 1 – M

cj – zj – M/5 – 2/5 7M/5 + 16/5 0 0 0↑

Since value of c2 – z2 > 0 (positive) in Table 2.24, the non-basic variable x2 is chosen to replace the artificial variable A1 in the basis. Thus, to get an improved basic feasible solution, apply the following row operations:

R1(new) → R1(old) × (5/7) (key element); R2(new) → R2(old) – (1/5) R1(new).R3(new) → R3(old) – (9/5) R1(new).The improved basic feasible solution is shown in Table 2.25.

Table 2.25. Improved Solutioncj → 1 2 3 – 1


cB

Basic Variables

B

Basic Variables

Value b(= xB )

x1 x2 x3 x4 Min Ratio xB /x1

2 x2 15/7 – 1/7 1 0 0 –3 x3 25/7 3/7 0 1 0 25/7 × 7/3 = 25/3

– 1 x4 15/7 6/7 0 0 1 15/7 × 7/6 = 15/6 →Z = 90/7 zj 1/7 2 3 –1

cj – zj 6/7 0 0 0↑

Once again, the solution shown in Table 2.25 is not optimal as c1 – z1 > 0 in x1-column. Thus, applying the following row operations for replacing non-basic variable x1 in the basis with basic variable x4:

R3(new) → R3(old) × (7/6) (key element) ; R1(new) → R1(old) + (1/7) R3(new).R2(new) → R2(old) – (3/7) R3(new).

NOTES


The improved basic feasible solution is shown in Table 2.26.


cj → 1 2 3 – 1


Basic Variables B Basic Variables Value b(= xB ) x1 x2 x3 x4

2 x2 15/6 0 1 0 1/6

3 x3 15/6 0 0 1 – 3/6

1 x1 15/6 1 0 0 7/6

Z = 15 zj 1 2 3 0

cj – zj 0 0 0 – 1

In Table 2.21, since all cj – zj ≤ 0, therefore, an optimal solution is arrived at with value of variables as: x1 = 15/6, x2 = 15/6, x3 = 15/6 and Max Z = 15.

Example 7. ABC Printing Company is facing a tight financial squeeze and is attempting to cut costs wherever possible. At present it has only one printing contract, and luckily, the book is selling well in both the hardcover and the paperback editions. It has just received a request to print more copies of this book in either the hardcover or the paperback form. The printing cost for the hardcover books is ` 600 per 100 books while that for paperback is only ` 500 per 100. Although the company is attempting to economize, it does not wish to lay off any employee. Therefore, it feels obliged to run its two printing presses – I and II, at least 80 and 60 hours per week, respectively. Press I can produce 100 hardcover books in 2 hours or 100 paperback books in 1 hour. Press II can produce 100 hardcover books in 1 hour or 100 paperbacks books in 2 hours. Determine how many books of each type should be printed in order to minimize costs.

Sol. Let x1 and x2 be the number of batches containing 100 hard cover and paperback books, to be printed respectively. The LP problem can then be formulated as follows.

Minimize Z = 600x1 + 500x2

subject to the constraints (i) 2x1 + x2 ≥ 80, (ii) x1 + 2x2 ≥ 60 (Printing press hours)and x1, x2 ≥ 0.

Standard form Adding surplus variables s1, s2 and artificial variables A1, A2 in the constraints, the standard form of the LP problem becomes

Minimize Z = 600x1 + 500x2 + 0s1 + 0s2 + MA1 + MA2

subject to the constraints (i) 2x1 + x2 – s1 + A1 = 80, (ii) x1 + 2x2 – s2 + A2 = 60and x1, x2, s1, s2, A1, A2 ≥ 0

Solution by simplex method. The initial basic feasible solution: A1 = 80, A2 = 60 and Min Z = 80M + 60M = 140M obtained by putting x1 = x2 = s1 = s2 = 0 is shown in Table 2.27.



NOTES

Operations Research Table 2.27. Initial Solution

cj → 600 500 0 0 M M


cB

Basic Variables

B

Basic Variables

Value b(= xB)

x1 x2 s1 s2 A1 A2

Min Ratio xB /x2

M A1 80 2 1 – 1 0 1 0 80/1

M A2 60 1 2 0 – 1 0 1 60/2 →

Z = 140M zj 3M 3M – M – M M M

cj – zj 600 – 3M 500 – 3M M M 0 0

↑

Since c2 – z2 value in x2-column of Table 2.27 is the largest negative, therefore non-basic variable x2 is chosen to replace basic variable A2 in the basis. For this, apply following row operations.

R2(new) = R2(old) ÷ 2(key element); R1(new) → R1(old) – R2(new)to get an improved basic feasible solution as shown in Table 2.28.


cj → 600 500 0 0 M

Basic Variables

Coefficient cB

Basic Variables

B

Basic Variables

Value b(= xB)

x1 x2 s1 s2 A1

Min Ratio xB /x1

M A1 50 3/2 0 – 1 1/2 1 100/3 →

500 x2 30 1/2 1 0 – 1/2 0 60

Z = 15,000 + 50M zj 3M/2 + 250 500 – M M/2 – 250 M

cj – zj 350 – 3M/2 0 M 250 – M/2 0

↑

Again, since value c1 – z1 of in x1-column of Table 2.28 is the largest negative, non-basic variable x1 is chosen to replace basic variable A1 in the basis. For this, apply following row operations:

R1(new) → R1(old) × (2/3) (key element) ; R2(new) → R2(old) – (1/2) R1(new)


NOTES


to get an improved basic feasible solution as shown in Table 2.29.

Table 2.29. Optimal Solutioncj → 600 500 0 0

Basic Variables

Coefficient cB

Basic Variables

B

Basic Variables

Value b(= xB )

x1 x2 s1 s2

600 x1 100/3 1 0 – 2/3 1/3500 x2 40/3 0 1 1/3 – 2/3

Z = 80,000/3 zj 600 500 – 700/3 – 400/3cj – zj 0 0 700/3 400/3

In Table 2.29, all cj – zj ≥ 0 and no artificial variable is present in the basis (solution mix). Hence, an optimum solution is arrived at with x1 = 100/3 batches of hardcover books, x2 = 40/3 batches of paperback books, at a total minimum cost, Z = ` 80,000/3.

Example 8. An advertising agency wishes to reach two types of audiences: Customers with annual income greater than 15,000 (target audience A) and customers with annual income less than 15,000 (target audience B). The total advertising budget is 2,00,000. One programme of TV advertising costs 50,000; one programme on radio advertising costs 20,000. For contract reasons, at least three programmes ought to be on TV, and the number of radio programmes must be limited to five. Surveys indicate that a single TV programme reaches 4,50,000 customers in target audience A and 50,000 in target audience B. One radio programme reaches 20,000 in target audience A and 80,000 in target audience B. Determine the media mix to maximize the total reach.

Sol. Let x1 and x2 be the number of insertions in TV and radio, respectively. The LP problem can then be formulated as follows:

Maximize (total reach) Z = (4,50,000 + 50,000) x1 + (20,000 + 80,000) x2

= 5,00,000 x1 + 1,00,000 x2 = 5x1 + x2subject to the constraints (i) 50,000 x1 + 20,000 x2 ≤ 2,00,000 or 5x1 + 2x2 ≤ 20 (Advt. budget) (ii) x1 ≥ 3 (Advt. on TV) (iii) x2 ≤ 5 (Advt. on Radio)and x1, x2 ≥ 0.

Standard form Adding slack/surplus and/or artificial variables in the constraints of LP problem. Then standard form of given LP problem becomes

Maximize Z = 5x1 + x2 + 0s1 + 0s2 + 0s3 – MA1

subject to the constraints (i) 5x1 + 2x2 + s1 = 20, (ii) x1 – s2 + A1 = 3, (iii) x2 + s3 = 5and x1, x2, s1, s2, s3, A1 ≥ 0


NOTES

Operations Research Solution by simplex method An initial basic feasible solution: s1 = 20, A1 = 3, s3 = 5 and Max Z = – 3M obtained by putting x1 = x2 = s2 = 0 is shown in simplex Table 2.30.


cj → 5 1 0 0 0 –M

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables

Value b(= xB)

x1 x2 s1 s2 s3 A1

Min Ratio xB /x1

0 s1 20 5 2 1 0 0 0 20/5 = 4

– M A2 3 1 0 0 – 1 0 1 3/1 = 3 →

0 s3 5 0 1 0 0 1 0 —

Z = – 3M zj – M 0 0 M 0 – M

cj – zj M + 5↑

1 0 – M 0 0

The c1 – z1 value in x1-column of Table 2.30 is the largest positive value. The non-basic variable x1 is chosen to replace basic variable A1 in the basis. For this apply following row operations

R2(new) → R2(old) ÷ 1(key element); R1(new) → R1(old) – 5R2(new)to get the improved basic feasible solution as shown in Table 2.31.


cj → 5 1 0 0 0

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables

Value b(= xB)x1 x2 s1 s2 s3

Min Ratio xB/s2

0 s1 5 0 2 1 5 0 5/5 = 1 →

5 x1 3 1 0 0 – 1 0 —

0 s3 5 0 1 0 0 1 —

Z = 15 zj 5 0 0 – 5 0

cj – zj 0 1 0 5 0

↑

Again in Table 2.31, c4 – z4 value in s2-column is the largest positive. Thus non-basic variable s2 is chosen to replace basic variable s1 into the basis. For this apply the following row operations:

R1(new) → R1(old) ÷ 5(key element); R2(new) → R2(old) + R1(new)


NOTES


to get the improved basic feasible solution as shown in Table 2.32.


cj → 5 1 0 0 0


Basic Variables B

Basic Variables Value b(= xB) x1 x2 s1 s2 s3

0 s2 1 0 2/5 1/5 1 0

5 x1 4 1 2/5 1/5 0 0

0 s3 5 0 1 0 0 1

Z = 20 zj 5 2 1 0 0

cj – zj 0 – 1 – 1 0 0

Since all cj – zj ≥ 0 in Table 2.32, the total reach of target audience cannot be increased further. Hence, the optimal solution is: x1 = 4 insertions in TV and x2 = 0 in radio with Max (total audience) Z = 20,00,000.

Example 9. An Air Force is experimenting with three types of bombs P, Q and R in which three kinds of explosives, viz., A, B and C will be used. Taking the various factors into account, it has been decided to use the maximum 600 kg of explosive A, at least 480 kg of explosive B and exactly 540 kg of explosive C. Bomb P requires 3, 2, 2 kg, bomb Q requires 1, 4, 3 kg and bomb R requires 4, 2, 3 kg of explosives A, B and C respectively. Bomb P is estimated to give the equivalent of a 2 ton explosion, bomb Q, a 3 ton explosion and bomb R, a 4 ton explosion respectively. Under what production schedule can the Air Force make the biggest bang?

Sol. Let x1, x2 and x3 be the number of bombs of type P, Q and R to be experimented, respectively.

Then the LP problem can be formulated as:Maximize Z = 2x1 + 3x2 + 4x3

subject to the constraints (i) Explosive A : 3x1 + x2 + 4x3 ≤ 600, (ii) Explosive B : 2x1 + 4x2 + 2x3 ≥ 480, (iii) Explosive C : 2x1 + 3x2 + 3x3 = 540,and x1, x2, x3 ≥ 0.

Standard form Adding slack, surplus and artificial variables in the constraints of LP problem, the standard form of LP problem becomes:

Maximize Z = 2x1 + 3x2 + 4x3, + 0s1 + 0s2 – MA1 – MA2

subject to the constraints (i) 3x1 + x2 + 4x3 + s1 = 600, (ii) 2x1 + 4x2 + 2x3 – s2 + A1 = 480, (iii) 2x1 + 3x2 + 3x3 + A2 = 540,and x1, x2, x3 s1, s2, A1, A2 ≥ 0.

Solution by simplex method The initial basic feasible solution: s1 = 600, A1 = 480, A2 = 540,


NOTES

Operations Research Max Z = –1,020 M obtained by putting basic variables x1 = x2 = s2 = 0 is shown in Table 2.33.


cj → 2 3 4 0 0 –M –M


cB

Basic Variables

B

Basic Variables

Value b(= xB)

x1 x2 x3 s1 s2 A1 A2

Min Ratio xB /x2

0 s1 600 3 1 4 1 0 0 0 600/1

–M A1 480 2 4 2 0 –1 1 0 → 480/4

–M A2 540 2 3 3 0 0 0 1 540/3

Z = 1,020M zj –4M –7M –5M 0 M –M –M

cj – zj 2 + 4M 3 + 7M↑

4 + 5M 0 –M 0 0

Since in Table 2.33, c2 – z2 value in x2-column is largest positive, non-basic variable x2 is chosen to replace basic variable A1 in the basis. For this, apply the following row operations:

R2(new) → R2(old) × (1/4) (key element); R1(new) → R1(old) – R2(new);R3(new) → R3(old) – 3R2(new)



cj → 2 3 4 0 0 M


Basic Variables

B

Basic Variables

Value b(= xB)

x1 x2 x3 s1 s2 A1

Min Ratio xB /x3

0 s1 480 5/2 0 7/2 1 1/4 0 960/7

3 x2 120 1/2 1 1/2 0 –1/4 0 240

–M A1 180 1/3 0 3/2 0 3/4 1 120 →

Z = 360 – 180M zj 3/2– M/2 3 3/2 – 3M/2 0 –3/4 – 3M/4 –M

cj – zj 1/2 + M/2 0 5/2 + 3M/2 0 3/4 + 3M/4 0

↑

In Table 2.34, c3 – z3 value in x3-column is largest positive, non-basic variable x3 is chosen to replace basic variable A2 into the basis. For this, apply following row operations:

R3(new) → R3(old) × (2/3) (key element); R1(new) → R1(old) – (7/2) R3(new);R2(new) → R2(old) – (1/2) R3(new).


NOTES



Table 2.35. Optimal Solutioncj → 2 3 4 0 0


Basic Variables B

Basic Variables Value b(= xB)

x1 x2 x3 s1 s2

0 s1 60 4/3 0 0 1 –3/23 x2 60 1/3 1 0 0 –1/24 x3 120 1/3 0 1 0 1/2

Z = 660 zj 7/3 3 4 0 1/2cj – zj –1/3 0 0 0 –1/2

In Table 2.35, all cj – zj ≤ 0 and artificial variables A1 and A2 have been removed from the basis (solution mix). Thus, an optimal solution is arrived at with x1 = 0 bombs of type P, x2 = 60 bombs of type Q, x3 = 120 bombs of type R, at largest benefit of Z = 660.

TEST YOUR KNOWLEDGE

1. A television company has three major departments for manufacturing two of its models – A and B. The monthly capacities of the departments are given as follows:

Per Unit Time Requirement (hour) Hours Available this MonthModel A Model B

Department I 4.0 2.0 1,600Department II 2.5 1.0 1,200Department III 4.5 1.5 1,600

The marginal profit per unit from model A is 400 and from model B is 100. Assuming that the company can sell any quantity of either product due to favourable market conditions, determine the optimum output for both the models, the highest possible profit for this month and the slack time in the three departments.

2. A manufacturer of leather belts makes three types of belts A, B and C which are processed on three machines M1, M2 and M3. Belt A requires 2 hours on machine M1 and 3 hours on machine M2 and 2 hours on machine M3. Belt B requires 3 hours on machine M1, 2 hours on machine M2 and 2 hours on machine M3 and Belt C requires 5 hours on machine M2 and 4 hours on machine M3. There are 8 hours of time per day available on machine M1, 10 hours of time per day available on machine M2 and 15 hours of time per day available on machine M3. The profit gained from belt A is ` 3.00 per unit, from Belt B is ` 5.00 per unit, from belt C is ` 4.00 per unit. What should be the daily production of each type of belt so that the products yield the maximum profit?

3. A company produces three products A, B and C. These products require three ores O1, O2 and O3. The maximum quantities of the ores O1, O2 and O3 available are 22 tonnes, 14 tonnes and 14 tonnes, respectively. For one tonne of each of these products, the ore requirements are:

A B CO1 3 – 3O2 1 2 3O3 3 2 3

Profit per tonne (` in thousand) 1 4 5


NOTES

Operations Research The company makes a profit of ` 1,000, 4,000 and 5,000 on each tonne of the products A, B and C, respectively. How many tonnes of each product should the company produce in order to maximize its profits.

4. A manufacturing firm has discontinued the production of a certain unprofitable product line. This has created considerable excess production capacity. Management is considering to devote this excess capacity to one or more of three products; call them product 1, 2 and 3. The available capacity on the machines that might limit output is summarized in the following table:

Machine Type Available Time (in Machine-hours per Week)

Milling Machine 250Lathe 150Grinder 50

The number of machine-hours required for each unit of the respective product is as follows:

Machine TypeProductivity

(in Machine-hours per Unit)Product 1 Product 2 Product 3

Milling Machine 8 2 3Lathe 4 3 0Grinder 2 – 1

The profit per unit would be ` 20, ` 6 and ` 8, respectively for product 1, 2 and 3. Find how much of each product the firm should produce in order to maximize its profit.

5. A farmer has 1,000 acres of land on which he can grow corn, wheat or soyabean. Each acre of corn costs ` 100 for preparation, requires 7 men-days of work and yields a profit of ` 30. An acre of wheat costs ` 120 to prepare, requires 10 men-days of work and yields a profit of ` 40. An acre of soyabean costs ` 70 to prepare, requires 8 men-days of work and yields a profit of 20. If the farmer has 1,00,000 for preparation and can count on 8,000 men-days of work, determine how many acres should be allocated to each crop in order to maximize profits?

6. The annual handmade furniture show and sale is supposed to take place next month and the school of vocational studies is also planning to make furniture for this sale. There are three wood-working classes – I year, II year and III year, at the school and they have decided to make styles of chairs – A, B and C. Each chair must receive work in each class. The time in hours required for each chair in each class is:

Chair I Year II Year III YearA 2 4 3B 3 3 2C 2 1 4

During the next month there will be 120 hours available to the I year class, 160 hours to the II year class, and 100 hours to the III year class for producing the chairs. The teacher of the wood-working classes feels that a maximum of 40 chairs can be sold at the show. The teacher has determined that the profit from each type of chair will be: A, ` 40; B, ` 35 and C, ` 30. How many chairs of each type should be made in order to maximize profits at the show and sale?

7. Mr Jain, the marketing manager of ABC Typewriter Company is trying to decide how he should allocate his salesmen to the company’s three primary markets. Market I is in the


NOTES


urban area and the salesman can sell, on the average, 40 typewriters a week. Salesmen in the other two markets, II and III can sell, on the average, 36 and 25 typewriters per week, respectively. For the coming week, three of the salesmen will be on vacation, leaving only 12 men available for duty. Also because of lack of company care, a maximum of 5 salesmen can be allocated to market area I. The selling expenses per week for salesmen in each area are 800 per week for area I, 700 per week for area II, and 500 per week for area III. The budget for the next week is ` 7,500. The profit margin per typewriter is ` 150. Determine how many salesmen should be assigned to each area in order to maximize profits?

8. An animal feed company must produce on a daily basis 200 kg of a mixture that consists ingredients x1 and x2 ingredient. x1 costs 3 per kg and x2 costs 8 kg. Not more than 80 kg of x1 can be used and at least 60 kg of x2 must be used. Find out how much of each ingredient should be used if the company wants to minimize costs.

9. A diet is to contain at least and 20 ounces of protein and 15 ounces of carbohydrate. There are three foods A, B and C available in the market, costing 2, 1 and 3 per unit, respectively. Each unit of A contains 2 ounces of protein and 4 ounces of carbohydrate; each unit of B contains 3 ounces of protein and 2 ounces of carbohydrate; and each unit of C contains 4 ounces of protein and 2 ounces of carbohydrate. How many units of each food should the diet contain so that the cost per unit diet is minimum?

10. A person requires 10, 12 and 12 units of chemicals A, B and C, respectively for his garden. A typical liquid product contains 5, 2 and 1 unit of A, B and C, respectively per jar. On the other hand a typical dry product contains 1, 2 and 4 units of A, B and C per unit. If the liquid product sells for ` 3 per jar and dry product for ` 2 per carton, how many of each should be purchased in order to minimize the cost and meet the requirement?

11. A scrap metal dealer has received an order from a customer for a minimum of 2,000 kg of scrap metal. The customer requires that at least 1,000 kg of the shipment of metal be of high quality copper that can be melted down and used to produce copper tubing. Furthermore, the customer will not accept delivery of the order if it contains more than 175 kg metal that he deems unfit for commercial use, i.e. metal that contains an excessive amount of impurities and cannot be melted down and defined profitably.

The dealer can purchase scrap metal from two different suppliers in unlimited quantities with following percentages (by weight) of high quality copper and unfit scrap.

Supplier A Supplier BCopper 25% 75%Unfit scrap 5% 10%

The cost per kg of metal purchased from supplier A and B are ` 1 and ` 4, respectively. Determine the optimal quantities of metal to be purchased for the dealer from each of the two suppliers.

12. A marketing manager wishes to allocate his annual advertising budget of 2,00,000 to two media vehicles – A and B. The unit cost of a message in media A is ` 1,000 and that of B is ` 1,500. Media A is a monthly magazine and not more than one insertion is desired in one issue, whereas at least five massages should appear in media B. The expected audience for unit messages in media A is 40,000 and that of media B is 55,000. Develop an LP model and solve it for maximizing the total effective audience.

13. A transistor radio company manufactures four models A, B, C and D. Models A, B and C, have profit contributions of 8, 15 and 25 respectively and has model D a loss of 1. Each type of radio requires a certain amount of time for the manufacturing of components, for assembling and for packing. A dozen units of model A require one hour for manufacturing, two hours for assembling and one hour for packing. The corresponding figures for a dozen units of model B are 2, 1 and 2, and for a dozen units of C are 3, 5 and 1. A dozen units of model D however, only require 1 hour of packing. During the forthcoming week, the company will be able to make available 15 hours of manufacturing, 20 hours of assembling and 10 hours of packing time. Determine the optimal production schedule for the company.


NOTES

Operations Research 14. A transport company is considering the purchase of new vehicles for providing transportation between the Delhi Airport and hotels in the city. There are three vehicles under consideration: Station wagons, minibuses and large buses. The purchase price would be 1,45,000 for each station wagon, ` 2,50,000 for each minibus and ` 4,00,000 for each large bus. The board of directors has authorized a maximum amount of ` 50,00,000 for these purchases. Because of the heavy air travel, the new vehicles would be utilized at maximum capacity, regardless of the type of vehicles purchased. The expected net annual profit would be ` 15,000 for the station wagon, ` 35,000 for the minibus and ` 45,000 for the large bus. The company has hired 30 new drivers for the new vehicles. They are qualified drivers for all three types of vehicles. The maintenance department has the capacity to handle an additional 80 station wagons. A minibus is equivalent to 1.67 station wagons and each large bus is equivalent to 2 station wagons in terms of their use of the maintenance department. Determine the number of each type of vehicle that should be purchased in order to maximize profit.

15. Omega Data Processing Company performs three types of activities: Payroll, accounts receivables, and inventories. The profit and time requirement for keypunch, computation and office printing, for a standard job, are shown in the following table:

JobProfit/

Standard Job (`)

Time Requirement (Min)

Keypunch Computation Printing

Payroll 275 1,200 20 100A/c Receivable 125 1,400 15 60Inventory 225 800 35 80

Omega guarantees overnight completion of the job. Any job scheduled during the day can be completed during the day or night. Any job scheduled during the night, however, must be completed during the night. The capacities for both day and night are shown in the following table:

Capacity (Min) Keypunch Computation PrintDay 4,200 150 400

Night 9,200 250 650

Determine the mixture of standard jobs that should be accepted during the day and night. 16. A furniture company can produce four types of chairs. Each chair is first made in the

carpentry shop and then furnished, waxed and polished in the finishing shop. The man-hours required in each are:

Chair Type1 2 3 4

Carpentry shop 4 9 7 10Finishing shop 1 1 3 40Contribution per chair (`) 12 20 18 40

The total number of man-hours available per month in carpentry and finishing shops are 6,000 and 4,000, respectively.

Assuming an abundant supply of raw material and an abundant demand for finished products, determine the number of each type of chairs that should be produced for profit maximization.

17. A company mined diamonds in three locations in the country. The three mines differed in terms of their capacities, number, weight of stones mined, and costs. These are shown in the table below:

Due to marketing considerations, a monthly production of exactly 1,48,000 stones was required. A similar requirement called for at least 1,30,000 carats (The average stone size was at least 130/148 = 0.88 carats). The capacity of each mine is measured in cubic meter.


NOTES


The mining costs are not included from the treatment costs and assume to be same at each mine. The problem for the company was to meet the marketing requirements at the least cost.

MineCapacity

(M3 of earth processed)

Treatment Costs

(` per M3)

Crude (Carats per

M3)

Stone Count (Number of

stone per M3)Plant 1 83,000 0.60 0.360 0.58Plant 2 3,10,000 0.36 0.220 0.26Plant 3 1,90,000 0.50 0.263 0.21

Formulate a linear programming model to determine how much should be mined at each location.

Hints and Answers 1. Let x1 and x2 = units of models A and B to be manufactured, respectively.

Max Z = 400x1 + 400x2

subject to 4x1 + 2x2 ≤ 1,600 5x1/2 + x2 ≤ 1,200 9x1/2 + 3x2/2 ≤ 1,600and x1, x2 ≥ 0Ans. x1 = 355.5, x2 = 0 and Max Z = 1,42,222.2.

2. Let x1, x2 and x3 = units of types A, B and C belt to be manufactured, respectively.Max Z = 3x1 + 5x2 + 4x3

subject to 2x1 + 3x2 ≤ 8 2x2 + 5x3 ≤ 10 3x1 + 2x2 + 4x3 ≤ 15and x1, x2, x3 ≥ 0

Ans. x1 = 89/41, x2 = 50/41, x3 = 64/41 and Max Z = 775/41. 3. Let x1, x2 and x3 = quantity of products A, B and C to be produced, respectively.

Max Z = x1 + 4x2 + 5x3

subject to 3x1 + 3x2 ≤ 22 x1 + 2x2 + 3x3 ≤ 14 3x1 + 2x2 ≤ 14and x1, x2, x3 ≥ 0

Ans. x1 = 0, x2 = 7, x3 = 0 and Max Z = ` 28,000. 4. Let x1, x2 and x3 = number of units of products 1, 2 and 3 to be produced per week, respectively.

Max Z = 20x1 + 6x2 + 8x3

subject to 8x1 + 2x2 + 3x3 ≤ 250 4x1 + 3x2 ≤ 150 2x1 + x3 ≤ 50and x1, x2, x3 ≥ 0

Ans. x1 = 0, x2 = 50, x3 = 50 and Max Z = 700. 5. Let x1, x2 and x3 = acreage of corn, wheat and soyabean, respectively.

Max Z = 30x1 + 40x2 + 20x3

subject to 10x1 + 12x2 + 7x3 ≤ 10,000


NOTES

Operations Research 7x1 + 10x2 + 8x3 ≤ 8,000 x1 + x2 + x3 ≤ 1,000and x1, x2, x3 ≥ 0

Ans. x1 = 250, x2 = 625, x3 = 0 and Max Z = ` 32,500. 6. Let x1, x2 and x3 = number of units of chair of styles A, B and C, respectively.

Max Z = 40x1 + 35x2 + 30x3

subject to 2x1 + 3x2 + 2x3 ≤ 120 4x1 + 3x2 + x3 ≤ 160 3x1 + 2x2 + 4x3 ≤ 100 x1 + x2 + x3 ≤ 40and x1, x2, x3 ≥ 0

Ans. x1 = 20, x2 = 20, x3 = 0 and Max Z = ` 1,500. 7. Let x1, x2 and x3 = salesman assigned to area, 1, 2 and 3, respectively.

Max. Z = 40 × 150x1 + 36 × 150x2 + 25 × 150x3 – (800x1 + 700x2 + 500x3)subject to

(i) x1 + x2 + x3 ≤ 12; (ii) x1 ≤ 5; (iii) 800x1 + 700x2 + 500x3 ≤ 7,500

and x1, x2, x3 ≥ 0 8. Let x and y = number of kg of ingredients x1, x2, respectively.

Min (total cost) Z = 3x + 8ysubject to

(i) x + y = 200; (ii) x ≤ 80; (iii) y ≥ 60

and x, y ≥ 0Ans. x = 80, y = 120 and Min Z = ` 1,200.

9. Let x1, x2 and x3 = number of units of food A, B and C, respectively which a diet must contain.Min (total cost ) Z = 2x1 + x2 + x3

subject to 2x1 + 2x2 + 4x3 ≥ 20 4x1 + 2x2 + 2x3 ≥ 15and x1, x2, x3 ≥ 0

10. Let x1 and x2 = number of units of liquid and dry product produced, respectively.Min (total cost) Z = 3x1 + 2x2

subject to (i) 5x1 + x2 ≥ 10; (ii) 2x1 + 2x2 ≥ 12; (iii) x1 + 4x2 ≥ 12

and x1, x2 ≥ 0 Ans. x1 = 1, x2 = 5 and Min Z = 13. 11. Let x1 and x2 = number of scrap (in kg) purchased from suppliers A and B, respectively.

Min (total cost) Z = x1 + 4x2

subject to 2.25x1 + 0.75x2 ≥ 1,000 0.05x1 + 0.10x2 ≥ 175 x1 + x2 ≥ 2,000and x1, x2 ≥ 0

Ans. x1 = 2,500, x2 = 500 and Min Z = 4,500.


NOTES


12. Let x1 and x2 = number of insertions messages for media A and B, respectively.Min (total effective audience) Z = 40,000x1 + 55,000x2subject to

(i) 1,000x1 + 1500x2 ≤ 2,00,000; (ii) x1 ≤ 12; (iii) x2 ≥ 5and x1, x2 ≥ 0

Ans. x1 = 12, x2 = 16/3 and Max Z = 77,3333.33. 13. Let x1, x2, x3 and x4 = unit of models A, B, C and D to be produce, respectively.

Max (total income) Z = 8x1 + 15x2 + 25x3 – x4

subject to (i) x1 + 2x2 + 3x3 = 15; (ii) 2x1 + x2 + 5x3 = 20 (iii) x1 + 2x2 + x3 + x4 = 10and x1, x2, x3, x4 ≥ 0

Ans. x1 = 5/2, x2 = 5/2, x3 = 5/2, x4 = 0 and Max Z = 120. 14. Let x1, x2 and x3 = number of station wagons, minibuses and large buses to be purchased,

respectively.Max Z = 15,000x1 + 35,000x2 + 45,000x3

subject to x1 + x2 + x3 ≤ 30 1,45,000x1 + 2,50,000x2 + 4,00,000x3 ≤ 50,00,000 x1 + 1.67x2 + 0.5x3 ≤ 80and x1, x2, x3 ≥ 0

15. Let xij represents ith job and jth activityMax Z = 275 (x11 + x12) + 125(x21 + x22) + 225 (x31 + x32)subject to 1,200 (x11 + x12) + 1,400 (x21 + x22) + 800 (x31 + x32) ≤ 13,400 20 (x11 + x12) + 15 (x21 + x22) + 35 (x31 + x32) ≤ 400 100 (x11 + x12) + 60 (x21 + x22) + 80 (x31 + x32) ≤ 1,050 1,200x12 + 1,400x22 + 800x32 ≤ 9,200 20x12 + 15x22 + 35x32 ≤ 250 100x12 + 60x22 + 80x32 ≤ 650and xij ≥ 0 for all i, j

16. Let x1, x2, x3 and x4 = chair types 1, 2, 3 and 4 to be produced, respectively.Max Z = 12x1 + 20x2 + 18x3 + 40x4

subject to 4x1 + 9x2 + 7x3 + 10x4 ≤ 6,000 x1 + x2 + 3x3 + 40x4 ≤ 4,000and x1, x2, x3, x4 ≥ 0

Ans. x1 = 4,000/3, x2 = x3 = 0, x4 = 200/3 and Max Z = ` 56,000/3. 17. Let x1, x2 and x3 = cubic metric of earth processed at plants 1, 2 and 3, respectively.

Min Z = 0.60x1 + 0.36x2 + 0.50x3

subject to0.58x1 + 0.26x2 + 0.21x3 = 1,48,000 (Stone count requirement)0.36x1 + 0.22x2 + 0.263x3 ≤ 1,30.000 (Carat requirement) x1 ≤ 83,000; x2 ≤ 3,10,000; x3 ≤ 1,90,000 (Capacity requirement)

Ans. x1 = 61,700; x2 = 3,10,000; x3 = 1,50,500 and Min Z = ` 2,23,880.


NOTES

Operations Research

2.5 TYPES OF LINEAR PROGRAMMING SOLUTIONSWhile solving any LP problem using simplex method, at the stage of optimal

solution, the following three types of solutions may exist:

2.5.1 Alternative (Multiple) Optimal SolutionsThe cj – zj values in the simplex table indicates the contribution in the objective

function value by each unit of a variable chosen to enter into the basis. Also, an optimal solution to a maximization LP problem is reached when all cj – zj ≤ 0. But, if cj – zj = 0 for a non-basic variable column in the optimal simplex table and such non-basic variable is chosen to enter into the basis, then another optimal solution so obtained will show no improvement in the value of objective function.

Example 1. Solve the following LP problem.Maximize Z = 6x1 + 4x2

subject to the constraints (i) 2x1 + 3x2 ≤ 30, (ii) 3x1 + 2x2 ≤ 24, (iii) x1 + x2 ≥ 3and x1, x2 ≥ 0.

Sol. Adding slack variables s1, s2, surplus variable s3 and artificial variable A1 in the constraint set, the LP problem becomes

Maximize Z = 6x1 + 4x2 + 0s1 + 0s2 + 0s3 – MA1

subject to the constraints (i) 2x1 + 3x2 + s1 = 30, (ii) 3x1 + 2x2 + s2 = 24 (iii) x1 + x2 – s3 + A1 = 3

and x1, x2, s1, s2, s3, A1 ≥ 0The optimal solution: x1 = 8, x2 = 0 and Max Z = 48 for this LP problem is shown

in Table 2.36.


cj → 6 4 0 0 0


Basic Variables

B

Basic Variables

Value b(= xB)

x1 x2 s1 s2 s3 Min Ratio xB /x2

0 s1 14 0 5/3 1 – 2/3 0 14/(15/3) = 42/5 →

0 s3 5 0 – 1/3 0 1/3 1 —

6 x1 8 1 2/3 0 1/3 0 8/(2/3) = 12

Z = 48 zj 6 4 0 2 0

cj – zj 0 0↑

0 – 2 0

In Table 2.36, c2 – z2 = 0 corresponds to a non-basic variable, x2. Thus, an alternative optimal solution can also be obtained by entering variable x2 into the basis and removing basic variable, s1 from the basis. The new solution is shown in Table 2.37.

Alternative optimal solutions arise when cj – zj = 0 for non-basic variable columns in the simplex table.


NOTES


Table 2.37. Alternative Solutioncj → 6 4 0 0 0


Basic Variables B

Basic Variables Value b(= xB ) x1 x2 s1 s2 s3

4 x2 42/5 0 1 3/5 – 2/5 00 s3 39/5 0 0 1/5 1/5 16 x1 12/5 1 0 – 2/5 3/5 0

Z = 48 zj 6 4 0 2 0cj – zj 0 0 0 – 2 0

The optimal solution shown in Table 2.37 is: x1 = 12/5, x2 = 42/5 and Max Z = 48. Since this optimal solution shows no change in the value of objective function, it is an alternative solution.

Once again, c3 – z3 = 0 corresponds to non-basic variable, s1. This again indicates that an alternative optimal solution exists. The infinite number of solutions that can be obtained for this LP problem are as follows:

VariablesSolution Values

General Solution1 2

x1 8 12/5 x1 = 8 λ + (12/5) (1 – λ)x2 0 42/5 x2 = 0 λ + (42/5) (1 – λ)s1 14 0 s1 = l4 λ + (0) (1 – λ)s3 5 39/5 s3 = 5 λ + (39/5) (1 – λ)

For each arbitrary value of λ, the value of objective function will remain same.

2.5.2 Unbounded SolutionIn a maximization LP problem, if cj – zj > 0 (cj – zj < 0 for a minimization case)

corresponds to a non-basic variable column in simplex table, and all aij values in this column are negative, then minimum ratio to decide basic variable to leave the basis cannot be calculated. It is because negative value in denominator would indicate the entry of a non-basic variable in the basis with a negative value (an infeasible solution). Also, a zero value in the denominator would result in a ratio having an infinite value and would indicate that the value of non-basic variable could be increased infinitely with any of the current basic variables being removed from the basis.

Example 2. Solve the following LP problem.Maximize Z = 3x1 + 5x2

subject to the constraints (i) x1 – 2x2 ≤ 6, (ii) x1 ≤ 10, (iii) x2 ≥ 1and x1, x2 ≥ 0

Sol. Adding slack variables s1, s2, surplus variable s3 and artificial variable A1 in the constraint set. Then the standard form of LP problem becomes

Maximize Z = 3x1 + 5x2 + 0s1 + 0s2 + 0s3 – MA1

Unbounded solution occurs when value of decision variables in the solution of LP problem becomes infinitely large without violating any given constraints.


NOTES

Operations Research subject to the constraints (i) x1 – 2x2 + s1 = 6, (ii) x1 + s2 = 10, (iii) x2 – s3 + A1 = 1and x1, x2, s1, s2, s3, A1 ≥ 0

The initial solution to this LP problem is shown in Table 2.38.

Table 2.38. Initial Solutioncj → 3 5 0 0 0 – M


Basic Variables

B

Basic Variables

Value b(= xB )x1 x2 s1 s2 s3 A1

Min Ratio xB /x2

0 s1 6 1 – 2 1 0 0 0 –0 s2 10 1 0 0 1 0 0 –

– M A1 1 0 1 0 0 – 1 1 1 →Z = – M zj 0 – M 0 0 M – M

cj – zj 3 5 + M 0 0 – M 0↑

Iteration 1: Since c2 – z2 ≥ 0, non-basic variable x2 is chosen to enter into the basis in place of basic variable A1. The new solution is shown in Table 2.39.

Table 2.39cj → 3 5 0 0 0 – M


Basic Variables B

Basic Variables

Value b(= xB)x1 x2 s1 s2 s3 A1

0 s1 8 1 0 1 0 – 2 20 s2 10 1 0 0 1 0 05 x2 1 0 1 0 0 – 1 1

Z = 5 zj 0 5 0 0 – 5 5cj – zj 3 0 0 0 5 – M – 5

In Table 2.39, c5 – z5 = 5 is largest positive, so variable s3 should enter into the basis. But, coefficients in the ‘s3’ column are all negative or zero. This indicates that s3 cannot be entered into the basis. However, the value of s3 can be increased infinitely without removing any one of the basic variables. Further, since s3 is associated with x2 in the third constraint, x2 will also be increased infinitely because it can be expressed as x2 = 1 + s3 – A1. Hence, the solution to the given problem is unbounded.

2.5.3 Infeasible SolutionIf LP problem solution does not satisfy all of the constraints, then such a solution

is called infeasible solution. Also, infeasible solution occurs when all cj – zj values satisfy optimal solution condition but at least one of artificial variables appears in the basis with a positive value. This situation may occur when an LP model is either improperly formulated or more than two of the constraints are incompatible.


NOTES


Example 3. Solve the following LP problemMaximize Z = 6x1 + 4x2

subject to the constraints (i) x1 + x2 ≤ 5, (ii) x2 ≥ 8and x1, x2 ≥ 0.

Sol. Adding slack, surplus and artificial variables, the standard form of LP problem becomes

Maximize Z = 6x1 + 4x2 + 0s1 + 0s2 – MA1

subject to the constraints (i) x1 + x2 + s1 = 5, (ii) x2 – s2 + A1 = 8and x1, x2, s1, s2, A1 ≥ 0

The initial solution to this LP problem is shown in Table 2.40.


cj → 6 4 0 0 – M


Basic Variables

B

Basic Variables

Value b(= xB)x1 x2 s1 s2 A1

Min Ratio xB /x2

6 s1 5 1 1 1 0 0 5/1 →

– M A1 8 0 1 0 – 1 1 8/1

Z = 30 – 8M zj 6 6 – M 0 M – M

cj – zj 0 – 2 + M 0 – M 0

↑

Iteration 1: Since c2 – z2 = M – 2 (≥ 0), non-basic variable x2 is chosen to enter into the basis to replace basic variable x1. The new solution is shown in Table 2.40.

Table 2.41. Optimal but Infeasible Solution

cj → 6 4 0 0 – M


Basic Variables

B

Basic Variables

Value b(= xB )

x1 x2 s1 s2 A1

4 x2 5 1 1 1 0 0– M A1 3 –1 0 –1 – 1 1

Z = 20 – 3M zj 4 + M 4 4 + M M – Mcj – zj 2 – M 0 – 4 – M – M 0

In Table 2.41, since all cj – zj ≤ 0, the current solution is optimal. But this solution is not feasible for the given LP problem because values of decision variables are: x1 = 0 and x2 = 5 violates second constraint, x2 ≥ 8. The presence of artificial variable A1 = 3 in the solution also indicates that the optimal solution violates the second constraint (x2 ≥ 8) by 3 units.


NOTES

Operations ResearchTEST YOUR KNOWLEDGE–A

1. Define slack and surplus variables in a linear programming problem. 2. Explain the various steps of the simplex method involved in the computation of an optimum

solution to a linear programming problem. 3. (a) Give outlines of the simplex method in linear programming. (b) What is simplex? Describe the simplex method of solving linear programming problem. (c) What do you understand by the term two-phase method of solving linear programming

problem? (d) Outline the simplex method in linear programming. Why is it called so? (e) Explain the purpose and procedure of the simplex method. 4. What do you mean by an optimal basic feasible solution to a linear programming problem? 5. Given a general linear programming problem, explain how you would test whether a basic

feasible solution is an optimal solution or not. How would you proceed to change the basic feasible solution in case it is not optimal?

6. Explain the meaning of basic feasible solution and degenerate solution in a linear programming problem.

7. Explain the term artificial variables and its use in linear programming. 8. What are artificial variables? Why do we need them? Describe the two-phase method of

solving an LP problem with artificial variables. 9. What is the significance of cj – zj numbers in the simplex table? Interpret their economic

significance in terms of marginal worth. 10. What is meant by the term opportunity cost? How is this concept used in designing a test

for optimality? 11. How do the graphical and simplex methods of solving LP problems differ from each other?

In what ways are they same? 12. How do maximization and minimization problems differ when applying the simplex method?

TEST YOUR KNOWLEDGE–B

1. Solve the following LP problems using the simplex method: (i) Max Z = 3x1 + 2x2

subject to x1 + x2 ≤ 4 x1 – x2 ≤ 2and x1, x2 ≥ 0

(ii) Max Z = 5x1 + 3x2

subject to x1 + x2 ≤ 2 5x1 + 2x2 ≤ 10 3x1 + 8x2 ≤ 12and x1, x2 ≥ 0

(iii) Max Z = 3x1 + 2x2 + 5x3

subject to x1 + 2x2 + x3 ≤ 430 3x1 + 2x3 ≤ 460 x1 + 4x3 ≤ 420and x1, x2, x3 ≥ 0


NOTES


2. Use the two-phase method to solve the following LP problems: (i) Max Z = 3x1 – x2

subject to 2x1 + x2 ≥ 2 x1 + 3x2 ≤ 2 x2 ≤ 4and x1, x2 ≥ 0

(ii) Min Z = 3x1 – x2 – x3

subject to x1 – 2x2 + x3 ≤ 11 – 4x1 + x2 + 2x3 ≥ 3 – 2x1 + x3 = 1and x1, x2, x3 ≥ 0

(iii) Min Z = 7.5x1 – 3x2

subject to 3x1 – x2 – x3 ≥ 3 x1 – x2 + x3 ≥ 2and x1, x2, x3 ≥ 0

3. Use penalty (Big-M) method to solve the following LP problems. (i) Min Z = 3x1 –x2

subject to 2x1 + x2 ≥ 2 x1 + 3x2 ≤ 3 x2 ≥ 4 x1, x2 ≥ 0

(ii) Min Z = 2x1 + x2

subject to 3x1 + x2 = 3 4x1 + 3x2 ≥ 6 x1 + 2x2 ≤ 4 x1, x2 ≥ 0

(iii) Max Z = 3x1 + 2x2 + 3x3

subject to 2x1 + x2 + x3 ≤ 2 3x1 + 4x2 + 2x3 ≤ 8and x1, x2, x3 ≥ 0

Hints and Answers 1. (i) x1 = 3, x2 = 1 and Max Z = 11 (ii) x1 = 2, x2 = 0 and Max Z = 10 (iii) x1 = 0, x2 = 100, x3 = 230 and Max Z = 1,350 2. (i) x1 = 1, x2 = 0, and Max Z = 6 (ii) All cj – zj ≤ 0 but Z = – 5/4 (< 0) and artificial variable A1 = 5/4 appears in the basis with

positive value. Thus the given LP problem has no feasible solution. (iii) x1 = 5/4, x2 = 0, x3 = 0 and Min Z = 75/8

3. (i) x1 = 3, x2 = 0 and Max Z = 9 (ii) x1 = 3/5, x2 = 6/5 and Min Z = 12/5 (iii) All cj – zj ≥ 0 artificial variable A1 = 0 appears in the basis with zero value. Thus an

optimal solution to the given LP problem exists.


NOTES

Operations Research 2.6 SLACK AND SURPLUS VARIABLESIn the general LP problem each constraint may take one of the three possible

forms, ≤, =, or ≥. Inequality constraints of the LP problems are converted into equalities by adding additional non-negative variables called slack and surplus (negative slack) variables:

Case 1: The constraints with ≤ inequality sign, i.e.

a xijj

n

j=∑

1 ≤ bi

can be converted to the equality,

a x sij j ij

n+

=∑

1 = bi, i = 1,2, …, m ...(4)

by adding non-negative variables si, called slack variables.Case 2: The constraints with ≥ inequality sign, i.e.

a xijj

n

j=∑

1 ≥ bi

can be converted to the equality

a x sij j ij

n−

=∑

1 = bi, i = 1, 2, …, m ...(5)

by subtracting non-negative variables si, called surplus variables.The general LP problem that involves mixed constraints can be stated as:

Optimize (Max or Min) Z = c xj jj

n

=∑

1 ...(6)

subject to the constraints

a xij jj

n

=∑

1 ≤ bi, i = 1, 2, . . ., r

a xij jj

n

=∑

1 ≥ bi, i = r + 1, r + 2, …, s ...(7)

a xij jj

n

=∑

1 = bi, i = s + 1, s + 2, …, m

and xj ≥ 0, j = 1, 2, …, nConstraints with ≤ inequality sign can be converted to form (4) and those with

≥ inequality sign to form (5). Those having an equality sign remain unchanged. The general LP problem can then be stated as:

Optimize Z = c x s sj jj

n

ii

r

ii r

s

= = = +∑ ∑ ∑+ +1 1 1

0 0. .


a x sij j ij

n+

=∑

1 = bi, i = 1, 2, …, r


NOTES

Linear Programming: The Simplex Method a x sij j i

j

n−

=∑

1 = bi, i = r + 1, r + 2, …, s (7)

a xij jj

n

=∑

1 = bi, i = s + 1, s + 2, …, m

and xj ≥ 0, j = 1, 2, . . ., nRemark. The coefficients of slack and surplus (negative slack) variables are zero in the

objective function due to the reason that they represent unused capacity (or resource).

2.6.1 Basic SolutionConstraints, Ax = b of an LP problem are written as a system of m simultaneous linear

equations in n (n > m) unknown, where A is an m × n matrix and rank (A) = m. Let B be an m × n non-singular submatrix of A obtained by reordering the column of A, and N an m × (n – m) matrix such that A = (B, N). Let x be partitioned as x = [xB, xN]T where xT

B ∈ Em and xTN ∈ En – m be the vector of variables associated with columns of matrix B

and N, respectively. Then constraints, Ax = b can be rewritten as:

[B, N]xxB

N

= b

or BxB + NxN = b or xB = B–1 b – B–1 N xN

If all the (n – m) variables not associated with the columns of matrix B are set equal zero, i.e. xN = 0, the solution to the resulting system of equations, i.e.

xxB

N

= B−

1

0; xB = [xB1, xB2, . . ., xBm]

is called the basic solution to the given system of equations. The m variables that can be different from zero are called the basic variables. If all these variables also satisfy the non-negativity conditions, x ≥ 0, then the basic solution constituted by them is called a basic feasible solution. Again if these satisfy all the constraints Ax = b, then the solution is known as a feasible solution. The remaining n – m variables, i.e. the components of xN are called non-basic variables. The matrix B is called the basic matrix (or simply the basic) having m linear independent columns selected from A. Let β1, β2, …, βm be the columns of basis matrix B that form a basis. Then we can write B = [β1, β2, …, βm]. The column aj of A can be expressed as a linear combination of columns of B as:

aj = y1j β1 + y2j β2 + . . . + ymj βm = (β1, β2, . . ., βm) (y1j, y2j, . . ., ymj) = Byj

or yj = B–1 aj

where yj = (y1j, y2j, . . ., ymj) are the scalars. Obviously the vector yj will change with the change in the columns of A that are part of basis matrix B.

If the value of the objective function Z can be increased or decreased with change in the values of basic variables, then such a solution is said to be unbounded.

2.6.2 Degenerate SolutionA basic feasible solution xB = B–1 b is said to be degenerate if at lest one component

of xB (basic variables) is zero. If all components of xB are non-zero (xB > 0), then it is called a non-degenerate basic feasible solution.


NOTES

Operations Research For a system of m equations with n variables (n > m) the total number of basic feasible solutions is less than or equal to the total number of combinations, nCm = n!/m! (n – m)!

2.6.3 Cost (or Price) VectorLet the cost vector c, associated with the variables in objective function Z, of an

LP problem be partitioned as c = (cB, cN), where cB and cN are the coefficients of basic variables xB and non-basic variables xN, respectively. The objective function can then be written as:

Z = cx = [cB, cN] xxB

N

= cB xB + cN xN

Since for the basic feasible solution the value of all the non-basic variables become zero, i.e. xN = 0, therefore the value of the objective function for basic feasible solution is given by Z = cB xB, where cB = (cB1, cB2, …, cBm). The vector, cB = (cB1, cB2, …, cBm) associated with the basic variable, xB = (xB1, xB2, . . ., xBm) is called cost (or price) vector.

Example 1. Find the all basic feasible solutions to the system of linear equations (i) x1 + 2x2 + x3 = 4, (ii) 2x1 + x2 + 5x3 = 5

Are the solutions degenerate?Sol. The given system of equations can be written in the Ax = b form as follows:

1 2 12 1 5

1

2

3

xxx

= 45

, where A =

1 2 12 1 5

; x =

xxx

1

2

3

and b = 45

Since, rank (A) = 2, there are two linearly independent columns of A. Therefore, any of the following 3[3C2 = 3] submatrices of order 2 can be considered as the basis matrix B because following determinants of order 2 are not equal to zero.

1 22 1

1 12 5

; and

2 11 5

Since each of these submatrices is non-singular, by putting variables not associated with the columns of B equal to zero, all possible basic feasible solution can be obtained. Let us consider the case where x2 = 0, i.e. it is not associated with the columns of B. We then have:

1 12 5

1

3

xx =

45

or xx1

3

=

1 12 5

45

1

−

[because xB = B–1 b]

= 13

5 12 1

45

13153

31

−−

=

−

=

−

since A–1 = adj A

A| |

or x1 = 3, and x3 = – 1


NOTES


We now set x1 = 0 and solve the system of equations. The resulting matrix is non-singular. Then,

2 11 5

2

3

xx =

45

or xx2

3

=

2 11 5

45

19

5 11 2

45

19156

51

=

−−

=

=

− / 332 3/

or x2 = 5/3, and x3 = 2/3Next we set x3 = 0 and solve the system of equations. Thus,

1 22 1

1

2

xx =

45

or xx1

2

=

1 22 1

45

13

1 22 1

45

13

63

1

= −

−−

= −

−−

−

==

21

or x1 = 2, and x2 = 1The summary of the solutions is given below:

Basic Vector Basic Variables Non-basic Variable

1. xB = −−

31

x1 = 3, x3 = –1

x1, x3 x2

2. xB = 5 32 3//

x2 = 5/3, x3 = 2/3

x2, x3 x1

3. xB = 21

x1 = 2, x2 = 1

x1, x2 x3

Here it may be noted that all these solutions are non-degenerate.

