IGNOU MBA Operations Research

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Table of contents :
UNIT 1 OPERATIONS RESEARCH-AN An Overview OVERVIEW......Page 1
UNIT 2 REVIEW OF PROBABILITY ANDSTATISTICS......Page 12
UNIT 3 LINEAR PROGRAMMING – Graphical Method GRAPHICAL METHOD......Page 25
UNIT 4 LINEAR PROGRAMMING -SIMPLEX METHOD......Page 50
UNIT 5 TRANSPORTATION PROBLEM......Page 75
UNIT 6 ASSIGNMENT PROBLEM......Page 93
UNIT 7 GOAL PROGRAMMING......Page 108
UNIT 8 INTEGER PROGRAMMING......Page 123
UNIT 9 DYNAMIC PROGRAMMING......Page 139
UNIT 10 NON – LINEAR PROGRAMMING......Page 155
UNIT 11 INVENTORY CONTROL - Models DETERMINISTIC MODELS......Page 174
UNIT 12 INVENTORY CONTROL :PROBABILISTIC MODELS......Page 203
UNIT 13 QUEUEING MODELS......Page 226
UNIT 14 COMPETITIVE SITUATIONS:GAME THEORY......Page 242
UNIT 15 SIMULATION......Page 258
CASE 1 INSULATOR INDIA LIMITED......Page 270
CASE 2 USE OF OPERATIONS RESEARCH TECHNIQUES:A CASE STUDY OF ECS CORPORATION......Page 273

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UNIT 1 OPERATIONS RESEARCH-AN OVERVIEW

Operation Research An Overview

Objectives After studying this unit, you should be able to: • • • •

Explain the meaning, purpose, and limitations of Operations Research Describe the historical background of Operations Research Discuss the approach and tools of Operations Research Describe the delicate relationship between Operations Research Specialist and Manager.

Structure 1.1

Introduction

1.2

History

1.3

Approach, Techniques and-Tools

1.4

Relationship between O.R. Specialist and Manager

1.5

Typical Applications of O.R

1.6

Phases and Processes of O.R. Study

1.7

Limitations of Operations Research

1.8

Summary

1.9

Key Words

1.10 Self Assessment Exercise 1.11 Further Readings

1.1

INTRODUCTION

Operations Research (O.R.) is a relatively a new discipline. Its content and boundaries are not yet fixed. Therefore, proposing a formal definition of the term OPERATIONS RESEARCH (Or OPERATIONAL RESEARCH as some people call it) is a difficult task. Operations Research begins when some mathematical and quantitative techniques are used to substantiate the decision being taken. We make decisions in our everyday life without even noticing them. Decision making is one of the main activities of a manager. In simple situations decisions are taken simply by common sense, sound judgment and expertise without using any mathematics. But here the decision we are concerned with are rather complex and heavily loaded with responsibility. Examples of such decision are finding the appropriate product mix when there are large number of products with different profit contributions and productional requirement or planning public transportation network in a town having its own layout of factories, apartment blocks etc. Certainly in such situations also decisions may well be arrived at intuitively from experience and common sense, yet they are more judicious if backed up by mathematical reasoning. The search of a decision may also be done by trial and error but such, a search may be cumbersome and costly. Preparative calculations may avoid long and costly research. Doing preparative calculations is the purpose of operations research. Operation Research does mathematical scoring of consequences of a decision with the aim of optimizing the use of time, efforts and resources and avoiding blunders. The tools of operations research are not from any one discipline, rather Mathematics, Statistics, Economics, Engineering, Psychology, etc. have contributed to this newer discipline of knowledge. Operations research takes tools from subjects like mathematics, statistics, economics, engineering, psychology etc. and combines them to make a new body of knowledge for decision making. Today, it has become a professional discipline that deals with the application of scientific methods for decision making, and especially to the allocation of scarce resources. O.R. worker tries to find order in apparent chaos by identifying the structure in complex situations. He develops the understanding of how the components of organizations interact, so as to explain and predict the effects of actions taken on

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Introduction to Operation Research

these components. Much of this work is done with the help of analytical and numerical techniques and by developing and manipulating mathematical and computer models of organizational systems. Such systems may be composed of people, machine and procedures. Complete information about such a system may not be available. A purpose of O.R. is to provide a rational basis for making decisions in the absence of complete informations. Operations Research can also be treated as science devoted to describing, understanding and predicting the behaviour of systems, particularly man-machine systems. Thus operations research workers are engaged in three classical aspects of science: a) Describing the behaviour of systems. b) Analyzing this behaviour by constructing appropriate models. c) Using these models to predict future behaviour, that is, the effects that will be produced by changes in the systems or in the methods of operations. The operating systems studied by operations research workers arise in a wide variety of practical, military, industry, and governmental environment. Thus, the results of their research frequently make important contributions to solutions of problems of choice, policy, and planning that arise in these environments. It is to be noted that operations research workers in an organization are not the decision makers themselves. They merely present their findings to the executives in-charge of operations, who is supposed to make decisions. Operations research function is a staff function. Operations research is assistance to executives in improving the operations under their control. What particularly distinguishes operations research from other research and engineering is its emphasis on analysis of operations as a whole. It is an interdisciplinary approach that provided useful solutions to problems of military operations in World War II, and has since been found equally successful in nonmilitary operations. Most present-day business applications are primarily concerned with mathematical and statistical analysis of the results of possible alternative actions. Often using techniques especially designed or refined for business problems, operations researchers have been able to provide remarkable and diverse benefits to companies. Some such benefits are improved inventory and reorder policies, minimum cost production schedules, optimum location and size of warehouses, and guidance in sales and advertising policies. The basic pattern applied is clarification of various courses of action open, estimation of the outcome to be expected from each, and evaluation of these in terms of the overall goal desired. As stated earlier defining O.R. is a difficult task. Salient aspects related to definition stressed by various experts on the subject are as follows: a) Pocock stresses that O.R. is an applied science: he states "OR is scientific methodology-analytical, experimental, quantitative-which by assessing the overall implication of various alternative courses of action in a management system, provides an improved basis for management decisions." b) Morse and Kimball have stressed the quantitative approach of O.R. and have described it as "a scientific method of providing executive departments with a quantitative basis for decisions regarding the operations under their control." c) Miller and Starr see O.R. as applied decision theory. They state, "O.R. 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 problem."

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d) Saaty considers O.R. as tool of improving the quality of answers to problems. He says, "O.R. is the art of giving bad answers to problems which otherwise have worse answers. All these definitions put together enable us to know what -O.R. is, and what it does. Activity 1 Define Operations Research. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

………………………………………………………………………………………… …………………………………………………………………………………………

1.2

Operation Research An Overview

HISTORY

O.R. has its beginning in World War II. The term, operations research, was coined by McClosky and Trefthen in 1940 in U.K. British scientists set up the first field installations of radars during the battle and observed air operations. Their analysis of these led to suggestions that greatly improved and increased the effectiveness of British fighters, and contributed to successful British defence. Operations research was then extended to antisubmarine warfare and to all phases of military, navel, and air operations, both in Britain and the United States, and was incorporated in the post-war military establishments of both the countries. The effectiveness of operations research in military spread interest in it to other governmental departments and industry. In the U.S.A. the National Research Council formed a committee on operations research in 1951, and the first book on the subject "Methods of Operations Research", by Morse and Kimball, was published. In 1952 the Operations Research Society of America came into being. Success of O.R. in military attracted the attention of industrial managers who were seeking solutions to their complex problems. Today, almost every large organisation or corporation in affluent nations has staff applying operations research, and in government the use of operations research has, spread from military to widely varied departments at all levels. This general acceptance to O.R. has come as managers have learned the advantage of the scientific approach on which O.R. is based. Availability of faster and flexible computing facilities and the number of qualified O.R. professionals enhanced the acceptance and popularity of the subject. The growth of O.R. has not been limited to the U.S.A. and the U.K. It has reached to many countries of the world: Indicative of this is that the International Federation of Operations Research Societies, founded in 1959, now comprises member societies from many countries of the world. India was one of the first few countries who started using O.R. In 1949, first O.R. unit was established in Regional Research Laboratory at Hyderabad. At about same time another group was set up in Defence Science Laboratory to solve the problems of stores, purchase and planning. In 1953, O.R. unit was established in Indian Statistical Institute, Calcutta, with the aim of using O.R. methods in national planning and survey. O.R. Society of India was formed in 1955. The society is one of the first members of International Federation of O.R. societies. The society started publishing OPSEARCH, a learned journal on the subject in 1963. Today O.R. is a popular subject in management institutes and schools of mathematics and is gaining currency in industrial establishments.

1.3

APPROACH, TECHNIQUES AND TOOLS

An O.R. team (in some companies this may be single individual) consists of trained researchers utilizing the skills and tools of applicable sciences, without any preconceived ideas of what the correct solution should be. The first step is to formulate the problem, after which a mathematical model is generally constructed to represent the system being studied. Mathematical models or conceptual models may be in the form of equations or formulae developed to relate important factors or variables of the operations under study. The model then can be tested and operated upon mathematically to determine the effects of changing the values of the variables. Optimization of some criterion may be purpose of such an exercise. Optimization means achieving the best-maximum or minimum value of the criterion. Sub-optimization occurs either when some criterion subordinate to the overall criterion is optimized or when the overall criterion is optimized over only a portion of the total of all possible alternative actions. A criterion is a measure by which results

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Introduction to Operation Research

can be evaluated. It measures the effectiveness of the operations under study in terms of the ultimate objective of the operations, and may be net profit, return on investment, cost or some other thing appropriate to a particular study. Operations research uses any suitable techniques or tools available. These frequently include common mathematical or statistical procedures, cost analysis, or electronic computation. However, O.R. analysts have given special impetus to the development and use of the techniques like, linear program, waiting line theory, game theory, inventory control models and simulation. In addition, some other common tools are non-linear programming, integer programming, dynamic programming, sequencing theory, Markov process, network scheduling-PERT and CPM, symbolic logic, information theory and utility/value theory. The list, of course, is not exhaustive. Detailed discussion on some of these techniques will be presented in appropriate units. But brief explanation of these term will not be out of place here. 1) Linear Programming Linear Programming is basically a constrained optimization technique which tries to optimize some criterion within some constraints. It consists of an objective function which is some measure of effectiveness like profit, loss or return on investment and several boundary conditions putting restriction on the use of resources. Objective function and boundary conditions are linear in nature. There are methods available to solve a linear programming problem. 2) Waiting Line Theory or Queuing Theory Waiting line theory or queuing theory deals with the situation in which queue is formed. Customers waiting for service, machines waiting for repairmen and aircraft's waiting for landing strips are some of the situations in which queue is formed. If we assume that there are costs associated with waiting in line, and if there are costs of adding more channels (service facilities), we want to minimize the sum of costs of waiting and the costs of providing service facilities. Waiting line theory helps to make calculations like expected number of people in the queue, expected waiting time in the queue, expected idle time for the server, etc. These calculations then-can be used to determine the desirable number of service facilities. 3) Game Theory Game theory is used for decision making under conflicting situations where there are one or more opponents. Opponents, in game theory, are called players. The motives of the players are dichotomized. The success of one player tends to be at the cost of others and hence they are in conflict. Game theory models a conflict situation and helps us to improve the decision process by formulating appropriate strategy. 4) Inventory Control Models When to buy, how much to buy and how much to keep in stores are some of the questions which production managers, purchase managers and material managers address themselves to. Inventory _ control models provide rational answer to these question in different situation of supply and demand for different kind of materials. Inventory control models help managers to decide reordering time, reordering level and optimal ordering quantity. The approach is to prepare a mathematical model of the situation that expresses total inventory costs in terms of demand, size of order, possible over or under stocking and other relevant factors and then to determine optimal order size, optimal order level etc. using calculus or some other technique. 5) Simulation

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Simulation is basically data generation technique. Sometimes it is very risky, cumbersome or time consuming to conduct real study or experiment to know more about a situation or problem. Analytical methods also cannot be used in all cases. Sometimes due to large number of variables or large number of interrelationships among the variables and the complexity of relationships, it is not possible to develop an analytical model representing the situation. Even if model building is possible, solving the model may not be possible. In such situations simulation is used. It is to be noted that simulation does not solve a problem; it only generates information or data needed for decision problem solving or decision making. Thus, simulation is a

data generation technique and is used when actual experimentation is not feasible, analytical model building or solution of model is not possible.

Operation Research An Overview

6) Non-Linear Programming Non-linear programming methods may be used when either the objective function or some of the constraints are not linear in nature. Non-linearity may be introduced by such factors as discount on price of purchase of large quantities and graduated income tax etc. Linear relationships may be employed to approximate non-linear conditions, but the approximation becomes poorer as the range is extended. Nonlinear methods may be used to determine the approximate area in which a solution lies and linear methods may then be used to obtain a more exact solution. 7) Integer Programming Integer programming method may be used when one or more of the variables can take only integral values. Examples are the number of trucks in a fleet, the number of generators in a powerhouse and so on. Approximate solutions can be obtained without using integer programming methods, but the approximation generally becomes poorer as the numbers become smaller. There are techniques .to obtain solution of integer programming problems. 8) Dynamic Programming Dynamic programming is a method of analyzing multistage decision processes, in which each elementary decision is dependent upon those preceding it as well as upon external factors. It drastically reduces the computational efforts otherwise necessary to analyze results of all possible combinations of elementary decisions. 9) Sequencing Theory Sequencing theory is related to Waiting Line Theory. It is applicable when the facilities are fixed, but the order of servicing may be controlled. The scheduling of service or the sequencing of jobs is done to minimize the relevant costs. 10) Markov Process Markov process for decision making is used in situations where various states are defined. The probability of going from one state to another is known and depends on the present state and is independent of how we have arrived at that state. Theory of Markov process helps us to calculate long run probability of being in a particular state (steady state probability). This steady state probability is used in decision making. 11) Network Scheduling-PERT AND CPM Network scheduling is technique used to plan, schedule and monitor large projects. Such large projects are very common in the field of construction, maintenance, computer system installation, research and development designs etc. The technique aims at minimizing trouble spots, such as, delays, interruptions and production bottlenecks by identifying critical factors and coordinating various parts of overall job. The whole projector job is diagrammatically represented with the help of network made of arrows representing different activities and interrelationships among them. Such a representation is used for identifying critical activities and critical path. Two basic techniques in network scheduling are Program Evaluation and Review Technique (PERT) and Critical Path Method (CPM). CPM is used when time taken by activities in the project are known for sure and PERT is used. when activities time is not known for sure-only probabilistic estimate of time is available. 12) Symbolic Logic Symbolic logic deals with substituting symbols for words, classes of things, or functional systems. It incorporates rules, algebra of logic and propositions. There have been only limited attempts to apply this technique to business problems; however it has had extensive application in the design of computing machinery. 13) Information Theory Information theory is-an analytical process transferred from the electrical communications field to operations research. It seeks to evaluate the effectiveness of information flow within a given system. Despite its application mainly to

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Introduction to Operation Research

communications networks, it has had an indirect influence in simulating the examination of business organizational structures with a view to improving information or communication flow. 14) Utility/Value Theory Utility/Value theory deals with assigning numerical significance to the worth of alternative choices. To date this has been only a concept, and is in the stage of elementary model formulation and experimentation. When developed this may be very helpful in the decision making process in assessing the worth of various possible outcomes. Activity 2 Describe some of the tools of Operations Research. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

1.4

RELATIONSHIP BETWEEN O.R. SPECIALIST AND MANAGER.

The role of the O.R. specialist is to help the manager make better decisions. Decision making is a key responsibility of managers. Managers now need and in the future will need Q.R. specialists because the size of organizations is increasing to the point where they are becoming unmanageable by traditional means and methods. Even without this problem, managers face enormously complicated and fast changing situations. Creative and workable solutions to these problems require the cooperative involvement of O.R. specialists and managers. Decision makers move towards O.R. a proach p when: a) They see the problem as complex involving many variables and relationships. b) They don't think they can develop a solution without O.R. methods. c) They view the problem as repetitive, thus quantitative solution which is repeated will save the time and money. d) They initially feel that the data in the problem are numeric. e) They know of similar situations in which decisions have been improved by using O.R. techniques. f) They had personal experience of applying O.R. methods to solve problems. g) The decision environment lends itself to a specification of goals. Skills in qualitative analysis are inherent in the manager and generally increase with experience. Skills in quantitative analysis can be acquired by study of mathematical tools such as those discussed above. Using these tools, "managers can improve their decision making effectiveness. They can compare and combine the qualitative and quantitative information at their disposal and thus make the best possible decisions. Activity 3 Comment on the relationship between O.R. specialist and manager.

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………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………….

1.5

TYPICAL APPLICATIONS OF O.R.

Operation Research An Overview

Although the complete list of O.R. techniques and their applications would fill volumes itself, the following is an abbreviated set of applications to show how widely these techniques are used today: 1)

Accounting: Cash flow planning. Credit policy analysis. Planning of delinquent account strategy.

2)

Construction: Allocation of resources to projects. Determination of proper work force. Deployment of work force. Project scheduling, monitoring and control.

3)

Facilities Planning: Factory size and location decision. Hospital planning. International logistics systems design. Estimation of number of facilities required. Transportation loading and unloading. Warehouse location decision.

4)

Finance: Dividend policy making. Investment analysis. Portfolio analysis.

5)

Manufacturing: Inventory control. Projection marketing balance. Production scheduling. Production smoothing.

6)

Marketing: Advertising budget allocation. Product introduction timing. Selection of product mix.

7)

Organizational behaviour: Personnel justification/planning. Scheduling of training programs. Skills balancing. Recruitment of Employees.

8)

Purchasing: Materials transfer. Optimal buying. Optimal reordering.

9)

Research and Development: Control of R & D projects. Product introduction planning.

A similar list can be prepared for any major field of human endeavour. Military activities alone would cover an entire book. Activity 4 Identify some areas of application of O.R. technique in your organization. …………………………………………………………………………………………

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Introduction to Operation Research

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………….

1.6

PHASES AND PROCESSES OF O.R. STUDY

O.R. is a logical and systematic approach to provide a rational basis for decision making. The phases and processes of O.R. study must also be quite logical and systematic. There are six important steps in O.R. study, but it is not necessary that in all the studies each and every step is invariably present. These steps are arranged in following logical order. Step I : Observe the Problem Environment Step I in the process of O.R. study is observing the problem environment. The activities that constitute this step are visits, conferences, observations, research etc. With the help of such activities, the O.R. scientist gets sufficient information and support to proceed and is better prepared to formulate the problem. Step II : Analyze and Define the Problem Step II is analyzing and defining the problem. In this step not only the problem is defined, but also uses, objectives and limitations of the study are stressed in the light of the problem. The end results of this step are clear grasp of need for a solution and understanding of its nature. Step III : Develop a Model Step III is to construct a model. A model is representation of some real or abstract situation. O.R. models are basically mathematical models representing systems, processes or environment in form of equations, relationships or formulae. The activities in this step defining interrelationships among variables, formulating equations, using known O.R. models or searching suitable alternate models. The proposed model may be field tested and modified in order to work under stated environmental constraints. A model may also be modified if the management is not satisfied with the answer that it gives. Step IV : Select Appropriate Data Input Garbage in and garbage out is a famous saying. No model will work appropriately if data input is not appropriate. Hence, taping right kind of data is a vital step in O.R. process. Important activities in this step are analyzing internal-external data and facts, collecting opinions and using computer data banks. The purpose of this step is to have sufficient input to operate and test the model. Step V : Provide a Solution and test Reasonableness Step V in O.R. process is to get a solution with the help of model and data input. Such a solution is not implemented immediately. First the solution is used to test the model and to find limitations if any. If the solution is not reasonable or if the model is not behaving properly, updating and modification of the model is considered at this stage. The end result of this step is solution that is desirable and supports current organizational objective. Step VI : Implement the Solution

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Implementation of the solutions obtained in previous step is the last step of O.R. process. In O.R. the decision making is scientific but implementation of decision involves so many behavioural issues. Therefore, the implementing authority has to resolve the behavioural issues. He has to sell the idea of use of O.R. not only to the workers but also to the superiors. Distance between management and O.R. scientist may offer a lot of resistance. The gap between one who provides a solution and one who wishes to use it should be eliminated. To achieve this O.R. scientist as well as

management should play a positive role. A properly implemented solution obtained through O.R. techniques results in improved working and wins the management support.

Operation Research An Overview

A brief summary of steps, process activities and process output is presented below:

1.7

LIMITATIONS OF OPERATIONS RESEARCH

Operations Research has certain limitations. However, these limitations are mostly related to the problems of model building and the time and money factors involved in its application rather than its practical utility. Some of them are as follows : 1) Magnitude of. Computations: O.R. tries to find out optimal solution taking into account all the factors. In the modern society these factors are enormous and expressing them in quantity and establishing relationships among these require voluminous calculations which can only be handled by machines. 2) Non-Quantifiable Factors: O.R. provides solution only when all elements related to a problem can be quantified. All relevant variables do not lend themselves to quantification. Factors which cannot be quantified, find no place in O.R. Models in O.R. do not take into account qualitative factors or emotional factors which may be quite important. 3) Distance between Manager and Operations Researcher: O.R. being specialist's job requires a mathematician or a statistician, who might not be aware of the business problems. Similarly, a manager fails to understand the complex working of O.R. Thus there is a gap between the two. Management itself may offer a lot of resistance due to conventional thinking. 4) Money and Time Costs: When the basic data are subjected to frequent changes, incorporating them into the O.R. models is a costly affair. Moreover, a fairly good solution at present may be more desirable than a prefect O.R. solution available after sometime.. 5) Implementation: Implementation of decisions is a delicate task. It must take into account the complexities of human relations and behaviour. Sometimes resistance is offered only due to psychological factors.

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Introduction to Operation Research

1.8

SUMMARY

O.R. is a relatively new academic discipline. It has its origin in World War II and soon became very popular throughout the world. O.R. is successfully used not only in military affairs but also in industry, government and business. India is one of the first few countries of the world who started using O.R. Defining O.R. is a difficult task as its boundaries and content are not yet fixed. O.R. can be regarded as use of mathematical and quantitative, techniques to substantiate the decision being taken. O.R. takes tools from subjects like mathematics, statistics, engineering, economics, psychology etc. and uses them to score the consequences of possible alternative actions. Today it has become a professional discipline that deals with the application of scientific methods to decision making-especially to the allocation of scarce resources. Developing appropriate mathematical models for situations, processes, systems or environment is basic essence of O.R. study. The model then can be tested, and operated on to determine the effects of changing the values of variables with particular reference to optimization of some criterion. Linear programming, nonlinear programming, dynamic programming, game theory, Markov process, waiting line theory and simulation are some of the models and techniques that are used in O.R. There is role for O.R. in almost all areas of business decisions. Operations Research has certain limitations also. These limitations are mostly related to the problems of model building and time and money factor involved in its applications rather than its practical utility. Magnitude of computation 'involved, lack of consideration for non-quantifiable factors and psychological issues involved in implementation are some of the limitations of O.R. Specialists in O.R. are not the decision makers themselves; they only provide rational basis for decision making to executives. There is a gap between who provides a solution and who wishes to use it. Due to this gap, management sometimes offers resistance to the use of O.R. The gap between O.R. scientists and managers is diminishing fast. O.R. techniques are gaining acceptance and respect day-by-day as they improve manager's decision making effectiveness.

1.9

KEY WORDS

Criterion is a measure by which results can be evaluated. Dynamic Programming is a method to analyze multistage decision process. Game Theory is a technique for decision making in situations of conflict. Information Theory is a theory to evaluate the effectiveness of information flow within a system. Integer Programming is a technique to ensure only best integral values of variables in programming problem Inventory Control Models are models used to express total inventory cost to calculate optimal ordering quantity and ordering time to minimize total inventory cost. Linear Programming is a constrained optimization technique to optimize a linear measure of effectiveness (objective function) under linear constraints. Non-Linear Programming is a constrained optimization technique with a non-linear measure of effectiveness or constraints or both. Optimization is achieving maximum or minimum (the best) value of some effectiveness criterion. Sub-optimization is said to occur either when some criterion subordinate to the overall criterion is optimized or when the overall criterion is optimized over only a portion of the total of all possible' alternative actions. 14

Waiting Line Theory deals with situations in which queues are formed: This is used to make calculations like expected number of items in the queue, expected waiting time, expected idle time, etc. which in turn are used to take decision regarding appropriate number of service channels.

1.10 SELF ASSESSMENT EXERCISE 1)

Define Operations Research.

2)

Discuss the limitations of operation research.

3)

Enumerate, with brief description, some of the techniques of O.R.

4)

Comment on the relationship between managers and O.R. specialists.

5)

Describe the various steps involved in O.R. study.

Operation Research An Overview

1.11 FURTHER READINGS Ackoff, Russell L., and Maurice W. Sasieni: "Fundamentals of Operations Research," John Wiley & Sons, Inc., New York, N.Y., 19681 Beer, Stafford, "Decision and Control," John Wiley & Sons, Inc., New York N.Y., 1966. Churchman, Charles W., Russell L. Ackoff, and L. Arnoff: "Introduction to Operations Research," John Wiley & Sons, Inc., New York, N.Y., 1957. Sasieni, Maurice, A. Yaspan, and L. Friedman: "Operations Research: Methods and Problems," John Wiley & Sons, Inc., New York, N.Y., 1959. Shamblin, James E., and G.T. Stevens, Jr.: "Operations Research, A Fundamental Approach," McGraw Hill Book Company, New York, N.Y., 1974. Sharma, J.K.,: "Mathematical Models in Operations Research," Tata McGraw Hill Publishing Company Ltd., New Delhi, 1989: Swarup, Kanti, Gupta, P.K., Manmohan: "'Tracts in Operations Research," Sultan Chand & Sons, New Delhi, 1989. Taha, H.A.: "Operations Research, An Introduction," Macmillan, Inc., New York, N.Y., 1971. Thierauf, Robert J., and Robert C. Klekamp: "Decision Making through Operations Research," 2nd. Ed., John Wiley & Sons, Inc., New York, N.Y., 1975. Wagner, Harvey M.: "Principles of Operations Research," 2nd. Ed., Prentice-Hall, Inc., Englewood Cliffs, N.J., 1975.

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Introduction to Operation Research

UNIT 2 REVIEW OF PROBABILITY AND STATISTICS Objectives After reading this unit, you should be able to: •

Discuss the relevance of probability and statistics in decision making



Explain the concepts of Random variable and Probability Distribution



Identify situations where different discrete probability distributions can be applied



Use suitable continuous distributions to various situations in decision making process



Summary measures of Probability distributions.

Structure 2.1

Introduction

2.2

Random Experiment and Probability

2.3

Random Variable: Discrete v/s Continuous

2.4

Probability Distribution and Summary Statistics

2.5

Some Important Discrete Probability Distributions

2.6

Some Important Continuous Probability Distributions

2.7

Summary

2.8

Key Words Appendices

2.1

16

INTRODUCTION

Uncertainty is part and parcel of business life. Demand of a product, product quality, stock market prices are but some of the areas, where, prediction about the future with certainty becomes impossible. So, decision making in such areas is facilitated through formal and precise expressions for the uncertainties involved. Formulation the market strategy of a company becomes more easy if the company is able to predict exact demand, product characteristics and competitors strategy for the future with certainty. For example, in order to decide exact time for company's product advertisement on a T.V., the executive needs to know the exact distribution of potential customers according to various socio-economic characteristics, such as income, education, rural/urban living etc. Probability and statistics provides us with the ways and means to attain the formal precise expressions for uncertainties involved in different situations that come across the decision making process. The objective of this unit is to review the techniques of probability and statistics you have already studied in your earlier course MS-8. Accordingly, the basic concepts: Probability, random variable and Probability distribution are first presented; followed by brief details regarding some specific probability distributions. A brief review of summary statistics for both discrete and continuous probability distribution has also been presented. Finally a few examples relating to business application of these distributions are given along with steps for computation. Usage of these concepts will be facilitated when you will study units relating to inventory management and waiting lines problems in operation research. Activity 1 Explain three situations relating to your experience with any organisation/firm you know, where you faced uncertainty in taking decisions. Also explain as to how you dealt with the uncertainty in each of the cases. 1) ................................................................................................................................. 2) ..................................................................................................................... 3) ................................................................................................................................

2.2

RANDOM EXPERIMENT AND PROBABILITY

Review of Probability and Statistical

The experiment whose outcomes are unpredictable before the actual happening is termed as random experiment and the outcomes are termed as its events. The chance of occurrence associated with each outcome of the random experiment is called its probability. Consider following examples for clear understanding of these terms: 1)

The demand for a product on any day which is unpredictable and can have three values, viz. high, medium and low demand.

2)

Toss of a coin with two outcomes either head or tail.

3)

The weight of a student selected at random from a college.

Here, in these examples it is clear that the experiment can be repeated large number of times and each time the set of out cases remains same but occurrence of any is unpredictable. You may recall that we have already given various methods of computation of probability and algebra of probability. It may be quickly pointed out that the probability associated with any event of the random experiment is a number P(A) such that i)

P(A) is always +ve and less than or equal to one

ii)

Sum of P(A) for all the events is always 1.

Thus if P(A) worked out using the assumption that all events of the random experiment are equally likely, it is termed as a Prior Probability. If P(A) is worked out as a limit of ratio of favourable trials to total number of trials, then it is called an empirical-probability. Here, it is assumed that experiment can be repeated indefinitely under identical conditions. These two definitions are also called objective probabilities. However, if the values to P(A) are assigned on subjective basis satisfying the above conditions. Then the probability is termed as Subjective Probability. This subjective probability plays very significant role in business decisions as the experts can assign subjective probabilities to various events on the basis of their expertise and experience. These concepts shall be more clear with following examples: Example 1 A coin is tossed two times. Describe its sample space along with probabilities associated with the events. Solution The sample space consist of following 4 points S = (HH, HT, TH, TT); Where H and T denote occurrence of head or tail. These four points are equally likely, mutually exclusive and exhaustive. So using prior probability definition each has probability of occurrence 1/4. Example 2 Following is record of demand of T.V. sets per day No. of T.V.s demanded 1 2 3 4 5 6 No. of days 8 12 10 5 3 2 Calculate the probability of demand for more than 3 T.V. sets. Solution Solution Here, the no. of experiments (No. of days) = 40 No. of favourable trials = 10 (It is sum of days when demand is 4 to 5 or 6 T.V. sets)

∴ P (Demand > 3) = 10/40 =

1 4

Activity 2 List all outcomes of throw of a pair of dice and write all possible outcomes having sum of the values on the face of the dice as 10. ………………………………………………………………………………………….

