Operation Research

Amity University

MBA IB The course aims to provide a thorough understanding of the essential features, relevance, application, tools and techniques of Operations Research. The objective of this course is to develop the understanding of models building and quantitative approach to decisions making in the functions of the management of any organization with special focus on International Business. It also aims to develop the understanding of the various optimization techniques used for decisions making in the functions of the management of any organization.

Semester Two

Ms. Sonia Singh

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Preface

This is an attempt to the integration of operation research with business practices for the

purpose of facilitating Decision Making and Forward Planning by the management. As

operation research provides as a set of concepts, these concepts furnish us the tools and

techniques of analysis. Scientific methods have been man’s outstanding asset to pursue a

number of activities. The application of O.R. methods helps in making decisions in

complicated situations. O.R. helps in making a better decision by studying the advantages

and disadvantages of alternative courses of actions.

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Course Contents:

Module 1: Introduction to Operations Research

Nature and Significance

Applications and Scope

Advantages, Opportunities and Shortcomings of OR Models

Module 2: Linear Programming

Introduction, Application, Advantages and Limitations of Linear Programming

Linear Programming Model Formulation

Graphical Solution Methods

Simplex Method

Big-M method

Module 3: Transportation and Assignment Problems

Introduction, Mathematical Model of Transportation problem

Methods for finding Initial Solution

Test for Optimality

Introduction to Assignment Problem

Methods of finding solution to Assignment Problem

Module 4: Theory of Games

Two-Person Zero-Sum Games

Pure Strategies

Mixed Strategies

Module 5: Network Analysis

Network Diagram

Critical Path Method

PERT

Probability in Network Analysis

Module 6: Inventory Theory

Introduction, Meaning of Inventory Control

Functional Role of Inventory

Factors Involved in Inventory Problem Analysis

Inventory Model Building: Concept of EOQ

Inventory Control Models Without Shortages

Inventory Control Models With Shortages

Module 7: Queuing Theory

Introduction, Essential Features of a Queuing System

Performance Measures of a Queuing System

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Probability Distributions in Queuing System

Classification of Queuing Models: Single Server Queuing Models , Multi-Server Queuing Models

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Unit 1 :Introduction to operation research

This chapter provides an overall view of subject of operation research. It covers some general idea on the subject, thus providing a perspective. 1.1 Development of operation research :

(i) Pre –world War II : The roots of operation extend to even early 1800s , it was in 1885 when Ferderick W Taylor emphasized the application of scientific analysis to method of production , that the real start took place. Taylor conducted experiments in connection with a simple shovel. His aim was to find that weight load of ore moved by shovel which would result in maximum of ore moved with minimum of fatigue. In 1917, A .K. Erlang , a Danish mathematician , published his work on the problem of congestion of telephone traffic. The difficulty was that during busy periods , telephone operators were unable to handle the calls they were made , resulting in delayed calls. The well known economic lot size model is attributed to F.W. Harris , who published his work on the area of inventory control in 1915. During the 1930s, H.C. Levinson , an American , applied scientific , analysis to the problems of merchandising . His work included scientific study of customers’ buying habits , response to advertising and relation of environment to the type of article sold. The industrial development , brought with it ,a new type of problems called executive –type problems. These problems are a direct consequences of functional division of labour in an organization. In an organization , each functional unit performs a part of the whole job and for its successful working , develops its own objectives. The production department wants to have maximum production , associated with the lowest possible cost. This can be achieved by producing one item continuously. The marketing department also wants a large but diverse inventory so that a customer may be provided immediate delivery over a wide variety of products. The finance department wants to minimize inventory so as to minimize the unproductive capital investments ‘tied up’ in it. Personnel department wants to hire good labour and to retain it. This is possible only when goods are produced continuously for inventory during the slack period. All the executive type problems can be solved by using OP techniques. The decision which is in the best interest of the organization as a whole is called Optimal decision and the one of the best interest of an individual department is called sub-optimal decision.

(ii) World War II : During World War II , the military management in England called on a term of scientists to study the strategic and tactical problems of air and land defence. Many of these problems were of executive type. The objective was to find out the most effective allocation of limited military resources to the various military operations and to the activities within each operations. The name operations research was apparently coined in 1940 because the team was carrying out research on operations.

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(iii) Post World War II : Immediately after world war , the success of military teams attracted the attention of industrial managers who were seeking solutions to there problems. In U.K. the critical economic situation required drastic increase in production efficiency and creation of new markets. In U.S.A. situation was different . Most of the war-experienced OR workers remained military services. Industrial executives did not call for much help because they were returning to the peace- time situation. OR has been known by a variety of names in that country such as operational analysis , operations evaluation ,system analysis ,decision analysis ,decision science, quantitative analysis and management science. To increase the impact of OR , the OR Society of America (ORSA) was formed in 1950. Today , the impact of operation research can be felt in many areas. This is shown by the ever increasing number of educational institutions offering this subject at degree level. The fast increasing number of management consulting firms speaks of the popularity of the subject. Some of the Indian organizations using OR techniques are : Indian Airlines , Railways , Defence Orgabizations , Fertilizer Corporation of India , Delhi Clioth Mills , Tata Iron and Steel .Co. etc.

1.2 Definitions of OR : Many definitions of OR has been suggested from time to time .Some of the definitions suggested are: 1) Or is a scientific method of providing executive departments with a

quantitative basis for decisions regarding the operations under their control. – Morse & Kimball

2) OR is a scientific approach to problem solving for executive management. – H.M.Wanger

3) OR is an experiment and applied science devoted to observing , understanding and predicting the behaviour of purposeful man-machine systems ; and OR workers are actively engaged in applying this knowledge to practical problems in business , government and society.- OR Society of American.

4) OR is the art of winning wars without actually fighting them. –Auther Clark It may be noted that most of the above definitions are not satisfactory because of the following reasons : i) They have been suggested at different times of development of

operations research and hence emphasize only its one or other aspect. ii) The interdisciplinary approach which is an important characteristic of

OR is not included in most of its definitions. iii) It is not easy to define OR precisely as it is not a science representing

ant well-defined social, biology or physical phenomenon. 1.3 Characteristic of OR : a) System or executive orientation of OR : This mean that an activity by any part of an organization has some effect on the activity of every other part. The optimum operation of one part of a system may not be the optimum operation for some other part. Therefore , to evaluate any decision , one must identify all possible interactions and determine their

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impact on the organization as a whole. When all factors affecting the system (organization) are known , a mathematical model can be prepared. A solution of this model will optimize the profit to the system as a whole. Such a solution is called an optimum (optimum) solution. b) The use of Interdisciplinary Teams : The second characteristic of OR study is that it is performed by a team of scientist whose individual members have been drawn from different scientific and engineering discipline . For example , one may find a mathematician , statistician , physician , psychologist , economist and an engineer working together on OR problems. Thus the OR team can look at the problem from many different angles in order to determine which one of the approach is the best.

1.4 NATURE OF OPERATIONS RESEARCH

As its name implies, O.R. involves research on (military) operations. This

indicates towards the approach as well as the area of application of the field.

Thus it is an approach to problems that concern how to co-ordinate and control

the operations or activities within an organization. Following is such example

which need further elaboration.

In order to run an organization effectively as a whole, the problem arises is of

co-ordination among the conflicting goals of its various functional departments.

For example, consider the problem of stocks of finished goods. The various

departments of the organizations would like to handle this problem differently. To

the marketing department, stock of a large variety of products are a means Of

supplying the company's customers with what they want, and when they want it.

Clearly, a fully stocked warehouse is of prime importance to the company. The

production department argues for long production runs preferably on a smaller

product range, particularly if there is a significant time lost when production is

switched from one variety to another. The result would again be a tendency to

increase the amount of stock carried but it is, of course, vital that the plant should

be kept running. On the other hand, the finance department sees stocks kept

capital tied up unproductively and argues strongly for their reduction. Finally,

there appears the personnel department who sees great advantage for labour

relations by having a steady level of production. All there are acting through their

specialization in what they would claim to be in the interests of their organization

that they may come up with contradictory solutions. To assimilate the whole

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system, the decision-maker must decide the best policy keeping in view the

relative importance of objectives and validity of conflicting claims of various

departments from the perspective of the whole organization.

Operations research seeks the optimal solution to a problem not merely one

which gives better solutions than one currents in use. The decision taken by the

decision-maker may not be acceptable to every department but it should be

optimal for a large portion of the total organization. In order to obtain such types

of solution, the decision-maker must follow up the effects and interactions of a

particular decision.

Operations research can be used to formulate the problem in terms of a

model, which on solving gives the required solution. While constructing the

model, intuition and judgement based on experience, which are valuable assets

for any practitioner, should also be incorporated.

1.5 NECESSITY/SIGNIFIANCE OF OPERATIONS RESEARCH IN INDUSTRY

After having studied as to what is operation research, we shall now try to answer

as to why study OR or what is its importance or why its need has been felt by the

industry.

As already pointed out, science of OR came into existence in connection with the

war operations, to decide the strategy by which enemy could be harmed to the

maximum possible extent with the help of the available warfare. War situation

required reliable decision-making. But its need' has beer. equally felt by the

industry due to the following reasons:

(a) Complexity: In a big industry, the number of factors influencing a decision

have increased. Situation has become big and complex because these factors

interact with each other in complicated fashion. There is, thus, great uncertainty

about the outcome of interaction of factors like technological, environmental,

competitive, etc. For instance, consider a factory production schedule which has

to take into account

(i) customer demand,

(ii) requirements of raw materials,

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(iii) equipment capacity and possibility of equipment failure, and

(iv) restrictions on manufacturing processes.

(b) Scattered responsibility and authority: In a big industry, responsibility and

authority of decision-making is scattered throughout the organization and thus

the organization, if it is not conscious, may be following inconsistent goals.

Mathematical quantification of OR overcomes this difficulty also to a great extent.

(c) Uncertainty: There is a great uncertainty about economic and general

environment. With economic growth, uncertainty is also increasing. This makes

each decision costlier and time consuming. OR is, thus, quite essential from

reliability point of view.

(d) Knowledge explosion: Knowledge is increasing at a very fast rate. Majority of

the industries are not up-to-date with the latest knowledge and are, therefore, at

a disadvantage. OR teams collect the latest information for analysis purposes

which is quite useful for the industries.

1.6 SCOPE OF OPERATIONS RESEARCH

Having-known-the definition of OR, it is easy to visualize the scope of operations

research. When we broaden the scope of OR, we find that it has really been

practised for hundreds of years even before World War II. Whenever there is a

problem of optimization, there is scope for the application of OR. Its techniques

have been used in a wide range of situations:

1. In Industry

In the field of industrial management, there is of chain of problems starting from

the purchase of raw materials to the dispatch of finished goods. The

management is interested in having an overall view of the method of optimizing

profits. OR study should also point out the possible changes in the overall

structure like installation of a new machine, introduction of more automation, etc.

OR has been successfully applied in industry in the fields of production, blending,

product mix, inventory control, demand forecast, sale and purchase,

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transportation, repair and maintenance, scheduling and sequencing, planning,

scheduling and control of projects and scores of other associated areas.

2. In Defence

OR has a wide scope for application in defence operations. In modern warfare

the defence operations are carried out by a number of different agencies, namely

airforce, army and navy. The activities performed by each of them can be further

divided into sub-activities viz. operations, intelligence, administration, training and

the like. There is thus a need to coordinate the various activities involved in order

to arrive at optimum strategy and to achieve consistent goals. Operations

research, conducted by team of experts from all the associated fields, can be

quite helpful to achieve the desired results.

3. Planning

In both developing and developed economies, OR approach is equally

applicable. In developing economies, there is a great scope of developing an OR

approach towards planning. The basic problem is to orient the planning so that

there is maximum growth of per capita income in the shortest possible time, by

taking into consideration the national goals and restrictions imposed by the

country. The basic problem in most of the countries in Asia and Africa is to

remove poverty and hunger as quickly as possible. There is, therefore, a great

scope for economists, statisticians, administrators, technicians, politicians and

agriculture experts working together to solve this problem with an OR approach.

4. Agriculture

OR approach needs to be equally developed in agriculture sector on national or

international basis. With population explosion and consequent shortage of food,

every country is facing- the problem of optimum allocation of land to various crops

in accordance with climatic conditions and available facilities. The problem of

optimal distribution of water from the various water resources is faced by each

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developing country and a good amount of scientific work can be done in this

direction.

5. Public Utilities

OR methods can also be applied in big hospitals to reduce waiting time of out-

door patients and to solve the administrative problems,

Monte Carlo methods can be applied in the area of transport to regulate train

arrivals and their running times. Queuing theory can be applied to minimize

congestion and passengers' waiting time.

OR is directly applicable to business and society. For instance, it is increasingly

being applied in LIC offices to decide the premium rates of various policies. It has

also been extensively used in petroleum, paper, chemical, metal processing,

aircraft, rubber, transport and distribution, mining and textile industries.

OR approach is equally applicable to big and small organizations. For example,

whenever a departmental store faces a problem like employing additional sales

girls, purchasing an additional van, etc., techniques of OR can be applied to

minimize cost and maximize benefit for each such decision.

Thus we find that OR has a diversified and wide scope in the social, economic

and industrial problems of today.

6 OPERATIONS RESEARCH AND DECISION-MAKING

Operations research or management science, as the name suggests, is the

science of managing. As is known, management is most of the time making

decisions. It is thus a decision science which helps management to wake better

decisions. Decision is, in fact, a pivotal word in managing. It is not only the

headache of management, rather all of us make decisions. We daily decide

about minor to major issues. We choose to be engineers, doctors, lawyers,

managers, etc. a vital decision which is going to affect us throughout our lives.

We choose to purchase at a particular shop a decision of relatively minor

importance.

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Decision-making can be improved and, in fact, there is a scope of large scale

improvement. The essential characteristics of all decisions are

(i) objectives,

(ii) alternatives,

(iii) influencing factors (constraints).

Once these characteristics are known, one car think of improving the

characteristics so as to improve upon the decision itself.

Let us consider a situation in which a decision has been taker to see a particular

movie and the problem is to decide the conveyance. Three alternatives are

available: rickshaw, autorickshaw and a local bus.

In the first level of decision-making, autorickshaw is chosen as the mode of

conveyance just by intuition, i.e., it is decided at random. Evidently, it is a highly

emotional and qualitative way of decision-making.