TEST YOUR KNOWLEDGE–A

1. What is meant by a basic solution of an LP problem? 2. What is meant by a basic solution to the system of m linear non-homogeneous equations in

n unknowns (m < n) Ax = b? 3. When is a basic solution to Ax = b said to be degenerate? 4. Explain the meaning of basic feasible solution. 5. Define: Basic feasible solution, optimum solution, optimum basic feasible solution. 6. Define (i) Feasible solution; (ii) Basic solution; (iii) Basic feasible solution; (iv) Unbounded

solution. 7. (a) Define a basic solution to a given system of m simultaneous linear questions in n

unknowns. (b) How many basic feasible solutions are there to a given system of 3 simultaneous linear

equations in 4 unknowns.


NOTES

Operations Research 8. What are slack and surplus variables? 9. What is the effect of converting the inequalities in the constraints into equalities by adding

slack and surplus variables in the objective function?


1. Find all the basic feasible solutions for the system of equations given below: (a) 2x1 + 6x2 + 2x3 + x4 = 3 6x1 + 4x2 + 4x3 + 6x4 = 2 xj ≥ 0, j = 1, 2, 3, 4 (b) (i) 3x1 + x2 – x3 = 8, (ii) x1 + x2 + x3 = 4 x1, x2, x3, ≥ 0 2. Show that the following system of linear equations has a degenerate solution: (i) 2x1 + x2 – x3 = 2, (ii) 3x1 + 2x2 – x3 = 3 3. Compute all the non-degenerate basic feasible solutions of the following equations: (i) x1 + 2x2 – x3 + x4 = 2, (ii) x1 + 2x1 + 0.5x3 + x5 = 2 Is the solution (1, 1/2, 0, 0, 0) a basic solution? 4. If x1 = 0, x2 = 0, x3 = 0, x4 = 7, x5 = 12, x6 = 10 is an extreme point solution of the set of

equations, (i) 3x1 – x2 + x4 = 7, (ii) 2x1 – 4x2 + x5 = 12 (iii) – 4x1 – 3x2 + 8x3 + x6 = 10

then find two more extreme point solutions of the above set of equations.

Hints and Answers 1. (a) (i) Basic: x1 = 0, x2 = 1/2 Non-basic: x3 = x4 = 0 (ii) Basic: x1 = 2, x3 = 7/2 Non-basic: x2 = x4 = 0 (infeasible also) (iii) Basic: x1 = 8/3, x4 = 7/3 Non-basic: x2 = x3 = 0 (infeasible also) (iv) Basic: x2 = 1/2, x3 = 0 Non-basic: x1 = x4 = 0 (v) Basic: x2 = 1/2, x3 = 0 Non-basic: x1 = x4 = 0 (vi) Basic: x3 = 0, x4 = 1 Non-basic: x1 = x2 = 0 (infeasible also) (b) (i) x1 = 0, x2 = 6, x3 = – 2 (ii) x1 = 3, x2 = 0, x3 = 1 (ii) x1 = 2, x2 = 2, x3 = 0 2. (i) Basic: x1 = 1, x2 = 0 Non-basic: x3 = 0 (degenerate solution) (ii) Basic: x1 = 5/3, x3 = 1/3 Non-basic: x1 = 0 (iii) Basic: x1 = 1, x3 = 0 Non-basic: x2 = 0 (degenerate solution) 3. The solution (x1 = 1, x2 = 1/2, x3 = 0, x4 = 0, x5 = 0) is not a basic solution.

2.7 BASIC FEASIBLE SOLUTION (Reduction From Feasible Solution)

Theorem 1. A collection of all feasible solutions (if they exist) of an LP problem constitute a convex set.

Proof. The LP problem in its standard form is written as:


NOTES


Optimize (Max. or Min.) Z = cxsubject to the constraints

Ax = b, and x ≥ 0Let S be the set of all feasible solutions of the system Ax = b, x ≥ 0. Now if S

contains only one element, then obviously S is a convex set and hence the statement of the theorem is true. However, if x′, x″ ∈ S such that x′ ≠ x″ then we have:

Ax′ = b, x′ ≥ 0, and Ax″ = b, x″ < 0Let there exists a point x″′ such that x″′ = λx′ + (1 – λ)x″, 0 ≤ λ ≤ 1. In order to

show that S is convex, we have to show that x″′ ∈ S. In other words, the point x″′ must satisfy the system Ax = b, x ≥ 0. Thus,

Ax″′ = A{λx′ + (1 – λ)x″} = λAx′ + (1 – λ)Ax″

= λb + (1 – λ)b = bSince x′, x″ ≥ 0 and 0 ≤ λ ≤ 1, we have x″′. Hence x″′ ∈ S and the set S is convex.Theorem 2. A necessary and sufficient condition for a vector x in a convex set S

to be an extreme point is that x is a basic feasible solution satisfying the system Ax = b, x ≥ 0. In other words, a point is a basic feasible solution to Ax = b if and only if it is an extreme point of the convex set of the feasible solution.

Proof. Consider the following LP problem:Minimize Z = cx

subject to the constraints Ax = b, and x ≥ 0

where A is an m × n matrix of rank m.(a) Basic Solution: Let x be an extreme point of the feasible region of the convex

polyhedron. Let the first p (≤ n) components xj (j = 1, 2, …, p) of x be positive. Then: a1 x1 + a2 x2 + … ap xp = b ...(6)

where aj (j = 1, 2, . . ., p) are the columns of A.To prove that x is a basic feasible solution, we should prove that the vectors

a1, a2, . . ., ap associated with the positive components of x are linearly independent. And we shall do so by contradiction. Suppose the vectors a1, a2, . . ., ap are not linearly independent, then there must exist scalars λ1, λ2, …, λp not all zero, such that:

λ1 a1 + λ2 a2 + . . . + λp ap = 0 ...(7)Let there exist two distinct feasible solution x′ and x″ such that x = x′ + x″, which

violate the assumption that x is an extreme point. Let x′ and x″ be defined as:

xj′ = x j p

j p p nj j+ = …

= + + …

δλ 1 20 1 2

, , ,, , ,

xj″ = x j p

j p p nj j+ = …

= + + …

δλ 1 20 1 2

, , ,, , ,

Since xj > 0 (j = 1, 2, …, p), it is possible to select an arbitrary δ > 0 such that: xj + δ λj ≥ 0 and xj – δ λj ≥ 0 ...(8)

Furthermore Ax′ = a aj jj

p

j j jj

px x′ = ±

= =∑ ∑1 1

( )δλ


NOTES

Operations Research = a aj j

j

p

j jj

px +

= =∑ ∑δ λ

1 1 = b + 0 = b

Similarly Ax″ = b. Thus, the two points x′ and x″ satisfy the system of equations Ax = b. Hence x′ and x″ are two different feasible solutions. It can also be seen that inequalities are true if:

0 < δ < Min x

j pj

jj| |

≠ = …

λλ: , , , ,0 1 2

That is, the first p-components of x′ and x″ will always be positive. Now, x′ + x″ = 2 (x1, x2, …, xp, 0, 0, …, 0)

or x = 1212

x x′ + ″

This shows that an extreme point x can be expressed as 0 linear combination of two distinct feasible points different from x, which cannot be true. Hence vectors a1, a2, …, ap are linearly independent. Since rank of A = m, therefore (m – p) additional column vectors from ap + 1, …, an of A can be added with their corresponding variables, together with a1, a2, …, ap, to form a linearly independent set of column vectors. If needed, after rearranging the columns, let the new column vectors be ap + 1, …, am. Let B = (a1, a2, …, ap, ap + 1, …, am) be the new basis matrix whose columns are linearly independent. Further, xN = 0 and xB = (x1, x2, …, xp, 0, 0, …, 0)T. Since Ax = b, x is a feasible solution. (b) Extreme-point Correspondence: Let x be a basic feasible solution to the system of equations Ax = b, x ≥ 0, such that x = (xB, 0)T, where xB = B–1 b, for non-singular matrix B called basis matrix (or basis).

Let there exist two distinct points x′ and x″ satisfying Ax = b (feasible solution) such that:

x = λ x′ + (1 – λ) x″; 0 < l < 1 ...(9)Now to prove that x is an extreme point it is sufficient to show that x′ = x″ = x. Let

x′ = xxB

N

′

′

and x″ = xxB

N

″

″

It may be noted that x′N ≥ 0 and x″ N ≥ 0. Then substituting in Eq. (9), we get

xB0

= λ λ

xx

xx

B

N

B

N

′

′

″

″

+ −

( )1

xB = λ x′B + (1 – λ) xB″ ...(10) 0 = λ x′N + (1 – λ) x″N ...(11)Since 0 < λ < 1 and x ′N, x″N ≥ 0, from Eq. (11) we have x′N = x″N = 0. Again since x′

and x″ satisfy Ax′ = b and Ax″ = b, we have b = Ax′ = BxB′ + NxN′ = B x′B

or x ′B = B–1 b = xB

Similarly x″B = xB. It follows that x = x′ = x″. This is a contradiction for x′ ≠ x″. Hence, x is an extreme point.


NOTES


Working Rule 1. Compute all xr/λr (j = 1, 2, . . ., p) for which λj > 0, and choose the minimum value. 2. Reduce to zero the value of the variable corresponding to the minimum ratio xr/λr

by using the relationship

ar = λλj

rj r

p

j

≠

∑ a , λr ≠ 0

Here at least one λj must be positive; however, if all λj ≤ 0, then multiply the equation: λ1 a1 + λ2 a2 + . . . + λp ap = 0by – 1 and obtain new values of λj ≥ 0.

3. The values of the new variables are given by xj′ = xj – xrj

jλλ .

Example 1. Let x1 = 2, x2 = 4 and x3 = 1 be feasible solution to the system of equations: (i) 2x1 – x2 + 2x3 = 2, (ii) x1 + 4x2 = 18

Reduce the given feasible solution to a basic feasible solution.Sol. We first write the system of equations in matrix notation as:

2 1 21 4 0

1

2

3

−

xxx

= 218

where A (a1, a2, a3) = 2 1 21 4 0

−

; b =

218

and x =

xxx

1

2

3

This system of equations can also be expressed as: a1 x1 + a2x2 + a3x3 = bSince it is given that x1 = 2, x2 = 4 and x3 = 1, we have 2a1 + 4a2 + a3 = b

where a1 = 21

, a2 =

−

14

and a3 = 20

are column vectors of A.

Since rank (A) = 2, only two out of three column vectors a1, a2 and a3 are linearly independent. Assuming that these vectors are linearly dependent, we express one of them as a linear combination of the remaining two as:

a3 = λ1 a1 + λ2 a2

20

= λ1 λ λ1 2

21

14

+

−

This gives 2 = 2 λ1 – λ2 and 0 = λ1 + 4λ2

where λ1 and λ2 are scalars, not all zero.On solving these two equations, we get λ1 = 8/9 and λ2 = – 2/9. Substituting values

of λ1 and λ2 in the linear combination, we get:

a3 = λ1 a1 + λ2 a2 = 89291 2a a−


NOTES

Operations Researchor 8

9291 2a a− – a3 = 0

where λ1 = 8/9, λ2 = – 2/9 and λ3 = – 1.To reduce the number of positive variables, the vector to be removed is chosen in

accordance with the Theorem 5, i.e.

xrrλ

= Min Minx x x xj

jjλ

λλ λ λ

, , ,>

=

0 1

1

2

2

3

3

= Min 28 9

42 9

11

94/

,/,

− −

=

Since xrrλ

= 94

corresponds to a vector a1, it should be removed in order to obtain a

new solution with two non-negative variables. The values of the new variables are given by:

xj′ = xj – xrr

jλλ

This gives x′1 = x1 – xrrλ

λ1 = 2 – 94

89

= 0

x2′ = x2 – xrrλ

λ2 = 4 – 94

29

−

= 92


λ3 = 1 – 94

(– 1) = 134

The new solution (0, 9/2, 13/4), so obtained, is also a feasible solution. As the two column vectors a2 and a3 associated with non-zero variables x2 and x3 are linearly independent, therefore, the required basic feasible solution is: x1 = 0, x2 = 9/2 and x3 = 13/4. This result can be verified by substituting the values of x1, x2 and x3 in the equation, a1 x1 + a2 x2 + a3 x3 = b.

Example 2. Show that the feasible solution: x1 = 1, x2 = 0, x3 = 1 and Z = 6 to the system of equations:

x1 + x2 + x3 = 2 and x1 – x2 + x3 = 2with Max Z = 2x1 + 3x2 + 4x3, is not basic feasible solution.

Sol. Writing the system of equations in matrix notation, we have:

1 1 11 1 1

1

2

3−

xxx

= 22

where A = (a1, a2, a3) = 1 1 11 1 1−

; b =

22

and x =

xxx

1

2

3

This system of equations can also be expressed as: a1 x1 + a2 x2 + a3 x3 = bSince it is given that x1 = 1, x2 = 0 and x3 = 1, we have: a1 + a3 = b


NOTES

Linear Programming: The Simplex Methodwhere a1 =

11

and a3 =

11

are the column vectors of A.

Since rank (A) = 2, only two out of three column vectors a1, a2 and a3 are linearly independent. We assume that these vectors are linearly dependent. Expressing one of them as a linear combination of the remaining two as:

a3 = λ1 a1 + λ2 a2

11

= λ λ1 2

11

11

+

−

This gives, 1 = λ1 + λ2 and 1 = λ1 – λ2

where λ1, λ2 are scalars, not all zero.On solving these two equations, we get λ1 = 1 and λ2 = 0. Substituting values of

λ1 and λ2 in the linear combination, we get: a3 = λ1 a1 + λ2 a2 = a1

(a) We take a1 – a3 = 0, where λ1 = 1, λ2 = 0, and λ3 = – 1. To reduce the number of positive variables, the vector to be removed is chosen in accordance with the Theorem 4, i.e.

xrrλ

= Min Min Minx x x xj

jjλ

λλ λ λ

, , , ,>

=

=−

0 11

11

1

1

2

2

3

3

= 1

Since xr/λr = 1 corresponds to vector a1, it should be removed to obtain a new solution with not more than two non-negative variables. The values of the new variables are given by:

xj′ = xj – xrr

jλλ

This gives, x′1 = x1 – xrrλ

λ1 = 1 – 1(1) = 0


λ2 = 0 – 1(0) = 0

x′3 = x3 – xrrλ

λ3 = 1 – 1(– 1) = 2

(b) By taking a3 – a1 = 0, we shall have λ1 = 1, λ2 = 0 and λ3 = 1. In this case:

xrrλ

= Minx jj

jλλ, >

0 = 1 (corresponds to a3)

Thus, the vector a3 can be removed in order to obtain a new solution with not more than two non-negative variables. The values of the new variables are given by:


λ1 = 1 – 1 (–1) = 2


λ2 = 0 – 1 (0) = 0


λ3 = 1 – 1 (1) = 0


NOTES

Operations Research The new solutions (0, 0, 2) and (2, 0, 0), so obtained, have two linearly independent column vectors, and hence both of these solutions are basic feasible. However, these two solutions are different from the given (1, 0, 1) solution. Thus, the given solution is not basic.

2.8 IMPROVING A BASIC FEASIBLE SOLUTIONTheorem 3. Given a non-degenerate basic feasible solution xB = B–1 b to the linear

programming problem, if a basis vector is replaced by a non-basis vector, then the new solution so obtained is also a basic feasible solution.

Proof. Suppose we have a feasible solution x = B–1 b of the following LP problem:Minimize Z = cx

subject to the constraints Ax = b, and x ≥ 0

where A is a m × n matrix of rank m. Let B be a basis matrix. Then any column vector say aj of A can be expressed as the linear combination of column (or basis) vectors βi of B, i.e.

aj = y1j β1 + y2j β2 + . . . + ymj βm = yij ii

mβ

=∑

1 bi = yj B ...(12)

or yj = B–1 aj

where yij are some scalars and yj = (y1j, y2j, . . ., ymj). By Theorem 1 column vector aj of A can replace any column vector say βr of B, for which yrj ≠ 0, and the new set of vectors will, still form basis matrix. Let us then select any vector aj, not in B for which at least one yrj ≠ 0, and insert this into the basis matrix B to replace its column vector βr. Then we have:

aj = y yij i rj ri r

mβ β+

≠∑

βr = ay

yy

j

rj i r

mij

rji−

≠∑ β , yrj ≠ 0 ...(13)

The original basic feasible solution xB B = b can also be written as:

i r

m

Bi ix=∑ β = b or

i r

m

Bi i Br rx x≠∑ +β β = b ...(14)

Substituting βr from Eq. (16) in Eq. (17), we get for a new basic solution:

i r

m

Bi Brij

rji

Br

rjjx x

yy

xya

≠∑ −

+β = b ...(15)

For a new basic solution to be feasible, it is required that:

x xyyBi Brij

rj− ≥ 0; i ≠ r, i = 1, 2, . . ., m ...(16)

and xyBr

rj ≥ 0; i = r


NOTES


From Eq. (19) we see that if xBr ≠ 0, then we must have yrj > 0. (The basic solution where xB = 0 is said to be degenerate. This will be discussed later.) In the case where all yij ≤ 0, i ≠ r, the new solution will be feasible. However, if not all yij ≤ 0, then we must carefully choose a vector to be removed from the basis. Otherwise we may get an infeasible solution. If yij > 0, then we shall select the column vector βr to be replaced by non-basic column vector in such a way that Eq. (19) holds true; that is:

xy

xy

Bi

ij

Br

rj− ≥ 0, yij > 0, i ≠ r

xy

xy

Bi

ij

Br

rj− ≤ 0, yij < 0, i ≠ r

Feasibility may be maintained in the new solution if we select the column vector βr (yrj ≠ 0) of B to be replaced by aj, not in B, by taking:

xyBi

rj = Min

xy

yBi

ijij, >

0

then the new basic solution will be feasible.Remark: We define the new non-singular matrix obtained from B by replacing βr with aj as:

B = 1 2ˆ ˆ ˆ[ , , , ]mβ β … β

where ˆrβ = βi (i ≠ r) and βr = aj. The corresponding new basic feasible solution is denoted

by

ˆ Bx = 1ˆ −B b

The new basic variables would then have the following form:

ˆBix = xBi – xBr ;ij

rj

yy

i ≠ r

ˆBrx = Br

ij

xy

; i = r

Theorem 4. Given a basic feasible solution xB = B–1 b to the set of constraints Ax = b, with the value of the objective function Z = cB xB. Then it is possible to obtain a new basic feasible solution ˆBx = –1B b with objective function Z by replacing one of the

columns in B by a column aj in A for which cj – zj > 0 holds and if at least one yij > 0 (i = 1, 2, …, m) such that Z > Z.

Proof. Suppose that we have a basic feasible solution xB = B–1 b. The value of the objective function for this solution is given by:

Z = –

1B bc0

= – 1

( , )B N

B bc c0

= cB B–1 b

= 1 1

m m

Bi i Bi Bii i

c b c x−

= =

=∑ ∑1B

Let the new value of the objective function Z ,obtained by replacing a column vector βr, in B by a vector aj in A but not in B, be given by:

Z = ˆ ˆ ˆ ˆm

Bi Bi Br Bri r

c x c x≠

+∑


NOTES

Operations Research where cBi = ˆBic (i ≠ r) and ˆBrc = cj, since only the cost or profit coefficient of the new variable entering the basis changes.

Substituting the value of ˆBix and ˆBrx from Eqs (19a) and (19b), we may find the new value of the objective function:

Z = –m ij Br

Bi BrBi jrji r rj

y xx xc cy y≠

+

∑ ...(20)

Adding the rth term to the summation does not change the value of Z because at i = r, we have

– ijBi BrBr

rj

yx xc

y

= 0

Thus we have, Z = 1 1

–m m

Br BrBi Bi Bi ij j

i rj i rj

x xc x c y cy y= =

+∑ ∑

= ( )Brj j

rj

xZ c zy

+ −

where zj = 1

m

Bi iji

c y=∑ = cB B–1 aj = cB yj for each non-basic variable.

By examining this expression we see that if cj – zj > 0, then an improved value of Z is obtained because Z has to be maximized. The value of cj – zj may be viewed as the change in Z due to per unit increase in xj. Since (xBr/yrj) ≥ 0, therefore an improved value of Z can be obtained only if cj – zj > 0 or zj – cj < 0.

Remarks: 1. In the absence of degeneracy we have xBr > 0 and hence (xBr/yrj) > 0. Since cj – zj > 0, it

follow that > Z. 2. If the given basic feasible solution is non-degenerate, then we select any column vector

aj for entering into the basis for which cj – zj > 0 and at least one yij > 0. In this case the new basic solution gives an improved value of Z. But if initial solution is degenerate and a vector aj in A, for which cj – zj > 0 and at least one yij > 0, is chosen for entering into the new basis, then an increase in Z will depend on whether (xBr/yrj) > 0 or not.

3. In any case, cj – zj > 0 indicates that the new value of Z will not be less than the old value of Z.

4. cj – zj may be interpreted as the change in the value of the objective function with per unit increase in the value of xj, when all other non-basic variables are zero, and the basic variables are allowed to change.

Example 3. Compute all basic feasible solutions to the LP problem:Maximize Z = 3x1 + 2x2 + x3

subject to the constraints (i) 3x1 + 2x2 + 2x3 + x4 = 8, (ii) 3x1 + 4x2 + x3 + x5 = 7and x1, x2, x3, x4, x5 ≥ 0


NOTES


Obtain the new basic feasible solution other than the initial solution and verify that the corresponding new value of Z is improved.

Sol. The given system of equations can be written in matrix from as follows:

1

2

3

4

5

3 2 2 1 03 4 1 0 1

xxxxx

= 87

where A = 3 2 2 1 0 ;3 4 1 0 1

x = (x1, x2, x3, x4, x5)T, b = [8, 7]T, B = 1 00 1

and here a1 = 33

, a2 = 24

, and a3 = 21

The y′j s for each column aj of A but not in B are:

y1 = B–1 a1 = 11

21

1 0 3 30 1 3 3

yy

= =

y2 = B–1 a2 = 12

22

24

yy

=

y3 = B–1 a3 = 13

23

21

yy

=

z1 = cB y1 = (0 0) 33

= 0, z2 = 0, z3 = 0

Since cB = (3, 2, 1, 0, 0) and xB = (0, 0, 0, 8, 7), we get Z = cB xB = 0. To obtain an improved basic feasible solution, we have to determine one of the column vector in B that can be replaced by a column vector in A but not in B. This is determined as follows:

Br

rj

xy

= 1 2

11 21, 0Min Min ,B BBi

ijrj

x xx yy y y

> =

, for j = 1

= 78 7Min ,33 3

=

That is, the second column β2 of B is replaced by the first column a1 and A not in B. Thus,

Z = Z + (cj – zj) Br

rj

xy

= Z + (cj – zj) 2

21

Bxy

= 0 + (3 – 0) 73 = 7 > Z (= 0)


NOTES

Operations Research 2.9 THE FUNDAMENTAL THEOREM OF LINEAR PROGRAMMING

LP in standard form:

min Tx

c x subject to

Ax = b, x ≥ 0, (b ≥ 0).Theorem 5. Let A be an m-by-n matrix of rank m.

(i) If there is a feasible solution then there is a basic feasible solution. (ii) If there is an optimal feasible solution then there is an optimal basic feasible

solution.Proof. Let x be a feasible solution. Then x ≥ 0, and we may write

Ax = i ii I

a x b∈

=∑ , I ⊂ {1, . . ., n}, xi > 0,

i.e., indices i such that xi = 0 are deleted. There are two cases: (a) {ai}i ∈ I are linearly independent. Then x is already a feasible basic solution.

Both (i) and (ii) proven! (b) {ai}i ∈ I are linearly dependent.

Then i ii I a y∈∑ = 0 for some yi′s not all vanishing. Can assume yp > 0 for at least

one p ∈ I. Define the vector y with components yi as above for i ∈ I and yi = 0 for i ∉ I. Define x e = x – ey. Then,

Axe = Ax – eAy = b – e i ii I

a y∈∑ = b.

That is, Axe = b and xe is feasible for small positive or negative e since

xie =

0

–0

i ix y i Ii I

> e ∈ ∉

At least for one p ∈ I, xpe = xp – eyp = 0 for e xp/yp. Choose the smallest such e and

associated p. The resulting xe is the feasible and

Axe = i ii I

a x∈

e

∈∑ , Ie = I / {p},

that is, we have found another feasible solution spanned by one less of the columns of A. Continue the same procedure until only linearly independent columns remain. Then, back to case (a) and part (i) is shown.

To show (ii), we also need to show that xe is optimal if x is optimal. Since xe is feasible for small positive or negative e, cT xe = cT x – e cT y < cT x for small e of the same sign as cT y if cT y ≠ 0. Since x is optimal, must have cT y = 0. Thus cT xe = cT x and xe is optimal.


NOTES


2.10 ALTERNATIVE OPTIMAL SOLUTIONThe optimal value of the objective function of an LP problem is always unique but

the set of basic variables yielding this optimal value need not be unique. There may be two or more basic feasible solutions that give the same value of the objective function.

Theorem 6. Given an optimal basic feasible solution to an LP problem and for some column aj of A but not in B, cj – zj = 0, yij ≤ 0 for all i = 1, 2, …, m or yij > 0 for at least one i, then aj may be inserted into the basis to yield an alternative optimal solution.

Proof. Let as an exercise for the reader.Remark: For selecting the column vector ar of A but not in B to enter into the basis, the

following formula may be used: (i) cr – zr = Max {cj – zj} for maximization problem. (ii) cr – zr = Min {cj – zj} for minimization problem.

2.11 UNBOUNDED SOLUTIONTheorem 7. Given any basic feasible solution to an LP problem. If for this solution

there is some column aj in A but not in B for which cj – zj > 0 and yij ≤ 0 (i = 1, 2, . . ., m), then the problem has an unbounded solution, if the objective function is to be maximized.

Proof. If we introduce any column vector aj of A for which all yij ≤ 0 (i = 1, 2, …, m) into the basis matrix B, then aj must enter into the basis either at a negative level or at the zero level. Thus, a new basic feasible solution will be infeasible, unless aj enters at a zero level. Let xB be the basic feasible solution to the given LP problem so that:

B xB = b or 1

m

Bi ii

x=

β =∑ b ...(18)

The value of the objective function at this solution is given by:

Z = cB xB = 1

m

Bi Bii

c x=∑

By adding and subtracting λ aj in Eq. (21), where λ be any scalar and aj the vector entering the basis, we have:

1

m

Bi ii

c=

β∑ + λaj – λaj = β

1 1

–m m

Bi i j ij ii i

x a y= =

β + λ λ β∑ ∑ = β, since aj = 1

m

ij ii

y=

β∑

1( – )

m

Bi ij i ji

x y a=

λ β + λ∑ = b ...(19)

Equation (19) represents a new solution and is given by: ˆBix = xBi – λ yij and

1ˆ

mBx+

= λ ; i = 1, 2, . . ., m

Since all yij ≤ 0, we get ˆiBx ≥ 0 when λ > 0. Thus Eq. (22) is a feasible solution in

which m + 1 variables may be at a positive level. But in general, this may not be the basic solution because there are more positive variables than constraints.


NOTES

Operations Research The new value of the objective function Z for this solution is given by:

Z = 1 1

ˆ ( – )m m

Bi Bi Bi Bi ij ji i

c x c x y c= =

+ λ + λ∑ ∑

= 1 1

( – )m m

Bi Bi j Bi iji i

c x c c y= =

+ λ λ∑ ∑ = Z + λ (cj – zj); λ = xBr/yrj

For a sufficiently large value of λ and cj – zj > 0, the value of Z can be increased up to infinity. Similarly, in the case of a minimization LP problem we make λ sufficiently small so that the value of Z can be decreased up to infinity, for cj – zj < 0. Such a solution is unbounded.

2.12 CONDITION OF OPTIMALITYIn this section we shall develop the criterion of optimality of an LP problem solution,

i.e., when the interactive procedure of solving an LP problem may be stopped.Theorem 8. Given a basic feasible solution to the LP problem, xB = B–1 b =

1 2( , , , )

mB B Bx x x… and Z = Z* such that cj – zj ≤ 0 for every column aj in A but not in B.

Then Z is maximum value of objective function Z and xB is an optimal basic feasible solution.

Proof. Let xB = B–1 b be feasible solution to the given LP problem and Z = cB xB be the corresponding value of objective function.

Let xj ≥ 0 (j = 1, 2, …, n) be any feasible solution to the same LP problem. Then the system Ax = b can be expressed in terms of column vectors of A as:

a1 x1 + a2 x2 + … + an xn = b ...(20)The value of the objective function at this solution is given by: Z* = c x = c1 x1 + c1 x2 + … + cn xn

Any column vector aj of A can be expressed as linear combination of column vectors βi of B. That is,

aj = 1

m

ij ii

y=

β∑

Substituting value of aj (j = 1, 2, . . ., n) in Eq. (23), we have:

1 1 2 21 1 1

m m m

i i i i n in ii i i

x y x y x y= = =

β + β + … + β∑ ∑ ∑ = b

1 21 21 1 1

m m m

j j j j j mj mi i i

x y x y x y= = =

β + β + … + β ∑ ∑ ∑ = b

Let for every column aj in A but not in B, cj – zj ≤ 0. Now we prove that Z is more than value of the objective function Z* for any other feasible solution. For all columns vector of A in B, i.e. aj ∈ B, we have:

yj = B–1 aj = B–1 βi = ei (unit vector)provided aj is in column i for B. Then:

cj – zj = cj – cB yj = cj – cB ei = cj – cj = 0


NOTES


Thus cj – zj = 0 for all columns of A in B. Applying the assumption that cj – zj ≤ 0 for all columns in A, then from Eq. (23), we have:

1( – )

m

j j ji

c z x=∑ ≤ 0

or 1 1

n n

j j j jj j

c x z x= =

≤∑ ∑ = 11

mn

Bi ijjij

c yx==

∑∑

= 1 21 2

1 1 1.

m

n n n

j j j j j mjB B Bj j j

x y x y x yc c c= = =

+ + …+ ∑ ∑ ∑

= 1 1 2 2

.m mB B B B B Bx c x c x c+ + + … + = Z

or Z* ≤ ZThis completes the proof of the theorem.


1. Prove that the set of all feasible solutions to an LP problem is a convex set. 2. Establish that every vertex of the convex set of feasible solutions is a basic feasible solution. 3. Show that every basic feasible solution to an LP problem corresponds to an extreme point

of the convex set of the feasible solutions. 4. Prove that the objective function of an LP problem assumes its optimum value at an extreme

point of the convex set generated by the set of all feasible solution. 5. Prove that if the objective function assumes its optimal value at more than one extreme

points, then every convex combination of these extreme point also gives the optimal value. 6. Show that any convex combination of k has different optimum solutions to an LP problem

is again an optimum solution to that problem. 7. Prove that if an LP problem admits of an optimal solution at least one basic feasible solution

must be optimal. 8. Find the condition under which a basic feasible solution constitutes a minimum feasible

solution. 9. Prove that the sufficient condition for a basic feasible solution of an LP problem to be

maximum is cj – zj ≤ 0, for all J for which the columns vectors aj ∈ A, but not in the basis. 10. State and prove the basic theorem that indicates how to improve a basic feasible solution

of an LP problem. 11. If there is an optimal basic feasible solution to an LP problem and for some aj not in basis

cj – zj = 0, yij ≤ 0 for all i, then prove that an alternative optimal solution also exists. 12. Prove that the extreme points of the feasible region of an LP problem correspond to the

feasible solutions of the problem.


1. Obtain all the basic feasible solution of the following system of linear equations: x1 + 2x2 + x3 = 4 and 2x1 + x2 + 5x3 = 5

2. What do you mean by an optimal basic feasible solution to an LP problem? Is the solution: x1 = 1, x2 = 1/2, x3 = x4 = x5 = 0, a basic solution of the equations. x1 + 2x2 + x3 + x4 = 2 and x1 + 2x2 + 0.5x3 + x5 = 2 ?


NOTES

Operations Research 3. The column vector (1, 1, 1) is a feasible solution of the system of equations: x1 + x2 + 2x3 = 4 and 2x1 + x2 + x3 = 2

4. Consider the system of equations: (i) x1 + 2x2 + 4x3 + x4 = 7; (ii) 2x1 – x2 + 3x1 – 2x4 = 4. Here, x1 = 1, x2 = 1, x3 = 1 and x4 = 0 is a feasible solutions. Reduce this feasible solution of two different basic feasible solution.

5. Show that the LP problem:Max Z = x1 + 2x2

subject to x1 – x2 + x3 = 4 x1 – 5x2 + x4 = 8and x1, x2, x3, x4 ≥ 0has an unbounded solution.

6. If x1 = 2, x2 = 3, x3 = 1 be a feasible solution of the following LP problem, then find the basic feasible solution:Max Z = x1 + 2x2 + 4x3

subject to 2x1 + x2 + 4x3 = 11 3x1 + x2 + 5x3 = 14and x1, x2, x3 ≥ 0

Hints and Answers 1. For each of the three possible submatrices of order 2, calculate xB = B–1 b. (a) x1 = 2, x2 = 1, x3 = 0 (b) x1 = 0, x2 = 5/3, x3 = 2/3 2. For each of the 10 possible submatrices of order 2, calculate xB = B–1 b. None of the 10 basic

feasible solution corresponds to the given solution. Hence the given solution is not basic. 3. Solve λ1 a1 + λ a2 + λ3 a3 = 0, where a1 = (1, 2)T, a2 = (1, 1)T, a3 = (2, 1)T, and b = (4, 2)T.

Taking λ3 = 1, we have λ1 = – 1 and λ2 = – 1. Also xr/λr = Min {xj/λj, λj > 0} = 1 (corresponds to a3); therefore, remove a3 and the new values of variables so obtained are x′1 = 2 and x′2 = 2. As vectors a1 and a2 associated with x′1 and x′2 are linearly independent, the given feasible solution is basic.

4. (i) For λ1 = 2, λ2 = 1, λ3 = – 1 and λ4 = 0, the solution is: (0, 1/2, 3/2, 0) (ii) For λ1 = – 2, λ2 = – 1, λ3 = 1 and λ4 = 0, the solution is: (3, 2, 0, 0)

5. Let B =

1 00 1 . Then BxB = b, where

xB =

48 ; B y2 = a2 y12 = – 1

c2 – z2 = c2 – =∑ 2

1

m

Bi ii

c y = 2 – (– 1) = 3 (> 0)

Z = Z + λ(c2 – z2) = Z + 3λ

for arbitrary large value of λ, Z can be made arbitrary large. 6. For λ = –1, λ2 = – 2 and λ3 = 1, the basic feasible solution is: x1 = 3, x2 = 5 and x3 = 0.


NOTES

Linear Programming: The Simplex MethodSUMMARY

• The simplex method is an interactive procedure for reaching the optimal solution to any LP problem. It consists of a series of rules that, in effect, algebraically examine corner (extreme) points of the solution space in a systematic way. Each step moves towards the optimal solution by increasing profit or decreasing cost, while maintaining feasibility. The simplex method consists of five steps: (i) identifying the pivot column, (ii) identifying the pivot row and key element, (iii) replacing the pivot row, (iv) computing new values for each remaining row; and (v) computing the zj and cj – zj values and examining for optimality. Each step of this iterative procedure is displayed and explained in a simplex table both for maximization and minimization LP problems.

• Finding and evaluating all basic feasible solution of an LP problem with more than two variables by using graphical method becomes difficult and complicated. Thus, an efficient method called the simplex method was developed by G.B. Dantzing in 1947 for solving a general class of LP model. This method is an iterative procedure of moving from one extreme point to another of the solution space. It leads to the optimal solution point and/or indicates that there exists an unbounded solution, in a finite number of steps.

• In the chapter, we discussed as to how an LP model can be stated in its standard form, along with certain important theorems that help to understand the procedure of (i) reducing a feasible solution to a basic feasible solution, (ii) improving a basic feasible solution, (iii) conditions of an alternative and unbounded solution and (iv) optimality condition.


NOTES

Operations Research

STRUCTURE

3.1 Introduction3.2 Formulation of Dual Linear Programming Problem3.3 Standard Results on Duality3.4 ManagerialSignificanceofDuality3.5 Advantages of Duality3.6 Duality Theorems3.7 Existence Theorem3.8 Dual-Simplex Method3.9 Integer Programming3.10 Types of Integer Programming Problems3.11 Enumeration and Cutting Plane Solution Concept3.12 Gomory’s all Integer Cutting Plane Method3.13 Gomory’s Mixed-Integer Cutting Plane Method3.14 Branch and Bound Method

U N I T

3DUALITY IN LINEAR

PROGRAMMING

3.1 INTRODUCTIONThe term ‘dual’, in general, implies two or double. The concept of duality is very

useful in Mathematics, Physics, Statistics, Engineering and Managerial decision-making. For example, in a two-person game theory, one competitor’s problem is the dual of the opponent’s problem.

In linear programming, duality implies that each linear programming problem can be analyzed in two different ways but would have equivalent solutions. Any LP problem (either maximization and minimization) can be stated in another equivalent form based on the same data. The new LP problem is called dual linear programming problem or in short dual. In general, it is immaterial which of the two problems is called primal or dual, since the dual of the dual is primal.


NOTES

Duality in Linear Programming

For example, consider the problem of production planning. The production manager attempts to determine quantities for each product to be produced with an objective tooptimizetheuseofavailableresourcessothatprofitismaximum.Butthroughadual LP problem approach, he may develop a production plan that optimizes resource utilization so that marginal opportunity cost of each unit of a resource is equal to its return (also called shadow price). The shadow price indicates an additional price to be paidtoobtainoneadditionalunitoftheresourcesinordertomaximizeprofitunderthe resource constraints. If a resource is not completely used, i.e. there is slack, then its marginal return is zero.

Theshadowpriceisalsodefinedastherateofchangeintheoptimalobjectivefunction value with respect to the unit change in the availability of a resource. To be more precise for any constraint, we have

Shadow price = Change in optimal objective function valueUnit change in the availabbility of resource

The interpretation of rate of change (increase or decrease) in the value of objective function depends on whether LP problem is of maximization or minimization type. The shadow price for a less than or equal to (≤) type constraint will always be positive. This is because increasing the right-hand side resource value cannot decrease the value of objective function. Similarly, the shadow price for a greater than or equal to (≥) type constraint will always be negative because increasing the right-hand side resonance value cannot increase the value of the objective function.

There is no need to solve both LP problems separately. Solving one LP problem is equivalent to solving the other simultaneously. Thus, if the optimal solution to one is known, the optimal solution of the other can also be read from the cj – zj row. In some cases, considerable computing time can be reduced by solving the dual LP problem.

3.2 FORMULATION OF DUAL LINEAR PROGRAMMING PROBLEM

There are two important forms of primal and dual LP problems, namely the symmetrical (canonical) form and the standard form.

3.2.1 Symmetrical FormSuppose the primal LP problem is given in the formMaximize Zx = c1 x1 + c2 x2 + ... + cn xn

subject to the constraints a11 x1 + a12 x2 + ... + a1n xn ≤ b1

a21 x1 + a22 x2 + ... + a2n xn ≤ b2

am1 x1 + am2 x2 + ... + amn xn ≤ bm

and x1, x2 ..., xn ≥ 0Then the corresponding dual LP problem is written as:Minimize Zy = b1 y1 + b2 y2 + ... + bm ym

Dual LP problemprovides useful economic information about worth of resources to be used.

Shadow price represents increase in the objective function value due to one-unit increase in the right hand side (resource) of any constraint.


NOTES

Operations Research subject to the constraints a11 y1 + a21 y2 + . . . + am1 ym ≤ c1

a12 y1 + a22 y2 + . . . + am2 ym ≤ c2

a1n y1 + a2n y2 + . . . + amn ym ≤ cn

and y1, y2, . . ., ym ≥ 0In general, the primal-dual relationship between a pair of LP problems can be

expressed as follows:

Primal Dual

Max Zx = j

n

j jc x=∑

1Min Zy =

i

m

i ib y=∑

1

subject to the constraints subject to the constraints

j

n

ij ja x=∑

1 ≤ bj ; i = 1, 2, ..., m

i

m

ji ia y=∑

1 ≥ cj; j = 1, 2, ..., n

and xj ≥ 0; j = 1, 2, ..., n and yi ≥ 0; i = 1, 2, ..., m

A summary of the general relationships between primal and dual LP problems is given in Table 3.1

Table 3.1. Primal-Dual RelationshipIf Primal Then Dual

(i) Objective is to maximize (i) Objective is to minimize(ii) jth primal variable, xj (ii) jth dual constraint(iii) ith primal constraint (iii) ith dual variable, yi

(iv) Primal variable xj unrestricted in sign (iv) Dual constraint j is = type(v) Primal constraint i is = type (v) Dual variable yi is unrestricted in sign(vi) Primal constraints ≤ type (vi) Dual constraints ≥ type

3.2.2 Economic Interpretation of Dual VariablesInthemaximizationprimalLPmodel,wemaydefineeachparameterasfollows: Z = return cj =profit(orreturn)perunitofvariable(activity)xj

xj = number units of variable j aij = units of resource, i consumed (required) bi = maximum units of resource, i available per unit of variable jThe new variables introduced in the dual problem are Zy and yi (dual variables).

When both the primal and the dual solutions are optimal, the value of objective function satisfiesthestrictequality,i.e. Zx = Zy.

The interpretation associated with the dual variables yi (i = 1, 2, . . ., m) is discussed below. Rewrite the primal LP problem as follows:

Dual variables represent the potential value of resources

aij = aji


NOTES


subject to the constraints a11 y1 + a21 y2 + . . . + am1 ym ≤ c1

a12 y1 + a22 y2 + . . . + am2 ym ≤ c2

a1n y1 + a2n y2 + . . . + amn ym ≤ cn

and y1, y2, . . ., ym ≥ 0In general, the primal-dual relationship between a pair of LP problems can be

expressed as follows:

Primal Dual

Max Zx = j

n

j jc x=∑

1Min Zy =

i

m

i ib y=∑

1

subject to the constraints subject to the constraints

j

n

ij ja x=∑

1 ≤ bj ; i = 1, 2, ..., m

i

m

ji ia y=∑

1 ≥ cj; j = 1, 2, ..., n

and xj ≥ 0; j = 1, 2, ..., n and yi ≥ 0; i = 1, 2, ..., m

A summary of the general relationships between primal and dual LP problems is given in Table 3.1

Table 3.1. Primal-Dual RelationshipIf Primal Then Dual

(i) Objective is to maximize (i) Objective is to minimize(ii) jth primal variable, xj (ii) jth dual constraint(iii) ith primal constraint (iii) ith dual variable, yi

(iv) Primal variable xj unrestricted in sign (iv) Dual constraint j is = type(v) Primal constraint i is = type (v) Dual variable yi is unrestricted in sign(vi) Primal constraints ≤ type (vi) Dual constraints ≥ type

3.2.2 Economic Interpretation of Dual VariablesInthemaximizationprimalLPmodel,wemaydefineeachparameterasfollows: Z = return cj =profit(orreturn)perunitofvariable(activity)xj

xj = number units of variable j aij = units of resource, i consumed (required) bi = maximum units of resource, i available per unit of variable jThe new variables introduced in the dual problem are Zy and yi (dual variables).

When both the primal and the dual solutions are optimal, the value of objective function satisfiesthestrictequality,i.e. Zx = Zy.

The interpretation associated with the dual variables yi (i = 1, 2, . . ., m) is discussed below. Rewrite the primal LP problem as follows:

Dual variables represent the potential value of resources

aij = aji

Primal LP Problem

Maximize (return) Zx = j

n

j jc x=∑

1

= j

n

=∑

1(Profitperunitofvariablexj) (Units of variable xj)


j

n

ij ja x=∑

1 ≤ bi

or j

n

=∑

1(Units of resource i, consumed per unit of variable, xj) (Units of variable xj)

≤ Units of resource, i availableand xj ≥ 0, for all j

Dual LP Problem

Minimize (cost) Zy = i

m

i ib y=∑

1

= i

m

=∑

1(Units of resource, i) (Cost per unit of resource, i)


i

m

ji ia y=∑

1 ≥ cj

or i

m

=∑

1(Units of a resource, i consumed per unit of variable yi )

(Cost per unit of resource, i) ≥Profitperunitforeachvariable,xj

and yi ≥ 0, for all iFrom these expressions of parameters of both primal and dual problems, it is

clear that for the unit of measurement to be consistent, the dual variable (yi) must be expressed in terms of return (or worth) per unit of resource i. This is called dual price (simplex multiplier or shadow price) of resource i. In other words, optimal value of a dual variable associated with a particular primal constraint indicates the marginal change (increase, if positive or decrease, if negative) in the optimal value of the primal objective function. For example, if y2 = 5, then this indicates that for every additional unit (up to a certain limit) of resource 2 (resource associated with constraint 2 in the primal), the objective function value will increase by 5 units. The value y2 = 5 is also called the marginal (or shadow or implicit) price of resource 2.

Similarly, for feasible solutions of both primal and dual LP problems, objective function value satisfy the inequality Zx ≤ Zy. This inequality is interpreted as: Profit ≤ Worth of resources.Thus,solongasthetotalprofit(return)fromallactivitiesislessthanthe worth of the resources, the feasible solution of both primal and dual is not optimal. Theoptimality(maximumprofitorreturn)isreachedonlywhentheresourceshavebeen completely utilized. This is only possible if the worth of the resources (i.e. input) isequaltoprofit(i.e. output).


NOTES

Operations Research 3.2.3 Economic Interpretation of Dual ConstraintsAs stated earlier, the dual constraints are expressed as:

i

m

ji i ja y c=∑ −

1 ≥ 0

Sincecoefficientsaji represents the amount of resource bi consumed by per unit of activity xj, and the dual variable yi represents shadow price per unit of resource bi, the quantity Σ aji yi (= zj) should be total shadow price of all resources required to produce one unit of activity xj.

For maximization LP problem, if cj – zj > 0 value corresponds to any non-basic variable, then the value of objective function can be increased. This implies that the value of variable, xj canbeincreasedfromzerotoapositivelevelprovideditsunitprofit(cj) is more than its shadow price, i.e.

cj ≥ i

m

ji ia y=∑

1

Profitperunitofactivityxj ≥ Shadow price of resources used per unit of activity, xj.

3.2.4 Rules for Constructing the Dual from PrimalThe rules for constructing the dual from the primal and vice-versa using the

symmetrical form of LP problem are: 1. AdualvariableisdefinedcorrespondstoeachconstraintintheprimalLPproblem

and vice versa. Thus, for a primal LP problem with m constraints and n variables, there exists a dual LP problem with m variables and n constraints and vice-versa.

2. The right-hand side constants b1, b2, . . ., bm of the primal LP problem becomes the coefficientsofthedualvariablesy1, y2, . . ., ym in the dual objective function Zy. Also thecoefficientsc1, c2, . . ., cn of the primal variables x1, x2, . . ., xn in the objective function become the right-hand side constants in the dual LP problem.

3. For a maximization primal LP problem with all ≤ (less than or equal to) type constraints, there exists a minimization dual LP problem with all ≥ (greater than or equal to) type constraints and vice versa. Thus, the inequality sign is reversed in all the constraints except the non-negativity conditions.

4. Thematrixofthecoefficientsofvariablesintheconstraintsofdualisthetransposeofthematrixofcoefficientsofvariablesintheconstraintsofprimalandviceversa.

5. If the objective function of a primal LP problem is to be maximized, the objective function of the dual is to be minimized and vice versa.

6. If the ith primal constraint is = (equality) type, then the ith dual variables is unrestricted in sign and vice versa.


NOTES


The primal-dual relationships may also be memorized by using the following table:

Dual VariablesPrimal Variables

Maximize Zxx1 x2 . . . xj . . . xn

y1 a11 a12 . . . a1j . . . a1n ≤ b1

y2 a21 a22 . . . a2j . . . a2n ≤ b2

ym am1 am2 . . . amj . . . amn ≤ bm

Minimize Zy ≥ c1 ≥ c2 . . . ≥ cj . . . ≥ cn ‘ Dual objective function coefficients

‘ jth dual constraintThe primal constraints should be read across the rows, and the dual constraints

should be read across the columns.

Example 1. Write the dual to the following LP problem.Maximize Z = x1 – x2 + 3x3

subject to the constraints (i) x1 + x2 + x3 ≤ 10, (ii) 2x1 – x2 – x3 ≤ 2, (iii) 2x1 – 2x2 – 3x3 ≤ 6and x1, x2, x3 ≥ 0

Sol. In the given LP problem there are m = 3 constraints and n = 3 variables. Thus, there must be m = 3 dual variables and n =3constraints.Further,thecoefficientsoftheprimal variables, c1 = 1, c2 = – 1, c3 = 3 become right-hand side constants of the dual. The right-hand side constants b1 = 10, b2 = 2, b3=6becomethecoefficientsinthedualobjective function. Finally, the dual must have a minimizing objective function with all ≥ type constraints. If y1, y2 and y3 are dual variables corresponding to three primal constraints in the given order, the resultant dual is

Minimize Zy = 10y1 + 2y2 + 6y3

subject to the constraints (i) y1 + 2y2 + 2y3 ≥ 1, (ii) y1 – y2 – 2y3 ≥ – 1, (iii) y1 – y2 – 3y3 ≥ 3and y1, y2, y3 ≥ 0

Example 2. Write the dual of the following LP problem.Minimize Zx = 3x1 – 2x2 + 4x3

subject to the constraints (i) 3x1 + 5x2 + 4x3 ≥ 7, (ii) 6x1 + x2 + 3x3 ≥ 4, (iii) 7x1 – 2x2 – x3 ≤ 10 (iv) x1 – 2x2 + 5x3 ≥ 3, (v) 4x1 + 7x2 – 2x3 ≥ 2and x1, x2, x3 ≥ 0

Sol. Since the objective function of the given LP problem is of minimization, the direction of each inequality has to be changed to ≥ type by multiplying both sides by – 1. The standard primal LP problem so obtained is:

Minimize Zx = 3x1 – 2x2 + 4x3


NOTES

Operations Research subject to the constraints (i) 3x1 + 5x2 + 4x3 ≥ 7, (ii) 6x1 + x2 + 3x3 ≥ 4, (iii) – 7x1 + 2x2 + x3 ≥ – 10 (iv) x1 – 2x2 + 5x3 ≥ 3, (v) 4x1 + 7x2 – 2x3 ≥ 2and x1, x2, x3 ≥ 0

If y1, y2, y3, y4 and y5aredualvariablescorrespondingtothefiveprimalconstraintsin the given order, the dual of this primal LP problem is stated as:

Maximize Zy = 7y1 + 4y2 – 10y3 + 3y4 + 2y5

subject to the constraints (i) 3y1 + 6y2 – 7y3 + y4 + 4y5 ≤ 3, (ii) 5y1 + y2 + 2y3 – 2y4 + 7y5 ≤ – 2 (iii) 4y1 + 3y2 + y3 + 5y4 – 2y5 ≤ 4and y1, y2, y3, y4, y5 ≥ 0

Example 3. Obtain the dual LP problem of the following primal LP problem:Minimize Z = x1 + 2x2

subject to the constraints (i) 2x1 + 4x2 ≤ 160, (ii) x1 – x2 = 30, (iii) x1 ≥ 10and x1, x2 ≥ 0

Sol. Since the objective function of the primal LP problem is of minimization, change all ≤ type constraints to ≥ type constraints by multiplying the constraint on both sides by – 1. Also write = type constraint equivalent to two constraints of the type ≥ and ≤. Then the given primal LP problem can be rewritten as:

Minimize Zx = x1 + 2x2

subject to the constraint (i) – 2x1 – 4x2 ≥ – 160, (ii) x1 – x2 ≥ 30 (iii) x1 – x2 ≤ 30 or – x1 + x2 ≥ – 30, (iv) x1 ≥ 10and x1, x2 ≥ 0

Let y1, y2, y3 and y4 be the dual variables corresponding to the four constraints in the given order. The dual of the given primal LP problem can then be formulated as follows:

Maximize Zy = – 160y1 + 30y2 – 30y3 + 10y4

subject to the constraints (i) – 2y1 + y2 – y3 + y4 ≤ 1, (ii) – 4y1 – y2 + y3 ≤ 2and y1, y2, y3, y4 ≥ 0

Let y = y2 – y3 (y2, y3 ≥ 0). The above dual problem then reduces to the formMaximize Zy = – 160y1 + 30y + 10y4

subject to the constraints (i) – 2y1 + y + y4 ≤ 1, (ii) – 4y1 – y ≤ 2and y1, y4 ≥ 0; y unrestricted in sign

Remark. Since second constraint in the primal LP problem is equality, therefore as per rule 6 corresponding second dual variable y (= y2 – y3) should be unrestricted in sign.