17

Introduction to Operation Research

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… Activity 3 Calculate the probability of getting a complete suit in a hand of Bridge. (A complete suit consist of all cards of the same color.) ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… Activity 4 Following is record of No. of defective nuts in packets of 150 nuts each

Calculate the probability that a packet selected at random will have i) 2 defectives ii) at least 2 defectives iii) atmost 2 defectives.

2.3

RANDOM VARIABLE

When all the outcomes of a random process can be identified through numerical values, then we can redefine the outcomes of the process by a variable called a random variable. For example the toss of two coins simultaneously will result into three distinct numerical values, namely two heads, one head and no head. Thus this experiment can be described with the help of random variable X taking values 2,1 and 0 respectively.; Again, if the random variable has finite possible values, like this example, (tossing of two coins), the variable is called a discrete random variable (d.r.v.) and if the variable has infinite possible values it is called continuous random variable (c.r.v.). For example, weight of a school student selected at random. It may be pointed out that the distinction between a continuous and discrete random variable is purely a nomenclature type as a continuous random variable can always be looked upon as a discrete random variable. The idea shall be more clear with this example. Suppose a marketeer regards the demand for sugar to be high, moderate or low according as-the demand is more than 10 quintals, between 10 to 4 quintals and less than 4 quintals respectively. Though the demand of sugar is a continuous random variable, but from the marketeer point of view it is discrete taking values 1, 2, 3 according as demand is low, moderate or high respectively.

2.4

PROBABILITY DISTRIBUTION

Let X be a discrete random variable taking values X1, X2…………,Xn. Now, if we can assign probabilities P(X1), p(X2),………..., p(Xn) to these values. Then P(X) is called 18

Review of Probability and Statistical

probability distribution function of X, if the following holds statistic

So, in case of toss of two coins simultaneously, we have No. of (X)

Probability of X

Heads

P(X)

0

1/4

1

1/2

2

1/4

Sum

1

From the table it is evident that P(X) defined above satisfy both the conditions, and so it is called Probability distribution of X. Cumulative Probability Distribution Function The concept and importance of cumulative probability distribution shall be clear from the following example. Example 3 Following is demand data for last 100 days of number of T.V. sets. Calculate the probability of the following events. i)

The demand on a particular day is 4 T.V. sets

ii) The demand on a particular day is 4 or less T.V. sets No. of T.V. Demanded

1

2

3

4

5

6

7

No. of days

5

15

20

30

18

9

3

Solution Table: Probability function and Cumulative Probability function for No. of T.V. demanded per day. No. of T.V. No, of Relative Cumulative Relative Cumulative demanded days frequency Frequency frequency = Cumulative D f P(D) (c.f.) Probability C (D) 1

5

0.05

5

0.05

2 15 0.15 20 0.20 3 20 0.20 40 0.40 4 30 0.30 70 0.70 5 18 0.18 88 0.88 6 9 0.09 97 0.97 7 3 0.03 100 1.00 So cumulative probability of the random variableX is sum of the probabilities for values of X less than and equal to a given value. Further, it can be seen that C(X) = C(X-1) + P(X) Using these expression the last column has been obtained for the table given above. P(4) = 0.30 and C(4) = 0.70. Expectation of Random Variable or Expected value of X denoted by E(X) is an expression for calculating, the value of X in the long run. It is also known as average value of X. Symbolically,

µ = E(X) =

∑ XP(X) = X P(X ) + X P(X 1

1

2

2

) + ......+ X n P(X n )

i.e. expected value of random variable is sum of the product of all possible values of the random variable with corresponding probabilities. So in the previous example average value of demand of T.V. sets/day is

19

Introduction to Operation Research

µ = E(X) = 1× 0.5+2 × 0.15+3 × 0.20+4 × 0.30+5 × 0.18+6 × 0.09+7 × 0.03 = 3.80 ≈ 4 T.V. sets Thus average or mean demand/day is 4 T.V. sets. It is also known as measure of central tendency. Standard deviation (S.D.) Standard deviation is a tool to describe the variability of the random variable X. Mathematically it is given as

σ = E[X-E(X)]2 =

∑ (X-µ) P(X) 2

=

∑ X p(X)-µ 2

2

Where µ = E(X) P(X) is Probability mass function of random variable X.



is sum over all possible values of discrete random variable X.

Activity 5 A flower seller has following distribution of demand for flowers per day: No. of flowers Demanded per day: Probability:

10

11

12

13

14

15

0.1

0.2

0.3

0.2

0.1

0.1

He knows from past experience that each flower sold gives him a profit of Rs. 3, where as each unsold flower gives a loss of rupees two. Estimate the number of flowers he would have every day with a view to maximize his earnings on an avera ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

2.5

STANDARD DISCRETE PROBABILITY DISTRIBUTIONS

You have already studied standard discrete probability distributions in MS-8 Block Unit 10. These are discussed below in brief alongwith their summary statistics. The usage of these distributions shall be explained with the help of examples. Binomial Distribution

20

Review of Probability and Statistical

An owner of a hotel having delux rooms has bought 2 Colour T.V. sets fitted with V.C.R. to rent to occupants of these rooms. He estimates from his past experience that about 1/3 of occupants would be willing to rent sets. Assuming 100% occupancy for all days. Calculate the probability that on a particular evening i) There are more requests than the T.V. sets ii) If the owner's cost per set per day is Rs. 100, what rent R must he charge in order to break even in the long run. Solution i) Daily request of T.V. sets on rent is a random variable, subject to the Binomial distribution with n = 6 and p = 1/3. We are interested to find the probability that the random variable takes value more than 2. This probability is P(A) = 1 - P(0) - P(l) - P(2)

21

Introduction to Operation Research

Now for Break even in the long run expected total revenue (T) is equal to total cost.



1138 145800 R = 200 ⇒ R = 729 1138

Example 5 Assume that cars arrive at a toll booth according to poisson distribution at a mean rate of 4 cars per 5-minute interval. Find the probability that during a random interval of 5 minutes. i)

Exactly 2 cars arrived.

ii) Atmost two cars arrived. iii) At least three cars arrived. Solution Here in this example λ = 4 and we are to calculate

Example 6 A and B are playing a certain game with odds in favour of A 2 to 3 for each game. Calculate the probability that a will win after loosing first two games. Also calculate mean and variance for No. of failures of A proceeding his first success. Solution It is a case of geometric distribution with p=3/5 and n=2

22

Review of Probability and Statistical

Activity 6 Incharge of an electronic section of departmental store in New Delhi knows from the past experience that the chance that the customer who is just browsing will buy something is 0.3. Suppose that 10 customers browse in his section every hour during the evening. Calculate the probability of following events during a specified hour in the evening. a) At least one browsing customer will buy something. …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… b) No more than 2 browsing customers will buy something. …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… Activity 7 Certain mass produced articles of which 0.5% are defective, are packed in cartons each containing 130 articles. What proportion of cartons are free from defective articles? Also find out the proportion of carton containing 2 or more defectives. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

2.6

CONTINUOUS PROBABILITY DISTRIBUTIONS

Continuous probability distributions are also known as probability, density functions (p.d.f.) for the continuous random variables. Some of the important p.d.f. have already been discussed in course MS=8 Block 3 Unit 11. However, some important continuous density functions are listed below along with their summary statistic for ready reference. Again a few examples are also discussed to explain the usage of these p.d.f. in decision making process. Rectangular Distribution Range :

0 to a

Density : f(x) =

1 a 23

Introduction to Operation Research

24

Review of Probability and Statistical

Example 7 I t is known from past experience that average life of a bulb (assumed to be a continuous random variable following exponential distribution) is 120 hours. Calculate the probability that the bulb will work for atmost 30 hours. Solution

Example 8 A company manufacturing plastic rope knows from the past experience that average load bearing capacity of the rope is 200 lbs with a S.D. of 20 lbs. Calculate the probability of the following events. i)

The roap fails to bear a load of 170 lbs.

ii) The roap bears a load of 240 N. Solution To calculate these probabilities following sequence of steps will be performed.

25

Introduction to Operation Research

1) Identify mean and S.D. of the normal distribution. In this case these are

µ = 200 lbs. σ = 20 lbs. 2) Sketch the graph to depict the desired area to be calculated as probability of the events. For (i) and (ii) they are given as Fig. 2.6.

Fig. 2.6: Required Area for (i) and (ii) under the normal curve.

Step (iii) Convert the X values to standard normal score (z)

Step (iv): Read area from 0 to Z under the standard normal curve using statistical table area under the normal curve. Here (i) Area from 0 to - 1.5is 0.4332 which is same as area from 0 to 1.5 (ii) Area from 0 to 2 is 0.4772 Step (v) Calculate required probabilities using the fact that area to the left of the mean is 0.5.

∴ (i) Probability that rope will fail to bear a load of 170 lbs = 0.5 - 0.4332 = 0.0668 (ii) Probability that rope will bear to a load of 240 lbs = 0.5 - 0.4772 = 0.0228 Activity 8 In an industrial complex, the average no. of total accidents per month is one-half. What is the probability that 4 months will pass without a fatal accident? Assume No. of accidents per month follows Poisson distribution. Further mean of exponential distribution is 1/A, where A is mean of Poisson distribution.

26

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

………………………………………………………………………………………… …………………………………………………………………………………………

Review of Probability and Statistical

Activity 9 Assume the mean height of soldiers to be 68.22 inches with a variance of 10.8 inches. How many soldiers in a regiment of 1,000 would you expect to have height more than six feet? Assume that height of soldiers follows normal distribution. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

2.7

SUMMARY

Probability in simple terms means the chance of occurrence of an event. The need to develop a formal and precise expression for uncertainty in decision making, has lead to different approaches to probability measurement namely, priori, relative frequency and subjective. However, these approaches follows the same axioms. The concept of random variable enables one to assign numerical values to all possible events of the random experiment. If the events are finite, then the random variable associated with it is called discrete random variable and corresponding probability function as probability mass function. Again, if the events are infinite; then the random variable is called continuous and probability function as the density function of this random variable. Binomial, Poisson, geometric and negative Binomial or Pascal are a few important discrete probability functions. Normal; uniform, exponential, Gamma and Beta probability density function are a few examples in case of continuous random variable. These probability functions enable us to calculate precise expression for uncertainties attached with day-to-day business decisions.

2.8

KEY WORDS

Random Experiment is an experiment whose outcomes are unpredictable before the actual happening. Probability is a precise expression for chance of occurrence of an event if the experiment is performed large number of times. Random Variable is a numerical valued function defined .on outcomes of the random experiment. Discrete Random Variable is a random variable which takes only finite values. Continuous Random Variable is a random variable which takes infinite possible values. Probability function is an expression to describe probability of occurrence for all possible values of the random variable. Probability mass function is a probability function defined for discrete random variable. Probability density function is a probability function defined for continuous random variable:

27

Introduction to Operation Research

Cumulative probability function is an expression for probability of the random variable taking a value equal to or less than a given value. Expectation is the average value of the random variable in the long run. Variance is an expression to measure the variability of random variable about its expectation. Standard deviation is an expression to measure the variability of random variable about its mean which has the same unit of measurement as that of random variable.

28

UNIT 3 LINEAR PROGRAMMING – GRAPHICAL METHOD

Linear Programming – Graphical Method

Objectives After studying this unit, you should be able to : •

Formulate management problem as a linear programming problem in suitable cases



identify the characteristics of a linear programming problem



make a graphical analysis of the problem



solve the problem graphically



identify various types of solutions



explain various applications of linear programming in business and industry.

Structure 3.1

Introduction

3.2

Formulation of a Linear Programming Problem

3.3

Formulation with Different Types of Constraints

3.4

Graphical Analysis

3.5

Graphical Solution

3.6

Multiple, Unbounded Solution and Infeasible Problems

3.7

Application of Linear Programming in Business and Industry

3.8

Summary

3.9

Key Words

3.10 Self-assessment Exercises 3.11 Answers 3.12 Further Readings

3.1

INTRODUCTION

Linear Programming is a versatile technique which can be applied to a variety of problems of management such as production, refinery operation, advertising, transportation, distribution and investment analysis. Over the years linear Programming has been found useful not only in business and industry but also in non-profit organisations such as government, hospitals, libraries and education. The technique is applicable in problems characterised by the presence of a number of decision variables, each of which can assume values within a certain range and affect their decision variables. The variables represent some physical or economic quantities which are of interest to the decision maker and whose domains are governed by a number of practical limitations or constraints. These may be due to availability of resources like men, material or money or may be due to a quality constraint or may arise from a variety of other reasons. The problem has a well defined objective. The common most objectives are maximisation of profit/contribution or minimisation of cost. Linear programming indicates the right combination of the various decision variables which can be best employed to achieve the objective taking full account of the practical limitations within which the problem must be solved. The most important feature of linear programming is the presence of linearity in the problem. This will enable you to convert the objective to a linear function of the decision variables and the constraints into linear inequalities. The problem thus reduce to maximising or minimising a linear function subject to a number of linear inequalities. Although only graphical methods of solution are presented in this unit, very efficient computational procedures known as algorithms are available to solve linear programming problems. The emergence of computers has been helpful to solve these problems with a large number of decision variables and constraints.

5

Programming Techniques – Linear Programming and Application

Activity 1 Fill in the blanks i)

A linear programming problem has a well defined objective function which is ........... and which is to be ...........or ..............................

ii)

The constraints in a linear programming problem arises due to limitation of These are linear ..........................or .............................

iii)

The solution of a linear programming problem indicates the right combination of .....................which ..................................... or ............................. the objective function satisfying the various .....: ...................................................................

3.2

FOMULATION OF A LINEAR PROGRAMMING PROBLEM

The formulation of a linear programming problem can be illustrated through what is known as a product mix problem. Typically, it occurs in a manufacturing industry where it is possible to manufacture a variety of products. Each of the products has a certain margin of profit per unit. These products use a common pool of resources whose availability is limited. The linear programming technique identifies the combination of the. products which will maximise the profit without violating the resource contraints. The formulation is illustrated with the help of following example. Example 1 Suppose an organisation is manufacturing two products P1and P2. The profit per tonne of the two products are Rs. 50 and Rs. 60 respectively. Both the products require processing in three types of machine. The following Table indicates the available machine hours per week and the time required on each machine for one tonne of P1 and P2. Formulate this product mix problem in the linear programming form. Table Showing the available machine capacities and machine hour requirement of the two products

Profit/tonne Machine 1 Machine 2 Machine 3

6

Product 1 Rs. 50

Product 2 Rs. 60 2 34 4

Total available Machine hours/weeks 300 509 812

Linear Programming – Graphical Method

This procedure is also commonly referred to as the formulation of the problem. Activity 2 Himalayan Orchards have canned apple and bottled juice as its products Rs. 2 and Rs. 1 respectively per unit. The following Table indicates the margin labour, equipment and material to produce each product per unit. Bottled Juice

Canned Apple

Total

Labour (man hours)

3.0

2.0

12.0

Equipment (machine hours)

1.0

2.3

6.9

Material (unit)

1.0

1.4

4.9

Formulate the product mix which will Formulate the linear programming problem specifying maximise profit without exceeding the various levels of resources. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………. Activity 3 A television manufacturer must decide how many black-and-white and how many colour sets he should produce for each day's sale so as to maximize his daily profit. Each day he has available the following supplies: TV Chasis 24 Production hours 160 Colour Tubes 10 Each black-and-white set requires 5 production hours and yields a profit of Rs. 60. Each colour set requites 10 production hours and yields a profit of Rs. 150. Formulate the manufacturer's problem as a linear programming problem. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………….

3.3

FORMULATION WITH DIFFERENT TYPES OF CONSTRAINTS

The formulation of a linear programming problem when the constraints have emerged due to resource limitation has been illustrated in the previous section. These constraints are of “less than equal to” type. However, there may be constraints of other types as well. The formulation of linear programming problems with different types of constraints is illustrated in the following examples. Example 2 Three nutrient components namely, thiamine, phosphorus and iron are found in a diet of two food items A and B. The amount of each nutrient (in milligrams per ounce i.e. mg/oz) is given below :

7

Programming Techniques – Linear Programming and Application

A

B

Thiamine

0.15 mg/oz

0.10 mg/oz

Phosphorus Iron

0.75 mg/oz 1.30 mg/oz

1.70 mg/oz 1.10 mg/oz

The cost of food A and B are Rs. 2 per oz. and Rs. 1.70 per oz. respectively. The minimum daily requirements of these nutrients are at least 1.00 mg. of thiamine, 7.50 mg of phosphorus and 10.00 mg of iron. Write the problem in the linear programming form. Solution Let us define by x1 and x2 the number of units (ounces) of A and B respectively purchased everyday. Since the purpose is to minimise the total cost of the food items and to satisfy the minimum daily requirement of nutrient the linear programming problem is given by

Example 3 An oil refinery can blend three grades of crude oil to produce quality R and quality S petrol. Two possible blending processes are available. For each production run the older process uses 5 units of crude A, 7 units of crude B and 2 units of crude C to produce 9 units of. R and 7 units of S. The newer process uses 3 units of crude A, 9 units of crude B and 4 units of crude C to produce 5 units of R and 9 units of S petrol. Because of prior contract commitments the refinery must produce at least 500 units of R and at least 300-units of S for the next month. It has available 1500 units of crude A, 1900 units of crude B and 1000 units of crude C. For each unit of R the refinery receives Rs. 60 while for each unit of S it receives'Rs. 90. Find out the linear programming formulation of the problem so as to maximise the revenue. Solution The decision variables in this case are

Fractional runs are permissible in blending processes. Then from the given conditions the problem can be formulated as

8

In the example 2 of this section the linear programming problem has only "greater than or equato" type constraints. In the example 3 the problem has both "greater

than or equal to" type and "less than or equal to" type constraints.

Linear Programming – Graphical Method

Activity 4 A company owns two flour mills, A and B, which have different production capacities, for high, medium and low grade flour. This company has entered a contract to supply flour to a firm every week with at least 12, 8 and 24 quintals of high, medium and. low grade respectively. It costs the company Rs. 1000 and Rs. 800 per day to run mill A and B respectively. On a day, mill A produces 6, 2 and 4 quintals of high, medium and low grade flour respectively. Mill B produces 2, 2 and 12 quintals of high, medium and low grade flour respectively. How many days per week should each mill be operated in order. to meet the contract order most economically. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… Activity 5 A scrap metal dealer has received a bulk order from a customer for a supply of at least 2000 kg of scrap metal. The customer has specified that at least 1000 kgs of the order must be of high quality copper that can be melted easily and can be used to produce tubes. Further, the customer has specified that the order should not contain more than 200 kg. of scrap which are unfit for commercial purposes. The scrap metal dealer purchases scrap from two different sources in an unlimited quantity with the following percentages (by weight) of high quality copper and unfit scrap.

Copper Unfit Scrap

Source A 40% 7.5%

Source B 75% 10%

The cost of metal purchased from Source A and Source B are Rs. 12.50 and Rs. 14.50 per kg. respectively. Determine the optimum quantities of metal to be purchased from the two Sources by the scrap metal dealer so as to minimise the total cost. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

3.4

GRAPHICAL ANALYSIS

Linear programming with two decision variables can be analysed graphically. The graphical analysis of a linear programming problem is illustrated with the help of the following example of product mix introduced in Section 3.2.

At first we draw the line 2x1 + x2 = 300 which passes through the points (0, 300) and (150, 0).

9

Programming Techniques – Linear Programming and Application

A linear inequality in two variables is known as a half plane. The corresponding equality or the line is known as the boundary of the half plane. The half plane along with its boundary is called a closed half plane. We must decide on which side of that line 2x 1 + x2 = 300 the half plane is located. An easy way is to solve the inequality for x2. For fixed x1, the ordinates satisfying this inequality are smaller than the corresponding ordinate on the line and thus the inequality is satisfied for all points below the line. This is the shaded region as indicated in Figure 3.1. Similarly, you may determine the closed half planes corresponding to the inequalities (Figures 3.2 and 3.3).

10

Since all the three constraints must be satisfied simultaneously we consider the intersection of these three closed half planes in Figure 3.4.

Linear Programming – Graphical Method

Feasible solution and feasible region Any non-negative value of (x1, x2) i.e is a feasible solution of the linear programming problem if it satisfies all the. constraints. The collection of all feasible solutions is known as the feasible region. The feasible region of the linear programming problem under discussion is indicated by the shaded part of Figure 3.4. Example 4 Consider the linear programming problem formulated in Section 3.3.

The feasible region of this problem is indicated in Figure 3.5. It may be noted that as the constraints are of "greater than or equal to" type of feasible region is unbounded on one side. 11

Programming Techniques – Linear Programming and Application

12

Example 5

Linear Programming – Graphical Method

Consider the linear programming problem formulated in Section 3.3.

The feasible region of this linear programming problem is indicated in Figure 3.6. The feasible region in this case presents another interesting feature. The critical region has been formed by the two constraints.

The remaining three constraints although present is not affecting the feasible region in any manner. Such constraints are known as redundant constraints. Activity 6 Find graphically the feasible region of the linear programming problem given in Activity 2. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. Activity 7 Find graphically the feasible region of the' linear programming problem given in Activity 3. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. Activity 8 Find graphically the feasible region corresponding to the linear programming problem given in Activity 5. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. ………………………………………………………………………………………….

13

Programming Techniques – Linear Programming and Application

Activity 9 A company is interested in the analysis of two products which can be made from the idle time of labour, machine and investment possible in this regard. It was found on investigation that the labour requirement for the first and the second products was 2 and 3 units respectively and the total available man hours was 24. Only product 1 required machine hour utilization of one hour per unit and at present only 9 spare machine hours are available, Product 2 requires one unit of a byproduct per unit and the daily availability of the byproduct is .6 units. According to the marketing department the sales potential of product 1 cannot exceed 5 units. In a competitive market, product 1 can be sold at a profit of Rs. 3 and product 2 at a profit of Rs. 5 per unit. Formulate the problem as a linear programming problem. Determine graphically the feasible region. Identify the redundant constraint. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. ………………………………………………………………………………………….

3.5

GRAPHICAL SOLUTION

Linear programming with two decision variables can be solved graphically, Although the method is quite simple the principle of solution is based on certain analytical concepts. Convex Set A region or a set R is convex if and only if for any two points on the set R the line segment connecting those points lies entirely in R. We refer to the feasible region presented in Figure 3.4. The points P (xi = 20, x2 = 25) and Q (x1 = 60, x2 = 75) are both feasible solutions of the corresponding linear programming problem. The line joining P and Q belongs entirely in R. Thus the collection of feasible solutions in a linear programming problem form a convex set. In fact, it is a special type of convex set known as convex polygon as this is formed by the intersection of a finite number of closed half planes. Extreme Point The extreme point E of a convex set R is a point such that it is not possible to locate two distinct points in or on R so that the line joining them will include E. Extreme points, are also referred to as vertices or corner points. We refer to Figure 3.4 for illustration. The point B (origin) is such that it is not 'possible to locate two distinct points in or on the convex set such that B belongs to the line joining them, Thus R is an extreme point of the convex set of feasible solutions. Other extreme points are F, H, J and C. The following result (Hadley, 1969), (Mittal, 1976) provides the solution of a linear programming problem: If the maximum or minimum value of a linear function defined over a convex polygon exists, then it must be on one of the extreme points. 14

We now illustrate the graphical solution of linear programming problems through the following examples.

Linear Programming – Graphical Method

Example 6

The feasible region which is a convex polygon is illustrated in Figure 3.7. The extreme points of this convex set are B, F, H, J and C. The objective function of this problem is 50x1 + 60x2. If we consider M as a parameter the graph 50x1 + 60x2 = M is a family of parallel straight lines with slope = − 1 . Some 6

of these lines will intersect the feasible region and contain many feasible solutions while the others will miss and contain no feasible solution. We wish to find the line of this family that intersects the feasible region and is farthest out from the origin as the problem is to maximise the objective function. The farthest is the line from the origin the greater will be the value of M.

In the Figure 3.7 we observe that the line 50x1 + 60x2 = M passes through the point J. The point J being the intersection of the lines 3x1 + 4x2 = 509 and 4x1 + 7x2=812 has . Since J is the only feasible solution on this line the coordinates the solution is unique. The corresponding value of M is 8326 which is the maximum value of the objective function. The fact that the objective function is maximised at J can also be ascertained from the values of the objective function at the various extreme points as shown in the Table below.

15

Programming Techniques – Linear Programming and Application

Each of the half planes lies above its boundary line (Figure 3.8). The feasible region is infinite at the upper side. It will not be possible to find the maximum in this case hut we are looking for a minimum. Let us introduce a parameter m in 2x1 + 1.7x2 = m and draw the lines for various values of m. We seek a line in this family of lines that intersects the feasible region and at the same time is as close as possible to the origin. As in the previous example, we compute the value of the objective function at the various extreme points in the following Table. Table Showing the Computation of Minimum Value of an Objective Function

16

Thus the value m is minimum when it passes through the point C. The, optimum values variables are x1 = 6.32 and x2 = 1.63; the minimum value of the objective

Linear Programming – Graphical Method

Example 8

As in the previous two examples the maximum value of the objective function is computed in the following Table. Table Showing the Computation of Maximum Value of an Objective Function

Thus the optimum values of the decision variables are x1 = 271.43, x2 = 0 and the maximum value of the objective function is 317573. Activity 10 Solve graphically the linear programming problem stated in Activity 2. ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... .........................................................................................................................................

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Programming Techniques – Linear Programming and Application

Activity 11 Solve graphically the linear programming problems stated in Activity 4. ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... Activity 12 Solve graphically the linear programming problem stated in Activity 5. ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... Activity 13 Solve graphically the linear programming problem stated in Activity 9. ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... .........................................................................................................................................

3.6

MULTIPLE, UNBOUNDED SOLUTION AND INFEASIBLE PROBLEMS

The linear programming problems discussed earlier possessed unique solutions. This was because the objective function passed only through the extreme point located at the intersection of two half planes. Linear programming problems where the objective function coincides with one of the half planes generated by a constraint will possess multiple solution. The presence of multiple solutions is illustrated through the following example. Example 9 A company buying scrap metal has two types of scrap metal available to him. The first type of scrap metal has 30% of metal A, 20% of metal B and 50% of metal C by weight. The second scrap has 40% of metal A, 10% of metal B and 30% of metal C. The company requires at least 240 kg. of metal A, 100 kg. of metal B and 290 kg. of metal C. The price per kg. of the two scraps are Rs. 120 and Rs. 160 respectively. 'Determine the optimum quantities of the two scraps to be purchased so that the requirements of the three metals are satisfied at a minimum cost. Solution 18

We introduce the decision variables x, and x, indicating the amount of scrap metal to be purchased respectively. Then the problem can be formulated as Minimise 120x1 + 160x,

Linear Programming – Graphical Method

The points A, B, C, D are the extreme points of the lower boundary of the convex set of feasible solutions. One of the members of the family of objective functions 120x1 + 160x2 = m coincides with the line CD with m = 96000. This is indicated by the fact that both the points C with co-ordinates xl = 400, x2 = 300 and ,D with co-ordinates x1 = 800, x2 = 0 are on the line 120x1 + 160x2 = 96000. Thus, every point on the line CD minimises the value of the objective function and the problem has multiple solutions. Unbounded Solution If the feasible region is unbounded it is possible to move the graph of the objective function arbitrarily far out from the origin while passing through the feasible points. In this case no maximum of the objective function exists. The solution of the problem is said to be unbounded. In the previous example the feasible region as shown in Figure 3.10 has no boundary for increasing values of x1 and x2. Thus, it is not possible to maximise the objective function in this case and the solution is unbounded. It may be pointed out that although it is possible to construct linear programming problems with unbounded solutions numerically no linear programming problem formulated from a real life situation can have unbounded solution.. An unbounded

19

Programming Techniques – Linear Programming and Application

solution in this case indicates a wrong formulation or presence of erroneous data. Infeasible Problem A linear programming problem is said to be infeasible if no feasible solution of the problem exists. An infeasible linear programming problem with two decision variables can be identified through its graph. This is illustrated in the following example. Example 10 A company buying scrap metal has two types of scrap available to them. The first type of scrap metal has 20% of metal A, 10% of impurity and 20% of metal by weight. The second type of scrap has 30% of metal A,10% of impurity and 15% of metal B by weight. The company requires at least 120 kg. of metal A, at most 40 kg. of impurity and at least 90 kg. of metal B. The price for the two scraps are Rs. 200 and Rs. 300 per kg. respectively. Determine the optimum quantities of the two scraps to be purchased so that the requirements of the two metals and the restriction on impurity are satisfied at minimum cost. Solution Introduce the decision variable x1 and x2 indicating the amount of scrap metal (in kg.) to be purchased. The problem can be formulated as

20

The region right of the boundary ACF includes all solutions which satisfy the first And the third constraints. The region located on the left of BD includes all solutions which satisfy the second constraint. Hence, there is no solution satisfying all the three constraints. Thus, the problem is infeasible.