In the second level of decision-making, the three conveyances are compared and

it is decided qualitatively that autorickshaw will be preferred since, though a little

costlier, it is timesaving and more comfortable.

In the third level of decision-making, the three alternatives are compared and it is

suggested that autorickshaw will be chosen, as it will be taking only !/?rd time

than an ordinary rickshaw and shall be only 10% costlier while more comfortable.

The local bus is rejected since it would not reach the theatre in time at all.

Though outcome of all these decisions is the same, still we can judge the quality

of each decision. We may brand the first decision as 'bad' since it is highly

emotional, while we may call the second decision as 'good' since it is scientific

though qualitative. The third decision is doubtlessly the best as it is scientific and

quantitative.

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It is the scientific quantification used :n OR, which helps management to make

better decisions. Thus in OR, the essential features of decisions, namely,

objectives, alternatives and influencing factors are expressed in terms of

scientific quantifications or mathematical equations. This gives rise to certain

mathematical relations, termed as a whole as mathematical model. Thus the

essence of OR is such mathematical models. For different situations different

models are used and this process is continuing since World War II when the term

OR was coined. However, with the advance of science and technology, decision-

making in business and industry has become highly complex and extremely

difficult. The decision-maker is not only faced with a large number of interacting

variables, which at times do not lend themselves to neat quantitative treatment

but also finds them too numerous and dynamic. Above all he has to take into

consideration the actions of the competitors over which he has no control. This

complexity of decision-making made the decision-makers look for various aids in

decision-making. It is in these situations that operations research comes to our

help. The managers today make full use of the OR techniques in various

functional areas. It has been realised beyond doubt that intuition alone has no

place in decision making since such a decision becomes highly questionable

when it involves the choice among several alternatives. OR provides the

management much needed tools for improving the various decisions.

7 SCOPE OF OPERATIONS RESEARCH IN MANAGEMENT

Operations research is a problem-solving and decision-making science. It is a kit

of scientific and programmable rules providing the management a ‘quantitative

basis' for decisions regarding the operations under its control. Some of the areas

of management where OR techniques have been successfully applied are:

Allocation and Distribution

(a) Optimal allocation of limited resources such as men, machines, materials,

time and money.

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(b) Location and size of warehouses, distribution centres, retail depots, etc.

(c) Distribution policy.

1.7 APPLICATIONS OF VARIOUS OR TECHNIQUES

Operations research at present finds extensive application in industry, business,

government, military and agriculture. Wide variety of industries namely, airlines,

automobiles, transportation, petroleum, coal, chemical, mining, paper,

communication, computer, electronics, etc. have made extensive use of OR

techniques. Some of the problems to which OR techniques have been

successfully applied are:

1. Linear programming has been used to solve problems involving assignment

of jobs to machines, blending, product mix, advertising media selection, least

cost diet, distribution, transportation, investment portfolio selection and many

others.

2. Dynamic programming has been applied to capital budgeting, selection of

advertising media, employment smoothening, cargo loading and optimal routing

problems.

3. Inventory control models have been used to determine economic order

quantities, safety stocks, reorder levels, minimum and maximum stock levels.

4. Queuing theory has been helpful to solve problems of traffic congestion,

repair and maintenance of broken-down machines, number of service facilities,

scheduling and control of air traffic, hospital operations, counters in banks and

railway booking agencies.

5. Decision theory has been helpful in controlling hurricanes, water pollution,

medicine, space exploration, research and development projects.

6. Network techniques of PERT and CPM have been used in planning,

scheduling and controlling construction of dams, bridges, roads, highways and

development and production of aircrafts, ships, computers, etc.

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7. Simulation has been helpful in a wide variety of probabilistic marketing

situations. It has been, for example, used to find NPV (Net Present Value)

distribution for the venture of market introduction of a new product.

8. Replacement theory has been extensively employed to determine the

optimum replacement interval for three types of replacement problems:

1.9 MODELS IN OR

A model, as used in operations research, is defined as an idealized

representation of the real life situation. It represents one or a few aspects of

reality. Diverse items such as a map, a multiple activity chart, an autobiography,

PERT network, break-even equation, balance sheet, etc. are all models because

each one of them represents a few aspects of the real life situation. A map, for

instance, represents the physical boundaries but normally ignores the heights of

the various places, above the sea level. The objective of the model is to provide

a means for analysing the behaviour of the system for the purpose of improving

its performance.

1.10 LASSIFICATION SCHEMES OF MODELS

The various schemes by which models can be classified are

1. By degree of abstraction

2. By function

3. By structure

4. By nature of the environment

5. By the extent of generality

6. By the time horizon

1. By Degree of Abstraction

Mathematical models (viz. linea programming formulation of the blending

problem or transportation problem) are the most abstract type since it requires

not only mathematical knowledge but also great concentration :o get .he idea of

the real-life situation they represent.

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Language models (cricket or hockey match commentary) are also abstract type.

Concrete models (model of earth, dam, building or plane) are the least abstract

since they instantaneously suggest the shape or characteristics of the modelled

entity.

2. By Function

Descriptive models explain the various operations in non-mathematical language

and try to define the functional relationships and interactions between various

operations. They simply describe some aspects of the system on the basis of

observation, survey or questionaire, etc. but do not predict its behaviour. The

organisational chart, pie diagram and layout plan describe the features of their

respective systems.

Predictive models explain or predict the behaviour of the system. Exponential

smoothing forecast model, for instance, predicts the future demand.

Normative or prescriptive models develop decision rules or criteria for optimal

solutions. They are applicable to repetitive problems, the solution process of

which can be programmed without managerial involvement. Linear programming

is a prescriptive or normative model as it prescribes what the managers must

follow.

3. By Structure

(a) Iconic or Physical Models

In iconic or physical models, properties of the real system are

represented by the properties themselves, frequently with a change of

scale. Thus, iconic models resemble the system they represent but differ

in size; they are images. For example, globes are used to represent the

orientation and shape or various continents, oceans and other

geographical features of the earth. A model of the solar system, likewise,

represents the sun and planets in space. Iconic models of atoms and

molecules are commonly used in physics, chemistry and other sciences.

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However, these models are usually scaled up or down. For example, in a

globe, the diameter of the earth is scaled down, but its shape, relative

sizes of continents, oceans, etc., are approximately correct. On the other

hand, a model of the atom is scaled up so as to make it visible to the

naked eye. Iconic models may be two-dimensional (photographs, maps,

blue prints, paintings, sketches of insects, etc.) or three-dimensional

(globes, automobiles, airplanes, etc.). Ordinarily it is easier to work with

the model of a building, earth, sun, atom, etc., than with the modelled

entity itself. Iconic models are quite specific and concrete but difficult to

manipulate for experimental purposes. They represent a static event.

Characteristics that are not relevant are not included. For instance, in the

models used for the study of atomic structure, the colour of the model is

irrelevant since it contributes no help in the study of the atom. Another

limitation of iconic model is that it is either two-dimensional or three-

dimensional. If a situation involves more than three dimensions, it cannot

be represented by an iconic model.

(b) Analogue or Schematic Models

Analogue models can represent dynamic situations and are used

more often than iconic models since they are analogous to the

characteristics of the system under study. They use one set of properties

to represent some other set of properties which the system under study

possesses. After the model is solved, the solution is re-interpreted in

terms of the original system.

For example, graphs are very simple analogues. They represent properties like

force, speed age, time, etc., in terms of distance. A graph is well suited for

representing quantitative relationship between any two properties and predicts

how a change in one property affects the other.

An organizational chart is a common schematic model. It represents the

relationships existing between the various members of the organization. A man-

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machine chart is also a schematic model. If represents a time varying interaction

of men and machines over a complete work cycle. A flow process chart is

another schematic model which represents the order of occurrence of various

events to make a product. Contour lines on a map are analogous of elevation.

Flow of water through pipes may be taken as an analogue of the `flow' of

electricity through wires. Similarly, demand curves and frequency distribution

curves used in statistics are examples of analogue models. In analogue

computers quantities are represented by voltages and they are, therefore, aptly

termed analogue.

Transformation of properties into analogous properties increases our ability to

make changes. Usually it is easier to change an analogue than to change an

iconic model and also lesser number of changes are required to get the same

results. For example, it is easier to change the contour lines on a two-

dimensional chart than to change the relief on a three-dimensional one. In

general, schematic models are less specific and concrete but easier to

manipulate than iconic models. They can represent dynamic situations and are

more commonly used than the iconic models.

(c) Symbolic or Mathematical Models

Symbolic models employ a set of mathematical symbols (letters,

numbers, etc.) to represent the decision variables of the system under

study. These variables are related together by mathematical

equation(s)/inequation(s) which describe the properties of the system. A

solution from the model is, then, obtained by applying well developed

mathematical techniques. The relationship between velocity, acceleration

and distance is an example of mathematical model. Similarly, cost-

volume-profit relation is a mathematical model used in investment

analysis.

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In many research projects, all the three types of models are used in

sequence; iconic and analogue models are used as initial approximations, which

are, then, refined into symbolic model. Mathematical models differ from those

traditionally used in physical sciences in two ways:

1. Since OR systems involve social and economic factors, these models use

probabilistic elements.

2. They consist of two types of variables; controllable and uncontrollable.

The objective is to select those values for controllable variables which optimize

some measure of effectiveness. Therefore, these models are used in decision

situations rather than in physical phenomena.

In OR, symbolic models are used wherever possible, not only because they are

easier to manipulate but also because they yield more accurate results. Most of

this text, therefore, is devoted to the formulation and solution of these

mathematical models.

4. By Nature of the Environment

(a) Deterministic Models

In deterministic models variables are completely defined and the

outcomes are certain. Certainty is the state of nature assumed in these

models. They represent completely closed systems and the parameters of

the system have a single value that does not change with time. For any

given set of input variables, the same output variables always result . EOQ

model is deterministic; here the effect of changes in the batch sizes on the

total cost is known. Similarly linear programming, transportation and

assignment models are deterministic models.

(b) Probabilistic Models

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They are the products of an environment of risk and uncertainty.

The input and/or output variables take the form of probability distributions.

They are semi-closed models and represent

the likelihood of occurrence of an event. Thus they represent, to an extent, the

complexity of the real world and the uncertainty prevailing in it. As an example,

the exponential smoothing model for forecasting demand is a probabilistic model.

5. By the Extent of Generality

(a) General Models

Linear programming model is known as a general model since it can be used for

a number of functions (viz. product mix, production scheduling, marketing, etc.)

of an organisation.

(b) Specific Models

Sales response curve or equation as a function of advertising is applicable in the

marketing function alone.

6. By the Time Horizon

(a) Static Models

They are one-time decision models. They represent the system at a specified

time and do not take into account the changes over time. In these models cause

and effect occur almost simultaneously and time lag between the two is zero.

They are easier to formulate, manipulate and solve. Economic order quantity

model is a static model.

(b) Dynamic Models

They are the models for situations in which time often plays an important role.

They are used for optimization of multistage decision problems which require a

series of decisions with the outcome of each depending upon the results of the

previous decisions in the series. Dynamic programming is a dynamic model.

1.11CHARACTERISTICS OF A GOOD MODEL

1. The number of simplifying assumptions should be as flew as possible.

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2. The number of relevant variables should be as few as possible. This means

the model should be simple yet close to reality.

3. It should assimilate the system environmental changes without change in its

framework. 4. It should be adaptable to parametric type of treatment.

5. It should be easy and economical to construct.

1.12 ADVANTAGES OF A MODEL

1. It provides a logical and systematic approach to the problem. 2. It indicates the

scope as well as limitations of a problem.

3. It helps in finding avenues for new research and improvements in a system.

4. It makes the overall structure of the problem more comprehensible and helps

in dealing with the problem in its entirety.

5. It permits experimentation and analysis of a complex system without directly

interfering in the working and environment of the system.

1.14 LIMITATIONS OF A MODEL

1. Models are only idealised representation of reality and should not 6e regarded

as absolute in any case.

2. The validity of a model for a particular situation can be ascertanied only by

conducting experiments on it.

1.15 CONSTRUCTING THE MODEL

It was pointed out in previous sections that formulation of the problem requires

analysis of :he system under study. This analysis shows the various phases of

the system and the way it can 7e controlled. With the formulation of the problem,

the first stage in model construction is over. The next step is to define a measure

of effectiveness, i.e., the next step is to construct a model in ,rhich effectiveness

of the system is expressed as a function of the variables cleftningg the system.

The general form of OR model is

E = f (xi, yj),

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where E = effectiveness of the system,

xi = variables of the system that can be controlled,

yi = variables of the system that cannot be controlled but do affect E.

Deriving of solution from such a model consists of determining those values of

control variables x;, for which the measure of effectiveness is optimized.

Optimization includes both maximization (in case of profits, production capacity,

etc.) and minimization (in case of losses, cost of production, etc.).

Various steps in the construction of a model are

1. Selecting components of the system

2. Pertinence of components

3. Combining the components

4. Substituting symbols

1.16-1 Selecting Components of the System

All the components of the system which contribute towards the effectiveness

measure of the system should be listed.

1.16-2 Pertinence of Components

Once a complete list of components is prepared, the next step is to find whether

or not to take each of these components into account. This is determined by

finding the effect of various alternative courses of action on each of these

components. Generally, one or more components (e.g., fixed costs) are

independent of the changes made among the various alternative courses of

action. Such components may be temporarily dropped from consideration.

1.16.3 Combining the Components

It may be convenient to group certain components of the system. For example,

the purchase price, freight charges and receiving cost of a raw material can be

combined together and called `raw material acquisition cost'. The next step is to

determine, for each component remaining on the modified list, whether its value

is fixed or variable. If a component is variable, various aspects of the system.

23

affecting its value must be determined. For instance manufacturing cost usually

consists of

(1) the number of units manufactured, and (ii) the cost of manufacturing a unit_

1.16-4 Substituting Symbols

Once each variable component in the modified list has been broken down like

this, symbols may be assigned to each of these sub-components.

The foregoing steps will be clear from the example considered below

A newsboy wants to decide the number of newspapers he should order to

maximize his expected profit. He purchases a certain number of newspapers

everyday and is able to sell some or all of them. He earns a profit on each paper

sold. He can return the unsold papers, but at a loss. The number of persons who

buy newspapers varies from day-to-day.