NOTES


Example 4. Obtain the dual LP problem of the following primal LP problem:Minimize Zx = x1 – 3x2 – 2x3

subject to the constraints (i) 3x1 – x2 + 2x3 ≤ 7, (ii) 2x1 – 4x2 ≥ 12, (iii) – 4x1 + 3x2 + 8x3 = 10and x1, x2 ≥ 0; x3 unrestricted in sign.

Sol. Let y1, y2 and y3 be the dual variables corresponding to three primal constraints in the given order. As the given LP problem is of minimization, all constraints can be converted to ≥ type by multiplying both sides by – 1, i.e., – 3x1 + x2 – 2x3 ≥ – 7. Since the third constraint of the primal is an equation, the third dual variable y3 will be unrestricted in sign. The dual of the given LP primal can be formulated as follows:

Maximize Zy = – 7y1 + 12y2 + 10y3

subject to the constraints (i) – 3y1 + 2y2 – 4y3 ≤ 1, (ii) y1 – 4y2 + 3y3 ≤ – 3, (iii) – 2y1 + 8y3 ≤ – 2and y1, y2 ≥ 0; y3 unrestricted in sign.

Example 5. Obtain the dual of the following primal LP problemMaximize Zx = x1 – 2x2 + 3x3

subject to the constraints (i) – 2x1 + x2 + 3x3 = 2, (ii) 2x1 + 3x2 + 4x3 = 1and x1, x2, x3 ≥ 0

Sol. Since both the primal constraints are of the equality type, the corresponding dual variables y1 and y2, will be unrestricted in sign. Following the rules of duality formulation, the dual of the given primal LP problem is

Minimize Zy = 2y1 + y2

subject to the constraints (i) – 2y1 + 2y2 ≥ 1, (ii) y1 + 3y2 ≥ – 2, (iii) 3y1 + 4y2 ≥ 3and y1, y2 unrestricted in sign.

Example 6. Write the dual of the following primal LP problemMaximize Z = 3x1 + x2 + 2x3 – x4

subject to the constraints (i) 2x1 – x2 + 3x3 + x4 = 1, (ii) x1 + x2 – x3 + x4 = 3and x1, x2 ≥ 0 and x3, x4 unrestricted in sign.

Sol. Here we may apply the following rules of forming a dual of given primal LP problem. (i) The x3 and x4 variables in the primal are unrestricted in sign, third and fourth

constraints in the dual shall be equalities. (ii)ThegivenprimalLPproblemisofmaximization;thefirsttwoconstraintsin

the dual LP problem will therefore be ≥ type constraints. (iii) Since both the constraints in the primal are equalities, the corresponding dual

variables y1 and y2 will be unrestricted in sign.


NOTES

Operations Research If y1 and y2 are dual variables corresponding to the two primal constraints in the given order, the dual of the given primal can be written as:

Minimize Zy = y1 + 3y2

subject to the constraints (i) 2y1 + y2 ≥ 3, (ii) – y1 + y2 ≥ 1, (iii) 3y1 – y2 = 2 (iv) y1 + y2 = – 1and y1, y2 unrestricted in sign.

TEST YOUR KNOWLEDGE 3.1

Write the dual of the following primal LP problems 1. Max Zx = 2x1 + 5x2 + 6x3

subject to (i) 5x1 + 6x2 – x3 ≤ 3 (ii) – 2x1 + x2 + 4x3 ≤ 4 (iii) x1 – 5x2 + 3x3 ≤ 1 (iv) – 3x1 – 3x2 + 7x3 ≤ 6

and x1, x2, x3 ≥ 0 2. Min Zx = 7x1 + 3x2 + 8x3

subject to (i) 8x1 + 2x2 + x3 ≥ 3 (ii) 3x1 + 6x2 + 4x3 ≥ 4 (iii) 4x1 + x2 + 5x3 ≥ 1 (iv) x1 + 5x2 + 2x3 ≥ 7and x1, x2, x3 ≥ 0

3. Max Zx = 2x1 + 3x2 + x3

subject to (i) 4x1 + 3x2 + x3 = 6, (ii) x1 + 2x2 + 5x3 = 4and x1, x2, x3 ≥ 0

4. Max Zx = 3x1 + x2 + 3x3 – x4

subject to (i) 2x1 – x2 + 3x3 + x4 = 1 (ii) x1 + x2 – x3 + x4 = 3and x1, x2, x3, x4 ≥ 0

5. Min Zx = 2x1 + 3x2 + 4x3

subject to (i) 2x1 + 3x2 + 5x3 ≥ 2 (ii) 3x1 + x2 + 7x3 = 3 (iii) x1 + 4x2 + 6x3 ≤ 5and x1, x2, ≥ 0, x3 is unrestricted

6. Min Zx = x1 + x2 + x3

subject to (i) x1 – 3x2 + 4x3 = 5, (ii) x1 – 2x2 ≤ 3 (iii) 2x2 – x3 ≥ 4and x1, x2 ≥ 0, x3 is unrestricted.

7. Min Zx = 8x1 + 3x2

subject to (i) x1 – 6x2 ≥ 2, (ii) 5x1 + 7x2 = – 4and x1, x2 ≥ 0.

8. Max Zx = 8x1 + 8x2 + 8x3 + 12x4

subject to (i) 30x1 + 20x2 + 25x3 + 40x4 ≤ 800 (ii) 25x1 + 10x2 + 7x3 + 15x4 ≤ 250 (iii) 4x1 – x2 = 0 (iv) x3 ≥ 5and x1, x2, x3, x4 ≥ 0


NOTES


9. Min Zx = 18x1 + 10x2 + 11x3subject to (i) 4x1 + 6x2 + 5x3 ≥ 480 (ii) 12x1 + 10x2 + 10x3 ≥ 1,200 (iii) 10x1 + 15x2 + 7x3 ≤ 1,500 (iv) x3 ≥ 50 (v) x1 – x2 ≤ 0and x1, x2, x3 ≥ 0

10. Min Zx = 2x1 – x2 + 3x3subject to (i) x1 + x2 + x3 ≥ 12 (ii) x2 – 2x3 ≥ – 6 (iii) 6 ≤ x1 + 2x2 + 4x3 ≥ 24and x1, x2 ≥ 0, x3 unrestricted.

Hints and Answers 1. Min Zy = 3y1 + 4y2 + y3 + 6y4

subject to (i) 5y1 – 2y2 + y3 – 3y4 ≥ 2 (ii) 6y1 + y2 – 5y3 – 3y4 ≥ 5 (iii) – y1 + 4y2 + 3y3 + 7y4 ≥ 6and y1, y2, y3, y4 ≥ 0

2. Max Zy = 3y1 + 4y2 + y3 + 7y4subject to (i) 8y1 + y2 + 4y3 + y4 ≤ 7 (ii) 2y1 + 6y2 + y3 + 5y4 ≤ 3 (iii) y1 + 4y2 + 5y3 + 2y4 ≤ 8and y1, y2, y3, y4 ≥ 0

3. Min Zy = 6y1 + 4y2subject to (i) 4y1 + y2 ≥ 2, (ii) 3y1 + 2y2 ≥ 3 (iii) y1 + 5y2 ≥ 1and y1, y2 unrestricted in sign.

4. Min Zy = y1 + 3y2subject to (i) 2y1 + y2 ≥ 3 (ii) – y1 + y2 ≥ 1 (iii) 3y1 – y2 ≥ 3 (iv) y1 + y2 ≥ – 1and y1, y2 unrestricted in sign.

5. Max Zy = 2y1 + 3y2 – 5y3subject to 2y1 + 3y2 – y3 ≤ 2 3y1 + 3y2 – 4y3 ≤ 3 5y1 + 7y2 – 6y3 = 4and y1, y3 ≥ 0 and y2 unrestricted.

6. Max Zy = – 5y1 – 3y2 + 4y3subject to (i) – y1 – y2 ≤ 1, (ii) – 3y1 + 2y2 + 2y3 ≤ 1 4y1 – y3 ≤ 1and y2, y3 ≥ 0 and y1 is unrestricted.

7. Max Zy = 2y1 – 4y2subject to (i) y1 + 5y2 ≤ 8, (ii) – 6y1 + 7y2 ≤ 3and y1 ≥ 0 and y2 is unrestricted.

8. Min Zy = 800y1 + 250y2 + y3 + 5y4subject to 30y1 + 25y2 + 4y3 ≥ 8 20y1 + 10y2 – y3 ≥ 8 25y1 + 7y2 + y4 ≥ 8 40y1 + 15y2 ≥ 12y1, y2 ≥ 0 and y3, y4 are unrestricted.


NOTES

Operations Research3.3 STANDARD RESULTS ON DUALITY

See appendix for detail proof of the following standard results: 1. The dual of the dual LP problem is again the primal problem. 2. If either the primal or the dual problem has an unbounded objective function

value, the other problem has no feasible solution. 3.Ifeithertheprimalordualproblemhasafiniteoptimalsolution,theotherone

also possesses the same, and the optimal value of the objective functions of the two problems are equal, i.e., Max Zx = Min Zy. This analytical result is known as the fundamental primal-dual relationship. These results are summarized as follows.

Dual Problem (Max)Primal Problem (Min)

Feasible Infeasible

Feasible Max Zy = Min Zx Max Zy → + ∞

Infeasible Min Zx → – ∞ Unbounded or infeasible

4. Complementary slackness property of primal-dual relationship states that for a positive basic variable in the primal, the corresponding dual variable will be equal to zero. Alternatively, for a non-basic variable in the primal (which is zero), the corresponding dual variable will be basic and positive.

3.4 MANAGERIAL SIGNIFICANCE OF DUALITYThe study of dual LP problem provides information about the value of the resources.

Thus, a thorough economic analysis is required to decide whether additional units of any resource are needed. If yes, then how much cost per unit is required to be paid for any resource.

Thesignificanceofthestudyofdualissummarizedasfollows: (i) The right-hand side of the constraint in primal LP problem represents the

amount of a resource available and the associated dual variable value is interpretated as the maximum amount likely to be paid for an additional unit of this resource.

(ii) The maximum amount that should be paid for one additional unit of a resource is called its shadow price (also called simplex multiplier).

(iii) The total shadow price (or marginal value) of the resources equals the optimal value of objective function.

(iv) The value of the ith dual variable represents the rate at which the primal objective function value will increase by increasing the right-hand side (resource value) of constraint i, assuming that all other data remain unchanged.

Remark: When the cost of a resource is sunk (cost that is incurred no matter irrespective to the value the dual variables). When the cost of a resource vary with the value of decision variables the shadow price represents the extra cost likely to be paid for one unit of the resource.


NOTES


Example 1. The optimal solution simplex table for the primal LP problemMaximize Z = 3x1 + 4x2 + x3

subject to the constraints (i) x1 + 2x2 + 3x3 ≤ 90 (time for operation 1), (ii) 2x1 + x2 + x3 ≤ 60 (time for operation 2) (iii) 3x1 + x2 + 2x3 ≤ 80 (time for operation 3)and x1, x2, x3 ≥ 0is given below:

cj → 3 4 1 0 0 0


cB

Basic Variables B

Basic Variables Value xB

(= b)

x1 x2 x3 s1 s2 s3

4 x2 40 0 1 10/6 4/6 – 1/3 0

3 x1 10 1 0 – 1/3 – 1/3 2/3 0

0 s3 10 0 0 8/6 8/6 –10/6 1

Z = 190 cj – zj 0 0 –28/6 –10/6 – 2/3 0

(a) Find the solution, maximum profit, idle capacity and the loss of the total contribution of every unit reduced from the right-hand side of the constraints.

(b) Write the dual of the given problem and give the initial simplex table.Sol. ConsideringthisLPprobleminthecontextofprofitmaximizationproduction

problem with constraints on time input resource.TheprimalLPproblemisconcernedwithmaximizingtheprofitcontributionfrom

three products say A, B and C, while the dual will be concerned with evaluating the time used in the three operations for producing the three products.

Theproductioncapacityofthethreeoperationsisaresourcetothefirm.ThisLPproblem can be solved along the following lines.

(a) The optimal solution of the primal LP problem can be read from the given table as:

x1 = 10, x2 = 40, x3=0andMax(profit)Zx = ` 190The value of slack variable, s3 = 10 units in the optimal solution is representing

idle capacity in operation 3.Interpretation of dual variables: The absolute values of cj – zj in the optimal

solution simplex table, under the slack variables columns represent the values of dual variables, i.e. marginal value (shadow price) or worth of one hour in each operation 1 and 2. The optimal solution of dual LP problem is:

y1 = 10/6, y2 = 2/3, y3 = 0 and Mix Zy = 190Thus optimal value of the dual variable y1 = 10/6 or 1.66 represents the per unit

price(worthormarginalprice)ofthefirstresource(i.e. time available in operation 1). Thismeansthatmarginalcontributionofoperation1tothetotalprofitis 1.66. In other


NOTES

Operations Research words,ifoneproductiveunitisremovedfromtherighthandsideofthefirstconstraint,the total contribution would reduce by ` 10/6, or ` 1.66.

However, if one productive unit is removed up to the extent of idle capacity from the right-hand side of the third constraint (time available in operation 1), the total contribution would not be affected.

(b) Let y1, y2 and y3 be the dual variables representing per unit price (worth) for time available in the operations 1, 2 and 3, respectively. Then the dual of the given problem is stated below:

Minimize Z = 90y1 + 60y2 + 80y3

subject to the constraints (i) y1 + 2y2 + 3y3 ≥ 3, (ii) 2y1 + y2 + y3 ≥ 4, (iii) 3y1 + y2 + 2y3 ≥ 1and y1, y2, y3 ≥ 0

Interpretation of dual constraints: Suppose company wants to continue production of product C. To see economic worth of C, examine c3 – z3 value under column-x3. Producing C will be economical only if c3 > z3. This is possible either by increasing the profit per unit c3 or decreasing the cost of the used resources z3 (= 3y1 + y2 + 2y3).An increase in theunit profitmaynot bepossible because ofcompetitive market. A decrease in z3 is possible by reducing the unit usage of time in operation 3. Let t1, t2 and t3 be the unit times of three operations to be reduced. Then to determine t1, t2 and t3 such that the new cost, z3 of three operations falls below the unitprofit,i.e.

3(1 – t1) y1 + (1 – t2)y2 + 2 (1 – t3) y3 < 1For given values of y1 = 5/3, y2 = 2/3 and y3 = 0, we get

3(1 – t1) (5/3) + (1 – t2) (2/3) < 1 or 5t1 + 2t2 > 16Thus, any values of t1 and t2 that satisfy 5t1 + 2t2 > 16 should make the product

profitable.However,thismaynotbepossiblebecausethisrequiresareductioninthetimes of operation 1 and 2.

Example 2. XYZ Company has three departments–Assembly, Painting and Packing. The company can make three types of almirahs. An almirah of type I requires one hour of assembly, 40 minutes of painting and 20 minutes of packing time, respectively. Similarly, an almirah of type II needs 80 minutes, 20 minutes and one hour, respectively. The almirah of type III requires 40 minutes each of assembly, painting and packing time. The total time available at assembly, painting and packing departments is 600 hours, 400 hours and 800 hours, respectively. Determine the number of each type ofalmirahsthatshouldbeproducedinordertomaximizetheprofit.Theunitprofitfortypes I, II and III is ` 40, 80 and 60, respectively.

Suppose that the manager of this XYZ company is thinking of renting the production capacities of the three departments to another almirah manufacturer– ABC company. ABC company is interested in minimizing the rental charges. On the other hand, the XYZ company would like to know the worth of production hours to them, in each of the departments, in order to determine the rental rates. (a) Formulate this problem as an LP problem and solve it to determine the number of each type of almirahs that should be produced by the XYZcompanyinordertomaximizeitsprofit.(b) Formulate the dual of the primal LP problem and interpret your results.

The shadow price is represented by the cj – zj value under slack variable columns in the optimal simplex table.


NOTES


words,ifoneproductiveunitisremovedfromtherighthandsideofthefirstconstraint,the total contribution would reduce by ` 10/6, or ` 1.66.

However, if one productive unit is removed up to the extent of idle capacity from the right-hand side of the third constraint (time available in operation 1), the total contribution would not be affected.

(b) Let y1, y2 and y3 be the dual variables representing per unit price (worth) for time available in the operations 1, 2 and 3, respectively. Then the dual of the given problem is stated below:

Minimize Z = 90y1 + 60y2 + 80y3

subject to the constraints (i) y1 + 2y2 + 3y3 ≥ 3, (ii) 2y1 + y2 + y3 ≥ 4, (iii) 3y1 + y2 + 2y3 ≥ 1and y1, y2, y3 ≥ 0

Interpretation of dual constraints: Suppose company wants to continue production of product C. To see economic worth of C, examine c3 – z3 value under column-x3. Producing C will be economical only if c3 > z3. This is possible either by increasing the profit per unit c3 or decreasing the cost of the used resources z3 (= 3y1 + y2 + 2y3).An increase in theunit profitmaynot bepossible because ofcompetitive market. A decrease in z3 is possible by reducing the unit usage of time in operation 3. Let t1, t2 and t3 be the unit times of three operations to be reduced. Then to determine t1, t2 and t3 such that the new cost, z3 of three operations falls below the unitprofit,i.e.

3(1 – t1) y1 + (1 – t2)y2 + 2 (1 – t3) y3 < 1For given values of y1 = 5/3, y2 = 2/3 and y3 = 0, we get

3(1 – t1) (5/3) + (1 – t2) (2/3) < 1 or 5t1 + 2t2 > 16Thus, any values of t1 and t2 that satisfy 5t1 + 2t2 > 16 should make the product

profitable.However,thismaynotbepossiblebecausethisrequiresareductioninthetimes of operation 1 and 2.

Example 2. XYZ Company has three departments–Assembly, Painting and Packing. The company can make three types of almirahs. An almirah of type I requires one hour of assembly, 40 minutes of painting and 20 minutes of packing time, respectively. Similarly, an almirah of type II needs 80 minutes, 20 minutes and one hour, respectively. The almirah of type III requires 40 minutes each of assembly, painting and packing time. The total time available at assembly, painting and packing departments is 600 hours, 400 hours and 800 hours, respectively. Determine the number of each type ofalmirahsthatshouldbeproducedinordertomaximizetheprofit.Theunitprofitfortypes I, II and III is ` 40, 80 and 60, respectively.

Suppose that the manager of this XYZ company is thinking of renting the production capacities of the three departments to another almirah manufacturer– ABC company. ABC company is interested in minimizing the rental charges. On the other hand, the XYZ company would like to know the worth of production hours to them, in each of the departments, in order to determine the rental rates. (a) Formulate this problem as an LP problem and solve it to determine the number of each type of almirahs that should be produced by the XYZcompanyinordertomaximizeitsprofit.(b) Formulate the dual of the primal LP problem and interpret your results.

The shadow price is represented by the cj – zj value under slack variable columns in the optimal simplex table.

Mathematical formulation The production data given in the problem may be summarized as below:

Types of AlmirahNumber of Hours Required

(per Unit) Profit per Unit of Almirah (`)

Assembly Painting PackingI 1 2/3 1/3 40II 4/3 1/3 1 80III 2/3 2/3 2/3 60

Total availability (hrs) 600 400 800

Let x1, x2 and x3 be the number of units of the three types of almirahs to be produced, respectively. The given problem can then be represented as an LP model:

Maximize Z = 40x1 + 80x2 + 60x3subject to the constraints (i) x1 + (4/3)x2 + (2/3)x3 ≤ 600, (ii) (2/3) x1 + (1/3) x2 + (2/3) x3 ≤ 400 (iii) (1/3)x1 + x2 + (2/3)x3 ≤ 800and x1, x2, x3 ≥ 0

Let y1, y2 and y3 be the per hour rental price for assembly, painting and packing departments, respectively to be paid by ABC company. Since company ABC desires to minimize total rental changes, then objective function becomes:

Min (total rental charges) Zy = 600y1 + 400y2 + 800y3The given data show that an almirah of type I requires 1 assembly hour, 2/3 painting

hour and 1/3 packing hour. The cost (i.e. rent) of time used for making almirah I is: y1 + (2/3) y2 + (1/3) y3. If ABC company used that time to make almirah I, it would earn ` c1=40incontributiontoprofit,andsoitwillnotrentoutthetimeunless:y1 + (2/3) y2 + (1/3) y3 ≥ 40. Similarly, the company will work for almirahs II and III.

Since the total rent of all the departments should be greater than or equal to the profitfromoneunitofthealmirah,thedualobjectivefunctionalongwiththeconstraintsthat determines for ABC, the value of the productive resources, can be written as:

Minimize (total rent) Zy = 600y1 + 400y2 + 800y3subject to the constraints

(i) y1 + (2/3) y2 + (1/3) y3 ≥ 40, (ii) (4/3) y1 + (1/3) y2 + y3 ≥ 80 (iii) (2/3) y1 + (2/3) y2 + (2/3) y3 ≥ 60and y1, y2, y3 ≥ 0

This dual objective function (minimum total rent acceptable by the ABC company) isequaltotheprimalobjectivefunction(maximumprofitthatcouldbeearnedbyXYZ company) from its resources.

Example 3. A firm manufactures two products A and B on machines I and II as shown below:

Machine Product Available Hours

A B

I 30 20 300

II 5 10 110

Profitperunit(`) 6 8


NOTES

Operations Research The total time available is 300 hours and 110 hours on machines I and II, respectively. Products A and Bcontributeaprofitof` 6 and ` 8 per unit, respectively. Determine the optimum product mix. Write the dual of this LP problem and give its economic interpretation.

Mathematical formulation. The primal and the dual LP problems of the given problem are

Primal problem Dual problem

x1 and x2 = number of units of A and B to be produced, respectively

Max Zx = 6x1 + 8x2

subject to the constraints 30x1 + 20x2 ≤ 300 5x1 + 10x2 ≤ 110and x1, x2 ≥ 0

y1 and y2 = cost of one hour on machines I and II, respectively

Min Zy = 300y1 + 110y2

subject to the constraints 30y1 + 5y2 ≥ 6 20y1 + 10y2 ≥ 8and y1, y2 ≥ 0

Solution of the primal problem. The optimal solution of the primal problem is given in Table 3.2.

Table 3.2. Optimal Solution of Primal Problem

cj → 6 8 0 0


Basic Variables B

Basic Variables Value xB (= b)

x1 x2 s1p s2p

6 x1 4 1 0 1/20 – 1/10

8 x2 9 0 1 – 1/10 3/20

Z = 96 cj – zj 0 0 – 1/10 – 6/10

Table 3.2 indicates that the optimal solution is to produce: x1 = 4 units of product A; x2 = 9 units of product B and Zx =totalmaximumprofit,` 96.

Solution of the dual problem. The optimal solution of the dual problem can be obtained by applying the Big-M method. The optimal solution is shown in Table 3.3.

Table 3.3. Optimal Solution of Dual Problem

bi → 300 110 0 0


Basic Variables B

Basic Variables Value yB

y1 y2 s1d s2d

300 y1 1/10 1 0 – 1/20 1/40110 y2 6/10 0 1 1/10 – 3/20

Z = 96 bi – zj 0 0 4 9

The optimal solution as given in Table 3.3 is: y1 = ` 1/10 per hour on machine I; y2 = ` 6/10 per hour on machine II and Zy = total minimum cost, ` 96.

The values of y1 = ` 1/10 and y2 = ` 3/5, indicate the worth of one hour of machine time of machines I and II, respectively. However, since the machines hours can be increasedbeyondaspecificlimit,thisresultholdstrueonlyforaspecificrangeofhoursavailable on machines I and II.


NOTES


Comparison of the solutions. For interpreting the optimal solution of the primal (or dual), its solution values can be read directly from the optimal simplex table of the dual (or primal). The method can be summarized in the following steps. 1. The slack variables in the primal, correspond to the dual basic variables in the

optimal solution and vice versa. For example, s1p is the slack variable of the firstprimalconstraint.Itcorrespondstothefirstdualvariabley1. Likewise s2p corresponds to the second dual variable y2. Moreover, s1d and s2d are the dual surplus variables and they correspond to x1 and x2, respectively of the primal problem. The correspondence between primal and dual variables is summarized in Table 3.4.

Primal Dual

Main variables xx1

2

ssd

d

1

2

Surplus variables

Slack variables ssp

p

1

2

yy1

2

Main variables

2. The value in the cj – zj row under columns of the slack/surplus variables, ignoring the negative sign, directly gives the optimal values of the dual/primal basic variables. For example, c3 – z3 = 1/10 and c4 – z4 = 6/10 values in Table 3.2 under primal slack s1p and s2p correspond to solution values of dual variables y1 and y2 in Table 3.3 and vice versa. The primal-dual relationship for these problems which are feasible can be summarized in Table 3.5.

Primal Dual• Values of the basic variables • (cj – zj) values in the non-basic

surplus variable columns• (cj – zj) values in the non-basic slack

variable columns • Value of the basic variables

3. The optimal value of the objective function is the same for primal and dual LP problems.

4. The dual variable yi represents the worth (dual price or shadow price) of one unit of resource i. For example, for y1 the worth of time on machine I, equals 1/10 and for y2, the worth of time on machine II, equals ` 6/10, as shown in Table 3.3.Example 4. A company wishes to get at least 160 million ‘audience exposures’

the number of times one of the advertisements is seen or heard by a person. Because of the nature of the product the company wants at least 60 million month and at least 80 million of the exposures to involve persons between 18 and 40 years of age. The relevant information pertaining to the two advertising media under consideration—magazine and television is given below:

Magazine Television • Cost per advertisement (` thousand) 40 200 • Audience per advertisement (million) 4 40 • Audience per advertisement with monthly income over 3 10

` 10,000 (million) • Audience (per advertisement) in the age group 18–40 (million) 8 10


NOTES

Operations Research The company wishes to determine the number of advertisements to be released each in magazine and television so as to keep the advertisement expenditure to the minimum. Formulate this problem as a LP problem.

What will be the minimum expenditure and its allocation among the two media? Write ‘dual’ of this problem. Solve the ‘dual’ problem to find answer to the problem.

Solution: The primal and dual LP problems of the given problem are:

Primal Problem Dual Problem

x1, x2 = number of advertisements in mag-azine and television, respectively.

Minimize Zx = 40x1 + 200x2

subject to 4x1 + 40x2 ≥ 160

3x1 + 10x2 ≥ 60

8x1 + 10x2 ≥ 80

and x1, x2 ≥ 0.

y1, y2, y3 = shadow price (or worth) of one unit of advertisement over audience characteristics, respectively.

Maximize Zy = 160y1 + 60y2 + 80y3

subject to 4y1 + 3y2 + 8y3 ≤ 40

40y1 + 10y2 + 10y3 ≤ 200

and y1, y2, y3 ≥ 0.

Example 5. XYZ manufacturing company operates a three-shift system at one of its plants. In a certain section of the plant, the number of operators required on each of the three shifts is as follows:

Shift Number of Operators

Day (6 a.m. to 2 p.m.) 50

Afternoon (2 p.m. to 10 p.m.) 24

Night (10 p.m. to 6 p.m.) 10

The company pays its operators at the basic rate of ` 10 per hour for those working on the day shift. For the afternoon and night shifts, the rates are one and a half times the basic rate and twice the basic rate, respectively. In agreement with each operator at the commencement of his employment, he is allocated to one of three schemes A, B or C. These are as follows:

A : Work (on average) one night shift, one afternoon shift, and two day shifts in every four shifts.

B : Work (one average) equal number of day and afternoon shifts.C : Work day shifts only.In schemes A and B, it is necessary to work strictly alternating sequences of specified

shifts, as long as the correct proportion of shifts is worked in the long run. (a) Formulate a linear programming model to obtain the required number of

operators at minimum cost. (b) By solving the dual of the problem, determine how many operators must be

employed under each of the three schemes. Does this result in over-provision of operators on any one of the three shifts?


NOTES


Sol. The primal and dual LP problems of the given problem are:

Primal Problem Dual Problemx1, x2 and x3 = number of operators employed under scheme A, B and C, respectively.

Minimize Zx = 20 × 14

x1 + 15 14

121 2x x+

+ 10 12

121 2 3x x x+ +

= 554

252

101 2 3x x x+ +

subject to the constraints (1/4)x1 ≥ 10, (1/4)x1 + (1/2)x2 ≥ 24 (1/2)x1 + (1/2)x2 + x3 ≥ 50and x1, x2, x3 ≥ 0.

y1, y2 and y3 = shadow price (or worth) per unit of resources—operators in three shifts, respectively.Maximize Zy = 10y1 + 24x2 + 50y3

subject to the constraints(1/4)y1 + (1/4)y2 + (1/2)y3 ≤ 55/4 (1/2)y2 + (1/2)y3 ≤ 25/2 y3 ≤ 10and y1, y2, y3 ≥ 0.

3.5 ADVANTAGES OF DUALITY 1. It is advantageous to solve the dual of a primal that has a less number of constraints

because the number of constraints usually equals the number of iterations required to solve the problem.

2. Thisavoidsthenecessityforaddingsurplusorartificialvariablesandsolvestheproblem quickly (the technique is known as the primal-dual method ). In economics, duality is useful in the formulation of the input and output systems. It is also useful in Physics, Engineering, Mathematics, etc.

3. The dual variables provide an important economic interpretation of the finalsolution of an LP problem.

4. It is quite useful when investigating changes in the parameters of an LP problem (the technique is known as the sensitivity analysis).

5. Duality is used to solve an LP problem by the simplex method in which the initial solution is infeasible (the technique is known as the dual simplex method).


1. Definethedualofalinearprogrammingproblem.Statethefunctionalpropertiesofduality. 2. Explain the primal-dual relationship. 3. Brieflydiscuss‘duality’inlinearprogramming. 4. What is the principle of duality in linear programming? Explain its advantages. 5. Whatisduality?WhatisthesignificanceofdualvariablesinanLPmodel? 6. State the general rules for formulating a dual LP problem from its primal. 7. What is a shadow price? How does the concept relate to the dual of an LP problem?


NOTES

Operations Research 8. How can the concept of duality be useful in managerial decision making? 9. State and prove the relationship between the feasible solutions of an LP problem and its

dual. 10. ProvethatthenecessaryandsufficientconditionforanyLPproblemanditsdual,inorder

to have optimal solutions is that both have feasible solutions.


Write the dual of the following primal LP problems 1. Max Zx = 2x1 + 5x2 + 6x3

subject to (i) 5x1 + 6x2 – x3 ≤ 3 (ii) – 2x1 + x2 + 4x3 ≤ 4 (iii) x1 – 5x2 + 3x3 ≤ 1 (iv) – 3x1 – 3x2 + 7x3 ≤ 6

and x1, x2, x3 ≥ 0. 2. Min Zx = 7x1 + 3x2 + 8x3

subject to (i) 8x1 + 2x2 + x3 ≥ 3 (ii) 3x1 + 6x2 + 4x3 ≥ 4 (iii) 4x1 + x2 + 5x3 ≥ 1 (iv) x1 + 5x2 + 2x3 ≥ 7and x1, x2, x3 ≥ 0.

3. Max Zx = 2x1 + 3x2 + x3

subject to (i) 4x1 + 3x2 + x3 = 6 (ii) x1 + 2x2 + 5x3 = 4and x1, x2, x3 ≥ 0.

4. Max Zx = 3x1 + x2 + 3x3 – x4

subject to (i) 2x1 – x2 + 3x3 + x4 = 1 (ii) x1 + x2 – x3 + x4 = 3and x1, x2, x3, x4 ≥ 0.

5. Min Zx = 2x1 + 3x2 + 4x3

subject to (i) 2x1 + 3x2 + 5x3 ≥ 2 (ii) 3x1 + x2 + 7x3 = 3 (iii) x1 + 4x2 + 6x3 ≤ 5and x1, x2 ≥ 0, x3 is unrestricted.

6. Min Zx = x1 + x2 + x3

subject to (i) x1 – 3x2 + 4x3 = 5 (ii) x1 – 2x2 ≤ 3 (iii) 2x2 – x3 ≥ 4and x1, x2 ≥ 0, x3 is unrestricted.

7. One unit of product A contributes ` 7 and requires 3 units of raw material and 2 hours of labour. One unit of product B contributes 5 and requires one unit of raw material and one hour of labour. Availability of raw material at present is 48 units and there are 40 hours of labour.

(a) Formulate this problem as a linear programming problem. (b) Write its dual. (c)Solvethedualbythesimplexmethodandfindtheoptimalproductmixandtheshadow

prices of the raw material and labour.


NOTES


8. A company makes three products: X, Y and Z out of three raw materials A, B and C. The raw material requirements are given below:

Raw MaterialsNumber of Units of Raw Material Required to Produce

One Unit of Product

X Y Z

A 1 2 1

B 2 1 4

C 2 5 1

Theunitprofitcontributionoftheproducts:X, Y and Z is ` 40, 25 and 50, respectively. The number of units of raw material available are 36, 60 and 45, respectively.

(a)Determinetheproductmixthatwillmaximizethetotalprofit. (b)Usingthefinalsimplextable,writethesolutiontothedualproblemandgiveitseco-

nomic interpretation. 9. Three food products are available at costs of ` 10, ` 36 and ` 24 per unit, respectively. They

contain 1,000, 4,000 and 2,000 calories per unit, respectively and 200, 900 and 500 protein unitsperunit,respectively.Itisrequiredtofindtheminimum-costdietcontainingatleast20,000 calories and 3,000 units of protein. Formulate and solve the given problem as an LP problem. Write the dual and use it to check the optimal solution of the given problem.

10. A company produces three products: P, Q and R from three raw materials A, B and C. One unit of product P requires 2 units of A and 3 units of B. A unit of product Q requires 2 units of B and 5 units of C and one unit of product R requires 3 units of A, 2 units of B and 4 units of C. The company has 8 units of material A, 10 units of material B and 15 units of material Cavailabletoit.ProfitsperunitofproductsP, Q and R are ` 3, ` 5 and ` 4, respectively.

(a) Formulate this problem as an LP problem. (b)Howmanyunitsofeachproductshouldbeproducedtomaximizeprofit? (c) Write the dual of this problem. 11. A diet conscious housewife wishes to ensure certain minimum intake of vitamins A, B

and C for the family. The minimum daily (quantity) needs of the vitamins A, B and C for the family are respectively 30, 20 and 16 units. For the supply of these minimum vitamin requirements,thehousewifereliesontwofreshfoodIandII.Thefirstoneprovides7,5and2 units of the three vitamins per gram, respectively and the second one provides 2, 4 and 8 unitsofthesamethreevitaminspergramofthefoodstuff,respectively.Thefirstfoodstuffcosts ` 3 per gram and the second ` 2 per gram. The problem is how many grams of each foodstuff should the housewife buy everyday to keep her food bill as low as possible?

(a) Formulate this problem as an LP model. (b) Write and then solve the dual problem. 12. A large distributing company buys coffee seeds from four different plantations. On these

plantations, the seeds are available only in a blend of two types A and B. The company wants to market a blend consisting of 30 per cent of type A and 70 per cent of type B. The percentage of each type used by each plantation and the selling prices per 10 kg of the blends of each plantation are as follows:

TypePlantation

Desired1 2 3 4

A 40% 20% 60% 80% 30%B 60% 80% 40% 20% 70%

Selling price per 5 kg ` 3 ` 2 ` 1.20 ` 1.50


NOTES

Operations Research What quantity of coffee seeds should the company buy from each plantation so that the total mixture will contain the desired percentages of A and B and at the same time keep the purchasing cost at a minimum? Also write the dual of the problem.

13. The XYZ Plastic Company has just received a government contract to produce three different plastic valves. These valves must be highly heat and pressure resistant and the company has developed a three-stage production process that will provide the valves with the necessary properties involving work in three different chambers. Chamber 1 provides the necessary pressure resistance and can process valves for 1,200 minutes each week. Chamber 2 provides heat resistance and can process valves for 900 minutes per week. Chamber 3 tests the valves and can work 1,300 minutes per week. The three valve types and the time in minutes required in each chamber are:

Valve TypeTime Required in

Chamber 1 Chamber 2 Chamber 3

A 5 7 4

B 3 2 10

C 2 4 5

The government will buy all the valves that can be produced and the company will receive thefollowingprofitmarginsoneachvalve:

A, ` 15; B, ` 13.50; and C, ` 10. How many valves of each type should the company produce each week in order to maximize

profits?WritethedualofthegivenLPproblemandgiveitseconomicinterpretation? 14. A medical scientist claims to have found a cure for the common cold that consists of three

drugs called K, S and H. His results indicate that the minimum daily adult dosage for effective treatment is 10 mg of drug K, 6 mg of drug S and 8 mg of drug H. Two substances are readily available for preparing pills for distribution to cold sufferers. Both substances contain all three of the required drugs. Each unit of substance A contains 6 mg, 1 mg and 2 mg of drugs K, S and H, respectively and each unit of substance B contains 2 mg, 3 mg and 2 mg of the same drugs. Substance A costs ` 3 per unit and substance B costs ` 5 per unit.

(a) Find the least-cost combination of the two substances that will yield a pill designed to contain the minimum daily recommended adult dosage.

(b) Suppose that the costs of the two substances are interchanged so that substance A costs ` 5 per unit and substance B costs ` 3 per unit. Find the new optimal solution.

15. The XYZ company has the option of producing two products during the period of slack activity. For the next period, production has been scheduled so that the milling machine is free for 10 hours and skilled labour will have 8 hours of time available.

Product Machine Time per Unit

Skilled Labour Time per Unit

Profit Contribution per Unit (`)

A 4 2 5

B 2 2 3

Solve the primal and dual LP problems and bring out the fact that the optimum solution of one can be obtained from the other. Also explain in the context of the example, what you understand by shadow prices (or dual prices or marginal value) of resource.

16. A company produces three products P, Q and R whose prices per unit are 3, 5 and 4 respectively. On unit of product P requires 2 units of m1 and 3 units of m2. A unit of product Q requires 2 units of m2 and 5 units of m3 and one unit of product R requires 3 units of m1,


NOTES


2 units of m2 and 4 units of m3. The company has 8 units of material m1, 10 units of material m2 and 15 units of material m3 available to it.

(a) Formulate the problem as on LP model. (b) How many units of each product should be produced to maximize revenue? (c) Write the dual problem. 17. A person consumes two types of food A and B everyday to obtain 8 units of proteins, 12 units

of carbohydrates and 9 units of fats which is his daily minimum requirements. 1 kg of food A contains 2, 6 and 1 units of protein, carbohydrates and fats, respectively. 1 kg of food B contains 1, 1 and 3 units of proteins, carbohydrates and fats respectively. Food A costs ` 8.50 per kg, while B costs ` 4 per kg. Determine how many kg of each food should he buy daily to minimize his cost of food and still meet the minimum requirements.

Formulate on LP problem mathematically. Write its dual and solve the dual by the simplex method.

18. AfirmproducesthreearticlesX, Y, Z at a total cost of 4 , 3, and 6 per item respectively. Total number of X and Z item produced should be at least 2 and number of Y and Z together beatleast5.Thefirmwantstominimizethecost.FormulatethisproblemasanLPproblem.Write its dual. Solve the dual by the simplex method. Can you point out the solution of the primal problem? If yes, what is it?

19. AfirmproducesthreetypesofbiscuitsA, B and C. It packs them in assortments of two sizes I and II. The size I contains 20 biscuits of type A, 50 of type B and 10 of type C. The size II contains 10 biscuits of type A, 80 of type B and 60 of type C. A buyer intends to buy at least 120 biscuits of type A, 740 of type B and 240 of type C. Determine the least number of packets he should buy. Write the dual LP problem and interpret your answer.

20. Consider the following product mix problem: Let x1 denote number of units of Product 1 to be produced daily and x2 the number of units of Product 2 to be produced daily. The production of Product 1 requires one hour of processing time in department D1. Production of 1 unit of Product 2 requires 2 hours of processing time in department D1 and one hour in department D2. The number of hours available in department D1 are 32 hours and in department D2, 8 hours. The contribution of one unit of Product 1 is ` 200 and of Product 2 is ` 300.

The solution to this LP model is given below:

cj → 200 300 0 0

cBBasic

VariablesSolution Values xB

x1 x2 s1 s2

200 x1 32 1 2 1 00 s2 8 0 1 0 1

Z = 6400 cj – zj –100 –200 0 6400

Given the dual of the primal model. Obtain the optimum solution to the dual LP model from the above table. Interpret the dual variables.

Hints and Answers 1. Min Zy = 3y1 + 4y2 + y3 + 6y4

subject to (i) 5y1 – 2y2 + 5y3 – 3y4 ≥ 2 (ii) 6y1 + y2 – 5y3 – 3y4 ≥ 5 (iii) – y1 + 4y2 + 3y3 + 7y4 ≥ 6 and y1, y2, y3, y4 ≥ 0.


NOTES

Operations Research 2. Max Zy = 3y1 + 4y2 + y3 + 7y4

subject to (i) 8y1 + 3y2 + 4y3 + y4 ≤ 7 (ii) 2y1 + 6y2 + y3 + 5y4 ≤ 3 (iii) y1 + 4y2 + 5y3 + 2y4 ≤ 8 and y1, y2, y3, y4 ≥ 0. 3. Min Zy = 6y1 + 4y2

subject to (i) 4y1 + y2 ≥ 2 (ii) 3y1 + 2y2 ≥ 3 (iii) y1 + 5y2 ≥ 1 and y1, y2 unrestricted in sign. 4. Min Zy = y1 + 3y2

subject to (i) 2y1 + y2 ≥ 3 (ii) – y1 + y2 ≥ 1 (iii) 3y1 – y2 ≥ 3 (iv) y1 + y2 ≥ – 1 and y1, y2, unrestricted in sign. 5. Max Zy = 2y1 + 3y2 – 5y3

subject to (i) 2y1 + y2 – y3 ≤ 2 (ii) 3y1 + y2 – 4y3 ≤ 3 (iii) 5y1 + 7y2 – 6y3 = 4 and y1, y3 ≥ 0 and y2 unrestricted. 6. Max Zy = – 5y1 – 3y2 + 4y3

subject to (i) – y1 – y2 ≤ 1 (ii) 3y1 + 2y2 + 2y3 ≤ 1 (iii) – 4y1 – y3 ≤ 1 and y2, y3 ≥ 0 and y1 is unrestricted. 7. Primal x1 and x2 = number of units of products A and B, respectively to be produced. Max Zx = 7x1 + 5x2

subject to (i) 3x1 + x2 ≤ 48 (ii) 2x1 + x2 ≤ 40 and x1, x2 ≥ 0 Ans. x1 = 0, x2 = 40, Max Zx = ` 200 Dual

y1 and y2 = worth of one unit of raw material and labour, respectively. Min Zy = 48y1 + 40y2

subject to (i) 3y1 + 2y2 ≥ 7 (ii) y1 + y2 ≥ 5 and y1, y2 ≥ 0 Ans. y1 = 0, y2 = 5 and Min Zy = ` 200 8. Primal

x1, x2 and x3 = units of the products X, Y and Z, respectively to be produced. Max Zx = 40x1 + 25x2 + 50x3

subject to (i) x1 + x2 + x3 ≤ 36 (ii) 2x1 + x2 + 4x3 ≤ 60 (iii) 2x1 + 5x2 + x3 ≤ 45 and x1, x2, x3 ≥ 0. Ans. x1 = 20, x2 = 0, x3 = 5; Max Zx = 1,050 Dual

y1, y2 and y3 = worth (or shadow price) per unit of raw materials A, B and C, respectively. Min Zx = 36y1 + 60y2 + 45y3


NOTES


subject to (i) y1 + 2y2 + 2y3 ≥ 40 (ii) y1 + y2 + 5y3 ≥ 25 (iii) y1 + 4y2 + y3 ≥ 50 and x1, x2, x3 ≥ 0. Ans. y1 = 0, y2 = 0, y3 = 10; Min Zy = 1,050 10. Primal

x1, x2 and x3 = units of the products P, Q and R to be produced, respectively. Max Z = 3x1 + 5x2 + 4x3

subject to (i) 2x1 + 3x3 ≤ 8 (ii) 5x1 + 2x2 + 2x3 ≤ 10 (iii) 5x2 + 4x3 ≤ 15 and x1, x2, x3 ≥ 0. Dual

Min Z = 8y1 + 10y2 + 15y3

subject to (i) 2y1 + 5y2 ≥ 3 (ii) 2y2 + 5y3 ≥ 5 (iii) 3y1 + 2y2 + 4y3 ≥ 4 and y1, y2, y3 ≥ 0. 11. Primal

x1 and x2 = units of fresh food I and II, to be bought, respectively Min Z = 3x1 + 2x2

subject to (i) 7x1 + 2x2 ≥ 30 (ii) 5x1 + 4x2 ≥ 20 (iii) 2x1 + 8x2 ≥ 16 and x1, x2 ≥ 0. Dual

y1, y2 and y3 = worth per unit of vitamins A, B and C, respectively to the body.Max Z = 30y1 + 20y2 + 16y3

subject to (i) 7y1 + 5y2 + 2y3 ≤ 3 (ii) 2y1 + 4y2 + 8y3 ≤ 2 and y1, y2, y3 ≥ 0. (b) The optimal solution to the dual problem is: y1 = 20/52, y2 = 0, y3 = 8/52 and Max Z = 14. 12. Primal

x1, x2, x3 and x4 = quantities of coffee seeds, respectively which the company buys from each plantation.

Min = 3x1 + 2x2 + 1.2x3 + 1.5x4

subject to (i) 0.4x1 + 0.2x2 + 0.6x3 + 0.8x4 ≥ 0.3 (ii) 0.6x1 + 0.8x2 + 0.4x3 + 0.2x4 ≥ 0.7 and x1, x2, x3, x4 ≥ 0. Dual

y1 and y2 = worth of coffee seeds A and B, respectively Max Zy = 0.3y1 + 0.7y2

subject to (i) 0.4y1 + 0.6y2 ≤ 3 (ii) 0.2y1 + 0.8y2 ≤ 2 (iii) 0.6y1 + 0.4y2 ≤ 1.2 (iv) 0.8y1 + 0.2y2 ≤ 1.5 and y1, y2 ≥ 0 13. Primal

x1, x2 and x3 = number of valves of the types A, B and C, respectively to be produced. Max Zx = 15x1 + 13.5x2 + 10x3


NOTES

Operations Research subject to (i) 5x1 + x2 + 2x3 ≤ 1,200 (ii) 7x1 + x2 + 4x3 ≤ 900 (iii) 4x1 + 10x2 + 5x3 ≤ 1,300 and x1, x2, x3 ≥ 0. Ans. x1 = 22,400/217, x2 = 5,500/62, x3 = 0 and Max Zx = ` 58,125/31 Dual

y1, y2 and y3 = work of Chambers 1, 2 and 3 production capacity, respectively. Min Zy = 1,200y1 + 900y2 + 1,300y3

subject to (i) 5y1 + 7y2 + y3 ≥ 15 (ii) 3y1 + 2y2 + 10y3 ≥ 13.5 (iii) 2y1 + 4y2 + y3 ≥ 10 and y1, y2, y3 ≥ 0. Ans. y1 = 0, y2 = – 41/31, y3 = – 645/62. 15. Primal

x1, x2 = number of units of product A and B respectively to be produced Max Zx = 5x1 + 3x2

subject to (i) 4x1 + 2x2 ≤ 10 (ii) 2x1 + 2x2 ≤ 8 and x1, x2 ≥ 0 Ans. x1 = 1, x2 = 3 and Max Zx = 14 Dual

y1, y2, y3 = worth (or shadow price) per unit of resource operators in three shifts respectively.

Min Zy = 10y1 + 8y2

subject to (i) 4y1 + 2y2 ≥ 5 (ii) 2y1 + 2y2 ≥ 3 and y1, y2 ≥ 0. Ans. y1 = 1, y2 = 1/2 and Min Zy = 14. 16. Primal

x1, x2, x3 = number of units of product P, Q and R to be produced, respectively. Max Zx = 3x1 + 5x2 + 4x3

subject to (i) 2x1 + 3x3 ≤ 8 (ii) 5x1 + 2x2 + 2x3 ≤ 10 (iii) 5x2 + 4x3 ≤ 15 and x1, x2, x3 ≥ 0. Ans. x1 = 44/41, x2 = 59/41, x3 = 80/41 Dual

y1, y2, y3 = worth per unit of material m1, m2 and m3 respectively. Min Zy = 8y1 + 10y2 + 15y3

subject to (i) 2y1 + 3y2 ≥ 3 (ii) 2y2 + 5y3 ≥ 5 (iii) 3y1 + 2y2 + 4y3 ≥ 4 and y1, y2, y3 ≥ 0. 17. Primal

x1, x2 = number of units of food A and B to be consumed, respectively. Max Zx = 8.50x1 + 4x2

subject to (i) 2x1 + x2 ≥ 8 (ii) 6x1 + x2 ≥ 12 (iii) x1 + 3x2 ≥ 9 and x1, x2 ≥ 0. Ans. x1 = 1, x2 = 6 and Min Zx = 65/2;


NOTES


Dual y1, y2, y3 = worth per unit of proteins, carbohydrates and fats respectively. Min Zy = 8y1 + 12y2 + 9y3

subject to (i) 2y1 + 6y2 + y3 ≤ 8.50 (ii) y1 + y2 + 3y3 ≤ 4 and y1, y2, y3 ≥ 0 Ans. y1 = 31/8, y2 = 1/8, y3 = 0 and Max Zy = 65/2; 18. Primal

x1, x2, x3 = number of units of articles X, Y and Z respectively. Max Zx = 4x1 + 3x2 + 6x3

subject to (i) x1 + x3 ≥ 2 (ii) x2 + x3 ≥ 5 and x1, x2, x3 ≥ 0. Ans. x1 = 0, x2 = 3, x3 = 2 with Min Z = 21 Dual

y1, y2 = worth per unit of resources to be used, respectively Min Zy = 2y1 + 5y2

subject to (i) y1 ≤ 4 (ii) y2 ≤ 3 (iii) y1 + y2 ≤ 6 and y1, y2 ≤ 0 Ans. y1 = 3, y2 = 3 with Max Z* = 21 19. Primal

x1, x2 = number of assortments of size I and II, respectively. Max Zx = x1 + x2

subject to (i) 20x1 + 10x2 ≥ 120 (ii) 50x1 + 80x2 ≥ 740 (iii) 10x1 + 60x2 ≥ 240 and x1, x2 ≥ 0 Ans. x1 = 2, x2 = 8 and Min Zx = 10 Dual

y1, y2, y3 = worth per unit of biscuits of type A, B and C, respectively. Min Zy = 120y1 + 740y2 + 240

subject to (i) 20y1 + 50y2 + 10y3 ≤ 1 (ii) 10y1 + 80y2 + 60y3 ≤ 1 and y1, y2, y3 ≥ 0 20. Primal

Max Zx = 200x1 + 300x2

subject to (i) x1 + 2x2 ≤ 32 (ii) 0x1 + x2 ≤ 8 and x1, x2 ≥ 0. Dual

Min Zy = 32y1 + 8y2

subject to (i) y1 + 0y2 ≥ 200 (ii) 2y1 + y2 ≥ 300 and y1, y2 ≥ 0 The optimum solution to the dual LP problem is: y1 = 200, y2 = 0s1 = 0s2 = 100, Z* = ` 600 Note that, y1 = marginal increase in profit for an addition 1 hour of capacity. Since

y1 = `200,profitbecomes` 6400 + 200 = ` 6600. Also y2=marginalincreaseinprofitgivenan addition hour of capacity in D2. Since y2=0,therewouldbenochangeintheprofit.