Linear Programming – Graphical Method

Activity 14 Solve graphically the following linear programming problem

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………. Activity 15 Solve graphically the following linear programming problem

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………. Activity 16 Solve graphically the following linear programming problem

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

3.7

APPLICATION OF LINEAR PROGRAMMING IN BUSINESS AND INDUSTRY

In the previous sections you have learnt the formulation and solution of a variety of

21

Programming Techniques – Linear Programming and Application

linear programming problems. These include the product mix problem, the problem of preparing a balance diet, the problem of blending in a oil refinery, and a problem of purchasing. You will be familiar with certain other types problems in business and industry which can be formulated and solved by linear programming technique. Example 11 (Advertising) The Consumer Product Corporation wishes to plan its advertising strategy. There are 'two magazines under consideration, magazine I and magazine II. Magazine I has a reach of 2000 potential customers per advertisement and magazine II has a reach of 3000 potential customers per advertisement, The cost per advertising is Rs. 6000 and Rs. 9000 in magazines I and II respectively and the firm has a monthly budget of Its, 1 lakh. There is an important requirement that the total reach for the income group under Its 20000 per annum should not exceed 3000 potential customers. The reach of magazine I and II for this income group is 300 and 150 potential customers respectively per advertisement. How many times the company should advertise in the two magazines to maximise the total reach? Solution Formulation of the problem Suppose x1 is the number of advertisement in magazine I and x2 is the number of advertisement in magazine II. Then the problem can be formulated as

Figure 3.12 : Solution of the Advertising Problem

The feasible region of the problem is indicated by the shaded part ABCD. You may also note that the objective function indicated by the line 2000x1 + 3000)(2 = M is 22

parallel to the half plane 6x1 + 9x2 = 100. Hence every point on the line AB including the two extreme points A and B are solution of the problem. From the graph it is easy to

Linear Programming – Graphical Method

Example 12 (Cost minimisation) The final product of a firm has a requirement that it must weigh exactly 150 kg. The two raw materials used in the manufacture of this product are A with a cost of Rs. 2 per unit and B with a cost of Rs. 8 per unit. Each unit of A weighs 5 kg. and each unit of B weighs 10 kg. At least 14 units of B and no more than 20 units of A must be used. How much of each type of raw material should be used for each unit of the final product if cost is to be minimised? Solution We introduce decision variables x1, x2 indicating the number of units 'of raw material A and raw material B respectively. Then the problem can be formulated as

Figure 3.13 : Solution of the Cost Minimisation Problem

The feasible region of the problem is the triangle ADB. The line 2x1 + 8x2 = 11 a passes through the extreme point B. Hence the optimum solution of the problem 1, ,, given by x1 = 2, x2 = 14 with a minimum cost of Rs. 116

23

Programming Techniques – Linear Programming and Application

Example 13 (Packaging) A manufacturer of packing material, manufactures two types of packing tins, round and flat. Major production facilities involved are cutting and joining. The cutting 'department can process 300 round tins or 500 flat tins per hour. The joining department can process 500 round tins or 300 flat tins per hour. If the contribution towards profit for a round tin is the same as that of a flat tin what is the optimum production level? Solution Let us introduce decision variables xi No. of round tins per hour, x2 = No. of flat tins per hour. Since the contribution towards profit is identical for both the products the objective function can be expressed as x1 + x2. Hence the problem can be formulated as

The feasible region ABCD is indicated by the shade. The coordinate of the extreme

24

Linear Programming – Graphical Method

Several other problems which can be formulated as two decision variable linear programming problems are given in the self-assessment exercises.

3.8

SUMMARY

Linear programming is a fascinating topic in operations research with wide applications in various problems of management. Regardless of the functional area a linear programming problem has a number of characteristics: We first identify the decision variables which are some economic or physical quantities whose values are of interest to the management. The problem must have a well defined objective function expressed in terms of the decision variables. The objective function may have to be maximised when it expresses the profit or contribution. In case the objective function indicates a cost, it has to be minimised. The decision variables interact with each other through some constraints. These constraints occur due to limited resources, stipulation on quality, technical, legal or a variety of other reasons. The objective function and the constraints are linear functions of the decision variables. When a problem of management is expressed in terms of the decision variables with appropriate objective function and constraints we say that the problem has been formulated. A linear programming problem with two decision variables can be solved graphically. Any non negative solution which satisfies all the constraints is known as a feasible solution of the problem. The collection of all feasible solutions is known as a feasible region. The feasible region of a linear programming problem is a convex set. The value of the decision variables which maximise or minimise the objective function is located on the extreme point of the convex set formed by the feasible solutions. This point and hence the solution of a linear programming problem with two decision variables can be identified graphically. In some problems, there may be more than one solution. It is also possible that a linear programming problem has no finite solution. Sometimes the problem may be infeasible indicating that no feasible solution of the problem exists. The diverse applicability of linear programming is illustrated in this unit.

3.9

KEY WORDS

Decision Variables are economic or physical quantities whose numerical values indicate the solution of the linear programming problem. The Objective Function of a linear programming problem is a linear function of the decision variables expressing the objective of the decision maker. Constraints of a linear programming problem are linear equations or inequalities arising out of practical limitations. A Closed Half Plane is a linear inequality in two variables which include the value of the variables for which equality is attained. A Feasible Solution of a linear programming problem is a solution which satisfies all the constraints including the non negativity constraints. The Feasible Region is the collection of all feasible solutions. A Redundant Constraint is a constraint which does not affect the feasible region. A Convex Set is a collection of points such that for any two points on the set, the line joining the points belongs to the set. A Convex Polygon is a convex set formed by the intersection of a finite number of closed half planes. An Extreme Point of a convex set is a point such that it is not possible to locate two distinct points in or on the set such that the line joining the latter two points will include the first point.

25

Programming Techniques – Linear Programming and Application

Multiple Solutions of a linear programming problem are solutions each of which maximise or minimise the objective function. An Unbounded Solution of a linear programme problem is a solution whose objective function is infinite. An Infeasible Linear Programming Problem has no feasible solution.

3.10 SELF-ASSESSMENT EXERCISES 1) Laxmi Furniture Mart (LFM) is in the business of manufacturing tables and chairs. In a day, LFM has 40 hours for assembly and 32 hours of finishing work. Manufacturing of a table requires 4 hours in assembly and 2 hours in finishing. A chair requires 2 hours in assembly and 4 hours in finishing. Profitability analysis indicates that every table would contribute Rs. 80, while a chair's contribution is Rs. 55. Because of the long standing business experience, LFM does.not find it difficult to sell their products. What should be daily production to maximise the contribution? 2) The India Manufacturing Corporation (IMC) has one plant located on the outskirts of the city. Its production is limited to two products naphtha and urea. The unit contribution for each product has been computed as Rs. 50 per unit of naphtha and Rs. 60 per unit of urea. The time requirements for each product and total time available in each department are as follows : Department Hours Required Available Hours in a Naphtha Urea Month 1 2 3 1500 2 3 2 1500 In addition, the demand for the products restrict the production to a maximum of 400 units of.each of these two products. What should be the daily production schedule so as to maximise the contribution? 3) A poultry farmer feeds his hens with an all-purpose grain and a balanced poultry feed. The all purpose grain costs Rs. 2 per kg. while the balanced feed costs Rs. 3 per kg. A kg. of grain provides 70 calorie units and 25 vitamin units while a kg. of balanced feed provides 75 calorie units and 50 vitamin units. For proper growth and egg production; the hens must be fed with a minimum of 100 calorie units and 75 vitamin units every week. What is the minimum weekly cost food programme and what is the extent of two ingredients in this food programme? 4) A company is making two different types of radio. The profit per unit of the first radio is Rs. 60 while the profit per unit of the second radio is Rs. 25. The first radio require 7 hours of assembly and 12 hours of body work. The second radio requires 9 hours of assembly work and 5 hours body work. The daily hours available for assembly work and body work are 63 hours and 60 hours respectively. Find the daily production schedule which maximises the profit. Give your comments about the solution. 5) Find the solution of the following linear programming problem if possible.

6)

26

Linear Programming – Graphical Method

9) The ABC Company has been a producer of picture tubes for television sets and certain printed circuits for radios. The company has just expanded into full scale production and marketing of AM and AM-FM radios. It has built a new plant that can operate 48 hours per week. Production of an AM radio in the new plant will require 2 hours and production of an AM-FM radio will require 3 hours. Each AM radio will contribute Rs. 40 to profits, while an AM-FM radio will contribute Rs. 80 to profits. The market department, after extensive research, has determined that a maximum of 15 AM radios and 10 AM-FM radios can be sold each week. Determine, using grapical method, the optimal production mix of AM and AM-FM radios that will maximise profits.

3.11 ANSWERS Activity 1 i)

linear, maximised, minimised.

ii)

resources, equations, inequalities.

iii) decision variables, maximise, minimise, constraints. Activity 2 Bottled juice x1 Canned apple x2 Maximise x1 + 2x2 Subject to : 3x1 + 2x2 x1 + 2.3x2 x1 + 1.4x2

12. 0 6.9 4.9

Activity 3

27

Programming Techniques – Linear Programming and Application

28

Linear Programming – Graphical Method

Activity 14 The problem has multiple solutions

Maximum value of the objective function = 3. Activity 15 The solution is unbounded. Activity 16 The problem is infeasible. Self-Assessment Exercises 1)

Tables 8, Chairs 4, Maximum Profit Rs, 860.

2)

Naphtha 300 units, Urea 300 units, Maximum profit Rs.33000.

3)

Balanced feed 1 1 units. Minimum Cost Rs. 4 1 .

4)

The problem has multiple solutions

2

2

First type of radio 5, second type of radio 0 First type of radio 5, second type of radio 73 Maximum Profit Rs. 300 5)

The solution is unbounded

6)

The problem is infeasible

7)

No feasible solution

8)

No feasible solution

9)

AM radio 9 AM-FM radio 10 Maximum profit Rs. 1160

3.12 FURTHER READINGS Hadley, G. 1969. Linear Programming, Addison Wesley, Reading, Massachusetts : USA. Mittal, K.V.1976. Optimization Methods in Operations Research and Systems Analysis, Wiley Eastern Limited New Delhi. Mustafi, C.K. 1988. Operations Research Methods and Practice, Wiley Eastern Limited : New Delhi.

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Programming Techniques – Linear Programming and Application

UNIT 4 LINEAR PROGRAMMING SIMPLEX METHOD Objectives After studying this unit, you should be able to : • describe the principle of simplex method • discuss the simplex computation • explain two phase and M-method of computation • work out the sensitivity analysis • formulate the dual linear programming problem and analyse the dual variables. Structure 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13

Introduction Principle of Simplex Method Computational aspect of Simplex Method Simplex Method with several Decision Variables Two Phase and M-method Multiple Solution, Unbounded Solution and Infeasible Problem Sensitivity Analysis Dual Linear Programming Problem Summary Key Words Self-assessment Exercises Answers Further Readings

4.1

INTRODUCTION

Although the graphical method of solving linear programming problem is an invaluable aid to understand its basic structure, the method is of limited application in industrial problems as the number of variables occurring there is substantially large. A more general method known as Simplex Method is suitable for solving linear programming problems with a larger number of variables. The method through an iterative process progressively approaches and ultimately reaches to the maximum.or minimum value of the objective function. The method also helps the decision maker to identify the redundant constraints, an unbounded solution, multiple solution and an infeasible problem. In industrial applications of linear programming, the coefficients of the objective function and the right hand side of the constraints are seldom known with complete certainty. In many problems the uncertainty is so great that the effect of inaccurate coefficients can be predominant. The effect of changes in the coefficients in the maximum.or minimum value of the objective function can be studied through a technique known as Sensitivity Analysis. Every linear programming problem has a dual problem associated with it. The solution of this problem is readily obtained from the solution of the original problem if simplex method is used for this purpose. The variables of dual problem are known as dual variables or shadow price of the. various resources. The solution of the dual problem can be used by the decision maker for augmenting the resources. The methodological aspects of the Simplex method is explained with a linear programming problem with two decision variables in the next section. 30

4.2

PRINCIPLE OF SIMPLEX METHOD

Linear Programming – Simplex Method

We explain the principle of the Simplex method with the help of the two variable linear programming problem introduced in Unit 3, Section 2. Example I Maximise 50x1 + 60x2

Solution We introduce variables x3 .>. 0, x4 equations

0, x5 r 0 So that the constraints become

The variables x3, x4, x5 are known as slack variables corresponding to the three constraints. The system of equations has five variables (including the slack variables) and three equations. Basic Solution In the system of equations as presented above we may equate any two variables to zero. The system then consists of three equations with three variables. If this system of three equations with three variables is solvable such a solution is known as a basic solution. In the example considered above suppose we take x, = 0, x2 = O. The solution of the system with remaining three variables is x3 = 300, x4 = 509, x5 = 812. This is a basic solution of the system. The variables x3, x4 and x5 are known as basic variables while the variables x1, x2 are known as non basic variables (variables which are equated to zero). Since there are three equations and five variables the two non basic variables can be chosen in 5c2, = 10 ways. Thus, the maximum number of basic solutions is 10, for in some cases the three variable three equation problem may not be solvable. In the general case, if the number of constraints of the linear programming problem is m and the number of variables (including the slack variables) is n then there are at most

basic solutions.

Basic Feasible Solution A basic solution of a linear programming problem is a basic feasible solution if it is feasible, i.e. all the variables are non negative. The solution x3 = 300, x4 = 509, x5 = 812 is a basic feasible solution of the problem. Again, if the number of constraints is m and the number of variables (including the slack variables) is n, the maximum number of basic feasible' solution is The following result (Hadley, 1969) will help you to identify the extreme points of the convex set of feasible solutions analytically. Every basic feasible solution of the problem is an extreme point of the convex set of feasible solutions and every extreme point is a basic feasible solution of the set of Constraints. When several variables are present in a linear programming problem it is not possible to identify the extreme points geometrically. But we can identify them through the 31

Programming Techniques – Linear Programming and Application

basic feasible solutions. Since one of the basic feasible solutions will maximise or minimise the objective function, we can carry out this search starting from one basic feasible solution to another. The simplex method provides a systematic search so that the objective function increases (in the case of maximisation) progressively until the basic feasible solution has been identified where the objective function is maximised. The computational aspect of the simplex method is presented in the next section. Activity 1 Fill up the blanks : i)

.................................variables are introduced to make …………… .......... type inequalities equations. ii) A system with m equations and n variables has at most ……………basic solutions. iii) A basic solution with m equations and n variables has .......................................... variables equal to zero. iv) A basic feasible solution is a basic solution whose variables are ........................... v) The maximum number of basic feasible solutions in a system with m equations and n variables is ……………….. vi) In a linear programming problem every ............................ point of the Convex set of feasible solutions is a .................................solution of the problem. vii) The objective function of a linear programming problem is maximised or minimised at a…………………solution.

4.3

COMPUTATIONAL ASPECT OF SIMPLEX METHOD

We again consider the linear programming problem Maximise 50x1 + 60x2

The slack variables provide a basic feasible solution to start the simplex computation. This is also known as initial basic feasible solution. If z denote the profit then z = 0 corresponding to this basic feasible solution. We denote by CB the coefficient of the basic variables in the objective function and by X13 the numerical values of the basic variables. The numerical values of the basic variables are

Table 1

The coefficients of the basic variables in the objective function are CBI = CB2 = CB3 = 0 The topmost row of Table 1 indicates the coefficient of the variables xi, x2; x3, x4 and x5 in the objective function respectively. The column under xi presents the coefficient of xi in the three equations. The remaining columns have also been formed in a similar manner. 32

On examining the profit equation z = 50x1 + 60x2 you may observe that if either x1

or x2 which is currently non basic is included as a basic variable the profit will increase. Since the coefficient of x2 is numerically higher we choose x2 to be included as a basic variable in the next iteration. An equivalent criterion of choosing a new basic variable can be obtained from the last row of Table 1 (corresponding to z). Since the entry corresponding to x2 is smaller between the two negative values x2 will be included as a basic variable in the next iteration. However with three constraints there can only be three basic variables. Thus by making x2 a basic variable one of the existing basic variables will become non basic. You may identify this variable using the following line of argument.

Linear Programming – Simplex Method

Referring back to Table 1, we obtain the elements of the next Table (Table 2) using the following rules : 1) In the z row we locate the quantities which are negative. If all the quantities are positive, the inclusion of any non basic variable will not increase the value of the objective function. Hence the present solution maximises the objective function. If there are more than one negative values we choose the variable as a basic variable corresponding to which the z value is least as this is likely to increase the profit most. 2) Let xi be the incoming basic variable and the corresponding elements of the jth column be denoted by ylj, Y2j and Y3j. If the present values of basic variables are XB1, XB2 and xB3 respectively, then we compute

3) Table 2 is computed from Table 1 using the following rules.

33

Programming Techniques – Linear Programming and Application

The element corresponding to x5 is

In this way we obtain the elements of the first and the second row in Table 2. The numerical values of the basic variables in Table 2 can also be computed in a similar manner.

3) Table 4 is computed from Table following the usual steps. 34

Linear Programming – Simplex Method

Activity 2 Himalayan Orchards have canned apple and bottled juice as its product with profit margin's of Rs. 2 and Rs. 1 respectively per unit. The following table indicates the labour, equipment and material to produce per unit of each product.

Labour (man hours) Equipment (machine hours) Material (unit)

Bottled Juice 3.0 1.0 1.0

Canned Apple 2.0 2.3 1.4

Total Resources 12.0 6.9 4.9

Find by simplex method the product mix which will maximise the profit. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………….

4.4

SIMPLEX METHOD WITH SEVERAL DECISION VARIABLES

The computational procedure explained in the previous section can be readily extended to linear programming problems with more than two decision variables. This is illustrated with the help of the following example. Example 2 The products A. B and C are produced in three machine centres X, Y and Z. Each product involves operation of each of the machine centres. The time required for each operation for unit amount of each product is given below. 100, 77 and 80 hours are available at machine centres X, Y and Z respectively. The profit per unit of A, B and C is Rs. 12, Rs. 3 and Rs. 1 respectively. Find out a suitable product mix so as to maximise the profit. Solution The linear programming formulation of the product mix problem is as follows : Maximise 12x1 + 3x2+x3 35

Programming Techniques – Linear Programming and Application

Table 1

1) Zl - C1= -12 is the smallest negative value. Hence x1 should be made a basic variable in the next iteration. 2) We compute minimum of the ratios

The variable x4 corresponding to which minimum occurs is made a non basic variable. 3) Table 2 is computed from Table 1 using the following rules a) The revised basic variables are x1i x5, x6. Accordingly we make CB1 = 12, CB2 = 0, CB3 = 0 b) As x1 is the incoming basic variable we make coefficient of x1 one by dividing each element of row 1 by 10. Thus the numerial value of the element corresponding to x2 is 2 , corresponding to x3 is 1 and so on in Table 2. 10

10

c) The incoming basic variable should appear only in the first row. So we multiply the first row of Table 2 by 7 and subtract it from the second row of Table I element by element. Thus the element corresponding to x1 in the second row of Table 2 is zero. The element corresponding to x2 is In this way we obtain the elements of the second and the third row in Table 2. The computation of the numerical values of basic variables in Table 2 is made in a similar manner.

36

1) Z2 -C2 = − 3 . Hence x2 should be made a basic variable at the next iteration. 5

Linear Programming – Simplex Method

2) We compute the minimum of the ratios

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………….

4.5

TWO PHASE AND M-METHOD

The simplex method illustrated in the last two sections was applied to linear programming problems with less than or equal to type constraints. As a result we could introduce slack variables which provided an initial basic feasible solution of the problem. Linear programming problems may also be characterised by the presence of both "less than or equal to" type or "greater than or equal to type" constraints. It may also contain some equations. Thus it is not always possible to obtain an initial basic feasible solution using slack variables. Two methods are available to solve linear programming by simplex method in such cases. These methods will be explained with the help of numerical examples. 37

Programming Techniques – Linear Programming and Application

Two Phase Method We illustrate the two phase method with the help of the problem presented in Activity 5 of Unit 3 Example 3

Solution Although the objective function 12.5x1 + 14.5x2 is to be minimised, the values of xi and x2 which minimised this objective function are also the values which maximise the revised objective function -12.5x1 - 14.5x2. The second and the third constraint are multiplied by 100 and 1000 respectively for computational convenience. Thus the linear programming problem can be expressed as

We convert the first two inequalities by introducing surplus variables x3 and x4 respectively. The third constraint is changed into an equation by introducing a slack variable x5. Thus the linear programming problem can be expressed as

Although surplus variables can convert greater than or equal to type constraints into equations they are unable to provide initial basic variables to start the simplex computation. We introduce two additional variables x6 and x7 known as artificial variables to facilitate the computation of an initial basic feasible solution. The computation is carried out in two phases. Phase I In this phase we consider the following linear programming problem

An initial basic feasible solution of the problem is given by x6 = 2000, x7=100000, 38

x5 200000. As the minimum value of the Phase I objective function is zero at the end of the Phase I computation both x, and x7 become zero. Phase II The basic feasible solution at the end of Phase I computation is used as the initial basic feasible of the problem. The original objective function is introduced in Phase II computation and the usual simplex procedure is used to solve the problem. Phase I Computation Table 1

Linear Programming – Simplex Method

x2 becomes a basic variable and x7 becomes a non basic variable in the l.ext iteration. It is no longer considered for re-entry. Table 2

x1 becomes a basic variable and x6 becomes a non basic variable in the next iteration. It is no longer considered for re-entry. Table 3

The Phase I computation is complete at this stage. Both the artificial variables have been removed from the basis. We have also found a basic feasible solution of the

Phase 11 Computation Table 1

39

Programming Techniques – Linear Programming and Application

The coefficients of the artificial variables in the objective function are large negative numbers. As the objective function is to 1 maximised in the optimum or optimal solution (where the objective function is maximised) the artificial variables will be zero. The basic variables of the optimal solution are therefore variables other than artificial variables and hence is a basic solution of the original problem. The successive simplex Tables are given below : Table 1

As M is a large positive number, the coefficient of M in the Zj Cj row would decide the incoming basic variable. As -76M < -41M, x2 becomes a basic variab16 in the next iteration replacing x7. The variable x7 being an artificial variable it is not considered for re-entry as a basic variable. Table 2

40

x1 becomes a basic variable replacing x6. The variable x6 being an artificial variable is not considered for re-entry as a basic variable.

Linear Programming – Simplex Method

Table 3

Activity 4 Solve the following linear programming problem by Two-Phase Method and. M-method using artificial variables corresponding to second and third constraints.

………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………… ………………………………………………………………………………....

4.6

MULTIPLE, UNBOUNDED SOLUTIONS AND INFEASIBLE PROBLEMS

The simplex method can identify multiple solutions of a linear programming problem. If a problem possesses an unbounded solution it is also located in course of simplex computation. If a linear programming problem is infeasible it is revealed by simplex computation. We illustrate these applications of simplex method with the help of a number of examples. Example 4 We consider the linear programming problem formulated in Unit 3, Section 7.

41

Programming Techniques – Linear Programming and Application

Solution After introducing slack variables x3 i 0, x4 0 the inequalities can be converted into equations as follows

The successive tables of simplex computation are shown below :

Since Zj - Cj ≥ 0 for all the variables, x1 = 0, x2 = 100/9 is an optimum solution of the problem. The maximum value of the objective function is 100000/3. However, the Z3 C3 value corresponding to the non basic variable x1 is also zero. This indicates that there is more than one optimum solution of the problem. In order to compute .the value 9f the alternative optimum solution we Introduce x1 as a basic variable replacing x4. The subsequent computation is presented in the next Table.

Thus x1 = 20/3, x2 = 20/3 also maximise the objective function. The maximum value as in the previous solution is 100000/3. Example 5 Consider the linear programming problem

Solution After introducing slack variables x3

0, x4

0 the corresponding equations are

The successive simplex iterations are shown below : 42

Linear Programming – Simplex Method

Z2 - C2 < 0

indicates x2 should be introduced as a basic variable in the next iteration.

However, both . Thus it is not possible to proceed with the simplex computation any further as you cannot decide which variable will be non basic at the next iteration. This is the criterion for unbounded solution. If in the course of simplex computation

Zj - Cj < 0

but

yij ≤ 0

for all i then the

problem has no finite solution. Intuitively, you may observe that the variable x2 in reality is unconstrained and can be increased arbitrarily. This is why the solution is unbounded. Example 6 We consider the .linear programming problem formulated in Unit 3, Section 6.

Solution After converting the minimisation problem into a maximisation problem and introducing slack, surplus, artificial variables the problem can be presented as

The variables x6 and x7 are artificial variables. We use two phase method to solve this problem. In phase I, we use the objective function : Maximise -x6 - x7 along with the constraints given above. The successive simplex computations are given below

43

Programming Techniques – Linear Programming and Application

If in course of simplex computation by two phase method one or more artificial variables remain basic variables at the end of Phase I computation, the problem has no feasible solution. Activity 5 Solve the linear programming problem by simplex Method and give your comments.

......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... Activity- 6 Solve the following linear programming problem by simplex method and give your comments.

......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... Activity 7 Solve the following linear programming problem by two phase method and give your

44

Linear Programming – Simplex Method

comments.

4.7

SENSITIVITY ANALYSIS

After a linear programming problem has been solved, it is useful to study the effect of changes in the parameters of the problem on the current optimal solution. Some typical situations are the impact, of the changes in the profit or cost in the objective function in the current solution or an increase or decrease in the resource level in the present composition of the product mix. Such an investigation can be carried out from the final simplex Table and is known as sensitivity analysis or post optimality analysis. We shall illustrate this method with the help of an example Example 7 We consider the linear programming problem introduced in Section 4.4 Maximise 12x1 + 3x2 + x3

The final simplex Table presenting the optimum solution '(x4, x5, x6 being the slack variables) is presented below : Table 1

Change in the Profit coefficient i)

Non Basic Variable The variable x3 non basic. If its profit level is increased to C3 then the solution remained unchanged so long as

If C3 > 1 16 then the present solution is no longer optimum; A new round of simplex computation is to be performed in which x3 becomes a basic variable;

45

Programming Techniques – Linear Programming and Application

However, if C3 < 1 16, the present solution remains optimum. ii) Basic Variable In this case, the changes can be both positive and negative, as in either case the current solution may become non-optimal. Let us consider the basic variable x1. Let us define by ∆1 = C1 − 12 as the change (both positive and negative) in the profit coefficient 12. We divide each Zj-Cj value of the non basic variable by the corresponding coefficient in the x1 row which is denote by Zj-Cj . Then the basic variables remain y1j

unchanged so long, as (Mustafi, 1988)

Thus the optimal solution is insensitive so long as the changed profit coefficient C1 is between Rs. 7 and Rs. 15 although the present profit coefficient is Rs. 12. Of course, the value of the objective function has to be revised after introducing the change value C1 . Activity 8

In the example considered in Section 4.7 find the range of the profit coefficient of x2 within which the present solution remains optimum. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………….

4.8

DUAL LINEAR PROGRAMMING PROBLEM

Every linear programming problem is associated with another linear programming problem known as its dual problem. The original problem in this context is known as the primal problem. The formulation of the dual linear programming problem (also sometimes referred to as the concept of duality) is substantially helpful to our 46

understanding of linear programming. The variables of the dual linear programming problem also known as dual variables have important economic interpretations which can be_ used by a decision maker for planning his resources. Under certain circumstances the dual problem is easier to solve than the primal problem. The solution of the dual problem leads to the solution of the primal problem and thus efficient computational techniques can be developed through the concept of duality. Finally in the problems of competitive strategy solution of both the primal and the dual problem is necessary to understand the problem fully. We introduce the concept of duality with the help of the product mix problem introduced in Section 4.4. Example 8 Three products A, B, C are produced in three machine centres x, y, z. Each product involves operation of each of the machine centres. The time required for each operation on various products is indicated in the following Table. 100, 77 and 80 hours are only available at machine centres x, y and z respectively. The profit per unit of A, B and C is Rs. 12, Rs. 3 and Re 1 respectively.

Linear Programming – Simplex Method

After introducing the slack variables x4 >=0, x5 >=0, x6>=0 and using simplex method we obtain the optimum solution as x1 = 73/8, x2 = 35/8 and the maximum value of the objective function as 981/8. The final Simplex Table is given below :

Suppose a prospective investor is planning to purchase the resources x, y, z. What offer is he going to make to manufacturer? Let us assume that W1, W2 and W3 are the offers made per hour of machine time x, y and z respectively. Then these prices W1, W2, W3 must satisfy the following conditions. a) b)

c)

Assuming that the prospective investor is behaving in a rational manner, he would try to bargain as much as possible and hence the total amount payable to the manufacturer would be as tittle as possible. This leads to the condition : Minimise 100W1 + 77W2 + 80W3 The total amount offered by the prospective investor to the three resources required to produce one unit of each product must be at least as high as the profit 47

Programming Techniques – Linear Programming and Application

earned by the manufacturer per unit. Since these resources enable the manufacturer to earn the specified profit corresponding to the product he would not like to sell it for anything less assuming he is behaving rationally. This leads to inequalities.

We have, thus, a linear programming problem to ascertain the values of the variables w1, w2, w3. The variables are known as dual variables. The primal problem presented in this example (i) considers maximisation of the objective function (ii) has less than or equal to type constraints and (iii) has non-negativity constraints on the variables. Such a problem is known as a primal problem in the standard form. Formulation of a dual problem

If the primal problem is in the standard form, the, dual problem can be formulated using the following 'rules. 1) The number of constraints in the primal problem is equal to the number of dual variables. The number of constraints in the dual problem is equal to the number of variables in the primal problem. The primal problem is a maximisation problem, the dual problem is a minimisation problem. 2) The profit coefficients of the primal problem appear on the right hand side of the constraints of the dual problem. 3) The primal problem has less than or equal to type constraints while the dual problem has greater than or equal to type constraints. 4) The coefficients of the constraints of the primal problem which appear from left to right are placed from top to bottom in the constraints of the dual problem and. vice versa. It is easy to verify these rules with respect to the example discussed before. Properties of the dual problem (Mustafi, 1988)

1) If the primal problem is in the standard form the solution of the dual problem can be obtained from the Zi - Cl values of the slack variables in the final simplex Table. Example

In the example discussed previously the variables x4, x5, x6 are slack variables. Hence the solution of the dual problem is w1 = z4 c4 = 15/16, w2 = 3/8, w3 = O. 2) The maximum value of the objective function of the primal problem is the minimum value of the objective function of the dual problem. Example

The maximum value of the objective of the primal problem is 981/8. The minimum value of the objective function of the dual problem is

The result has an important practical implication, If the problem is analysed by both the manufacturer and the investor then neither of the two can outmanoeuver the other. Shadow price

The shadow price of a resource is the unit price that is equal to the increase in profit to be realised by one additional unit0 of the resource. 48

Linear Programming – Simplex Method

Example

The maximum value of the objective function can be expressed as

If the first type of resource is increased by one unit the maximum profit will increase by 15/16 which is the value of the first dual variable in the optimum solution. Thus the dual variable is also referred to as the shadow price or imputed price of a resource. This is the highest price the manufacturer would be willing to pay for the resource. The shadow price of the third resource is zero as there is already an unutilised amount; profit is not increased by more of it until the current supply is totally exhausted. 3)

Suppose the number of constraints and variables in the primal problem is m and n respectively. The number of constraints and variables in the dual problem is, therefore, n and m respectively. Suppose the slack variables in the primal are denoted by y1, y2, ….., yn and the surplus variables in the dual problem are denoted by z1, z2, ….., zn

4)

If the primal problem is in the non standard form, the stricture he dual problem remains unchanged. However, if a constraint is greater than equal to type, the corresponding dual variable is negative or zero. If a constraint in the primal problem is equal to type, the corresponding dual variable is unrestricted in sign (may be positive or negative).