To construct the model for this problem, we identify the various relevant

components (variables) and then assign symbols to them.

Let N = number of newspapers ordered per day,

A = profit earned on each newspaper sold,

B = loss on each newpaper returned,

D = demand i.e. number of newspapers sold per day,

p(D) = probability that the demand will be equal to D

on any randomly selected day,

P = net profit per day.

If D > N i.e., demand is more than the number of newspapers ordered, the profit

to the newsboy is

P(D > N) = NA.

If on the other hand, demand is less than the number ordered, the profit is

P(D < N) = DA - (N - D)B.

:. Net expected profit per day, P can be expressed as

24

This is a decision model of the risk type. Here, P is the measure of performance,

N is the controlled variable, D is an uncontrollable variable, while A and B are

uncontrollable constants. Solution of this model consists of finding that value of N

which maximizes P.

1.16 APPROXIMATIONS (SIMPLIFICATIONS) IN OR MODELS

While constructing a model one comes across two conflicting objectives:

(i) the model should be as easy to solve as possible,

(ii) it should be as accurate as possible.

Moreover, the management must be able to understand the solution of the model

and must be capable of using it. Obviously, one must pay due care to the

mathematical complexity of the solution. Therefore, while constructing the model,

the reality (problem under study) should be simplified but only to the point that

there is no significant loss of accuracy. Some of the common simplifications

include

1. Omitting certain variables

2. Aggregating variables

3. Changing the nature of variables

4. Changing the relationship between variables, and 5. Modifying constraints

1.17-1 Omitting Certain Variables

Clearly, variables having a large effect on system's performance cannot be

omitted. It requires a lot of study to decide which variables have and which do not

have large effects. For instance, in production and inventory control models, the

effect of production-run sizes on in process inventory costs is usually negligible

as compared to effect of other variables and is, therefore, neglected.

1.17-2 Aggregating Variables

Most problems involve a large number of decision variables. For instance,

some inventory problems involve the purchase of more than a million items. For

solving such problems, the controlled variables are grouped into `families'. A

25

family is, then, supposed to consist of all identical members. One principle of

`family' formation is

1. Low usage, low cost

2. Low usage, high cost

3. High usage, low cost

4. High usage, high cost

1.18-3 Changing the Nature of Variables

The nature of variables may be changed in three ways:

(i) by treating a variable as constant,

(ii) by treating a discrete variable as continuous, and

(iii) by treating a continuous variable as discrete.

A variable may be treated as constant with its value equal to the mean of the

variable's distribution. For example, in most production quantity models setup

cost is treated as constant. From both analytical and computational viewpoints it

is easier to treat a discrete variable as continuous. Most of OR techniques deal

with continuous variables. Even if the discrete variables are few in number, the

computational difficulties become quite large. For instance, in inventory control

models, withdrawals of items from stock that are actually discrete are assumed

as continuous at a constant rate, over a planning period.

However, for processes in which time between events is a relevant variable,

considerable simplification may be obtained by assuming that events occurring

within a certain period occur instantaneously at the beginning or end of the

period.

1.17-4 Changing the Relationship between Variables

Models can be simplified by modifying the functional form of the model. Non-

linear functions require a complex solution method. The most powerful

computational techniques are applicable only to models having linear functions.

Therefore, non-linear functions are usually approximated to linear functions (e.g.,

in linear programming). Many times, a curve is approximated to a series of

26

straight lines (e.g., in non-linear programming). Quadratic functions are used as

approximations since their derivatives are linear (e.;.. in quadratic programming).

Discrete functions (e.g., binomial and Poisson) are sometimes approximated to

continuous normal functions.

1.17-5 Modifying Constraints

Constraints can be deleted, added or modified to simplify the model. If it is not

possible to solve a model with all the constraints, some of them may be

temporarily ignored and a `solution' obtained. If this `solution' happens to satisfy

these constraints too, it is accepted. If it does not, constraints are added, one at a

time, with increasing complexity, until a solution satisfying these constraints is

obtained. A general rule regarding constraints is that when they are dropped the

solution derived from the model becomes optimistic (it gives better performance

than the `true' solution). On the other hand, adding of constraints makes the

solution pessimistic.

1.18 TYPES OF MATHEMATICAL MODELS

Many OR models have been developed and applied to problems in business and

industry. Some of these models are:

1. Mathematical techniques

2. Statistical techniques

3. Inventory models

4. Allocation models

5. Sequencing models

6. Project scheduling by PERT and CPM

7. Routing models

8. Competitive models

9. Queuing models

10. Simulation techniques

27

Unit –2 LINEAR PROGRAMMING

2.1 INTRODUCTION

28

All organizations, whether large or small, need optimal utilization of their scarce

or limited resources to achieve certain objectives. Scarce resources may be

money, manpower, materials, machine capacity technology, time, etc. In order to

achieve best possible result(s) with the available resources, the decision-maker

must understand all facts about the organization activities and the relationships

governing among chosen activities and its outcome. The desired outcome may

be measured in terms of profits, time, return on investment, costs, etc.

Of all the well known operations research models, linear programming is the

most popular and most widely applied technique of mathematical programming.

Basically, linear programming is e deterministic mathematical technique which

involves the allocation of scarce or limited resources in an optimal manner on the

basis of a given criterion of optimality. Frequently, the criterion of optimality is

either profits, costs, return on investment, time, distance, etc.

George B. Dantzig, while working with U.S. Air Force during World War II, in

1947, developed. the technique of linear programming. Linear programming was

developed as a technique to achieve the best plan out of different plans for

achieving the goal through various activities like procurement recruitment,

maintenance, etc.

During its early stage of development, it was applied primarily to military

logistics problems such as transportation, assignment and deployment decisions.

However, after the war, became e popular technique in business and industry.

Today linear programming has found applications in government, hospitals,

libraries, education and almost in all functional areas of management Production

scheduling and inventory control, transportation of goods and services, capita

investments, advertising and promotion planning, personnel assignment and

development, etc., are few examples of the type of problems that can be solved

by linear programming.

In linear programming decisions are made under certainty, i.e., information

on available resources and relationships between variables are known. Therefore

actions chosen will invariably lead to optimal or nearly optimal results. It also

29

helps in verifying the results arrived at by intuitive decision making and indicate

the errors involved in the selection of optimal course of actions. In sum, we can

say that linear programming provides a quantitative basis to assist a

decisionmaker in the selection of the most effective and desirable course of

action from a given number of available alternatives to achieve the result in an

optimal manner.

2:2 BASIC TERMINOLOGY, REQUIREMENTS, ASSUMPTIONS,

ADVANTAGES AND LIMITATIONS

Basic Terminology: The word 'linear' used to describe the relationships

among two or more variables which are directly or precisely proportional. For

example, doubling (or tripling) the production of a product will exactly double (or

tripling) the profit and required resources.

The word 'programming' means that the decisions are taken systematically

by adopting various alternative courses of actions.

A program is 'optimal' if it maximize or minimize some measure or criterion of

effectiveness such as profit, cost, or sales. The term 'limited' refers to the

availability of resources during planning horizon.

2.3 Basic Requirements: Regardless of the way one defines linear

programming, certain basic requirements are necessary before it can be used for

optimization problems.

(i) Decision Variables and their Relationship: The decision variable refers to any

activity (product, service, project etc.) that is competing with other decision

variables (activities) for limited resources. These variables are usually

interrelated in terms of utilization of resources and need simultaneous

solutions. The relationship among these variables should be linear.

(ii) Objective Function: The linear programming problem must have a well

defined (explicit) objective function to optimize. For example, maximization of

profits, minimization of cost or elapsed time of the system being studied. It

should be expressed as linear function of decision variables. The single-

objective optimization is an important requirement of linear programming.

30

(iii) Constraints: There must be limitations on resources, which are to be

allocated among various competing activities. These resources may be

production capacity, manpower, money, time, space or technology. These

must be capable of being expressed as linear equalities or inequalities in

terms of decision variables. These impose restrictions on the activities

(decision variables) in optimizing the objective function.

(iv) Alternative Courses of Action: There must be alternative courses of action.

For example, there may be many processes open to a firm for producing a

commodity and one process can be substituted for another.

(c•) Non-negative Restriction: All variables must assume non-negative values

as negative values of physical quantities is meaningless. If any of the

variables is unrestricted in sign, a trick can be employed which will enforce

the non-negativity without changing the original information of the problem.

(vi) Linearity: All relationships among decision variables in the objective function

and constraints must exhibit linearity, that is, relationship among decision

variables must be directly proportional. For example, if our resource increase

by some percentage, than it should result increase in the outcome by the

same percentage.

2.4 Basic Assumptions: In all linear programming models, there are certain

assumptions which must be met in order for such models to be applicable.

(i) Proportionality: The amount of each resource used and associated

contribution to profit (or cost) in the objective function must be proportional to

the value of each decision variable. For example, if the number of units of an

item produced were doubled, then the total amount of each resource

required in the manufacture of the item would also be doubled, as would the

total profit contribution from items. In other words marginal measure of

profitability and marginal usage of each resource are being considered as

constant over the entire range of productive activity.

(ii) Divisibility (or Continuity): It is assumed that the solution value of the

decision variables and the amount of resources used need not be integer

31

values, i.e. continuous values of the decision variables and resources must

be permissible W obtaining an optimal solution.

(iii) Additivity: It is required that the total profitability, and total amount of each

resource utilized must be equal to the sum of the respective individual

amounts. For example, the production of an item can not affect the

profitability associated with the production of any other item and vice-versa,

i.e., total profitability and total amount of each resource untilized that results

from the joint production of two items, must be equal to the sum of the

quantities resulting from the items being produced individually.

(iv) Deterministic Coefficients (or Parameters): It is assumed that all coefficients

(or parameters e.g. profit or cost associated with each product; amount of

resources required per unit each product, and the amount of input-output or

technological coefficients in linear programming model are known with

certainty (i.e. constant). Such type of data, usually obtained from marketing,

production or accounting data. However, this may not be always true. Then

in such cases, the decision-maker is required to obtain a set of coefficients

that will allow a reasonable decision to be made.

2.5 Advantages of Linear Programming: Following are certain advantages

of linear programming

1. Linear programming helps in attaining the optimum use of productive factors. It

also indicates how a decision-maker can employ his productive factors

effectively by selecting and distributing these elements.

2. Linear programming techniques improve the quality of decisions. User of this

technique becomes more objective and less subjective.

3. Linear programming gives possible and practical solutions since there might

be other constraint operating outside of the problem which must be taken into

account. Just because we can produce so many units, does not mean that

they can be sold. It allows modification of its mathematical solution for the

sake of convenience to the decision-maker.

32

4. Highlighting of bottlenecks in the production processes is the most significant

advantage of this technique. For example, when bottleneck occurs, some

machines cannot meet demand while other remain idle for sometime.

5. Linear programming also help in re-evaluation of a basic plan for changing

conditions. If conditions change when the plan is party carried out, they can be

determined so as to adjust the remainder of the plan for best result.

2.6 Limitations of Linear Programming: Inspite of having many

advantages there are some limitations associated with it, which are given below :

1. Linear programming treats a11 relationship as linear. However, generally,

neither the objective function no the constraints in real life situations

concerning business and industrial problems are linearly related to the

variables.

2. There is no guarantee that it will give integer valued solutions. For example, W

finding out how many men and machines would be required to perform a

particular job a non-integer valued solution will be meaningless. Rounding off

the solution to the nearest integer will not yield an optimal solution. In such

cases other methods would be used.

3. Linear programming model does not take into consideration the effect of time

and uncertainty. Thus it shall be defined in such a way that any change due to

internal as well as external factors can be incorporated.

4. Sometimes large-scale problems can not be solved with linear programming

technique even when assistance of computer is available. The main problem

can be decomposed into several small problems and solved separately.

5. Parameters appear in the model are assumed to be constant but in real-life

situation, they are frequently neither known nor constants.

6. It deals with only single objective, where as in real life situations we may come

across more than one objective.

2.7 APPLICATION AREAS OF LINEAR PROGRAMMING

33

Linear programming is the most widely used technique of decision making in

business and industry and in various other fields. In this section, we will discuss a

few of the broad application areas it linear programming.

Agricultural Applications

These applications fall into two categories, farm economics and farm

management. The former deals with agricultural economy of a nation or region,

while the later is concerned with the problems o# the individual farm.

The study of farm economics deals with interregional competition and

optimum spartial allocation of crop production. Efficient production patterns were

specified by linear programming model under regional land resources and

national demand constraints.

Linear Programming can be applied in agricultural planning, e.g., allocation

of limited resources such as acreage, labour, water supply and working capital,

etc in such a way so as to maximize net revenue.

Military Applications

Military applications include the problem of selecting an air weapon system

against gurillas so as §o keep them pinned down and at the same time minimize

the amount of aviation gasoline used, a variation of transportation problem that

maximizes the total tonnage of bombs dropped on a set of targets and the

problem of community defence against disaster, the solution of which yields the

number of defence units that should be used in a given attack in order to provide

the required level of protection at the lowest possible cost.

Production Management

Product mix: A company can produce several different products each of which

require the use of limited production resources. In such cases it is essential to

determine the quantity of each product to be produced knowing their marginal

contribution and amount of available resource used by each of them. The

objective is to maximize the total contribution subject to all constraints.

Production planning: This deals with the determination of minimum cost

production plan over a planning period of an item with a fluctuating demand

34

considering the initial number of units in inventory, production capacity,

constraints on production, manpower, and all relevant cost factors. The objective

is to minimize total operation costs.

Assembly-line balancing: This problem is likely to arise when an item can be

made by assembling different components. The process of assembling requires

some specified sequence(s). The objective is to minimize the total elapse time.

Blending problems: These problems arise when a product can be made from a

variety of available raw materials, each of which has a particular composition and

price. The objective, here is to determine the minimum cost blend subject to

availability of the raw materials and the minimum and maximum constraints on

certain product constituents.

Trim loss: When an item is made in standard size (e.g. glass, paper, sheet), the

problem that arises is to determine which combination of requirements should be

produced from standard materials in order to minimize the trim loss.

Financial Management

• Portfolio selection: This deals with the selection of specific investment activity

among several other activities. The objective is to find the allocation which

maximizes the total expected return or minimizes risk under certain limitations.

• Profit planning: This deals with the maximization of profit margin from

investment in plant facilities and equipment, cash on hand and inventory.