NOTES

Operations Research In primal LP problem solution, s2 = 8 indicates that there is an excess capacity of 8 hours in D2. Thus, in dual solution we have y2 = 0. Dual constraint (i) is written as: y1 + 0y2 ≥ 200 or

total value of hours in department D and Drequired to produce1 unit

1 2oof Product 1

≥ unit profit received from manufacturingand selling 1 unit of Productt 1

3.6 DUALITY THEOREMSTheorem 1. The dual of the dual is the primal.Proof. Let the primal problem beMaximise z = CX, C ∈Rn

subject to AX ≤ b, bT ∈Rm ...(1)

and X ≥ 0Then, the dual of (3, 1, 4) isMinimise w = bTY, bT ∈Rm

subject to ATY ≥ CT, C ∈Rn ...(2)

and Y ≥ 0The dual of (3, 1, 5) isMinimise s = (CT)T, U, C ∈Rn

subject to (AT)T ≤ (bT)T, bT ∈Rm ...(3)

and U ≥ 0By observing (CT)T = C and (AT)T = A, (3, 1, 6) becomesMaximise s = CU, C ∈Rn

subject to AU ≤ b, bT ∈Rn ...(4)

and U ≥ 0Theorem 2. Let X0 be a feasible solution to the primal problem.Maximise z = CX subject to AX ≤ b, X ≥0, where XT and C ∈ Rn , bT ∈ Rm and A

is an m × n real matrix. If Y0 be a feasible solution to the dual of the primal, namely Minimise w = bT Y subject to AT Y ≥ CT, Y ≥ 0, YT ∈Rm then CX0 ≤ bTY0.

Proof. Since X0 and Y0 are the feasible solutions to the primal and its dual, respectively, we have

AX0 ≤ b, X0 ≥ 0 and AT Y ≥ CT, Y0 ≥ 0.Hence, C ≤ Y0

T A or CX0 ≤ Y0T AX0 ≤ Y0

T bor, CX0 ≤ bT Y0 [ Y0

T b = bTY0]Theorem 3. Let X0 be a feasible solution to the primal problem. Maximise

Z = CX subject to AX ≤ b, X ≥ 0 and Y0 be a feasible solution to its dual: Maximise w = bT Y subject to ATY ≥ CT, Y ≥ 0 where C, XT ∈ Rn and YT, bT ∈ Rm and A is a m × n real matrix. If CX0 = bTW0 then both X0 and W0 are optimum solution to the primal and dual respectively.


NOTES


Proof. Let X0′ be any other feasible solution to the primal problem. Then, theorem 2 gives,

CX0′ ≤ bTY0

Thus, CX0′ ≤ CX0 [ CX0 = bT Y0]and hence X0 is an optimum solution to the primal problem, because primal is of maximisation type.

Similarly, if Y0′ is any other feasible solution to the dual problem, then bT Y0 ≤ bT Y0′ and thus Y0 is an optimum solution to the dual problem, because dual is a minimisation problem.

3.6.1 Basic Duality TheoremLet a primal problem be Maximise z = CX subject to AX ≤ b, X ≥ 0, C, XT ∈ Rn and

the associated dual be minimise w = bT Y subject to AT Y ≥ CT, Y ≥0, YT, bT ∈ Rm. If X0 (Y0) is an optimum solution to the primal (dual) then there exists a feasible solution Y0(X0) to the dual (primal) such that CX0 = bT Y0.

Proof: The standard primal can be written as Maximise z = CX subject to AX + IXs = b, where Xs

T ∈ Rm is the slack vector and I is the associated m × m identity matrix.

Let X0 = [XB, 0] be an optimum solution to the primal, where XBT ∈ Rm is optimum

basic feasible solution given by XB = B–1 b, B being the optimal basis of A. Then, the optimal primal objective function is

z = CX0 = CBXB where CB is the cost vector associated with XB.Now, the net evaluations in the optimal simplex table are given by

zj – cj = CB Pj – cj = C B a c a A

C B e e IB j j j

B j j

−

−

− ∈

− ∈

1

1 0

for all

for all

Since XB is optimal, we must have zj – cj ≥ 0 for all j.This gives CB B–1 aj ≥ cj and CB B–1 ej ≥ 0 for all j.

or, CBB–1 A ≥ C and CB B–1 ≥ 0 in matrix form.or, AT B–1 CB

T ≥ CT and B–1 CBT ≥ 0.

Now, if we let B–1 CBT = Y0, the above become

ATW0 ≥ CT and W0 ≥ 0, W0T ∈ Rm. This means that W0 is a feasible solution to the

dual problem. Moreover, the corresponding dual objective function value is bT Y0 = Y0T b

= CBB–1 b = CBXB = CX0. Thus, given an optimal solution X0 to the primal, there exists a feasible solution Y0 to the dual such that CX0 = bTY0.

Similarly, starting with Y0, the existence of X0 can be proved.Corollar. If X0 is an optimal solution to the primal, an optimal solution to the

dual is given by W0 = B–1 CB, where B is the primal optimal basis.Theorem. Fundamental Theorem of DualityIf the primal or the dual has a finite optimum solution, then the other problem also

pocesses a finite optimum solution and the optimum values of the objective functions of the two problems are equal.


NOTES

Operations Research Proof: Consider the primal-dual pair.Primal: Maximise z = CX, subject to AX ≤ b and X ≥ 0.Dual: Minimise w = bT Y, subject to AT Y ≥ CT and W ≥ 0.Necessary condition: Let a feasible solution X0 (Y0) be an optimum solution to

the primal (dual) problem. It then follows from Theorem 3.6.1 that there exists a feasible solution Y0(X0) to the dual (primal) problem, such that CX0 = bT Y0.

It now follows from Theorem 3 that Y0 must be optimal. This proves the necessity ofthecondition.SufficiencyfollowsfromTheorem3.

3.7 EXISTENCE THEOREMIf either the primal or the dual problem has an unbounded objective function value,

then the other problem has no feasible solution.Proof: Let the given primal problem have unbounded solution. Then, for any value

of the objective function, say + ∞, there exists a feasible solution say X yielding this solution, that is, CX → ∞. If possible let the dual problem have a feasible solution. Then from Theorem 2 for each feasible solution Y0 of the dual, there exists a feasible solution. X0 to the primal such that CX0 ≤ bT Y0. That is bT Y → ∞. Now, as b is a constant and W0 has to satisfy the constraint AT Y0 ≥ CT, therefore the dual objective function bT Y0 mustbefinite.ThiscontradictstheresultbT Y0 →∞. Hence, the dual problem has no feasible solution.

By a similar argument it can be shown that when the dual has an unbounded solution, the primal has no solution.

Remark. The above result shows the following relation between the solutions of the primal and dual.

Dual problemPrimal problem

Feasible solution No feasible solutionFeasible solution Optimum Dual unboundedNo feasible solution Primal unbounded Unbounded or

No feasible solution

3.8 DUAL-SIMPLEX METHODThe simplex method is an algorithm that always deals with a basic feasible solution

and this algorithm is terminated as soon as an optimal solution is achieved. That is, the procedure should be stopped when all cj – zj ≤ 0 for maximization problem and cj – zj ≥ 0 for minimization problem. However, if one or more solution values (i.e., xBi) are negative and the optimality condition cj – zj both for maximization and minimization is satisfied,thenthecurrentoptimalsolutionmaynotbefeasible(becausecj – zj = cj – cB B–1 aj is completely independent of the vector b).Insuchcases,itispossibletofinda starting basic, but not a feasible solution that is dual feasible, i.e., all cj – zj ≤ 0 for a miximization problem. In all such cases a variant of the simplex method called the dual-simplex method would be used. In the dual simplex method we always attempt to retain optimality while bringing the primal back to feasibility (i.e., xBi ≥ 0 for all i).


NOTES


3.8.1 Dual-Simplex AlgorithmThe steps of a dual-simplex algorithm may be summarized as follows:Step 1: Determine an initial solution. Convert the given LP problem into the

standardformbyaddingslack,surplusandartificialvariablesandobtainaninitialbasic feasible solution. Display this solution in the initial dual-simplex table.

Step 2: Test optimality of the solution. If all solution values are positive (i.e., xBi ≥ 0 for all i), then there is no need of applying a dual-simplex method because the improved solution can be obtained by simplex method itself. Otherwise go to step 3.

Step 3: Test feasibility of the solution. If there exists a row, say r, for which the solution value is negative (i.e., xBr < 0) and all elements in row r and column j are positive (i.e., yrj ≥ 0 for all j), then the current solution is infeasible; hence go to Step 4.

Step 4: Obtain improved solution (i) Select a basic variable associated with the row (called key row) that has the largest negative solution value, i.e.,

xBr = Min {xBi; xBi < 0} (ii) Determine the minimum ratios only for those columns that have a negative element in row r. Then, select a non-basic variable for entering into the basic associated with the column for which:

c cyk k

rk

− = Min

c zy

yj j

rjrj

−<

; 0 , for all j

The element (i.e., yrk) at the intersection of key row and key column is called key element. The improved solution can then be obtained by making yrk as 1 and all other element of the key column as zero. Here, it may be noted that the key element is always positive.

Step 5: Revise the solution. Repeat steps 2 to 4 until either an optimal solution is reached or there exists no feasible solution.

Aflowchartofsolutionprocedureofthedual-simplexmethodisshowninFig.3.1.Example 1. Use the dual simplex method to solve the LP problem:Maximize Z = – 3x1 – 2x2

subject to the constraints (i) x1 + x2 ≥ 1, (ii) x1 + x2 ≤ 7 (iii) x1 + 2x2 ≥ 10, (iv) x2 ≤ 3and x1, x2 ≥ 0

Sol. In order to apply the dual-simplex method, make all the constraints of the type ≤ by multiplying by – 1. Then add slack variables in the constraints of the given LP problem. Thus, the problem becomes:


NOTES

Operations Research

Fig. 3.1. Flow Chart of Dual-Simplex Method

Maximize Z = – 3x1 – 2x2 + 0s1 + 0s3 + 0s4


(i) – x1 – x2 + s1 = – 1, (ii) x1 + x2 + s2 = 7,

(iii) – x1 – 2x2 + s3 = – 10, (iv) x2 + s4 = 3

and x1, x2, s1, s2, s3, s4 ≥ 0

An initial basic solution (infeasible) is obtained by setting x1 = x2 = 0, as shown in Table 3.7. This gives the solution values as: s1 = – 1, s2 = 7, s3 = – 10, s4 = 3 and Max Z = 0.


cj → – 3 – 2 0 0 0 0

cBBasic

Variables BSolution Values

b (= xB) x1 x2 s1 s2 s3 s4

0 s1 – 1 – 1 – 1 1 0 0 00 s2 7 1 1 0 1 0 00 s3 – 10 – 1 – 2 0 0 1 0 →0 s4 3 0 1 0 0 0 1

Z = 0 zj 0 0 0 0 0 0cj – zj – 3 – 2 0 0 0 0

↑

Since all cj – zj ≤ 0 and all solution values (i.e., xBi) are not non-negative, an optimal but infeasible solution has been obtained. Now in order to obtain a feasible solution,


NOTES


we select a basic variable to leave the basis and a non-basic variable to enter the basis, as follows:

Key row (variable to leave the basis) = Min {xBi : xBi < 0} = Min {– 1, – 10} = – 10 (= xB3

)That is, the basic variable s3 leaves the basis.

Key column (variable to enter the basis) = Min c zy

yj j

rjrj

−<

; 0

= Min {(– 3/– 1), (– 2/– 2)}= Min {3, 1} = 1That is, variable x2 enters the basis.Iteration 1: The new solution is obtained after introducing x2 into the basis and

dropping s3 from the basis, as shown in Table 3.8.Table 3.8

cj → – 3 – 2 0 0 0 0

cBBasic

Variables BSolution

Values b(= xB) x1 x2 s1 s2 s3 s4

0 s1 4 – 1/2 0 1 0 – 1/2 0

0 s2 2 1/2 0 0 1 1/2 0

– 2 x2 5 1/2 1 0 0 – 1/2 0

0 s4 – 2 – 1/2 0 0 0 1/2 1 →

Z = – 10 zj 0 – 2 0 0 1 0

cj – zj – 2 0 0 0 – 1 0↑

Table 3.8 shows that the solution is still infeasible (because s4 = – 2) but is optimal. Thus, we proceed to iteration 2.

Iteration 2: Since s4 (= xB4) = – 2 is the only row with negative solution value,

therefore, the variable s4 leaves the basis. Also in the key row y41 is the only element having negative value, thus x1-column is the key column and variable x1 enters the basis. The new solution is shown in Table 3.9.

Table 3.9

cj → – 3 – 2 0 0 0 0

cBBasic

Variables BSolution

Values b(= xB) x1 x2 s1 s2 s3 s4

0 s1 6 0 0 1 0 – 1 – 10 s2 0 0 0 0 1 1 1

– 2 x2 3 0 1 0 0 0 1– 3 x1 4 1 0 0 0 – 1 – 2

Z = – 18 zj – 3 2 0 0 3 4cj – zj 0 0 0 0 – 3 – 4

As shown in Table 3.9 all cj – zj ≤ 0 as well as all solution values are also positive (i.e., xBi ≥ 0), therefore, the current solution is the optimal solution. Hence the optimal basic feasible solution to the given LP problem is: x1 = 4, x2 = 3 and Max Z = – 18.


NOTES

Operations Research Example 2. Use the dual simplex method to solve the following LP problem:

Maximize Z = – 2x1 – x3


(i) x1 + x2 – x3 ≥ 5, (ii) x1 – 2x2 + 4x3 ≥ 8

and x1, x2, x3 ≥ 0

Sol. Since the objective function of the LP problem is that of maximization, therefore all the constraints should be of ≤ type. Thus, convert the constraints to the ≤ type by multiplying both sides by – 1 and rewrite the LP problem.

Maximize Z = – 2x1 + 0 . x2 – x3

subject to the constraints (i) – x1 – x2 + x3 ≤ – 5, (ii) – x1 + 2x2 – 4x3 ≤ – 8and x1, x2, x3 ≥ 0

Convert this problem into the standard form by adding slack variables s1 and s2 in the constraints. The given problem can then be written as:

Maximize Z = – 2x1 + 0 . x2 – x3 + 0 . s1 + 0 . s2

subject to the constraints (i) – x1 – x2 + x3 + s1 = – 5, (ii) – x1 + 2x2 – 4x3 + s2 = – 8and x1, x2, x3, s1, s2 ≥ 0

An initial solution to this LP problem is shown in Table 3.10.


cj → – 3 0 – 1 0 0

cBBasic Variables

BSolution Values

b(= xB) x1 x2 x3 s1 s2

0 s1 – 5 – 1 – 1 1 1 0

0 s2 – 8 – 1 2 – 4 0 1 →

Z = 0 zj 0 0 0 0 0

cj – zj – 2 0 – 1 0 0

↑

In Table 3.10, we noticed that the initial solution is infeasible because s1 = – 5 and s2 = – 8, but it is optimal, since all cj – zj ≤ 0. Thus we need to apply the dual simplex method in order to get both the feasible as well as the optimal solution.

Iteration 1: In order to obtain a feasible solution we select a basic variable to leave the basis and a non-basis variable to enter into the basis. This is done as follows:

Key row (variable to leave the basis) = Min {xBi : xBi < 0} = Min {– 5, – 8} = – 8 (corresponds to xB2

)That is, basic variable s2 in row 2 leaves the basis


NOTES


Key column (variable to enter the basis)

= Min c zy

yj j

rjrj

−<

; 0 = Min −

−−−

21

14

,

= 14

(corresponds to column 3)

That is, non-basis variable x3 enters the basis. The key element has been marked in the circle in Table 3.10.

The new solution is obtained after entering x3 into the basis and removing s2 from the basis. This is shown in Table 3.11.

Table 3.11

cj → – 2 0 – 1 0 0

cBBasic

Variables BSolution Values

b(= xB) x1 x2 x3 s1 s2

0 s1 – 7 – 5/4 – 1/2 0 1 1/4 →

– 1 x3 2 1/4 – 1/2 1 0 – 1/4

Z = – 2 zj – 1/4 1/2 – 1

0 1/4

cj – zj – 7/4 – 1/2 0 0 – 1/4↑

Table 3.11 shows that the solution is still infeasible (because s1 = – 7) but is optimal. So, we proceed to the second iteration.

Iteration 2: Since variable s1 = – 7 is the only variable having negative value, we should select variable s1 to leave the basis. Now, the variable to enter into the basis is determined as:

Key column (variable to enter the basis)

= Min c zy

yj j

rjrj

−<

; 0

= Min −−

−−

7 45 4

1 21 2

//; /

/ = 1 (corresponds to column = 2)

That is, the variable x2 enters into the basis. The new solution is shown in Table 3.12.

Table 3.12

cj → – 2 0 – 1 0 0

cB Basic Variables B

Solution Values b(= xB)

x1 x2 x3 s1 s2

0 x2 14 5/2 1 0 – 2 – 1/2– 1 x3 9 3/2 0 1 – 1 – 1/2

Z = – 9 zj – 3/2 0 – 1 1 1/2cj – zj – 1/2 0 0 – 1 – 1/2


NOTES

Operations Research As shown in Table 3.12, all cj – zj ≤ 0 and all solution values are also positive (i.e., xBi > 0 for all i), therefore, the current solution is the optimal solution. Hence, the optimal basic feasible solution to the given LP problem is: x1 = 0, x2 = 14, x3 = 9 and Max Z = – 9.

Example 3. Use dual simplex method to solve the following LP problem:Minimize Z = 3x1 + x2

subject to the constraints (i) x1 + x2 ≥ 1, (ii) 2x1 + 3x2 ≥ 2and x1, x2 ≥ 0

Sol. Convert the given minimization problem into maximization problem, and convert the constraints of ≥ type into ≤ type by multiplying them by – 1. Thus, the given LP problem becomes:

Maximize Z* = – 3x1 – x2

subject to the constraints (i) – x1 – x2 ≤ – 1, (ii) – 2x1 – 3x2 ≤ – 2and x1, x2 ≥ 0where Z* = – Z.

Convert this problem into a standard form by adding slack variables s1 and s2 in the constraints. The given problem can then be expressed as:

Maximize Z * = – 3x1 – x2 + 0s1 + 0s2

subject to the constraints (i) – x1 – x2 + s1 = – 1, (ii) – 2x1 – 3x2 + s2 = – 2and x1, x2, s1, s2 ≥ 0

An initial solution of this LP problem is shown in Table 3.13.


cj → – 3 – 1 0 0

cBBasic

Variables BSolution

Values b(= xB) x1 x2 s1 s2

0 s1 – 1 – 1 – 1 1 00 s2 – 2 – 2 – 3 0 1 →

Z* = 0 zj 0 0 0 0cj – zj – 3 – 1 0 0

↑

In Table 3.13, we noticed that the initial solution is infeasible because s1 = – 1, and s2 = – 2 but it is optimal since all cj – zj ≤ 0. Thus, we need to apply the dual simplex method in order to get both the feasible as well as the optimal solution.

Iteration 1: In order to obtain a feasible solution we select a basic variable to leave the basis and a non-basic variable to enter into the basis as follows:

Key row (variable to leave the basis) = Min {xBi : xBi < 0} = Min {– 1, – 2} = – 2 (corresponds to xB2

)


NOTES


That is, basic variable s2 leaves the basis.Key column (variable to enter the basis)

= Min c zy

yj j

rjrj

−<

; 0

= Min −−

−−

=32

13

13

, (corresponds to column 2)

That is, non-basic variable x2 enters into the basis.The new solution obtained after introducing x2 into the basis and removing s2 from

the basis is shown in Table 3.14.

Table 3.14

cj → – 3 – 1 0 0



x1 x2 s1 s2

0 s1 – 1/3 – 1/3 0 1 – 1/3

– 1 x2 2/3 2/3 1 0 – 1/3

Z* = – 2/3 zj – 2/3 – 1 0 1/3

cj – zj – 7/3 0 0 – 1/3

Table 3.14 shows that solution is still infeasible (because s1 = – 1/3) but optimal. So we proceed to second iteration.

Iteration 2: Variable s1 = – 1/3 is the only basic variable that has a negative value; it should, therefore, be removed from the basis. Variable to enter into the basis is selected as:

Key row (variable to enter the basis)

= Min c zy

yj j

rjrj

−<

; 0 = Min −

−−−

7 31 3

1 31 3

//, /

/

= 1 (corresponds to column 4)That is, the non-basic variable s2 enters into the basis. The new solution is shown

in Table 3.15

Table 3.15cj → – 3 – 1 0 0



x1 x2 s1 s2

0 s2 1 1 0 – 3 1– 1 x2 1 1 1 – 1 0

Z* = – 1 zj – 1 – 1 1 0cj – zj – 2 0 – 1 0

As shown in Table 3.15, all cj – zj ≤ 0 and all solution values are also positive (i.e., xBi > 0 for all i), therefore, the current solution is the optimal solution. Hence the optimal basic feasible solution to the given LP problem is: x1 = 0, x2 = 1 and Min Z = 1.


NOTES

Operations ResearchTEST YOUR KNOWLEDGE–A

1. What is dual simplex algorithm? State the various steps involved in the dual simplex algorithm.

2. What is the essential difference between regular simplex method and dual simplex method? 3. Show that the value of the objective function of the dual for any feasible solution is never

less than the value of the objective function of the primal corresponding to any feasible solution.

4. Show, with the help of an example, how when one solves an LP problem by the simplex method, going through infeasible but better than optimal solution, one directly goes through infeasible better than optimal solution of the dual LP problem. How is this fact utilized in the solution of the dual?

5. State the duality theorem for an LP problem and hence develop the computational algorithm for solving it by the dual simplex method.


Use dual simplex method to solve the following LP problems 1. Min Z = x1 + x2

subject to (i) 2x1 + x2 ≥ 2, (ii) – x1 – x2 ≥ 1 and x1, x2 ≥ 0 2. Max Z = 2x1 + 2x2

subject to (i) x1 + 2x2 ≥ 1, (ii) 2x1 + x2 ≥ 1 and x1, x2 ≥ 0 3. Max Z = – 2x1 – 2x2 – 4x3

subject to 2x1 + 3x2 + 5x3 ≥ 2 3x1 + x2 + 7x3 ≤ 3 x1 + 4x2 + 6x3 ≤ 5 and x1, x2, x3 ≥ 0 4. Min Z = x1 + 2x2 + 3x3

subject to x1 – x2 + x3 ≥ 4 x1 + x2 + 2x3 ≤ 8 x2 – x3 ≥ 2 and x1, x2, x3 ≥ 0 5. Min Z = 3x1 + 2x2 + x3 + 4x4

subject to 2x1 + 4x2 + 5x3 + x4 ≥ 10 3x1 – x2 + 7x3 – 2x4 ≤ 2 5x1 + 2x2 + x3 + 6x4 ≥ 15 and x1, x2, x3, x4 ≥ 0


NOTES


7. Min Z = 2x1 + x2

subject to (i) 3x1 + x2 ≥ 3, (ii) 4x1 + 3x2 ≥ 6 (iii) x1 + 2x2 ≥ 3 and x1, x2 ≥ 0 8. Min Z = x1 + 2x2

subject to (i) 2x1 + x2 ≥ 4, (ii) x1 + 2x2 ≤ 7 and x1, x2 ≥ 0 9. Min Z = 6x1 + 7x2 + 3x2 + 5x4

subject to 5x1 + 6x2 – 3x3 + 4x4 ≥ 12 x2 + 5x3 – 6x4 ≥ 10 2x1 + 5x2 + x3 + x4 ≥ 8 and x1, x2, x3, x4 ≥ 0 10. Use the dual simplex method to solve the LP problem: Min Z = 3x1 + 5x2 + 4x3

subject to – 2x1 – x2 + 5x3 ≥ 2 3x1 + 2x2 + 4x3 ≥ 16 and x1, x2, x3 ≥ 0 Write complementarybasis corresponding tofirst iteration.Write the simplex

multiplierswithrespecttothebasisoffirstiteration.Verifytheseresults.

Hints and Answers 1. Write Min Z = Max (– Z ) and convert all the constraints of the ≤ type ; x1 = 1, x2 = 0 and

Min Z = – 1. 2. x1 = 1/3, x2 = 1/3 and Min Z = 4/3 3. x1 = 0, x2 = 2/3, x3 = 0 and Max Z = – 4/3 4. x1 = 6, x2 = 2, x3 = 0 and Min Z = 10 5. x1 = 65/23, x2 = 1, x3 = 20/23 and Min Z = 215/23 7. x1 = 3/5, x2 = 6/5, and Min Z = 12/5 8. x1 = 0, x2 = 2 and Min Z = 4 9. x1 = 16/13, x2 = 6/13, x3 = 8/13 and Min Z = 2,280/13 10. x1 = x2 = 0, x3 = 4 and Min Z = 16

3.9 INTEGER PROGRAMMINGIn linear programming, each decision variable, slack and/or surplus variable is

allowed to take any discrete or fractional value. However, there are certain real-life problemsinwhichthefractionalvalueofthedecisionvariableshasnosignificance.Forexample, it does not make sense to say that 1.5 men will be working on a project or 1.6 machines will be used in a workshop. The integer solution to a problem can, however, be obtained by rounding off the optimum value of the variables to the nearest integer value. This approach can be easy in terms of economy of effort, time, and the cost that might be required to derive an integer solution. This solution, however, may not satisfy


NOTES

Operations Research all the given constraints. Secondly, the value of the objective function so obtained may notbetheoptimalvalue.Allsuchdifficultiescanbeavoidedifthegivenproblem,wherean integer solution is required, is solved by integer programming techniques.

Integer LP problems are those in which some or all of the variables are restricted to integer (or discrete) values. An integer LP problem has important applications. Capital budgeting, construction scheduling, plant location and size, routing and shipping schedule, batch size, capacity expansion, fixed charge, etc., are fewproblems thatdemonstrate the areas of application of integer programming.

3.10 TYPES OF INTEGER PROGRAMMING PROBLEMSLinearintegerprogrammingproblemscanbeclassifiedintothreecategories:

(i) Pure (all) integer programming problems in which all decision variables are restricted to integer values.

(ii) Mixed integer programming problems in which some, but not all, of the decision variables are restricted to integer values.

(iii) Zero-one integer programming problems in which all decision variables are restricted to integer values of either 0 or 1.

Thebroader classification of integerLPproblemsand their solutionmethodsare summarized in Fig. 3.2. In this chapter, we shall discuss two methods: (i) Gomory’s cutting plane method and (ii) Branch and Bound method, for solving integer programming problems.

Integer Programming Problems

Linear Integer Programming Non-linear Integer Programming

Problems Problems

Pure Integer Mixed nteger 0–1 Polynomial GeneralI

Problems Problems Problems Programming Non-linear

Problems

(i) Cutting Plane (i) Cutting Plane Pure Integer Mixed Integer

Method Method Problem Problem

(ii) Enumerative (ii) Enumerative

or Branch and Method

Bound) Method (iii) Balas Additive

MethodGeneralized Penalty

Function Method

Problems

Fig. 3.2. ClassificationofIntegerLPProblemsandtheirSolutionMethods

The pure integer linear programming problem in its standard form can be stated as follows:

Maximize Z = c1 x1 + c2 x2 + . . . + cn xnsubject to the constraints

a11 x1 + a12 x2 + . . . + a1n xn = b1 a21 x1 + a22 x2 + . . . + a2n xn = b2 am1 x1 + am2 x2 + . . . + amn xn = bm

and x1, x2, . . ., xn ≥ 0 and are integers.


NOTES

Duality in Linear ProgrammingFRACTIONAL CUT METHOD

3.11 ENUMERATION AND CUTTING PLANE SOLUTION CONCEPT

The fractional cut (cutting-plane) method to solve integer LP problems was developed by R.E. Gomory in 1956. This method is based on creating a sequence of linear inequalities called cuts. Such a cut reduces a part of the feasible region of the given LP problem, leaving out a feasible region of the integer LP problem. The hyperplane boundary of a cut is called the cutting plane.

Illustration. Consider the following linear integer programming (LIP) problemMaximize Z = 14x1 + 16x2

subject to the constraints (i) 4x1 + 3x2 ≤ 12, (ii) 6x1 + 8x2 ≤ 24and x1, x2 ≥ 0 and are integers.

Relaxing the integer requirement, the problem is solved graphically as shown in Fig. 3.3. The optimal solution to this LP problem is: x1 = 1.71, x2 = 1.71 and Max Z = 51.42. This solution does not satisfy the integer requirement of variables x1 and x2.

Rounding off this solution to x1 = 2, x2 = 2 does not satisfy both the constraints and therefore, the solution is infeasible. The dots in Fig. 3.3, also referred to as lattice points, represent all of the integer solutions that lie within the feasible solution space of the LPproblem.However,itisdifficulttoevaluateeverysuchpointinordertodeterminethe value of the objective function.

Fig. 3.3. Concept of Cutting Plane

In Fig. 3.3, it may be noted that the optimal lattice point C, lies at the corner of the solution space OABC, obtained by cutting away the small portion above the dotted line. This suggests a solution procedure that successively reduces the feasible solution space until an integer-valued corner is found.

A cut is the linear constraint added to the given LP problem constraints.

Hyperplane boundary of a cut is called the cutting plane.


NOTES

Operations Research The optimal integer solution is: x1 = 0, x2 = 3 and Max Z = 48. The lattice point, C is not even adjacent to the most desirable LP problem solution corner, B.

Remark. Reducing the feasible region by adding extra constraints (cut) can never give an improved objective function value. If Z1P represents the maximum value of objective function in an ILP problem and ZLP the maximum value of objective function in an LP problem, then Z1P ≤ ZLP.

3.12 GOMORY’S ALL INTEGER CUTTING PLANE METHOD

In this section, a procedure called Gomory’s all-integer algorithm will be discussed for generating ‘cuts’ (additional linear constraints) so as to ensure an integer solution to thegivenLPprobleminafinitenumberofsteps.Gomory’salgorithmhasthefollowingproperties. (i) Additional linear constraints never cutoff that portion of the original feasible

solution space that contains a feasible integer solution to the original problem. (ii) Each new additional constraint (or hyperplane) cuts off the current non-integer

optimal solution to the linear programming problem.

3.12.1 Method for Constructing Additional Constraint (Cut)Gomory’s method begins by solving an LP problem ignoring the integer value

requirement of the decision variables. If the solution so obtained is an integer, i.e., all variables in the ‘xB’-column (also called basis) of the simplex table assume non-negative integer values, the current solution is the optimal solution to the given ILP problem. However, if some of the basic variables do not have non-negative integer value, an additional linear constraint called the Gomory constraint (or cut) is generated. After having generated a linear constraint (or cutting plane), it is added to the bottom of the optimal simplex table. The new problem is then solved by using the dual simplex method. If the optimal solution, so obtained, is again a non-integer, then another cutting plane is generated. The procedure is repeated until all basic variables assume non-negative integer values.

3.12.2 Steps of Gomory’s All Integer Programming AlgorithmAn iterative procedure for the solution of an all integer programming problem by

Gomory’s cutting plane method can be summarized in the following steps.Step 1: Initialization. Formulate the standard integer LP problem. If there

areanynon-integercoefficientsintheconstraintequations,convertthemintointegercoefficients. Solve the problemby the simplexmethod, ignoring the integer valuerequirement of the variables.

Step 2: Test the optimality (a) Examine the optimal solution. If all basic variables (i.e. xBi = bi ≥ 0) have integer

values, then the integer optimal solution has been obtained and the procedure is terminated.

(b) If one or more basic variables with integer value requirement have non-integer solution values, then go to Step 3.


NOTES


Step 3: Generate cutting plane Choose a row r corresponding to a variable xr that has the largest fractional value fr and follow the procedure to develop a ‘cut’ (a Gomory constraint) as explained in Eq. (5):

– fr = sg – j r

rj jf x≠∑ , where 0 ≤ frj < 1 and 0 < fr < 1

If there are more than one variables with the same largest fraction, then choose the onethathasthesmallestprofit/unitcoefficientintheobjectivefunctionofmaximizationLPproblemorthelargestcost/unitcoefficientintheobjectivefunctionofminimizationLP problem.

Step 4: Obtain the new solution. Add this additional constraint (cut) generated in Step 3 to the bottom of the optimal simplex table. Find a new optimal solution by using the dual simplex method, i.e., choose a variable that is to be entered into the new solution having the smallest ratio: {(cj – zj)/yij ; yij < 0} and return to Step 2. The process is repeated until all basic variables with integer value requirement assume non-negative integer values.

TheprocedureforsolvinganILPproblemissummarizedinaflowchartshownin Fig. 3.4.

Fig. 3.4. Flow Chart for Solving Integer LP Problem

Example 1. Solve the following Integer LP problem using Gomory’s cutting plane method.

Maximize Z = x1 + x2

subject to the constraints (i) 3x1 + 2x2 ≤ 5, (ii) x2 ≤ 2and x1, x2 ≥ 0 and are integers.


NOTES

Operations Research Sol. Step 1: Obtain the optimal solution to the LP problem ignoring the integer value restriction by the simplex method.

Table 3.16. Optimal Non-integer Solution

cj → 1 1 0 0


Basic Variables B

Basic Variables Value b (= xB ) x1 x2 s1 s2

1 x1 1/3 1 0 1/3 – 2/3

1 x2 2 0 1 0 1

Z = 7/2 cj – zj 0 0 – 1/3 – 1/3

In Table 3.16, since all cj – zj ≤ 0, the optimal solution of LP problem is: x1 = 1/3, x2 = 2 and Max Z = 7/2.

Step 2: In the current optimal solution, shown in Table 3.16 all basic variables in the basis (xB-column) did not assume integer value. Thus solution is not desirable. To obtain an optimal solution satisfying integer value requirement, go to step 3.

Step 3: Since x1 is the only basic variable whose value is a non-negative fractional value, therefore considerfirst row (x1-row) as source row in Table 3.16 to generate Gomory cut as follows:

13

= x1 + 0.x2 + 13

s1 – 23

s2 (x1-source row)

The factoring of numbers (integer plus fractional) in the x1-source row gives

0 13

+

= (1 + 0)x1 + 0 1

31 1

31 2+

+ − +

s s

Eachofthenon-integercoefficientsisfactoredintointegerandfractionalpartsinsuch a manner that the fractional part is strictly positive.

Rearrangingalloftheintegercoefficientsontheleft-handside,weget

13

+ (s2 – x1) = 13

131 2s s+

Since value of variables x1 and s2 is assumed to be non-negative integer, left-hand side must satisfy

13

≤ 13

s1 + 13

s2 (Ref. Eq. 4)

13 1

+ sg = 13

131 2s s+ or s s sg1

13

131 2− − = – 1

3 (Cut I)

where sg1 is the new non-negative (integer) slack variable.

Adding this equation (also called Gomory cut) at the bottom of Table 3.16, the new values so obtained is shown in Table 3.17.


NOTES



cj → 1 1 0 0 0

Basic Variables

Coefficient cB

Basic Variables BBasic

Variables Value b (= xB )

x1 x2 s1 s2 sg1

1 x1 1/3 1 0 1/3 – 2/3 0

1 x2 2 0 1 0 10 0

0 sg1– 1/3 0 0 – 1/3 – 1/3 1 →

Z = 7/2 cj – zj 0 0 – 1/3 – 1/3 0

Ratio: Min (cj – zj)/y3j(< 0) – – 1 1 –

↑

Step 4: Since the solution shown in Table 3.17 is infeasible, apply the dual simplex methodtofindafeasibleaswellasanoptimalsolution.Thekeyrowandkeycolumnare marked in Table 3.17. The new solution is obtained by applying the following row operations.

R3(new) → R3(old) × – 3; R1(new) → R1(old) – (1/3) R3(new)The new solution is shown in Table 3.18.

Table 3.18. Optimal Integer Solution

cj → 1 1 0 0 0

Basic Variables

Coefficient cB

Basic Variables BBasic

Variables Value b (= xB)

x1 x2 s1 s2 sg1

1 x1 0 1 0 0 – 1 1

1 x2 2 0 1 0 1 0

0 sg11 0 0 1 1 – 3

Z = 2 cj – zj 0 0 0 0 –1

Since all cj – zj ≤ 0 and value of basic variables shown in xB-column of Table 3.18 is integer, the solution: x1 = 0, x2 = 2, sg1

= 1 and Max Z = 2, is an optimal basic feasible solution of the given ILP problem.

Example 2. Solve the following Integer LP problem using the cutting plane method.Maximize Z = 2x1 + 20x2 – 10x3

subject to the constraints (i) 2x1 + 20x2 + 4x3 ≤ 15, (ii) 6x1 + 20x2 + 4x3 = 20and x1, x2, x3 ≥ 0 and are integers.

Also show that it is not possible to obtain a feasible integer solution by simple rounding off method.


NOTES

Operations Research Sol. Adding slack variable s1inthefirstconstraintandartificialvariableinthesecond constraint, the LP problem is stated in the standard form as:

Maximize Z = 2x1 + 20x2 – 10x3 + 0s1 – MA1

subject to the constraints (i) 2x1 + 20x2 + 4x3 + s1 = 15, (ii) 6x1 + 20x2 + 4x3 + A1 = 20and x1, x2, x3, s1, A1 ≥ 0 and are integers.

The optimal solution of the LP problem, ignoring the integer value requirement using the simplex method is shown in Table 3.19.

Table 3.19. Optimal Non-integer Solutioncj → 2 20 – 10 0

Basic Variables

Coefficient cB

Variables in Basic B

Basic Variables

Value b (= xB)x1 x2 s3 s1

20 x2 5/8 0 1 1/5 3/402 x1 5/4 1 0 0 – 1/4

Z = 15 cj – zj 0 0 – 14 – 1

The non-integer optimal solution shown in Table 3.19 is: x1 = 5/4, x2 = 5/8, x3 = 0 and Max Z = 15.

To obtain an optimal solution satisfying integer value requirement, we proceed to construct Gomory’s constraint. In this solution, the value of both basic variables x1 and x2 are non-integer. Since the fractional part of the value of basic variable x2 = (0 + 5/8) is more than that of basic variable x1 (= 1 + 1/4), the x2-row is selected for constructing Gomary cut as follows:

58

= 0 . x1 + x2 + 15

3403 1x s+ (x2-source row)

The factoring of the x2-source row yields

0 58

+

= (1 + 0)x2 + 0 15

0 3403 1+

+ +

x s

58

– x2 = 15

3403 1x s+ or 5

8 ≤ 15

3403 1x s+

On adding a slack variable sg1, the Gomory’s fractional cut becomes:

58

+ sg1 = 15

3403 1x s+ or s x sg1

15

3403 1− − = − 5

8 (Cut I)

Adding this additional constraint at the bottom of optimal simplex Table 3.19, the new values so obtained are shown in Table 3.20.

Iteration 1: Remove the variable sg1 from the basis and enter variable s1 into the

basis by applying the dual simplex method. The new solution is shown in Table 3.21.


NOTES



cj → 2 20 – 10 0 0

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables

Value b (= xB)x1 x2 x3 s1 sg1

20 x2 5/8 0 1 1/5 3/40 0

2 x1 5/4 1 0 0 – 1/4 0

0 sg1– 5/8 0 0 – 1/5 – 3/40 1 →

Z = 15 cj – zj 0 0 –14 –1 0

Ratio: Min (cj – zj)/y3j (< 0) – – 70 40/3 –↑

Table 3.21

cj → 2 20 – 10 0 0

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables

Value b (= xB )x1 x2 x3 s1 sg1

20 x2 0 0 1 0 0 – 1

2 x1 10/3 1 0 2/3 0 – 10/3

0 s1 25/3 0 0 8/3 1 – 40/3

Z = 20/3 cj – zj 0 0 – 34/3 0 – 40/3

The optimal solution shown in Table 3.21 is still non-integer. Therefore, one more fractional cut needs to be generated. Since x1 is the only basic variable whose value is a non-negative fractional value, consider the x1-row (because of largest fractional part) for constructing the cut:

103

= x x sg1 323

103 1

+ − (x1-source row)

The factoring of the x1-source row yields

3 13

+

= (1 + 0) x1 + 0 23

4 233 1

+

+ − +

x sg

13

+ (3 + x1 + 4sg1) = 2

3233 1

x sg+ or 13

≤ 23

233 1

x sg+

On adding another Gomory slack variable sg2, the second Gomory’s fractional cut

becomes:

13 2

+ sg = 23

233 1

x sg+ or s x sg g2 1

23

233− − = − 1

3 (Cut II)

Adding this cut to the optimal simplex Table 3.21, the new table so obtained is shown in Table 3.22.


NOTES

Operations Research Table 3.22. Optimal but Infeasible Solutioncj → 2 20 – 10 0 0 0

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables

Value b (= xB )

x1 x2 x3 s1 sg1sg2

20 x2 0 0 1 0 0 – 1 02 x1 10/3 1 0 2/3 0 – 10/3 00 s1 25/3 0 0 8/3 1 – 40/3 00 – 1/3 0 0 – 2/3 0 – 2/3 1 →

Z = 20/3 cj – zj 0 0 – 34/3 0 – 40/3 0Ratio: Min (cj – zj)/y4j (< 0) — — 17 — 20 —

↑

Iteration 2: Enter non-basic variable x3 into the basis to replace basic variable sg2

by applying the dual simplex method. The new solution is shown in Table 3.23.

Table 3.23. Optimal but Non-Integer Solutioncj → 2 20 – 10 0 0 0

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables

Value b (= xB )x1 x2 x3 s1 sg1

sg2

20 x2 0 0 1 0 0 1 02 x1 3 1 0 0 0 – 4 00 s1 7 0 0 0 1 – 16 4

– 10 x3 1/2 0 0 1 0 1 – 3/2Z = 1 cj – zj 0 0 0 0 – 2 – 17

The optimal solution shown in Table 3.23 is still non-integer because variable x3 does not assume integer value. Thus, a third fractional cut needs to be constructed with the help of the x3-row:

12

= x3 + s sg g1 2

32

− (x3–source row)

0 12

+

= (1 + 0)x3 + (1 + 0) s sg g1 22 1

2+ − +

12

2 22 13+ − −( )s x sg g = 1

2 2sg or 1

212 2

≤ sg

The required Gomory’s fractional cut obtained by adding slack variable sg3 is:

12 3

+ sg = 12 2sg or s sg g3 2

12

− = − 12

(Cut III)

Adding this cut to the bottom of the optimal simplex Table 3.23, the new table so obtained is shown in Table 3.24.


NOTES


Table 3.24

cj → 2 20 – 10 0 0 0 0

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables


sg2sg3

20 x2 0 0 1 0 0 1 0 0

2 x1 3 1 0 0 0 –4 1 0

0 s1 7 0 0 0 1 –16 4 0

– 10 x3 1/2 0 0 1 0 1 –3/2 0

0 sg3–1/2 0 0 0 0 0 –1/2 1 →

Z = 1 cj – zj 0 0 0 0 – 2 – 17 0

Ratio: Min (cj – zj)/y5j (< 0) – – – – – 34 –↑

Iteration 3: Remove the variable sg3 from the basis and enter variable sg2

into the basis by applying the dual simplex method. The new solution is shown in Table 3.25.


cj → 2 20 – 10 0 0 0 0

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables


sg2sg3

20 x2 0 0 1 0 0 1 0 0

2 x1 2 1 0 0 0 – 4 0 2

0 s1 3 0 0 0 1 – 16 0 8

– 10 x3 2 0 0 1 0 1 0 – 3

0 sg21 0 0 0 0 0 1 – 2

Z = – 16 cj – zj 0 0 0 0 – 2 0 – 34

In Table 3.25, since value of all basic variables is an integer value and all cj – zj ≤ 0, the current solution is an integer optimal solution: x1 = 2, x2 = 0, x3 = 2 and Max Z = – 16.

Example 3. The owner of a readymade garments store sells two types of shirts –Zee-shirtsandButton-downshirts.Hemakesaprofitof` 3 and ` 12 per shirt on Zee-shirts and Button-down shirts, respectively. He has two tailors, A and B, at his disposal, for stitching the shirts. Tailors A and B can devote at the most 7 hours and 15 hours per day, respectively. Both these shirts are to be stitched by both the tailors. Tailors A and B spend 2 hours and 5 hours, respectively in stitching one Zee-shirt, and 4 hours and 3 hours, respectively on stitching a Button-down shirt. How many shirts ofbothtypesshouldbestitchedinordertomaximizethedailyprofit? (a) Formulate and solve this problem as an LP problem. (b) If the optimal solution is not integer-valued, use Gomory technique to derive

the optimal integer solution.


NOTES

Operations Research Mathematical formulation. Let x1 and x2 = number of Zee-shirts and Button down shirts to be stitched daily, respectively.

Then the mathematical model of the LP problem is stated as:Maximize Z = 3x1 + 12x2

subject to the constraints (i) Time with tailor A: 2x1 + 4x2 ≤ 7 (ii) Time with tailor B: 5x1 + 3x2 ≤ 15

and x1, x2 ≥ 0 and are integers.Sol. (a) Adding slack variables s1 and s2, the given LP problem is stated into its

standard form as:Maximize Z = 3x1 + 12x2 + 0s1 + 0s2

subject to the constraints (i) 2x1 + 4x2 + s1 = 7, (ii) 5x1 + 3x2 + s2 = 15and x1, x2, s1, s2 ≥ 0 and are integers

The optimal solution of the LP problem, obtained by using the simplex method is given in Table 3.26.


cj → 3 12 0 0

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables

Value b (= xB)x1 x2 s1 s2

12 x2 7/4 1/2 1 1/4 0

0 s2 39/4 7/2 0 – 3/4 1

Z = 21 cj – zj – 3 0 – 3 0

The non-integer optimal solution shown in Table 3.26 is: x1 = 0, x2 = 7/4 and Max Z = 21. (b) To obtain the integer-valued solution, we proceed to construct Gomory’s fractional cut, with the help of x2-row (because of largest fraction value) as follows:

74

= 12

141 2 1x x s+ + (x2-source row)

or 1 34

+

= 0 12 1+

x + (1 + 0) x2 + 0 1

4 1+

s

34

+ (1 – x2) = 12

141 1x s+ or 3

4 ≤ 12

141 1x s+

On adding Gomory slack variable sg1, the required Gomory’s fractional cut becomes:

34 1

+ sg = 12

141 1x s+

or s x sg112

141 1− − = – 3

4 (Cut I)


NOTES


Adding this additional constraint to the bottom of the optimal simplex Table 3.26, the new table so obtained is shown in Table 3.27.


cj → 3 12 0 0 0

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables

Value b (= xB )x1 x2 s1 s2 sg1

12 x2 7/4 1/2 1 1/4 0 0

0 s2 39/4 7/2 0 – 3/4 1 0

0 sg1– 3/4 – 1/2 0 – 1/4 0 1 →

Z = 21 cj – zj – 3 0 – 3 0 0

Ratio: Min (cj – zj)/y3j (< 0) 6 – 12 – –↑

Iteration 1: Remove variable sg1 from the basis and enter variables x1 into the

basis by applying dual simplex method. The new solution is shown in Table 3.28.

Table 3.28. Optimal but Non-integer Solution

cj → 3 12 0 0 0

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables

Value b (= xB)x1 x2 s1 s2 sg1

12 x2 1 0 1 0 0 1

0 s2 9/2 0 0 – 5/2 1 7

3 x1 3/2 1 0 1/2 0 – 2

Z = 33/2 cj – zj 0 0 – 3/2 0 – 6

The optimal solution shown in Table 3.28 is still non-integer. Therefore, by adding one more fractional cut, with the help of the x1-row, we get:

32

= x s sg1 112

21

+ − (x1-source row)

1 12

+

= (1 + 0 ) x1 + 0 1

2 1+

s + (– 2 + 0) sg1

12

1 21 1+ − +( )x sg = 1

2 1s or 12

≤ 12 1s

On adding Gomory slack variable sg2, the required Gomory’s fractional cut becomes:

12 2

+ sg = 12 1s or s sg2

12 1− = − 1

2 (Cut II)

Adding this cut to the optimal simplex Table 3.28, the new table so obtained is shown in Table 3.29.


NOTES

Operations Research Table 3.29. Optimal but Infeasible Solutioncj → 3 12 0 0 0 0


cB

Basic Variables B

Basic Variables

Value b (= xB)

x1 x2 s1 s2 sg1sg2

12 x2 1 0 1 0 0 1 00 s2 9/2 0 0 – 5/2 1 7 03 x1 3/2 1 0 1/2 0 2 00 – 1/2 0 0 – 1/2 0 0 1 →

Z = 33/2 cj – zj 0 0 – 3/2 0 – 6 0Ratio: Min (cj – zj)/y4j (< 0) 0 0 3 0 0 0

↑

Iteration 2: Remove variable sg2 from the basis and enter variable s1 into the

basis by applying the dual simplex method. The new solution is shown in Table 3.30.

Table 3.30. Optimal Solutioncj → 3 12 0 0 0 0


Basic Variables B

Basic Variables

Value b (= xB )x1 x2 s1 s2 sg1

sg2

12 x2 1 0 1 0 0 0 00 s2 7 0 0 0 1 7 – 53 x1 1 1 0 0 0 2 10 s1 1 0 0 1 0 0 – 2

Z = 15 cj – zj 0 0 0 0 – 6 – 3

In Table 3.30, since all basic variables have assumed integer value and all cj – zj ≤ 0, the current solution is an integer optimal solution. Thus, the company should produce x1 = 1 Zee-shirt, x2=1Button-downshirtinordertoyieldMax(profit)Z = 15.


1. Solve the following all integer programming problems, using Gomory’s cutting plane algorithm:

(i) Max Z = x1 + 2x2

subject to (a) 2x2 ≤ 7, (b) x1 + x2 ≤ 7

(c) 2x1 ≤ 11

and x1, x2 ≥ 0 and integers.

(ii) Max Z = 2x1 + 1.7x2

subject to (a) 4x1 + 3x2 ≤ 7, (b) x1 + x2 ≤ 4



NOTES


(iii) Max Z = 3x1 + 2x2 + 5x3

subject to (a) 5x1 + 3x2 + 7x3 ≤ 28 (b) 4x1 + 5x2 + 5x3 ≤ 30

and x1, x2, x3 ≥ 0 and integers.

(iv) Max Z = 3x1 + 4x2

subject to (a) 3x1 + 2x2 ≤ 8, (b) x1 + 4x2 ≥ 10


(v) Max Z = 4x1 + 3x2

subject to (a) x1 + 2x2 ≤ 4, (b) 2x1 + x2 ≤ 6


(vi) Max Z = 7x1 + 9x2

subject to (a) – x1 + 3x2 ≤ 6, (b) 7x1 + x2 ≤ 35


2. AnairlineownsanageingfleetofBoeing 737jetairplanes.It isconsideringamajorpurchase of up to 17 new Boeing models 757 and 767 jets. The decision must take into accounts several cost and capacity factors including the following: (i)theairlinecanfinanceup to ` 4,000 million in purchases; (ii) each Boeing 757 will cost ` 350 million, while each Boeing 767 will cost ` 220 million; (iii) at least one-third of the planes purchased should be the longer-ranged 757; (iv) the annual maintenance budget is to be no more than ` 80 million (v) the annual maintenance cost per 757 is estimated to be ` 8,00,000 and it is ` 5,00,000 for each 767 purchased; (vi) each 757 can carry 1,25,000 passengers per year, whileeach767canfly81,000passengersannually.Formulatethisproblemasintegerprogramming problem in order in order to maximize the annual passenger carrying capacity and solve the problem by using the cutting plane method.