Example 9

Consider the primal linear programming problem Maximise 12x1 + 15x2 + 9x3

49

Programming Techniques – Linear Programming and Application

Activity 9

A garment manufacturer has a production line making two styles of shirts. Style I requires 200 grams of cotton thread, 300 grams of dacron thread, and 300 grams of linen thread. Style II requires 200 grams of cotton thread, 200 grams of dacron thread and 100 grams of linen thread. The manufacturer makes a net profit of Rs. 19.50 on Style I, Rs. 15.90 on Style II. He has in hand an inventory of 24 kg of cotton thread, 26 kg of dacron thread and 22 kg of linen thread. His immediate problem is to determine, a production schedule, given the current inventory to make a maximum profit. Then he would like to know at what price it would be profitable to buy thread Solve the problem and explain how the concept of duality can be helpful to find out the right price for various kinds of thread. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………….

4.9

SUMMARY

The simplex method is the appropriate method for solving a linear programming problem with more than two decision variables. For less than or equal to type constraints slack variables are introduced to make inequalities equations. A particular type of solution known as a basic feasible solution is important for simplex computation." Every basic feasible solution is an extreme point of the convex set of feasible solutions and vice versa. A basic feasible solution of a system with m equations and n variables has m non negative variables known as basic variables and n-m variables with value zero known as non-basic variables, We can always find a basic feasible solution with the help of the slack variables. The objective function is maximised or minimised at one of the basic feasible solutions. Starting with the initial basic feasible solution obtained from the slack variables the simplex method improves the value of the objective function step by step by bringing in a new basic variable and making one of the present basic variables non basic. The selection of the new basic variable and the omission of a current basic variable are performed following certain rules so that the revised basic feasible solution improves the value of the objective function. The iterative procedure stops when it is no longer possible to obtain a better value of the objective function than the present one. The existing basic feasible solution is the optimum solution of the problem which maximises or minimises the objective function as the case may be. When one or more of the constraints are greater than or equal to type surplus variables are introduced to make inequalities equations. Surplus variables, however. cannot be used to obtain an initial basic feasible solution. If some of the constraints are greater than or equal to type or equations artificial variables are used to initiate the simplex computation Two methods, namely, Two Phase Method and M-method are available to solve linear programming problems in these cases. The simplex method can identify multiple or unbounded solutions and infeasible problems. The simplex method also provides a mean for carrying out sensitivity or post optimality analysis of the problem. It is possible to study the effect of change in profit contribution for a particular product without solving the problem all over again. The effect of change in various resource levels can also be ascertained by making a few additional calculations. Every linear programming problem has an accompanying linear programming problem known as a dual problem. The variables of the dual problem are known as dual variables. The dual variables have an economic interpretation which can be used by management: for planning its resources. The solution of the dual problem can be 50

obtained from the simplex computation of the original problem. The solution has a number of important properties which can also. be helpful for computational purposes.

Linear Programming – Simplex Method

4.10 KEY WORDS A Slack Variable corresponding to a less than or equal to type constraint is a non negative variable introduced to convert the constraint into an equation. A Basic Solution of a system of m equations and n variables (m < n) is a solution where at least n-m variables are zero.

A Basic Feasible Solution of a system of m equations and n variables (m < n) is a solution where m variables are non negative and n-m variables are zero. A Basic Variable of a basic feasible solution has a non negative value. A Non Basic Variable of a basic feasible solution has a value equal to zero. A Surplus Variable corresponding to a greater than or equal to type constraint is a non negative variable introduced to convert the constraint into an equation. An Artificial Variable is a non negative variable introduced to facilitate the computation of an initial basic feasible solution. The Optimum Solution of a linear programming problem is the solution where the objective function is maximised or minimised. The Sensitivity Analysis of a linear programming problem is a study of the effect of changes of the profit or resource level on the solution. The Dual Problem corresponding' to a linear programming problem is another linear programming problem formulated from the parameters of the original problem. The Primal Problem is the original linear programming problem. The Dual Variables are the variables of the dual linear programming problem. The Shadow Price of a resource is the change in the optimum value of the objective function per unit increase of the resource.

4.11 SELF-ASSESSMENT EXERCISES 1) A manufacturer has production facilities for assembling two different types of television sets. These facilities can be used to assemble both black and white and coloured sets. At the present time the firm is producing only one model of each type of set. The black and white set contributes Rs. 150 towards profit while a coloured set contributes Rs. 450 towards profit. The number of coloured television sets manufactured everyday cannot exceed 50 as the number of coloured picture tubes available everyday is 50. Each black and white set requires 6 man-hours of chassis assembly time, whereas each coloured set requires 18 man hours. The daily available man hours for the chassis assembly line is 1800. A black and white set must spend one man hour on. the set assembly line whereas a coloured set must spend 1.6 man hours on the set assembly line. The daily available man hours on this line is 240. A black and white television set requires 0.5 man hours of testingand final inspection whereas a coloured set requires 2 man hours. The total available man hours per day for testing and inspection is 162. Formulate and solve the problem using simplex method so that the profit is maximised. 2) A small scale unit is in a position to manufacture three products A, B and C. Raw material required per piece of product A, B and C is 2 kg, 1 kg, and 2 kg respectively while the total daily availability is 50 kg. The raw material is processed on a machine by the labour force and on a day the availability of machine hours is 30 while the availability of labour hour is 26. The time required per unit production of the three products is given below 51

Programming Techniques – Linear Programming and Application

Product A

Machine Hour 1

Labour Hour 1

2 B 3 2 C 1 1 The net per unit profit from products A, B, C respectively are Rs. 25, Rs. 30 and Rs. 40. Find the linear programming formulation of the problem. Solve the problem by simplex method to obtain the maximum profit per day. 3) A small electronics dealer buys various components to assemble them into transistors, tape recorders and small stereo sets. In a week the dealer has time to assemble at the most 500 units of any one or, the combined items. Transistors and tape recorders have a weekly combined order of at least 150 units. Transistors being very popular, the number of these units assembled must exceed the number of ' tape recorders and stereos combined exactly by 100 units. The contribution towards profit made by each transistor, tape recorder and stereo is Rs.75, Rs. 125 and Rs. 150 respectively. How many units of these musical items be assembled each week to maximise the total profit? 4) Solve the following linear programming problem and give your comments

5) Solve the following problem by simplex method and give your comments.

6) Solve the linear programming problem and give your comments

7) In Activity 3 find the range within which the profit coefficients of the three decision variables can vary without changing the optimum solution. 8) Formulate the dual linear programming problem of Exercise 2 and find the optimum values of the dual variables. 9) Write down the primal problem corresponding to the dual problem given below and hence find its solution

52

4.12 ANSWERS

Linear Programming – Simplex Method

53

Programming Techniques – Linear Programming and Application

4.13 FURTHER READINGS Hadly, G. 1969. Linear Programming, Addison Wesley, Reading Massachusetts : USA. Mittal, K.V. 1976. Optimisation Methods in Operations Research and Systems Analysis, Wiley Eastern Limited : New Delhi. Mustafi, C.K. 1988. Operations Research Methods and Practice, Wiley Eastern Limited : New Delhi.

54

Transportation Problem

UNIT 5 TRANSPORTATION PROBLEM Objectives After studying this unit, you should be able to : • formulate a transportation problem • locate a basic feasible solution of a transportation problem by various methods • ascertain minimum transportation cost schedule by Modified Distribution (MODI) method • explain Stepping Stone Method to find the minimum transportation cost schedule • discuss appropriate method to make unbalanced transportation problems balanced • deal with degenerate, transportation problems • formulate and solve transhipment problems . • discuss suitable method when the problem is to maximise the objective function instead of minimising it. Structure 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.11 5.12

Introduction Basic Feasible Solution of a Transportation Problem Modified Distribution Method Stepping Stone Method Unbalanced Transportation Problem Degenerate Transportation Problem Transhipment Problem Maximisation in a Transportation Problem Summary Key Words Self-assessment Exercises Answers Further Readings

5.1

INTRODUCTION

The transportation problem is a special type of linear programming problem where the objective is to minimise the cost of distributing a product from a number of sources or origins to a number of destinations. Because of a special structure present the usual simplex method is unsuitable for solving transportation problems. These. problems require a special method of solution. The special features of a transportation problem are illustrated with the help of the following example. Example 1 Consider a manufacturer who operates three factories and despatches his products to five different retail shops. The Table below indicates the capacities of the three factories, the quantity of products required at the various retail shops and the cost of shipping one unit of the product from each of three factories to each of the five retail shops.

55

Programming Techniques – Linear Programming and Application

The Table usually referred to as Transportation Table provides the basic data regarding the transportation problem. The capacity of factories 1, 2, 3, is 50, 100 and 150 respectively. The requirement at retail shops 1, 2, 3, 4,, 5 is 100, 70, 50, 40 and 40 respectively. The quantities inside the bordered rectangle are known as unit transportation cost. The cost of transportation of one unit from factory 1 to retail shop 1 is 1, factory 1 to retail shop 2 is 9 and so on. A transportation problem can be formulated as a linear programming problem using variables with two subscripts. Let

The problem thus has 8 constraints and 15 variables. It will be unwise if not impossible to solve such a problem using simplex method. This is why a special computational procedure is necessary to solve transportation-problem. In the next section a number procedures will be presented to derive an initial basic feasible solution of the problem. Activity 1 Fill up the blanks i)

The adjective of a transportation problem is to ............................. transportation cost.

ii)

The constraints of a transportation problem are ..............................

the

iii) If a transportation problem has m factories and n retail shops the number variables is and the number of constraints is

5.2

BASIC FEASIBLE SOLUTION OF A TRANSPORTATION PROBLEM

We illustrate with the example introduced in the last section the computation of an initial feasible solution of a transportation problem. Although the problem has eight constraints and fifteen variables one of the constraints can be eliminated since al + a2 + a3 = bt + b2 + b3 + b4 + b5. Thus the problem has in fact seven constraints and fifteen variables. Any basic feasible solution thus has at most seven non zero 56

In general, any basic feasible solution of a transportation problem with m origins (such as factories) and n destinations (such as retail shops) has at most m + n - 1 non

Zero Xij The following methods are available for the computation of an initial basic feasible solution. 1) The North West Corner Rule In the North West corner rule allocations are made starting from the North West (upper left) corner completely disregarding the transportation cost. By applying North West corner rule to the transportation problem of Section 5.1, we obtain x11 = 50 as the capacity of factory 1 is 50. Eliminating the first row as the first' factory is unable to supply any more the reduced transportation Table becomes

Transportation Problem

The value of b1 is reduced to 50 in the revised transportation Table as 50 units have already been supplied in retail shop 1 from factory 1. We now allocate 50 units to the north west corner of the revised transportation Table. Thus x21= 50. Prodeeding in this way we obtain x22=50, x32 = 20, x33 = 50, x34 = 40, x35 = 40. The corresponding transportation cost is given by 1 × 50 + 24 × 50 + 12 × 50 + 35 × 20 + 1 × 50 + 23 × 40 + 26 × 40 = 4520. It is clear mat as soon as a value of xij is determined a row or a column is eliminated from further consideration. The last value of xij eliminates both a row and a column. Hence a feasible solution computed by the north-west corner rule can have at most m + n - 1 positive xij if the transportation problem has m origins and n destinations. Thus the solution is a basic feasible solution. In the present problem m = 3, n = 5. Hence using the north west corner rule we have derived a basic feasible solution with seven non zero xij. 2) Matrix Minimum Method We look for the row and the column corresponding to which Cij is minimum in the entire transportation Table. If there are two or more minimum costs then we should select the row and the column corresponding to the lower numbered row. If they appear in the same row we should select the lower numbered column. We choose the value of the corresponding xij as much as possible subject to capacity and requirement constraints. A row or a column is dropped and the same procedure is repeated with the reduced transportation cost matrix. The method is illustrated with the help of the transportation problem presented in Section 5.1. We observe that C11 = 1 which is the minimum transportation cost in the entire transportation Table. Hence x11 = 50 and the first row is eliminated from any further allocation. The reduced Transportation matrix is

the minimum transportation cost in the reduced transportation table. So x25 = 40. Proceeding in this way we observe that x33 = 50, x22 = 60, x3l = 50, x32 = 10, x34 = 40. The basic feasible solution developed by the matrix minimum method has a transportation cost 1 × 50 + 1 × 40 + 1 × 50 + 12 × 60 + 14 × 50 + 33 × 10 + 23 × 40 = 2810 The minimum transportation cost obtained by using matrix minimum method is much lower than the corresponding cost of the solution derived by using north west corner rule. This is to be expected as the matrix minimum method takes into account the unit transportation cost while choosing the values of the basic variables. C25 = 1 is

57

Programming Techniques – Linear Programming and Application

3) Vogel Approximation Method (VAM)

Vogel Approximation method for finding a basic feasible solution involves the following steps. i)

From the transportation Table we determine the penalty for each row and column. The penalties are calculated for each row (column) by subtracting the lowest cost element in that row (column) from the next lowest cost element in the same row (column).

ii) We identify the row or column with the largest penalty among all the rows and columns. If the penalties corresponding to two or more rows or columns are equal we select the topmost row and the extreme left column. iii) We select Xij as a basic variable if Cii is the minimum cost in the row or column with largest penalty. We choose the numerical value of xi, as high as possible subject to the row and the column constraints. Depending upon whether ai or bl is the smaller of the two, ith row or jth column is eliminated. iv) The step (ii) is now performed on the reduced matrix until all the basic variables have been identified.

58

Transportation Problem

Find an initial basic feasible solution of the transportation problem by using i)

North west corner rule

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………. ii) Matrix minimum method ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………. iii) Vogel approximation method ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………….

5.3

MODIFIED DISTRIBUTION (MODI) METHOD

The modified distribution method, also known as MODI method or a-v method provides a minimum cost solution to the transportation problem. The steps involved in the Modified distribution method are as follows : 1) Find out a basic feasible solution of the transportation problem using one of the 'three methods described in the previous section. 2) We introduce dual variables corresponding to the row constraints and the column constraints. If there are in origins and n destinations then there will be m+n dual variables. The dual variables corresponding to the row constraints are denoted by ui (I = 1, 2, ….., m) while the dual variables corresponding to column constraints are denoted by vi (j=1, 2, …….., n). The values of the dual variables should be determined from the following equations. ui + vj = cij if xij > 0. 3) Any basic feasible solution of the transportation problem has m+n - 1 Xij > 0. Thus there will be m+n 1 equations to determine m+n dual variables. One of the dual variables can be chosen arbitrarily. It is to be also noted that as the primal constraints are equations, the dual variables are unrestricted in sign.

59

Programming Techniques – Linear Programming and Application

4) If xii = 0, the dual variables computed in 3 are compared with the cij values of this allocation as

cij - u i - v j If all cij - u i - v j ≥ 0 , then by an application of complementary slackness theorem (see Unit 4) it can be shown that the corresponding solution of the transportation problem is optimum. If one or more of cij - u i - v j < 0 , we choose

the cell with least value of cij - u i - v j and allocate as much as possible subject to the row and the column constraints. The allocation of a number of adjacent cell are adjusted so that a basic variable becomes non basic. 5) A fresh set of dual variables are computed and entire procedure is repeated. Let us consider the following transportation problem given in Example I with a basic feasible solution computed by Matrix Minimum method,

60

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………….

5.4

Transportation Problem

STEPPING STONE METHOD

Stepping Stone Method is another method for finding the optimum solution of the transportation problem. The various steps necessary in the stepping stone method are given below : 1) Find an initial basic feasic feasible solution of the transportation problem. 2) Next check for degeneracy. A basic feasible solution with m origins and n destinations is said to be degenerate if the number of non zero basic variables is less than m + n - 1. When •a transportation problem is degenerate it has to be properly modified. This is included in Section 5.6'. 3) Each empty (non allocated) cell is now examined for a possible decrease in the transportation cast. One unit is allocated to an. empty cell. A number of adjacent cells are balanced so that the row and the column constraints are not violated. If the net result of such readjustment is a decrease in the transportation cost we include as many units as possible in the selected empty cell and carry out the necessary readjustment with other cells. 4) Step 3 is performed with all the empty cells till no further reduction in the transportation cost is possible. If there-is another allocation with zero increase or decrease in the transportation cost than the transportation problem has multiple solutions.

The stepping stone method is illustrated with the help of the following example. Example 3

Consider the following transportation problem (cost in rupees)

The figures in the parenthesis indicate the allocation in the corresponding cells. 1) The solution is not degenerate as the number of non zero basic variables is m+n-1 = 6. 2) The cell BD is empty. The result of allocating one unit along with the necessary adjustment in the adjacent cells is indicated in Table 2.

61

Programming Techniques – Linear Programming and Application

The increase in the transportation cost per unit quantity of reallocation is +3+6-4-5=0. This indicates that every unit allocated to route BD will neither increase nor .decrease the transportation cost. Thus, such a reallocation is unnecessary. 4) The result of reallocating one unit to cell' CD is indicated in Table 3.

The net increase in the transportation cost per unit of reallocation is +3+6+2-4-5-6 = -4. Thus the new route would be beneficial to the company. The maximum amount that can be allocated in CD is 100 and this will make the current basic variable corresponding to cell CF non basic. Table 4 shows the transportation table after the reallocation.

This procedure was repeated with remaining empty cells CE, AF, CF, AG, BG. The results are summarised in the following Table.

Since reallocation in any other unoccupied cell can not decrease the transportation cost the present allocation is optimum. The minimum transportation cost is thus 62

Transportation Problem

The transportation schedule is, however, not unique as there are a number of unoccupied cells with zero increase in transportation cost. Activity 4

Solve the •following transportation problem by stepping stone method

......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... .........................................................................................................................................

5.5

UNBALANCED TRANSPORTATION PROBLEM

We solved the various transportation problems with the assumption that the total supply at the origins is equal to the total requirement at the destinations. If they are unequal the problem is known as an unbalanced transportation problem. If the total supply is more than the total demand we introduce an additional column which will indicate the surplus supply with transportation cost zero. Likewise, if the total demand is more than the total supply an additional row is introduced in the Table which represents unsatisfied demand with transportation cost zero. The balancing of an unbalanced transportation problem is illustrated in the following example. Example 4

The total requirement is 1100 whereas the total supply 800. Thus we introduce an additional row with transportation cost zero indicating the unsatisfied demand.

63 Now the problem can be worked out as discussed in previous sections.

Programming Techniques – Linear Programming and Application

5.6

DEGENERATE TRANSPORTATION PROBLEM

If a basic feasible solution of a transportation problem with an origins and n destinations has fewer than in + n - 1 positive xij (occupied cells) the problem is said to be a degenerate transportation problem. While in the simple computation degeneracy does not cause any serious difficulty, it can cause computational problem in a transportation problem. If we apply modified distribution method, then the dual variables ui and vj are obtained from the cij values of the basic variables. If the problem is degenerate, you will be unable to locate one or more cij value which should be equated to corresponding ui + vj. Computational difficulty will also arise while applying stepping stone method to a degenerate transportation problem. It is thus necessary to identify a degenerate transportation problem at the very beginning and take appropriate step to avoid any computational difficulty. The degeneracy in a transportation problem can be identified through the following result (Mustafi, 1988) : A degenerate basic feasible solution in a transportation problem exists if and only if some partial sum of availabilities (row) is equal to a partial sum of requirements (column).

As for illustration the transportation problem presented in Section5.5 3 degenerate as a1, = 300 = b1 a2 + a3 = 800 = b2 + b3. Perturbation Technique

The degenerate basic solutions of the transportation problem can be avoided if we ensure that no partial sum of ai and bj are the same. We set up a new problem where *

a i = ai + d

i = 1, 2, &., m

b j = bj

j = 1, 2, &., n-1

b n = bn+ md

d>0

*

*

This modified problem has been constructed in such a manner that no partial sum of *

*

a i is equal to a partial sum of b j . After the problem is solved, we put d = 0 leading to the optimum solution of the original problem. For illustration, consider the transportation problem presented in Section 5.5 (after the addition of the additional row). Here m=3, n=3. The perturbed problem is given by

The problem can be solved using any of the methods described before. Then we take d = 0 to obtain the solution of the original problem. Activity 5

Solve the perturbed problem given above. 64

………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

………………………………………………………………………………………….

5.7

Transportation Problem

TRANSHIPMENT PROBLEM

In a transportation problem consignments are always transported from an origin to a destination. There could be situation where it might be economical to transport items in several stages : First within certain origins and destinations and finally the reform to the ultimate receipt points. It is not uncommon to maintain dumps for central storage of certain bulk material. Similarly, movement of material involving two different modes of transport - road and railways or between stations connected by broad gauge and metre gauge lines will necessarily require transhipment. Titus for the purpose of transhipment the distinction between an origin and destination is dropped so that from a transportation problem with m origins and n destinations we obtain a transhipment problem with m+n origins and m+n destinations. The formulation and solution of a transhipment problem is illustrated with the help of the following example. Example 5

Consider a transportation problem where the origins are plants and destinations are depots. The unit transportation costs, capacity at the plants and the requirements at the depots are indicated below :

When each plant is also considered a destination and each depot is also considered an origin, there are altogether five origins and five destinations. Some additional cost data are also necessary. There are presented in the following Tables.

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Programming Techniques – Linear Programming and Application

From Table 1, Table 2, Table 3 and Table 4 we obtain the transportation formulation of the transhipment problem

A bluffer stock of 450 which is the total capacity and total requirement in the original transportation problem is added to each row and column of the transhipment problem, The resulting transportation problem has m + n = 5 origins and m+n = 5 destinations, On solving the transportation problem presented in Table 5

1) Transport x21 = 300 units from plant B to plant A. 'his increase the availability at plant A to 450 units including the 150 units originally available from A. 2) From plant A transport x13 = 300 ,to depot X and X14 = 150 to depot Y. 3) From 300 units available at depot X transport x35 = 150 units to depot Z. The total transhipment cost is

1 × 300 + 3 × 150 + 1 × 300 + 1 × 150 = 1200 If, however, the, consignments are transported from plants A, B to depots X, Y, Z only according to the transportation Table 1, the minimum transportation cost schedule is x13 = 150 x21 = 150 x22 = 150 with a minimum cost of 3450. Thus transhipment reduces the cost of cargo movement in this case. Activity 6

Solve using modified distribution method the transportation problem given in Table 5 (Transhipment problem). ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

5.8

66

MAXIMISATION IN A TRANSPORTATION PROBLEM

There are certain types of transportation problems where the objective function is to be maximised instead of being minimised. These problems can be solved by converting the maximisation problem into a minimisation problem. The formulation and solution of this class of problems ate illustrated with the help of the following example.

Example 6

Transportation Problem

A firm has three factories located in City A, City B and City C and supplies goods to four dealers 'spr'ead all over the. country. The production capacities of these factories are 1000, 700 and 900 units per month respectively. The net return per unit product varies for different combinations of dealers and factories which is given in Table 1.

Determine a suitable allocation to maximise the total net return. If xij denotes the number of units to be despatched from the ith city to the jth dealer q be the corresponding return, then the objective function is

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5.9

SUMMARY

Transportation Problem is a special type of linear programming problem. Simplex method is not suitable for the solution of a transportation problem. On the other hand the transportation problem has a special structure which may be utilised to develop efficient computational techniques for its solution.

In the most general form, a transportation problem has a number of origins and a number of destinations. A certain amount of a particular consignment is available in each origin. Likewise, each destination has a certain requirement. The transportation problem indicates the amount of consignment to be transported from various origins to different destinations so that the total transportation cost is minimised without violating the availability constraints and the requirement constraints. The decision variables xia of a transportation problem indicate the amount to be transported from the ith origin to the jth destination. Two subscripts are necessary to describe these decision variables. A number of techniques are available for computing an initial basic feasible solution of a transportation problem. These are the North West Corner rule, Matrix Minimum method and Vogel's Approximation Method (YAM). Optimum solution of a transportation problem can obtained from Modified Distribution (MODI) Method or Stepping Stone Method. Sometimes the total available consignment at the origins is different from the total requirement at the destinations. Such a transportation problem is said to be unbalanced. An unbalanced transportation problem can be made balanced where the total available consignment at the origins is equal to the total requirement at the destinations by introducing an additional row or column with zero transportation; cost. The basic feasible solutions of a transportation problem with m origins and n destinations should have m+n - 1 positive basic variables. However, if one or more basic variables are zero the solution is said to be degenerate. A degenerate transportation problem can be modified by perturbation method so that the problem can be solved without any difficulty. A problemhaving a structure similar to that of a transportation problem where the objective function is to be maximised can also be solved by the techniques developed in this unit with slight modification. Transportation problem can be generalised into a transhipment problem where

68

shipment is possible from origin to origin or destination as well as from destination to origin or destination. This may result in an economy of transportation in some cases. A transhipment problem can be formulated as a transportation problem with an increased number of origin and destinations.

Transportation Problem

5.10 KEY WORDS The Origin of a transportation problem is .the' location from which shipments are despatched. The Destination of, a transportation problem is the location to which shipments are transported. The Unit Transportation Cost is the cost of transporting one unit of the consignment from an origin to a destination. The North West Corner Rule is a method of computing a basic feasible solution of a transportation problem where basic variables are selected from the North West Corner, i.e. top left corner. The Matric Minimum Method is a method of computing a basic feasible solution of a transportation problem where the basic variables are chosen according to the unit cost of transportation. The Vogel's Approximation Method (VAM) is an iterative procedure of computing a basic feasible solution of the transportation problem. The Modified Distribution Method (MODI) is a method of computing optimum solution of a transportation problem. The Stepping Stone Method is a method of computing optimum solution of a transportation problem. An Unbalanced Transportation Problem is a transportation problem where the total availability at the origins is different from the total requirement at the destinations.

A Degenerate Transportation Problem with in origins and n destinations has a basic feasible solution with fewer than m+n - 1 positive basic variables. The Perturbation Technique is a method of modifying a degenerate transportation problem so that the degeneracy can be resolved. The Transhipment Problem is a transportation problem where shipment is possible from an origin to an origin or a destination as well as from a destination to an origin or a destination.

5.11 SELF-ASSESSMENT EXERCISES 1) Consider the following transportation problem with the following unit transportation costs, availability and requirement.

Find a basic feasible solution of problem by (i) North West Corner Rule. (ii) Matrix Minimum Method (iii) VAM. Compute the corresponding transportation costs.

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Programming Techniques – Linear Programming and Application

2) Find the optimum solution of the transportation problem given in exercise 1 by (i) Modified Distribution Method (ii) Stepping Stone Method. 3) High Speed Transport Corporation has the following requirement and availability of goods from I, II and III to A, B and C with the following unit cost matrix.

Find the minimum cost transportation schedule. 4) A company has three plants at locations A, B and C which supply to warehouses located at D, E, F, G and Monthly plant capacities are 800, 500 and 900 units respectively. Monthly warehouse requirements are 400, 400, 500, 400, and 800 units respectively. The unit transportation costs are given below :

Determine an optimum distribution for the company. in order to minimise the total transportation cost. 5) A transportation problem has two origins and three destinations.. The unit costs of transportation, availability at the origins and the requirement at the destinations are given below.

Find a minimum cost transhipment solution of the problem. Compare this cost with the corresponding minimum cost transportation solution.

70

Tenders are submitted by four different manufacturers who undertake to supply not more than the quantities indicated below : Manufacturer W X Y Z Total Dress Quantity 300 250 15 200

The store estimates that its profit (in suitable units) per dress will vary according to the manufacture as shown in the following Table. .

Transportation Problem

How should the orders for dresses be placed so as to maximise the profit?

5.12 ANSWERS

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Programming Techniques – Linear Programming and Application

5.13 FURTHER READINGS Hadley, G. 1969. Linear Programming, Addison Wesley, Reading, Massachusetts USA. Mittal, K.V. 1976. Optimisation Methods in Operations Research and Systems Analysis, Wiley Eastern Limited : New Delhi.

Mustafi, C.K. 1988. Operations Research Methods and Practice, Wiley Eastern Limited : New Delhi.

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Assignment Problem

UNIT 6

ASSIGNMENT PROBLEM

Objectives After studying this unit, you should be able to : • formulate an assignment problem • use Hungarian method for solving an assignment problem • choose appropriate method for balancing and solving an unbalanced assignment problem • introduce appropriate modification when some of the assignments are infeasible • modify the assignment problem when the objective is to maximise the objective function • formulate and solve the crew assignment problem. Structure 6.1 Introduction 6.2 Solution of the Assignment Problem 6.3 Unbalanced Assignment Problem 6.4 Problem with some Infeasible Assignments 6.5 Maximisation in an Assignment Problem 6.6 Crew Assignment Problem 6.7 Summary 6.8 Key Words 6.9 Self-assessment Exercises 6.10 Answers 6.11 Further Readings

6.1

INTRODUCTION

The assignment problem in the general form can be stated as follows : Given n facilities, n jobs and the effectiveness of each facility for each job, the problem is to assign each facility to one and only one job in such a way that the measure of effectiveness is optimised (Maximised or Minimised). Several problems of management has a structure identical with the assignment problem. A departmental head may have five people available for assignment and five jobs to fill. He may like to know which job should be assigned to which person so that-all these tasks can be accomplished in the shortest possible time. Likewise a truck company may have an empty truck in each of the cities 1, 2, 3, 4, 5, 6 and needs an empty truck in each of the cities 7, 8, 9, 10, 11, 12. It would like to ascertain the assignment of trucks to various cities so as to minimise the total distance covered. in a marketing set up by making an estimate of sales performance for different salesmen as well as for different territories one could assign a particular salesman to a particular territory with a view to maximise overall sales. It may be noted that with n facilities and n jobs there are n! possible assignments. One way of finding an optimum assignment is to write all the n! possible arrangements, evaluate their total cost (in terms of the given measure of effectiveness) and select the assignment with minimum cost. The method leads to a computational problem of formidable size even when the value of n is moderate. Even for n 10 the possible number of arrangements is 3628800. It is thus necessary to develop a suitable computation procedure to solve an assignment problem.