Marketing Management

• Media selection: Linear programming technique helps in determining the

advertising media mix so as to maximize the effective exposure, subject to

limitation of budget, specified exposure rates to different market segments,

specified minimum and maximum number of advertisements in various

media.

• Traveling salesman problem: The problem of salesman is to find the shortest

route starting from a given city, visiting each of the specified cities and then

returning to the original point of departure, provided no city shall be visited

35

twice during the tour. Such type of problems can be solved with the help of

the modified assignment technique.

• Physical distribution: Linear programming determines the most economic and

efficient manner of ~locating manufacturing plants and distribution centres for

physical distribution.

Personnel Management

• Staffing problem: Linear programming is used to allocate optimum manpower

to a particular job so as to minimize the total overtime cost or total

manpower.

• Determination of equitable salaries: Linear programming technique has been

used in determining equitable salaries and sales incentives.

• Job evaluation and selection: Selection of suitable man for a specified job and

evaluation of jot in organizations has been done with the help of linear

programming technique.

Other applications of linear programming include in the area of

administration, education health care, fleet utilization, awarding contracts and

capital budgeting, etc.

2.8 FORMULATION OF LINEAR PROGRAMMING MODELS

The usefulness of linear programming as a tool for optimal decision making and

resource allocation is based on its applicability to many diversified decision

problems. The effective use and application require, as a first step, the

formulation of the model when the problem is presented.

The three basic steps in formulating a linear programming model are as follows:

Step 1: Identify the decision variables to be determined and express them in

terms of algebraic symbols.

Step 2: Identify all the limitations or constraints in the given problem and

then express them as linear inequalities or equations in terms of above

defined decision variables.

36

Step 3: Identify the objective (criterion) which is to be optimized (maximize or

minimize) am express it as a linear function of the above defined decision

variables.

FORMULATION OF LINEAR PROGRAMMING PROBLEMS

First, the given problem must be presented in linear programming form. This

requires defining the variables of the problem, establishing inter-relationships

between them and formulating the objective function and constraints. A model,

which approximates as closely as possible to the given problem, is then to be

developed. If some constraints happen to be nonlinear, they are approximated to

appropriate linear functions to fit the linear programming format. In case it is not

possible, other techniques may be used to formulate and then solve the model.

EXAMPLE (Production Allocation Problem)

A firm produces three products. These products are processed on three different

machines. The time required to manufacture one unit of each of the three

products and the daily capacity of the three machines are given in the table

below.

TABLE 2.1

Machine Time per unit (minutes) Machine

capacity Product l Product 2 Product 3 (minuteslday)

M1 2 3 2 440

M2 4 - 3 470 M3 2 5 - 430

It is required to determine the daily number of units to be manufactured for each

product. The profit per unit for product 1, 2 and 9 is Rs. 4, Rs. 3 and Rs. 6

respectively. It is assumed that a11 the amounts produced are consumed in the

37

market. Formulate the mathematical (L.P) model that will maximize the daily

profit.

Formulation of Linear Programming Model

Step 1:

From the study of the situation find the key-decision to be made. It this

connection, looking for variables helps considerably. In the given situation key

decision is to decide the extent of products 1, 2 and 3, as the extents are

permitted to vary.

Step 2:

Assume symbols for variable quantities noticed in step 1. Let the extents

(amounts) of products, 1, 2 and 3 manufactured daily be xr, xZ and x3 units

respectively.

Step 3:

Express the feasible alternatives mathematically in terms of variables. Feasible

alternative s are those which are physically, economically and financially

possible. In the given situation feasible alternatives are sets of values of st, x=

and x3,

where c,, x=, x3 ? 0,

since negative production has no meaning and is not feasible.

Step 4:

Mention the objective quantitatively and express it as a linear function of

variables. In the present situation, objective is to maximize the profit.

i.e., maximize Z = 4x1 + 3x2, + 6x3.

Step 5:

Put into words the influencing factors or constraints. These occur generally

because of constraints on availability (resources) or requirements (demands).

Express these constraints also as linear equations/inequalities in terms of

variables.

Here, constraints are on the machine capacities and can be mathematically

expressed as

38

2x1 + 3x2, + 2x3 < 440,

4x1 + 0x2, + 3x3 < 470,

2x1 + 5x2, + 0x3 < 430,

2.9 GRAPHICAL METHOD OF SOLUTION

Once a problem is formulated as mathematical model, the next step is to solve

the problem the optimal solution. A linear programming problem with only two

variables presents a simple case, for which the solution can be derived using a

graphical or geometrical method. Though, in actual practice such small problems

are rarely encountered, the graphical method provides a pictorial representation

of the solution process and a great deal of insight into the basic concepts used in

solving large L.P. problems. This method consists of the following steps:

1. Represent the-given problem in mathematical form i.e., formulate the

mathematical model for the given problem.

2. Draw the xi and X2-axes. The non-negativity restrictions xi 0 -and xZ ? 0 imply

that the values of the variables x, and xz can lie only in the first quadrant. This

eliminates a number of infeasible alternatives that lie in 2nd, 3rd and 4th

quadrants.

3. Plot each of the constraint on the graph. The constraints, 'whether equations

or inequalities are plotted as equations. For each constraint, assign any arbitrary

value to one variable and get the value of the other variable. Similarly, assign

another arbitrary value to the other variable and find the value of the first

variable. Plot these two points and connect them by a straight line. Thus each

constraint is plotted as line in the first quadrant.

4. identify the feasible region (or solution space) that satisfies all the constraints

simultaneously. For type constraint, the area on or above the constraint line i.e.,

away from the origin and for < type constraint, the area on or below the constraint

39

line i. e., towards origin will be considered. The area common to all the

constraints is called feasible region and is shown shaded. Any point on or within

the shaded region represents a feasible solution to the given problem. Though a

number of infeasible points are eliminated, the feasible region still contains a

large number of feasible points.

5. Use iso-profit (cost) junction line approach. For this, plot the objective function

by assuming Z = 0. This will be a line passing through the origin. As the value of

Z is increased from zero, the line starts moving to the right, parallel to itself. Draw

lines parallel to this line till the line is farthest way from the origin (for a

maximization problem). For a minimization problem, the line will be nearest to the

origin. The point of the feasible region through which this line passes will be the

optimal point. [t is possible that this line may coincide with one of the edges of

the feasible region. In that case, every point on that edge will give the same

maximum/minimum value of the objective function and will be the optimal point.

Alternatively use extreme point enumeration approach. For this, find the co-

ordinates of each extreme point (or corner point or vertex) of the feasible region.

Find the value of the objective function at each extreme point. The point at which

objective function is maximum/minimum is the optimal point and its co-ordinates

give the optimal solution.

40

EXAMPLE

Find the maximum value of

Z = 2x1 + 3x2,

subject to x1 + x2 < 30,

x2 > 3,

x2 < 12,

x1 –x2 > 0,

0 < x1 < 20

Solution

The solution space satisfying the given constraints and meeting the non-

negativity restrictions x1 > 0 and x2 > 0 is shown shaded in Fig. 2.4. Any point in

this shaded region is a feasible solution to the given problem.

The co-ordinates of the five vertices of the convex region ABCDE are A(3, 3),

B(12, 12), C(18, 12), D(20, 10) and E(20, 3).

Fig.

Values of the objective function Z = 2x[ + 3xZ at these vertices are Z(A) = 15, Z(B)

= 60, Z(C) = 72, Z(D) = 70 and Z(E) = 49.

Since the maximum value of Z is 72, which occurs at the point C(18,12), the

solution to the given problem is x, = 18, x2 = 12, Zmax= 72.

41

2.10 Simplex method : EXAMPLE

Solution

Maximize Z = 3X1 + 3X2 (Objective function) . (2.16)

subject to X1 + X2 < 450, (Machine M1, time constraint) 2.17

2X1 + X2 < 600, (Machine M1, time constraint)

Where X1, X2 2.18

Step 1. Express the problem in standard form

The given problem is said to be expressed in standard form if the given

(decision) variables are non-negative, right-hand side of the constraints are

non-negative and the constraints are expressed as equations. Since the first

two conditions are met with in the problem, non-negative slack variables s1 and

s2 are added to the left-hand side of the first and second constraints

respectively to convert them into equations. Values of s, and s2 vary with the

values that x, and x= take in any solution. Slack variables represent unutilised

capacity or resources. In the current problem s, denotes the time (in minutes)

for which machine M, remains unutilised or id1e; similarly s, denotes the idle

time for machine M,. Since slack variables represent idle resources, they

contribute zero to the objective function. Accordingly, they are associated with

zero coefficients in the objective function. Accordingly, the problem in standard

form, can be written as

maximize Z = 3x1, + 4x2 + Os1, + Os2, (2.19)

subject to x1, + x2 + s1 = 450 2.20

2x1, + x2 + s2 = 600

where x1, x2, s1 , s2 > 0 2.21

42

43

2.12 ARTIFICIAL VARIABLES TECHNIQUES

In the earlier problems, the constraints were of (5) type (with non-negative right-

hand sides). The introduction of slack variables readily provided the initial basic

feasible solution. There are, however, many linear programming problems where

slack variables cannot provide such a solution. In these problems at least one of

the constraints is of (?) or (=) type. In such cases, we introduce another type of

variables called artificial variables. These variables are fictitious and have no

physical meaning. They assume the role of slack variables in the first iteration,

only to be replaced at a later iteration. Thus they are merely a device to get the

starting basic feasible solution so that simplex algorithm be applied as usual to

get optimal solution. There are two (closely related) techniques available to solve

such problems. They are

1. The big M-method or M-technique or method of penalties due to A.

Charnes. 2. The two-phase method due to Dantzig, Orden and Wolfe.

2.12 The Big M-Method

This method consists of the following basic steps :

Step 1. Express the linear programming problem in standard form by introducing

slack variables. These variables are added to the left-hand sides of the

constraints of (<) type and subtracted from the constraints of (?) type.

Step 2. Add non-negative variables to the left-hand sides of all the constraints of

initially (>) or (<) type. These variables are called artificial variables. The

purpose of introducing the

Linear Programming 1 16b

artificial variables is just to obtain an initial basic feasible solution. They have,

however, two drawbacks:

(i) They are fictitious, have no physical meaning or economic significance and

have no relevance to the problem.

(ii) Their introduction (addition) violates the equality of constraints that has been

already established in step 1.

44

They are, therefore, rightly termed as artificial variables as opposed to other real

decision variables in the problem. Therefore, we must get rid of these variables

and must not allow them to appear in the final solution. To achieve this, these

variables are assigned a very large per unit penalty in the objective function. This

penalty is designated by - M far maximization problems and + M for minimization

problems, where M > 0. Value of M is much higher than the cost coefficients of

other variables and for hand calculations it is not necessary to assign any

specific value to it.

Step 3. Solve the modified linear programming problem by the simplex method.

The artificial variables are a computational device. They keep the starting

equations in balance and provide a mathematical trick for getting a starting

solution. By having a high penalty cost it is ensured that they will not appear in

the final solution i.e., they will be driven to zero when the objective function is

optimized by using the simplex method.

While making iterations, using the simplex method, one of the following three

cases may

arise:

1. If no artificial variable remains in the basis and the optimality condition is

satisfied, then the solution is an optimal feasible solution to the given problem.

Also, the original constraints are consistent and none of them is redundant.

2. u at ieam one artificial variable appears in the basis at zero level (with zero

value in bcolumn) and the optimality condition is satisfied, then the solution is

optimal feasible (though degenerate) solution to the given problem. The

constraints are consistent though redundancy may exist in them. By redundancy

is meant that the problem has more than the required number of constraints.

3. If at least one artificial variable appears in the basis at a non-zero level (with

positive value in b-column) and the optimality condition is satisfied, then the

original problem has no feasible solution; for if a feasible solution existed, the

artificial variables could be driven to zero, yielding an improved value of the

objective function. The problem has no feasible solution either because the

constraints are inconsistent or because there are solutions, but none is feasible.

45

In economic terms this means that the resources of the system are not sufficient

to meet the expected demands. The final solution to the problem is not optimal

since the objective function contains an unknown quantity M. Such a solution

satisfies the constraints but does not optimize the objective function and is also

called pseudo-optimal solution.

Remarks: I. Slack variables are added to (the left-hand sides) the constraints of

(S) type and subtracted from the constraints of (?) type.

2. Artificial variables are added to the constraints of (?) and (_) type. Equality

constraints require neither slack nor surplus variables.

3. Variables, other than the artificial variables, once driven out in an iteration,

may re-enter in a subsequent iteration. But, an artificial variable, once driver, can

never re-enter, because of the large penalty coefficient M associated with it in the

objective function. Advantage can be taken of this fact by not computing its

column in iterations subsequent to the one from which it was driven out.

4. For computer solutions, some specific value has to be assigned to M. Usually,

the largest value that can be represented in the computer is used.

EXAMPLE

Food X contains 6 units of vitamin A per gram and 7 units of vitamin B per

gram and costs !2 paise per gram. Food Y contains 8 units of vitamin A per

gram and ll units of vitamin B per gram and costs 20 poise per gram. The

daily minimum requirement of vitamin A and vitamin B is 100 units and 120

units respectively. Find the minimum cost of product mix by the simplex

method. [RU. B. Com. April, 20117; Meerut M.Com. 19701 Solution. Let

xi and x, be the grams of food X and Y to be purchased. Then the problem can

be formulated as follows :

Minimize Z = 12x1 + 20x2

subject to 6x1 + 8x2 > 100,

7x1 + 12x2 > 120,

x1 + x2 > 0

46

Step 1. Express the problem in standard form

Slack variables s1 and S2 are subtracted from the left-hand sides of the

constraints to convert them to equations. These variables are also called

negative slack variables or surplus variables. , Variable si represents units of

vitamin A in product mix In--excess of the minimum requirement

of 100, s, represents units of vitamin B in produc ~nix in excess of

requirement of 120. Sir.. ° they represent `free' foods, the cost coefficients

ssociated with them in the objective function are zeros. The problem,

therefore, can be writte as follows :

Step 2. Find initial basic feasible solution

Putting x, = xZ = 0, we get s, _ - 100, s2 = - 120 as the first basic solution but it

is not feasible as s, and sZ have negative values that do not satisfy the non-

negativity restrictions. Therefore, we introduce artificial variables A, and AZ in

the constraints, which take the form 6x, + 8xZ -- s, + A, = 100,

7x, + 12x2 - sZ + Az = 120, x,, xz, s,, sZ, Ar, Az >- 0.