3. An air conditioning and refrigeration company has been awarded a contract for the air conditioning of a new computer installation. The company has to make a choice between two alternatives: (a) Hire one or more refrigeration technicians for 6 hours a day, or (b) hire one or more part-time refrigeration apprentice technicians for 4 hours a day. The wage rate of refrigeration technicians is ` 400 per day, while the corresponding rate for apprentice technicians is ` 160 per day. The company does not want to engage the technicians on work for more than 25 man hours per day. It also wants to limit the charges of technicians to ` 4,800. The company estimates that the productivity of a refrigeration technician is 8 units and that of part time apprentice technician is 3 units. Formulate and solve this problem as an integer LP problem to enable the company to select the optimal number of technicians and apprentices.

4. A manufacturer of toys makes two types of toys, A and B. Processing of these two toys is done on two machines X and Y. The toy A requires two hours on machine X and six hours on machine Y. Toy B requires four hours on machine XandfivehoursonmachineY. There are sixteen hours of time per day available on machine X and thirty hours on machine Y.Theprofitobtainedonboththetoysisthesame,i.e., ` 5 per toy. Formulate and solve this problem as an integer LP problem to determine the daily production of each of the two toys?

5. The ABC Electric Appliances Company produces two products: Refrigerators and ranges. The production of these takes place in two separate departments. Refrigerators are produced in department I and ranges in department II. Both these are sold on a weekly basis. Due to the limited facilities in the departments the weekly production cannot exceed 25 refrigerators in department I and 35 ranges in department II. The company regularly employs a total of 50 workers in the departments. The production of one refrigerator requires two man-weeks of labour and of one range one man-week. A refrigerator contributesaprofitof` 300 and a range of ` 200. Formulate and solve this problem as an


NOTES

Operations Research integer LP problem to determine the units of refrigerators and ranges that the company shouldproducetorealizethemaximumprofit?

6. XYZ company produces two types of tape recorders: Reel-to-reel model and cassette model, on two assembly lines. The company must process each tape recorder on each assembly line. It has found that the this whole process requires the following amount of time:

Assembly Line Reel-to-reel Cassette1 6 hours 2 hours2 3 hours 2 hours

The production manager says that line 1 will be available for 40 hours per week and line 2 for only 30 hours per week. After the mentioned hours of operation each line must be checkedforrepairs.Thecompanyrealizesaprofitof 300 on each reel-to-reel tape recorder and ` 120 on each cassette recorder. Formulate and solve this problem as an integer LP problem in order to determine the number of recorders of each type to be produced each weeksoastomaximizeprofit.

7. A company that manufactures metal products is planning to buy any of the following three types of lathe machines–manual, semi-automatic and fully-automatic. The manual lathe machine costs 1,500, while the semi-automatic and fully-automatic lathe machines cost ` 4,000 and ` 6,000 respectively. The company’s budget for buying new machines is `1,20,000.Theestimatedcontributiontowardsprofitfromthemanual,semi-automaticand fully-automatic lathe machines are: ` 16, ` 20 and ` 22, respectively. The available floorspaceallowsfortheinstallationofonly36newlathemachines.Themaintenancerequired on fully automatic machines is low and the maintenance department can maintain 50 fully-automatic machines in a year. The maintenance of a semi-automatic machine takes 20 per cent more time than that of a fully automatic machine and a manually operated machine takes 50 per cent more time for maintenance of a fully-automatic machine. Formulate and solve this problem as an integer LP problem in order to determine the optimal number of machines to be bought.

8. A stereo equipment manufacturer can produce two models–A and B–of 40 and 80 watts of total power, each. Each model passes through three manufacturing divisions 5 namely 1, 2 and 3, where model A takes 4, 2.5 and 4.5 hours each and model B takes 2, 1 and 1.5 hours each. The three divisions have a maximum of 1,600, 1,200 and 1,600 hours every month,respectively.ModelAgivesaprofitcontributionof 400 each and B of 100 each. Assuming abundant product demand, formulate and solve this problem as an integer LP problem, to determine the optimal product mix and the maximum contribution.

9. A manufacturing company produces two types of screws–metal and wooden. Each screw has to pass through the slotting and threading machines. The maximum time that each machine can be run is 150 hours per month. A batch of 50 wooden screws requires 2 minutes on the threading machine and 3 minutes on the slotting machine. Metal screws of the same batch size require 8 minutes on the threading machine and 2 minutes on the slottingmachine.Theprofitcontributionforeachbatchofwoodenandmetalscrewsis ` 1 and ` 2.50, respectively. Formulate and solve this problem as an integer LP problem inordertodeterminetheoptimalproductmixformaximumprofitcontribution.

10. A diectician for a hospital is considering a new breakfast menu that includes oranges and cereal. This breakfast must meet the minimum requirements for the Vitamins A and B. The number of milligrams of each of these vitamins contained in a purchasing unit, for each of these foods, is as follows:

VitaminMilligrams per Purchasing

Unit of Food Minimum Requirement (mg)

Oranges (doz) Cereal (box)A 1 2 20B 3 2 50


NOTES


The cost of the food ingredients is 15 per dozen for oranges and 12.50 per box for cereal. For dietary reasons, at least one unit of each food type must be used in the menu plan. Formulate and solve this problem as an integer programming problem.

11. The dietician at a local hospital is planning the breakfast menu for the maternity ward patients. She is primarily concerned with Vitamin E and iron requirements, for planning the breakfast. According to the State Medical Association (SMA) new mothers must get at least 12 milligrams of Vitamin E and 24 milligrams of iron from breakfast. The SMA handbook reports that a scoop of scrambled egg contains 2 milligrams of Vitamin E and 8 milligrams of iron. The handbook also recommends that new mothers should eat at least two scoops of cottage cheese for their breakfast. The dietician considers this as one of the model constraints. The hospital’s accounting department estimates that one scoop of cottage cheese costs ` 2 and one scoop of scrambled egg also costs ` 2. The dietician is attemptingtodeterminetheoptimumbreakfastmenuthatsatisfiesalltherequirementsand minimizes the total cost. The cook insists that he can serve foods by only full scoop, thus necessitating an integer solution. Determine the optimum integer solution to the problem.

12. A building contractor has just won a contract to build a municipal library building. His present labour workforce is inadequate to immediately take up this work as the force is already involved with other jobs on hand. The convertor must therefore immediately decide whether to hire one or more labourers on a full-time basis (eight hours a day each) ortoallowovertimetooneormoreoftheexistinglabourforce(fivehoursadayeach).Extra labourers can be hired for ` 40 per day (for eight hours) while overtime costs ` 43 perday(forfivehoursperday).Thecontractorwantstolimithisextrapaymentto` 400 per day and to use no more than twenty labourers (both full-time and overtime) because of limited supervision. He estimates that the new labour employed on a full-time basis will generate `15adayinprofits,whileovertimelabour` 20 a day. Formulate and solve this problem as an integer LP problem to help the building contractor to decide the optimum labour force.

13. The ABC company requires an output of at least 200 units of a particular product per day. To accomplish this target it can buy machines A or B or both. Machine A costs ` 20,000 and B ` 15,000. The company has a budget of ` 2,00,000 for the same. Machines A and B will be able to produce 24 and 20 units, respectively of this product per day. However, machine Awillrequireafloorspaceof12squarefeetwhilemachineB will require 18squarefeet.Thecompanyonlyhasatotalfloorspaceof180squarefeet.Formulateand solve this problem as an integer LP problem to determine the minimum number of machines that should be purchased.

14. A company produces two products A and B. Each unit of product A requires one hour of engineeringservicesandfivehoursofmachinetime.ToproduceoneunitofproductB, two hours of engineering and 8 hours of machine time are needed. A total of 100 hours of engineering and 400 hours of machine time is available. The cost of production is a non-linear function of the quantity produced as given in the following table:

Product A Product BProduction (units) Unit Cost (`) Production (units) Unit Cost (`)

0–50 10 0–40 750–100 8 40–100 3

The unit selling price of product A is ` 12 and of product B is ` 14. The company would like a production plan that gives the number of units of A and the number of units of B to beproducedthatwouldmaximizeprofit.Formulateandsolvethisproblemasanintegerlinear programming problem to help the company maximize its total revenue.


NOTES

Operations Research 15. XYZCorporationmanufacturesanelectricdevice,finalassemblyofwhichisaccomplishedby a small group of trained workers operating simultaneously on different devices. Due to spacelimitations,theworkinggroupmaynotexceedteninnumber.Thefirm’soperatingbudget allows 5,400 per month as salary for the group. A certain amount of discrimination isevidencedbythefactthatthefirmpaysmeninthegroup 700 per month, while women doing the same work receive ` 400. However, previous experience has indicated that a man will produce about ` 1,000 in value added per month, while a woman worker adds `900.Ifthefirmwishestomaximizethevalueaddedbythegroup,howmanymenandwomen should be included? (A non-integer solution for this problem will not be accepted).

16. A manufacturer of baby dolls makes two types of dolls. One is sold under the brand name ‘Molina’ and the other under ‘Suzie’. These two dolls are processed on two machines –A and B. The processing time for each ‘Molina’ is 2 hours and 6 hours on machines A and B, respectively and that for each ‘Suzie’ is 5 hours and 5 hours on machines A and B, respectively. There is 16 hours of time available per day on machine A and 30 hours on machine B.Theprofitcontributionfroma‘Molina’is` 6 and that from a ‘Suzie’ is ` 18. Formulate and solve this problem as an integer LP problem to determine the optimal weekly production schedule of the two dolls.

Hints and Answers 1. (i) x1 = 4, x2 = 3 and Max Z = 10 (ii) x1 = 1, x2 = 2, and Max Z = 3.7 (iii) x1 = 0, x2 = 0, x3 = 4 and Max Z = 20 (iv) x1 = 0, x2 = 4 and Max Z = 16 (v) x1 = 3, x2 = 0 and Max Z = 12 (vi) x1 = 4, x2 = 3 and Max Z = 55 4. x1 and x2 = number of units of toy A and toy B, respectively, to be produced Max Z = 5x1 + 5x2

subject to (i) 2x1 + 5x2 ≤ 16, (ii) 6x1 + 5x2 ≤ 30and x1, x2 ≥ 0 and integers.Ans: x1 = 3, x2 = 2 and Max Z = ` 25

5. x1 and x2 = number of units of refrigerator and range, respectively, to be produced Max Z = 300x1 + 200x2

subject to (i) 2x1 + x2 ≤ 60, (ii) x1 ≤ 25 (iii) x2 ≤ 35and x1, x2 ≥ 0 and integers.Ans: x1 = 13, x2 = 34 and Max Z = ` 10,700

14. x1 and x2 = number of units of product A and product B, respectively, to be produced Max Z = 12x1 – {10x1 – 2(x1 – 50)} + 14x2 – {7x2 – 4(x2 – 40)} subject to (i) x1 + 2x2 ≤ 100; (ii) 5x1 + 8x2 ≤ 400 (iii) x1 + 2x2 ≤ 100; (iv) 5x1 + 8x2 ≤ 100 and x1, x2 ≥ 100 and integers. 16. x1, and x2 = number of units of shirt ‘Molina’ and ‘Suzie’; respectively, to be produced Max Z = 6x1 + 18x2

subject to (i) 2x1 + 5x2 ≤ 16 (ii) 6x1 + 5x2 ≤ 30 and x1, x2 ≥ 0 and integers Ans: x1 = 3, x2 = 2 and Max Z = 36


NOTES


3.13 GOMORY’S MIXED-INTEGER CUTTING PLANE METHOD

In the previous section, Gomary cutting method was discussed for LP problems where all decision variables, slack variables and surplus variables were assumed to have integer values. This implies that this method is applicable only when all variables in the LP problem assume integer values. For example, consider the following constraint:

12

x1 + x2 ≤ 113

or 12

x1 + x2 + s1 = 113

x1, x2, s1 ≥ 0 and integers.In this equation decision variables x1 and x2 can assume integer values only if s1

is non-integer. This situation can be avoided in two ways: 1. The non-integer coefficients in the constraint equation can be removed by

multiplying both sides of the constraint with a proper constant. For example, the above constraint is multiplied by 6 to obtain: 3x1 + 6x2 ≤ 22. However, this type of conversionispossibleonlywhenthemagnitudeoftheintegercoefficientsissmall.

2. Use a special cut called Gomory’s mixed-integer cut or simply mixed-cut, where only a subset of variables may assume integer values and the remaining variables (including slack and surplus variables) remain continuous. The details of developing this cut are presented below.

3.13.1 Method for Constructing Additional Constraint (Cut)Consider the following mixed integer programming problem:Maximize Z = c1 x1 + c2 x2 + . . . + cn xn

subject to the constraints a11 x1 + a12 x2 + . . . + a1n xn = b1

a21 x1 + a22 x2 + . . . + a2n xn = b2

am1 x1 + am2 x2 + . . . + amn xn = bm

and xj are integers; j = 1, 2, . . ., k (k < n).Suppose that the basic variable, xr is restricted to be integer and has a largest

fractional value among all those basic variables that are restricted to take integer values. Then rewrite the rth constraint (row) from the optimal simplex table as follows:

xBr = xr + j r

rj ja x≠∑ ...(6)

where xj represents all the non-basic variables in the rth row except variable, xr and xBr is the non-integer value of variable xr.

Decomposecoefficientsofxj, xr variables and xBr into integer and non-negative fractional parts, as shown below:

xBr = [xBr] + fr

and R+ = { j : arj ≥ 0}, columns in simplex table for which arj ≥ 0 R– = { j : arj < 0}, columns in simplex table for which arj < 0


NOTES

Operations Research Then Eq. (6) can be rewritten as: [xBr] + fr = (1 + 0) xr +

j Rrj j

j Rrj ja x a x

∈ ∈+ −

∑ ∑+ ...(7)

RearrangethetermsinEq.(7)sothatalloftheintegercoefficientsappearontheright-hand side.

This gives: j R

rj jj R

rj ja x a x∈ ∈+ −

∑ ∑+ = fr + {[xBr] – xr} = fr + I ...(8)

where fr is a strictly positive fraction number (i.e., 0 < fr < 1), and I is the integer value.Since the terms in the bracket on the right-hand side of Eq. (8) are integers, left-

hand side in Eq. (8) is either positive or negative according as fr + I is positive or negative.Case 1: Let fr + I be positive. Then it must be fr , 1+ fr, 2 + fr , . . . and we shall

then have

j Rrj j

j Rrj ja x a x

∈ ∈+ −

∑ ∑+ ≥ fr ...(9)

Since arj ∈ R– are non-positive and xj ≥ 0.

j Rrj ja x

∈ +

∑ ≥ j R

rj jj R


∑ ∑+

and hence j R

rj ja x∈ +

∑ ≥ fr

Case 2: Let fr + I be negative. Then it must be fr , – 1 + fr , – 2 + fr , . . . and we shall have

j R

rj ja x∈ −

∑ ≤ j R

rj jj R


∑ ∑+ ≤ – 1 + fr ...(10)

Multiplying both sides of Eq. (10) by the negative number ffr

r −

1 , we get

ff

a xr

r j Rrj j−

∈ −

∑1 ≥ fr ...(11)

Either of inequalities (9) and (11) holds, since in both the cases the left-hand side is non-negative and one of these is greater than or equal to fr. Thus, any feasible solution to mixed-integer programming must satisfy the following inequality:

j R

rj jr

r j Rrj ja x f

fa x

∈ ∈+ −

∑ ∑+−

1 ≥ fr ...(12)

Inequality(12)isnotsatisfiedbytheoptimalsolutionoftheLPproblemwithoutinteger requirement. This is because by putting xj = 0 for all j, the left-hand side becomes zero,andright-handsidebecomespositive.Thus,inequality(12)definesacut.

Adding a non-negative slack variable we can rewrite Eq. (12) as

sg = – fr + j R

rj jr

r j Rrj ja x f

fa x

∈ ∈+ −

∑ ∑+−

1 ...(13)

Equation (13) represents the required Gomory’s cut.For generating the cut (13) it was assumed that the basic variable xr should take

integer value. But if one or more xj, j ∈ R are restricted to be integers to improve a


NOTES


cut shown as Eq. (12) to a better cut, i.e.,thecoefficientsarj, j ∈ R+ and arj { fr/( fr – 1)}, j ∈ R– are desired to be as small as possible, we can proceed as follows. The value of the coefficientsofxr can be increased or decreased by an integral amount in Eq. (8) so as to getatermwithsmallestcoefficientsinEq.(12).Inordertoreducethefeasibleregionasmuchaspossiblethroughcuttingplanes,thecoefficientsofintegervariablexr must beassmallaspossible.Thesmallestpositivecoefficientforxr in Eq. (8) is:

f ff

frjr

rrj; ( )

11

−−

The smaller of the two coefficientswould be considered tomake the cut (13)penetrate deeper into the original feasible region. A cut is said to be deep if the intercepts of the hyperplane represented by a cut with the x-axis are larger. Obviously,

fr ≤ ffr

r1 − (1 – frj); frj ≤ fr and fr >

ffr

r1 −(1 – frj); frj > fr

Thus, the new cut can be expressed as

sg = – fr + j R

rj jf x∈∑ * ...(14)

where frj* =

a a x

ff

a a x

rj rj j

r

rrj rj j

,

,

≥

−

<

0

10

and non-integer

and non-inteeger

and integer

and

f f f x

ff

f f f x

rj rj r j

r

rrj rj r j

,

( ) ,

≤

−

− >1

1 iinteger

...(15)

3.13.2 Steps of Gomory’s Mixed-Integer Programming AlgorithmGomory’s mixed-integer cutting plane method can be summarized in the following

steps:Step 1: Initialization. Formulate the standard integer LP problem. Solve it by

simplex method, ignoring integer requirement of variables.Step 2: Test of optimality

(a) Examine the optimal solution. If all integer restricted basic variables have integer values, then terminate the procedure. The current optimal solution is the optimal basic feasible solution to the integer LP problem.

(b) If all integer restricted basic variables are not integers, then go to Step 3.Step 3: Generate cutting plane. Choose a row r corresponding to a basic variable

xr that has the highest fractional value fr and generate a cutting plane as explained earlier in the form [Ref. Eq. 13]:

sg = – fr + j R

rj jr

r j Rrj ja x f

fa x

∈ ∈+ −

∑ ∑+−

1 , where 0 < fr < 1.

Step 4: Obtain the new solution. Add the cutting plane generated in Step 3 to the bottom of the optimal simplex table. Find a new optimal solution by using the dual simplex method and return to Step 2. The process is repeated until all restricted basic variables are integers.


NOTES

Operations Research Example 1. Solve the following mixed-integer programming problem:Maximize Z = – 3x1 + x2 + 3x3

subject to the constraints (i) – x1 + 2x2 + x3 ≤ 4, (ii) 2x2 – (3/2) x3 ≤ 1, (iii) x1 – 3x2 + 2x3 ≤ 3and x1, x2 ≥ 0, x3 non-negative integer.

Sol. Expressing LP problem in its standard form as follows:Maximize Z = – 3x1 + x2 + 3x3 + 0s1 + 0s2 + 0s3

subject to the constraints (i) – x1 + 2x2 + x3 + s1 = 4, (ii) 2x2 – (3/2)x3 + s2 = 1, (iii) x1 – 3x2 + 2x3 + s3 = 3and x1, x2, s1, s2, s3 ≥ 0, x3 non-negative integer.

Step 1: Ignoring the integer requirement, the optimal solution of the LP problem using the simplex method is given in Table 3.31.


cj → – 3 1 3 0 0 0

Basic Variables

Coefficient cB

Basic Variables B

Basic Variables

Value b (= xB)x1 x2 x3 s1 s2 s3

1 x2 5/7 – 3/7 1 0 2/7 0 – 1/7

0 s2 48/7 9/7 0 0 1/7 1 10/7

3 x3 13/7 5/14 0 1 3/7 0 2/7

Z = 44/7 cj – zj – 51/14 0 0 – 11/7 0 – 5/7

The non-integer optimal solution shown in Table 3.32 is: x1 = 0, x2 = 5/7, x3 = 13/7 and Max Z = 44/7.

Step 2: Since basic variable, x3 is required to be integer, then considering x3-row to develop Gomory’s mixed-integer cut as follows:

137

= 514

x1 + x3 + 37

271 3s s+ (x3-source row)

Sincethecoefficientsofnon-basicvariablesx1, s1 and s3 are positive, therefore factorizingthecoefficientsusingrule(15),weget:

1 67

+

= 0 514

+

x1 + (1 + 0)x3 + 0 37

0 271 3+

+ +

s s

67

+ (1 – x3) = 514

37

271 1 3x s s+ + , i.e., 6

7 ≤ 514

37

271 1 3x s s+ +

On adding slack variable sg1, we obtain Gomory’s mixed integer cut as follows:

67 1

+ sg = 1 1 35 3 2

14 7 7x s s+ +

or − − − +514

37

271 1 3 1

x s s sg = − 67

(Mixed integer cut I)


NOTES


Adding this constraint to the bottom of the Table 3.31. The new values so obtained are shown in Table 3.32.


cj → – 3 1 3 0 0 0 0


cB

Basic Variables

B

Basic Variables

Value b (= xB)

x1 x2 x3 s1 s2 s3 sg1

1 x2 5/7 – 3/7 1 0 2/7 0 – 1/7 0

0 s2 48/7 9/7 0 0 1/7 1 10/7 0

3 x3 13/7 5/14 0 1 3/7 0 2/7 0

0 sg1– 6/7 – 5/14 0 0 – 3/7 0 – 2/7 1 →

Z = 44/7 cj – zj – 51/14 0 0 – 11/7 0 – 5/7 0

Ratio: Min(cj – zj)/y4j (< 0) 51/5 — — 11/3 — 5/2 —↑

Iteration 1: Enter the non-basic variable, s3 into the basis to replace with basic variable, sg1

using dual simplex method. The new solution so obtained is shown in Table 3.33.

Table 3.33. Optimal Solutioncj → – 3 1 3 0 0 0 0


cB

Basic Variables

B

Basic Variables

Value b(= xB)

x1 x2 x3 s1 s2 s3 sg1

1 x2 8/7 – 1/4 1 0 1/2 0 0 – 1/20 s2 18/7 – 1/2 0 0 – 2 1 0 53 x3 1 0 0 1 0 0 0 10 s3 3 5/4 0 0 3/2 0 1 – 7/2

Z = 29/7 cj – zj – 11/4 0 0 – 1/2 0 0 – 5/2

Since variable x3 has assumed integer value and all cj – zj ≤ 0, the optimal mixed integer solution is:

x1 = 0, x2 = 8/7, x3 = 1 and Max Z = 29/7.

3.14 BRANCH AND BOUND METHODTheBranchandBoundmethoddevelopedfirstbyAHLandandAGDoigisused

to solve all-integer, mixed-integer and zero-one linear programming problems. The concept behind this method is to divide the feasible solution space of an LP problem into smaller parts called subproblems and then evaluate corner (extreme) points of each subproblem for an optimal solution.

The branch and bound method starts by imposing bounds on the value of objective function that help to determine the subproblem to be eliminated from consideration when the optimal solution has been found. If the solution to a subproblem does not yield an

Branch and Bound method breaks the feasible solution region into smaller regions until an optimal solution is obtained.


NOTES

Operations Research optimal integer solution, a new subproblem is selected for branching. At a point where no more subproblem can be created, an optimal solution is arrived at.

Thebranchandboundmethodfortheprofit-maximizationintegerLPproblemcanbe summarized in the following steps:

The ProcedureStep 1: Initialization. Consider the following all integer programming problemMaximize Z = c1 x1 + c2 x2 + . . . + cn xn

subject to the constraints a11 x1 + a12 x2 + . . . + a1n xn = b1 a21 x1 + a22 x2 + . . . + a2n xn = b2 ...(LP-A)

am1 x1 + am2 x2 + . . . + amn xn = bmand xj ≥ 0 and non-negative integers.

Obtain the optimal solution of the given LP problem ignoring integer restriction on the variables. (i) If the solution to this LP problem (say LP-A) is infeasible or unbounded, the

solution to the given all-integer programming problem is also infeasible or unbounded, as the case may be.

(ii)Ifthesolutionsatisfiestheintegerrestrictions,theoptimalintegersolutionhas been obtained. If one or more basic variables do not satisfy integer require-ment, then go to Step 2. Let the optimal value of objective function of LP-A be Z1. This value provides an initial upper bound on objective function value and is denoted by ZU.

(iii) Find a feasible solution by rounding off each variable value. The value of objec-tive function so obtained is used as a lower bound and is denoted by ZL.

Step 2: Branching step (i) Let xk be one basic variable which does not have an integer value and also has

the largest fractional value. (ii) Branch (or partition) the LP-A into two new LP subproblems (also called

nodes) based on integer values of xk that are immediately above and below its non-integer value. That is, it is partitioned by adding two mutually exclusive constraints:

xk ≤ [xk] and xk ≥ [xk] + 1 to the original LP problem. Here [xk] is the integer portion of the current

non-integer value of the variable xk. This is obviously done to exclude the non-integer value of the variable xk. The two new LP subproblems are as follows:

LP Subproblem B LP Subproblem C

Max Z = j

n

j jc x=∑

1Max Z =

j

n

j jc x=∑

1

subject to j

n

ij ja x=∑

1 = bi subject to

j

n

ij ja x=∑

1 = bi

xk ≤ [xk] xk ≥ [xk] + 1and xj ≥ 0. and xj ≥ 0.


NOTES


Step 3: Bound step. Obtain the optimal solution of subproblems B and C. Let the optimal value of the objective function of LP-B be Z2 and that of LP-C be Z3. The best integer solution value becomes the lower bound on the integer LP problem objective function value (Initially this is the rounded off value). Let the lower bound be denoted by ZL.

Step 4: Fathoming step. Examine the solution of both LP-B and LP-C (i) If a subproblem yields an infeasible solution, then terminate the branch. (ii) If a subproblem yields a feasible solution but not an integer solution, then

return to Step 2. (iii) If a subproblem yields a feasible integer solution, examine the value of the

objective function. If this value is equal to the upper bound, an optimal solu-tion has been reached. But if it is not equal to the upper bound but exceeds the lower bound, this value is considered as new upper bound and return to Step 2. Finally, if it is less than the lower bound, terminate this branch.

Step 5: Termination. The procedure of branching and bounding continues until no further sub-problem remains to be examined. At this stage, the integer solution corresponding to the current lower bound is the optimal all-integer programming problem solution.

Remark. The above algorithm can be represented by an enumeration tree. Each node in the tree represents a subproblem to be evaluated. Each branch of the tree creates a new constraint that is added to the original problem.

Example 2. Solve the following all integer programming problem using the branch and bound method.

Maximize Z = 2x1 + 3x2

subject to the constraints (i) 6x1 + 5x2 ≤ 25, (ii) x1 + 3x2 ≤ 10and x1, x2 ≥ 0 and integers.

Sol. Relaxing the integer conditions, the optimal non-integer solution to the given integer LP problem obtained by graphical method as shown in Fig. 3.5 is: x1 = 1.92, x2 = 2.69 and max Z1 = 11.91.

Fig. 3.5. Optimal Noninteger Solution

The value of Z1 represents initial upper bound as: ZL = 11.91. Since value of variable x2 is non-integer, therefore selecting it to decompose (branching) the given problem into


NOTES

Operations Research two sub-problems by adding two new constraints x2 ≤ 2 and x2 ≥ 3 to the constraints of original LP problem as follows:

LP Sub-problem B LP Sub-problem C

Max Z = 2x1 + 3x2 Max Z = 2x1 + 3x2

subject to (i) 6x1 + 5x2 ≤ 25, (ii) x1 + 3x2 ≤ 10, subject to (i) 6x1 + 5x2 ≤ 25, (ii) x1 + 3x2 ≤ 10

(iii) x2 ≤ 2, (iii) x2 ≥ 3,

and x1, x2 ≥ 0 integers. and x1, x2 ≥ 0 and integers.

Sub-problem B and C are solved graphically as shown in Fig. 3.6. The feasible solutions are:

Sub-problem B: x1 = 2.5, x2 = 2 and Max Z2 = 11Sub-problem C : x1 = 1, x2 = 3 and Max Z3 = 11

Fig. 3.6. Solution of LP Sub-problems B and C

The value of decision variables at one of the extreme point of feasible region (solution space) of LP sub-problem C, are: x1 = 1 and x2 = 3. Since these are integer values, so there is no need to further decompose (branching) this sub-problem. The value of objective function, Max ZL = 11 becomes lower bound on the maximum value of objective function, Z for future solutions.

Fig. 3.7. Solution of LP Sub-problems D and E


NOTES


LP sub-problem B is further subdivided into two LP sub-problems D and E (shown in Fig. 3.7) by taking variable x1 = 2.5. Adding two new constraints x1 ≤ 2 and x1 ≥ 3 to sub-problem B. Also Max Z = 11 is also not inferior to the ZL = 11.

LP Sub-problem D LP Sub-problem E

Max Z = 2x1 + 3x2 Max Z = 2x1 + 3x2

subject to (i) 6x1 + 5x2 ≤ 25, (ii) x1 + 3x2 ≤ 10 subject to (i) 6x1 + 5z2 ≤ 25, (ii) x1 + 3x2 ≤ 10,

(iii) x2 ≤ 2, (iv) x1 ≤ 2 (iii) x2 ≤ 2, (iv) x1 ≥ 3

and x1, x2 ≥ 0 and integers. and x1, x2 ≥ 0 and integers.

Sub-problems D and E are solved graphically as shown in Fig. 3.7. The feasible solutions are:

Sub-problem D: x1 = 2, x2 = 2 and max Z4 = 10Sub-problem E: x1 = 3, x2 = 1.4 and max Z5 = 10.2The solution of LP sub-problem D is satisfying integer value requirement of

variables but is inferior to the solution of LP sub-problem E in terms of value of objective function, Z5 = 10.2. Hence the value of lower bound ZL = 11 remains unchanged and sub-problem D is not considered for further decomposition.

Since the solution of sub-problem E is non-integer, it can be further decomposed into two sub-problems by considering variable, x2. But the value of objective function (Z5 = 10.2) is inferior to the lower bound and hence this does not give a solution better than the one already obtained. The sub-problem E is also not considered for further branching. Hence, the best available solution corresponding to sub-problem C is the integer optimal solution: x1 = 1, x2 = 3 and Max Z = 11 of the given integer LP problem. The entire branch and bound procedure for the given Integer LP problem is shown in Fig. 3.8.

Fig. 3.8. Branch and Bound Solution

Example 3. Solve the following all-integer programming problem using the branch and bound method

Minimize Z = 3x1 + 2.5x2subject to the constraints (i) x1 + 2x2 ≥ 20, (ii) 3x1 + 2x2 ≥ 50and x1, x2 ≥ 0 and integers.


NOTES

Operations ResearchSol. Relaxing the integer requirements, the optimal non-integer solution of the

given integer LP problem, obtained by the graphical method, is: x1 = 15, x2 = 2.5 and Z1 = 51.25. This value of Z1 represents the initial lower bound, ZL = 51.25 on the value of the objective function, i.e., the value of the objective function in the subsequent steps cannot be less than 51.25.

The variable x2 (= 2.5) is the only non-integer solution value and is therefore is selected for dividing the given problem into two subproblems: B and C. In order to eliminate the fractional part of x2 = 2.5, two new constraints x2 ≤ 2 and x2 ≥ 3 are created by adding in the given set of constraints as shown below:

LP Subproblem B LP Subproblem C

Max Z = 3x1 + 2.5x2 Max Z = 3x1 + 2.5x2

subject to (i) x1 + 2x2 ≥ 20, (ii) 3x1 + 2x2 ≥ 50 (iii) x2 ≤ 2

subject to (i) x1 + 2x2 ≥ 20, (ii) 3x1 + 2x2 ≥ 50 (iii) x2 ≥ 3


Subproblems B and C are solved graphically. The feasible solutions are:Subproblem B: x1 = 16, x2 = 2 and Min Z2 = 53.Subproblem C: x1 = 14.66, x2 = 3 and Min Z3 = 51.5.Since the solution of subproblem B is all-integer, therefore no further decomposition

(branching) of this subproblem is required. The value of Z2 = 53 becomes the new lower bound. A non-integer solution of subproblem C and also Z3 < Z2 indicates that further decomposition of this problem need to be done in order to search for a desired integer solution. However, if Z3 ≥ Z2, then no further branching was needed from sub-problem C. The second lower bound takes on the value ZL = 51.5 instead of ZL = 51.25 at node A.

Dividing subproblem C into two new subproblems: D and E by adding constraints x1 ≤ 14 and x1 ≥ 15, as follows:

LP Subproblem D LP Subproblem EMax Z = 3x1 + 2.5x2 Max Z = 3x1 + 2.5x2

subject to (i) x1 + 2x2 ≥ 20, (ii) 3x1 + 2x2 ≥ 50 (iii) x2 ≥ 3, (iv) x1 ≤ 14

subject to (i) x1 + 2x2 ≥ 20, (ii) 3x1 + 2x2 ≥ 50 (iii) x2 ≥ 3, (iv) x1 ≥ 15


Subproblems D and E are solved graphically. The feasible solutions are:Subproblem D: x1 = 14, x2 = 4 and Min Z4 = 52.Subproblem E: x1 = 15, x2 = 3 and Min Z5 = 52.5.The feasible solutions of both subproblems D and E are all-integer and therefore

branch and bound procedure is terminated. The feasible solution of subproblem D is considered as optimal basic feasible solution because this solution is all-integer and the value of the objective function is the lowest amongst all such values.


NOTES


The branch and bound procedure for the given problem is shown in Fig. 3.9.

1

1

2

2

21

1 2

4

1 2

3

1

5

2

Fig. 3.9. Branch and Bound Solution


1. What is integer linear programming? How does the optimal solution of an integer programming problem compare with that of the linear programming problem?

2. What is integer linear programming? Explain the merits and demerits of ‘rounding-off’ a continuous optimal solution to an LP problem in order to obtain an integer solution.

3. What is the effect of the ‘integer’ restriction of all the variables on the feasible space of integer programming problem?

4. Explain how Gomory’s cutting plane algorithm works. 5. Addition of a cut makes the previous non-integer optimal solution infeasible. Explain. 6. What is the meaning and the role of the lower bound and upper bound in the branch and

bound method? 7. Describe any one method of solving a mixed-integer programming problem. 8. Sketch the branch and bound method in integer programming. 9. Discuss the advantages of the branch and bound method. 10. Discuss the advantages and disadvantages of solving integer programming problems by

(a) the cutting plane method and (b) the branch and bound method.


Solve the following integer programming problems using Gomory’s cutting plane algorithm or by branch or bound method: 1. Max Z = 1.5x1 + 3x2 + 4x3 subject to (i) 2.5x1 + 2x2 + 4x3 ≤ 12 (ii) 2x1 + 4x2 – x3 ≤ 7 and x1, x2, x3 ≥ 0 2. Max Z = 2x1 + 3x2 subject to (i) x1 + 3x2 ≤ 9, (ii) 3x1 + x2 ≤ 7 (iii) x1 – x2 ≤ 1 and x1, x2 ≥ 0 and integers. 3. Max Z = 7x1 + 6x2

subject to (i) 2x1 + 3x2 ≤ 12, (ii) 6x1 + 5x2 ≤ 30 and x1, x2 ≥ 0 and integers.


NOTES

Operations Research 4. Max Z = 5x1 + 4x2 subject to (i) x1 + x2 ≥ 2, (ii) 5x1 + 3x2 ≤ 15 (iii) 3x1 + 5x2 ≤ 15 and x1, x2 ≥ 0 and integers. 5. Max Z = – 3x1 + x2 + 3x3 subject to (i) – x1 + 2x2 + x3 ≤ 4, (ii) 2x2 – 1.5x3 ≤ 1 (iii) x1 – 3x2 + 2x3 ≤ 3 and x1, x2 ≥ 0; x3 non-negative integer. 6. Max Z = x1 + x2 subject to (i) 2x1 + 5x2 ≥ 16, (ii) 6x1 + 5x2 ≤ 30 and x2 ≥ 0, x1 non-negative integer. 7. Min Z = 4x1 + 3x2 + 5x3

subject to (i) 2x1 – 2x2 + 4x3 ≥ 7, (ii) 2x1 + 6x2 – 2x3 ≥ 5 and x2 ≥ 0; x1, x3 non-negative integers. 8. Max Z = 110x1 + 100x2 subject to (i) 6x1 + 5x2 ≤ 29 (ii) 4x1 + 14x2 ≤ 48 and x1, x2 ≥ 0 and integers. 9. Afirmisconsideringinvestinginplantmodernizationandplantexpansion.Allofthese

proposed projects would be completed within 2 years, with varying requirements of money and plant engineering. The management is willing to use the following data in selecting the best set of proposals. Three resource limitations are:

First year expenditure : ` 4,20,000 Second year expenditure : ` 4,40,000 Engineering hours : ` 15,000 hours

Project Description

Expenditure (‘000 `)

Net Present Value

(‘000s `)

Engineer-ing Hours

(00)1st year 2nd year

• Modernizeshopefloor 220 0 70 50• Buildnewshopfloor 95 270 95 80• Equipment for new production

line0 170 60 43

• Modernize maintenance shop 60 100 80 90• Processing sub-contract raw

material60 220 145 45

• Install new raw material processing plant

195 0 70 30

• Buy trucks and containers 80 32 45 0

Thesituationrequiresthatanewormodernizedshopfloorbeprovided.Theequipmentforproductionlineisapplicableonlytothenewshopfloor.Thecompanymaynotwanttobuy or build raw material processing facilities. Formulate the given problem as an integer linear programming problem in order to maximize the net present value of the money.

10. ABC manufacturing company faces the following problem: should they make or buy each of their several products? The company policies specify that they will either make or buy the whole lot of each product in a complete lot. The company has four products to make or buy with six machines involved in making these products, if they are made in the shop. The time per unit (in hours) reqired are as follows:


NOTES


Forty hours are available on each machine. One hundred ten units of each product are needed. The costs of making the products are listed below:

Product : 1 2 3 4Cost/unit (in `) : 2.25 2.22 4.50 1.90

The cost to buy the products are:

Product : 1 2 3 4Cost/unit (in `) : 3.10 2.60 4.75 2.25

Formulate this problem as a zero-one integer programming problem.

ProductMachine

A B C D E F1 0.04 0.02 0.02 0 0.03 0.062 0 0.01 0.05 0.15 0.09 0.063 0.02 0.06 0 0.06 0.02 0.024 0.06 0.04 0.15 0 0 0.05

Hints And Answers 1. x1 = 0, x2 = 2, x3 = 2 and Max Z = 14. 2. x1 = 0, x2 = 3 and Max Z = 9 3. x1 = 5, x2 = 0 and Max Z = 35 4. x1 = 3, x2 = 0 and Max Z = 15 5. x1 = 0, x2 = 8/7, x3 = 1 and Max Z = 29/7 6. x1 = 4, x2 = 6/5 and Max Z = 26/5 7. x1 = 3, x2 = 1.25, x3 = 1 and Min Z = 20.75 8. x1 = 4, x2 = 1, Max Z = 540

SUMMARY

• Here we have discussed the relationship between an LP problem and its dual. The method of formulating a dual LP problem of the given LP problem was explained with several examples. The main focus, while solving a dual LP problem, was to findforeachresourceitsbestmarginalvalue,alsoknownasdualorshadowprice.Thisvaluereflectsthemaximumadditionalpricetobepaidinordertoobtainoneadditionalunitofanyresourceinordertomaximizeprofit(orminimizecost)under resource constraints.

• If one or more solution values are negative and the optimality condition cj – zj bothformaximizationandminimizationLPproblemissatisfied,thenthecurrentoptimal solution may not be feasible (because cj – zj = cj – cB B–1 aj is completely independent of the vector b).Insuchcases,itispossibletofindastartingbasicsolution but not a feasible solution that is dual feasible. In all such cases a variant of the simplex method called the dual-simplex method would be used. In the dual simplex method we always attempt to retain optimality while bringing the primal back to feasibility.

• Here, an extension of linear programming, referred to as integer linear programming, was introduced where few or all variables must be an integer. If all variables of a problem are integers, then such problems are referred to as all-integer linear programming problems. If some, but not necessarily all, variables are integers, then such problems are referred to as mixed integer linear programming problems. Most integer programming applications involve 0-1 variables.

• The number of applications of integer linear programming continues to grow rapidly due to the availability of integer linear programming software packages.


NOTES

Operations Research U N I T

4DYNAMIC PROGRAMMING AND

THEORY OF GAMES

A. DYNAMIC PROGRAMMING

4.1 INTRODUCTIONThe decision-making process often involves several decisions to be taken at

different times. For example, problems of inventory control, evaluation of investment opportunities, long-term corporate planning, and so on, require sequential decision-making. The mathematical technique of optimizing a sequence of interrelated decisions over a period of time is called dynamic programming. The dynamic programming approach uses the idea of recursion to solve a complex problem, broken into a series of interrelated (sequential) decision stages (also called subproblems) where the outcome of a decision at one stage affects the decision at each of the following stages. The word dynamic has been used because time is explicitly taken into consideration.

Dynamic programming (DP) differs from linear programming in two ways:(i) In DP, there is no set procedure (algorithm) as in LP to solve any decision-

problem. The DP technique allows to break the given problem into a sequence of smaller subproblems, which are then solved in a sequential order (stage).

Dynamic programming is the mathematical technique of optimizing a sequence of interrelated decision over a period of time.

STRUCTURE

4.1 Introduction4.2 Dynamic Programming Terminology4.3 Characteristics of Dynamic Programming4.4 Optimal Sub-Divison Problem4.5 Dynamic Programming Approach for Solving Linear Programming

Problem4.6 Two-Person Zero-Sum Games4.7 Minimax and Maximin Principles: Games with Saddle Point4.8 Game without Saddle Point4.9 Dominance Properties4.10 Solution Methods for Games without Saddle Point


NOTES

Dynamic Programming and Theory of Games

(ii) LP approach provides one-time period (single stage) solution to a problem whereas DP approach is useful for decision-making over time and solves each subproblem optimally.

4.2 DYNAMIC PROGRAMMING TERMINOLOGYRegardless of the type or size of a decision problem, there are certain terms and

concepts that are commonly used in dynamic programming approach of solving such problems.

Stage: The dynamic programming problem can be decomposed or divided into a sequence of smaller subproblems called stages. At each stage there are a number of decision alternatives (courses of action) and a decision is made by selecting the most suitable alternative. Stages represent different time periods in the planning period. For example, in the replacement problem each year is a stage, in the salesman allocation problem each territory represents a stage.

State: Each stage in a dynamic programming problem is associated with a certain number of states. These states represent various conditions of the decision process at a stage. The variables that specify the condition of the decision process or describe the status of the system at a particular stage are called state variables. These variables provide information for analyzing the possible effects that the current decision could have upon future courses of action. At any stage of the decision-making process there could be a finite or infinite number of states. For example, a specific city is referred to as state variable, in any stage of the shortest route problem.

Return function: At each stage, a decision is made that can affect the state of the system at the next stage and help in arriving at the optimal solution at the current stage. Every decision that is made has its own worth or benefit associated and can be described in an algebraic equation form, called a return function. This return function, in general, depends upon the state variable as well as the decision made at a particular stage. An optimal policy or decision at a stage yields optimal (maximum or minimum) return for a given value of the state variable.

Figure 4.1 depicts the decision alternatives known at each stage for their evaluation. The range of such decision alternatives and their associated returns at a particular stage is a function of the state input to the stage itself. The state input to a stage is the output from the previous (larger number) stage and the previous stage output is a function of the state input to itself, and the decision taken at that stage. Thus, to evaluate any stage we need to know the values of the state input to it (there may be more than one state inputs to a stage) and the decision alternatives and their associated returns at the stage.

State n

Input to

first stage

(stage) n

Stage n

(Decision

problem)

State n – 1

Input to second

stage (stage – 1)

�n

� A function of input

and decision at

stage n

Stage n – 1

(Decision

problem)

State n – 2

Input to third

stage (stage – 2)

�n

� A function of

input and decision

at stage – 1n

Fig 4.1. Information Flow between Stages

For a multistage decision process, functional relationship among state, stage and decision may be described as shown in Fig. 4.2.

Stages are the sequence of smaller sub-problems of the main problem to be analyzed.


NOTES

Operations Research

Stagen

State

s = (s , d )

(Output)n–1 n n

Stagen – 1

States

(Input)n–2

States

(Input)1

Stage1

States

(Output)0

Decisiondn

Decisiondn – 1

Decisiond1

States

(Input)n

Returnfunction

f = r (s , d )n n n n

Returnfunction

f = r (s , d )n – 1 n – 1 n – 1 n – 1

Returnfunction

f = r (s , d )1 1 1 1

Fig 4.2. Functional Relationship among Components of a DP Model

where n = stage number sn = state input to stage n from stage n + 1. Its value is the status of the system

resulting from the previous (n + 1) stage decision. dn = decision variable at stage n (independent of previous stages). This represents

the range of alternatives available when making a decision at stage n. fn = rn(sn, dn) = return (or objective) function for stage n.

Further suppose that there are n stages at which a decision is to be made. These n stages are all interconnected by a relationship (called transition function):that is, Output at stage n = (Input to state n) * (Decision at stage n)

sn–1 = sn * dn

where * represents any mathematical operation, namely addition, subtraction, division or multiplication. The units of sn, dn and sn–1 must be homogeneous.

It can be seen that at each stage of the problem, there are two input variables: state variable, sn and decision variable, dn. The state variable (state input) relates the present stage back to the previous stage. For example, the current state sn provides complete information about various possible conditions in which the problem is to be solved when there are n stages to go. The decision dn is made at stage n for optimizing the total return over the remaining n – 1 stages. The decision dn, which optimizes the output at stage n, produces two outputs: (i) the return function rn(sn, dn) and (ii) the new state variable sn–1.

The return function is expressed as function of the state variable, sn. The decision (variable), dn indicates the state of the process at the beginning of the next stage (stage n – 1), and is denoted by transition function (state transformation)

sn–1 = tn(sn, dn),where tn represents a state transformation function and its form depends on the particular problem to be solved. This formula allows the transition from one stage to another.

4.3 CHARACTERISTICS OF DYNAMIC PROGRAMMINGThe basic features which characterize dynamic programming problem are as follows:

1. The problem can be sub divided into stages with a policy decision required at each stage.

2. Each stage has a number of states associated with it.

States at each stage represents conditions of decision-making at that stage.

Return function is an algebraic equation that represents worth or benefit associated with a decision taken at each stages.


NOTES


3. The effect of the policy decision at each stage is to transform the current state into a state associated with the next stage.

4. The state of the system at a stage is described by a set of variables, called state variables.

5. Remaining decisions constitutes an optimal policy regardless of the policy adopted in previous stage, i.e., given the current stage, an optimal policy for the remaining stages is independent of the policy adopted in previous stage.

6. A recursive equation is formed that connect the optimal solution of the previous stage and the contribution of the current stage, to identify the optimal policy for each state with (n – 1) stages.

7. Using this recursive equation, the solution procedure moves backward stage by stage to get the optimal policy for each state of that stage until the optimal policy for the initial stage.

4.4 OPTIMAL SUB-DIVISION PROBLEMThe decision problems where conditions (constraints) at each stage, (i.e., state

variables) are known with certainty, can be solved by dynamic programming.

Model I: Shortest Route ProblemExample 1. A salesman located in a city A decided to travel to city B. He knew the

distances of alternative routes from city A to city B. He then drew a highway network map as shown in the Fig 4.3. The city of origin A, is city 1. The destination city B, is city 10. Other cities through which the salesman will have to pass through are numbered 2 to 9. The arrow representing routes between cities and distances in kilometres are indicated on each route. The salesman’s problem is to find the shortest route that covers all the selected cities from A to B.

2 5

3

4

6

7

1

8

10

9

4

6

3

City A

7

310

5

8

6 10

5

4 8

4

7

3

8

4

7

9City B

Stage 4Stage 5 Stage 3 Stage 2 Stage 1

Fig 4.3. Network of Routes

Sol. To solve the problem, we need to define problem stages, decision variables, state variables, return function and transition function. For this particular problem, the following definitions will be used to denote various the state and decision variables.


NOTES

Operations Research dn = decision variables that define the immediate destinations when there are n(n = 1, 2, 3, 4) stages to go.

sn = state variables describe a specific city at any stage. Dsn

,dn = distance associated with the state variable, sn, and the decision variable, dn for the current nth stage.

fn(sn, dn) = minimum total distance for the last n stages, given that salesman is in state sn and selects dn as immediate destination.

fn*(sn) = optimal path (minimum distance) when the salesman is in state sn with

n more stages to go for reaching the final stage (destination).We start calculating distances between a pair of cities from destination city

10(= x1) and work backwards x5 → x4 → x3 → x2 → x1 to find the optimal path. The recursion relationship for this problem can be stated as follows:

fn*(sn) = Minnd

{Dsn, dn + f*n–1(dn)}; n = 1, 2, 3, 4

where f*n–1(dn) is the optimal distance for the previous stages.Working backward in stages from city B to city A, we determine the shortest

distance to city B (node 10) in stage 1, from state s1 = 8 (node 8) and state s1 = 9 (node 9) in stage 2. Since the distances associated with entering stage 2 from state s1 = 8 and s1 = 9 are D8,10 = 7 and D9,10 = 9, respectively, the optimal value of f1*(s1) is the minimum value between D8,10 and D9,10. These results are shown in Table 4.1.

Table 4.1. Stage 2 Decision, d1 → ( , ) ,=

1 11 1 1 s df s d D10

Minimum Distance f1*(s1)

Optimal Decision d1

States, s1 8 7 7 109 9 9 10

We move backward to stage 3. Suppose that the salesman is at state s2 = 5 (node 5). Here he has to decide whether he should go to either d2 = 8(node 8) or d2 = 9(node 9). For this he must evaluate two sums:

D5,8 + f1*(8) = 4 + 7 = 11 (to state s1 = 8) D5,9 + f1*(9) = 8 + 9 = 17 (to state s1 = 9)This distance function for travelling from state s2 = 5 is the smallest of these two

sums: f2(s2) =

2 8,9{11,17}Min

=d = 11 (to state s1 = 8)

Similarly, the calculation of distance function for travelling from state s2 = 6 and s2 = 7 can be completed as follows:

For state, s2 = 6 f2(6) = 2

*6,8 1

*8,96,9 1

(8) 3 7 10

(9) 7 9 16=

+ = + =

+ = + =Mind

D f

D f

= 10 (to state s1 = 8)

For state, s2 = 7 f2(7) = 2

*7,8 1

*8,97,9 1

(8) 8 7 15

(9) 4 9 13Min=

+ = + =

+ = + =d

D f

D f

= 13 (to state s1 = 9)


NOTES


These results are entered into the two-stage table as shown in Table 4.2.

Table 4.2. Stage 3 Decision, d2 →

f2(s2 , d2) = Ds2,d2

+ f1*(d2)Minimum Distance

f *2(s2)

Optimal Decision

d28 9States, s1 5 11 17 11 8

6 10 16 10 87 15 13 13 9

The results that we obtain by continuing the same process for stages 4 and 5, are shown in Tables 4.3 and 4.4.

Table 4.3. Stage 4 Decision, d3 → f3(s3, d3) = Ds3

,d3 + f2*(d3) Minimum

Distance f3*(s3)

Optimal Decision d35 6 7

States, s3 2 18 20 18 18 5 or 7

3 14 18 17 14 54 17 20 18 17 5

Table 4.4. Stage 5 Decision, d4 →

f4(s4, d4) = Ds4,d4

+ f3*(d4) Minimum Distance f4*(s4)

Optimal Decision d4

2 3 4States, s4 1 22 20 20 20 3 or 4

The above optimal results at various stages can be summarized as below:Entering states (nodes)

Sequence 10 8 5 3 110 8 5 4 1

Distances 7 4 3 6 207 4 6 3 20

=

=

From the above, it is clear that there are two alternative shortest routes for this problem, both having a minimum distance of 20 kilometres.