6.2

SOLUTION OF AN ASSIGNMENT PROBLEM

Suppose is the measure o effectiveness when ith person is assigned jth job. It is

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Programming Techniques – Linear Programming and Application

74

also assumed that the overall measure of effectiveness is to be minimised (such as total time taken to accomplish all the jobs). As in the case of transportation problem we introduce decision variables xij with double subscripts so that xij is the number of ith individuals assigned to jth job. Since ith person can be assigned only one job and jth job can be assigned to only one person we have

The assignment problem is thus a special case of transportation problem where m = n and ai = bi = 1. It may however be easily observed that any basic feasible solution of an assignment problem contains (2n-1) variables of which (n--1) variables are zero. Due to this high degree of degeneracy the usual computational techniques of a transportation problem become very inefficient A separate computational device is required to solve assignment problem. The basic result on which the solution of an assignment problem is based can be stated as follows (Mustafi, 1988) : If a constant is added to every element of a row and/or a column of the cost matrix of an assignment problem the resulting assignment problem has the same optimum solution as the original problem and vice versa. This result can be used in two different ways to solve the assignment problem. If in an assignment problem some cost elements are negative, we limy convert them into an equivalent assignment problem where all the cost elements are non negative by adding a suitably large constant to the elements of the relevant row or column. Next we look for a feasible solution which has zero assignment cost after adding suitable constants to the elements of various rows and columns. Since it has been assumed that all the cost elements are non negative, this assignment must be optimum. Based on this principle a computational technique known as Hungarian Method is developed which is discussed below. Hungarian Method : The method is listed below in the form of a series of computational steps, when the objective function is that of minimisation type. Step 1 : Find out the cost table from the given problem. If the number of origins are not equal to the number of destinations, a dummy origin or destination must be added [For details, please refer Section 6.3]: Step 2 : Find the smallest cost in each row of the cost table. Subtract this smallest

cost element from each element in that row. Therefore, there will be atleast one zero in each row of this new table, called the first Reduced Cost Table. Find the smallest element in, each column of the reduced cost table. Subtract this smallest cost element from each element in that column.• As a result of this, each row and column now has atleast one zero value in the second reduced cost table. Step 3 : Determine an assignment as follows : i) For each, row or column with a single zero value cell that has not be assigned or eliminated, box that zero value as an assigned cell. ii) For every zero that becomes assigned, cross out all other zeros in the same row and for column. iii) If for a row and for column there are two or more zero and one cannot be chosen by inspection, choose the assigned zero cell arbitrarily. . iv) The above process may be continued until every zero cell is either assigned (boxed) or crossed out. Step 4 : An optimal assignment is found, if the number of assigned cells equals the number of rows (and columns). In case you had chosen a zero cell arbitrarily, there may be an alternate optimum. If no optimum solution is found (some rows or columns without an assignment), please go to step 5. Step 5 Draw a set of lines equal to the number of assignments made in Step 3, covering all the zeros in the following way. Mark check (V) to those rows where no assignment has been made, i) ii) Examine the checked (V) rows, If any zero cell occurs in those rows, check (V) the respective columns that contain those zeros. iii) Examine the checked (V) columns. If any assigned zero occurs in those columns, check (V) the respective rows that contain those assigned zeros. iv) The process may be repeated until no more rows or column can be checked. v) Draw lines through all unchecked rows and through all checked columns. Step 6 Examine those elements that are not covered by a line. Choose the smallest of these elements and subtract this smallest from all the elements that do not have a line through them, Add this smallest element to every element that lies at the intersection of two lines. The resulting matrix is a new revised cost tableau. Example 1 A job shop has four men available for work on four separate jobs. Only one man can work on any one job. The cost of assigning each man to each job is given in Table 1 below. The objective is to assign men to jobs such that the total cost of assignment is a minimum.

Solution Step 1 : The ost tableau (Table

Assignment Problem

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Programming Techniques – Linear Programming and Application

Step 2 : Find the first and second reduced cost tableau (Table '2 & 3). Table 2 First Reduced Cost Tableau

Step 3 : Determine an Assignment. Examine row A of Table 3. You will find that it has only one zero (Al). Box this zero. Cross-out all other zeros in the boxed column. This way you can eliminate cell B1. Now examine row C. You find that it has one zero (C2) Box this zero. Eliminate all the zeros in the boxed column. This is how cell D2 gets eliminated. There is one zero in column 3. Therefore, D3 gets boxed and this enables us to eliminate cell D4. Therefore, we can box (assign) or eliminate all zeros. (Refer Table 4).

76

Step 4 : The solution obtained in Step 3 is not optimal. This is because we were able to make three assignments when four were required. Step 5 : Cover all the zeros of Table 4 with three lines, since three assignments were made. Check (V) row B since it has no assignment. Please note that row B has a zero in column 1, therefore, we check (V) column 1. We then check (V) row A, since column 1 has an assigned zero in row A. Please note that no other rows or columns can be checked. You may draw three lines

through unchecked rows C & D ad column 1, the checked column. This is shown in Table 5.

Assignment Problem

Step 6 : Develop the new revised tableau. Examine those elements that are not covered by a line in Table 5. Take the smallest element. This is 1(one) in our case. By subtracting 1 from the uncovered cells and adding 1 to elements (Cl & D1) that lie at the intersection of two. lines, we get the new revised cost tableau as given in Table 6 below.

Step 7 : Go to Step 3 and repeat the procedure until you arrive at an optimal assignment. Step 8 : Determine an assignment. By examining each of the four rows in Table 6, we find that it is only row C which has got only one columns have two zeros. Choose a zero arbitrarily, say Al and box this cell. Thus cell A3 and B1 get eliminated. Therefore, row B(B4) and column 3 (D4) has one zero. These are boxed and cell D4 is eliminated. Thus all zeros are either boxed or eliminated in Table 7.

Since the number of assignments equal the number of rows (columns), the assignment in Table 7 is optimal. The total cost of this assignment is 78.

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Programming Techniques – Linear Programming and Application

Remember that we had chosen a zero cell arbitrarily, an alternate optimum solution exists and is given by A3, B1, C2 and D4. You may please verify it yourself. Activity 1 A tourist car rental firm has one car in each of the five depots DI, D2, D3, D4, D5 and a customer in each of the five cities Cl, C2, C3, C4, C5. The distances in Kilometers between the depots and the cities are given in the following matrix. How should be cars be assigned to the customers so as to minimise the total distance covered?

…………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………………………...

6.3

UNBALANCED ASSIGNMENT PROBLEM

The number of persons to be assigned and the number of jobs were assumed to be the same in the previous section. Such an assignment problem is known as a balanced assignment problem. If the number of persons is different from the number of jobs the assignment problem is said to be unbalanced. If the number of jobs is less than the number of persons, some of them cannot be assigned any job. We introduce one or more dummy jobs of zero duration to make the assignment problem balanced. This balanced problem can be solved using the method developed in the previous section. The persons to whom the dummy jobs are assigned are left out of assignment. Likewise, if the number of persons is less than the number of jobs we introduce one or more dummy persons with duration time zero to make the assignment problem balanced. We solve this balanced assignment problem and the jobs assigned to the dummy persons are left out. Two examples have been presented to illustrate the solution of unbalanced assignment problems. Example 2 Solve the following unbalanced assignment problem of minimising total time for performing all the jobs.

78

Solution Since the number of jobs is less than the number of operators we introduce a dummy job with duration zero. The revised assignment problem is given below.

Assignment Problem

Using the Hungarian Method described in the previous section the assignment leading to minimum cost is : Operator 1 to job 4, Operator 2 to job 1, Operator 3 to dummy 6, Operator 4 to job 5, Operator 5 to job 2, Operator 6 to job 3. The total minimum completion time is 16. Operator 3, therefore, can not be assigned.

Activity 2 Work out the various steps of the solution of the example 2 …………………………………………………………………………………………… …………………………………………………………………………………………… Activity 3 Work out the various steps of the Solution of the example 3 …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………………………

6.4

PROBLEM WITH SOME INFEASIBLE ASSIGNMENTS

It is sometimes possible that a particular person is incapable of doing certain work or a specific job cannot be performed on a particular machine. The solution of the assignment problem should take into account these restrictions so that the infeasible

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Programming Techniques – Linear Programming and Application

assignments can be avoided. This can be achieved by assigning a very high cost to the cells where assignments are prohibited. The method of solution is illustrated using the following example. Example 4 A metal shop has five jobs to be done and has five machines to do them. The cost matrix gives the cost of processing each job on any machine. Because of specific job requirement and machine specifications certain jobs cannot be done on certain machines. These have been shown by X in the cost matrix. The assignment of jobs to machines must be done on a one to one basis. The objective is to assign the jobs to the machines so as to minimize the total cost within the restrictions mentioned above.

Because certain jobs cannot be done on certain machines we assign a high cost (Say 500) to these cells and modify the cost matrix before solution. The revised assignment problem is .given below

Solving this revised assignment problem we allot Job 1 to Machine 3, Job 2 to Machine 5, Job 3 to Machine 4, Job 4 to Machine 2 and Job 5 to Machine 1. The minimum cost of assignment is 220. Activity 4 Give the step by step solution of the example given in this section …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………………………

6.5

MAXIMISATION IN AN ASSIGNMENT PROBLEM

There are problems where certain facilities have to be assigned to a number of jobs so as to maximise the overall performance of the assignment. The problem can be converted into a minimisation problem and Hungarian Method can be used for its solution. Using notations as in Section 6.2 this problem can be formulated as

80

Assignment Problem

Activity 5 Verify the solution of the assignment problem presented in Examples 5. …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………………………

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6.6

CREW ASSIGNMENT PROBLEM

The method of solution discussed in this unit can be utilised to plan the assignment of crew members in different locations by a transport company. The technique is illustrated with the help of the following example Example 6 A trip from Madras to Bangalore takes six hours by bus. A typical time table of the bus service in both directions is given below :

The cost of providing this service by the transport company depends upon the time spent by the bus crew (driver and conductor) away from their places in addition to service times. There are five crews. There is a constraint that every crew should be provided with more than 4 hours of rest before the return trip again and should not wait for more than 24 hours for the return trip. The company has residential facilities for the crew of Madras as well as at Bangalore. Find Which line of service be connected with which other line so as to reduce the waiting time to the minimum. 1) As the service time is constant for• each line it does not appear directly in the computation. If the entire crew resides at Madras then the waiting times in hours at Bangalore for different route connections are given in the following Table :

If Route a is combined with Route 1, the crew after arriving at Bangalore at 12 Noon start in at 5.30 next morning. Thus the waiting time is 17.5 hours. Some of the assign meats are infeasible. Route 3 leaves Bangalore at 15.00 hours. Thus the crew of Route a reaching Bangalore at 12 Noon are unable to take the minimum stipulated rest of four hours if they are asked to leave by Route 3. Hence a3 is an infeasible assignment. Its cost is thus M, a large positive number. 2) Similarly, if the crew are assumed to reside at Bangalore then the waiting times of the crew in hours at Madras for different route combinations are given in

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3) As the crew can be asked to reside either at Madras or at Bangalore, minimum

waiting time from the above operation can be computed for different route combination by choosing the minimum of the two waiting tins. This is presented in Table 5. The asterisk marked waiting times indicates that the crew are based at Madras, otherwise they are based at Bangalore.

Assignment Problem

4) We now use Hungarian Method to solve assignment problem in Table 3. Solution Step .1 : The cost tableau in terms of waning time (Table 3) Step 2 : Find the first and second reduced cost tableau (Tables 4 and 5).

Step 3 : Determine an Assignment Examine row a of Table 5. You will find that it has Only one zero (a4). Box this zero. Cross out all other zeros in the boxed column. This way you can eliminate cell M. Now examine row C. It has only one zero (C5). Box this zero. There is no other zero in the boxed column. Now examine row d. It has only one zero (d1). Box this zero. There is no other zero in the boxed column. Now examine row e. It has two zero (e2 and e3), We box one of the two zeros arbitrarily. Suppose we box e2. There is no other zero in the boxed column. However, the other zero in the same row i.e. e3 gets crossed out, This way, all the zeros of Table 5 get boxed or eliminated (Refer Table 6).

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84

Since the number of assignments equal the number of rows (columns), the assignment in Table 9 is optimal.

Assignment Problem

Therefore the routes to be paired to achieve the minimum waiting time are a - 3, b - 3, c 5, d - 1 and e - 2. Referring back to Table 3 we can now obtain the waiting times of these assignments as well as the, residence of the crew. Thus is indicated in Table 10.

The minimum total waiting time is thus 33.5 hours.

6.6

SUMMARY

The Assignment Problem considers the allocation of a number of jobs to a number of persons so that the total completion time is minimized. If the number of persons is-the same as the number of jobs, the assignment problem is said to be balanced. If the number of jobs is different from the number of persons the assignment problem is said to be unbalanced. An unbalanced assignment problem can be converted into a balanced assignment problem by introducing a dummy person or a dummy job with completion time zero. Although an assignment problem can be formulated as a linear programming problem, it is solved by a special method known as Hungarian Method because of its special structure. If the times of completion or the costs corresponding to every assignment is written down in a matrix form, it is referred to as a Cost matrix. The Hungarian Method ,is based on the principle that if a constant is added to every element of a row and/or a column of cost matrix the optimum solution of the resulting assignment problem is the same as the original problem and vice versa. The original cost matrix can be reduced to another cost matrix by adding constants to the elements of rows. and columns where the total cost or the total completion time of an assignment is zero. Since the optimum solution remains unchanged after this reduction this assignment is also the optimum solution of the original problem. Various ramifications of the assignment problem are possible. If a person is unable to carry out a particular job the corresponding-cost or completion time is taken as very large which automatically prevents such an assignment. If the objective is to maximise a performance index through assignment, Hungarian Method can be applied to a revised cost matrix obtained from the original cost matrix. The method of solution can also be utilised for allocating crew members to various stations of a transport organisation.

6.8

KEY WORDS

Assignment Problem is a special type of linear programming problem where the objective is to minimise the cost or time of completing a number of jobs by a number of persons. Balanced Assignment Problem is an assignment problem where the number of persons is equal to the number of jobs. Unbalanced Assignment Problem is an assignment problem where the number of

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Programming Techniques – Linear Programming and Application

persons is not equal to the number of jobs. Hungarian Method is a technique of solving assignment problems. A Dummy Job is an imaginary job with cost or time zero introduced to make an unbalanced assignment problem balanced An Infeasible Assignment occurs in the cell (i, j) of the assignment cost matrix if ith person is unable to perform jth job.

6.9

SELF-ASSESSMENT EXERCISES

1) Six contractors have submitted tenders to take up six projects advertised. It is noted that one contractor can be assigned one job as otherwise the time for completion and the quality of workmanship will be affected. The estimates of cost in thousand rupees given by each of them are indicated' below :

Find out the assignment such that the total cost of completing the six projects is minimum. What is the minimum cost? 2) A freight terminal can accommodate six trucks simultaneously. There is cost of sorting and transferring of loads associated with parking of each truck on each of the six sports. On a certain day, four trucks are to be simultaneously parked at the terminal. -The cost matrix. is given below.

Find out the assignment which minimises the total cost of parking. 3) A sales manager has to assign salesman to four territories. He has four candidates of varying experience and capabilities and assesses the possible profit in suitable units for each salesman in each territory as given below :

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Find an assignment that maximises the profit. 4) An airline that operates seven days a week has a time table shown below. Crews 86 must have minimum rest of six hours between flights. Obtain the pairing of flights

that minimises waiting time away from the city. For any given pairing the crew will be based at the city that results in the smaller waiting time.

Assignment Problem

For each pair also mention the city where the crew should be based.

6.10 ANSWERS

6.11 FURTHER READINGS Cooper, L and D. Steinberg, 1974. Methods and Applications of Linear Programmings, Saunders, Philadelphia : USA. Mustafi, C.K. 1988. Operations Research Methods and Practice, Wiley Eastern Limited : New Delhi. Sasieni, M.A. Yaspan and L. Friedman 1959. Operations Research Methods and Problems, J. Wiley and Sons, New York : USA.

87

Goal Programming

UNIT 7 GOAL PROGRAMMING Objectives After completion of this unit, you should be able-to: • • •

Explain the concept of goal programming, Formulate business problems involving multiple goals as goal programming problems, Describe the simplex procedure used in modified form to solve the goal programming problem.

Structure 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9

Introduction Concepts of goal programming Goal programming model formulation Graphical method of goal programming The simplex method of goal programming Application areas of goal programming Summary Self-assessment exercises Further Readings

7.1

INTRODUCTION

Organizational objectives vary according to the characteristics, types, philosophy of management, and particular environmental conditions of the organization. There is no single universal goal for all organizations. Profit maximization, which is regarded as the sole objective of the business firm in classical economic theory, is one of the most widely accepted goals of management. However, in today's dynamic business environment, profit maximization is not the only objective of management. In fact, business firms do on quite a few occasions place higher priorities on goals other than profit maximization. We have seen, for example, firms place great emphasis on social responsibilities, social contributions, public relations, industrial and labour relations, etc. Such goals are sought because of outside pressure or voluntary management decisions. Non economic goals exist and are gaining greater significance. Thus, we see management has multiple conflicting objectives to achieve in the present day business scenario. This implies that the decision criteria should be multidimensional and the decision involves multiple goals. In this context, goal programming assumes greater importance as a powerful quantitative technique capable of handling multiple decision criteria.

7.2

CONCEPTS OF GOAL PROGRAMMING

Linear programming technique that has been applied quite extensively to various decision problems has a limited value for problems involving multiple goals. The difficulty in using linear programming lies in the unidimensionality of the objective function, which requires maximization or minimization of one objective function - an over simplification of reality! However, in real life situations multiple goals and subgoals must be taken into account. Various goals may be expressed in different units of measurement such as rupees, hours, tons, etc. Often multiple goals of management are in conflict or are achievable only at the expense of other goals. Furthermore these goals are incommensurable. Thus, the solution of the problem requires an establishment of a hierarchy of importance among these incompatible goals so that the low-order goals are considered only after the higher-order goals are satisfied or have reached the point beyond which no further improvement are desired. If management can provide an ordinal ranking of goals in terms of their contributions or importance to the organization and all goal constraints are in linear relationships, the problem can be solved by goal programming.

5

Programming Techniques – Further Applications

How is an objective function to be determined and expressed in algebraic form when there exist multiple, incommensurable, and conflicting goals? Goal programming provides an answer to- this. In 8oul programming, instead of trying to maximize or minimize the objective criterion directly as in linear programming, deviations between goals and what can be achieved within the given set of constraints are to be minimized. In goal programming, the objective function contains primarily the deviational variables that represent each goal or subgoal. The deviational variable represented in two dimensions in the objective function, a positive and a negative deviation from each subgoal and/or constraint. Then, the objective function becomes the minimization of these deviations, based on the relative importance or priority assigned to them. The solution of linear programming is limited by quantification. Unless the management can accurately quantify the relati9nship of the variables in cardinal numbers, the solution is only as good as the inputs. The distinguishing characteristic of goal programming is that it allows for ordinal solution. 'Stated differently management may be unable to specify the cost or utility of u goal or a subgoal, but often upper or lower limits may be stated for each subgoal. The manager is usually called upon to specify the priority of the desired attainment of each goal or subgoal and rank them in ordinal sequence for decision analysis. The true value of goal programming is, therefore, the solution of problems involving multiple, conflicting goals according to the manager's priority structure. Activity 1 Describe the specialities of goal programming ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... .........................................................................................................................................

7.3

GOAL PROGRAMMING MODEL FORMULATION

It is important to consider formulation of the model before getting into the details of goal programming solution. Infact, the most difficult problem in the application of management science to decision analysis is the formulation of the model for the practical problem in question. Model formulation is the process of transforming a real-world decision problem into a management science model. While the advent of task, micro-computers the powerful problem formulation requires a great deal of efforts in terms of conceptualization. In order to provide some experience and insight into formulating and analyzing a goal programming problem, typical examples will be provided in this unit. One key to successful application of goal programming is the ability to recognize when a problem can be solved by goal programming and to formulate the corresponding model. Example 1 The manager of the only record shop in a town has a decision problem that involves multiple goals. The record shop employs five full-time and four part-time salesmen. The normal working hours per month for a full-time salesman and a part-time salesman are 160 hours and 80 hours respectively. According to the performance records of salesmen, the average sales has been five records per hour for full-time salesmen and two records per hour for part-time salesmen. The average hourly wage rates are Rs. 3 for full-time salesmen and Rs. 2 for part-time salesmen, Average profit from the sales of a record is Rs. 1.50.

6

In view of past sales record and increased enrollment at the local college, the manager feels that the sales goal for the next month should be 5500 records. Since the shop is open six days a week, overtime is often required of salesmen (not necessarily overtime but extra hours for the part-time salesmen). The manager believes that a good employer-employee relationship is an essential factor of business success.

Therefore, he feels that a stable employment level with occasional overtime jai Prognmming requirement is a better practice than an unstable employment level with no overtime. However he feels that overtime of more than 100 hours among the fulltime salesmen should be avoided because of the declining sales effectiveness caused by fatigue. The manager has set the following goals: 1) The first goal is to achieve a sales goal of 5500 records for the next month. 2) The second goal is to limit the overtime of full-time salesmen to 100 hours. 3) The third goal is to provide job security to salesmen. The manager feels that full utilization of employees' regular working hours (no layoffs) is an important factor for a good employer-employee relationship. However, he is twice as concerned with the full utilization of full-time salesmen as with the full utilization of part-time salesmen. 4) The last goal is to minimize the sum of overtime for both full-time and part-time salesmen. The manager desires to assign differential weights to the minimization of overtime according to the net marginal profit ratio between the full-time and part-time salesmen. Based on the problem stated above, the following constraints can be formulated. 1. Sales Goal

Goal Programming

Achievement of the sales goal, which is set at 5500, is a function of total working hours of the full-time and part-time salesmen and their productivity (sales per hour) rates.

2. Regular Working Hours Salesmen hours are determined by the regular working hours for each type of salesman and the number of full-tune and part-time salesmen employed. With the full-time salesmen the total regular working hours per month will be 5 x 160 = 800. For part-time salesmen, the total salesmen hours/month will be 4 x 80 = 320. Thus we have:

3.

Overtime

In this record shop example, the manager’s second goal is to limit the overtime of full-time salesmen to 100 hours. we do not have a deviational variable to minimize in order to achieve this goal in the above formulated constraints . Therefore, we must introduce a new constraint. Note that the manager listed the minimization of overtime for full-time salesmen as a part of the fourth goal in essence , he has two separate goals regarding the overtime work of full-time salesmen. To limit the

7

Programming Techniques – Further Applications

overtime for full-time salesmen as a part of the fourth goal, In essence„ he has two separate goals regarding the overtime work of full-time salesmen. To limit the

We introduced both the negative and positive deviation from the allowed 100 hours of overtime because the actual overtime can be less than, equal to, or even more than 100 hours. Now, we have a deviational variable to minimize to achieve the second goal, i.e., d+21.

8

In the above model, the differential weight of 3 is assigned to d3 at the P4 level on the basis of net marginal profit ratio per hour between the full-time and part-time salesmen. The productivity ratio (sales per hour) between the full-time and part-time salesmen is 5 to 2, while the hourly wage rate for overtime is Rs. 4.50 and Rs. 2.00. (We have assumed here that a full-time saleman will charge 50% of the normal wage rate/hour as extra charges for overtime, i.e., Rs. 3 + Rs. 1.50 = 4.50). The marginal profit per hour of overtime is Rs. 3 hour the full-time salesmen and Re. 1 for the parttime salesmen. The relative cost of an hour of overtime for the part-time salesmen is three times that of the full-time salesmen. Therefore, 3P4 is assigned to d3 where as P4 is assigned to d2. Example 2 A textile company produces two types of materials A and B. The material A is produced according to direct orders from furniture manufacturers. The material B, is distributed to retail fabric stores. The average production rates for the material A and B are identical at 1000 metres/hour. By running two shifts the operational capacity of the plant is 80 hours per week. The marketing department reports that the maximum estimated sales for the following week is 70000 metres of material A and 45000 metres of material B. According to the accounting department the profit from a metre of material A is Rs. 2.50 and from a metre of material B is Rs. 1.50. The management of the company decides that a stable employment level is a primary goal for the firm. Therefore, whenever there is demand exceeding normal production capacity, the management simply expands production capacity by providing overtime. However, management feels that overtime operation of the plant of more than 10 hours per week should be avoided because of the accelerating costs. The management has the following goals in the order of importance: 1) The first goal into avoid any underutilization of production capacity (i.e., to maintain stable employment at normal capacity). 2) The second goal is to limit the overtime operation of the plant to 10 hours. 3) The third goal is to achieve the sales goals of 70000 metres of material A and 45000 metres of material B. 4) The last goal is to minimize the overtime operation of the plant as much as possible. Formulate this as a goal programming problem to help the management for the best decision.

Production Capacity The production capacity is limited to 80 hours at present by running two- shifts. However, since overtime of the plant is allowed to a certain extent, the constraint may be written as:

Goal Programming

Sales Constraints In this case, the maximum sales for material A and material B are set at 70,060 and 45,000 metres, respectively. Hence, it is assumed that overachievements of sales beyond the maximum limits are impossible. Then, the sales constraints will be (as before, k, and X2 are expressed in thousands):

Overtime Operation Constraint From the case itself, only production and sales constraints can be formulated. However, the analysis of goals indicates that overtime operation of the plant is to be minimized to 10 hours or less. To solve the problem by goal programming, we need a deviational variable that represents the overtime operation of the plant beyond 10 hours. By minimizing this deviational variable to zero we can achieve the goal. Since there is no such deviational variable in the three constraints presented above, we must create a new constraint. The overtime operation of the plant d1+, should be limited to 10 hours or less. However, it may not be possible to limit the overtime operation to 10 hours or less in order to meet higher order goals. Therefore, d+ can be smaller than, equal to or even greater than 10 hours. By introducing some new deviational variables, a constraint regarding overtime can be expressed as: d +1 + d -12 - d +12, = 10 where d-12 = negative deviation of overtime operation from 10 hours. d +12 =overtime operation beyond 10 hours. One metre of material A gives a profit of Rs.,2.50 and one metre of B gives a profit of Pa. 110, Since the production rate is same for both A an d B namely 1000 metres per hours, the hourly profits of A and B are in the ratio of 2.50: 1.50 or 5 : 3. Hence it is appropriate to assign these as differential weights in goal 3. The differential weights imply that management is relatively more concerned with the achievement of the sales goal for material A than that for material B. Now, the model can be formulated as:

Example 3 XYZ computer company produces three. different types of computers: Epic, Galaxie, and Utopia. The production of all computers is conducted in a complex and modern assembly line. The production of an Epic requires five hours in the assembly line, a Galaxie requires 8 hours, and a Utopia requires 12 hours, The normal operation line

9

Programming Techniques – Further Applications

of the assembly are 170 hours per month. The marketing and accounting departments have estimated that profits per unit for the three types of computers are Rs. 100000 for the Epic, Rs. 144000 for the Galaxie, and Rs. ,252000 for the Utopia. The marketing department further reports that the demand is such that the firm can expect to sell all the computers it produces in the month. The president of the firm has established the following goals according to their importance: 1) Avoid underutilization of capacity in terms of regular hours of operation of the assembly line. 2) Meet the demand of the north eastern sales district for five Epics, five Galaxies, and eight Utopias (differential weights should be assigned according to the net profit ratios among the three. types of computers). 3) Limit overtime operation of the assembly line to 20 hours. 4) Meet the sales goal for each type of computer: Epic, 10; Galaxie, 12; and Utopia, 10 (again assign weights according to the relative profit function for each computer). 5) Minimize the total overtime operation of the assembly line. With the practice and experience gained from the previous two examples, we can easily set up a goal programming model for the above problem. 1. Normal Operation Capacity of the Assembly Line . The normal capacity of the assembly line for the month is 170 hours. With this capacity, the firm produces three types of computers. The total operation hours required to produce the computers will simply be a function of production rate (in number of hours) for a unit of each type of computer. Now, we can formulate the normal operation capacity of the assembly line as:

2. Sales Constraint

3. Overtime Operation of the Assembly Line We have already mentioned that frequently we have to introduce new constraints in order to define deviational variables that we must minimize to achieve certain goals. The president listed the minimization of -overtime operation of the assembly line to 20 hours. As we do not have a deviational- variable to minimize in order to achieve this goal, we introduce this following constraints:

10

Goal Programming

In the objective function of the above model, we note differential weights assigned to the second and fourth priority factors. We remember that the criterion used for the weights was the net profit ratio among the three types of computers. Here, we have an assumption that the cost of operating the assembly line is proportional to which computer the line is producing. The net profit ratio, therefore, will simply be determined by dividing the profit by operation hours required to produce each type of computer. For the Epic, the profit is Rs. 100,000 per unit and it requires five hours of assembly operation. Hence, the profit per hour of assembly operation for Epic is Rs. 20,000. Similarly, profits per hour of assembly operation for the Galaxie and Utopia will be Rs. 18,000 and Rs. 21,006 respectively. The differential weights are based on these figures. Activity 2 In company XYZ, the management specifies overtime to be restricted to a maximum of 25 hours. Write an appropriate goal constraint incorporating the deviational variables. ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………. Activity 3 Company XYZ, produces two products. The maximum sales potential for product 1 and product 2 are 30 units and 40 units respectively. Write the goal constraints for achieving the sales goal by incorporating the deviational variables. ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………. Activity 4 In example 3 under formulation, if the hourly profits are changed into hourly costs for the products, how will the objective function change? Write the changed objective function. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………. Steps of Goal Programming Model Formulation Thus far-we have illustrated model formation of goal programming problems with relatively simple examples. The steps we have taken in the model formulation can be briefly summarized as follows: 1. Define Variables and Constants The first step id model formation is the definition of'c4oice variables and the right hand side constants. The right hand side -constants may bi either available resources or specified goal levels. It requires a careful analysis of the problem in order to

11

Programming Techniques – Further Applications

identify all relevant variables that have some effect on the set of goals stated by the decision maker. 2. Formulate Constraints Through an analysis of the relationships among choice variables and their relationships to the goal , a set of constraints should be formulated .A constraint may be either a system constraint that define the relationship between choice variables and the goals . it should be remembered that if there is no deviational variable to minimize in order to achieve in a certain goal , a new constraint must be created . Also , if further refinement of goals and priorities is required , it may be facilitated by decomposing certain deviational variables. 3. Develop the Objective Function Through the analysis of the decision marker’s goal structure ,the objective function must be developed . First , the preemptive priority factors should be assigned to certain deviational variables that are relevant to goal attainment . Second , if necessary differential weights must be assigned to deviational variables at the same priority level . It is imperative that goals at the same priority level be commensurable.