Now artificial variables with values greater than zero violate the equality in

constraints established in step I. Therefore, A, and AZ should not appear in the

final solution. To achieve this, they are assigned a large unit penalty (a large

positive value, + M) in the objective function, which can be written as

minimize Z = 12x1 + 20x2 + Os1 + Os2 + MA1 + MA2.

Problem, now, has six variables and two constraints. Four of the variables

have to be zeroised to get initial basic feasible solution to the `artificial

system'. Putting x, = x= = s, = s, = 0, we get

A, = 100, A2 = 120, Z = 220 M.

Note that we are starting with a very heavy cost (compare it with zero profit in

maximization problem) which we shall minimize during the solution procedure.

47

Unit -3

The Transportation Model

3.1 INTRODUCTION TO THE MODEL

In the previous chapter the general nature of the linear programming problem

and its solution by the graphical, simplex and other methods was discussed. It

was stated that the simplex algorithm could be used to solve any linear

programming problem for which the solution exists. However, as the number of

variables and constraints increase, the computation by this method becomes

more and more laborious. Therefore, where-ever possible, we try to simplify the

calculations. One such model requiring simplified calculations is the distribution

model or the transportation model It deals with the transportation of a product

available at several sources to a number of different destinations. The name

"transportation model" is, however, misleading. 'this model can be used for a

value variety of situations such as scheduling, production, investment, plant

location, inventory control, employment scheduling, personnel assignment,

product mix problems and many others, so that the model is really not confined

to transportation or distribution only.

The origin of transportation models dates back to 1941 when F.L. Hitchcock

presented a study entitled `The Distribution of a Product from Several Sources

to Numerous Localities.' The presentation is regarded as the first important

contribution to the solution of transportation problems. In 1947, T.C. Koopmans

presented a study called `Optimum Utilization of the Transportation System'.

These two contributions are mainly responsible for the development of

transportation models which involve a number of shipping sources and A.

number of destinations. Each shipping source has a certain capacity and each

destination has a certain requirement associated with a certain c9st of shipping

from the sources to the destinations. The objective is to minimize the cost of

transportation while meeting the requirements at the destinations.

48

Transportation problems may also involve movement of a product from plants to

warehouses, warehouses to wholesalers, wholesalers to retailers and retailers

to customers.

3.2 ASSUMPTIONS IN THE TRANSPORTATION MODEL

1. Total quantity of the item available at different sources is equal to the total

requirement at different destinations.

2. Item can be transported conveniently from all sources to destinations.

3. The unit transportation cost of the item from all sources to destinations is

certainly and pecisely known.

4. The transportation cost on a given route is directly proportional to the number

of units shipped on that route.

5. "f he objective is to minimize the total transportation cost for the organisation

as a whole and not for individual supply and distribution centres.

3.3 DEFINITION OF THE TRANSPORTATION MODEL

Transportation models deal with problems concerning as to what happens to

the effectiveness function when we associate each of a number of origins

(sources) with each of a possibly different number of destinations (jobs). The

total movement from each origin and the total movement to each destination is

given and it is desired to find how the associations be made subject to the

limitations on totals. In such problems, sources can be divided among the jobs

or jobs may be done with a combination of sources. The distinct feature of

transportation problems is that sources and jobs must be expressed in terms of

only one kind of unit.

Suppose that there are m sources and n destinations. Let a; be the number of

supply units available at source i(i = l, 2, 3, ..., m) and let bj be the number of

demand units required at destination j(i = 1, 2, 3, ..., n). Let cij represent the unit

49

transportation cost for transportating the units from source i to destination j. The

objective is to determine the number of units to be transported from source i to

destination j so that the total transportation cost is minimum. In addition, the

supply limits at the sources and the demand requirements at the destinations

must be satisfied exactly.

If ij(xij > 0) is the number of units shipped from source i to destination j, then the

equivalent linear programming model will be

Find xij (i = 1,2,3,…m: j = 1,2,3,……n) in order to

The two sets of constraints will be consistent i.e.; the system will be in balance

if

Equality sign of the constraints causes one of the constraints to be redundant

(and hence it can be deleted) so that the problem will have (m + n - I)

constraints and (m x n) unknowns. Note that a transportation problem will have

a feasible solution only if the above restriction

is satisfied. Thus, is necessary as well as a sufficient condition

for a

50

transportation problem to have a feasible solution. Problems that satisfy this

condition are called balanced transportation problems. Techniques have been

developed for solving balanced or standard transportation problems only. It

follows that any non-standard problem in which the supplies and demands do

not balance, must be converted to a standard tansportation problem before it

can be solved. This conversion can be achieved by the use of a dummy source/

destination.

The above information can be put in the form of a general matrix shown below:

51

In table 3.1, c„, i = I, 2, ..., m; j = l, 2, ..., n, is the unit shipping cost from the dh

origin to jth destination, x,; is the quantity shipped from the ith origin to jth

destination, a; is the supply available at origin i and b, is the demand at

destination j.

Definitions

A few terms used in connection with transportation models are defined below.

1. Feasible Solution. A feasible solution to a transportation problem is a set of

nonnegative allocations, x,i that satisfies the rim (row and column) restrictions.

2. Basic Feasible Solution. A feasible solution to a transportation problem is

said to be a basic feasible solution if it contains no more than m + n - 1 non-

negative allocations, where m is the number of rows and n is the number of

columns of the transportation problem.

3. Optimal Solution. A feasible solution (not necessarily basic) that minimizes

(maximizes) the transportation cost (profit) is called an optimal solution.

4. Non-degenerate Basic Feasible Solution. A basic feasible solution to a (m x

n) transportation problem is said to be non-degenerate if,

(a) the total number of non-negative allocations is exactly m + n - I (i.e.,

number of independent constraint equations), and

(b) these m + n - I allocations are in independent positions.

5. Degenerate Basic Feasible Solution: A basic feasible solution in which the

total number of non-negatives allocations is less than m + n - 1 is called

degenerate basic feasible solution.-. .

3.4 MATRIX TERMINOLOGY

The matrix used in the transportation models consists of squares called `cells',

which when stacked form `columns' vertically and `rows' horizontally.

The cell located at the intersection of a row and a column is designated by its

row and column headings. Thus the cell located at the intersection of row A and

column 3 is called cell (A, 3). Unit costs are placed in each cell.

52

3.5 FORMULATION AND SOLUTION OF TRANSPORTATION MODELS

In this section we shall consider a few examples which will make clear the

technique of formulation and solution of transportation models.

EXAMPLE 3.5-1 (Transportation Problem)

A dairy firm has three plants located throughout a state. Daily milk production at

each plant is as follows:

Plant 1 ... 6 million litres, plant 2 ... I million litres, and plant 3 ... 10 million litres.

Each day the firm must fulfil the needs of its four distribution centres. Milk

requirement at each centre is as follows:

Distribution centre 1 ... 7 million litres, distribution centre 2 ... 5 million litres,

distribution centre 3 ... 3 million litres, and distribution centre 4 ... 2 million litres.

Cost of shipping one million litres of milk from each plant to each distribution

centre is given in the following table in hundreds of rupees:

Table 3.3 Distribution centres

(i) Show that the problem represents a network situation. (ii) Formulate the

mathematical model forYhe problem.

(iii) The dairy firm wishes to determine as to how much should 6e the shipment

from which milk plant to which distribution centre so that the total cost of

shipment is the minimum. Determine the optimal transportation policy. -

53

Solution. (t) Let us represent the example graphically:

We find that the above situation takes the shape of a network. (it)

Formulation of Model

Step 1:

Key decision to be made is to find how much quantity of milk from which

plant to which distribution centre be shipped so as to satisfy the

constraints and minimize the cost. Thus the variables in the situation are:

x11, x12, x13, x14, x21, x22, x23, x24, x31, x32, x33, and x34. These variables

represent the quantities of milk to be shipped from different plants to

different distribution centres and can be represented in the form of a

matrix shown below:

54

In general, we can say that the key decision to be made is to find the

quantity of units to be transported from each origin to each destination.

Thus, if there are in origins and r: destinations, then .xy are the decision

variables (quantities to be found), where

i = 1, 2,…..m,

and j = 1, 2, ..., n.

Step 2:

Feasible alternatives are sets of values of xij, where xij > 0.

Step 3:

Objective is to minimize the cost of transportation.

In general, we can say that if c, i is the unit cost of shipping from ith source

to jib destination, the objective is

Step 4: Constraints are

(i) because of availability or supply:

Thus, in all, there are 3 constraints (equal to the number of plants).

In general, there will be m constraints if number of origins is m, which can

be expressed as

55

(ii) because of requirement or demand

In general, there are n constraints if the number of destinations is n, which can

be expressed as

Thus we find that the given situation involves (3 x 4 = 12) variables and (3 + 4

= 7) constraints. In general, such a situation will involve (In x n) variables and

(m + n) constraints. However, because the transportation model is always

balanced, one of these constraints must be redundant. Thus, the model has or

+ n - 1 independent constraint equations, which means that the starting basic

feasible solution consists of m + n - l basic variables.

It can be easily seen that in this model the objective function as well as the

constraints are linear functions of the variables and, therefore, the model can

be solved by simplex method. However, as a large number of variables are

involved, computations required will be much more. The following points may

be noted in a transportation model:

1. All supply as well as demand constraints are of equality type.

2. They are expressed in terms of only one kind of unit.

3. Each variable occurs only once in the supply constraints and only once in the

demand constaints.

4. Each variable in the constraints has unit coefficient only.

Therefore, the transportation model is a special case of general L.R model

where in the above four conditions hold good and can be solved by a special

56

technique called the transportation technique which is easier and shorter than

the simples technique.

(iii) Solution of the Transportation Model

The solution involves making a transportation model (in the form of a matrix),

finding a feasible solution, performing optimality test and iterating towards

optimal solution if required.

Step 1: Make a Transportation Model

This consists in expressing supply from origins, requirements at destinations

and cost of shipping from origins to destinations in the form of a matrix shown

below.

A check is made to find if total supply and demand are equal. If yes, the

problem is said to be a balanced or self contained or standard problena. If

not, a dummy origin or destination (as the case may be) is added to balance the

supply and demand. Table 3.5 represents the transportation table for the given

problem.

Step II: Find a Basic Feasible Solution

This can be easily obtained by applying a technique which has' been developed

by Dantzig and which Charnes and Cooper refer to as "the north-west comer

rule". Other methods for finding the initial feasible solutions are also described. In

all these techniques it is assumed at the beginning that the transportation table is

blank i.e., initially all x; = 0.

57

The difference among these methods is the "quality" of the initial basic feasible

solution they produce, in the sense that a better starting solution will involve a

smaller objective value (minimization problem). In general, the Vogel's

approximation method yields the best starting solution and the north-west corner

method yields the worst. However, the latter is easier, quick and involves the

least computations to get the initial solution.

North-West Corner Rule or North-West Corner Method (NWCM)

This rule may be stated as follows:

(i) Start in the north-west (upper left) corner of the requirements table i.e., the

transportation matrix framed in step I and compare the supply of plant 1 (call it

S1) with the requirement of distribution centre 1 (call it DI).

(a) If Di < S, i.e., if the amount required at Di is less than the number of units

available at S,, set xii equal to DI, find the balance supply and demand and

proceed to cell (l, 2) (i.e., proceed horizontally).

(b) If D1, = S1, set xi i equal to D1, compute the balance supply and demand and

proceed to cell (2, 2) (i.e., proceed diagonally). Also make a zero allocation to the

least cost cell in S1 /D1.

(c) If D1, > S1, set.ri; equal to S1, compute the balance supply and demand and

proceed to cell (2, 1) (i.e., proceed vertically).

(ii) Continue in this manner, step by step, away from the north-west corner until,

finally, a value is reached in the south-east corner.

Thus in the present example (see table 3.6), one proceeds as follows:

58

(i) set x, I equal to 6, namely, the smaller of the amounts avaiblable at S, (6) and

that needed at D, (7) and

(ii) proceed to cell (2, 1) (rule c). Compare the number of units available at S2

(namely 1) with the amount required at D i (1) and accordingly set x2 i = 1. Also

set X22 = 0 as per rule (b) above.

(iii) proceed to cell (3, 2) (rule b). Now supply from plant S3 is 10 units while the

demand for D2 is 5 units; accordingly set x3Z equal to 5.

(iv) proceed to cell (3, 3) (rule a) and allocate 3 there.

(v) proceed to cell (3, 4) (rule a) and allocate 2 there.

It can be easily seen that the proposed solution is a feasible solution since all

supply and demand constraints are fully satisfied.

The following points may be noted in connection with this method:

(i) The quantities allocated are put in parenthesis and they represent the values

of the corresponding decision variables. These cells are called basic or allocated

or occupied or loaded cells. Cells without allocations are called non-basic or

vacant or empty or unoccupied or unloaded cells. Values of the corresponding

variables are all zero in these cells.

(ii) This method of allocation does not take into account the transportation cost

and, therefore, may not yield a good (most economical) initial solution. The

transportation cost associated with this solution is

e = Rs. [2 x 6 + 1 x 1 + 8 x 5 + 15 x 3 + 9 x 2] x 100 = Rs. 11,600.

(2) Row Minima Method

This method consists in allocating as much as possible in the lowest cost cell of

the first centre so that either the capacity of the first plant is exhausted or the

requirement at jth distribution centre is satisfied or both. In case of tie among the

cost, select arbitrarily. Three cases arise: (i) if the capacity of the first plant is

completely exhausted, cross off the first row and proceed to the second row.

59

(ii) if the requirement at jth distribution centre is satisfied, cross off the jth column

and reconsider the first row with the remaining capacity.

(iii) if the capacity of the first plant as well as the requirement at jth distribution

centre are completely satisfied, make a zero allocation in the second lowest cost

cell of the first row. Cross off the row as well as the jth column and move down to

the second row.

Continue the process for the resulting reduced transportation table until all the

rim conditions (supply and requirement conditions) are satisfied.