Model II. Multiplicative Separable Return Function and Single Additive Constraint

Consider the general form of the recursive equation involving multiplicative separable return function and single additive constraints as follows:

Maximize Z = {f1(d1) . f2(d2),…,fn(dn)}subject to the constraints


NOTES

Operations Research a1d1+ a2d2 + … + andn = b,and dj, aj, b ≥ 0 for all j = 1, 2, …, nwhere j = jth number of stage (j = 1, 2, …, n)

dj = decision variable at jth stage; aj = constant.Defining state variables, s1, s2 , …, sn such that: sn = a1d1 + a2d2 + … + andn = b sn–1 = a1d1 + a2d2 + … + an–1dn–1 = sn – andn

sj–1 = sj – ajdj

s1 = s2 – a2d2

In general, the state transition function takes the form: sj–1 = tj (sj, dj); j = 1, 2, …, n,i.e., a function of next state and decision variables.At the nth stage, sn is expressed as the function of the decision variables. Thus,

the maximum value of Z, denoted by f*n(sn) for any feasible value of sn is given by: f*n(sn) =

0Maximum

jd >{f1(d1) · f2(d2) · …· fn(dn)}

subject to the constraint sn = bBy keeping a particular value of dn constant, the maximum value of Z is given by: fn(dn) ∗

0Max

jd >{f1(d1) · f2(d2) ·…· fn–1(dn–1)} = fn(dn) ∗ f *n–1(sn–1); j = 1, 2, …, n – 1

The maximum value f*n–1(dn–1) of Z due to decision variables dj(j = 1, 2, …, n – 1) depends upon the state variable sn–1 = tn(sn, dn). The maximum of Z for any feasible value of all decision variables will be:

fj*(sj) = 0

Maxjd >

[fj(dj) ∗ f *j–1(sj–1)]; j = n, n – 1, …, 2

f1(s1) = f1(d1)where sj–1 = tj(sj, dj).

The value of f *j(sj) represents the general recursive equation.

Example 2. (Optimal subdivision problems) Divide quantity b into n parts so as to maximize their product. Let fn(b) be the maximum value. Then, show that:

f1(b) = b and fn(b) = 10{ ( )}nz b

Max z f b z−≤ ≤

−

Hence find fn(b) and the division that maximizes it.Sol. Let xj be the jth part of the quantity b(j = 1, 2, …, n). The problem then becomes: Maximize fn(b) = x1 · x2 · … · xn

subject to the constraints x1 + x2 + … + xn = b; xj > 0; j = 1, 2, …, nConsidering each part xj(j = 1, 2, …, n) of b as a stage. Since xj may assume any

positive value, satisfying the condition: x1 + x2 + … + xn = b, alternatives at each stage are infinite. Thus xj’s may be considered continuous variables. The recursive equation of the problem for all values of n can be obtained as follows:


NOTES


For n = 1, the above result becomes f1(b) = x or b (initially true).For n = 2, (i.e., two stage problem), the quantity b is divided into two parts, say

x1 = z and x2 = b – z.Then: f2(b) = Max{x1, x2} =

0Max { ( )}

z bz b z

< ≤−

= 10Max { ( )}

z bzf b z

< ≤− , since f1(b – z) = b – z

Similarly, for n = 3, the maximum product of b is divided into three parts, given the initial choice of z which leaves (b – z) to be further divided into two parts. We denote the maximum possible product for (b – z) into two parts by f2(b – z). Thus, using the principle of optimality, we have:

f3(b) = 20Max { ( )}

z bzf b z

< ≤−

Continuing in a similar manner, the recursive equation for general value of n is given by:

fn(b) = 10Max { ( )}nz b

zf b z−< ≤

− …(1)

Solution to the recursive equation: The solution to recursive Eq. (1), for getting the optimal policy, can be obtained with the help of differential calculus.

For n = 2, the function z(b – z) attains its maximum value for z = b/2, satisfying the condition 0 < z ≤ b.

Hence, the functional equation becomes:

f2(b) = 2

0Max ·

2 2 2z b

b b bb< ≤

− =

Then for n = 2, we have: Optimal policy: ,

2 2b b

and f2(b) = 2

2b

For n = 3, the functional equation becomes: f3(b) = 20

Max { ( )}z b

zf b x< ≤

−

= 2 32

0 0

( )Max Max2 4 3z b z b

b z b z bz z< ≤ < ≤

− − = =

The maximum value of 2

2b zz −

is attained for z = 3b , satisfying the condition

0 < z ≤ b because

f2(b – z) = 2 2

2 22 1 2 1

3 3 2 3 3bf b f b b b − = = =

.

Thus, for n = 3, we have:

Optimal policy: , ,3 3 3b b b

and f3(b) = 2

3b

Hence, in general, for an n-stage problem, we assume that:

Optimal policy: , , ,b b bn n n

…


NOTES

Operations Researchand fn(b) =

nbn

for n = 1, 2, …, m.

Now, by induction it can also be shown that the result holds good for n = m + 1. This method is discussed below.

For n = m + 1, the functional equation becomes: fm+1(b) =

0Max { ( )}mz b

zf b z< ≤

−

= 1

0 0

( )Max Max1( )

mm m

mz b z b

b z b z bz zm mm

+

< ≤ < ≤

− − = = +

The maximum value of mb zz

m−

is attained for z = 1b

m +, that is, the result is

also true for n = m + 1.

Hence, the required optimal policy is: , , ,b b bn n n

…

for fn(b) = nb

n

Remark: The maximum value z(b – z)2 was obtained by using the concept of maximum and minimum. For example, if:

f(z) = 2

2b zz −

,

then { ( )}d f zdz = 1

4 {2z(b – z) (– 1) + (b – z)2}

But for calculating maximum or minimum of f(z), we have to equate:

{ ( )}d f zdz = 0; i.e., – 2z(b – z) + (b – z)2 = 0

This gives z = b/3. Further, the second derivative of f(z), i.e., 2

2 { ( )}d f zdz

is negative,

at z = b/3. Hence, the maximum value of f(z) is obtained at z = b/3.

Example 3. Determine the value of u1, u2 and u3, so as to:Maximize Z = u1 · u2 · u3

subject to the constraints. u1 + u2 + u3 = 10

and u1, u2, u3 ≥ 0Sol. Let us define state variable xj(j = 1, 2, 3) such that: x3 = u1 + u2 + u3 = 10, at stage 3 x2 = x3 – u3 = u1 + u2, at stage 2 x1 = x2 – u2 = u1, at stage 1The maximum value of Z for any feasible value of state variable is given by: f3(x3) = Max

uu f x

33 2 2{ · ( )}

f2(x2) = Maxu

u f x2

2 1 1{ · ( )}

f1(x1) = u1 = x2 – u2


NOTES

Dynamic Programming and Theory of GamesThus, f2(x2) = Max Max

u uu x u u x u

2 22 2 2 2 2 2

2{ · ( )}− = −{ }Differentiating f2(x2) with respect to u2 and equating to zero (necessary conidtion

for maximum or minimum value of a function), we have: x2 – 2u2 = 0 or u2 = x2/2Now using Bellman’s principle of optimality, we get: f2(x2) = (x2/2) · x2 – (x2/2)2 = x2

2/4

and f3(x3) = Max Maxu u

u f x u x3 3

3 2 2 3 22 4{ · ( )} { · ( / )}=

= Max u x u3

3 32

4· ( )−

Again differentiating f3(x3) with respect to u3 and equating to zero we get:

14

{u3 · 2(x3 – u3)} (– 1) + (x3 – u3)2} = 0

(x3 – u3) (– 2u3 + x3 – u3) = 0 (x3 – u3) (x3 – 3u3) = 0Now either u3 = x3, which is trivial as u1 + u2 + u3 = x3 or x3 – 3u3 = 0, or u3 = x3/3

= 10/3. Therefore,

u2 = x x u2 3 32 2

1210 10

3103

=−

= −

=

u1 = x2 – u2 = 203

103

103

− =

Thus u1 = u2 = u3 = 10/3 and hence Max {u1 · u2 · u3} = (10/3)3 = 1,000/27.

Example 4. A company has decided to introduce a product in three phases. Phase 1 will feature making a special offer at a greatly reduced rate to attract the first-time buyers. Phase 2 will involve intensive advertising to persuade the buyers to continue purchasing at a regular price. Phase 3 will involve a follow up advertising and promotional campaign.

A total of ` 5 million has been budgeted for this marketing campaign. If m is the market share captured in Phase 1, fraction f2 of m is retained in Phase 2, and fraction f3 of market share in Phase 2 is retained in Phase 3. The expected values of m, f2 and f3 at different levels of money expended are given below. How should the money be allocated to the three phases in order to maximize the final share?

Money Spent (` Millions)Effect of Market Share

m percent f2 f3

0 0 0.30 0.501 10 0.50 0.702 15 0.70 0.853 22 0.80 0.904 27 0.85 0.935 30 0.90 0.95

Sol. This problem can be treated as a three-stage problem, taking each phase as a stage and amount of money spent as the state of the system. Let us adopt the following notations:


NOTES

Operations Research s = millions of rupees spent on marketing campaign at a particular stage xj = amount of money allocated to phase (stage) j; (j = 1, 2, 3) pj(xj) = market share captured in phase j; (j = 1, 2, 3)The problem can now be mathematically expressed as:Maximize Z = p1(x1) · p2(x2) · p3(x3)

subject to the constraint x1 + x2 + x3 = 5; x1, x2, x3 ≥ 0Using the forward induction approach, the recursive equation calculations are as

follows:

f *j(s) = Max0 1≤ ≤ −× −{ }

x s j n j jj

p s f s x( ) ( )*

Phase 1 (j = 1): If the whole amount is spent in Phase 1, that is x1 ≤ 5, the optimal return will be:

x*1 : 0 1 2 3 4 5

m = f*1(x1) : 0 10 15 22 27 30

Phase 2 (j = 2): If the whole amount is spent in Phase 2, then:

f* 2(s) = Max0 2 2 1 2

2≤ ≤× −{ }

x sf x f s x( ) ( )*

= Max0 5 2 2 1 2

2

5≤ ≤

× −{ }x

f x f x( ) ( )*

The computations for stages 2 and 3 are shown in Table 4.5 and 4.6 respectively.

Table 4.5. Phase 2 (j = 2)

Decision x2

States sf2(s, x2) = f2(x2) × f1*(s – x2) Optimal

Return f*2(s)

Optimal Decision

x*20 1 2 3 4 5

0 0 – – – – – 0 0

1 3.0 – – – – – 3.0 0

2 4.5 5.0 – – – – 5.0 1

3 6.6 7.5 7.0 – – – 7.5 1

4 8.1 11.0 10.5 8.0 – – 11.0 1

5 9.0 13.5 15.4 12.0 8.5 – 15.4 2

Table 4.6. Phase 3 (j = 3)

Decision x3

States s

f3(s, x3) = f3(x3) × f2*(s – x3) Optimal Return

f*3(s)

Optimal Decision x*30 1 2 3 4 5

5 7.7 7.7 6.37 4.5 2.79 – 7.7 0, 1

In Table 4.6, when x3 = 0, s = x1 + x2 = 5, and Max f*2(s) = 15.4 for which x2 = 2 and, therefore, x1 = 3. But for x3 = 1, s = x1 + x2 = 4, and Max f*2(s) = 11.0 for which x2 = 1 and, therefore, x1 = 3.


NOTES


Hence, the optimal policy to obtain a maximum market share of 7.7 per cent can be any one of the following:

(i) spend 3, 2, 0 million rupees in first, second and third phases (stage), respectively or,

(ii) spend 3, 1, 1 million rupees in first, second and third phases (stage), respectively.

Example 5. An electronic device consists of four components, each of which must be functional for the system to function. The system reliability can be improved by installing parallel units in one or more components. The reliability of components R, with one, two, or three parallel units, and the corresponding cost, C are given below. The maximum amount available for this device is 100. The problem is to determine the number of parallel units in each component.

Number of Parallel

units

Components

1 2 3 4

R C R C R C R C

1 0.70 10 0.50 20 0.70 10 0.60 20

2 0.80 20 0.70 40 0.90 30 0.70 30

3 0.90 30 0.80 50 0.95 40 0.90 40

Sol. The reliability of the given electronic device is the product of the reliabilities of each of its four components. If Rj and uj represent the reliability of the component j and units in parallel in component i, then the reliability of whole system, consisting of n components in series, will be:

R1u1 × R2u2 × … × Rnun.Since the objective is to maximize the reliability of the system, the problem can

be stated as:Maximize Z = R1u1 × R2u2 × … × Rnun

subject to the constraint c1u1 + c2u2 + … + cnun ≤ C

where cjuj = cost of component j C = total capital availableSince the electronic device consists of n = 4 components, for solving this problem

consider each components as a stage. The state at any stage will be the capital to be allocated. Let us adopt the following notations:

xj = capital allocated to stage j, through first stage inclusive. fj(xj) = return when available capital cj is allocated optimally over n stages

(components) Rjuj = reliability of component j(j = 1, 2, 3, 4)Now the recursive equation can be expressed as: fj(xj) = Max

0 1 1 2 2 1< < −× × …× −c u x n n j j j jj j j

R u R u R u f x c u[{ }· ( )]; j = 1, 2, 3, 4.


NOTES

Operations Research The electronic device in this problem will consist of at least one unit in each component. Thus, the range of investment, xj(j = 1, 2, 3, 4) in each case will be as follows:

c11 ≤ x1 ≤ C – c21 – c31 – c41 or 10 ≤ x1 ≤ 50 c11 + c12 ≤ x2 ≤ C – c31 – c41 or 30 ≤ x2 ≤ 70 c11 + c12 + c13 ≤ x3 ≤ C – c41 or 40 ≤ x3 ≤ 80 c11 + c12 + c13 + c14 ≤ x4 ≤ C or 60 ≤ x4 ≤ 100The computation of return at each stage, using the forward induction approach,

is shown in Table 4.7 to 4.10.

Table 4.7. Stage 1(j = 1)

Decision u1

States x1

f1(x1) = R1u1 Optimal Return f*1(x1)

Optimal Decision

u*1u

R C1 1

0 70 10=

= =. ;u

R C1 2

0 80 20=

= =. ;u

R C1 3

0 90 30=

= =. ;

10 0.70 – – 0.70 1

20 0.70 0.80 – 0.80 2

30 0.70 0.80 0.90 0.90 3

40 0.70 0.80 0.90 0.90 3

50 0.70 0.80 0.90 0.90 3

Table 4.8. Stage 2(j = 2) Decision u2

States x2

f2(x2) = R2u2 × f1*(x2 – c2x2) Optimal Return f*2(x2)

Optimal Decision

u*2

uR C

2 10 50 20

== =. ;

uR C

2 20 70 40

== =. ;

uR C

2 30 8 50

== =. ;0

30 0.5 × 0.7 = 0.35 – – 0.35 1

40 0.5 × 0.8 = 0.40 – – 0.40 1

50 0.5 × 0.9 = 0.45 0.7 × 0.7 = 0.49 – 0.49 2

60 0.5 × 0.9 = 0.45 0.7 × 0.8 = 0.56 0.8 × 0.7 = 0.56 0.56 2, 3

70 0.5 × 0.9 = 0.45 0.7 × 0.9 = 0.63 0.8 × 0.8 = 0.64 0.64 3


States x3

f3(x3) = R3u3 × f2*(x3 – c3x3) Optimal Return f *3(x3)

Optimal Decision

u*3u

R C3 1

0 70 10=

= =. ;u

R C3 2

0 90 30=

= =. ;u

R C3 3

0 95 40=

= =. ;

40 0.7 × 0.35 = 0.245 – – 0.245 1

50 0.7 × 0.40 = 0.280 – – 0.280 1

60 0.7 × 0.49 = 0.343 0.9 × 0.35 = 0.315 – 0.343 1

70 0.7 × 0.56 = 0.392 0.9 × 0.40 = 0.360 0.95 × 0.35 = 0.3325 0.392 1

80 0.7 × 0.64 = 0.448 0.9 × 0.49 = 0.441 0.95 × 0.40 = 0.3800 0.448 1


NOTES



States x4

f4(x4) = R4u4 × f3*(x4 – c4u4 ) Optimal Return f *4(x4)

Optimal Decision

u*4u

R C4 1

0 60 20=

= =. ;u

R C4 2

0 70 30=

= =. ;u

R C4 3

0 90 40=

= =. ;

60 0.6 × 0.245 = 0.147

– – 0.147 1

70 0.6 × 0.280 = 0.168

0.7 × 0.245 = 0.171

– 0.171 2

80 0.6 × 0.343 = 0.205

0.7 × 0.280 = 0.196

0.9 × 0.245 = 0.220

0.220 3

90 0.6 × 0.392 = 0.235

0.7 × 0.343 = 0.240

0.9 × 0.280 = 0.252

0.252 3

100 0.6 × 0.448 = 0.268

0.7 × 0.392 = 0.274

0.9 × 0.343 = 0.308

0.308 3

In Table 4.10, at stage 4, the value of return function f4(x4) is the maximum i.e., 0.308 at x4 = 100, and u4 = 3.

This makes x3 = 100 – 40 = 60. In Table 4.9, x3 = 60 corresponds to u3 = 1, and we are left with x2 = 60 – 10 = 50. In table 4.8, x2 = 50 corresponds to u2 = 2, and we are left with x1 = 50 – 40 = 10. In table 4.7, x1 = 10 corresponds to u1 = 1.

Hence, for maximum reliability, the device must have 1, 2, 1, and 3 units in components 1, 2, 3, and 4, respectively, in order to attain a maximum reliability of 0.308 or 30.8 per cent.

Example 6. Use dynamic programming to show that:p1 log p1 + p2 log p2 + … + pn log pn

subject to the constraintp1 + p2 + … + pn and pi ≥ 0, for all i

is minimum when p1 = p2 = … = pn = 1/n. Sol. The problem is to divide unity into n parts, p1, p2, …, pn such that the quantity

fn(1) = p pi ii

nlog

=∑1

is minimum, where fn(1) denotes the minimum attainable sum of pi log pi (i = 1, 2, …, n), when 1 is divided into n parts

For n = 1 (Stage 1) we have,f1(1) = p1 log p1 = 1 log 1, as unity is divided only into p1 = 1 part.

For n = 2 (Stage 2) the unity is to be divided into two parts p1 and p2 such that p1 + p2 = 1. If p1 = z and p2 = 1 – z, then:

f2(1) = Min0 1 1 1 2 2< ≤

+z

p p p p{ log log }

= Min0 1

1 1< ≤

+ − −z

z z z z{ log ( ) log ( )}

= Min0 1 1 1< ≤

+ −z

z z f z{ log ( )}


NOTES

Operations Research Similarly, in general, for an n-stage problem, where unity is to be divided into n parts, the recursive equation is:

fn(1) = Min0 1 1 1 2 2< ≤

+ +…+z n np p p p p p{ log log log }

= Min0 1 1 1< ≤ −+ −

z nz z f z{ log ( )}

Solution of recursive equation: The solution to the above recursive equation to get the optimal policy can be obtained with the help of differential calculus.

The minimum value of recursive equation of f(z) = z log(z) + (1 – z) log (1 – z) is attained at z = 1/2, satisfying the condition 0 < z ≤ 1. Hence, for stage 2, the optimal policy is p1 = p2 = 1/2 and

f2(1) = 12

12

1 12

1 12

2 12

12

log log log+ −

−

=

Similarly, for stage 3, the minimum value of the recursive equation f3(1) = Min

0 1 2 1< ≤+ −{ }

zz z f zlog ( )

= Min0 1

2 12

12< ≤

+−

−

z

z z z zlog log

is attained at z = 1/3, satisfying the condition 0 < z ≤ 1. Hence, for stage 3, the optimal policy is, p1 = p2 = 1/3 and

f3(1) = 13

13

21 1

32

1 132

log log

+

−

−

= 13

13

2 13

13

3 13

13

log log · log

+

=

Thus, in general, the optimal policy is: p1 = p2 = … = pn = 1/n

and fn(1) = nn n1 1log

The above result can be proved valid for any value of n by using mathematical induction. For n = m + 1, the recursive equation is:

fm+1(1) = Min0 1

1< ≤

+ −z mz z f z{ log ( )}

= Min0 1

1 1< ≤

+− −

z

z z m zm

zm

log log

Since the minimum of the function

F(z) = z log z + 1 1− −

zm

zm

log

is attained for z = 1/(m + 1), therefore using the concept of maximum and minimum, we have:

fm+1(1) = 11

11

11

11m m

mm m+ +

+ + +

log log


NOTES

Dynamic Programming and Theory of Games = ( ) logm

m m+

+ +

1 1

111

Hence, the result is also true for n = m + 1.

Example 7. Use dynamic programming to solve the following problem:

Minimize Z = y12 + y2

2 + y32

subject to the constraint. y1 + y2 + y3 = 10

and y1, y2, y3 ≥ 0Sol. Let the state variables s1, s2 and s3 such that: s3 = y1 + y2 + y3 = 15; s2 = y1 + y2 = s3 – y3; s1 = y1 = s2 – y2

The recursive equations can now be written as:

f3(s3) = Miny

y f s3

32

2 2{ ( )}+

f2(s2) = Miny

y f s2

22

1 1{ ( )}+

f1(s1) = Miny

y1

12{ } = y1

2 = (s2 – y2)2

The recursive equation f2(s2) can also be expressed as:

f2(s2) = Miny

y s y2

22

2 22{ ( ) }+ − ; f1(s1) = (s2 – y2)2

Now, by using the concept of maxima and minima in differential calculus, the minimum value of f2(s2) can be obtained as explained below:

Differentiating f2(s2) with respect to y2 and equating to zero, we get 2y2 – 2(s2 – y2) = 0 or y2 = s2/2

Thus, f1(s2) = s s s s2

2

22

222

2 2 2

+ −

=

f3(s3) = Min Miny y

y f s y s3 3

32

2 2 32 2

2

2+{ } = +

( )

= Miny

y s y3

32 2 3

2

2+

−

( )

Differentiating f3(s3) with respect to y3 and equating to zero to get its minimum value,

2y3 – (s3 – y3) = 0 or y3 = s3/3

Thus, f3(s3) = s s s s3

2

333

32

312 3 3

+ −

= .

But for minimum value of f3(s3) we must have y1 + y2 + y3 = 10. Therefore, f3(s3) = (10)2/3 = 100/3 and


NOTES

Operations Research y3 = s3/3 = 10/3 y2 = s2/2 = (s3 – y3)/2 = {10 – (10/3)}/2 = 10/3 y1 = s1/2 = (s3 – y3) – y2 = {10 – (10/3)}/2 – 10/3 = 10/3Hence, the minimum value of y1

2 + y22 + y3

2 is 100/3 for y1 = y2 = y3 = 10/3.

Example 8. Suppose there are n machines that can perform two jobs. If x number of them do the first job, then they produce goods worth g(x) = 3x, and if y number of them perform the second job, then they produce goods worth h(y)= 2.5y. Machines are subject to depreciation so that after performing the first job only a(x) = x/3 machines remain available and after performing the second job b(y) = 2y/3 machines remain available in the beginning of the second year. The process is repeated with the remaining machines. Obtain the maximum total return after three years and also find the optimal policy in each year.

Sol. Let us define the following notations n = stages, each year being viewed as a stage; n = 1, 2, 3 xn = number of machines devoted to job 1 in the year n yn = number of machines devoted to job 2 in the year n s = state variable, the number of machines at any stage (year) fj(s) = maximum return function when initial available machines are s with

n more stages (years) to goStage 1 (n = 1) In this problem, there are three stages (years). Using the backward

induction approach, i.e., n = 1 (third year), we have s = s3 number of machines at the beginning of the year. Then the return function can be expressed as:

f1(s3) = Maxx y

x y3 3

3 2 53 3,{ . }+

subject to the constraint x3 + y3 ≤ s3 and x3, y3 ≥ 0

where x3, y3 = number of machines devoted to jobs 1 and 2, respectively in the third year.Since function f1(s3) is a linear function in x3 and y3, its maximum can be obtained

by using the concept of extreme point solutions of LP problem as shown in Fig. 4.4.As shown in Fig 4.4, the maximum value of return function occurs at B(s3, 0), where f1(s3) = 3x3 + 2.5y3 = 3s3 + 2.5 × 0 = 3s3.Hence, the optimal decision at this stage is: x*3 = s3, y*3 = 0 and f*1(s3) = 3s3.

A(0, s )3

x + y = s33 3

(s , 0)3

0

y3

x3

Fig 4.4. Extreme Point Solution (Stage 1)


NOTES


Stage 2 (n = 2) Now consider the second year as the second stage. The return function at this stage is:

f2(s2) = Max Immediate return + Maximum known returnfromStage 2 fromStage1

= Maxx y

x y f x y2 2

3 2 53

232 2 1

2 2,

*( . )+ + +

subject to the constraint x2 + y2 ≤ s2 and x2, y2 ≥ 0

where x2 and y2 = number of machines devoted to jobs 1 and 2, respectively, in the second year

x2/3 and 2y2/3 = machines that will be available at the beginning of the next year.By the definition of f *1(s3) = 3s3 (Stage 1) the return function of this stage becomes:

f2(s2) = Maxx y

x y x y2 2

3 2 5 33

232 2

2 2,

.+ + +

= Maxx y

x y2 2

4 4 52 2,{ . }+

This again is a linear function in x2 and y2, and its maximum value occurs at corner point A(0, s2), as shown in Fig 4.5.

The maximum value of the return function at A(0, s2) is: f2(s2) = 4 × 0 + 4.5 × s2 = 4.5s2. Hence, the optimal decision at this stage is: x*2 = 0, y*2 = s2 and f*2(s2) = 4.5s2.

A (0, s )2

x + y = s22 2

B(s , 0)2

0

y2

x2


Stage 3 (n = 3) Now consider the first year as the third stage. The return function at this stage is:

f3(s1) = Immediate return + Maximum known returnfromStage3 fromStage 2

= Maxx y

x y f x y1 1

3 2 53

231 1 2

1 1,

*( . )+ + +

= Maxx y

x y x y1 1

3 2 5 4 53

231 1

1 1,

. .+ + +

; f2(s2) = 4.5s

= Maxx y

x y1 1

4 5 5 51 1,. .+{ }


NOTES

Operations Research subject to the constraint x1 + y1 ≤ s1 and x1, y1 ≥ 0This return function is also a linear function in x1 and y1, and its maximum value

occurs at corner point A(0, s1), as shown in Fig. 4.6. The maximum value of return function at A(0, s1) is:

f3(s1) = 4.5 × 0 + 5.5 × s1 = 5.5s1

A (0, s )1

x + y = s11 1

B(s , 0)1

0

y1

x1


Hence, the optimal decision at this stage is: x*1 = 0, y*1 = s1(= m machines, say) and f *3(s1) = 5.5m

The values of x2, x3 and y2, y3 can also be obtained in terms of m, as follows: y2 = s2 = 2y1/3 = 2m/3 and x2 = 0 x3 = s3 = 2y2/3 = 2/3(2m/3) = 4m/9 and y3 = 0Table 4.11 summarizes the optimal decision at each stage (year).

Table 4.11. Summary of the Optimal Decision

Stage Values of Decision Variable

3 (First year) x1 = 0, y1 = m2 (Second year) x2 = 0, y2 = 2m/31 (Third year) x3 = 4m/9, y3 = 0

The maximum possible return is 5.5 m.

Example 9. A company has five salesmen who have to be allocated to four marketing zones. The return (or profit) from each zone depends upon the number of salesmen working in that zone. The expected returns for different number of salesmen in different zones, as estimated from the past records, are given in the following table. Determine the optimal allocation policy.

Number of SalesmenMarketing Zones

1 2 30 45 30 351 58 45 452 70 60 523 82 70 644 93 79 725 101 90 82


NOTES


Sol. This problem can be treated as a three-stage problem, taking each marketing zone as a stage and the number of salesmen allocated to a particular zone as a state variable. Let us adopt the following notations:

s = total number of salesmen available xj = number of salesmen allocated to marketing zone (j = 1, 2, 3) pj(xj) = return from zone j when xj salesmen are allocated, j = 1, 2, 3Now, the given problem can be mathematically stated as: Maximize Z = p1(x1) + p2(x2) + p3(x3)

subject to the constrain x1 + x2 + x3 ≤ 5; and x1, x2, x3 ≥ 0 and integersTo obtain the recurrence relation, let there be s salesmen available for allocation to

all marketing zones. then fj(s) represents the optimal allocation of salesmen to marketing zone j, and xj represents the initial allocation to marketing zone j. Then the general recursive equation becomes:

fj(s) = Max0 1≤ ≤

++ −{ }x s j j j jj

p x f s x( ) ( )* ; j = 1, 2, 3

Using backward induction, we start by optimizing the last stage, stage 3 – marketing zone 3. Computations at each stage are shown below:

Stage 2(j = 3)

s: 0 1 2 3 4 5f *

3(s) : 35 45 52 64 72 82

x*3 : 0 1 2 3 4 5

Stage 3(j = 2) For this stage the recursive equation will become:

f2(s) = Max0 2 2 3 2

2≤ ≤+ −{ }

x sp x f s x( ) ( )*

The computations for stages 2 and 1 are shown in Tables 4.12 and 4.13 respectively.

Table 4.12. Computations for Stage 2

Decision x2

States s

f *2 (s) = p2(x2) + f2(s – x2)

Optimal Return

Optimal Decision x*

2

0 1 2 3 4 50 65 – – – – – 65 01 75 80 – – – – 80 12 82 90 95 – – – 95 23 94 97 105 105 – – 105 2, 34 102 109 112 115 114 – 115 35 112 117 124 122 124 125 125 5

Stage 1 (j = 1)Table 4.13. Computations for Stage 1

Decision x1

States s

f *1 (s) = p1(x1) + f*

2(s – x1)Optimal Return

Optimal Decision x*

1

0 1 2 3 4 55 170 173 175 177 173 166 177 3


NOTES

Operations Research In Table 4.13, the maximum return ` 177, corresponds to x1 = 3 salesmen, which leaves, s = 5 – 3 = 2 salesmen for two other stages. From Table 4.11, for s = 2, we have x2 = 2. This leaves s = 2 – 2 = 0, i.e., no salesman for stage 1 (marketing zone 3). Hence, the optimal solution, so obtained, is:

Stage (Marketing zone)

Value of decision Variable (Salesman)

3 (Zone 3) 02 (Zone 2) 21 (Zone 1) 3

Example 10. Find the minimum value of Z = y1

2 + y22 + … + yn

2

subject to the constraint y1 · y2 · … yn = c(c ≠ 0)

and yj ≥ 0; j = 1, 2, …, nSol. Let fn(c) be the minimum attainable value of Z when c is divided into n factors.

For n = 1, we have y1 = c and therefore, f1(c) = Miny c

y c1

12 2

={ } =

For n = 2, let y1 = x and y2 = c/x. Then

f2(c) = Min Min0 1

222

02 2

< ≤ < ≤+{ } = +

x c x cy y x c x{ ( / ) }

= Min0

21

2< ≤

+x c

x f c x{ ( / ) }

Since f1(c) = c2, we have f1(c/x) = (c/x)2.For n = 3, let y1 = x, y2 · y3 = c/x. Then,

f3(c) = Min Min0 1

222

32

02

2< ≤ < ≤+ +{ } = +

x c x cy y y x f c x{ ( / )}

Proceeding in the same way and using the principle of optimality, the recursive equation will become:

fn(c) = Min0

21< ≤ −+

x c nx f c x{ ( / )}; for all n

Solution of recursive equation: The solution of the recursive equation to get the optimal policy can be obtained with the help of differential calculus. Let f(x) = x2 + (c/x)2. Then,

dfdx

= 2x – 2 2

2cx

= 0

which gives x = (c)1/2; therefore, y1 = (c)1/2 and y2 = c/x = (c)1/2.Since the second derivative of f(x), with respect to x, is positive, f(x) is minimum.

Also,

f2(c) = Min0

2 2 1 2 21 2

2

< ≤+ = +

x c

x c x c cc

{ ( / ) } ( )/ / = 2c

Thus, f2(c/x) = 2(c/x)


NOTES


Hence the optimal policy is (c1/2, c1/2) and f2(c) = 2c. Again,

f3(c) = Min Min0

22 0

2 2< ≤ < ≤

+ = +x c x c

x f c x x c x{ ( / )} { ( / )}

= ( )/ /c cc

1 3 21 32+

= 3c2/3.

since minimum of f(x) = x2 + 2c/x was obtained at x = c1/3. Hence the optimal policy is: {c1/3, c1/3, c1/3} and f3(c) = 3c1/3.

Continuing in this manner, the optimal policy for an n stage problem will be: {c1/n, c1/n, …, c1/n} and fn(c) = nc2/n

Example 11. Use dynamic programming to solve the problem:Maximize Z = u1

2 + u22 + u3

2

subject to the constraint u1 · u2 · u3 = 6

and u1, u2, u3 ≥ 0 and integersSol. This problem can be considered as a three-stage problem. Let x3 = u1 · u2 · u3 = 6,

x2 = u1 · u2 = 63u , and x1 = u1 = x

u

2

2

Stage 1: For x1 = u1, where u1 can vary from 1 to 6, we have

f1(x1) = Min Min0 6 1

20 6 1

2

1≤ ≤ ≤ ≤{ } = { }

x ux u , where u1 integer.

u1 : 1 2 3 4 5 6

f *1(x1) : 1 4 9 16 25 36

Stage 2: For x2 = u1u2 = 6 where u1, u2 integers, and u1 can vary from 1 to 6, we have

u2 = 61u = 6 3 2 … 1 (depending on the value of u1)

f2(x2) = Max[u22 + f *

1(x1)] = Max uxu2

2 2

2

2

+

u1 → 1 2 3 6f1(x1) → 1 4 9 36

u2 u22

1 1 2* 5* 10 372 4 5* 8 133 9 10* 136 36 37*

The optimal values of f2(x2) are:

x2 : 1 2 3 6

f *2(x2) : 2 5 10 37


NOTES

Operations ResearchStage 3: For x3 = u1u2u3 = 6, and u3 = 6

1 2 2u uux

= , we have

f3(x3) = Max ux3

2

2

26+

= Max[u32 + f *2(x2)]

x3 → 1 2 3 6f *

1(x1) → 2 5 10 37u3 u3

2

1 1 3* 6* 11* 38*2 4 6* 9* 143 9 11* 146 36 38*

The optimal values of f3(x3) are:

x3 : 1 2 3 6f *

3(x3) : 3 6 11 38

Hence, for x3 = 6, Max, Z = 38 with u3 = 1 or 6.Proceeding backwards, following sets of values for u1, u2, u3 can be obtained

(u1, u2, u3) = (1, 6, 1); (1, 1, 6) (6, 1, 1).Example 12. A firm has divided its marketing area into three zones. The amount

of sales depends upon the number of salesmen in each zone. The firm has been collecting the data regarding sales and salesmen in each area over a number of past years.

Profit (in 1000’s ) is summarized in the table below. Next year only 9 salesmen are available, and the problem is to allocate these salesmen to three different zones so that the total sales are maximum.

No. of SalesmenZone

1 2 30 30 35 421 45 45 542 60 52 603 70 64 704 79 72 825 90 82 956 98 93 1027 105 98 1108 100 100 1109 90 100 110

Sol. Let three zones represent the three stages and the number of salesmen represent the state variables.

Stage 1 (zone 1): The amount of sales corresponding to different number of salesmen allocated to zone 1 are reproduced as follows:

No. of salesmen: 0 1 2 3 4 5 6 7 8 9Profit (000’ of `): 30 45 60 70 79 90 98 105 100 90


NOTES


Stage 2 (zone 1 and 2): Nine salesmen can be divided among two zones in 10 different ways: Each combination generates certain returns. The returns for all the salesmen are shown in Table 4.14. For a particular salesman, the profit for all possible combinations can be read along the diagonal. Maximum profit is marked by a star (*)

Table 4.14. Total ReturnsZone 1 → x1: 0 1 2 3 4 5 6 7 8 9

f1(x1): 30 45 60 70 79 80 98 105 100 90Zone 2

x2 f2(x2)

0 35 65* 80* 95* 105* 114 125* 133 140 135 1251 45 75 90 105* 115* 125 135* 143* 150 1452 52 82 97 112 122 131 142 150 1573 64 94 109 124 134 143* 154* 1624 72 102 117 132 142 151 1625 82 112 127 142 152 1616 93 123 138 153 163*7 98 128 143 1588 100 130 1459 100 130

Stage 3: This stage considers distribution of 9 salesmen in zones 1, 2 and 3, so as to decide allocation of certain number of salesmen to zone 3 and the remaining to zone 2 and 1 combined. Table 4.15 shows the returns obtained by allocating all to the three zones.

Table 4.15. Total Returns

Salesmen: 0 1 2 3 4 5 6 7 8 9

Total profit f2(x2) + f1(x1):

65 80 95 105 115 125 135 143 154 163

Salesmen in Zone 2 + Zone 1:

0 + 0 0 + 1 0 + 2 0 + 3 1 + 2

1 + 3 0 + 5 1 + 5 3 + 4 1 + 6

3 + 5 6 + 3

Salesmen in Zone 3: 9 8 7 6 5 4 3 2 1 0

Profit f3(x3): 110 110 110 102 95 82 70 60 54 42

Total profit, f3(x3) + f2(x2) + f1(x1):

175 190 205 207 210* 207 205 203 208 205

Since maximum profit for 9 salesmen is 2,10,000 provided 5 salesmen are allocated to zone 3 and from the remaining four, 1 is allocated to zone 2 and 3 to zone 1.

4.5 DYNAMIC PROGRAMMING APPROACH FOR SOLVING LINEAR PROGRAMMING PROBLEM

A linear programming problem in n decision variables and m constraints can be converted into an n stage dynamic programming problem with m states. Consider a general linear programming problem:

Maximize Z = j

n

j jc x=∑

1


NOTES

Operations Research subject to the constraints

j

n

ij ja x=∑

1≤ bi; i = 1, 2,…, m

and xj ≥ 0; j = 1, 2, …, nTo solve and LP problem using dynamic programming, the value of the decision

variable xj is determined at stage j(j = 1, 2, …, n). The value of xj at several stages can be obtained either by the forward or the backward induction method. The state variables at each stage are the amount resources available for allocation to the current stage and the succeeding stages.

Let a1j, a2j, …, amj be the amount (in units) of resources i(i = 1, 2,…,m) respectively, allocated to an activity cj at jth stage, and let fn(a1j, a2j, …, amj) be the optimum value of the objective function of a general LP problem for stages j, j + 1, …, n, and for states a1j, a2j, …, amj. Thus, the LP problem may be defined by a sequence of function as:

fn(a1j, a2j, …, amj) = Maxj

n

j jc x=∑1

This maximization is taken over the decision variable xj such that:

j

n

ij ja x=∑

1 ≤ bi; xj ≥ 0

The recursive relations for optimization are:fn(a1n, a2n, …, amn) or fn(b1, b2, …, bm)

= Max0 1 1 1 2 2≤ ≤ −+ − − −

a x b n n n n n n n m mn nin n i

c x f b a x b a x b a x{ ( , , ..., )}

The maximum value b that a variable xm can assume is:

b = Minba

ba

ban n

m

mn

1

1

2

2, , ,…

because the minimum value satisfies the set of a constraints simultaneously.Example 13. Use dynamic programming to solve the following linear programming

problem.Maximize Z = 3x1 + 5x2

subject to the constraints (i) x1 ≤ 4 (ii) x2 ≤ 6 (iii) 3x1 + 2x2 ≤ 18

and x1, x2 ≥ 0.Sol. This linear programming problem can be considered as a two-stage , three-

stage problem because there are two decision variables and three constraints with available resources.

The optimal value of f1(b1, b2, b3) at the first stage is given by: f1(b1, b2, b3) = Max

0 11

3≤ ≤

{ }x b

x , where b1 = 4, b2 = 6 and b3 = 18.

The feasible value of x1 is a non-negative value that satisfies all the given constraints x1 < b1 (= 4), 3x1 ≤ b3(= 18). Thus, the maximum value of b that x1 can assume is b = Min (4, 18/3) = 4. Therefore,

f1(4, 6, 18) = Max Min0 4 1 2

1

3 3 4 6 23≤ ≤

{ } = −

x

x x, ·


NOTES

Dynamic Programming and Theory of Gamesand x*

1 = Min 4 6 23 2, ·−

x

The recursive relation for optimization of this two-stage problem is:

f2(b1, b2, b3) = Max0 2 1 1 2 2 3 2

2

5 2≤ ≤

+ − −{ }x b

x f b b x b x*( , , )

where the maximization of x2, satisfying the conditions of x2 ≤ b2 (= 6) and 2x2 ≤ b3(= 18), is the minimum of b = min (6, 9) = 6. Therefore, the recurrence relationship can be expressed as:

f2(4, 6, 18) = Max0 6 2 1 2 2

2

5 4 6 18 2≤ ≤

+ − −x

x f x x{ ( , , )}*

= Max 3Min0 2 2

2

5 4 6 23≤ ≤

+ −

x b

x x,

Since, Min 4 6 23 2, ·−

x = 4 0 3

6 23

3 6

2

2 2

;

;

≤ ≤

− ≤ ≤

x

x x

we get Max 3Min5 4 6 232 2x x+ −

, = 5 12 0 318 3 3 6

2 2

2 2

x xx x

+ ≤ ≤+ ≤ ≤

;;

Now the maximum value of 5x2 + 12 = 27 at x2 = 3 and maximum value of 18 + 3x2 = 36 at x2 = 6.

Therefore, the optimal value of f *2 = (4 ,6, 18) = 36, is obtained at x2 = 6. Since,

x*1 = Min Min4 6 2

34 6 2

362, ,−

= − ×

x = 2

The optimum solution to the given LP problem is: x1 = 2, x2 = 6 and Max Z = 36.


1. State Bellman’s “principle of optimally” and explain with the help of an illustrative example how it can be used to solve a multistage decision problem.

2. Discuss dynamic programming with suitable examples. 3. Define the following dynamic programming terms: (a) Stage (b) State variable (c) Decision variable (d) Immediate return (e) Optimal return (f) State transformation function. 4. (a) What is the dynamic recursive relation? Describe the general process of backward

recursion. (b) Explain the recursive nature of computations in dynamic programming. 5. Why is it frequently desirable to solve a problem with a number of decision variables by

dividing it into a series of subproblems? 6. (a) Explain the concept of dynamic programming and the relation between dynamic and

linear programming approach. (b) Illustrate, with example, the method for solving a linear programming problem by the

dynamic programming approach.


NOTES

Operations Research 7. (a) Discuss dynamic programming with suitable examples. (b) Define a standard warehouse problem. Outline a procedure to solve it. 8. Discuss briefly: (a) The general similarities between dynamic programming and linear programming. (b) How does dynamic programming conceptually differ from linear programming? 9. In which areas of an organization can dynamic programming be applied successfully? Discuss.


Models I and II 1. Use dynamic programming to find the value of:

Max Z = y1 · y2 · y3

subject to the constraint y1 + y2 + y3 = 5 and y1, y2, y3 ≥ 0

2. Show how the functional equation technique of dynamic programming can be used to determine the shortest route when it is constrained to pass through a set of specified nodes, which is a definite subset of the set of nodes of a given network.

3. State Bellman’s principle of optimality and apply it to solve the following problem Max Z = x1 · x2 … xn

subject to the constraint x1 + x2 + … + xn = c; x1, x2, …, xn ≥ 0

4. A government space project is researching on a certain engineering problem that must be solved before a man can safely fly to the moon.

The research teams are currently trying three different approaches for solving the problems. An estimate has been made that, under present circumstances, the probability that the respective teams; say A, B and C will not succeed are 0.40, 0.60 and 0.80, respectively. Thus, the current probability that all three teams will fail is (0.40) (0.60) (0.80) = 0.192. Since the objective is to minimize this probability, the decision has been made to assign two or more top scientists among the three teams in order to reduce the probability of failure as much as possible.

The following table gives the estimated probability that the respective teams will fail when 0, 1 or 2 additional scientists are added to that team.

Number of new scientists

Team

. . .

. . .

. . .

A B C012

0 40 0 60 0 800 20 0 40 0 500 15 0 20 0 30

How should the additional scientists be allocated to the team? 5. A truck can carry a total 10 tonnes of a product. Three types of products are available for

shipment. Their weights and values are tabulated. Assuming that at least one of each type must be shipped, determine the type of loading that will maximize the total value.

Product Type Value (`) Weight (tonnes)A 20 1B 50 2C 50 3


NOTES


6. Consider the problem of designing an electronic device that consists of three main components. The components are arranged in series so that the failure of one of them will result in the failure of the whole device. Therefore, it is decided that the reliability (probability of failure) of the device should be increased by installing parallel units on each component. Each component may be installed in, at the most, three parallel units. The total capital (in thousand `) available for the device is 10. The following data is available:

Number of parallel Units, mi

Components

1 2 3

r1 c1 r2 c2 r3 c3

1 0.50 2 0.70 3 0.60 1

2 0.70 4 0.80 5 0.80 2

3 0.90 5 0.90 6 0.90 3

where ri, ci (i = 1, 2, 3) is the reliability and the cost of the ith component, respectively. Determine the number of parallel units that will maximize the total reliability of the system, without exceeding the given capital.

Model III 7. Use the principle of optimality to find the maximum value of:

Max Z = b1x1 + b2x2 + … + bnxn

subject to the constraint x1 + x2 + … + xn = c; x1, x2, …, xn ≥ 0

8. A student has to take an examination in three courses x, y and z. He has three days for studying. He feels that it would be better to devote a whole day to study one single course. So he may study a course for one day, two days or three days or not at all. His estimates of grades he may get according to days of study he puts in are as follows:

Study DaysCourse

x y z

0 1 2 1

1 2 2 2

2 2 4 4

3 4 5 4

How many days should he allocate to each course so that he gets the best results. 9. A chairman of a certain political party is planning his election to the Parliament. He has

engaged the services of six volunteer workers and wishes to assign them four districts in such a way as to maximize their effectiveness. He feels that it would be inefficient to assign one worker or more than one district but he is also willing to assign no worker to any one of the districts, judging by what the workers can accomplish in other districts.

The following table gives the estimated increase in the number of votes in favour of the party’s candidate if it were allocated to the various workers:


NOTES

Operations ResearchNumber of workers

Districts

1 2 3

0 0 0 0

1 25 20 33

2 42 38 43

3 55 54 47

4 63 65 50

5 69 73 52

6 74 80 53

How many workers should be assigned to each of the three districts in order to maximize the total number of votes in his favour?

10. Min Z = y21 + y2

2 + y23 subject to the constraint

y1 + y2 + y3 ≥ 15; y1, y2, y3 ≥ 0 11. Min Z = x2

1 + 2x22 + 4x3

subject to the constraint x1 + 2x2 + x3 ≤ 8; x1, x2, x3 ≥ 0

12. Develop the functional equation to determine m1, m2, …, mn so as to:

Max Z = m p mi i ni

n( / )α

=∑1

subject to the constraint m1 + m2 + … + mn = M

13. We have a machine that deteriorates with age and so we need to formulate a replacement policy for the same. We have to own such a machine during each of the next years. The operating cost c(i) of a machine i years old at the beginning of the year, trade in value t(i) received when such a machine is traded for a new machine at the start of the year and s(i), the salvage value received for a machine that has just turned age i at the end of 5 years are given below:

i 0 1 2 3 4 5 6c(i) 10 13 20 40 70 100 100t(i) – 32 21 11 5 0 0s(i) – 25 17 8 0 0 0

If a new machine costs ` 50 and at present we have a machine that is two years old, what is the optimum policy of replacement? Solve the problem using dynamic programming.

14. The following table provides the monthly demand for an item of a company during the winter season. Virtually all direct labour is casual and temporary so any number of items up to the plant capacity of 400 items can be produced each month. The direct labour cost is ` 40 per item.

Month November December January FebruaryDemand 100 200 300 400

The plant is completely shut down during a month where there is no production, and supervisory personnel are given an unpaid holiday to save a further ` 2,000 in monthly payroll costs. Demands may be filled either from current production or from inventory. Each item held in inventory at a beginning of the month costs the company ` 5. The production lot size may be 0, 100, 200, 300 or 400 items. Determine the monthly production levels in such a way that the total cost is minimized.


NOTES


Hints and Answers 1. Recursive equation

f *j(sj) = Max{zj × fj–1(sj – zj)}; j = 1, 2, 3where zj = jth part of quantity 5 (stage j), and sj = state variable at jth stageOptimal policy: y1 = y2 = y3 = 5/3 and f3(5) = (5/3)3

4. Recursive equation

f *j = Min

x s j j j jj

p x f s x≤ −× −{ }( ) ( )*

1 ; j = 1, 2, 3

where xj = number of additional scientists allocated to team j (stage j) s = number of new scientists available for assignment pj(xj) = probability of failure for team j if it is assigned xj additional

scientistsOptimal policy: x1 = 1, x2 = 0, x3 = 1, and probability that three teams will fail is 0.060.


f *j = Min

x s j j j jj

p d f s d≤ +× −{ }( ) ( )*

1 ; j = 1, 2, 3

where dj = weight (in tonnes) to be loaded on jth of product pj(dj) = expected value obtained from allocation of dj tonnes of weight s = total weight (= 10 tonnes) which a truck can carry.Optimal policy: d1 = 3(product A); d2 = 1 (product B) d3 = 2(product C) with maximum return, f *

3(s) = ` 270 6. Recursive equation

f *n(xn) = Max{rn(cn) × fn–1(xn – xn)}; n = 1, 2, 3

where c = total capital available cn = capital allocated at stage n(state) cj = capital cost of jth component (stage)Optimal policy: m1 = 3, m2 = 1 and m3 = 2, and maximum reliability is 0.504.


fj(c) = Max0 0 1≤ ≤ −+ −{ }

z j jj

b z f c z( ) ; j = 1, 2, …,n

where xj = z, i.e., z is the jth part of c.Apply forward process to get optimal policy: x1 = x2 = … = xn–1 = 0, xn = c, and fn(c) = bnc.

8. Recursive equation fj(sj) = Max{fj(dj)} + fj–1(sj–1); j = 2, 3i.e., f1(s1) = f1(d1)where fj(dj) = grades earned by devoting dj number of days; j = 1, 2, 3 sj = state variable, number of days available for studyThe state variable is defined as s3 = d1 + d2 + d3 ≤ 3; s2 = d1 + d2

= s3 – d3; s1 = d1 = s2 – d2

Apply backward process to arrive at an optimum policy:d1 = 1, d2 = 0, d3 = 2 and maximum return of 8.


NOTES

Operations Research 9. Recursive equation fj(s) = Max

0 1≤ ≤ ++ −{ }x s j j j jj

E x f s x( ) ( ) ; j = 1, 2, 3

f1(s) = E1(s1)where xj = state variable, the number of workers available for district (stage)

j (j = 1, 2, 3) Ej(xj) = expected number of votes when xj workers are assigned to district

j(j = 1, 2, 3) s = total number of workers, i.e., x1 + x2 + x3 = 6

Apply the backward process (i.e., start from district 3) to arrive at the optimal policy: x1 = 2, x2 = 3 and x3 = 1; maximum increase in number of votes is 129.

10. Recursive equation: fj(sj) = Max

02

1≤ ≤ −+ −{ }y s j j j jj j

y f s y( ) ; j = 1, 2, 3

f1(s1) = y12

where s3 = y1 + y2 + y3 ≥ 15; s3 = y1 + y2 = s3 – y3; s3 = y1 = s2 – y3;Apply the forward induction method to get the optimal policy: y1 = y2 = y3 = 5 and Min Z = 75.