7.4

GRAPHICAL METHOD OF GOAL PROGRAMMING

The graphical method of solving goal programming problem is quite similar to the graphical method of linear programming. In linear programming, the method is used to maximize or minimize an objective function with one goal, whereas in goal programming, it is used to minimize the total deviation from a set of multiple goals. The deviation from the goal with the highest priority is minimized to the fullest extent possible before deviation from the next goal is minimized. The graphical method is illustrated with the help of the following example. Example A manufacturing firm produces two types of product - A and B. According to past experience, production of either Product A or Product B requires an average of one hour in the plant.. The plant has a normal production capacity of 400 hours a month. The marketing department of the firm reports that because of limited market, the maximum number of Product A and Product B that can be sold in a month are 240 and 300 respectively. The net profit from the sale of Product A and Product B are Rs. 800 and lbs. 400 respectively. The manager of the firm has set the following goals arranged in the order of importance (preemptive priority factors). Pl : He wants to avoid any underutilization of normal production capacity. P2 : He wants to sell maximum possible units of Product A and Product B. Since the net profit from the sale of Product A is twice the amount from that of Product B, the manager has twice, as much desire to achieve sales for Product A as for Product B. P3: He wants to minimize the overtime operation of the plant as much as possible. Solution Let x1and x2 be the numbers of units of Product A and Product B to be produced respectively. Since overtime operation is allowed, the plant capacity constraint can be expressed as: x1. +x2+dl - - d1+ =400 where d1- is the underutilization of production capacity and d t is the overtime operation of the normal production capacity. Since the sales goals given are the maximum possible sales volume, positive deviations will not appear in the sales constraints. The sales constraints care be expressed as:

12

Goal Programming

The first step in solving the problem by graphical method is to plot all the goal constraints on a graph as displayed in figure 7.1 since the underutilization and overutilization of the plant capacity is permissible, the deviational variable d1-and d1+ , are indicated by arrows in figure 7.1 . Similarly d2- and d3- , are also denoted by arrows in the diagram. Once all the goal constraints have been plotted on the graph the feasible region is confined to the shaded portion OABC. The next step is to analyse each goal in the objective function in search of the optimal solution in the feasible region. The first goal is completely achieved at line EF in the reduced feasible region EBF. Now, the manager would like to move to priority level 2, where the differential weight for product A is two times that of B (assigned in the ratio of the profits). The sales goal of 'product A is completely achieved at line EB in the feasible region E3F. The sales goal of product B is fully met at line FB in the feasible region EBF. So, we see that the first two goals are completely achieved at point B in the feasible region EBF. The third goal on overtime operation cannot The best way to explain the simplex method of goal programming is through an example. So, let us take example 2 under goal programming model formulation (Refer 7.3). 13

Programming Techniques – Further Applications

be achieved without dispensing with the first two goals in the priority order . The optimal solutions to the problem is given as under:

7.5

THE SIMPLEX METHOD OF GOAL PROGRAMMING

The best way to explain the simplex method of goal programming is through an example. So, let us take example 2 under goal programming model formulation (Refer 7.3)

Before the solution by the simplex method is presented for the goal programming problem, a few points to be observed are given below: First, in goal programming the purpose is to minimize the unattained portion of goal as much as possible. This is achieved by minimizing the deviational variables through the use of certain preemptive priority factors and differential weights.' Since there is no profit maximization or cost minimization in the objective function, the preemptive factors and differential weights represent q values. Second, it should be remembered that preemptive priority factors are oridinal weights and they are not commensurable. Consequently, Zj or (Zj - Cj) cannot be expressed by a single row as in linear programming .Rather the simplex criterion becomes a matrix of (m x n) size, where m represents the number of preemptive priority factors and n is the number of variables (inclusive of decision and deviational variables). Third, since the simplex criterion Zj - Cj is expressed-as a matrix rather than a row, a new procedure must be devised for identifying the key column. Again since Pj>> Pj+1the selection procedure of the column must be initiated from Pj and move gradually to the lower priority levels. The solution to this problem by simplex method is explained below: Table 5.3 Cj

0

P1

5P3

3P3

0

P4

P2

CB

XB

b

x1

x2

d1-

d2-

d3-

d12-

d1+

d12+

Pt

d1-

80

1.

1

1

0

0

0

-1

0

5P3 3P3 0

d2d3d12-

70 45 10

1 0 0

0 1 0

0 0 0

1 0 0

0 1 0

0 0 1

0 0 1

0 0 -1

P4

0

0

0

0

0

0

0

-1

0

P3 P2 P1

485 0 80

5 0 1

3 0 1

0 0 0

0 0 0

0 0 0

0 0 0'

0 0 -1

0 -1 0

Zj-Cj

14

0

The initial simplex table is presented in Table 5.3. The first four rows of the table are set up in the same way as for L.P. problem, with the coefficients of the associated variables placed in the appropriated entries. Below the thick line which separates the constraints from the objective function, there are four rows and each row stands for a priority goal level.

Note that the optimal criterion (Z, - Q is a 4 x 8 matrix because there are four priority factors and eight variables (two decision .variables and six deviational variables) in the model. Preemptive priority goals are written in basic variables column XB below the thick line from the lowest at the top to the highest at the bottom.

Goal Programming

It should be apparent that the selection of the key column is based on the per unit contribution rate of each variable in achieving the goals. When the first goal is completely attained, then the key column selection criterion moves on to the second goal, and so on. This is why the preemptive priority factors are listed from the lowest to the highest so that. the key column can be easily identified at the bottom of the table. To make the simplex table relatively simple, the Z' matrix is omitted. In goal programming the Zi values'(P4 = 0, P3 = 485, P2 = 0 and Pl = 80) in the resources column (XB) represents the unattained portion of each goal. The key column, would be determined by selecting the largest positive element in Zj Cj row at the Pl level as there exists an unattained portion of this highest goal (Pi = 80). There are two identical positive Zj - C; values in the Xl and X2 columns. In order to break this tie, we check the next flower priority levels. Since at the priority 3, the largest element is 5 in a row, therefore xl becomes the key column. The values of Zj Cj are computed as follows: Zj - Cj = (Elements in CB column)* (corresponding elements in X' columns) -.CJ (priority factors assigned to deviational variables).

Since PI, P2i P3 and P4 are not commensurable, we must list their coefficients separately in their rows in the simplex criterion (Zj - Cj) as shown in. Table 5.3. The key row is determined by selecting the minimum positive or zero value when values in the resources column (XB)) are divided by the coefficients in the key column. In this problem the key row refers to d2- row as shown in Table 5:3. If there is a tie, then select a row that has the variable with the highest priority factor. Coefficient lying at the intersection of key row and key columns is called key element and in Table 5.3 the key element is circled. By utilizing the usual simplex procedure, The table 5.3 is revised to obtained Table 5.4. Again, Table 5.4 does not give the optimal solution as the resources column (XB) indicates unattained portions of goals. Proceeding in the usual manner, Table 5.4 suggests that an improved solution can be obtained if negative deviational variable d1 is driven out and decision variable x2 enters into the solution. The new improved solution is shown in Table 5.5. The solution in Table 5.5 indicates that production of 70,000 metres of material A and 10,000 metres of material B is sufficient to achieve the first, second and fourth goals and the value of d 3 = 35 suggest that 35,000 metres of material is not achieved. It is also observed that all the elements in Pl and P2 are either zero or negative which indicates that the first two priorities are achieved. Therefore, to improve the solution, the selection of the key column is done at P3 level. Since the only positive element 3 occurs at P3 level which lies in d column. Thus d 1 enters into the solution and d-is driven out as shown in Table 5.5. Finally, Table 5.6 presents the optimal solution. The solution is optimal in the sense that it enables -the decision maker to attain his goals as closely as possible within the given decision environment and priority structure.

15

Programming Techniques – Further Applications

16

Activity 5

Goal Programming

What are the difference in the application of simplex procedure between goal programming and linear programming? ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

7.6

APPLICATION AREA OF GOAL PROGRAMMING

The salient feature of goal programming is its capability to handle managerial problems that involve multiple incompatible goals according to their importance. If management is capable of establishing ordinal importance of goals in a linear decision system, the goal programming model provides management with the opportunity to analyse the soundness of their goal structure. In general, a goal programming model performs three types of analysis: (1) it-determines the input requirements to achieve a set of goals; (2) it determines the degree of attainment of defined goals with given resources; and (3) it provides the optimum solution under the varying inputs and goal structures. The. goal programming approach to be taken should be carefully examined by the decision maker before he employs the technique. The most important advantage of goal programming is its great flexibility, which allows model simulation with numerous variation of constraints and goal priorities. Every quantitative technique has some limitations. Some of these are inherent to all quantitative tools and some are attributed to the particular characteristics of technique. The most important limitation, of goal programming belongs to the first category. The goal programming model simply provides the best solution under the given set of constraints and priority structure. Therefore, if the decision maker's goal priorities are not on accordance with the organization objectives, the solution will not be the global optimum for the organization. For an effective application of goal programming - and for the matter of all mathematical techniques -, a clear understanding of the assumptions and limitations of the technique is a prerequisite. The application of goal programming for managerial decision analysis forces the detision maker to think of goals and constraints in terms of their importance to the organization. It may be mentioned in the passing that the limitations of linear programming technique in terms of assumptions, namely proportionality, additivity, divisibility, and deterministic are attributable to goal programming also. Goal programming can be applied to almost unlimited managerial decision areas. In the area of marketing, it can be applied to media planning and product mix decisions. In finance, it can be applied to portfolio selection, capital budgeting, and financial planning. In production, it can be applied to aggregate production planning and scheduling: In, the academic field, it can be used in assigning faculty teaching schedules and for university admissions planning. In HRD area, it can be used for manpower planning. In public systems area, it can be applied to transportation systems and medical care planning. It maybe pointed out that this list is not an exhaustive one but only indicative of the typical potential areas in which goal programming can be effectively applied. -

7.7

SUMMARY

In this unit, at the outset, we have provided a brief introduction to goal programming as a powerful tool to tackle multiple and incompatible goals of any enterprise some of which may be non-economic in nature. The next point in discussion has been on the concepts of goal programming. The distinction between goal programming and linear programming have been brought

17

Programming Techniques – Further Applications

out. The distinguishing features of goal programming revolve around its ability to use the ordinal principle of preemptive priority structure of the goals of management which may be incommensurable. The formulation of goal programming models with its steps have been covered with three typical and comprehensive examples. You would be in a position to conceptualize complex business problems after carefully going through these examples. In the graphical method of solving the goal programming problem, one problem was formulated and solved graphically for a meaningful appreciation. The modified simplex method employed for solving goal programming problem with its fine distinctions over linear programming situation has been elaborately dealt with. One practical example has been solved step by step to reach the final solution using the simple algorithm. Then the application areas of goal programming in management problems have been briefly covered.

7.8

SELF-ASSESSMENT EXERCISES

1)

What is goal programming? State clearly its assumptions.

2)

Identify the major differences between linear programming and goal programming.

3)

Explain the following terms: i) Deviational variables ii) Preemptive priority structure iii) Differential weights iv) Cardinal value and ordinal value Identify the important areas where goal programming can be effectively applied.

4) 5)

An office equipment manufacturer produces two kinds of products, chairs and lamps. Production of either a chair or lamp requires 1 hour of production capacity in the plant. The plant has a maximum production capacity of 110 hours per week. Because of the limited sales capacity, the maximum number of chairs and lamps that can be sold are 6 and 8 per week, respectively. The gross margin from the sale of a chair is Rs. 80 and Rs. 40 for a lamp. The plant manager has set the following goals arranged in the order of importance: 1) He wants to avoid any underutilization of production capacity. 2) He wants to sell as many chairs and lamps as possible. Since the gross margin from the sale of chair is set at twice the amount of profit from a lamp, he has twice as much desire to achieve the sales goal for chairs as for lamps. 3) He wants to minimize the overtime operation of the plant as much as possible. Formulate this as a goal programming problem and then solve both by graphical and simplex method.

6)

18

The production manager faces the problem of job allocation among three of hit teams. The processing rate of the three teams are 5, 6, and 8 units per hour respectively. The normal working hours for each team are 8 hours per day. The production manager has the following goals for the next day in order of priority: 1) The manager wants to avoid any underachievement of production level, which is set at 180 units of product. 2) Any overtime operation of team 2 beyond 2 hours and team 3 beyond3 hours should be avoided. 3) Minimize the sum of overtime.

Formulate this as a goal programming problem and solve the same by simplex method.

7.8 FURTHER RKABINGS

Goal Programming

Lee S.M. and L.J. Moore. Introduction to Decision Science. New York; Petrocelli/ Charter Inc. 1975. Lee S.M. Goal Programming for Decision Analysis. Philadelphia: Auerbach Publishers. 1972. M.P. Gupta and J.K. Sharma. Operations Researh for Management. National Publishing House. New Delhi 1984.

19

Programming Techniques – Further Applications

UNIT 8

INTEGER PROGRAMMING

Objectives After reading this unit, you should be able to: • Discuss the importance of integer programming models in supporting decisions. • Explain the use of integer variables for providing modelling capabilities beyond those available in linear programming. • Describe the rationale behind Cutting Plane and Branch and Bound methods used for solving integer programming models. Structure 8.1 Introduction. 8.2 Some Integer Programming Formulation Techniques 8.3 Unimodularity 8.4 Cutting Plane Method 8.5 Branch and Bound Method 8.6 Summary 8.7 Self-assessment Exercises 8.8 Further Readings

8.1 INTRODUCTION

20

In practice ,we quite often face situations where the variables of interest have to be integers. Consider for example, the product-mix problem, where a Company, operating within the existing departmental capacities, has to decide on the number of units of each product to be manufactured, m0 as to maximize the profit. Under certain assumptions,, it is possible to formulate the situation as a Linear program and to solve it by using any linear programming (L.P.) technique. The optimal solution in such cases may result in fractional values of the decision variables. It is obvious that such fractional values do not make any sense in practice, and as such, one is . tempted to round-off these values to the nearest integers and use them for action. Rounding-off, however, may result in sub-optimal or infeasible solutions. It is possible to take care of such eventualities in the initial L.P. formulation by specifying that the decision variables can take only positive integer values. Such a linear program with decision variables restricted to integer values is called a linear integer program. While the integer restrictions on decision variables in a L.P. framework may be inherent in the problem situation, there are also situations, where these restrictions may be imposed or engineered by the analyst. Integer variables, specifically those which can take a value of zero or one, have been used in many, problem situations to provide modelling capabilities beyond those available in L.P. Consider for example, the capital budgeting problem, where a company is to choose m among n projects. As each project has an associated return and uses a specific quantity of one or more resources, the problem is to choose the projects so as to maximize return within the given resource constraint. Analysts have used integer variables to formulate this problem. The decision variables are defined as xi corresponding to each project j and xj is allowed to take a value of one or zero depending on whether the project j is chosen or not. Once again, if the integer restrictions are relaxed and the problem is solved by L.P. technique, fractional values of xjs' may result, which will be meaningless in this case. In this unit we will be interested in looking at some formulations and solution techniques for linear integer programming models; and in all our future references to such models we will use the term "integer programming (I.P.) models'', Accordingly, in the next section, we present some I.P. formulations to help you appreciate the application of such models. Before discussing the methods of solving, I.P. models, it is necessary to understand the conditions under which L.P. techniques give integer solutions of I.P. problems. Unimodularity property of a matrix is useful in this context and is presented in the subsequent section In the last two sections, we present two methods for solving I.P. models.

Activity I

Integer Programming

Give three problem situations where integer restrictions are inherent in the problems. ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... Activity 2 Solve the L.P. given below and round-off the values of all the decision variables to the nearest integers. Check as to whether the integer solution thus obtained is a feasible solution to the problem. What inference would you draw from the results of this check?

8.2 SOME LP. FORMULATION TECHNIQUES I.P. formulations of problem situations do not pose problem when the variables inherently are discrete in nature. Problems arise in situations where direct L.P. or I.P. formulations are not possible. Codification of variables are helpful in such situations. We first present a direct I.P. formulation of a problem situation. This is followed by two other situations that require codification and transformation. Situation 1 You have landed in a treasure island full of three types of valuable stones, emerald(E), ruby(R) and topaz(T). Each piece of E, R, and T weighs 3,2,2 kg., and is known to have a value of 4,3,1 crore respectively. You have got a bag with you that can carry a maximum of 11 kgs. Your problem is to decide on how many pieces of each type to carry, within the capacity of your bag, so as to maximize the total value carried. The stones cannot be broken. We start the formulation exercise by defining the decision variables. From the problem situation it is apparent that we want to determine the number of each type of stone to be carried. Hence we define variables corresponding to each type of stone, as follows: X1= Number of emeralds to be carried X2 = Number of rubies to be carried X3 = Number of topaz to be carried. You may note that the pieces of a type of stone are uniform so far as their weights and values are concerned. We now express the Objective function and the constraint in terms of the decision variables. The total value carried as a consequence of X1 of E, X2 of R and X3 of T, under the assumption of linearity, is 4 X1 + 3 X2 + X3, and as per the problem this has to be maximized.

21

Programming Techniques – Further Applications

Under similar assumption, we can express the total weight to be carried, as 3 X1 + 2 X2 + 2 X3, and the capacity of the bag being 11 kgs., the weight carried has to be less than 11, Finally, we note the fact that stones cannot be broken, suggests that the variables have to take discrete values. Thus we have the following formulation:

The above model with a single constraint is known as the Knapsack problem in O.R. literature, and is an example of a direct formulation as the decision variables here are integers by definition. Solution of the above I.P. is given in a later section; meanwhile, you may try your, own method to solve the problem. Explore for example, by relaxing the integer restriction and solving the resulting L.P. Then try to generate different feasible solutions for the I.P. Situation 2 A Company has to decide on the production plan of an item for the next three periods, so as to meet the demands in different periods at minimum cost. The relevant costs are the production and inventory carrying costs. In any period, only if production is undertaken then a cost of Rs. 5 is incurred. The variable cost of production is Rs. 5 per unit. The inventory carrying cost is Rs. 1 per unit per period, and is levied on the inventory at the end of any period. Assume that the demands for the upcoming periods are d1, d2, d3, and the beginning inventory is known to be I,,. Here, our interest is to find the number of units of the item to be manufactured in different periods. We also note, that as demands are known figures, the inventories ,get determined simultaneously with the production. Hence, we have the following decision variables: X1 = number of units to be produced in period 1 X2 = number of units to be produced in period 2 X3 = number of units to be produced in period 3 I1 = number of units in inventory at the end of period 1 I2 number of units in inventory at the end of period 2 I3 = number of units in inventory at the end of period 3 In general, we can write Xt = number of units to be produced in period t and It = end inventory of period t, where t = 1,2,3. The problem in formulation becomes evident once we try to write the production cost function in period t.

Clearly, the function as above cannot be incorporated in the objective function; hence we take recourse to codification. We define three binary variables Y1, Y2, Y3 for each period, where Y, (t = 1, 2, 3) in general, will take a value of 1 i if there is production in that period (i.e. X, > 0), and a value of 0 otherwise.

22

The above completes the formulation, however, the conditions in (6) is still not in an acceptable I.P. format. Standard transformation of (6) in the form of a linear constraint is given below:

Integer Programming

Situation Consider a tailor facing the problem of sequencing four stitching jobs given by different people. All jobs have arrived simultaneously; and he has only one stitching machine. From his experience, it is possible for him to make a fairly accurate estimate of the stitching time of the jobs. Each customer has specified a date by which the respective job is to be delivered. The tailor wants to determine a sequence so as to minimize the average tardiness of all the jobs, where, tardiness of any job is given by Maximum (0, completion time-delivery time). It is also agreed that once. a job is started on a machine, it cannot he taken out before completion. As the problem of sequencing is solved by specifying the starting time for each job, we define the decision variables accordingly. Assuming a datum of zero time for starting the work, and an arbitrary numbering of the jobs as 1 to 4, let XI denote the starting time for job I from the datum (I = 1 to 4). It the processing time for job I (known) is denoted by PI, then the completion time for job I = X1 + Pl. If D1 denote the given delivery date for job I from the same datum, then by definition, tardiness for Job I = Max [0, Xi + PI - DI]. Hence, we can write down the objective functions as:

The constraints of the problem are implicit, and is given by the fact that two jobs cannot be undertaken simultaneously. This would imply that for any pair of job I and J, either I precedes J, or J precedes I. To express this in terms of the decision variables, we note that if,l precedes J, then completion time of I has to be less than the starting time for J, i.e.,

Example I Consider a Company which would like to take up three projects. However, because of budget limitations, not all the projects can be selected. It is known that project j has a present value of C3,;and would require an investment of ait in period t. The capital available in period t is ht. The problem is to choose projects so as to maximize present value, without violating the budget constraints. Formulate the problem as an I.P.

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Programming Techniques – Further Applications

Solution For choice situations of "yes-no", "go-no go" type, where the objective is to determine whether or not a particular activity is to be undertaken, integer binary variables that can take a value of 0 or 1, can be used to represent the decision variables. Here, we find that for each project, we want to find out whether it should be taken up or not, as such we define three decision variables Xj (j =1, 2, 3) corresponding to each project, and Xj

= 1, if project j is selected = 0, otherwise

Then, the objective function and the constraints can be expressed in terms of the decision variables, to give the required formulation :

Activity 3 Consider the example given above, suppose the projects are arbitrarily numbered as 1, 2 and 3. Also, over and above the budget constraints for different periods, it has been specified that between projects 1 and 2 only one can be selected. Show, how you will incorporate this constraints in the above formulation. ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... .........................................................................................................................................

24

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Integer Programming

8.3 UNIMODULARITY We have noted in the introduction that in general, an I.P. model cannot be solved as an L.P., because if fractional values are obtained in the solution, rounding-off may not be possible or desirable. It is worthwhile examining in this context, conditions, it any, under which I.P. models can be solved as L.Ps, to yield integer solutions. Existence of such conditions in a problem situation or formulation would make our task simpler as we do not have to apply any new methods for solution. The objective of this section is to examine such conditions.

Without loss of generality, we can assume that all the technological coefficient a11, a12, ... amn, and .all the right hand side constraints b1, b2 ….. bm are integers. (Even if they are not integers, note that it is always possible to convert them to integers' by suitable multiplication on both the sides of the constraints.) If the integer restrictions on the variables are relaxed, and we specify that the variables can take any value greater than or equal to zero, then the above becomes a L.P. problem. From L.P. theory we know that, given m equations and n unknowns (m less than n), we can set (n-m)'variables to zero and solve for the rest of the m unknowns, provided a solution exists. The (n-m) variables that are set to zero are called nonbasic variables, and the rest m variables are called basic variables. For example, without loss of generality, the first m variables xi, x2 ,..., xm can be the basic variables, and the remaining x m+i + …... Xn, the nonbasic variables. If we now set x m+i…. xn to zero, the constraint set of the above problem can be expressed as follows :

It may be noted here that m equations in m variables now constitute the constraint set of P2. The number of different ways in which the m basic variables can be chosen. from among n variables is known to be nCm. This implies that a total of nCm combinations of constraint sets as in P2 can arise. However, a solution may not exist for all such sets. When a solution exists it is referred to as basic solution. The clads of basic solutions with all non-negative values of the m variables constitute a basic feasible solution (bfs). Recall that L.P. technique essentially gives us a systematic) approach of exploring the different basic feasible solutions to arrive at the optimum. It does not test for or guarantee integer solution. To understand when the bfs will be integer, we first note that the nonbasic variables being all set to zero, are integers and as such do not concern us. Thus, we are interested in finding out the condition under which all the basic variables are integers. For our purpose we first state the conditions in the context of the problem Pl. The proof of the condition requires familiarity with matrix algebra and is presented in the appendix of this section for those who are interested.

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Programming Techniques – Further Applications

Condition: If the matrix of the technological coefficients denoted by A with elements ate, is totally unimodular, then every basic solution of P1 is integer. We note that in P1, the matrix A is given by

We also note that as all the coefficients a12, ... amn are integers, A is integer. We now specify the rule for detecting unimodularity of A. Rule : An integer matrix A with all elements = 0, 1, - 1 is totally unimodular if : a)

Each column does not have more than two nonzeros.

b)

The rows can be classified into two subsets such that: i)

if a column contains two nonzero elements with the same sign, one element is in each of the subsets.

ii)

if a column contains two nonzero elements of different sign, both elements are in the same subset.

It is apparent from the above condition that if we detect unimodularity in the A matrix of an I.P., then we can use any L.P. technique to arrive at a solution. This is because of the fact that all basic solutions being integers, all basic feasible solutions if they exist will also be integers and hence the optimal solution, if it exists will have to be integer.

From Xm = B-1 b we can see that as b is an integer matrix, a sufficient condition for Xm to be an integer vector is that B-1 has to be integer. The condition that B-1 is integer is not a necessary condition because it is possible to have Xm integer even if B-1 is not integer. Now, B-1 is given by: B-1=B*/(det B), where, B*, the adjoint matrix, is known to be integer as B is integer, and (det B) denoting the determinant of B is not equal to zero. Hence, from the expression of B-1 it is clear that if determinant of B is equal to 1 then B-1 is integer, which in turn gives us the condition for a bfs to be integer. The definition of Unimodularity is relevant in this context and is presented below:

26

A square integer matrix B is said to be =modular if determinant of B is equal to 1. An (m x n) integer matrix (as the matrix A in P1) is said to be totally unimodular if every square, nonsingular submatrix of A is unimodular.

Integer Programming

Comment on the solution procedure that can be adopted for the problem.

1

1

1

0

0

0

0

0

0

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1

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l

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0•

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1

From the rule given for detection of unimodularity, it is apparent that each column of A has two nonzeros, also rows I to 3 can be one set and 4 to 6 another set , as given the rule, hence A is totally unimodular. The problem thus can be solved by any L.P. procedure to yield integer solution. This problem is known as the Assignment problem in the O.R. I iterate -e. ' Activity Consider the standard transportation pr6blem. Check whether the matrix A is unimodular. What inference would you draw from the above check.

27

Programming Techniques – Further Applications

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8.4 CUTTING PLANE METHOD Historically, the first method for solving I.P. problems was the cutting plane method. The basic steps involved in the method are as follows i)

Solve a L.P.; to start with take the I.P. problem with integer restriction relaxed.

ii)

Check whether the optimal solution is integer. If yes, stop. The resulting solution is the required one. If no, go to step iii.

iii) Generate a new constraint from the fractional optimal solution obtained in i. The constraint is created so as to rule out the fractional values obtained, without ruling out any integer solutions to the problem. Add the constraint to the L.P. in i; and go back to step i. Thus, the method involves solving a L. P. at every iteration. The approach is to reduce or cut the solution space in every successive iteration, ruling out the current fractional solution, while ensuring that no integer solutions are excluded in the process. The process is continued till an integer solution is found or infeasibility occurs. As the constraints are essentially used to cut the solution space, these are referred to as cuts, and the method is referred to as the cutting plane method. The original method as stated above has been found to. be inefficient computationally, however, it has helped in developing valuable insights to I.P. solution procedure, and subsequently has helped in generating other methods. For our purpose, we present here the original method, useful for solving an I.P. with all integer variables. Generation of a cut, the rationale and the mechanics, is central to the method and is presented below. Activity 6 Read carefully the steps given above. The steps do not mention anything regarding infeasibility. Modify the steps to, account for infeasibility. (Hint :. If at any iteration the L.P. is infeasible, what can you say about the I.P. solution?) ......................................................................................................................................... ....................................................................................………………………………… ………………………………………………………..................................................... ........................................................................................................................................ We now consider the following problem :

28

Hence, (4) is an example of a, cut as described earlier. The example, however does not give us a systematic way of generating a cut. We take this up in the following paragraphs.

Integer Programming

Assume that as given in the first step in cutting plane method, we solve the above as a L.P., and some of the variables in the optimal solution are not integers. Obviously, such variables have to be among the basic variables. We now need to generate a cut. We take any one of the non integer basic variables, say the Ith one. Let the equation .corresponding to this variable in the final simplex tableau be represented as:

Where, AIJ ‘s are the technological coefficients in the final tableau. bI is the optimal value of the Ith basic variable, and is assumed to be non integer. It may be noted that though the summation in (4) has been written over all variables, actually in the optimal tableau, it will be for all the nonbasic variables, and only one basic variable X, will have a coefficient of 1.

Equation (7) has been derived with the assumption that Xj ,s are integers, as such, it gives us the condition to be imposed, if we are to get integer solution. The linear constraint so derived is what we have called earlier a cut. At every iteration such a cut is to be generated from the optimal tableau, by taking any one equation corresponding to a basic variable that is not an integer. We further note that in equation (7), the summation is over all nonbasic variables, hence if this is added to the original L.P., SI = -fl, renders the current optimal solution infeasible. Under such situations the Dual Simplex method can be applied. Example 4 Use the cutting plane method for solving the I.P. problem as given by equations (1) to (3). Solution Step 1: The optimal tableau obtained by solving the relaxed problem can be verified to be : X1

X2

RHS

X1

1

21

5/3

Obj.fn. Z

0

2

10

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Programming Techniques – Further Applications

X1 S1 Z

X1

X2

S1

RHS

1 0 0

2/3 -2/3 2

0 1 0

5/3 -2/3 10

Note from the above tableau that the second equation corresponds to the cut, and the current solution is optimal but infeasible. Dual Simplex can now be applied; the steps are as shown below. i)

Determine the leaving variable: The basic variable with the most negative value is chosen as the leaving variable. In our example, the basic variables are X1 = 5/3, and S1 = -2/3. Hence, S1 is the leaving variable.

ii)

Determine the entering variable: The ratios of the L.H.S. coefficients of the Z equation to the corresponding coefficients of the leaving variable equation is taken, and neglecting the ratios associated with positive or zero denominators, the nonbasic variable corresponding to the smallest ratio is chosen as the entering variable, if the problem is of minimization type; for a maximization problem, the variable corresponding to the smallest absolute value is chosen. For the given problem, the respective equations and the ratios are `shown below: X1 0 0 -

Z Equation S1 Equation Ratios

X2 2 -2/3 -3

S1 0 1 -

Thus X2 is the entering variable. The optimal' solution is shown in the tableau below X1

X2

S1

RHS

X1

1

0

1

1

X2

0

1

-3/2

1

Z

0

0

3

8

At the end of the second iteration, we find that the optimal solution is X, = 1, and X2 = 1. The solution being integer, this is the required answer. Activity 7 Assume that the following equation is taken from the optimal tableau of a L.P.: X5 + 3/4 X6 + 1/4 X7 = 3.5 Which basic variable in the final tableau does the equation correspond to? Generate a cut from the above. 30

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Integer Programming

8.5 BRANCH AND BOUND Branch and Bound is a generic method for solving a class of optimization problems. The optimal solution is selected by successive partitioning of the set of all possible solutions into two mutually exclusive and exhaustive subsets, and establishing limit for each set such that all solutions in the set are worse than the limit. The first part of the procedure involving partitioning is called branching, while the second part of establishing limit is referred to as bounding. In this section we will be restricting ourselves to the application of the procedure in solving I.P. problems. The method is illustrated with the help of an example.