In this problem, we first allocate to cell (1, 1) in the first row as it contains the

minimum cost 2. We allocate min. (6, 7) (6) in this cell. This exhausts the supply

capacity of plant I and thus the first row is crossed off. The next allocation, in the

resulting reduced matrix is made in cell (2, 2) of row 2 as it contains the minimum

cost 0 in that row. We allocate min. (1, 5) (1) in this cell. This exhausts the supply

capacity of plant 2 and thus the second row is crossed off. The next allocation, in

the resulting reduced matrix is made in cell (3, 1) of row 3 as it contains the

minimum cost 5 in that row. We allocate min. (1, 10) (1) in this cell. This exhausts

the requirement condition of distribution centre 1 and hence the first column is

crossed off. Proceeding in this way we allocate (4), (2) and (3) units to cells (3,

60

2), (3, 4) and (3, 3) till all the rim conditions are met with. The resulting matrix is

shown in table 3.7.

The transportation cost associated with this solution is

Z = Rs. [2 x 6 + 0 x 1 + 5 x 1 + 8 x 4 + 15 x 3 + 9 x 2] x 100 = Rs. 11,200, which

is less than the cost associated with solution obtained by N-W corner method.

(3) Column Minima Method

This method consists in allocating as much as possible in the lowest cost cell of

the first column so that either the demand of the first distribution centre is

satisfied or the capacity of the ith plant is exhausted or both. In case of tie

among the lowest cost cells in the column, select arbitrarily. Three cases arise:

(i) if the requirement of the first distribution centre is satisfied, cross off the first

column and move right to the second column. -

(ii) if the capacity of ith plant is satisfied, cross off ith row and reconsider the

first column with the remaining requirement.

(iii) if the requirement of the first distribution centre as well as the capacity of

the ith plant are completely satisfied, make a zero allocation in the second

lowest cost cell of the first column. Cross off the column as well as the ith row

and move right to the second column.

Continue the process for the resulting reduced transportation table until all the

rim conditions are satisfied.

61

In the given problem we allocate first to cell (2, 1) in the first column as it

contains the minimum cost 1. We allocate min. (1, 7) = (1) in this cell. This

exhausts the supply capacity of plant 2 and thus the second row is crossed off.

The next allocation in the resulting reduced matrix

is made in cell (1, 1) of column 1 as it contains the second lowest cost 2 in that

column. We allocate min. (6, 6) = (6) in this cell. This exhausts the supply

capacity of plant 1 as well as the requirement of distribution centre 1. Therefore,

we allocate zero in cell (3, 1) of the first column, cross off first row and first

column and move on to the second column. Proceeding in this way we allocate

(5), (3) and (2) to cells (3, 2), (3, 3) and (3, 4) till all the rim conditions are met

with. The resulting matrix is shown in table 3.8.

The transportation cost associated. with this solution is

Z= Rs. [2 x 6 + 1 x 1 + s x 0 + 8 x 5 + IS x 3 + 9 x 2] x 100 = Rs. 11,600, which

is same as the cost associated with solution obtained by N-W corner method.

(4) Least-Cost Method (or Matrix Minima Method or Lowest Cost Entry

Method)

This method consists in allocating as much as possible in the lowest cost

cell/cells and then further allocation is done in the cell/cells with second lowest

62

cost and so on. In case of tie among the cost, select the cell where

allocation of more number of units can be made. Consider the matrix for

the problem under study.

Here, the lowest cost cell is (2, 2) and maximum possible allocation (meeting

supply and requirement positions) is made here. Evidently, maximum

feasible allocation in cell (2, 2) is (1). This meets the supply position of plant

2. Therefore, row 2 is crossed out, indicating that no allocations are to be

made in cells (2, 1), (2, 3) and (2, 4).

The next lowest cost cell (excluding the cells in row 2) is (1, 1); maximum

possible allocation of (6) is made here and row I is crossed out. Next lowest

cost cell in row 3 is (3, 1) and allocation of (1) is done here. Likewise,

allocations of (4), (2) and (3) are done in cells (3, 2), (3, 4) and (3, 3)

respectively. The transportation cost associated with this solution is

Z =Rs.(2x6+0x1+5x1+8x4+15x3+9x2)x100

=Rs.(12+0+5-'-32+45+18)x100=Rs.11,200,

which is less than the cost associated with the solution obtained by N-W

corner method.

(5) Vogel's Approximation Method (VAM) or Penalty Method or Regret

Method

Vogel's approximation method is a heuristic method and is preferred to the

methods described above. In the transportation matrix if an allocation is

made in the second lowest cost cell instead of the lowest, then this allocation

will have associated with it a penalty corresponding to the difference of these

two costs due to `loss of advantage'. That is to say, if we compute the

63

difference between the two lowest costs for each row and column, we find

the opportunity cost relevant to each row and column. It would be most

economical to make allocation against the row or column with the highest

opportunity cost. For a given row or column, the allocation should obviously

be made in the least cost cell of that row or column. Vogel's approximation

method, therefore, makes effective use of the cost information and yields a

better initial solution than obtained by the other methods. This method

consists of the following substeps:

Substep Write down the cost matrix as shown below.

Enter the difference between the smallest and second smallest element in

each column below the corresponding column and the difference between

the smallest and second smallest element in each row to the right of 'he row.

Put these numbers in brackets as shown. For example, in column 1, the two

lowest elements are 1 and 2 and their difference is 1 which is entered as [1]

below column l. Similarly, the two smallest elements in row 2 are 0 and 1 and

their difference 1 is entered as [1] to the right of row 2. A row or column

"difference" can be thought of a penalty for making allocation in second

smallest cost cell instead of smallest cost cell. In other words this difference

indicates the unit penalty incurred by failing to make an allocation to the

smallest cost cell in that row or column. In case the smallest and second

smallest elements in a row/column are equal, the penalty should be taken as

zero.

64

Substep 1: Select the row or column with the greatest difference and allocate

as much as possible within the restrictions of the rim conditions to the lowest

cost cell in the row or column selected.

In case of tie among the highest penalties, select the row or column having

minimum cost. In case of tie in the minimum cost also, select the cell which can

have maximum allocation. If there is tie among maximum allocation cells also,

select the cell arbitrarily for allocation. Following these rules yields the best

possible initial basic feasible solution and reduces the number of iterations

required to reach the optimal solution.

Thus since [6] is the greatest number in brackets, we choose column 4 and

allocate as much as possible to the cell (2, 4) as it has the lowest cost 1 in

column 4. Since supply is 1 while the requirement is 2, maximum possible

allocation is (1).

Substep 3: Cross out of the row or column completely satisfied by the allocation

just made. For the assignment just made at (2, 4), supply of plant 2 is

completely satisfied. So, row 2 is crossed out and the shrunken matrix is written

below.

This matrix consists of the rows and columns where allocations have not yet

been made, including revised row and column totals which take the already

made allocation into account. Substep 4: Repeat steps 1 to 3 until all

assignments have been made.

(a) Column 2 exhibits the greatest difference of [5]. Therefore, we allocate (5)

units to cell (1, 2), since it has the smallest transportation cost in column 2.

65

Since requirements of column 2 are completely satisfied, this column is crossed

out and the reduced matrix is written again as fable 3.12.

(b) Differences are recalculated. The maximum difference is [5]. Therefore, we

allocate (1) to the cell (1, I) since it has the lowest cost in row 1. Since

requirements of row 1 are fully satisfied, it is crossed out and the reduced

matrix is written below.

In table 3.13, it is, possible to find row differences but it is not possible to find

column differences. Therefore, remaining allocations in this table are made by

following the least cost method,

(c) As cell (3, 1) has the lowest cost 5, maximum possible allocation of (6) is

made here. Likewise, next allocation of (1) is made in cell (3, 4) and (3) in cell

(3, 3) as shown.

All allocations made during the above procedure are shown below in

thelallocation matrix.

66

The above repetitions can be made in a single matrix as shown in table 3.15.

Table 3.15

The cost of transportation associated with the above solution is

Z=Rs.(2x1+3x5+1 x 1+5x6+15x3+9x1) x100

= Rs. (2 + 15 + I + 30 + 45 + 9) x 100 = Rs. 10, 200,

which is evidently the least of all the values of transportation cost found by

different methods. Since Vogel's approximation method results in the most

economical initial feasible solution, we shall use this method for finding such a

solution for all transportation problems henceforth.

Step III: Perform Optimality Test

Make an optimality test to find whether the obtained feasible solution is optimal

or not. An optimality test can, of course, be performed only on that feasible

'solution in which

(a) number of allocations is m + n - I, where m is the number of rows and n is

the number of columns. [n,the given situation, m = 3 and n = 4 and number of

67

allocations is 6 which is equal to (m + n - 1) (3 + 4 - 1 = 6). Hence optimality

test can be performed.

(b) these (m + n - I ) allocations should be in independent positions.

A look at the feasible solution of the problem under consideration indicates that

all the allocations are in independent positions as it is impossible to increase or

decrease any allocation without either changing the position of the allocations

or violating the row and column restrictions. For example, if the allocation in cell

(l, 1) is changed from (I) to (3), the allocation in cell (1, 2) must be changed

from (5) to (3) in order to satisfy the row restriction. Similarly, the allocation in

cell (3, 1) must be changed from (6) to (4) in order to meet the column

restriction. This will, in turn, require changes in the allocations of cell (3, 3)

and/or cell (3, 4).

A simple rule for allocations to be in independent positions is that it is

impossible to travel from any allocation, back to itself by a series of horizontal

and vertical jumps from -o&' occupied cell to another, without a direct reversal

of route. For instance, the occupied cells in table 3.16 are not in independent

positions because the cells (2, 2), (2, 3), (3, 3) and (3, 2) from a closed loop.

68

Now test procedure for optimality involves examination of each vacant cell to

find whether or not making an allocation in it reduces the total transportation

cost. The two methods commonly used for this purpose are the stepping-stone

method and the modified distribution (MODI) method.

1. The stepping-Stone Method

Consider the matrix giving the initial feasible solution for the problem under

consideration. Let us start with any arbitraty empty cell (a cell without

allocation), say (3, 2) and allocate + 1 unit to this cell. As already discussed, in

order to keep up the column 2 restriction, - 1 must be allocated to cell (1, 2) and

to keep up the row i restriction, + 1 must be allocated to cell (1,.1) and

consequently- 1 must be allocated to cell (3, 1); this is shown in the matrix

below.

The net change in transportation cost as a result of this perturbation is called

the evaluation of the empty cell in question.

Evaluation of cell (3, 2) = Rs. 100 x (8 x 1 - 5 x 1 + 2 x 1 - 5 x 1)

= Rs. (0 x 100) = Rs. 0.

Thus the total transportation cost increases by Rs. 0 for each unit allocated to

cell (3, 2). Likewise, the net evaluation (also called opportunity cost) is

calculated for every empty cell. For this the following simple procedure may be

adopted.

Starting from the chosen empty cell, trace a path m the matrix consisting of a

series of alternate horizontal and vertical lines. The path begins and

69

terminates in the chosen cell. All other corners of the path lie in the cells for

which allocations have been made. The path rnay skip over any number of

occupied or vacant cells. Mark the corner of the path in the chosen vacant cell

as positive and other corners of the path alternately -ve, +ve, -ve and so on.

Allocate I unit to the chosen cell; subtract and add I unit from the cells at the

comers of the path, maintaining the row and column requirements. The net

cnange in the total cost resulting from this adjustment is called the evaluation

of the chosen empty cell. Evaluations of the various empty cells (in hundreds

of rupees) are:

If any cell evaluation is negative, the cost can be reduced so that the solution

under consideration can be improved i.e., it is not optimal. On the other hand,

if all cell evaluations are positive or zero, the solution in question will be

optimal. Since evaluations of cells (1, 3) and (2, 3) are -ve, initial basic

feasible solution given in table 3.15 is not optimal.

Now in a transportation problem involving m rows and n columns, the total

number. of empty cells will be m.n - (m + n - 1) = (m - 1)(n - 1). Therefore,

there are (m - 1)(n - 1) such cell evaluations which must be calculated and for

large problems, the method can be quite inefficient. This method is named

'stepping-stone' since only occupied cells or `stepping stones' are used in the

evaluation of vacant cells.

2. The Modified Distribution (MODI) Method or the u-v Method

70

In the stepping-stone method, a closed path is traced for each unoccupied

cell. Cell evaluations are found and the cell with the most negative evaluation

becomes the basic cell. In the modified distribution method, cell evaluations of

all the unoccupied cells are calculated simultaneously and only one closed

path for the most negative cell is traced. Thus it provides considerable time

saving over the stepping-stone method. This method consists of the following

substeps:

Substep 1: Set-up a cost matrix containing the unit costs associated with the

cells for which allocations have been made. This matrix for the present

example is

Substep 2: Introduce dual variables corresponding to the supply and demand

constraints. If there are m origins and n destinations, there will be m + n dual

variables. Let u;(i = 1, 2, ..., m) and v j (j = I, 2, ..., n) be the dual variables

corresponding to supply and demand constraints. Variables ui and vj are such

that ui+vj = cij for all occupied cells. . .

Therefore, enter a set of numbers u; (i = 1, 2, 3) along the left of the matrix

and vj (j = 1, 2, 3, 4) across the top of the matrix so that their sums equal the

costs entered in substep 1.

Thus,

ui + vj =2

ul + v2 = 3,

u2 + v4 = 1,

u3 + v1 = 5,

u3 + v3 = 15,

and u3 + v4 = 9.

71

Since number of dual variables are m + n (3 + 4 = 7 in the present problem)

and number of allocations (in a non-degenerate solution) are m + n - 1 (3 + 4

- 1 = 6 in the present problem), one variable is assumed arbitrarily. Let v i =

0. Therefore, from the above equations

u1=2,v2= 1, u3=5, v3=10, v4 = 4, u2 = -3.

The values of these dual variables satisfy the complementary slackness

theorem which states that if primal constraints are equations, dual variables

are unrestricted in sign (Refer section 6.1.3). Therefore, the matrix may be

written

Now for any vacant (empty) cell, u i + vj is called the implicit cost, whereas c ij;

is called the actual cost of the cell. The two costs are compared and c ij (ui +

vj) are calculated for each empty cell. If all c ij - (ui + vj) > 0, then by the

application of complementary slackness theorem it can

be shown that the corresponding solution is optimum. If any c ij - (u; + v,) c0,

the solution is not optimal. c i (ui + vj) is called the evaluation of the cell (i, j)

or opportunity cost of cell (i, j). Thus we have the following three substeps:

Substep 3: Fill the vacant cells with the sums of u; and y. This is shown in

table 3.20.