14. The recursive equation: fn(s) = Min{c(s, xn) + fn+1(s + xn – dn)}

where c(s, xn) = 55 2 000 40 055 0

+ + >=

, ;;

x xx

n n

n

ifif

s + xn – dn = number of items on hand at the end of month n; xn = production quantity;

s = stock of itemOptimal policy the value of the cost function for stage 1, f1(0) = ` 47,000 corresponds to the optimal policy: x1 = 300, x2 = 0, x3 = 300 and x4 = 400.

B. THEORY OF GAMES

4.6 TWO-PERSON ZERO-SUM GAMESA game with only two players, say A and B, is called a two-person zero-sum game,

only if one player’s gain is equal to the loss of other player, so that total sum is zero.Payoff matrix The payoffs (a quantitative measure of satisfaction that a player

gets at the end of the play) in terms of gains or losses, when players select their particular strategies (courses of action), can be represented in the form of a matrix, called the payoff matrix. Since the game is zero-sum, the gain of one player is equal to the loss of other and vice versa. In other words, one player’s payoff table would contain the same amounts in payoff table of other player, with the sign changed. Thus, it is sufficient to construct a payoff table only for one of the players.

If player A has m strategies represented by the letters: A1, A2, . . ., Am and player B has n strategies represented by the letters: B1, B2, . . ., Bn. The numbers m and n need


NOTES


not be equal. The total number of possible outcomes is therefore m × n. It is assumed that each player not only knows his own list of possible strategies but also of his competitor. For convenience, it is assumed that player A is always a gainer whereas player B a loser. Let aij be the payoff that player A gains from player B if player A chooses strategy i and player B chooses strategy j. Then the payoff matrix is shown in the Table 4.16.

Table 4.16. Payoff Matrix

Player A’s StrategiesPlayer B’s Strategies

B1 B2 … Bn

A1 a11 a12 … a1n

A2 a21 a22 … a2n

Am am1 am2 … amn

Since player A is assumed to be the gainer, therefore he wishes to gain as large a payoff aij as possible, player B on the other hand would do his best to reach as small a value of aij as possible. Of course, the gain to player B and loss to A must be – aij.

Various methods discussed in this chapter to find value of the game under decision-making environment of certainty are as follows:

Assumptions of the Game 1. Each player has available to him a finite number of possible strategies (courses of

action). The list may not be the same for each player. 2. Players act rationally and intelligently. 3. List of strategies of each player and the amount of gain or loss on an individual’s

choice of strategy is known to each player in advance. 4. One player attempts to maximize gains and the other attempts to minimize losses. 5. Both players make their decisions individually, prior to the play, without direct

communication between them.


NOTES

Operations Research 6. Both players select and announce their strategies simultaneously so that neither player has an advantage resulting from direct knowledge of the other player’s decision.

7. The payoff is fixed and determined in advance.

4.7 MINIMAX AND MAXIMIN PRINCIPLES: GAMES WITH SADDLE POINT

The selection of an optimal strategy by each player, without the knowledge of the competitor’s strategy, is the basic problem of playing games. Since the payoffs for either player provides all the essential information, therefore, only one player’s payoff table is required to evaluate the decisions. By convention, the payoff table for the player whose strategies are represented by rows (say player A) is constructed. The objective of the study is to know how these players must select their respective strategies so that they are able to optimize their payoff. Such a decision-making criterion is referred to as the minimax-maximin principle. Such principle in pure strategies game always leads to the best possible selection of a strategy for both players.

Maximin principle For player A the minimum value in each row represents the least gain (payoff) to him, if he chooses his particular strategy. These are written in the matrix by row minima. He will then select the strategy that gives the largest gain among the row minimum values. This choice of player A is called the maximin principle, and the corresponding gain is called the maximin value of the game.

Minimax principle For player B, who is assumed to be the looser, the maximum value in each column represents the maximum loss to him, if he chooses his particular strategy. These are written in the payoff matrix by column maxima. He will then select the strategy that gives the minimum loss among the column maximum values. This choice of player B is called the minimax principle, and the corresponding loss is the minimax value of the game.

Optimal strategy A course of action that puts any player in the most preferred position, irrespective of the course of action his competitor(s) adopt, is called as optimal strategy. In other words, if the maximin value equals the minimax value, then the game is said to have a saddle (equilibrium) point and the corresponding strategies are called optimal strategies.

Value of the game This is the expected payoff at the end of the game, when each player uses his optimal strategy, i.e., the amount of payoff, V, at an equilibrium point. A game may have more than one saddle points. A game with no saddle point is solved by choosing strategies with fixed probabilities.Remarks 1. The value of the game, in general, satisfies the equation: maximin value ≤ V ≤

minimax value. 2. A game is said to be a fair game if the lower (maximin) and upper (minimax) values

of the game are equal and both equals zero. 3. A game is said to be strictly determinable if the lower (maximin) and upper

(minimax) values of the game are equal and both equals the value of the game.

Maximin principle maximize the player’s minimum gains.

Value of the game is the expected gain or loss in a game when a game is played a large number of times.


NOTES


4.7.1 Rules to Determine Saddle PointThe reader is advised to follow the following three steps, in this order, to determine

the saddle point in the payoff matrix. 1. Select the minimum (lowest) element in each row of the payoff matrix and write

them under ‘row minima’ heading. Then, select the largest element among these elements and enclose it in a rectangle, .

2. Select the maximum (largest) element in each column of the payoff matrix and write them under ‘column maxima’ heading. Then select the lowest element among these elements and enclose it in a circle, .

3. Find out the element(s) that is same in the circle as the well as rectangle and mark the position of such element(s) in the matrix. This element represents the value of the game and is called the saddle (or equilibrium) point.

Example 1. For the game with payoff matrix:

Player APlayer B

B1 B2 B3

A1 – 1 2 – 2A2 6 4 – 6

determine the optimal strategies for players A and B. Also determine the value of game. Is this game (i) fair? (ii) strictly determinable?

Sol. In this example, gains to player A or losses to player B are represented by the positive quantities, whereas, losses to A and gains to B are represented by negative quantities. It is assumed that A wants to maximize his minimum gains from B. Since the payoffs given in the matrix are what A receives, therefore, he is concerned with the quantities that represent the row minimums. Now A can do no worse than receive one of these values. The best of these values occurs when he chooses strategy A1. This choice provides a payoff of – 2 to A when B chooses strategy B3. This refers to A’s choice of A1 as his maximum payoff strategy because this row contains the maximum of A’s minimum possible payoffs from his competitor B.

Table 4.17.

Player APlayer B

Row minimumB1 B2 B3

A1 – 1 2 – 2 – 2 ← Maximin

A2 6 4 – 6 – 6

Column maximum 6 4 – 2 ← Minimax

Similarly, it is assumed that B wants to minimize his losses and wishes that his losses to A be as small as possible. The column maximums also represent the greatest payments B might have to make to A. The smallest of these losses is – 2, which occurs when A chooses his course of action, A1 and B chooses his course of action, B3. This choice of B3 by B is his minimax loss strategy because the amount of this column is the minimum of the maximum possible losses.

Saddle point is the payoff value that represents both minimax and maximin value of the game.


NOTES

Operations Research In Table 4.17, the quantity – 2 in row A1 and column B3 is enclosed both in the box and the circle. That is, it is both the minimum of the column maxima and the maximum of the row minima. This value is referred to as saddle point.

The payoff amount in the saddle-point position is also called value of the game. For this game, value of the game is, V = – 2, for player A. The value of game is always expressed from the point of view of the player whose strategies are listed in the rows.

The game is strictly determinable. Also since the value of the game is not zero, the game is not fair.

Example 2. Find the range of values of p and q that will render the entry (2, 2) a saddle point for the game.

Player APlayer B

B1 B2 B3

A1 2 4 5A2 10 7 qA3 4 p 6

Sol. First ignore the values of p and q in the payoff matrix, and then determine the maximin and minimax values in the usual manner, as shown in Table 4.19.

Table 4.19.

Player APlayer B

Row minimumB1 B2 B3

A1 2 4 5 2

A2 10 7 q 7 ← Maximin

A3 4 p 6 4

Column maximum 10 7 6↑ Minimax

As shown in Table 4.19, since there exists no unique saddle point, therefore, the saddle point will exist at the position (2, 2) only when p ≤ 7 and q > 7.


1. Explain: Minimax and Maximin principle used in the theory of games. 2. What is a game in game theory? What are the properties of a game? Explain the ‘best

strategy’ on the basis of minimax criterion of optimality. 3. Explain the two-person zero-sum game, giving a suitable example. 4. Define ‘saddle point’. Is it necessary that a game should always possess a saddle point? 5. Let A = [aij] be the payoff matrix for a two-person zero-sum game. If V denotes the minimax

value of the game, then show that V ≥ V. 6. Define: (i) Competitive game; (ii) Pure strategies; (iii) Mixed strategies (iv) Two-person

zero-sum (or rectangular) game, (v) Payoff matrix. 7. State the major limitations of the game theory. 8. Explain the difference between pure strategy and mixed strategy.


NOTES


9. Which competitive situation is called a game? What is the maximin criterion of optimality? 10. What are the assumptions made in the theory of games? 11. Describe the maximin principle of game theory. What do you understand by pure strategies,

and ‘saddle point’. 12. Game theory provides a systematic quantitative approach for analysing competitive

situations in which the competitors make use of logical processes and techniques in order to determine an optimal strategy for winning: Comment.


1. Consider the game with the following payoff table:

Player APlayer B

B1 B2

A1 2 6A2 – 2 λ

(a) Show that the game is strictly determinable, whatever λ may be. (b) Determine the value of the game. 2. Determine which of the following two-person zero-sum games are strictly determinable and

fair. Give the optimum strategies for each player in the case of it being strictly determinable.

(a)Player A

Player B (b)Player A

Player B

B1 B2 B1 B2

A1 1 2 A1 – 5 2

A2 4 – 3 A2 – 7 – 4

3. Solve the following games by using maximin (minimax) principle, whose payoff matrix are given below: Include in your answer: (i) strategy selection for each player, (ii) the value of the game to each player. Does the game have a saddle point?

(a)Player A

Player B

B1 B2 B3 B4

A1 1 7 3 4

A2 5 6 4 5

A3 7 2 0 3

(b)Player A

Player B

B1 B2 B3 B4 B5

A1 – 2 0 0 5 3

A2 3 2 1 2 2

A3 – 4 – 3 0 – 2 6

A4 5 3 – 4 2 6


NOTES

Operations Research (c)Player A

Player B

B1 B2 B3 B4

A1 3 – 5 0 6

A2 – 4 – 2 1 2

A3 5 4 2 3

4. (a)Player A

Player B

B1 B2 B3

A1 – 2 15 – 2

A2 – 5 – 6 – 4

A3 – 5 20 – 8

(b)Player A

Player B

B1 B2 B3 B4

A1 – 5 3 1 10

A2 5 5 4 6

A3 4 – 2 0 –5

5. Firm AFirm B

B1 B2 B3 B4 B5

A1 3 – 1 4 6 7

A2 – 1 8 2 4 12

A3 16 8 6 14 12

A4 1 11 – 4 2 1

6. Two competitive manufacturers are producing a new toy under licence from a patent holder. In order to meet the demand they have the option of running the plant for 8, 16 or 24 hours a day. As the length of production increases so does the cost. One of the manufacturers, say A, has set up the matrix given below. He uses the matrix to estimate the percentage of the market that he could capture and maintain the different production schedules:

Manufacturer AManufacturer B

C1 : 8 hrs C2 : 16 hrs C3 : 24 hrs

S1 : 8 hrs 60% 56% 34%

S2 : 16 hrs 63% 60% 55%

S3 : 24 hrs 83% 72% 60%

(i) At which level should each produce? (ii) What percentage of the market will B have? 7. Two computer manufacturers A and B are attempting to sell computer systems to two

banks 1 and 2. Company A has 4 salesmen, company B only has 3 salesmen available. The


NOTES


computer companies must decide upon how many salesmen to assign to sell computer to each bank. Thus company A can assign 4 salesmen to bank 1 and none to bank 2 or three to bank 1 and one to bank 2, etc.

Each bank will buy one computer system. The probability that a bank will buy from a particular computer company is directly related to the number of salesmen calling from that company, relative to the total salesmen calling. Thus, if company A assigns three salesmen to bank 1 and company B assigns two salesmen, the odds would be three out of five that bank 1 would purchase company A’s computer system. (If none calls from either company the odds are one-half for buying either computer.)

Let the payoff be the expected number of computer systems that company A sells. (2 minus this payoff is the expected number company B sells).

What strategy would company A use in allocating its salesmen? What strategy should company B use? What is the value of the game to company A? What is the meaning of the value of the game in this problem?

8. Assume that two firms are competing for the market share for a particular product. Each firm is considering what promotional strategy to employ for the coming period. Assume that the following payoff matrix describes the increase in market share of Firm A and the decrease in market share for Firm B. Determine the optimal strategies for each firm.

Firm AFirm B

Much PromotionNo Promotion Moderate

PromotionNo Promotion 5 0 –10Moderate Promotion 10 6 2Much Promotion 20 15 10

(i) Which firm would be the winner, in terms of market share? (ii) Would the solution strategies necessarily maximize the profits for either of the firms?

(iii) What might the two firms do to maximize their profits?

Hints and Answers 1. (a) Ignoring the value of λ, the maximin value = 2 = minimax. Thus the game is strictly

determinable. (b) Value of game is 2 with optimal strategy for player A and B is A1 and B1, respectively. 2. (i) Not fair, value of game = 1 (ii) Not fair, value of game = – 5 3. (a) Optimal strategy, A : A2 ; B : B3, Value of game = 4 (b) Optimal strategy, A : A2 ; B : B3, Value of game = 1 (c) Optimal strategy, A : A3 ; B : B3, Value of game = 2 4. (a) Two points at (A1, B1), (A1, B3), Value of game = – 2 (b) Optimal strategies: (A2, B3), Value of game = 4 5. Optimal strategy: (A3, B3), Value of game = 6 6. (i) Optimal strategy: (S3, C3), i.e., both should produce at the level of 24 hours per day. (ii) At the optimal level of 24 hours per day, B will have 60 per cent of the market share.


NOTES

Operations Research 4.9 DOMINANCE PROPERTIESThe rules of dominance are used to reduce the size of the payoff matrix. These rules

help in deleting certain rows and/or columns of the payoff matrix that are inferior (less attractive) to at least one of the remaining rows and/or columns (strategies), in terms of payoffs to both the players. Rows and/or columns once deleted can never be used for determining the optimum strategy for both the players.

The rules of dominance are especially used for the evaluation of two-person zero-sum games without a saddle (equilibrium) point. Certain dominance principles are stated as follows: 1. For player B, who is assumed to be the loser, if each element in a column, say Cr

is greater than or equal to the corresponding element in another column, say Cs in the payoff matrix, then the column Cr is said to be dominated by column Cs and therefore, column Cr can be deleted from the payoff matrix. In other words, player B will never use the strategy that corresponds to column Cr because he will loose more by choosing such strategy.

2. For player A, who is assumed to be the gainer, if each element in a row, say Rr, is less than or equal to the corresponding element in another row, say Rs, in the payoff matrix, then the row Rr is said to be dominated by row Rs and therefore, row Rr can be deleted from the payoff matrix. In other words, player A will never use the strategy corresponding to row Rr, because he will gain less by choosing such a strategy.

3. A strategy say, k can also be dominated if it is inferior (less attractive) to an average of two or more other pure strategies. In this case, if the domination is strict, then strategy k can be deleted. If strategy k dominates the convex linear combination of some other pure strategies, then one of the pure strategies involved in the combination may be deleted. The domination would be decided as per rules 1 and 2 above.

Remark Rules (principles) of dominance discussed are used when the payoff matrix is a profit matrix for the player A and a loss matrix for player B. Otherwise the principle gets reversed.

Example 1. Players A and B play a game in which each has three coins, a 5p, 10p and a 20p. Each selects a coin without the knowledge of the other’s choice. If the sum of the coins is an odd amount, then A wins B’s coin. But, if the sum is even, then B wins A’s coin. Find the best strategy for each player and the values of the game.

Sol. The payoff matrix for player A is

Player APlayer B

5p : B1 10p : B2 20p : B3

5p : A1 – 5 10 2010p : A2 5 – 10 – 1020p : A3 5 – 20 – 20

It is clear that this game has no saddle point. Therefore, further we must try to reduce the size of the given payoff matrix as further as possible. Note that every element

Dominance Property is the procedure to reduce the size of the payoff matrix according to the tendency of players.

7. Payoff matrix for company A

Company ACompany B

B1 B2 B3 B4

A1 1/2 0 0 0

A2 1 1/2 1/3 1/4

A3 1 2/3 2/4 2/5

A4 1 3/5 3/5 3/6

A5 1 4/5 4/6 4/7

Optimal strategy for A is A5 and for B is B4. Value of game = 4/7, i.e., probability of success for A is 4/7 or 51 per cent supply.

8. Optimal strategy for both A and B is ‘Much promotion’; Value of game = 10.

4.8 GAME WITHOUT SADDLE POINTIn certain cases, no saddle point exists, i.e., maximin value ≠ minimax value. In

all such cases, players must choose the mixture of strategies to find the value of game and an optimal strategy.

The value of game obtained by the use of mixed strategies represents the least payoff, which player A can expect to win and the least which player B can expect to lose. The expected payoff to a player in a game with payoff matrix [aij] of order m × n is defined as:

E( p, q) = p a qi ij jj

n

i

m

==∑∑

11 = PT AQ (in matrix notation)

where P = ( p1, p2, ..., pm) and Q = (q1, q2, ..., qn) denote probabilities (or relative frequency with which a strategy is chosen from the list of strategies) associated with m strategies of player A and n strategies of player B respectively, where p1 + p2 + … + pm = 1 and q1 + q2 + … + qn = 1.

A mixed strategy game can be solved by using following methods: • Algebraic method • Analytical or calculus method • Matrix method • Graphical method, and • Linear programming method.

Remark For solving a 2 × 2 game, without a saddle point, the following formula is also used. If payoff matrix for player A is given by:

Player A

Player Ba aa a11 12

21 22

then the following formulae are used to find the value of game and optimal strategies:

p1 = a aa a a a

22 21

11 22 12 21

−+ − +( )

; q1 = a aa a a a

22 12

11 22 12 21

−+ − +( )

where p2 = 1 – p1; q2 = 1 – q1

and V = a a a aa a a a

11 22 21 12

11 22 12 21

−+ − +( )


NOTES


4.9 DOMINANCE PROPERTIESThe rules of dominance are used to reduce the size of the payoff matrix. These rules

help in deleting certain rows and/or columns of the payoff matrix that are inferior (less attractive) to at least one of the remaining rows and/or columns (strategies), in terms of payoffs to both the players. Rows and/or columns once deleted can never be used for determining the optimum strategy for both the players.

The rules of dominance are especially used for the evaluation of two-person zero-sum games without a saddle (equilibrium) point. Certain dominance principles are stated as follows: 1. For player B, who is assumed to be the loser, if each element in a column, say Cr

is greater than or equal to the corresponding element in another column, say Cs in the payoff matrix, then the column Cr is said to be dominated by column Cs and therefore, column Cr can be deleted from the payoff matrix. In other words, player B will never use the strategy that corresponds to column Cr because he will loose more by choosing such strategy.

2. For player A, who is assumed to be the gainer, if each element in a row, say Rr, is less than or equal to the corresponding element in another row, say Rs, in the payoff matrix, then the row Rr is said to be dominated by row Rs and therefore, row Rr can be deleted from the payoff matrix. In other words, player A will never use the strategy corresponding to row Rr, because he will gain less by choosing such a strategy.

3. A strategy say, k can also be dominated if it is inferior (less attractive) to an average of two or more other pure strategies. In this case, if the domination is strict, then strategy k can be deleted. If strategy k dominates the convex linear combination of some other pure strategies, then one of the pure strategies involved in the combination may be deleted. The domination would be decided as per rules 1 and 2 above.

Remark Rules (principles) of dominance discussed are used when the payoff matrix is a profit matrix for the player A and a loss matrix for player B. Otherwise the principle gets reversed.

Example 1. Players A and B play a game in which each has three coins, a 5p, 10p and a 20p. Each selects a coin without the knowledge of the other’s choice. If the sum of the coins is an odd amount, then A wins B’s coin. But, if the sum is even, then B wins A’s coin. Find the best strategy for each player and the values of the game.

Sol. The payoff matrix for player A is

Player APlayer B

5p : B1 10p : B2 20p : B3

5p : A1 – 5 10 2010p : A2 5 – 10 – 1020p : A3 5 – 20 – 20

It is clear that this game has no saddle point. Therefore, further we must try to reduce the size of the given payoff matrix as further as possible. Note that every element

Dominance Property is the procedure to reduce the size of the payoff matrix according to the tendency of players.


NOTES

Operations Research Since player A is the gainer player and expects at least V, therefore, we must have

Player APlayer B

ProbabilityB1 B2 … Bn

A1 a11 a12 … a1n p1

A2 a21 a22 … a2n p2

Am am1 am2 … amn pm

Probability q1 q2 … qn

a11 p1 + a21 p2 + … + am1 pm ≥ V a12 p1 + a22 p2 + … + am2 pm ≥ V …(1) a1n p1 + a2n p2 + . . . + amn pm ≥ V

where p1 + p2 + … + pm = 1 and pi ≥ 0 for all iSimilarly, the expected loss to player B, when player A selects strategies

A1, A2, ..., Am, one by one, can also be determined. Since player B is the loser player, therefore, he must have:

a11 q1 + a12 q2 + … + a1n qn ≤ V a21 q1 + a22 q2 + … + a2n qn ≤ V …(2) am1 q1 + am2 q2 + … + amn qn ≤ V

where q1 + q2 + … + qn = 1 and qj ≥ 0 for all jTo get the values of pi’s and qj’s, the above inequalities are considered as equations

and are then solved for given unknowns. However, if the system of equations, so obtained, is inconsistent, then at least one of the inequalities must hold as a strict inequality. The solution can now be obtained only by applying the trial and error method.

Example 2. A company is currently involved in negotiations with its union on the upcoming wage contract. Positive signs in table represent wage increase while negative sign represents wage reduction. What are the optimal strategies for the company as well as the union? What is the game value?

Conditional costs to the company (` in lakhs)Union Strategies

U1 U2 U3 U4

C1 0.25 0.27 0.35 – 0.02Union C2 0.20 0.06 0.08 0.08Strategies C3 0.14 0.12 0.05 0.03

C4 0.30 0.14 0.19 0.00

Sol. Suppose, Company is the gainer player and Union is the looser player. Transposing payoff matrix because company’s interest is to minimize the wage increase while union’s interest is to get the maximum wage increase.

Algebraic method is used to determine the probability of using different strategies by players A and B.

of column B3 (strategy B3 for player B) is more than or equal to every corresponding element of row B2 (strategy B2 for player B). Evidently, the choice of strategy B3, by the player B, will always result in more losses as compared to that of selecting the strategy B2. Thus, strategy B3 is inferior to B2. Hence, delete the B3 strategy from the payoff matrix. The reduced payoff matrix is shown below:

Player APlayer B

B1 B2 B3

A1 –5 10 20A2 5 –10 –10A3 5 –20 –20

After column B3 is deleted, it may be noted that strategy A2 of player A is dominated by his A3 strategy, since the profit due to strategy A2 is greater than or equal to the profit due to strategy A3, regardless of which strategy player B selects. Hence, strategy A3 (row 3) can be deleted from further consideration. Thus, the reduced payoff matrix becomes:

Player APlayer B

Row minimumB1 B2

A1 – 5 10 – 5 ← Maximin

A2 5 –10 –10

Column maximum 5 10

↑ Minimax

As shown in the reduced 2 × 2 matrix, the maximin value is not equal to the minimax value. Hence, there is no saddle point and one cannot determine the point of equilibrium. For this type of game situation, it is possible to obtain a solution by applying the concept of mixed strategies.

The solution to this game can now be obtained by applying any of the methods used for mixed-strategy games that were discussed later.

4.10 SOLUTION METHODS FOR GAMES WITHOUT SADDLE POINT

4.10.1 Algebraic MethodThis method is used to determine the probability of using different strategies by

players A and B. This method becomes quite lengthy when a number of strategies for both the players are more than two.

Consider a game where the payoff matrix is: [aij]m × n. Let ( p1, p2, …, pm) and (q1, q2, …, qn) be the probabilities with which players A and B select their strategies (A1, A2, …, Am) and (B1, B2, …, Bn), respectively. If V is the value of game, then the expected gain to player A, when player B selects strategies B1, B2, …, Bn, one by one, is given by left-hand side of the following simultaneous equations, respectively.


NOTES


Since player A is the gainer player and expects at least V, therefore, we must have

Player APlayer B


A1 a11 a12 … a1n p1

A2 a21 a22 … a2n p2

Am am1 am2 … amn pm


a11 p1 + a21 p2 + … + am1 pm ≥ V a12 p1 + a22 p2 + … + am2 pm ≥ V …(1) a1n p1 + a2n p2 + . . . + amn pm ≥ V

where p1 + p2 + … + pm = 1 and pi ≥ 0 for all iSimilarly, the expected loss to player B, when player A selects strategies

A1, A2, ..., Am, one by one, can also be determined. Since player B is the loser player, therefore, he must have:

a11 q1 + a12 q2 + … + a1n qn ≤ V a21 q1 + a22 q2 + … + a2n qn ≤ V …(2) am1 q1 + am2 q2 + … + amn qn ≤ V

where q1 + q2 + … + qn = 1 and qj ≥ 0 for all jTo get the values of pi’s and qj’s, the above inequalities are considered as equations

and are then solved for given unknowns. However, if the system of equations, so obtained, is inconsistent, then at least one of the inequalities must hold as a strict inequality. The solution can now be obtained only by applying the trial and error method.

Example 2. A company is currently involved in negotiations with its union on the upcoming wage contract. Positive signs in table represent wage increase while negative sign represents wage reduction. What are the optimal strategies for the company as well as the union? What is the game value?

Conditional costs to the company (` in lakhs)Union Strategies

U1 U2 U3 U4

C1 0.25 0.27 0.35 – 0.02Union C2 0.20 0.06 0.08 0.08Strategies C3 0.14 0.12 0.05 0.03

C4 0.30 0.14 0.19 0.00

Sol. Suppose, Company is the gainer player and Union is the looser player. Transposing payoff matrix because company’s interest is to minimize the wage increase while union’s interest is to get the maximum wage increase.

Algebraic method is used to determine the probability of using different strategies by players A and B.


NOTES

Operations Research Company StrategiesC1 C2 C3 C4

U1 0.25 0.20 0.14 0.30Union U2 0.27 0.16 0.12 0.14Strategies U3 0.35 0.08 0.15 0.19

U4 – 0.02 0.08 0.13 0.00

In this payoff matrix strategy U4 is dominated by strategy U1 as well as U3. After deleting this strategy, we get

Company StrategiesC1 C2 C3 C4

U1 0.25 0.20 0.14 0.30Union U2 0.27 0.16 0.12 0.14Strategies U3 0.35 0.08 0.15 0.19

Company’s point of view, strategy C1 is dominated by C2 as well as C3, while C4 is dominated by C3. Deleting strategies C1 and C4 we get

Company StrategiesC2 C3

U1 0.20 0.14Union U2 0.16 0.12Strategies U3 0.08 0.15

Again strategy U2 is dominated by U1 and is, therefore, deleted to give

Company StrategiesC2 C3 Probability

Union U1 0.20 0.14 0.07/0.13 = 0.538Strategies U3 0.08 0.15 0.06/0.13 = 0.461Probability 0.01/0.13 0.12/0.13

= 0.076 = 0.923

Optimal strategy for the company : (0, 0.076, 0.923, 0)Optimal strategy for the union : (0.538, 0, 0.461, 0)Value of the game, V : 0.538 × 0.20 + 0.461 × 0.08 = ` 14360

Example 3. In a game of matching coins with two players, suppose A wins one unit of value when there are two heads, wins nothing when there are two tails and losses 1/2 unit of value when there is one head and one tail. Determine the payoff matrix, the best strategies for each player and the value of the game to A.

Sol. The payoff matrix for the given matching coin games is given by:

Player APlayer B

B1 B2A1 1 – 1/2A2 – 1/2 0


NOTES


As the payoff matrix does not have a saddle point, the game will be solved by the algebraic method.

For Player A Let p1 and p2 be probabilities of selecting strategy A1 and A2, respectively. Then the expected gain to player A, when player B uses its B1 and B2 strategies, respectively, is given by:

p1 – (1/2) p2 ≥ V; B selects B1 strategy …(1) – (1/2) p2 + 0.p2 ≥ V; B selects B2 strategy …(2)

where p1 + p2 = 1 …(3)For obtaining value of p1 and p2, we consider inequalities (1) and (2) as equations

and then with the help of Eq. (3), we get p1 = – 2V and p2 = – 6V.Substituting these values of p1 and p2 in Eq. (3) we get V = –1/8. Thus, p1 = 0.25

and p2 = 0.75.For Player B Let q1 and q2 be the probabilities of selecting strategies B1 and B2,

respectively. Then the expected loss to player B, when player A uses its A1 and A2 strategies, respectively is given by:

q1 – (1/2).q2 ≤ V ; A selects A1 strategy …(4) – (1/2).q1 + 0.q2 ≤ V ; A selects A2 strategy …(5)

and q1 + q2 = 1 …(6)Consider inequalities (4) and (5) as equations and then, with the help of Eq. (6),

we get q1 = 2V and q2 = – 6V.Substituting values of q1 and q2 in Eq. (6), we get V = – 1/8. Thus, q1 = 0.25 and

q2 = 0.75.Hence, the probability of selecting strategies optimally for players A and B are

(0.25, 0.75) and (0.25, 0.75), respectively, and the value of the game is V = – 1/8.Alternative methods

(a) Using the following results, the values of p1, p2, q1, q2 and value of game, V can be calculated as follows:

For Player A p1 = a a

a a a a22 21

11 22 12 21

0 1 21 0 1 2 1 2

14

−+ − +

= − −+ − − −

=( )

( / )( / / )

, and

p2 = 1 – p1 = 1 14

34

−

=

For Player B q1 = a a

a a a a22 12

11 22 12 21

0 1 21 0 1 2 1 2

14

−+ − +

= − −+ − − −

=( )

( / )( / / )

, and

q2 = 1 – q1 = 1 14

34

−

=

Expected value of game,

V = a a a a

a a a a11 22 21 12

11 22 12 21

1 0 1 2 1 21 0 1

. .( )

( / ) ( / )( /

−+ − +

= × − − × −+ − − 22 1 2

18−

= −/ )

(b) Expected payoff function E(p, q) = {(Expected gain to player A when player B uses strategy B1)

× Probability of player B using strategy B1} + {(Expected gain to player A when player B uses strategy B2) × Probability of player B using strategy B2}


NOTES

Operations Research = p1 q1 – (1/2) p2 q1 – (1/2)p1 q2 + (0)p2 q2

= p1 q1 – (1/2) (1 – p1) q1 – (1/2) p1 (1 – q1) = p1 q1 – (1/2)q1 + (1/2). p1 q1 – (1/2)p1 + (1/2)p1 q1

Differentiating partially E( p, q) with respect to p1 and q1 and by equating with zero, we get

∂∂Ep1

= q1 – (1/2)q1 – (1/2) + (1/2)q1 = 0,

i.e., q1 = (1/4) and q2 = 1 – q1 = (3/4)

∂∂Eq1

= p1 – (1/2) + (1/2)p1 + (1/2)p1 = 0,

i.e., p1 = (1/4) and p2 = 1 – p1 = (3/4)Substituting value of p1 and q1 in E( p, q), we get: Expected value of game V = (1/4) (1/4) – 1/2 (3/4) (1/4) – 1/2 (1/4) (3/4) = (1/16) – (3/32) – (3/32) = – 1/8

Example 4. In a small town, there are only two stores, ABC and XYZ, that handle sundry goods. The total number of customers is equally divided between the two, because the price and the quality of goods sold are equal. Both stores have good reputation in the community, and they render equally good customer service. Assume that a gain of customers by ABC is a loss to XYZ and vice versa. Both stores plan to run annual pre-Diwali sales during the first week of November. Sales are advertised through a local newspaper, and through radio and television. With the aid of an advertising firm store, ABC constructed the game matrix given below. (Figures in the matrix represent a gain or loss of customers).

Strategy of ABCStrategy of XYZ

Newspaper Radio TelevisionNewspaper 30 40 – 80Radio 0 15 – 20Television 90 20 50

Determine the optimal strategies and the worth of such strategies for both ABC and XYZ.

Sol. The first step is to search for a saddle point. There is no saddle point in the problem.

The second step is to observe whether the payoff matrix can be reduced by rules of dominance. Yes, each element in first column is more than the corresponding element in the third column. Thus, for store XYZ, the newspaper as a media of advertisement is less attractive as compared to television. Therefore we remove column 1 from the payoff matrix. We now get:

ABCXYZ

B2 B3

A1 40 – 80A2 15 – 20A3 20 50


NOTES


In the reduced payoff matrix, each element in the second row is less than the corresponding element in third row. So, store ABC will never like to choose radio as a media of advertisement because it would be earning less profit than it would if it advertised through television. Thus, by deleting the second row from the reduced matrix, we get the further reduced 2 × 2 payoff matrix as shown below:

ABCXYZ

ProbabilityB2 B3

A1 40 – 80 p1

A3 20 50 p2

Probability q1 q2

Even the reduced 2 × 2 payoff matrix does not have the saddle point. Thus, both the stores use mixed strategies.

For Store ABC : Let p1 and p2 be probabilities of selecting strategy A1 (newspaper) and A3 (television), respectively. Then, the expected gain to store ABC when store XYZ uses its B2 and B3 strategies is given by:

40p1 + 20p2 and – 80p1 + 50p2 ; p1 + p2 = 1For store ABC the probability p1 and p2 should be such that the expected gains

under both conditions are equal. That is 40p1 + 20p2 = – 80p1 + 50p2

40p1 + 20(1 – p1) = – 80p1 + 50(1 – p1), where p1 + p2 = 1 150p1 = 30 or p1 = 1/5, and p2 = 1 – p1 = 4/5Thus, the store ABC should apply strategy A1 (newspaper) with a probability of

1/5 and strategy A3 (television), with a probability of 4/5.For Store XYZ : Let q1 and q2 be the probabilities of selecting strategy B2 (radio)

and B3 (television), respectively. Then, the expected loss to store XYZ, when store ABC uses its strategies A1 and A3, should be

40q1 – 80q2 = 20q1 + 50q2 where q1 + q2 = 1 40q1 – 80(1 – q1) = 20q1 + 50(1 – q1) 150q1 = 130 or q1 = 13/15, and q2 = 1 – q1 = 2/15.Thus, store XYZ should apply strategy B2 (radio) with a probability of 13/15 and

strategy B3 (television) with a probability of 2/15.Substituting the values of p1, p2 or q1, q2 in any of the gain or loss equations, we

shall get the expected value of the game (i.e. 24). This is shown below:Expected Gain to Store ABC (i) 40p1 + 20p2 = 40 × (1/5) + 20 × (4/5) = 24 (ii) – 80p1 + 50p2 = – 80 × (1/5) + 50 × (4/5) = 24Expected Loss to Store XYZ (i) 40q1 – 80q2 = 40 × (13/15) – 80 × (2/15) = 24 (ii) 20q1 + 50q2 = 20 × (13/15) + 20 × (2/15) = 24

Here it may be noted that the expected loss to one store is the same as the expected gain to the other store.


NOTES

Operations Research Example 5. Two Firms A and B have, for years, been selling a competing product that forms a part of both firms’ total sales. The marketing executive of Firm A raised the question: ‘What should be the firm’s strategies in terms of advertising for the product in question’. The market research team of Firm A developed the following data for varying degree of advertising: (i) No advertising, medium advertising, and large advertising for both firms will result in equal market shares. (ii) Firm A with no advertising: 40 per cent of the market with medium advertising by Firm B and 28 per cent of the market with large advertising by Firm B. (iii) Firm A using medium advertising: 70 per cent of the market with no advertising by Firm B and 45 per cent of the market with large advertising by Firm B. (iv) Firm A using large advertising: 75 per cent of the market with no advertising by Firm B and 47.5 per cent of the market with medium advertising by Firm B.

Based upon the above information, answer the marketing executive’s questions. What advertising policy should Firm A pursue when consideration is given to the above factors: selling price, ` 4.00 per unit; variable cost of product, ` 2.50 per unit; annual volume of 30,000 units for Firm A; cost of annual medium advertising ` 5,000 and cost of annual large advertising ` 15,000? What contribution, before other fixed costs, is available to the firm?

Sol. The payoff matrix of the game between Firms A and B is as follows:

Firm AFirm B Row

MinimumNo Advt., B1 Medium Advt., B2 Large Advt., B3

No Advt., A1 50 40 28 28Medium Advt., A2 70 50 45 45Large Advt., A3 75 47.5 50 47.5Column Maximum

75 50 50

From the payoff matrix it is observed that there is no saddle point in the problem.Applying rules of dominance, delete the first row (dominated by third row) and

then the first column (dominated by both columns 2 and 3) from the payoff matrix. The reduced payoff matrix, so obtained, is shown below:

Firm AFirm B

ProbabilityB2 B3

A2 50 45 p1

A3 47.5 50 p2

Probability q1 q2

Even the reduced 2 × 2 payoff matrix does not have the saddle point. Thus, both the firms use mixed strategies. Adopting the same procedure, as discussed in earlier examples, the expected gain to Firm A can be calculated as follows:

Expected Gain to Firm A 50p1 + 47.5p2 = 45p1 + 50p2; p1 + p2 = 1 50p1 + 47.5(1 – p1) = 45p1 + 50(1 – p1)


NOTES


7.5p1 = 2.5 or p1 = 1/3 and p2 = 1 – p1 = 2/3. Expected gain = 50p1 + 47.5p2

= 50(1/3) + 47.5(2/3) = 145/3.Thus, the optimal policy for Firm A is to apply strategy A2 (medium advertising)

with probability 0.33 and strategy A3 (large advertising) with probability 0.67 on any one play of the game. With this policy, the firm may expect to gain 145/3 = 48.3 per cent of the market share.

Table 4.20. Market Share of Firm A

Firm AFirm B

Large Advt.No Advt. Medium Advt.

No Advt. 0.50 × 30,000 = 15,000 0.40 × 30,000 = 12,000 0.28 × 30,000 = 8,400Medium Advt. 0.70 × 30,000 = 21,000 0.50 × 30,000 = 15,000 0.45 × 30,000 = 13,500Large Advt. 0.75 × 30,000 = 22,500 0.475 × 30,000 = 14,250 0.50 × 30,000 = 15,000

Given that the expenditure on medium and large advertisements is ` 5,000 and ` 15,000, respectively, net profit to Firm A can be calculated by using following equation:

Net profit = (Sales price – Cost price) × Sales volume – Advertising expenditureThe net profit to Firm A is shown in Table 4.21

Table 4.21. Profit to Firm A

Firm AFirm B

No Advt. Medium Advt. Large Advt.No advt. 1.5 × 15,000

= 22,5001.5 ×12,000

= 18,0001.5 × 8,400 = 12,600

Medium advt. 1.5 × 21,000 – 5,000 = 26,500

1.5 × 15,000 – 5,000 = 17,500

1.5 × 13,500 – 5,000 = 15,250

Large advt. 1.5 × 22,500 – 15,000 = 18,750

1.5 × 14,250 – 15,000 = 6,375

1.5 × 15,000 – 15,000 = 7,500

Observations 1. If Firm A chooses the strategy of No advertising, then the minimum profit is

` 12,600, because Firm B can adopt its strategy Large advertising. 2. If Firm A chooses the strategy of Medium advertising, then the minimum profit

is ` 15,250 because Firm B can again adopt its strategy Large advertising. 3. If Firm A chooses the strategy of Large advertising, then the minimum profit is

` 6,375 because firm B can adopt its strategy Medium advertising.Based on these observations, the Firm A must adopt the policy of ‘Medium

Advertising’ in order to gain the maximum profit of ` 15,250 among these three alternatives and must spend ` 5,000 for advertising.

4.10.2 Arithmetic MethodThe arithmetic method (also known as short-cut method) provides an easy method

for finding optimal strategies for each player in a payoff matrix of size 2 × 2, without saddle point. The steps of this method are as follows:


NOTES

Operations Research 1. Find the differences between the two values in the first row and put it against the second row of the matrix, neglecting the negative sign (if any).

2. Find the difference between the two values in the second row and put it against first row of the matrix, neglecting the negative sign (if any).

3. Repeat Steps 1 and 2 for the two columns also.The values obtained by ‘swapping the differences’ represent the optimal relative

frequencies of play for both players strategies.These may be converted to probabilities by dividing each of them by their sum.

The value of the game can be obtained by applying any of the methods discussed earlier.Remark The arithmetic method should not be used to solve a 2 × 2 game that has a saddle

point because the method yields an incorrect answer.

4.10.3 Matrix MethodIf the game matrix is in the form of a square matrix, then the optimal strategy

mix as well as value of the game may be obtained by the matrix method. The solution of a two-person zero-sum game with mixed strategies with a square payoff matrix may be obtained by using the following formulae:

Player A’s optimal strategy = [ ]

[ ]

1 1

1 111

P

P

adj

adj

Player B’s optimal strategy = [ ]

[ ]

1 1

1 111

P

P

cof

adj

Value of the game = (Player A’s optimal strategies) × (Payoff matrix pij) × (Player B’s optimal strategies) where Padj = adjoint matrix, Pcof = cofactor matrix. Player A’s optimal strategies are in the form of a row vector and B’s optimal strategies are in the form of a column vector.

This method can be used to find a solution of a game with size of more than 2 × 2. However, in rare cases, the solution violates the non-negative condition of probabilities, i.e., pi ≥ 0, qj ≥ 0, although the requirement p1 + p2 + … + pm = 1 or q1 + q2 + … + qn = 1 is met.

Example 7. Solve the following game after reducing it to a 2 × 2 game

Player APlayer B

B1 B2 B3

A1 1 7 2A2 6 2 7A3 5 1 6

Sol. In the given game matrix, the third row is dominated by the second row and in the reduced matrix third column is dominated by the first column. So, after elimination of the third row and the third column the game matrix becomes:


NOTES

Dynamic Programming and Theory of GamesPlayer A

Player BB1 B2

A1 1 7A2 6 2

For this reduced matrix, let us calculate Padj and Pcof as given below:

Padj = 2 76 1

−−

and Pcof =

2 67 1

−−

Player A’s optimal strategies = 1 1

2 76 1

1 12 76 1

11

4 610

4 610

[ ] −−

[ ] −−

= − −−

=[ ] [ ]

This solution can be broken down into the optimal strategy mix for player A as p1 = 4/10 = 2/5 and p2 = 6/10 = 3/5, where p1 and p2 represent the probabilities of player A’s using his strategies A1 and A2, respectively.

Similarly, the optimal strategy mixture for player B is obtained as:

Player B’s optimal strategies = 1 1

2 67 1

1 12 76 1

11

5 510

5 510

[ ] −−

[ ] −−

= − −−

=[ ] [ ]

This solution can also be broken down into the optimal strategy mixture for player B as q1 = 5/10 = 1/2 and q2 = 5/10 = 1/2, where q1 and q2 represent the probabilities of player B’s using his strategies B1 and B2, respectively. Hence:

Value of the game, V = 25

35

1 76 2

1 21 2

//

= 4.

4.10.4 Graphical Solution of 2 × n and m × 2 Games The graphical method is useful for the game where the payoff matrix is of the size

2 × n or m × 2, i.e., the game with mixed strategies that has only two undominated pure strategies for one of the players in the two-person zero-sum game.

Optimal strategies for both the players assign non-zero probabilities to the same number of pure strategies. Therefore, if one player has only two strategies, the other will also use the same number of strategies. Hence, this method is useful in finding out which of the two strategies can be used.

Consider the following 2 × n payoff matrix of a game, without saddle point.

Player APlayer B


A1 a11 a12 … a1n p1

A2 a21 a22 … a2n p2



NOTES

Operations Research Player A has two strategies A1 and A2 with probability of their selection p1 and p2, respectively, such that p1 + p2 = 1 and p1, p2 ≥ 0. Now for each of the pure strategies available to player B, the expected pay off for player A would be as follows:

B’s Pure Strategies A’s Expected Payoff

B1 a11p1 + a21 p2

B2 a12p1 + a22 p2

Bn a1np1 + a2np2

According to the maximin criterion for mixed strategy games, player A should select the value of probability p1 and p2 so as to maximize his minimum expected payoffs. This may be done by plotting the straight lines representing player A’s expected payoff values.

The highest point on the lower boundary of these lines will give the maximum expected payoff among the minimum expected payoffs and the optimum value of probability p1 and p2.

Now, the two strategies of player B corresponding to those lines which pass through the maximin point can be determined. This helps in reducing the size of the game to (2 × 2), which can be easily solved by any of the methods discussed earlier.

The (m × 2) games are also treated in the same way except that the upper boundary of the straight lines corresponding to B’s expected payoff will give the maximum expected payoff to player B and the lowest point on this boundary will then give the minimum expected payoff (minimax value) and the optimum value of probability q1 and q2.

Example 7. Use the graphical method for solving the following game and find the value of the game.

Player APlayer B

B1 B2 B3 B4

A1 2 2 3 –2

A2 4 3 2 6

Sol. The game does not have a saddle point. If the probability of player A’s playing A1 and A2 in the strategy mixture is denoted by p1 and p2, respectively, where p2 = 1 – p1, then the expected payoff (gain) to player A will be

B’s Pure Strategies A’s Expected PayoffB1 2 p1 + 4 p2

B2 2 p1 + 3 p2

B3 3 p1 + 2 p2

B4 – 2 p1 + 6 p2

These four expected payoff lines can be plotted on a graph to solve the game.The graph for player A: A graphic solution is shown in Fig. 4.7. Here, the

probability of player A’s playing A1, i.e. p1 is measured on the x-axis. Since p1 cannot exceed 1, the x-axis is cutoff at p1 = 1. The expected payoff of player A is measured along y-axis. From the game matrix, if player B plays B1, the expected payoff of player A is


NOTES


2 when A plays A1 with p1 = 1 and 4 when A plays A2 with p1 = 0. These two extreme points are connected by a straight line, which shows the expected payoff of A when B plays B1. Three other straight lines are similarly drawn for B2, B3 and B4.

Fig. 4.7. Graph for Player A

It is assumed that player B will always play his best possible strategies yielding the worst result to player A. Thus, the payoffs (gains) to A are represented by the lower boundary when he is faced with the most unfavourable situation in the game. Since player A must choose his best possible strategies in order to realize a maximum expected gain, the highest expected gain is found at point P, where the two straight lines meet.

E3 = 3p1 + 2p2

= 3p1 + 2(1 – p1) E4 = – 2p1 + 6p2

= – 2p1 + 6(1 – p1)In this manner the solution of the original (2 × 4) game reduces to that of the game

with payoff matrix of size (2 × 2) as given below:

Player APlayer B

B3 B4

A1 3 – 2A2 2 6

The optimum payoff to player A can now be obtained by setting E3 and E4 equal and solving for p1, i.e.,

3p1 + 2(1 – p1) = – 2p1 + 6(1 – p1)or p1 = 4/9; p2 = 1 – p1 = 5/9

Substituting the value of p1 and p2 in the equation for E3 (or E4) we have:Value of the game, V = 3 × 4/9 + 2 × 5/9 = 22/9The optimal strategy mix of player B can also be found in the same manner as for

player A. If the probabilities of B’s selecting strategy B3 and B4 are denoted by q3 and q4, respectively, then the expected loss to B will be:


NOTES

Operations Research L3 = 3q3 – 2q4 = 3q3 – 2(1 – q3) (if A selects A1) L4 = 2q3 + 6q4 = 2q3 + 6(1 – q3) (if A selects A2)To solve for q3, equate the two equations: 3q3 – 2(1 – q3) = 2q3 + 6(1 – q3)

or q3 = 8/9; q4 = 1 – q3 = 1/9Substituting the value of q3 and q4 in the equation for L3 (or L4), we haveValue of the game, V = 3 × 8/9 – 2 × 1/9 = 22/9

Example 9. Two firms A and B make colour and black & white television sets. Firm A can make either 150 colour sets in a week or an equal number of black & white sets, and make a profit of ` 400 per colour set, or 150 colour and 150 black & white sets, or 300 black & white sets per week. It also has the same profit margin on the two sets as A. Each week there is a market of 150 colour sets and 300 black & white sets and the manufacturers would share market in the proportion in which they manufacture a particular type of set.

Write the pay-off matrix of A per week. Obtain graphically A’s and B’s optimum strategies and value of the game.

Sol. For firm A, the strategies are:A1 : make 150 colour sets, A2 : make 150 black & white sets.For firm B, the strategies are:B1 : make 300 colour sets, B2 : make 150 colour and 150 black & white sets.B3 : make 300 black & white sets.For the combination A1B1, the profit to firm A would be: {150/(150 + 300)}

× 150 × 400 = ` 20,000 wherein 150/(150 + 300) represents share of market for A, 150 is the total market for colour television sets and 400 is the profit per set. In a similar manner, other profit figures may be obtained as shown in the following pay-off matrix:

A’s StrategyB’s Strategy

B1 B2 B3

A1 20,000 30,000 60,000A2 45,000 45,000 30,000

This pay-off table has no saddle point. Thus to determine optimum mixed strategy, the data are plotted on graph as shown in Fig. 4.8.

Fig. 4.8. Graph for Player A


NOTES


Lines joining the pay-offs on axis A1 with the pay-offs on axis A2 represents each of B’s strategies. Since firm A wishes to maximize his minimum expected pay-off, we consider the highest point of intersection, P on the lower envelope of A’s expected pay-off equation. This point P represents the maximum expected value of the game. The lines B1 and B3 passing through P, define the strategies which firm B needs to adopt. The solution to the original 2 × 3 game, therefore, reduces to that of the simpler game with 2 × 2 pay-off matrix as follows:

A’s StrategyB’s Strategy

ProbabilityB1 B3

A1 20,000 60,000 p1

A2 45,000 30,000 p2

Probability q1 q2

The optimal mixed strategies of player A are: A1 = 3/11, A2 = 8/11. Similarly, the optimal mixed strategies for B are: B1 = 6/11, B2 = 0, B3 = 5/11. The value of the game is V = 38,182.

4.10.5 Solution of a Game by Linear-Programming MethodThe two-person zero-sum games can also be solved by linear programming. The

major advantage of using linear programming technique is that it helps to solve the mixed-strategy games of larger dimension payoff matrix.

To illustrate the transformation of a game problem to a linear programming problem, consider a payoff matrix of size m × n. Let aij be the element in the ith row and jth column of game payoff matrix, and letting pi be the probabilities of m strategies (i = 1, 2, …, m) for player A. Then, the expected gains for player A, for each of player B’s strategies will be:

V = p ai iji

m

=∑

1, j = 1, 2, …, n

The aim of player A is to select a set of strategies with probability pi (i = 1, 2, …, m) on any play of game such that he can maximize his minimum expected gains.

Now to obtain values of probability pi, the value of the game to player A for all strategies by player B must be at least equal to V. Thus to maximize the minimum expected gains, it is necessary that:

a11 p1 + a21 p2 + … + am1 pm ≥ V a12 p1 + a22 p2 + … + am2 pm ≥ V

a1n p1 + a2n p2 + … + amn pm ≥ Vwhere p1 + p2 + … + pm = 1, and pi ≥ 0 for all i.