X1, X2, X3 are all non-negative integers. Suppose we relax the integer restriction and solve the resulting L.P. By now, you must have discovered that the solution is trivial. You have to only find the maximum of the ratios of objective function coefficients to the constraint coefficients, for each of the variables; and you should fill up your bag with the variable associated with the maximum ratio. Thus, Max [4/3, 3/2, 1/2] = 3/2, and, as the maximum occurs for X2i the bag should be filled up with this only, i.e., X2 = 11/2, X1 = X3 = 0 is the L.P. optimal, with the value of the objective function, Z = 33/2. It is easy to verify here, that objective function value of any integer solution to the problem have to be less than 33/2. The observation is summarized below If an I.P. problem of maximization type is solved as an L.P., the optimal objective function value so obtained will have to be an upper bound of any optimal integer solution. Activity 8 With reference to the above example, give the rationale behind the following statements a)

The optimal solution to the L.P. is to fill up the bag with X2 (Ruby) only. (Note that L.P. assumes that the stones can be broken).

b)

Any integer optimal solution cannot have Z > 33/2.

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………. The above illustrates the "bound" aspect of Branch and Bound. In general, it may be noted, the integer restriction can only lead to a worsening of the objective function value. As such, for a maximization problem, the L.P. optimal is an upper bound, while for a minimization problem it gives the lower bound. We next inspect the optimal solution of the L.P. obtained above. X2 is the only non integer variable in the solution with a value of 5.5. (In case more than one variables are non integer, we choose any one among these.) For an integer solution, a value of greater than 5: and less than 6 being ruled out, we can specify that either

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Programming Techniques – Further Applications

X2 is less than equal to -5, or it is greater than equal to 6. These can be used to partition the solution space of the original L. P. into two distinct problems as follows

It may be noted that by the above partitioning, we have not ruled out any integer solution to the original problem. Suppose we now solve both the problems. The various possible outcomes are listed below: i) Both the problems have feasible solution. Let the objective function values obtained for Pl and P2 be Zl and Z2 respectively. (Note that both these values have to be less than equal to Z, as the objective function value can never increase because of an additional constraint). The solutions in turn may indicate any of the three outcomes : a) Both PI and P2 have integer solution. b) Only one of the problems yields an integer solution. c) None of the problems yield integer solution. It (a) occurs, then we can find max (Z1, Z2), and the optimal solution is the one associated with the problem for which the maximum occurs. This is true, as the two partitions between them covers the total solution space. If (b) occurs, the integer solution gives us a feasible solution to the I. P. problem. Without loss of generally, let us assume that P1 gives us an integer solution. Thus, no further branching from Pl arises. To determine branching from P2, we first note that, the maximum objective function value, that any problem arising from P2 can have is Z2. Thus, if Zl. is already greater than or equal to Z2, we do not need to explore P2 further, as there cannot be any gain in doing so. In fact, the objective function value of any feasible integer solution obtained at any stage of solving, is a lower bound of the required optimal (an upper bound for a minimization problem), and can be used to discard inferior branches. The solution corresponding to P1 is the required optimal solution. However, if Z1 < Z2, then we have to branch from P2, as we cannon rule out the possibility of reaching a feasible integer solution with a objective function value > Zl Any one of the non integer variable may be chosen, and branching done based on it, as shown earlier. If (c) occurs, we need to branch from both. The method is exactly same as shown earlier. The new upper bounds for the branches from Pl and P2 are Zl and Z2 respectively. ii) Both the problems are infeasible. As the whole solution space has been covered, we can say that the I.P. does not have any feasible solution. iii) Only one of the problems is feasible. If the solution is integer, then it is the required optimal solution. If the solution is not integer, then we branch from that point, with the new upper bound as the objective function value of the feasible problem (P1 to P2). In our case, P1 and P2 solutions are given below P1 Objective function 16.33 X2 5.00 X1 0.33 P2 Infeasible 32

Integer Programming

Both the problems are feasible, with P4 giving an integer solution. This is equivalent to the outcome i (b) as listed earlier. As the objective function value of P4 is greater than P3, we do not stand to gain by branching any further from P3. This is because of the fact (as argued earlier) that any problem arising from P3 can have an objective function value less than equal to'15.5. Thus P4 gives us the required optimal solution with 1 emerald and 4 rubies to be carried, giving a total value of 16 crore. It is useful to smmarize the above procedure in terms of a -diagram. The problems are represented by nodes, with the solutions written inside, and the objective function values denoting the bounds, written beside the nodes. Denoting the first L.P. (the relaxed version of the given I.P.) by PO, we have the following diagram:

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Programming Techniques – Further Applications

If after branching at any node, we find that both the problems are feasible and non integer, it may be useful to take any one node and branch. This way we go exploring depth-wise to identify a feasible integer solution, if it exists. Once a feasible integer solution with objective function value of Z has been found, the unexplored* nodes that are having their objective function values less than or equal to Z need not be explored further. From the rest of the nodes, any one can be chosen and the same strategy may be followed. Thus all nodes are not required to be solved. In our case we did not have to explore problem P3.

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8.6 SUMMARY A linear program with decision variables restricted to integer values 2 called a linear integer program. For our purpose, we have referred to this as "integer program". Such integer restrictions on decision variables may either be inherent in the problem situation, or these may be imposed by the analyst. In the former type, the decision variables, by their very nature, have to take discrete values, The latter type, on the other hand, results from the analyst's ingenuity in modelling problem situations. 0-1 integer variables used for modelling "go-no go" decisions is but one example of this type. In either of the Aftes, relaxing the integer restrictions and hence solving a L.P. may not yield an integer optimal solution. Rounding-off to the nearest integer may not be possible, or may be even meaningless. This justifies the study of integer programming, for developing an understanding of the modelling capabilities available in 0-1 variables, and learning the methods for solving. Some I.P. formulations has been presented first, to help. you develop an insight to the use of 0-1 variables. The unimodularity property of a matrix has been introduced subsequently, to help you detect the I.P. formulations that can be solved by using L.P. techniques only. Finally, the cutting plane and the branch and bound Methods for solving an I.P. have been presented.

8.7 SELF-ASSESSMENT EXERCISES

34

Integer Programming

8.8 FURTHER READINGS Bradley, S.P. 1977. Applied Mathematical Programming, Addison-Wesley Publishing Co., Inc.: Philippines. Garfinkel, R.S. and Nemhauser, G.L. 1972. Integer Programming, John Wiley & Sons Inc. : New York. Greenberg, H. 1971. Integer Programming, Academic Press: New York. Mustafi, C.K. 1988. Operations Research: Methods and Practice, Wiley Eastern Limited, New Delhi. Taha, H.A. 1976. Operations Research: An Introducltion, Macmillan Publishing Co. Inc.: New York.

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Programming Techniques – Further Applications

UNIT 9 DYNAMIC PROGRAMMING Objectives After reading this unit, you should be able to : • Explain the relevance of dynamic programming in decision making. • Discuss the rationale behind dynamic programming methodology. Formulate and solve some standard problems using dynamic programming. • Structure 9.1 Introduction 9.2 Dynamic Programming Methodology: An Example 9.3 Definitions and Notations 9.4 Dynamic Programming Applications 9.5 Summary 9.6 Self-Assessment Exercises 9.7 Further Readings

9.1

36

INTRODUCTION

In practice, many problems involve taking decisions over several stages in a sequence. Consider, for example, a production planning situation, where a Company has to .decide on the production plan of an item for the next three months, so as to meet the demands in different months at minimum cost. Assume that it does not have any stock to start with. The different months for which the production is to be decided, constitute the stages. Taking the best decision, month by month, may not be the optimal policy in such cases. Say, the relevant costs are the fixed cost of production in any period, which is incurred only if production is undertaken in that period and the inventory carrying cost per unit per period on any inventory left at the end of a period. Then, if the fixed production cost is very high compared to the inventory carrying cost, it is always better to produce all the demands in the first period itself. A period by period decision, in such cases, can never suggest carrying inventory to be optimal, and hence can never lead to the optimal plan. Linear Programming formulations of such problems are possible, provided the objective function and the constraints can be expressed as linear functions. In the given situation, the objective function is not linear; and in Unit 8, we have shown how it can be formulated as an integer program. Dynamic Programming (D.P.) provides us with an alternative methodology for solving such multistage problems, involving decisions over several stages in a sequence. By the very nature of any planning problem, decisions are called for in every period, so that the "multistage" aspect is essentially inherent in such problems. Associated with each stage or period are input data, decision variables(s), output parameter(s), and objective function. The stages can be thought of to be connected in series, with the output of a stage forming one of the inputs to the immediately succeeding stage. Finally, for determining the effectiveness of the decisions taken, the objective function of the problem is arrived at from the individual objective functions of the stages. In the example above, for each month, the input data are the beginning inventory and the demand, the decision variable is the production quantity, the output is the end inventory, and the objective function is the cost expressed in terms of the decision variable. The stages are essentially linked in series, as the end inventory of any period is the beginning inventory of the immediately succeeding period. Once such characterization is achieved for a problem situation, it is possible to apply D.P. for formulation and solution. Other situations also exist, where different decisions are to be taken, not necessarily in a sequence. Imposing an arbitrary sequence in such a situation, does not chang6 the original problem, and helps us in applying D.P. methodology. Consider, for example, a problem of allocating some resource among three competing activities Obviously, there being no time dimension, the allocation is to be done simultaneously. If we arbitrarily now, number the activities as 1,2 and 3, and assume that we will be allocating in that sequence, it does not make any difference so far as

the original problem is concerned. In each stage now, we can conceive of allocating to each activity. The input data to each stage would be the resource requirements per unit of each activity and the amount of each resource available for allocation at the start of the allocation process. The decision variable is obviously the amount that is allocated at that stage, and the output will be the amount left for allocation to any stage or activity, is nothing but the amount available for allocation beginning of the next stage. These provide the linkages between two stages. The objective functions for each stage could be the return one would get by the allocation in the particular stage or activity. Thus, a variety or situations may be amenable to D.P. methodology by being characterized is multistage problems.

Dynamic Programming

The objective of this unit is to present the D.P. methodology. First, an example is presented to demonstrate the methodology. The definitions of stage, state, and the principle of optimality, are then presented, and the notations specific to D.P. introduced. Finally, some more applications are presented. Activity 1 Consider a L.P. with n variables and m constraints. Is it possible to characterize it as a multistage problem? Justify your answer. ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ........................................................................................................................................

9.2

D.P.METHODOLOGY -- AN EXAMPLE

Consider the following diagram where circles denote cities, and lines between two such circles represent highways connecting the cities. The numbers inside the circles represent city numbers, and those given beside the lines denote the distances between the cities connected by the lines. Suppose you start from city 1 and would like to go to 10. You would like to know the shortest route from city 1 to 10.

The above problem can definitely be solved by enumerating all possible routes from city 1 to 10, and calculating the total distance for each route. The route that gives the minimum distance would be the required answer. However, this involves too much of computation, hence it may be useful to look for other methods. We attempt this in the following paragraphs. We first note from the diagram that in order to travel from 1 to 10, one has to necessarily cover four highways, i.e., any route has to contain four lines. Thus, the problem may be thought of as one of deciding on these four lines or highways that

37

Programming Techniques – Further Applications

should constitute the shortest route. The first line has to be chosen from among 1-2, 1-3, 1-4, as 1 is the starting city. Similarly, it is apparent from the diagram that the second highway has to emanate from 2,3 or 4, the third has to be from 5, 6 or 7, and the fourth from 8 or 9. If we now define a stage as 4 point where a decision is called for, thee, city 1 can be conceived to be in stage 1, cities, 2, 3, 4 in stage, 2, cities 5, 6. 7 in stage 3, dies 8, 9 in stage 4, with city 10 as the destination. 'The stages are connected. At the beginning of stage 1, we are in city l, and as soon as the decision at this stage is taken, it marks the end of stage 1, and it implies beginning of stage 2, and so on. We also realize that beginning of any stage, we can be in any one of the cities that constitute the stage. We refer to these cities as states. Thus, beginning of stage 2, we can be in city 2, 3 or 4, which form the different states at stage 2. In our effort to reduce computation, we now look at the problem stage by stage, starting from the last stage. Beginning of stage 4, we can be in either 8 or 9 (states). We note that if during the course of our travel, we are at 8 ever, the only highway we can choose is 8-10. If we would have reached 9 instead, then the only choice would have been highway 9-10. We summarize this information in the format below

In the above table, N = 1, denotes one more stage to go, i.e., we are at stage 4, The states we can be in at sage 4, are 8 and 9, as shown in column L the decision implies the destination from the states, which is 10 here, Finally, the distances from the states to the destination are listed in the last column. We now go back one stage. Beginning of stage 3, there are two more stages to go (i.e., N = 2), and we can be in states 5, 6 or 7. Suppose, we now try to answer the following question: If, at any stage of our travel, we are at state 5, what will be the best course of action for reaching the destination? Unlike the earlier case, we find that for any state in stage 3, it is possible for us to go at 8 or 9. From 5, for example, if we choose 5-8, then the distance covered would be 6, and from 8, the only way to the destination requires a distance of 9 units to be covered, as already summarized in the table 1. Thus, if the decision is to go to 8, then the corresponding distance would be 15 (= 6 + 9). Instead, if we would have chosento go to 9, then the length of the highway 5-9 added to the distance corresponding to state 9 in the table, would give us the total distance to be covered. This works out to 14 (= 8 + 6). Thus, in the course of our travel if we are ever at 5, out of the two possible decisions at the stage, namely, going to 8 or to 9, it is better to choose 9, asthis will involve lesser total distance. This answers our question. Similar exercise can be done for the other states in the stage, and the information can be summarized as shown in the table below. Decision State

8

Table 2 N = 2 9

5 6 7

6+9=15 4+9= 13 3+9=12

8+6=14 9+6= 15 7+6=13

Best. Decision

Best Distance

9 8 8

14 13 12

The above gives us the best courses of actions to be followed, if we are at any of the states 5, 6 or 7 in the course of our travel. Together with the best decision, the distance corresponding to the best action is also listed. Activity 2 38

Suppose, at the beginning of stage 3, you are at state 6. Comment on. the following alternative argument, for determining the best course of action : "The possible

decisions are either to go to 8, or to go to 9. The distances of 6-8 and 6-9 are 4 and 9 respectively. As 61 is the shorter of the two, it is best to go to 8, if you are ever at 6, in the course of your travel." You may note that we have reached the same answer.

Dynamic Programming

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………. It is worthwhile noting at this point that, if our problem consisted of only two stages, namely 3 and 4, then we have found all the possible shortest routes, corresponding to all the possible starting points 5, 6 or 7. In the process so far, we have in fact enumerated all routes, and not saved on computation, We now proceed backward to stage 2. Beginning of stage 2, there are three states possible, 2, 3 or 4. We are interested in finding and, tabulating the best possible decision for reaching the destination (10), if we ever are at any of these states in the course of our travel. It should be obvious by now, that the best decision that w are looking for, corresponds to all the stages that we may have to travel through to reach 10, and doe; not merely refer to the best for this stage. From 2, there are three possible decisions, namely, to go*to 5, 6 or 7. If we decide to go to 5, then tile minimum distance we have to cover to reach 10, will be the sum of the distance of 25 and the best distance from 5 onwards, The. former is given in the diagram as 6, while the latter information as already summarized in the table 2, is 14. The overall effect of choosing 5 9 thus 19. The effect of the other decisions from 2 can be found out in the same way, and the best decision would be the one that gives the minimum overall distance. The exercise can be repeated for the other states as well. The outcome is summarized in the table below. Table 3 N = 3. Decision State

5

6

7

Best Decision

Best Distance

2

6+14=20

8+13=21

9+12=21

5

20

3 4

5+14=19 6+14=19

4+13=17 5+13=18

1+12=18 7+12=19

6 6

17 18

Proceeding in the same fashion, we move to the beginning of stage 1. At this stage, there is only one state (city 1), and three possible decisions. Using the same method for creating a table, we create table 4 as shown below. Table 4 N = 4 Decision State 1

2

3

4

4+20=24

6+17=23

6+18=24

Best Decision 3

Best Distance 23

The above tells us that, if we are at 1, then the best decision for reaching the destination would be to go to city3, and the overall distance to be covered would be 23. The value 23 thus gives us the total distance corresponding to the shortest route. To obtain the highways that lead to this shortest route, we note from table 4, that 1-3 is to be included. We now go back to table 3 and note that if ever we reach city 3, then the best decision is to go to city 6, hence 3-6 is to be included. Similarly, table 2 tells us that the best decision from 6 is to go to 8, therefore 6-8 is also there on the shortest route. Finally, we know that from 8, we can go only to 10, which is shown in table 1 as well. Thus, we have the following answer to the example. To shortest route from 1 to 10 is given by 1-3-6-8-10, and the distance to be covered is 23. While solving the problem, we have used the concepts of stage and state. Moreover, the problem has been solved stage by stage, and to ensure that suboptimal solution

39

Programming Techniques – Further Applications

does not result, we have cumulated the objective function value in a particular way. Working backwards, for every stage, we have found the decisions in that stage that will allow us to reach the final destination optimally, starting from each of the states of the stage. It may be noted that, these decisions could be taken optimally, without the knowledge of how we actually reach the different states. This has been stated as the "principle of optimality" in the D.P. literature. The concepts and the notations are formalized in the next section. Activity 3 Consider the above example. We have noted at the end of table 2 that there has been no saving in computation. What is your inference on savings in computation at the end of table 4? You may like to enumerate all the routes for the problem and see for yourself, where, if at all, savings occur. ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

9.3

DEFINI'T'IONS NOTATIONS

We have noted in the introduction that D.P. provides us with an alternative methodology for solving multistage problems. Identification of stage and state corresponding to a problem situation forms the first step in this direction. The method is based on the principle of optimality. The principle essentially recognizes the fact that period by period optimization is myopic. It gives a systematic and efficient way of reaching the optimal, and also reveals the conditions under which a multistage problem can be solved by D.P. In the last section, the methodology has been illustrated through an example. Here, the objective is to formalize the procedure, to help you in gaining insight into the D.P. methodology. Definitions and notations are presented first, and are followed by discussion. Stage The points at which decisions are called for are referred to as stages. Each stage can be thought of having a beginning and an end. The different stages come. in a sequence, with the ending of a stage marking the beginning of the immediately succeeding stage. State The variable that links up two stages is called a state variable. At any stage, the status of the problem can be described by the values the state variable can take. These values are referred to as states. Principle of Optimality The principle states that the optimal decision from any state in a stage to the end, is , ,independent of how one actually arrives at that state. The above have been used in the example given in the preceding section. Notations specific to D.P. are now introduced. Let fN(S) denote the overall optimal objective function value with N more stages to go and S as the starting state. Let J denote the decisions or possible alternatives at S. Let dsj denote the effect of starting from state S and deciding on alternative J. This is the objective function corresponding to the stage we are in. Let fO(D) denote the overall optimal objective function value with no more stages to go and D, the destination as the starting stage; fO(D) = 0. 40

For a. better understanding of the notations, we have the help of the example of the preceding section. A look at the tables created there tells us that the fN(S) values as defined here are stored in the last column of every table.

Dynamic Programming

The objective function fN(S) is thus computed in D.P. through an equation that is a function of at least two stages. This equation is referred to as recursive equation of simply recursion. D.P. thus involves a stage by stage solution procedure. Starting from the last stage, given the associated objective function of that stage, we can write it down as a function of the decision variable pertaining to that stage and the statge variable beginning of the stage. This function gives the effect of starting with a particular state and taking a decision, if the problem consisted only of the last stage. Next we go to the last but one stage. Here also the associated stage objective function is found in terms of the decision variable, and the beginning state variable. This gives the effect of the decision pertaining to this stage only. We now note that, once we start with a beginning state and take a decision, we attain an ending state. This ending state is nothing but the beginning state of the next stage, for which we have already found the effect. Thus, in order to ascertain the overall effect of starting this period with a particular state and taking a decision, we need to add the former and the latter. It should be apparent that in the given example once we know the particular state we are in, we can always find the optimal from that point to the final destination. It is irrelevant how we reach the state, so far as optimality is concerned. This has been enunciated as the principle of optimality. It should be noted however, that the principle applies only if certain conditions are fulfilled. We have seen that the procedure involves finding the decision at each stage that optimises some function of the individual stage returns. The example solved in the preceding section can be formulated as a mathematical program with Xij = 1, if the fine between city I and J is in the shortest route. = 0, otherwise. The objective function f(.) may be expressed as 11 dij, Xij. This function is an addition of the effects of the individual stages. The nature of this function is important for applying the stage by stage procedure. The function is required to be separable and monotonous. Additive functions are separable, hence we could apply D.P. in the given example. Multiplicative functions are also separable, thus, if our problem would have been to find the route from I to 10 that maximizes or minimizes the product of all the distances of the lines in the route, then also D.P. could have been applied. If the individual effect or objective function of a stage i is denoted in general by E,(.), then, while Ei and El- are separable, f(.) = [E1 + E2] [E3 + E4] is not separable. Monotonicity, on the other hand, implies that if the effect in any individual stage i for a decision Xi = a (say) in that stage, is more compared to a decision of Xi = b. then the same should be presented in the function f(.), i.e., the value of 4E1, E-, ... Ei (Xi = a), ... EN] is -more than f (E1, E2,…..Ei (Xi = b), ... EN]. Activity 4 Suppose you are playing a game, and you have been given the figure shown in the example of the preceding section. The objective is once again to start from l and reach 10, so as to maximize your total points. The figures beside the lines now denote the points you collect by traversing the lines. The total points of a route is given by

41

Programming Techniques – Further Applications

the product of the points of the lines in the route. Find the required route. Suppose, the total points in a route is given as ,the addition of the points of the lanes in a route, will the answer change? Give justifications for your answer. ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... Example 1 Consider the problem of finding the shortest route as given in the example in the preceding section. Suppose you are interested now to find the longest route, is it possible to apply D.P. for solution? If your answer is yes, then solve the problem. If the answer is no, then give your reasons for the answer. Solution It is apparent that the only change in the overall objective function is that, instead of is minimisation problem, it now becomes a maximization problem. The overall function still is an additive function of the individual stage functions, and as such is separable. The monotonicity property is also preserved, hence D.P. can be applied. Going exactly in the same way as earlier, the general recursion can be written as follows: fN(S) = Maximum [dsj + fN-1(1)], N = 1 to 4, where, f0(10) = 0. The recursion can be applied for N = 1 to 4, to create the four tables. The final route is then obtained as usual by working back from table 4 to table 1, as demonstrated earlier. The tables and the result are presented below. Table 1 N = 1 States

Decision

Distance

8

10

9

9

10

6

Table 2 N = 2 Decision State

8

9

Best Decision

Best Distance

5

6.+9= 15

8+6= 14

8

15

6

4+9=13

9+6=15

9

15

7

3+9= 12

7+6= 13

9

13

Table 3 N = 3 Decision State

5

6

7

Best Decision

Best Distance

2

6+15=21

8+15=23

9+13=22

6

23

3

5+15=20

4+15=29

6+13=19

5

20

4

5+15=20

5+15=20

7+13=20

5,6,7

20

Table 4 N = 4

42

Decision State

2

3

4

Best Decision

Best Distance

1

4+23=27

6+20=26

6+20=26

2

27

The required route is 1-2-6-9-10 with a total distance of 27.

Dynamic Programming

Example 2 A company has to decide on the production of an item for the next four periods, so as to meet the demands at minimum cost. The relevant costs are the production and inventory carrying costs. In any period, only if production is undertaken then a cost of Rs. 10 is incurred. The variable cost of production is Rs. 2 per unit. The inventory carrying cost is Rs. 2 per unit per period, and is levied on the inventory at the end of any period. Assume that the demands for the upcoming periods are 4, 1,1 and 3 units respectively,, and the beginning inventory is 0. Solution Every period a decision is called for, hence periods are equivalent to stages. The beginning or entering inventory of any stage is the same as the end4nventory of the preceding stage, as such, it constitutes the state variable for our problem. We now proceed in the usual way, starting from the last stage. One more period to go, i.e., N 7 1, implies that we are at the beginning of the last stage. As the demand for the last period is 3, the maximum number of units with which we can start this period is also 3. If we start with more than 3, then obviously we will be left with some stock at the end of the planning horizon, which is undesirable because of the associated inventory cost. Thus, the entering inventory or the state variable, in the last stage can take values of 0, 1, 2 and 3. There is only one possible decision for each of these values. For example, if we start with no inventory, as demand has to be met, and as no inventory should be there at the end of the period, the best decision is to produce 3 units. Similarly, if we start with I unit, then the only decision is to produce 2.units. This information together with the cost corresponding to a decision can be tabulated as follows States: entering inventory

Table 1 N = 1 Decision: Production

Cost

0 3 16 1 2 14 2 1 12 3 0 0 We go backwards by one stage. Beginning of the third period implies N = 2. As the tool demand for the last two periods is 4 units, the beginning inventory of this stage can take any value between 0 to 4. If we start the stage with no stock, it is possible for us to decide on a production ranging from 1 to 4 units. Less than I unit will be infeasible as the demand for the current period will not be met, and more than 4 is not desirable as stock will be left at the end of the planning horizon. Thus, unlike the last stage, here there are several decisions possible with respect to a particular state. This can also be verified by the following procedure. Let the decision in a stage (the production quantity for that period) be denoted by x. Let the state variable be denoted by the symbol i. For any period, we know that the beginning inventory plus the production, minus the demand should be equal to the end inventory, i.e., end inventory = i + x - d. As demand for the current stage is 1 unit, we have, x = end inventory + I - i. Thus, for different values of i, and the different end inventories possible, as given in table 1, it is possible to find the range of x. For example, for i = 0, x = i + end inventory. As end inventory is equivalent to the beginning inventory of the next stage, which in this case can take values from 0 to 3. Hence x can take values between 1 and 4 as we have found earlier. We now try to answer the following question If at any stage, we are at state i, what will be the best course of action to minimise the total cost? We found that for any state in stage 3, there are more than one course of action. Starting with no stock (i = 0), for "ample, if we choose to produce 1 unit, then the

43

Programming Techniques – Further Applications

end inventory is zero and the cost or effect corresponding to this stage is 12 (production cost of 1 unit + inventory carrying cost =10 + V 1 + 0). End inventory of 0 implies that we start the last stage with i = 0, for which we know from table 1, that the only decision is to produce 3 units at a cost of 16. Hence, if we start stage 3 with 0 units, the total cost of the decision of producing 1 unit is 28. If we would have chosen to produce 2 units, then the end inventory would have been 1. The cost for this stage and the last stage would have been 16 and 14 respectively, giving us a total cost of 30 for the decision. Similarly, the cost corresponding to other decisions can also be found. The best course of action is given by the decision associated with the minimum cost. The above can be expressed in the form of a recursion. Let fN(i) denote the overall optimal objective function value with N more stages to go and i as the starting state. Let, C(x) = cost of producing x units = 10 + 2* x. With N-periods to go, if we start with entering inventory i, and produce x units, then, given a demand of d units, the inventory at the end of the stage is i + i - d. This is nothing but the beginning state with N-1 periods to go, and the best decision from that point onwards can be expressed as f N-1 (i + x - d). The inventory carrying cost for the period is on the end inventory, and can be expressed as : 2 * (i + x - d). Hence, we have the following recursion

The recursion can now be applied to create table 2 to 4. Table 2 N = 2 x0 i 0 1 2 3 4

1 --

12+0+16 =28 0+0+16 12+2+14 =16 =28 0+14+2 12+4+12 =16 =28 0+12+4 12+6+0 =16 1-8 0+6+0 =6

2

3

4

Min Cost

14+2+14 =30 14+4+12 =30 14+0+6 =20 -

16+4+12 =32 16+6+0 =22 -

18+6+0 ==24 --

24

Best Decision 4

16

0

-

16

0

-

-

16

0

-

-

-

6

0

Table 3 N = 3

44

X

6

1

2

3

4

5 34

Min Cost 32

Best Decision 2

0

-

36

32

36

40

1

24

30

34

38

32

-

24-

0

2

18

32

36

30

--

-

18

0

20

34

28

-

-

-

20

0

4

22

26

--

-

-

-

22

0

5

19

-

-

-

-

-

19

0

Dynamic Programming

The required production plan (found by working backwards from table 4), is to produce 6 units in the first period and 3 units in the last period, and the resulting cost is Rs. 44. Activity 5 In the example solved above, give the break-up of the total cost for each cell of table 3. Also describe the procedure for arriving at the production plan by working backwards from table 4. ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... ......................................................................................................................................... .........................................................................................................................................