72

Snbstep J: Subtract the cell values of the matrix of substep 3 from the

original cost matrix.

The resulting matrix is called cell evaluation matrix.

Substep S: Signs of the values in the cell evaluation matrix indicate whether

optimal solution has been obtained or not. The signs have the following

significance:

(a) A negative value in an unoccupied cell indicates that a better solution can

be obtained by allocating units to this cell.

(6) A positive value in an unoccupied cell indicates that a poorer solution will

result by allocating units to the cell.

(c) A zero value in an unoccupied cell indicates that another solution of the

same total value can be obtained by allocating units to this cell. In the present

example since two cell evaluations are negative. it is possible to obtain a better

solution by making these cells as basic cells.

Step IV: Iterate Towards an Optimal Solution

This involves the following substeps:

73

Substep'1 From the cell evaluation matrix, identify the cell with the most

negative cell evaluation. This is the rate by which total transportation cost can

be reduced if one unit is allocated this cell; in case more units are allocated, the

cost will come down proportionately. Therefore,-as many units as possible

(keeping in mind the rim conditions) will be allocated to this cell to bring down

the cost by maximum amount. !n ease of tie in the cell evaluation, the cell

wherein maximum allocation can be made is selected. This cell is now called

the identified cell. With reference to the simplex method, this identified cell is

currently the non-basic cell that has been decided to be made basic (decided to

enter the solution) by making allocation in it. [n the present problem both the

tied cells will have the same maximum allocation of 1 unit. Hence cell (1, 3) is

selected arbitrarily.

Substep 2: Write down again the initial basic feasible solution that is to be

improved. Check mark (V) the identified cell. This is shown in table 3.23.

Having decided the vacant cell that is to be made basic, the next thing is to

decide which basic cell should be made non-basic by changing its present

allocation to zero. For this we go to substep 3.

Substep 3: Trace a closed path in the matrix. This closed path has the following

characteristics:

(i) It begins and terminates in the identified cell.

(ii) It consists of 9 series of alternate horizontal and vertical lines only (no

diagonals). (iii) It can be traced clockwise or anticlockwise.

74

(iv) All other corners of the path lie in the allocated cells only.

(v) The path may skip over any number of allocated or vacant cells.

(vi) There will always be one and only one closed path, which may be traced.

The closed path has even number of corners (4, 6, 8, ...) and any allocated cell

can be considered only once. The closed path may or may not be square or

rectangular in shape; it may have a peculiar configuration and the lines may

even cross over.

Substep 4: Mark the identified cell as positive and each occupied cell at the

corners of the path alternately -ve, +ve, -ve and so on.

Substep 5: Make a new allocation in the identified cell by entering the smallest

allocation on the path that has been assigned a-ve sign. Add and subtract this

new allocation from the cells at the corners of the path, maintaining the row and

column requirements. This causes one basic cell to become zero and other

cells remain non-negative. The basic cell whose allocation has been made

zero, leaves the solution.

Since cell evaluation (in hundreds of rupees) is - 1 and 1 unit has been

reallocated, the total transportation cost should come down by Rs. (100 x 1) =

Rs. 100: This can be vertified by actually calculating the total cost for table

3.25.

The total cost of transportation for this 2nd feasible solution is

= Rs.(3x5+11 xl+l x 1+7x5+2x 15+1 x9)x 100

= Rs. (15 + 11 + 1 + 35 + 30 + 9) x 100

75

= Rs. 10,100, which is less than for the first (starting) feasible solution by Rs.

100.

Step V: Check for Optimality

Let us check whether the solution obtained above is optimal or not. This shall

be checked by repeating the steps under `check for optimality' already made. In

the above feasible solution, (a) number of allocations is (m + n - 1) i.e., 6,

(b) these (m + n - 1) allocations are in independent positions. -

Above conditions being satisfied, an optimality test can be performed as

follows:

Substep l: Set.up the cost matrix containing the costs associated with the cells

for which allocations have been made.

Substep 2: Enter a set of numbers vj along the top of the matrix and a set of

number: u; at the left side se that their sum is equal to costs entered in matrix of

substep 1, shown below:-

These values are shown entered in matrix 3.26.

76

Substep 3: Fill the vacant cells with the sums of u i and vj.

Substep 4: Subtract the cell values of this matrix from the original cost

matrix. Table 3.28

This matrix 3.29 is called cell evaluation matrix.

Substep S: Since one cell value is -ve, the 2nd feasible solution is not

optimal.

Step VI: Iterate Towards an Optimal Solution

This involves the following substeps:

Substep 1: In the cell evaluation matrix, identify the cell with the most

negative entry. It is the cell (2, 3).

Substep 2: Write down again the feasible solution in question.

77

Mark the empty cell (J) for which the evaluation is negative, This is called

identified cell. Substep 3: Trace the path shown in the matrix.

Substep 4'. Mark the identified cell as +ve and others alternately -ve and

+ve.

Substep 5: Make the new allocation in the identified cell by entering the

smallest allocation on the path which has been assigned negative sign.

Subtract and add this amount from other cells. Tables 3.31 and 3.32 result.

Table 3.31

For this allocation matrix the transportation cost is

Z=Rs.(5x3+1x11+1x6+1x15+2x9+7x5)x100=Rs.10,000.

Thus it is a better solution. Let us see if it is an optimal solution.

Step VII: Test for Optimality

In the above feasible solution

(a) number of allocations is m + -n - 1 i.e., 6.

(b) These m + n - 1 allocations are in independent positions. Hence repeat

the following substeps:

Substep 1: Set-up the cost matrix containing costs associated with cells for

which allocations have been made. This is table 3.33.

78

Substep 3: Fill up the vacant cells also as shown above.

Substep 4: Subtract the cell values of the above matrix from the original

cost matrix. Tables 3.35 and 3.36 result.

Substep 5: Since all the cell values are positive, the third feasible solution given

by table 3.37 is the optimal solution.

79

Therefore the optimal solution is:

Milk plant Distribut1 2 3

ion

centre

2

No. of units

transported

5

Transportation

Cost/unit (Rs.)

300

Total transportation

cost (Rs.)

1,500

3 1 1,100 1,100

3 1 600 600

1 7 ;00 3,500

3 1 1,500 1,500

4 2 900 1,800

Rs. 10,000

80

UNIT 4 :

Assignment Problem

4.1 DEFINITION OF THE ASSIGNMENT MODEL

An assignment problem concerns as to what happens to the effectiveness

function when we associate each of a number of `origins' with each of the

same number of `destinations'. Each resource or facility (origin) is to be

associated with one and only one job (destination) and associations are to

be made in such a way so as to maximize (or minimize) the total

effectiveness. Resources are not divisible among jobs, nor are jobs divisible

among resources.

The assignment problem may be defined as follows:

Given n facilities and n jobs and given the effectiveness of each facility for

each job, the problem is to assign each facility to one and only one job so as

to optimize the given measure of effectiveness.

Table 4.1 represents the assignment of n facilities (machines) to njobs c ij is

cost of assigning ith facility to jth job and c ij represents the assignment of ith

facility to jib job. If ith facility can be assigned to jib job, xij = 1 otherwise

zero. The objective is to make assignments that minimize the total

assignment cost or maximize the total associated gain.,

Thus an assignment problem can be represented by n x n matrix which

constitutes n: possible ways of making assignments. One obvious way to

find the optimal solution is to write all the n! possible arrangements, evaluate

the cost of each and select the one involving the minimum cost. However, this

enumeration method is extremely slow and time consuming even for small

81

values of n. For example, for n = 10, a common situation, the number of

possible arrangements is 10! = 3,628,800. Evaluation of so large a number of

arrangements will take a prohibitively large time. This confirms the need for an

efficient computational technique for solving such problems.

4.2 MATHEMATICAL REPRESENTATION OF THE ASSIGNMENT MODEL

Mathematically, the assignment model can be expressed as follows:

Let xij denote the assignment of facility i to job j such that

We see that if the last condition is replaced by xij ? 0, we have transportation

model with all requirements and available resources equal to 1.

4.3 COMPARISON WITH THE TRANSPORTATION MODEL

An assignment model may be regarr:,5d as a special case of the transportation

model. Here, (refer table 4.1) facilities represent the 'sources' and jobs

represent the `destinations'. Number of sources is equal to the number of

destinations, supply at each source is unity (a; = 1 for all a) and demand at

each destination is also unity (b, = 1, for all j). The cost of `transporting'

(assigning) facility i to job j is c;; and the number of units allocated to a cell can

be either one or zero, i. e. they are non-negative quantities.

However the transportation algorithm is not very useful to solve this model

because of degeneracy. In this model, when an assignment is made, the row as

82

well as column requirements are satisfied simultaneously (rim conditions being

always unity), resulting in degeneracy. Thus the assignment problem is a

completely degenerate form of the transportation problem. In n x n problem,

there will be n assignments instead of n + n - I or 2n - 1 and we will have to fill

in 2n - I - n = n - 1 epsilons which will make the computations quite

combersome. However, the special structure of the assignment model allows a

more convenient and simple method of solution.

The technique used for solving assignment model makes use of two theorems:

4.4 SOLUTION OF THE ASSIGNMENT MODELS

The technique of solution of the assignment models will be made clear now.

Since the solution applies the concept of opportunity costs, a brief description

of this concept may be useful. The cost of any action consists of opportunities

that are sacrificed in taking that action. Consider the following table which

contains the cost in rupees of processing each ofjobs A, e and Con machines

Suppose it is decided to process job A on machine X. The table shows that the

cost of this assignment is Rs. 25. Since machine Y could just as well process job

A for Rs. 15, clearly assigning job A to machine X is not the best decision.

Therefore, when job A is arbitrarily assigned to machine X, it is done by

sacrificing the opportunity to save Rs. 10 (Rs. 25 - Rs. 15). The sacrifice is

referred to as an opportunity cost. The decision to process job A on machine X

precludes the assignment of this job to machine Y, given the constraint that one

and only one job can be assigned to a machine. Thus opportunity cost of

83

assignment of job A to machine X is Rs. 10 with respect to the lowest cost

assignment for job A. Likewise, a decision to assign job A to machine Z would

involve an opportunity cost of Rs. 7 (Rs. 22 - Rs. 15). Finally, since assignment

of job A to machine Y is the best assignment, the opportunity cost of this

assignment is zero (Rs. 15 - Rs. 15). More precisely these costs can be called

the machine-opportunity costs with regard to job A. Similarly, if the lowest cost of

row B is subtracted from all the costs in this row, we would have the machine-

opportunity costs with regard to jab B. The same step in row C would give the

machine-opportunity costs for job C. This is represented in the following table:

In addition to these machine-opportunity costs, there are job-opportunity costs

also. Job A, B or C, for instance, could be assigned to machine X. The

assignment of job B to machine X involves a cost of Rs. 31, while the assignment

of job A to machine X costs only Rs. 25. Therefore, the opportunity cost of

assigning job B to machine X is Rs. 6 (Rs. 31 - Rs. 25). Similarly, the opportunity

cost of assigning job C to machine X is Rs. 10 (Rs. 35 - Rs. 25). A zero

opportunity cost is involved in the assignment of job A to machine X, since this is

the best assignment for machine X (column X). Hence job-opportunity costs for

each column (each machine) are obtained by subtracting the lowest cost entry in

each column from all the cost entries in that column. If the lowest entry in each

column of table 4.3 is subtracted from all the cost entries of that column, the

resulting table is called total opportunity cost table.

84

It may be recalled that the objective is to assign the jobs to the machines so as to

minimize total costs. With the total opportunity cost table this objective will be

achieved if the jobs are assigned to the machines in such a way as to obtain a

total opportunity cost of zero. The total opportunity cost table contains four cells

with zeros, each indicating a zero opportunity cost for ±at cell (assignment).

Hence job A could be assigned to machine X or Y and job B to machine Z all

assignments having zero opportunity costs. This way job C, however, could not

be assigned to any machine with a zero opportunity cost since assignment of job

B to machine Z precludes the assignment of job C to this machine. Clearly, to

make an optimal assignment of the three jobs to the three machines, there must

be three zero cells in the table such that a complete assignment to these cells

can be made with a total opportunity cost of zero.

There is, in fact, a convenient method for determining whether an optimal

assignment can be made. This method consists of drawing minimum number of

lines covering all zero cells in the total opportunity cost table. If the minimum

number of lines equals the number of rows (or columns) in the table, an optimal

assignment can be made and the problem is solved. If, however, the minimum

number of lines is less than the number of rows (or columns), an optimal

assignment cannot be made. In this case there is need to develop a new total

opportunity cost table. In the present example, since it requires only two lines to

cross (cover) all zeros, and there are three rows, an optimal assignment is not

possible. Clearly, there is a need to modify the total opportunity cost table by

including some assignment not in the rows and columns covered by the lines. Of

85

course, the assignment chosen should have the least opportunity cost. In the

present case it is the assignment of job B to machine Y with an opportunity cost

of 1. In other words, we would like to change the opportunity cost for this

assignment from 1 to zero.

To accomplish this we (a) choose the smallest element in the table not covered

by a straight line and subtract this element from all other elements not having a

line through them (b) add this smallest elemenf to all elements lying at the

intersection of any two lines. The revised total opportunity cost table is shown

below.

The test for optimal assignment described above is applied again to the revised

opportunity cost table. As the minimum number of lines covering all zeros is three

and there are three rows (or columns), an optimal assignment can be made. The

optimal assignments are A to X, B to Y and C to Z.

In larger problems, however, the assignments may not be readily apparent and

there is need for more systematic procedure.

86

4.5 THE HUNGARIAN METHOD FOR SOLUTION OF THE ASSIGNMENT

PROBLEMS

The Hungarian method suggested by Mr. Koning of Hungary or the Reduced

matrix method or the Flood's technique is used for solving assignment problems

since it is quite efficient and results in subtantial time saving over the other

techniques. It involves a rapid reduction of the original matrix and finding of a set

of n independent zeros, one in each row and column, which results in an optimal

solution. The method consists of the following steps:

1. Prepare a square matrix. This step will not be required for n x n assignment

problems. For m x n (m # n) problems, a dummy column or a dummy row, as the

case may be, is added to make the matrix square.