Dividing both sides of the m inequalities and equation by V the division is valid as long as V > 0. In case V < 0, the direction of inequality constraints must be reversed. But if V = 0, the division would be meaningless. In this case a constant can be added to all entries of the matrix, ensuring that the value of the game (V ) for the revised matrix


NOTES

Operations Research becomes more than zero. After optimal solution is obtained, the true value of the game is obtained by subtracting the same constant value. Let pi/V = xi, (≥ 0), we then have

a pV

a pV

a pVmm

111

212

1+ + … + ≥ 1

a pV

a pV

a pVmm

121

222

2+ + … + ≥ 1

a pV

a pV

a pVn n mnm

11

22+ + … + ≥ 1

pV

pV

pVm1 2+ + … + = 1

Since the objective of player A is to maximize the value of the game, V, which is equivalent to minimizing 1/V, the resulting linear programming problem can be stated as:

Minimize Zp ( = 1/V ) = x1 + x2 + … + xm

subject to the constraints a11 x1 + a21 x2 + . . . + am1 xm ≥ 1 a12 x1 + a22 x2 + … + am2 xm ≥ 1

a1n x1 + a2n x2 + … + amn xm ≥ 1and x1, x2, . . ., xm ≥ 0where xi = pi /V ≥ 0; i = 1, 2, …, m

Similarly, Player B has a similar problem with the inequalities of the constraints reversed, i.e., minimize the expected loss. Since minimizing V is equivalent to maximizing 1/V, therefore, the resulting linear programming problem can be stated as:

Maximize Zq (= 1/V) = y1 + y2 + … + yn

subject to the constraints a11 y1 + a12 y2 + … + a1n yn ≤ 1 a21 y1 + a22 y2 + … + a2n yn ≤1

am1 y1 + am2 y2 + … + amn yn ≤ 1 y1, y2, . . ., yn ≥0

where, yj = qj/V ≥ 0; j = 1, 2, . . ., nIt may be noted that the LP problem for player B is the dual of LP problem for

player A and vice versa. Therefore, the solution of the dual problem can be obtained from the primal simplex table. Since for both the players Zp = Zq, the expected gain to player A in the game will be exactly equal to expected loss to player B.

Remark Linear programming technique requires all variables to be non-negative and therefore to obtain a non-negative of value V of the game, the data to the problem, i.e., aij in the payoff table should all be non-negative. If there are some negative elements in the payoff table, a constant to every element in the payoff table must be added so as to make the smallest element zero; the solution to this new game will give an optimal mixed strategy for the original game. The value of the original game then equals the value of the new game minus the constant.


NOTES


Example 8. For the following payoff matrix, transform the zero-sum game into an equivalent linear programming problem and solve it by using the simplex method.

Player APlayer B

B1 B2 B3

A1 1 –1 3

A2 3 5 –3

A3 6 2 –2

Sol. The first step is to find out the saddle point (if any) in the payoff matrix as shown below:

Player APlayer B

B1 B2 B3 Row minimum

A1 1 –1 3 – 1 ← Maximin

A2 3 5 – 3 – 3

A3 6 2 – 2 – 2

Column maximum 6 5 3 ← Minimax

The given game payoff matrix does not have a saddle point. Since the maximin value is –1, therefore, it is possible that the value of game (V) may be negative or zero because –1 < V < 1. Thus, a constant that is at least equal to the negative of maximin value, i.e., more than –1 is added to all the elements of the payoff matrix. Thus, adding a constant number 4 to all the elements of the payoff matrix, the payoff matrix becomes:

Player APlayer B

ProbabilityB1 B2 B3

A1 5 3 7 p1

A2 7 9 1 p2

A3 10 6 2 p3

Probability q1 q2 q3

Let pi ( i = 1, 2, 3) and qj ( j = 1, 2, 3) be the probabilities of selecting strategies Ai (i = 1, 2, 3) and Bj (j = 1, 2, 3) by players A and B, respectively.

The expected gain for player A will be as follows: 5p1 + 7p2 + 10p3 ≥ V (if B uses strategy B1) 3p1 + 9p2 + 6p3 ≥ V (if B uses strategy B2) 7p1 + p2 + 2p3 ≥ V (if B uses strategy B3) p1 + p2 + p3 = 1

and p1, p2, p3 ≥ 0Dividing each inequality and equality by V, we get,

5 7 101 2 3pV

pV

pV

+ + ≥ 1, 3 9 61 2 3pV

pV

pV

+ + ≥ 1

7 21 2 3pV

pV

pV

+ + ≥ 1, pV

pV

pV

1 2 3+ + = 1V


NOTES

Operations Research In order to simplify, we define new variables: x1 = p1/V, x2 = p2/V and x3 = p3/VThe problem for player A, therefore becomes,Minimize Zp ( = 1/V ) = x1 + x2 + x3

subject to the constraints (i) 5x1 + 7x2 + 10x3 ≥ 1, (ii) 3x1 + 9x2 + 6x3 ≥ 1, (iii) 7x1 + x2 + 2x3 ≥ 1and x1, x2, x3 ≥ 0

Player B’s objective is to minimize his expected losses that can be reduced to minimizing the value of the game V. Hence, the problem of player B can be expressed as follows:

Maximize Zq (= 1/V) = y1 + y2 + y3

subject to the constraints (i) 5y1 + 3y2 + 7y3 ≤ 1, (ii) 7y1 + 9y2 + y3 ≤ 1, (iii) 10y1 + 6y2 + 2y3 ≤ 1and y1, y2, y3 ≥ 0where y1 = q1/V; y2 = q2/V and y3 = q3/V.

It may be noted here the that problem of player A is the dual of the problem of player B. Therefore, the solution of the dual problem can be obtained from the optimal simplex table of primal.

To solve the problem of player B, introduce slack variables to convert the three inequalities to equalities. The problem now becomes:

Maximize Zq = y1 + y2 + y3 + 0s1 + 0s2 + 0s3

subject to the constraints (i) 5y1 + 3y2 + 7y3 + s1 = 1, (ii) 7y1 + 9y2 + y3 + s2 = 1 (iii) 10y1 + 6y2 + 2y3 + s3 = 1and y1, y2, y3, s1, s2, s3 ≥ 0

The initial solution is shown in Table 4.22.Table 4.22. Initial Solution

cj → 1 1 1 0 0 0

Basic Variables

Coefficient cB

Basic Variables

B

Basic Variables Solution yB(= b) y1 y2 y3 s1 s2 s3

Min. Ratio yB /y1

0 s1 1 5 3 7 1 0 0 1/5

0 s2 1 7 9 1 0 1 0 1/7

0 s3 1 10 6 2 0 0 1 1/10 →

Z = 0 zj 0 0 0 0 0 0cj – zj 1 1 1 0 0 0

↑


NOTES


Proceeding with the usual simplex method, the optimal solution is shown in Table 4.23.


cj → 1 1 1 0 0 0


cB

Basic Variables

B

Basic Variables Solution yB(= b)

y1 y2 y3 s1 s2 s3

1 y3 1/10 2/5 0 1 3/20 – 1/10 0

1 y2 1/10 11/15 1 0 – 1/60 7/60 0

0 s3 1/5 24/5 0 0 – 1/5 – 3/5 1

Z = 1/5 zj 17/15 0 0 2/15 1/15 0

cj – zj – 2/5 0 0 – 2/15 – 1/15 0

The optimal solution (mixed strategies) for B is: y1 = 0; y2 = 1/10 and y3 = 1/10 and the expected value of the game is: Z = 1/V – constraint (= 4) = 5 – 4 = 1.

These solution values are now converted back into the original variables: If 1/V = 1/5 then V = 5

y1 = q1/V, then q1 = y1 × V = 0 y2 = q2/V, then q2 = y2 × V = 1/10 × 5 = 1/2 y3 = q3/V, then q3 = y3 × V = 1/10 × 5 = 1/2The optimal strategies for player A are obtained from the cj – zj row of the Table 4.23. x1 = 2/15, x2 = 1/15 and x3 = 0Then p1 = x1 × V = (2/15) × 5 = 2/3; p2 = x2 × V = (1/15) × 5 = 1/3; p3 = x3 × V = 0Hence, the probabilities of using strategies by both the players are: Player A = (2/3, 1/3, 0); Player B = (0, 1/2, 1/2) and, Value of the game V = 1.


1. Explain the following terms: (i) Two-person zero-sum game, (ii) Principles of dominance, (iii) Pure strategy in game theory 2. How is the concept of dominance used in simplifying the solution of a rectangular game? 3. Explain the theory of dominance in the solution of rectangular games. 4. Explain the graphical method of solving 2 × n and m × 2 games.


NOTES

Operations ResearchTEST YOUR KNOWLEDGE–B

1. Consider a modified form of ‘matching coins’ game problem. The matching player is paid ` 8, if the two coins turn both heads and ` 1 if the coins turn both tails. The non-matching player is paid ` 3 when the two coins do not match. Given the choice of being a matching or non-matching player, which one would you choose and what would be your strategy?

2. Use dominance rules to reduce the size of the following payoff matrix to (2 × 2) size and hence, find the optimal strategies and value of the game.

(a)Player A


Player B

B1 B2 B3 B1 B2 B3 B4

A1 3 – 2 4 A1 3 2 4 0

A2 – 1 4 2 A2 3 4 2 4

A3 2 2 6 A3 4 2 4 0

A4 0 4 0 8

3. Players A and B, each take out one or two matches and guess how many matches the opponent has taken. If one of the players guesses correctly, then the loser has to pay him the sum of the number held by both players. Otherwise, the pay out is zero. Write down the payoff matrix and obtain the optimal strategies of both players.

4. Explain the principle of dominance in game theory and solve the following game:

Player APlayer B

B1 B2 B3

A1 1 7 2A2 6 2 7A3 5 2 6

5. Player A

Player B

B1 B2 B3 B4

A1 8 10 9 14

A2 10 11 8 12

A3 13 12 14 13

6. Player APlayer B

B1 B2 B3 B4 B5

A1 2 4 3 8 4

A2 5 6 3 7 8

A3 6 7 9 8 7

A4 4 2 8 4 3


NOTES


7. In an election campaign, the strategies adopted by the ruling and opposition parties, along with payoffs, (ruling party’s per cent share in votes polled) are given below:

Ruling Party’s Strategies

Opposition Party’s StrategiesCampaign one day in each city

Campaign two days in large towns

Spend two days in large rural sectors

Campaign one day in each city 55 40 35Campaign two days in large towns 70 70 55Spend two days in large rural sectors 75 55 65

Assume a zero-sum game. Find the optimum strategies for both parties and expected payoff to ruling party.

8. Player A is paid ` 8 if two coins turn heads at the same time and ` 10 if two coins turn tails at the same time. Player B is paid ` 3 when the two coins do not match. Given the choice of being A or B, which one would you choose and what would be your strategy?

9. Even though there are several manufacturers of scooters, two firms with brand names Janata and Praja, control their market in Western India. If both manufacturers make changes in the model in the same year, their respective market shares would remain constant. Likewise, if neither makes model changes, even then their market shares would remain constant. The payoff matrix, in terms of increased/decreased percentage market share, under different possible conditions is given below:

JanataPraja

No change Minor change Major changeNo Change 0 – 4 – 10Minor Change 3 0 5Major Change 8 1 0

(a) Find the value of the game. (b) What change should Janata consider if this information is available only to itself? 10. A steel company is negotiating with its union for revising wages of its employees. The

management, with the help of a mediator, has prepared a payoff matrix shown below. Plus sign, represents wage increase, while negative sign stands for wage decrease. The union has also constructed a table which is comparable to that developed by the management. The management does not have the specific knowledge of game theory to select the best strategy or strategies for the firm. You have to assist the management on the problem. What game value and strategies are available to the opposing group?

Additional Costs to Steel Company (`)

Company StrategiesUnion Strategies

U1 U2 U3 U4

C1 2.50 2.70 3.50 0.20

C2 2.00 1.60 0.80 0.80

C3 1.40 1.20 1.50 1.30

C4 3.00 1.40 1.90 0

11. In zero-sum two-person children’s game of stone, paper and scissors, both players simultaneously call out stone, paper or scissors. The combination of paper and stone is a win of one unit for player calling paper (paper covers stone); stone and scissors is a win for stone (stone breaks scissors), and scissors and paper is a win for scissors (scissors cut paper).


NOTES

Operations Research A call of the same item represents no payoff. Write the payoff matrix and the equivalent linear programming problem to the above game. Find the optimal strategy for both the players and the value of the game.

12. Two candidates, X and Y, are competing for the councillor’s seat in a city municipal corporation. X is attempting to increase his total votes at the expense of Y. The strategies available to each candidate involve personal contacts, newspaper insertions, speeches or television appearance, and advertising. The increase in votes available to X, given various combinations of strategies, are given below. (Assume that this is a zero-sum game, i.e., any gain of X is equal to the votes lost by Y). Determine the optimal strategies that should be adopted by X during his election campaign. How many votes should X gain by adopting the optimal strategy?

Candidate XCandidate Y

Personal contacts Newspapers TelevisionPersonal contacts 30,000 20,000 10,000Newspapers 60,000 50,000 25,000Television 20,000 40,000 30,000

13. Vishal who has an amount of ` 1 lakh, is planning to invest it among three companies. He wants to buy equity shares in company A, B and C. The payoffs in terms of (i) growth in capital, and (ii) returns on the capital, are known for each of the investments under each of the three economic conditions which may prevail – recession, growth and stability. Assuming that Vishal must make his choice among the three portfolios for a period of one year in advance, his expectations of the net earning (in ‘000 `) of his ` 1 lakh portfolio after one year is represented by the following matrix:

CompanyEconomic Condition

Recession Stability GrowthA –15 6 10B 4 7.5 8C 6.5 6 15

Determine the optimal strategies for investment and the expected present return for the investor under such a policy.

14. Two firms F1 and F2 make colour and black & white television sets. F1 can make either 300 colour sets in a month or an equal number of black & white sets. He makes a profit of 200 on a colour set and 150 on a black & white set. F2 can, on the other hand, make either 600 colour sets or 300 colour and 300 black & white sets or 600 black & white sets per month. It also has the same profit margin on the two sets as F1. Each month there is a market of 300 colour sets and 600 black & white sets and the manufacturers would share the market depending upon the proportion in which they manufacture a particular type of sets.

Write the payoff matrix of F1 and F2 per month. Obtain F1 and F2’s optimal strategies and the value of the game.

15. Obtain the strategies for both players and the value of the game for two-person zero-sum game whose payoff matrix is given as follows:

(a)Player A


Player BB1 B2 B3 B1 B2 B3 B4 B5 B6

A1 1 3 11 A1 1 3 – 1 4 2 – 5A2 8 5 2 A2 – 3 5 6 1 2 0


NOTES


16. Obtain the optimal strategies for both players and the value of the game for two-person zero-sum game whose payoff matrix is given as follows:

Player APlayer B

B1 B2

A1 – 6 7A2 4 –5A3 – 1 – 2A4 – 2 5A5 7 –6

17. For the following payoff tables, transform the zero-sum game into an equivalent linear programming problem and solve it using the simplex method.

(a)Player A

Player B (b)Company B

Company AB1 B2 B3 A1 A2 A3

A1 9 1 4 B1 2 –2 3A2 0 6 3 B2 – 3 5 – 1A3 5 2 8

18. A soft drink company calculated the market share of two of its products against its major competitor, which has three products. The company found out the impact of additional advertisement in any one of its products against the other.

Company ACompany B

B1 B2 B3

A1 6 7 15A2 20 12 10

What is the best strategy for the company as well as the competitor? What is the payoff obtained by the company and the competitor in the long run? Use the graphical method to obtain the solution.

Hints and Answers 1. Matching player = Non-matching player = (4/15, 11/15) and V = – 1/15. 2. (a) Third column is dominated by the first column; in the reduced matrix third row

dominates an average of first and second row. Reduced matrix has the saddle point. Optimal strategy for player A is A3 and for player

is B2 or B3. Value of game = – 2. (b) Player A: (0, 0, 2/3, 1/3); Player B: (0, 0, 2/3, 1/3) and V = 8/3. 3. The payoff matrix is:

Player B2 0

Player A 0 4

Player A: (2/3, 1/3); Player B: (2/3, l/3) and V = 4/3. 4. Row 3 is dominated by row 2; Column 3 in the reduced matrix is dominated by column 1. Player A: (2/5, 3/5, 0), Player B: (1/2, 1/2, 0) and V = 4.


NOTES

Operations Research 5. Column 4 is dominated by column 1. In reduced matrix, row 1 is dominated by row 3. In reduced matrix, average of columns 2 and 3 dominate column 1. The reduced matrix has

a saddle point. Optimal strategy for player A is A3 (row 3) and for player B is B1 (column 1). The value of game is 12.

6. Optimal strategy for player A is A3 and for player B is B1. The value of game is 6.

7. B1 B2 B3 For ruling party: (0, 2/5, 3/5)A1 55 40 30 For opposition party: (0, 2/5, 3/5)A2 70 70 55 Expected value of game = 61A3 75 55 65

8. Player A

Player B

B1(H) B2(T) For player A : (4/15, 11/15)A1 (H) 8 – 3 For player B : (4/15, 1/15)A2 (T) – 3 1 Expected value of game = 1/15

9. (a)Janata

PrajaB1 B2 For Janata: (0, 1/6, 5/6)

A2 0 5 For Praja: (0, 5/6, 1/6)A3 1 0 Expected value of game = 5/6

(b) Janata may consider to have minor change with probability 5/6 and of major change with probability 1/6.

10. First interchange rows into columns to recast payoff matrix because ‘Union’ is considered as ‘maximizing’ player.

The game has no saddle point. Delete row 4 (dominated by row 3) and then delete column 1 (dominated by columns 2 and 3) and column 4 (dominated by column 3). The reduced payoff matrix is

C2 C3

U1 2.00 1.40 1.50 – 0.80 = 0.70, p(U1) = 7/13U3 0.80 1.50 2.00 – 1.40 = 0.6, p(U3) = 6/13

1.50 – 1.40 = 0.10

2.00 – 0.80 = 1.20

p(C2) = 1/13, p(C3) = 12/13; Value of the game = 18.8/13. 15. (a) Player A: (3/11, 8/11), Player B: (0, 2/11, 9/11) and V = 49/11 (b) Player A: (4/5, 1/5), Player B: (0, 3/5, 0, 2/5, 0, 0,) and V = 17/5 16. Player A: (0, 0, 0, 13/20, 7/20), Player B: (11/20, 9/20) and V = 23/20. 17. (a) Player A: (3/8, 13/24, 1/12), Player B: (7/24, 5/9, 11/72) and V = 91/24 (b) A constant, C = 4 is added to all the elements of the payoff matrix. 18. Company A: ( 2/3, 1/3, 0), Company B: (7/12, 5/12) and V = 1/3

SUMMARY • Dynamic programming is an approach in which the problem is broken down into a

number of smaller subproblems called stages. These subproblems are then solved sequentially until the original problem is finally solved. A particular sequence of alternatives (courses of action) adopted by the decision-maker in a multistage decision problem is called a policy. The optimal policy, therefore, is the sequence


NOTES


of alternatives that achieves the decision-maker’s objective. The solution of a dynamic programming problem is based upon Bellman’s principle of optimality (recursive optimization technique), which states:

The optimal policy must be one such that, regardless of how a particular state is reached, all later decisions (choices) proceeding from that state must be optimal.

Based on this principle of optimality, the best policy is derived by solving one stage at a time, and then sequentially adding a series of one-stage-problems are solved until the overall optimum of the initial problem is obtained. The solution procedure is based on a backward induction process and forward induction process. In the first process, the problem is solved by solving the problem in the last stage and working backwards towards the first stage, making optimal decisions at each stage of the problem. In certain cases, the second process is used to solve a problem by first solving the initial stage of the problem and working towards the last stage, making an optimal decision at each stage of the problem.

• The theory of games provides mathematical models that are useful in explaining interactive decision-making process, where two or more competitions (players) are involved under condition of conflict and competition.

A two-person game allows only two competitors to be involved in the game. In such a game if the sum of gains of one player is equal to the sum of losses of another, then the game is said to be two-person zero-sum game.

For a payoff matrix without saddle point, a number of solution methods can be used to find the value of the game and probabilities of using strategies by both players.


NOTES

Operations Research

STRUCTURE

5.1 Introduction5.2 The General Non-Linear Programming Problem5.3 Graphical Solution Method5.4 Quadratic Programming5.5 Lagrange Multipliers5.6 Conditions for a General Non-Linear Programming Problems5.7 Convex Non-Linear Programming Problem (CNLPP)

U N I T

5NON-LINEAR

PROGRAMMING METHODS

5.1 INTRODUCTIONLinear programming is useful for solving decision problems that involves linear

relationship among decision variables. Any non-linear change in the input variable values either in objective function or constraints, restrict the use of usual simplex method to solve the decision problem. Hence, decision-makers use non-linear programming methods to solve such decision problems.

5.2 THE GENERAL NON-LINEAR PROGRAMMING PROBLEM

The general non-linear programming problem can be stated in the following form:Optimize (max or min) Z = f(x1, x2, ..., xn)Subject to the constraints gi(x1, x2, ..., xn) {≤, =, ≥} bi; i = 1, 2, ..., m

and xj ≥ 0 for all j = 1, 2, ..., nwhere f (x1, x2, ..., xn) and gi (x1, x2, ..., xn) are real valued function of n decision variables, and at least one of these is non-linear.

Several methods have been developed for solving non-linear programming problems. In this chapter we will discuss the methods for solving quadratic programming problems, separable programming problems, geometric programming problems and stochastic programming problems.


NOTES

Non-Linear Programming Methods5.3 GRAPHICAL SOLUTION METHOD

As we know, the optimal solution of an LP problem is obtained at one of the extreme points of the feasible solution space. However, in case of non-linear programming, the optimal solution may not be obtained at the extreme point of its feasible region. This is illustrated through Examples 1 and 2.

Example 1. Solve graphically the following NLP problemMaximize Z = 2x1 + 3x2Subject to the constraints

(i) x12 + x2

2 ≤ 20 (ii) x1 . x2 ≤ 8and x1, x2 ≥ 0

Sol. In the given NLP problem, the objective function is linear, and constraints are non-linear. Plot the given constraints on the graph by the usual method, as shown in Fig. 5.1.

The constraint x12 + x2

2 = 20 represents a circle whose radius and centre are: a = 20; (h, k) = (0, 0) respectively and x1 · x2 = 8 represents a rectangular hyperbola whose asymptotes are represented by the x-axis and y-axis.

Solving the two equations: x12 + x2

2 = 20 and x1 . x2 = 8, we get (x1, x2) = (4, 2) and (x1, x2) = (2, 4). These solution points, which also satisfy both the constraints, may be obtained within the shaded non-convex region OABCD, also called the feasible region.

Now we need to find a point (x1, x2) within the convex region OABCD where the value of the given objective function Z = 2x1 + 3x2 is maximum. Such a point can be located by the iso-profit function approach. That is, draw parallel objective function 2x1 + 3x2 = k lines for different constant values of k, and stop the process when a line touches the extreme boundary point of the feasible region for some value of k. Starting with k = 6 and so on we find that the iso-profit line with k = 16 touches the extreme boundary point C(2, 4) where the value of Z is maximum. Hence the optimal solution is: x1 = 2, x2 = 4 and Max Z = 16.

x2

D

O

2

3

1

C (2, 4)

B (4, 2)

2

A

Z = 6Z = 12

Z = 16x1

�20

x ·x = 81 2

x + x = 201 22 2

1

2

Fig. 5.1. Graphical Solution

Example 2. Graphically solve the following NLP problemMaximize Z = 8x1 – x1

2 + 8x2 – x22

subject to the constraints (i) x1 + x2 ≤ 12, (ii) x1 – x2 ≥ 4and x1, x2 ≥ 0


NOTES

Operations Research Sol. In this NLP problem, the objective function is non-linear whereas the constraints are linear. Plot the given constraints on the graph by the usual method, as shown in the Fig. 5.2.

The feasible region is shown by the shaded region in Fig. 5.2. Thus, in the feasible region the optimal point (x1, x2) must be that at which a side of the convex region is tangent to the circle, Z = 8x1 – x1

2 + 8x2 – x22.

The gradient of the tangent to the circle can be obtained by differentiating the equation of the circle, Z with respect to x1 as follows:

Z = 8x1 – x12 + 8x2 – x2

2

8 – 2x1 + 8 22

12

2

1

dxdx

x dxdx

− = 0

or dxdx

2

1 = 2 88 2

1

2

xx

−−

...(1)

The gradient of the line x1 + x2 = 12 and x1 – x2 = 4 is:

and dx dx dx

dx

dx dx dxdx

1 22

1

1 22

1

0 1

0 1

+ = = −

− = =

or

or ...(2)

respectively.x2

12

10

8

6

4

2

O 2 4 6 8 10 12

C

x1

Z = 24B

2

FeasibleRegion

1

D

1

2

x + x = 121 2

x – x = 41 2

A


If the line x1 + x2 = 12 is the tangent to the circle, then substituting dxdx

2

1 = – 1 from

Eqn. (2) in Eqn. (1), we get:

2 88 2

1

2

xx

−−

= – 1, i.e. x1 = x2

and hence for x1 = x2, the equation x1 + x2 = 12 gives (x1, x2) = (6, 6). This means the tangent of the line x1 + x2 = 12 is at (6, 6). But this does not satisfy all the constraints.

Similarly, if the line x1 – x2 = 4 is the tangent to the circle, then substituting dxdx

2

1

= 1 from Eqn. (2) in Eqn. (1), we get


NOTES

Non-Linear Programming Methods

2 88 2

1 81

21 2

xx

i e x x−−

= + =, . .,

and hence for x1 + x2 = 8, the equation x1 – x2 = 4 gives (x1, x2) = (6, 2). This means the tangent of the circle to the line – x2 = 4 is at (6, 2). This point lies in the feasible region and also satisfies both constraints. Hence, the optimal solution is: x1 = 6, x2 = 2 and Max Z = 24.

Example 3. Solve graphically the following NLP problemMinimize Z = x1

2 + x22

subject to the constraints (i) x1 + x2 ≥ 8, (ii) x1 + 2x2 ≥ 10, (iii) 2x1 + x2 ≥ 10and x1, x2 ≥ 0.

Sol. Since in the given NLP problem all constraints are linear, plotting them on the graphs as usual. The shaded solution space bounded by convex region ABCD is shown in Fig. 5.3. The objective function is non-linear and represents a circle. If r is the radius of the circle, Z = (r) = x1

2 + x22. Then the objective is to determine the minimum value

of r, so that the circle with center (0, 0) and radius, r just touches the solution space. As shown in Fig. 5.3 the solution point (4, 4) lies on the line x1 + x2 = 8, and the line is tangent to the circle at this point.

Since the circle touches one of the sides of the convex region, one of the side the convex solution space would be tangent to the circle. Thus the solution can also be obtained by differentiating the equation: Z = x1

2 + x22 with respect to x1, i.e.

2x1dx1 + 2x2dx2 = 0 or dxdx

2

1 = −

xx1

2.

Differentiate the constraint equation which form the sides of the convex space as follows:

dx1 + dx2 = 0 or dxdx

2

1 = – 1,

dx1 + 2dx2 = 0 or dxdx

2

1 = − 1

2,

and 2dx1 + dx2 = 0 or dxdx

2

1 = – 2.

10

A

8

6

4

2

00 2 4 6 8 10

x2

1

2

B

3 CD

Optimal point

x1

x + x = 8,1 2

x + 2x = 10,1 2

2x + x = 10,1 2

1

2

3



NOTES

Operations Research Three alternative solutions which can now be obtained are:(i) Taking equations first and second and the constraint x1 + x2 = 8, we have

dxdx

2

1 = −

xx1

2= – 1 or x1 = x2, which gives x1 = x2 = 4.

This solution satisfies all the constraint equations and so is feasible.(ii) Taking equations first and third and the constraint x1 + 2x2 = 10, we have

dxdx

2

1 = −

xx1

2= − 1

2, which gives x1 = 2 and x2 = 4.

This solution does not satisfy all the constraints and is discarded. (iii) Taking equations first and third and the constraint 2x1 + x2 = 10, we have

dxdx

2

1 = −

xx1

2 = – 2, or x2 =

x12

, which gives x1 = 4 and x2 = 2.

This solution does not satisfies all the constraints and is discarded.Hence optimal solution is: x1 = 4, x2 = 4, and Min Z = 32.Example 4. Solve graphically the following NLP problem:Maximize Z = x1 + x2

subject to the constraints (i) x1x2 – 2x2 ≥ 3, (ii) 3x1 + 2x2 ≤ 24,and x1, x2 ≥ 0.

Sol. In this NLP problem the objective function is linear, while one of the constraints is non-linear. Plotting the constraint: x1x2 – 2x2 ≥ 3 on the graph assuming that it is an equation: x1x2 – 2x2 = 3 or x2(x1 – 2) = 3. Thus, for x2 ≥ 0, the value of x1 cannot be less than 2.

For different values of x1, the corresponding values of x2 which satisfy the equation x1x2 – 2x2 = 3 are given below:

x1 : 2.1 2.2 2.4 2.6 3.0 3.5 4.0 5.0 6.0 7.0 8.0 12.0x2 : 30 15 7.5 5.0 3 2 1.5 1.0 0.75 0.6 0.5 0.3

When these points are plotted as usual, the graph of the line x1x2 – 2x2 = 3, is shown in Fig. 5.4. Also plotting the constraint, 3x1 + 2x2 = 24, on the graph. The different values of x1 and x2 are, x1 = 0, x2 = 12 and x1 = 8, x2 = 0. The solution space bounded by two lines is shown by shaded area is Fig. 5.4.

The objective function line, Z = x1 + x2 inclined at 45°, when moved away from the origin. The farthest point through which it passes gives optimal solution: x2 = 8.45, x1 = 2.35 and Max Z = 10.81.

Alternative Approach: If 3x1 + 2x2 = 24, then x1 = 24 2

32− x

. Substituting this

value of x1 in other constraint equation, we get

24 2

32

2−

x x – 2x2 = 3 or 24x2 – 2x22 – 6x2 = 9


NOTES

Non-Linear Programming Methods16

14

12

10

8

6

4

2

0 2 4 6 8 10 12

B

2

A

1

1

2

x x – 2x = 31 2 2

3x + 2x = 241 2


or 2x22 – 18x2 + 9 = 0, which gives x2 =

18 324 724

± − = 8.46 or 0.53.

For x2 = 8.46, we get x1 = 24 2 8 463

− ( . ) = 2.34 and Max Z = 10.81.

x2 = 0.53, we get x1 = 24 2 0 533

− ( . ) = 7.64 and Max Z = 8.17.

Hence, the optimal solution is: x1 = 2.34, x2 = 8.46 and Max Z = 10.81.

TEST YOUR KNOWLEDGE

1. Solve the following non-linear programming problems graphically. (a) Max Z = x1

subject to (1 – x1)2 – x2 ≥ 0and x1, x2 ≥ 0

(b) Max Z = x1

subject to (3 – x1)3 – (x2 – 2) ≥ 0 (3 – x1)2 + (x2 – 2) ≥ 0and x1, x2 ≥ 0

Also show that the Kuhn-Tucker necessary conditions for maxima do not hold. What do you conclude?

2. Max Z = x1 + 2x2 subject to x1

2 + x22 ≤ 1

2x1 + x2 ≤ 2and x1, x2 ≥ 0


NOTES

Operations Research 3. (a) Min Z = x12 + x2

2

subject to x1 + x2 ≥ 4 2x1 + x2 ≥ 5and x1, x2 ≥ 0

(b) Min Z = (x1 – 1)2 + (x2 – 2)2

subject to x1 ≤ 2 x2 ≤ 1and x1, x2 ≥ 0

4. (a) Max Z = 100x1 – x12 + 100x2 – x2

subject to x1 + x2 ≥ 80 x1 + 2x2 ≤ 100and x1, x2 ≥ 0

(b) Min Z = (x1 – 2)2 + (x2 – 1)2

subject to – x1 + x2 ≥ 0 x1 + x2 ≤ 2and x1, x2 ≥ 0

5. The operations research team of ABC company has come up with the mathematical data needed for two products that the firm manufactures. It also has determined that this is a non-linear programming problem, having linear constraints and objective function that is the sum of a linear and a quadratic form. The pertinent data gathered by the OR team is:

Max (contribution) Z = 12x + 21y + 2xy – 2x2 – 2y2 subject to the constraints (i) 8 – y ≥ 0, (ii) 10 – x – y ≥ 0, and x, y ≥ 0 Find the maximum contribution and the number of units that can be expected from these

two products, which are a part of the firm’s total output, where x and y represent the number of units of the two products.

6. A company manufactures two products, A and B. It takes 30 minutes to process one unit of product A and 15 minutes to process each unit of B. The maximum machine time available is 35 hours per week. Products A and B require 2 kg and 3 kg of raw material per unit, respectively. The available quantity of raw material is envisaged to be 180 kg per week.

The product A and B, which have unlimited market potential, sell for ` 200 and ` 500 per unit, respectively. If the manufacturing costs for products A and B are 2x2 and 3y2, respectively, find how much of each product should be produced per week, where x and y represent the quantity of products A and B to be produced, respectively.

7. The manufacturing and raw material costs for manufacturing each of the two products N A and B N is proportional to the square of the quantity made. The products are made from a limited supply of a particular raw material and both the processed on the same machine.

It takes 40 minutes to process one unit of product A and 30 minutes to process each unit of B. The machine operates for a maximum of 40 hours a week. Product A needs 2 kg and product B needs 3 kg of raw material, per unit, which is in limited supply of 200 kg per week.

If the net incomes from the products are ` 250 and ` 650 per unit and the manufacturing costs are 2x2 and 3y2 respectively, find how much of each product should be produced, where x and y represent the number of units of product A and B, respectively, to be produced.

8. ABC investment Company’s manager is trying to create a portfolio of two stocks that promise annual returns of 16 and 20 percent, respectively. Additionally, he estimates that the risk of each investment will vary directly with twice the square of the size of investment. Each lot (100 shares) of the first stock costs ` 3000 and each lot (100 shares) of the second stock cost ` 2000. The total amount of cash available for investment is ` 6000. Formulate and solve this problem as a NLP problem to maximize return and minimize risk.


NOTES


9. The total product of a restaurant was found to depend mostly on the amount of money spent of money spent on advertising and the quality of the preparation of the food (measured in terms of the salaries paid to the chefs). In fact the manager of the restaurant found that if he pays his chefs x per hour and spends y a week on advertising, the restaurant’s weekly profit (in rupees) will be: Z = 412x + 806y – x2 – xy.

What hourly wages should the manager pay this chefs and how much should he spend on advertising so as to maximize the restaurant’s profit?

10. A factory is faced with a decision regarding the number of units of a product it should produce during the months of January and February respectively. At the end of January sufficient units must be on hand so as to supply regular customers with a total of at least 100 units. Furthermore at the end of February, the required quantity will be 200 units. Assume the factory ceases production at the end of February. The production cost C is a simple function of output x and is given by C = 2x2. In addition to the production cost, units produced in January, that are not sold until February incur an inventory cost of ` 8 per unit. Assume the initial inventory to be zero. Formulate the problem as an NLP problem and show that the minimum cost solution is to produce 149 units is January and 151 units in February. The number of units produced must be equal to the number demanded and distributed.

Hints and Answers 1. (a) x1 = 1, x2 = 20 and Max Z = 1 (b) x1 = 3, x2 = 2 and Max Z = 3 3. (a) x1 = 2, x2 = 2 and Min Z = 8 (b) x1 = 0, x2 = 1 and Min Z = 2 4. (a) x1 = 60, x2 = 20 and Max Z = 400 (b) x1 = 1, x2 = 1 and Min Z = 1 6. Let x and y = quantity of product A and B to be produced, respectively. Max Z = (200 – 2x2) + (500 – 2y2) subject to (i) 0.5 x + 0.25y ≤ 35, (ii) 2x + 3y ≤ 80; and x, y ≥ 0.

5.4 QUADRATIC PROGRAMMINGAmong several non-linear programming methods available for solving NLP

problems, we shall discuss in this section, an NLP problem with non-linear objective function and linear constraints. Such an NLP problem is called quadratic programming problem. The general mathematical model of quadratic programming problem is as follows:

Optimize (Max or Min) Z = c x x d xj jj

n

j

n

k

n

j jk k= = =∑ ∑ ∑+

1 1 1

12


a xij jj

n

=∑

1 ≤ bi

and xj ≥ 0 for all i and j.In matrix notations, QP problem is written as:

Optimize (Max or Min) Z = cx + 12x DxT


NOTES

Operations Research subject to the constraints Ax ≤ b

and x ≥ 0 ...(3)where x = (x1, x2, ..., xn)T; c = (c1, c2, ..., cn); b = (b1, b2, ..., bm)T ...(4)

D = [djk] is an n × n symmetric matrix, i.e., djk = dkj; A = [aij] is an m × n matrix.If the objective function in QP problem is of minimization, then matrix D is

symmetric and positive definite (i.e., the quadratic term xT Dx in x is positive for all values of x except at x = 0) and objective function is strictly convex in x. But, if the objective function is of maximization, then matrix D is symmetric and negative-definite (i.e., xTDx < 0 for all values of x except for x = 0) and objective function is strictly concave in x. If matrix, D is null, then the QP problem reduces to the standard LP problem.

5.4.1 Kuhn-Tucker ConditionsThe necessary and sufficient Kuhn-Tucker conditions to get an optimal solution

to the maximization QP problem subject to linear constraints can be derived as follows:Step 1: Introducing slack variables si

2 and rj2 to constraints, the QP problem

becomes:

Max f(x) = c x x d xj jj

n

j

n

k

n

j jk k= = =∑ ∑ ∑−1 1 1

12


a xij jj

n

=∑

1 + si

2 = bi; i = 1, 2, ..., m

– xj + rj2 = 0; j = 1, 2, ..., n

Step 2: Forming the Lagrange function as follows:

L(x, s, r, l, µ) = f(x) – j

n

i ij j i ia x s b=∑ + −1

2λ { } – j

n

j j jx r=∑ − +1

2µ { }

Step 3: Differentiate L(x, s, r, l, µ) partially with respect to the components of x, s, r, l and µ. Then equate these derivatives with zero in order to get the required Kuhn-Tucker necessary conditions. That is,

(i) c – 12

(2xT D) – lA + µ = 0, or

cj – k

n

k jk i ij ji

mx d a

= =∑ ∑− +

1 1λ µ = 0; j = 1, 2, ..., n

(ii) – 2ls = 0 or li si2 = 0, or

λi ij ji

n

ia x b=∑ −

1 = 0, i = 1, 2, ..., m

(iii) – 2µ r = 0 or µjrj = 0, j = 1, 2, ..., n µj xj = 0, j = 1, 2,..., n


NOTES

Non-Linear Programming Methods (iv) Ax + s2 – b = 0; i.e. Ax ≤ b, or

a xij jj

n

=∑

1 ≤ bi, i = 1, 2,..., m

(v) – x + r2 = 0, i.e. x ≥ 0, or xj ≥ 0, j = 1, 2, ..., n

(vi) li, µj, xj, si, rj ≥ 0These conditions, except (ii) and (iii), are linear constraints involving 2(n + m)

variables. The condition µj xj = li si = 0 implies that both xj and µj as well as si and li cannot be basic variables at a time in a non-degenerate basic feasible solution. The conditions µj xj = 0 and li si = 0 are also called complementary slackness conditions.

TEST YOUR KNOWLEDGE

1. What is meant by quadratic programming? How does a quadratic programming problem differ from a linear programming problem? Give an example.

2. Derive the Kuhn-Tucker necessary conditions for an optimal solution to a quadratic programming problem.

3. Obtain the Kuhn-Tucker conditions for a solution of the problem.

Max Z = cx + 12

xT dx,

subject to Ax = b and x ≥ 0.

5.5 LAGRANGE MULTIPLIERSHere we discussed the optimization problem of continuous functions. The non-linear

programming problem is composed of some differentiable objective function and equality side constraints, the optimization may be achieved by the use of Lagrange multipliers. A Lagrange multiplier measures the sensitivity of the optimal value of the objective function to change in the given constraints bi in the problem. Consider the problem of determining the global optimum of

Z = f(x1, x2,..., xn)subject to the

gi(x1, x2, ..., xn) = bi, i = 1, 2, ..., m.Let us first formulate the Lagrange function L defined by: L (x1, x2, ..., xn; l1, l2, ..., lm) = f(x1, x2, ..., xn) + l1 g1(x1, x2, ..., xn) + l2 g2 (x1, x2, ..., xn) + ... + lmgm (x1, x2, ..., xn)

where l1, l2, ..., lm are Lagrange Multipliers.For the stationary points

∂∂Lx j

= 0, ∂∂Liλ

= 0 ∀ j = 1(1)n ∀i = 1(1)m

Solving the above equation to get stationary points.


NOTES

Operations Research ObservationAlthough the Lagrangian method is often very useful in application yet the

drawback is that we cannot determine the nature of the stationary point. This can sometimes be determined from physical consideration of the problem.

Example 1. A rectangular box open at top is to have volume of 32 cubic meters. Find the dimensions of the box requiring the least material for its construction.

Sol. Let x1, x2, x3 be the sides of the rectangular face with x1, x3 in the sides of the bottom faces and s be its surface then

s = 2x1x2 + 2x2x3 + x1x3

subject to x1x2x3 = 32 and x1, x2, x3 > 0Form the Lagrangian function L(x1, x2, x3, l) = 2x1x2 + 2x2x3 + x1x3 + l(x1x2x3 – 32)The stationary points are the solutions of the following:

∂∂Lx1

= 2x2 + x3 + lx2x3 = 0 ...(1)

∂∂Lx2

= 2x1 + 2x3 + lx1x3 = 0 ...(2)

∂∂Lx3

= 2x2 + x1 + lx1 x2 = 0 ...(3)

∂∂Lλ

= x1x2x3 – 32 = 0 ...(4)

From equations (1) and (2), we get x1 = 2x2

and from equations (1) and (3), we get x2 = x3

Putting these value in equation (4), we get x1 = 4, x2 = 2, x3 = 4 Min. S = 48

5.6 CONDITIONS FOR A GENERAL NON-LINEAR PROGRAMMING PROBLEMS

5.6.1 Necessary and Sufficient Conditions for OptimalityOne can find many necessary conditions with different strong assumptions in

the literature, but only a few of them are directly applicable to practical optimization problems. The two fundamental results, which have an impact on modern non-linear programming algorithms, are the Fritz John and the Karush-Kuhn-Tucker conditions.

A prerequisite for stating the necessary conditions for the problem is the definition of the Lagrangian function, named after the mathematician Lagrange who treated equality-constrained optimization problem in the second half of the eighteenth century.


NOTES


Definition. (Lagrangian Function (Nocedal and Wright) The mapping :

N N Ny g h× × × → is defined by

(y, l0, l, µ): = l0 f(y) + i

N

i ii

N

i i

g h

g y h y= =∑ ∑+1 1

λ µ( ) ( ) ...(1)

and is called the Lagrangian function for the problem. l0 and the components of the vectors

lT = [l1, ..., λNg] and µT : = [ , ..., ]µ µ1 Nh

...(2)

are called Lagrange multipliers.Theorem. (First-Order Necessary Conditions, Fritz John Conditions) Suppose

that the functions f(· ), g(· ) and h (· ) are continuously differentiable with respect to y. Then, the following Fritz John conditions are satisfied at a local minimum (y*, l, µ): 1. sign condition:

l0 ≥ 0; ...(3) 2. optimality condition:

∇y (y*, l0, l, µ) = l0 ∇f(y*) + ∇gT(y*)l + ∇hT (y*)µ = 0. ...(4)

where the Jacobian matrices of g(· ) and h(· ) are of size ∇g(y*) ∈ N Ng y. and ∇h(y*) ∈ N Nh y. ;

3. complementarity conditions: l ≥ 0, g(y*) ≤ 0, lT g(y*) = 0; ...(5)

4. equality constraints: h(y*) = 0. ...(6)Proof: Proofs can be found in John, Mangasarian, and Bertsekas.

Each vector (y*, l0, l, µ) ∈ N N Ny g h+ + +1 which satisfies the conditions is called a Fritz John stationary point. It can be shown that for a local minimum y* a point (l0, l, µ) ≠ 0 with l0 > 0 only exists if an additional condition, called a regularity condition or a constraint qualification, is satisfied. If such a constraint qualification holds, l0 = 1 can be chosen without loss of generality.

In textbooks on optimization one can find numerous different constraint qualifications for non-convex problems. In general, it is desirable to use the weakest constraint qualification in order to restrict the allowed constraints as little as possible. But it is a major drawback of the weaker constraint qualifications, e.g., the quasiregularity condition, that they cannot be evaluated in a simple manner. Therefore, stronger but much easier evaluable constraint qualifications like the Mangasarian-Fromowitz constraint qualification or the even stronger linear independence constraint qualification are often used. They are defined as follows.

Definition. (Mangasarian-Fromowitz Constraint Qualification) Let a point y ∈ S be feasible and the active set I(y) be defined. Then, the Mangasarian-Fromowitz constraint qualification (MFCQ) holds if the gradients of the equality constraints ∇hj (y), j = 1, ...., Nh are linearly independent (i.e., the constraints in this case are said to be regular) and there exists a vector b ∈ Ny such that

∇hjT(y)b = 0 (j = 1, ..., Nh) and ∇gi

T(y)b < 0 (∀i ∈ (y)).


NOTES

Operations Research Definition. (Linear Independence Constraint Qualification) Let a point y ∈ S be feasible and the active set I(y) be defined. Then, the linear independence constraint qualification (LICQ) holds if the gradient ∇ gi (y), ∀i ∈ I(y) and ∇ hj (y), j = 1, ..., Nh are linearly independent.

Remark. One can realize that these constraint qualifications are strongly restrictive. One can even show that these constraint qualifications are over restrictive just with a simple test. If an equality constraint hi(y) = 0 is replaced by two equivalent inequality constraints, e.g., with gi(y) < = 0 and gi (y) > = 0, neither the MFCQ nor the LICQ holds.

Theorem. (Karush–Kuhn–Tucker conditions) Assume that the functions f(· ), g(· ), and h(· ) are continuously differentiable with respect to y and a constraint qualification holds. Then the following conditions at a local minimum (y*, l, µ)

∇y (y*, 1, l, µ) = 0 ...(7) ∇µ (y*, 1, l, µ) = h(y*) = 0 ...(8) ∇l (y*, 1, l, µ) = g(y*) ≤ 0 ...(9) lT g(y*) = 0 ...(10) l ≥ 0 ...(11)

are satisfied and are known as Karush–Kuhn–Tucker (KKT) conditions.Definition. (Strict Complementarity Condition) The strict complementarity

condition is satisfied, if l – g(y) > 0 ...(12)

holds.In order to ensure that any stationary KKT point (y*, l, µ) is indeed an optimal

solution of problem, second-order sufficient conditions (SOSC) are needed.

5.7 CONVEX NON-LINEAR PROGRAMMING PROBLEM (CNLPP)

Now we define a special class of NLPP, namely, CNLPP i.e., convex Non-Linear Programming Problem.

An NLPP is called a CNLPP, if it is of the formMin f(x)

subject to gi(X) ≤ 0 ∀ i = 1, 2, ..., mand, X ≥ 0where, f(X) and gi(X) are all convex functions in En.

It should be noted that a CNLPP is always taken as minimisation problem.we first show that the set of the BFS of CNLPP is a convex set.

Theorem. The set SF of BFS of a CNLPP is convex.Proof. Let CNLPP beMin f(X)

subject to gi(X) ≤ 0, i = 1, 2, ..., m X ≥ 0Let X1 , X2 ∈ SF and X = (1 – l)X1 + lX2, 0 ≤ l ≤ 1.Then, gi(X) = gi ((1 – l)X1 + lX2) ≤ (1 – l) gi(Xi) + lgi(X2)

because gi(X) are convex functions.


NOTES


gi(X) ≤ 0as g1 (X1), gi(X2) ≤ 0 and l(1 – l) ≥ 0

Hence, X satisfies the constraints. Since X1, X2 ≥ 0 and l, (1 – l) ≥ 0, we have X ≥ 0. Therefore, X ∈ SF . Hence, SF is convex.

If all the constraints of CNLPP are strictly equality then we can employ Lagrange’s Multiplier method to solve it. If one or more constraints are inequalities, then the Lagrange’s multiplier method cannot be used.

In this case we give another method, which is Kuhn-Tucker Method and is an extension of Lagrange’s Multiple Method.

We construct the function L(X, l) = f(X) + l1g1(X) + l2g2(X) + ... + lmgm (x) = f(X) + lT G(X)Where, l = (l1, l2, ..., lm)T

and, G(X) = (g1(x), g2(x), ..., gm(x))T

As usual li’s are called Lagrange’s Multipliers.We now first define a point which is going to play an important role, namely,

saddle point and then prove a result showing the relation between a saddle point and the optimal solution of CNLPP.

5.7.1 Non-negative Saddle Point and its ConditionDefinition: The point (X*, l*) is called a non-negative saddle point of L(X, l), if L(X*, l) ≤ L(X*, l*) ≤ L(X*, l*) ∀ X, l ≥ 0

i.e., it is a non-negative minimum point with respect to X and non-negative maximum point with respect to l of L(X, l*) and L(X*, l) respectively. Clearly, X*, l* ≥ 0.

Theorem. If (X*, l*) is the non-negative saddle point of L(X, l) then X* is the optimal solution of the CNLPP whose Lagrangian function is L(X, l).

Proof. Let the CNLPP beMin f(x)

subject to g1(X) ≤ 0, X ≥ 0We need to show that X* satisfies the constraints i.e., gi(X*) ≤ 0, X* ≥ 0

and, f(X*) ≤ f(X) ∀ X ∈ S, set of feasible solution.Given that (X*, l*) is a non-negative saddle point of L(X , l). Therefore, L(X*, l) ≤ L(X*, l*) ≤ L(X, l*) ∀ X, l ≥ 0.

i.e., f(X*) + lT G(X*) ≤ f(X*) + l*T G(X*) ≤ f(X) + l*TG(X)which gives from the left inequality,

lTG(X*) ≤ l*TG(X*) ∀ l ≥ 0or (l – l*)T G(X*) < 0 ∀ l > 0 L(X*, l) ≤ L(X*, l*) ≤ L (X, l*) ∀ X, l ≥ 0i.e., f(X*) + lTG(X*) ≤ f(X*) + l*TG(X*) ≤ f(X) + l*T G(X)which gives from the left inequality.

lT G(X*) ≤ l*T G(X*) ∀ l ≥ 0


NOTES

Operations Research or, (l – l*)T G(X*) ≤ 0 ∀ l ≥ 0which means G(X*) ≤ 0, i.e., g(X*) ≤ 0 ∀ i = 1, 2, ..., m because it is.

Already done that X* ≥ 0. Hence, X* satisfies the constraints.From the right hand inequality, we have f(X*) + l*T G(X*) ≤ f(x) + l*TG(X).Now X ∈ S, therefore G(X) ≤ 0, X ≥ 0. Also l* ≥ 0.∴ l*TG (X) ≤ 0Since, X* ∈ S, we get l*T G(X*) ≤ 0.

TEST YOUR KNOWLEDGE

1. A company has determined that its function is the Cobb-Douglas function f(x, y) = x2/3y1/3 where x is the number of labor hours and y is the number of capital units. The budget constraint for the company is given by 100x + 100y = 400000. Then find the maximum production using Lagrange multiplier. [Ans. 2116 units; x = 2666, y = 1333]

2. The equation P(x, y) = – 2x2 + 60x – 3y2 + 72y + 100 model profit when x represents the number of handmade chairs and y is the number of handmade rockers produced per week. The maximum situation occurs when x = 15 and y = 12. However due to an insufficient labor force they can only make a total of 20 chairs and rockers per week (x + y = 20). So, how many chair and how many rockers will give the realistic maximum profit?

[Ans. 10 chairs, 9 rockers] 3. What is meant by non negative saddle point? Write down the necessary condition for non-

negative saddle point. 4. Derive the necessary conditions for a general non-linear programming problem. 5. What is Karush–Kuhn–Tucker condition for non-linear programming problem? 6. Discuss about convex non-linear programming problem.

SUMMARY

• Linear programming requires the objective functions and constraints to be linear. However, if either of these are not linear, then non-linear programming methods are used to find optimal value of the objective function with or without constraints. In the more general procedure, conditions necessary for an optimum value of a function subject to inequality constraints, are known as Kuhn-Tucker conditions.

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