9.4

D.P. APPLICATIONS

In the foregoing sections we have illustrated how Dynamic Programming can be useful for solving a class of multistage problems. Our presentation has been based on two example situations. The first situation consisted in finding the shortest route, and is known as the Stage Coach problem in Operations Research literature. The second situation involved finding the production plan, given a nonlinear post function. In both the situations, decisions were called for in a sequence, and this sequence was built-in in the problem structure. Many applications of D.P. however exist where such sequences are not apparent.-In this section, we present one situation and an example, -and the reader is referred to the readings cited in the last section, for more such applications. Situation We take up the familiar example on Emeralds, Rubies and Topaz as described in Unit 8 on Integer. Programming: You have landed in a treasure island full of three types of valuable stones, Emerald(E), ruby(R) and topaz(T). Each piece of E, R and T weighs 3, 2, 2 kg., and is known to have a value of " 4,3,1 crore respectively. You have got a bag with you that can carry a maximum of 11 kgs. Your problem is to decide on how many pieces of each type to carry, within the capacity of your bag, so as to maximize the total value carried. The stones cannot be broken. The above does not involve decisions to be made in a sequence. However, without changing the problem, it is possible, to impose an arbitrary sequence. For example, we may assume-that decision on the number of Emeralds are to be taken first, then the decision on the number of Rubies, and finally, the decision on the number of Topaz. These then will form the three stages of the problem. As we have already seen, if we decide independently in each stage, we may not get a feasible or an optimal solution. Such stage-wise decisions are possible, provided we link up properly the stages, to preserve the overall optimality. The linkage between a stage and all the subsequent stages, is provided by the total amount of weight that is left to be allocated among these stages. Specifically, the state -variable corresponding to stage i may be identified as the amount of weight that is left to be allocated for stage i to stage n. Finally, recall that the situation was formulated as an I.P. in the last unit :

45

Programming Techniques – Further Applications

The overall objective function is an additive function of the individual stage objective, as can be seen from the formulation. Thus we can apply D.P. for solution. As usual, we star + t from the last stage. Beginning of last stage (N = 1), we are faced with the decision on the, number of Topaz to carry. Given that a maximum of 11 kg. can be carried, it is possible that at the starting of this stage, 0, 1, 2, 3, ... 11 kg. are still left to be filled up by Topaz. These are the possible states we can be in at N = 1. Suppose we start the stage with a weight W left to be allocated for carrying Topaz, then as each Topaz weight 2 kg., the maximum number of Topaz (X3) We will be able to carry is, [w/2], where [a] denotes the largest integer less than or equal to a. As each Topaz has a value of I crore, the effect of the decision of carrying'[W/2] Topaz will be Rs. [W/2]. This constitute the best decision if we start the last stage with Wk&still left to be filled up Following our usual notation, with fN(W) denoting o the overall maximum with N more to go, and the starting state as W, we construct the table I below : Table 1 N = 1 W 3 fl (W) X

0 0 0

1 0 0

2 1 1

3 1 1

4 2 2

5 2 2

6 3 3

7 3 3

8 4 4

9 4 4

10 5 5

11 5 5

In the above table, VV denotes the state variable, X3 denotes the best decision, and fl(W) denotes the overall objective function value corresponding to the starting state and the best decision. As already explained, X3 values corresponding to each W are found as the largest integer less than equal to W/2. Also note that the second, row and the, third row values are same because of the fact that each Topaz has a value of .1 crore. We now go back by one stage, Beginning of this stage, we are yet to decide on the number of Rubies and Topaz to be carried. In this stage we are going to decide on the number of Rubies only. The question we need to answer, may be formulated as follows: If W k& is still to be filled up, how many Rubies should we carry so as to maximize the overall value? Suppose that beginning of this stage, 11 kg. are yet to be filled up. We are interested in finding the over all maximum value that is possible to achieve, and the decision in this stage that gives the value. By our notations, these are f2(11), and X2 respectively. X2 denotes the number of Rubies; as each Ruby weighs 2 Kg.; if 11 Kg. is the amount that is yet to be filled up, X2 can take any integer value between 0 to [11/2], Le, between 0 to 5. Moreover, if we decide to carry X2 Rubies, then the weight that will remain to be filled up at the beginning of the last stage is (11 - 2 X2).. By definition; this will be the state value with one more -stage to go, and the overall best from the state till end has already been found and stored in table I as fl(11 - 2X2). To find the best decision, we have to add to this, the effect of the decision of this stage *(carrying X2 Rubies), which is given by 3 X2. Thus, we have the following recursion

This implies that if we start sage 2 with 11 k& still to be filled up, then the best, decision is to fill up the bag with 5 Rubies. Activity 6 46

Consider a modified version of the above problem. Suppose there were only two types of stones, Ruby and Topaz, instead of the three types mentioned above. All

other data remaining unchanged, how many of each kind you will carry in your bag so as to maximize the total value carried?

Dynamic Programming

......................................................................................................................................... ......................................................................................................................................... ...................................................................................................................................... In general, with 2 stages to go and W as the starting state, the maximum value from this point to the end can be expressed as :

Using the recursion we can create table 2 as shown below. Table 2 N = 2 W 2

0 0

1 0

2 1

3 1

4 2

5 2

6 3

7 3

8 4

9 4

10 5

11 5

f2(W)

0

0

3

3

6

6

9

9

12

12

15

15

X

We now go back by one more stage. This implies that we are at the beginning of the entire problem, and we have not decided on the number to be carried for any of the stones. As such the whole bag is to be filled up with a load of 11 Kg. Thus, with N = 3, we can be in only one state, that of W = 11. We are now interested in finding out the best decision (No. of Emeralds to be carried) pertaining to this stage that will maximize the overall value carried. The recursion can be written exactly in the same way as in the last stage f3(l 1) = Maximum [4 X 1 - f2(11 - 3 X1)] X, = 0, 1 ... [11/3] It may be noted that the effect corresponding to this stage is 4 X,, where X, represents the decision on the number of Emeralds, and the coefficient of 4 represent the value per Emerald. If 11 Kg. is to be filled up, then the maximum number of Emeralds that can be taken are [11/3] = 3, as each Emerald weighs 3 Kg. Once again, the best decision of this stage need not be the best overall decision, hence we allow X, to take values from 0 to 3. The balance, left to be filled up is equal to (11-3 Xj) and it constitutes the state of the system with N = 2. In table 2 we have already found out the maximum value corresponding to all the states the system can be in, beginning of stage 2. Hence for each value of X1, this is added with the effect of this stage to arrive at the overall optimum. f3(11) =Max [0 + f2(11), 4 + f2(8), 8 + f2(5), 12 + f2(2)] =Max [0'+ 15, 4 + 12, 8 + 6, 12 + 3] = 16 (corresponding to X1=1) Thus, the maximum value that can be carried is 16 crores, and the number of Emeralds to be carried is 1. The decisions on other stones can be found by working backwards. One emerald weighs 3 kg, hence beginning -of stage 2, we will have (113) = 8 kg. still to be filled up. Thus, with N = 2, we have W = 8. The corresponding best decision is tabulated in table 2, and is equal to 4. Thus, number of Rubies to be carried is 4. Finally, we are left with no more capacity, and obviously the best decision with N = 1 and W = 0 is listed in table 1 as 0, implying that no topaz is to be carried. You may check that we obtained the same optimal solution in Unit 8. Activity 7 In tie above example create the table corresponding to N 3. .........................................................................................................................................

47

Programming Techniques – Further Applications

......................................................................................................................................... ......................................................................................................................................... .........................................................................................................................................

Solution It is useful to compare the given problem with the one on Emerald etc. that we have just solved. Note that the earlier problem was an integer programming (I.P.) problem. The differences between the two can be identified as follows: i)

The number of variables are two and three respectively in the L.P. and the Y.P.

ii)

There are two constraints in the L.P., whereas there is only one constraint in the I.P.

iii) The variables are not restricted to integer in the L.P., whereas they are restricted to integers in the I.P. In the I.P., there were three items and decision for each were taken in stages, in an aribitrary sequence. In the L.P., as there are only two variables, there can be only two stages. Assume that the decision on X corresponds to stage 1 and that of Y to stage 2. Thus, i, above is advantageous as it will involve lesser computation. In the I. P., state is specified by the amount still left to be filled up in that stage and all the subsequent stages. The state variable is thus associated with the constraint of the problem. In the L.P. there are two constraints, as such, two state, variables will be needed to describe the state of the system. For a better understanding, assume that the I.P. is now modified to accommodate an extra constraint specifying that the total number of stones (XI + X2 + X3) that can be carried is less than or equal to 5. This implies that we need to keep track of not only the balance weight capacity W, be filled up, but also the balance number of stones that can be carried at the beginning of every stage. The state is now represented by a pair of numbers, corresponding to each state variable. The computation also increases as we have to now find the maximum over all the values of both the state variables. The third difference brings out a point which has so far not arisen in our' exposition on the D.P. methodology. In all the examples that we have considered, the variables were restricted to nonnegative integers. L.P. allows for continuous values of variables and as such it will be useful to find how this is solved by D.P. methodology. A physical representation of the L.P. formulation may be useful at this stage. Assume that we want to decide on the number of litres of two items, A and B to be carried, . so as to maximize the value. The value per litre of A and B are Rs.5 and Rs.9 and the weight per unit of A and B are.5 and 3 Kg. respectively. The total weight one is allowed to carry is 27 Kg. It is also known that each of item B requires 5 units of a special solid for preserving. A does not require this. However, for every litre of A carried, we will get one such solid free. Three units of the solids are in hand. It is apparent that X and Y denote the quantities of A and B that are to be carried. The first constraint specifies that the requirement of the special solid should be within the available capacity. The requirement is 5Y for Y litres of B carried, and the availability is 3 units in hand plus X units we get free for carrying X litres of A. Thus. 5YD, otherwise, if P is less than or equal to D, the item will be used as fast as it is produced (see Figure 11.13).

Figure 11.13: Gradual Replacement Inventory Situation

Cycle time t is the sum of the production time t1 plus the depletion time, t2 of maximum inventory level BD. Production starts at point A, and stops at point E as soon as the level of inventory becomes BE. Production time, tl = 20

Q Q and cycle time, t = P D

Maximum Inventory level BE = (P-D) x t1 = (P-D) Minimum Inventory Level = 0

Q ; P

Inventory Control – Deterministic Models

Example 4 An unit is used at the rate of 100 per day and can be manufactured at a rate of 600 per day. It costs Rs. 2000 to set up the manufacturing process and Rs. 0.1 per unit per day held in inventory based on the actual inventory any time. Shortage is not allowed. Find the minimum cost and the optimum number of units per manufacturing run.

4) Price Discounts Model (Instantaneous Supply with No Shortages) When items are purchased in bulk some discount in price is usually offered by the supplier. When discount is applicable for all the units purchased, it is known as all units discount. If discounts are offered only for items which are in excess of the specified amount, it is known as incremental discount. In the incremental discount all the prices offered in different slabs are applicable in finding the total cost. While, in all units discount only one price at any one slab is applicable for finding total cost. We will discuss here only all units discount case. Buying in large quantities may result in the following advantages: Lower unit price, Lower ordering costs, cheaper transportation, Fewer stock-outs, Mass display by retailers, preferential treatment by sellers. At the same time, large quantity buying may involve these disadvantages. Higher carrying costs, older stocks, lower-stock turnover, more capital required, less flexibility, Heavier deterioration and depreciation. Whenever quantity discounts are available in the case of purchased items, the price, C may vary according to the following scheme:

where pj-1 is greater than pj, for j = 1,2, ........... n, pj being the price per unit for the jth lot size. If shortages are not allowed, the total cost per year is explained by the set of relations:

and Ch = ipi, i being percentage change for j = 0, 1, 2,..., n. Since price or unit cost varies with size of purchase, Q, the fixed cost term CD in. equation (21) cannot be ignored for minimizing the total cost. Equations (21) for quantity discounts are illustrated in Figure 11.14. The heavy curves on the various price discounts show feasible part of the total cost which is a step function. For finding the overall optimum, the following procedure is adopted:

21

Inventory and Waiting Line Models

Figure 11.14: Quantity Discounts

22

Since 894 is greater than 750, optimum purchase quantity is 894 units.

Inventory Control – Deterministic Models

Example 6 Consider example 5 with the ordering cost of Rs. 100 only. Answer

5) Dynamic Demand Models In these models, it is assumed that demand, although known with certainty, may vary from one period to the next. a) Production Inventory Model (Incremental Cost Method) The situation is explained with the help of an example. Example 7 A manufacturing concern has a fixed weekly cyclic demand as follows:

Policy is to maintain constant daily production seven days a week. Shortage cost is Rs. 4 per unit per day and storage cost depends upon the size of Q, the quantity carried, as follows

The charges are based on the end of the day situation. Determine the optimal starting stock level. Solution Let the manufacturing rate be the average of the total sales which is

We find the total weekly cost by tabulation for various starting stocks. Tables 1, 2 and 3 are for starting stocks of 8, 9 and 10 units respectively. 23

Inventory and Waiting Line Models

Table 1 : Cost Analysis with 8 Initial Stock

Table 2 : Cost Analysis with 9 Initial Stock

Table 3 : Cost Analysis with 10 Initial Stock

The optimal solution is for starting stock of 9 units on Monday. Minimum total cost for this is Rs. 82. Procedure for solving such problems is incremental cost analysis which is self explanatory through this example. b) Dynamic Inventory Model (Prescribed Rule Method) The company dealing with inventory may prescribe some rule of procurement of items, say procuring every three months, or every month. Example 8 A company estimates the demand of an item as follows:

Ordering Cost = Rs. 54, Carrying charge per unit per month = 2% at the end of each month. 24

Unit cost Rs. 20. Supply is instantaneous. There is no lead time and no stock-outs. Only full month requirement is ordered.

Solution Tabular method is used for finding the total cost for the policy of ordering quarterly.

Inventory Control – Deterministic Models

Total replenishment cost = 4x54 = Rs. 216 Total carrying cast = 1058 x (0.02) (20) = Rs. 423.20 Total annual cost = 216+423.20 = Rs. 639.20 c) Dynamic Inventory Model (Fixed EOQ Method) We consider the same example 8. The average monthly demand =

1200 = 100 12

units/month Therefore, EOQ =

2 × 54 × 100 = 164 approx. (0.02)(20)

Since full months' requirement is to be ordered, 164 lies between 130 and 218 units. Since 164 is closer to 130 than 218, we order one month requirement at the beginning of January. Similarly, at the start of February, we order two months requirement. The detail results are as follows: Table 5 : Total Cost Analysis for EOQ Method

d) Dynamic Inventory Model (The Silver-Meal Heuristic) In this method, the relevant costs per unit time for the duration of the replenishment quantity are minimized. If the replenishment arrives at the beginning of the first period and it covers the requirements upto the end of Tth period, than, we find the average cost per period as :

25

Inventory and Waiting Line Models

This is illustrated in Table 6(a) to (e) for Example 8.

The ordering schedule has been obtained. Total cost for this schedule is 130.8+122.0+105.6+54+70.4=Rs. 482.8. This is further better solution than obtained under EOQ method. e) Dynamic Inventory Model (Wagner-Whitin Method) This is a dynamic programming method by which an optimal result can be obtained for the dynamic, inventory problem. We consider Example 8 here again. In this example, the planning period is 12 months and starting inventory for January as well as ending inventory for December is zero. Dynamic programming method is based on a principle of optimality which optimizes the total of the relevant cost for the current period plus the optimum cost for the remaining periods''. We assume in Example 8 that a replenishment takes place only when the inventory level is zero. There is an upper limit to how far before a period j we would include its requirement D(j) in a replenishment quantity. When, the carrying costs become higher than the ordering costs for a period j, that is

then the replenishment should be at the beginning of period j. For our problem, Co/Ch = 54/0.4=135 units. This implies that if inventory to be carried for any period exceeds 135 units, it is preferable to order that much inventory at the start of that period, The algorithm works backward in time asking the following type of question : "If we are now at the beginning of month j needing to replenish what is the best size to make the replenishment?" To start with the month of December, the replenishment needed is 41 to meet the demand and the associated cost is Rs. 54. There is no carrying cost. At the start of November, there are two options: Option 1: Order 238 units to cover the November's demand, then proceed in the best way from the start of Dec. with zero inventory. The cost associated with this option is 54+54=Rs. 108. Option 2: Order 279 units to cover the demand for both November and December. The associated cost is Rs. 54+41 x 0.4x1 =Rs. 70.40 26

Therefore, the best strategy is option 2 for Rs. 70.40

At the start of October, there are three options.

Inventory Control – Deterministic Models

Option 1 : Order 160 units for October's demand and use the best way for November with zero inventory. The related cost = Rs. 54+70.40=Rs. 124.40 Option 2 : Order 398 units to cover October + November demands and use the best way for December with zero inventory. The related cost = Rs. 54+238x0.4+Rs. 54 = Rs. 203.20 Option 3 : Replenish three months requirement to cover upto December. The related cost = Rs. 54+238x0.4+41x 0.4x2 = Rs. 182.00 The best strategy is, therefore, option 1 for the cost of Rs. 124.40. Similarly, we carry on further and get the following best strategies for each month with their related costs in Table 7.

Total ordering cost = 6x54 324 Total carrying cost 392 X0.4= 156.8 Annual total cost = Rs. 480.80, which is less than that obtained by heuristic.

11.9

DETERMINISTIC ULTI4 TEM INVENTORY MODELS

When the inventories consist of many items, the control requires special type of attention. Such inventory problems may have different types of limitations such as finance, cost structure, space and purchasing work load etc. As the number of restrictions increase the problem becomes more and more complicated. We discuss some of these problems. 1. Model with unknown Cost Structures Many organisations in India do not maintain the proper records which may provide a sufficient cost information to generate the two basic parameters of inventory control carrying cost and the procurement cost. In some situations, some organisations which have not developed cost structure related to inventory control, still wish to minimize total cost of inventory management. In some critical situations, an organisation may need to take immediate actions to improve the situation without considering the cost structure. Although, it may first appear that use of inventory models without cost information is impossible, but we will show here that it is possible to get many of benefits of inventory techniques even when ordering and carrying costs are not known. There are two approaches to such a problem: (i) To minimize the total carrying cost while keeping the number of orders per year fixed or (ii) To minimize the total number of orders per year while keeping the same level of inventory.

27

Inventory and Waiting Line Models

Model I We develop a model to minimize total inventory holding cost and keeping the total number of orders per year unchanged. From EOQ model, we know that

where K =

2C0 /Ch , a constant, because ordering cost and carrying cost are

deterministic values. The above equation says that EOQ is proportional to the square root of demand for any item of control. For this equation, we get

Since K is constant for any single item, we take K as the constant for all the inventory items. Thus, we take K

K=



∑(

D D/Q*

)

or

Sum of square roots of demand of each item in inventory Sum of the number of orders per year for each item

An example will illustrate this Model. Example 9 A company has the following procurement pattern of five items irrespective of their level of demand. Reduce the inventory levels while keeping total number of orders per year the same.

28

According to the policy of the company ordering four times a year each item, total average inventory comes out to be Rs. 216762.5, while in the new schedule as obtained in Table 9, the average inventory is Rs. 119902.50 which is much less. At the same time total number of orders almost remain the same. Thus, substantial savings can still be obtained when cost information is not known. Model II Here, we develop the model to minimize the number of orders per year (or purchasing workload) while keeping the same level of inventory. We know that

From here, we take up K=

∑ Q* ∑ D

Inventory Control – Deterministic Models

for all the inventory items.

In case of Example 9, we explain to minimize the purchasing work load in Table 10.

Constant K is obtained as, K =

433525 = 197.95662. Thus from Table 10, it is 2190

obvious that total purchasing workload has reduced by 45%. There is, therefore, a definite cost saving by applying these methods for multiple items even when cost information is not known. 2) Models with Known Cost Structures When the cost structure is known completely, we discuss the following models. In these models, we consider symbolic notations Coi, Chi, Di for ordering cost, carrying cost and demand for ith item respectively. Model Without Limitations If there are no restrictions for storing items, the items may be purchased according to their individual economic order quantities. The total variable cost per annum for nitems can be expressed as

The optimum order size for each item is

Model With Limitations The constraints on the inventories may be of floor space restrictions, capital availability, number of orders per year etc. We will discuss a model with single restriction. Floor space restriction and capital availability restrictions are interchangeable. If Qi is the order quantity for item i, and fi is the floor space covered by one unit of item i (or fi being the capital requirement for one-unit of item i), and total availability of floor space (capital) is f, then the restriction is as follows:

29

Inventory and Waiting Line Models

The objective is to minimize the total inventory management cost expressed by (22). This problem is converted to unconstrained minimization problem first and, then, optimal result is obtained. The problem is to minimize the function known as Lagrangian function,

The values of optimal order sizes are

Example 10 A retailer purchases three items. He works on the 1imitation that he is not able to invest more than Rs. 20000 at any time. Other relevant information is as follows:

Solution: The optimal lot sizes in the absence of constraints are:

With these optimal sizes, the maximum investment = 500x20+75x 100+200x 50=10,000+7500+10,000=Rs. 27500. This is greater than the maximum allowable investment in inventory. Therefore, equation (24) is used with the following changes.

If limitations are not imposed on the purchase of quantities, the optimal total cost of Inventory management is 2000+1500+2000=Rs. 5500. Under the limitations the total cost is Rs. 6792.96 which is higher than the total cost without limitations. Many a times the application of equation (24) to find optimal order quantities under limitation does not help in getting the result. Therefore, we apply trial and error procedure in the following manner; 30

a) First find the EOQs for all type items without considering the limitation i.e. taking p =o, find Qi. If these values satisfy the constraint (23), then this solution becomes optimal because the constraint is not active.

b) If the constraint is not satisfied by the values obtained under (a) above, we give some value to p (arbitrarily but intuitively) say p = h and solve for Qis. Q,s satisfy the constraint, these are optimal quantities. Otherwise, we interpolate or extrapolate the value p in between 0 and h or beyond if. With this value of p, the order sizes obtained will be approximately optimal.

Inventory Control – Deterministic Models

11.10 SUMMARY This unit has illustrated the introductory concepts of inventory and inventory control. Objectives, functions of inventory, classification of inventories and various factors that affect the maintaining of inventories are described. Deterministic Inventory models have been developed for various operating conditions. Firstly single item inventory control models have been discussed followed by the multiple item inventory models. Illustrations have been given by way of graphs, figures, tables and examples and their solutions.

11.11 KEY WORDS ABC-Classification : Classification of inventory items in terms of annual usage value in the categories of high value (A), medium value (B), and low value (C). Backlog : Accumulation of unfilled demands. Carrying (Holding) Cost : Cost of maintaining one unit of an item in the stock per unit of time (or one year). Decoupling : Use of inventories to break-apart operations so that one operation's supply is independent of another's supply. Delivery lag : Time taken to receive the material for use since when its need was felt. This is also known as lead time. Deterministic : Having known-conditions (or factors). EOQ : The order quantity which minimizes the total inventory cost. Holding Cost : Same as carrying cost. Inventory : Stores of goods and stocks. Inventory control : Technique of maintaining stock-items at desired levels. Lead time : Same as delivery lag. Lot size : Size of the purchase or size of the production during a cycle period. Multi-echlon Inventories : Products stored at various levels (factory, warehouse, retailer, customer) in a distribution system. Ordering cost : Cost associated with placing an order. It includes postage, stationery, telephone calls to vendor, labour and computer costs etc. Procurement Cost : Same as ordering cost. Replenishment Cost : Same as ordering cost. Safety Stock : Extra Stock needed to absorb variation in demand and supply to provide cushion. Set up cost : The cost related to setting up a machine to start production. Shortage (Stock out) Costs : Costs associated with demand when stocks have been depleted generally known as lost sales or back-order costs, which includes goodwill losses, idle labour and idle machine costs, loss of sales etc. VED classification : A subjective division of inventory items into vital, essential and desirable categories.

11.12 SELF-ASSESSMENT EXERCISES 1) What constitutes inventory ordering costs, carrying costs and stock out costs? Explain the behavior and relationship of these costs.

31

Inventory and Waiting Line Models

2) Define inventory. Some businessmen consider inventory as necessary evil while others-think inventory as an asset. What is your point of view? 3) An item is required at a rate. of 18000 units per year. Storage cost is Rs. 0.10 per unit per month. If the cost of placing an order is Rs. 400, find (a) EOQ (b) Number of orders per year (c) Cycle period (d) Total annual cost if per unit cost is Rs. 2. 4) An item is produced at. the rate of 50 units per day and is consumed at the rate of 25 units per day. If the set up cost is Rs. 100 per production run and holding cost in stock is Rs. 365 per unit per year, find the economic lot size per run, number of runs per year and the total related cost. 5) From the following find out the EOQ and the total inventory, cost. Annual demand = 800 units, set up cost = Rs. 20, set up Carrying Cost = 2% per month of purchase price. Minimum Stock = 200 units. Purchase price per unit = Re. 1 for order below 2000 units = Re. 0.95 for order of 2000 units = Re. 0.80 for single order 6) The following thirty numbers represent the annual value in thousands of rupees of some thirty items of materials selected at random. Carry out an ABC analysis and list out the values of `A', `B' and `C' items.

7) A small shop produces three machine parts in lots. The shop has only 650 sq. ft. storage space. The appropriate data for the three items is presented below:

Inventory carrying charges are 20% of average inventory value per year for each of the three items. Determine the optimal lot sizes. 8) A company maintains inventories of 10 items. The company is not able to determine its carrying cost or ordering cost with sufficient reliability to support the use of EOQ approaches, but wants to manage as effectively- as possible. The following information is available:

32

Answer the following: a) Without increasing the purchasing workload, what percentage reduction can be made in the average inventory carried? b) Without increasing the average inventory carried, what percentage reduction can be made in the no. of orders per year. c) If the company is willing to increase its average inventory by 10 per cent, what percentage reduction in purchasing workload can be achieved? d) If the company is willing to increase the purchasing workload by 25 per cent, which is the minimum average inventory it can achieve? 9) The demand for an item is 10,000 units per year. The cost per unit is Re 1.00. The set up cost is Rs. 10. The inventory carrying cost is 20 per cent per year. The

cost of back order is Rs. 5 per unit per year. Find the optimal size of order, the maximum inventory and the total number of units back ordered per year.

Inventory Control – Deterministic Models

10) If carrying cost is 20% per, unit/year, unit price is Rs. 2, and ordering cost is Rs. 20 per order for an item used in a company in the following pattern: Find the ordering schedule and the total inventory cost using a) Fixed EOQ method b) Quarterly replenishment Policy c) Silver Meal Heuristic d) Dynamic Programming.

11.13 FURTHER READINGS Buffa; E.S. ,(1990): Modern Production/Operations Management, Wiley Eastern Limited. Everett E. Adam, Jr and Ronald J. Ebert (1986) : Productions and Operations Management: Concepts, Models and Behaviour, Prentice Hall International. Hadley G. and Whitin, T.M.(1963): Analysis of Inventory System, Prentice Hall, N.J., U.S.A. Levin, R & Kirkpatrick,. C.A., (1978) : Quantitative Approaches to Management, McGraw Hill, Kogakusha Ltd., International Students Edition. Mustafi, C K. (1988) : Operations Research, Methods and Practice, Wiley Eastern . Limited. . Peterson R and Silver, E.A. (1979) : Decision Systems for Inventory Management and Production Planning, Wiley, New York. Taha, H.A. (1976) : Operations Research : An Introduction, MacMillan Publishing Co. Inc. New York.

33

Inventory and Waiting Line Models

UNIT 12 INVENTORY CONTROL : PROBABILISTIC MODELS Objectives After studying this unit, you should be able to •

discuss various probabilistic models.



explain various approaches to the probabilistic problems,



set various inventory levels under various probabilistic conditions.



describe the role of simulation study for inventory problems.



clarify the problems in terms of variability in demand and lead time,

Structure 12.1

Introduction

12.2

Inventory Models with Probabilistic Demand

12.3

Single Period Probabilistic Models

12.4

Multi-Period Probabilistic Models

12.5

Inventory Control Systems

12.6

Fixed Order Quantity System

12.7

Periodic Review System

12.8

Other Variants of Probabilistic Models

12.9

Summary

12.10

Key Words

12.11

Self-assessment Exercises

12.12

Further Readings

12.1 INTRODUCTION In previous unit, we have discussed Simple deterministic inventory models where each and every influencing factor is known completely. But, in actual business life complete certainty never occurs. Therefore, we will discuss here some practical situations of inventory problems by relaxing the condition of certainty for some of the factors. The major influencing factors for the inventory problems are price, demand and lead time. Other factors such as carrying cost, ordering and stock-out costs are also affecting the inventory problems, but their nature is not so much disturbing. This is because their estimation provides almost, on the average, as known values. Even price can also be averaged out to reflect the condition of certainty. But sometimes, price fluctuations are too much in the market and hence they influence the inventory decisions. Similarly, variability in demand or consumption of an item as well as the variability in lead time influences the overall inventory policy.

12.2

INVENTORY MODELS WITH PROBABILISTIC DEMAND

Inventory models where only demand is probabilistic or random will be discussed here. Demand pattern may be having discrete probability distribution or continuous probability distribution as explained in Figure 12.1.

34

12.3

SINGLE PERIOD PROBABILISTIC MODELS

Inventory Control – Probabilistic Models

These models deal with the inventory situation of the items-such as perishable goods, spare parts and seasonal goods requiring one time purchase only. The demand for such items may be discrete or continuous. Since purchases are made only once, the lead time factor is least important in these models. In single period models, the problem is studied using marginal (or incremental) analysis and the decision procedure consists of a sequence of steps. In such cases, there are two types of costs involved, namely (a) Over-stocking cost, and (b) Under-stocking cost. These two costs represent opportunity losses incurred when the number of units stocked is not exactly equal to the number of units actually demanded. The following symbolic notations are to be used : D = demand of an item in units (a random variable) Q = the number of units stocked (or to be purchased) C1 = Over-stocking cost (also known as over-ordering cost). This is an opportunity loss associated with each unit left unsold. = C + Ch - V C2 = Under-stocking cost (also known as under-ordering cost). This is an opportunity loss due to not meeting the demand. = S-C-Ch/2+Cs where C is the unit cost price; Ch, the unit carrying cost for the entire period; Cs, the shortage cost; S, the unit selling price and, V, the salvage value. 1) Single Period Discrete Probabilistic Demand Model (Incremental Analysis Method) The cost equation for this type of problem may be developed as follows. For any quantity in stock Q, only D units are consumed. Then for specified period of time, the cost associated with Q units in stock is either: a) (Q-D)C1, where D, the number of units used or demanded is less than or equal to the number of units Q, in stock, i.e., D ≤ Q b) (D-Q)C2, where the number of units required is greater than the number of units in stock, i.e. D>Q. Since, the demand D is random variable, its probability distribution of demand is known. p(D) denotes the probability that the demand is D units, such that total probability is one, i.e.,

If Q* is the optimal quantity stocked, then the total expected cost f(Q*) will be minimum. Thus, if we stock one unit more or less than the optimal quantity, the total expected cost will be higher than the optimal. Thus,

35

Inventory and Waiting Line Models

Therefore, the optimal stock level Q* satisfies the relationship (6). For practical application of (6), the three step procedure is as follows: Step 1:From the data, prepare a table showing p(D), the probability and the cumulative probability P(D ≤ Q) for each reasonable value of D. Step 2 :Compute the ratio

C2 which is known as service level. C1 + C2

Step 3 :Find the value of Q which satisfies the inequality (6). Example 1 A trader stocks a particular seasonal product at the beginning of the season and cannot re-order. The item costs him Rs. 25 each and he sells at Rs. 50 each. For any item that cannot be met on demand, the trader has estimated a goodwill cost of Rs. 15. Any item unsold will have a salvage value of Rs. 10. Holding cost during the period is estimated to be 10 per cent of the price. The probability distribution of demand is as follows:

Determine the optimal number of items to be stocked. Solution As per step 1, we put the data regarding demand distribution in the Table 1 below: Table 1 : Probability Distribution of Demand

Looking at Table 1, this ratio lies between cumulative probabilities of 0.60 and 0.80 which in turn reflect `the values of Q as 3 and 4. That is, P(D ≤ 3)=0.60