2. Reduce the matrix. Subtract the smallest element of each row from all the

elements of the row. So there will be at least one zero in each row. Examine if

there is at least one zero in each column. If not, subtract the minimum element of

the column(s) not containing zero from all the elements of that column(s). This

step reduces the elements of the matrix until zeros, called zero opportunity costs,

are obtained in each column.

3. Cheek whether an optimal assignment can be made in the reduced matrix or

not. For this (a) Examine rows successively until a row with exactly one

unmarked zero is obtained. Make an assignment to this single zero by making

square (0) around it. Cross (x) all other zeros in the same column as they will not

be considered for making any more assignment S in that column. Proceed in

this way until all rows have been examined.

(b) Now examine columns successively until a column with exactly one

unmarked zero is found. Make an assignment there by making a square (E])

around it and cross (x) any other zeros in the same row.

In case there is no row or column containing single unmarked zero (they contain

more than one unmarked zero), mark square (0) around any unmarked zero

arbitrarily and cross (x) all other zeros in its row and column Proceed in this

manner till there is no unmarked zero left in the cost matrix.

Repeat sub-steps (a) and (b) till one of the following two things occur:

87

(i) There is one assignment in each row and in each column. In this case the

optimal assignment can be made in the current solution, i.e. the current feasible

solution is an optimal solution. The minimum number of lines crossing all zeros is

n, the order of the matrix.

(ii) There is some row and/or column without assignment. In this case optimal

assignment cannot be made in the current solution. The minimum number of

lines crossing all zeros have to be obtained in this case by following step 4.

4. Find the minimum number of lines crossing all zeros. This consists of the

following sub

steps:

(a) Mark (√) the rows that do not have assignments.

(b) Mark (√) the columns (not already marked) that have zeros in marked rows.

(c) Mark (√) the rows (not already marked) that have assignments in marked

columns.

(d) Repeat sub-steps (b) and (c) till no more rows or columns can be marked.

(e) Draw straight lines through all unmarked rows and marked columns. This

gives the minimum number of lines crossing all zeros. If this number is equal to

the order of the matrix, then it is an optimal solution, otherwise go to step 5.

5. Iterate towards the optimal solution. Examine the uncovered elements. Select

the smallest dement and subtract it from all the uncovered elements. Add this

smallest element to every dewient that lies at the intersection of two lines. Leave

the remaining elements of the matrix as =wk This yields second basic feasible

solution.

6. Repeat steps 3 through 5 successively until the number of lines crossing all

zeros becomes equal to the order of the matrix. In such a case every row and

column will have one assignment. This indicates that an optimal solution has

been obtained. The total cost associated with this solution is obtained by adding

the original costs in the assigned cells.

88

Flow chart of these steps is shown in Fig. 4.1 below.

89

4.6 FORMULATION AND SOLUTION OF THE ASSIGNMENT MODELS

In this section we shall consider a few examples which will make clear the

techniques of formulation and solution of the assignment models.

The Assignment Model ) 329

EXAMPLE 4.6-1 (Assignment Problem)

A machine tool company decides to make four subassemblies through four

contractors. Each contractor is to receive only one subassembly. The cost of

each subassembly is determined by the bids submitted by each contractor and is

shown in table 4.7 in hundreds of rupees.

(i) Formulate the mathematical model for the problem.

(ii) Show that the assignment model is a special case of the transportation model.

(iii) Assign the different subassemblies to contractors so as to minimize the total

cost. [P.UB.Ii(Elect) Oct., 1993; ]WIFTMohali, 2000]

(i) Formulation of the Model

Step I

Key decision is what to whom i.e., which subassembly be assigned to which

contractor or what are the `n' optimum assignments on 1-1 basis. ,

Feasible alternatives are n! possible arrangements for n x n assignment situation.

In the given situation there are 4! different arrangements.

Step III

Objective is to minimize the total cost involved,

90

Step IV

Constraints' (a) Constraints on subassemblies are xiI +x1z +xi3 +x1a = 1,

(b) Constraints on contractors are

(ii) Comparing this model with the transportation model, we find that ai = 1 and bi

= 1. Thus, the assignment model can be represented as in table 4.8.

Therefore, the assignment model is a special case of the transportation model in

which

a) all right-hand side constants in the constraints are unity i.e., a, = l , bi = l.

(b) all coefficients of x; in the constraints are unity.

(c) m = n.

91

(iii) Solution of the Model

We shall apply the Flood's Technique for solving the assignment problems. This

technique also known as the Hungarian Method or the Reduced Matrix Method

consists of the following steps:

Step I

Prepare a Square Matrix: Since the situation involves a square matrix, this step is

not necessary. '

Step If

Reduce the Matrix: This involves the following substeps:

Substep 1: In the effectiveness matrix, subtract the minimum element of each

row from all the elements of that row. The resulting reduced matrix will have at

least one zero element in each row. Check if there is at least one zero element in

each column also. If so, stop here. If not, proceed to substep 2.

Substep 2: Mark the columns that do not have zero element. Now subtract the

minimum element of each such coulmn from all the elements of that column.

In the given situation, the minimum element in first row is 13. So, we subtract 13

from all the elements of the first row. Similarly we subtract 11, 10 and 14 from all

the elements of row 2, 3 and 4 respectively. This gives at least one zero in each

row as shown in table 4.9.

In table 4.9 column 4 has no zero element. We go to substep 2 and subtract the

minimum element 1 from all its elements. Table 4.10 represents the resulting

reduced matrix that contains at least one zero element in each row and in each

column.

92

Step III

Check if Optimal Assignment can be made in the Current Solution or not

Basis for making this check is that if the minimum number of lines crossing all

zeros is less than n (in our example n = 4), then an optimal assignment cannot

be made in the current solution. If it is equal to n (= 4), then optimal assignment

can be made in the current solution.

Approach for obtaining minimum number of lines crossing all zeros consists of

the following substeps:

Substep 1: Examine rows successively until a row with exactly one unmarked

zero is found. Make a square (p) around this zero, indicating that an assignment

will be made there. Mark (x) all other zeros in the same column showing that they

cannot be used for making other assignments. Proceed in this manner until all

rows have been examined.

In the given problem, row 1 has a single unmarked zero in column 2. Make an

assignment there by enclosing this zero by a square []. It means subassembly 1

is assigned to contractor 2. Since contractor 2 has been assigned sub assembly

1 and as a contractor can be assigned only one subassembly, any other zero in

column 2 is crossed. Since there is no other zero in this column, crossing is not

required. Next, row 2 has a single unmarked zero in column 1, make an

assignment. Row 4 has a single unmarked zero in column 3, make an

assignment and cross the 2nd zero in column 3. Now, row 3 has a single

unmarked zero in column 4, make an assignment here. This is shown in the

matrix below.

93

Substep 2: Next examine columns for single unmarked zeros, making them (E])

and also marking (x) any other zeros in their rows.

In case there is no row or column containing single unmarked zero (there are

more than one unmarked zeros), mark (0) one of the unmarked zeros arbitraily

and (x) all other zeros in its row and column. Repeat the process till no unmarked

zero is left in the cost matrix.

Substep 3: Repeat substeps I and 2 successively till one of the two. things

occurs:

(a) there may be no row and no column without assignment i.e., there is one

assignment in each row and in each column. In such a case the optimal

assignment can be made in the current solution i.e., the current feasible solution

is an optimal solution. The minimum number of lines crossing all zeros will be

equal to `n'.

(b) there may be some row and/or column without assignment. Hence optimal

assignment cannot be made in the current solution. The minimum number of

lines crossing all zeros have to be obtained in this case.

In the present example, substeps 2 and 3 are not necessary since there is no

column left unmarked. Since there is one assignment in each row and in each

column, the optimal assignment can be made in the current solution. Thus

minimum total cost is

=Rs (13 x I+11 x 1+11 x 1+14x 1)x 100=Rs.4,900, and the optimal assignment

policy is

94

Subassembly 1- Contractor 2,

Subassembly 2- Contractor 1,

Subassembly 3- Contractor 4,

Subassembly 4- Contractor 3,

The minimal cost of Rs. 4,900 can also be determined by summing up all the

elements that were subtracted during the solution procedure i. e., [(13 + 11 +

10 + 14) + 1 ] x 100 = Rs. 4,900.

95

UNIT : 5 THE THEORY OF GAMES

5.1 Introduction :

The theory of games (or game theory or competitive strategies) is a

mathematical theory that deals with the general features. of competitive

situations. This theory is helpful when two or more individuals or organisations

with conflicting objectives try to make decisions. In such situations,

a decision made by one decision-maker affects the decision made by one or

more of the remaining decision-makers and the final outcome depends upon the

decision of all the parties. Such situations often arise in the fields of business,

industry, economics, sociology and military training. This theory is applicable to a

wide variety of situations such as two players struggling to win at chess,

candidates fighting an election, two enemies planning war tactics, firms

struggling to maintain their market shares, launching advertisement campaigns

by companies marketing competing product, negotiations between organisations

and unions, etc. These situations differ from the ones we have discussed so far

wherein nature was viewed as a harmless opponent.

The theory of games is based on the. minimax principle put forward by J. von

Neumann which implies that each competitor will act so as to minimize has

miximum loss (or maximize his minimum gain) or achieve best of the worst. So

far only simple competitive problems have been analysed by this mathematical

theory. The theory does not describe how a game should be played; it describes

only the procedure and principles by which plays should be selected.

Though the theory of games was developed by von Neumann (called father of

game theory) in 1928', it was only after 1944 when he and MorgensYern

published their work named `Theory of Games and Economic Behaviour', that

the theory received its proper attention. Since, so far the theory has been

capable of analysing very simple situations only, there has remained a wide gap

between what the theory can handle and the most actual situations in business

and industry. So, the primary contribution of game theory has been its concepts

rather than its formal application to the solution of real problems.

96

5.2 CHARACTERISTICS OF GAMES

A competitive game has the following characteristics :

(i) There are finite number of participants or competitors. If the number of

participants is 2,

the game is called two-person game ; for number greater than two, it is called n-

person game.

(ii) Each participant has available to him a list of finite number of possible

courses of action. The list may not be same for each participant.

Decision Theory, Games, Investment Analysis and Annuity) 795

(iii) Each participant knows all the possible choices available to others but

does not know which of them is going to be chosen by them.

(iv) A play is said to occur when each of the participants chooses one of the

courses of action available to hikn. The choices are assumed to be made

simultaneously so that no participant knows the choices made by others

until he has decided his own.

(v) Every combination of courses of action determines an outcome which

results in gains to the participants. The gain may be positive, negative or

zero. Negative gain is called a loss.

(vi) The gain of a participant depends not oniy on his own actions but also

those of others.

(vii) The gains (payoffs) for each and every play are fixed and specified in

advance and are . known to each player. Thus each player knows fully the

information contained in the payoff matrix.

(viii) The players make individual decisions without direct communication.

5.3 GAME MODELS

There are various types of game models. They are based on the factors like

the number of players participating-, the sum of gains or losses and the

number of strategies available, etc.

97

1. Number of persons : If a game involves only two players, it is called two-

person game; if there are more than two players, it is named n-person

game. An n-person game does not imply that exactly n players are involved

in it. Rather it means that the participants can be classified into n mutually

exclusive groups, with all members in a group having identical interests.

2. Sum of payoffs : If the sum of payoffs (gains and losses) to the players is

zero, the game is called zero-sum or constant-sum game, otherwise non

zero-sum game.

3. Number of strategies : If the number of strategies (moves or choices) is

finite, the game is called a finite game; if not, it is called infinite game.

5.4 DEFINITIONS

1. Game : It is an activity, between two or more persons, involving actions

by each one of them according to a set of rules, which results in some gain

(+ve, -ve or zero) for each. If in a game the actions are determined by

skills, it is called a game of strategy, if they are determined by chance, it is

termed as a game of chance. Further a game may be finite or infinite. A

finite game has a finite number of moves and choices, while an infinite

game contains an infinite number of them. -

2. Player : Each participant or competitor playing a game is called a player.

Each player is equally intelligent and rational in approach.

3. Play : A play of the game is said to occur when each player chooses one

of his courses of action. •

4. Strategy : It is the predetermined rule by which a player decides his

course of action from his list of courses of actions during the game. To

decide a particular strategy, the player need not know the other's strategy.

98

5. Pure strategy : It is the decision rule to always select a particular course

of action. It is usually represented by a number with which the course of

action is associated.

6, Mixed strategy : It is decision, in advance of all plays, to choose a course

of action for each play in accordance with some probability distribution.

Thus, a mixed strategy is a selection among pure strategies with some fixed

probabilities (proportions). The advantage of a mixed strategy, after the

pattern of the game has become evident, is that the opponents are kept

guessing as to which course of action will be adopted by a player.

Mathematically; a mixed strategy of a playerwith m possible courses of actions

is a set X of m non-negative numbers whose sum is unity, where each number

represents the probability with which each course of action (pure strategy) is

chosen. Thus if x, is the probability of choosing course i, then

where

Evidently a pure strategy is a special case of mixed strategy in which all but

one x, are zero. A player may be able to choose only m pure strategies but he

has an infinite number of mixed strategies to choose from.

7. Optimal strategy : The strategy that puts the player in the most preferred

position irrespective of the strategy of his opponents is called an optimal

strategy. Any deviation from this strategy would reduce his payoff.

8. Zero-sum game : It is a game in which the sum of payments to all the

players, after the play of the game, is zero. In such a game, the gain of

players that win is exactly equal to the loss of players that lose e.g., two

99

candidates fighting elections, wherein the gain of votes by one is the loss of

votes to the other.

9. Two-person zero-sum game: It is a game involving only two players,. in

which the gain of one player equals the loss to the other. It is also called a

rectangular game or matrix game because the payoff matrix is rectangular in

form. If there are n players and the sum of the game is zero, it is called n-

person zero-sum game. The characteristics of a two person zero-sum game

are

(a) only two players are involved,

(b) each player has a finite number of strategies to use, (c) each specific

strategy results in a payoff,

(c) total payoff to the two players at the end of each play is zero.

10. Nonzero-sum game : Here a third party (e.g. the `house' or a `kitty')

receives or makes some payment. A payoff matrix for such a game is shown

below. The left-hand entry in each cell is the payoff to A,

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and the right-hand entry is the payoff to B. Note that for play combination (1,1)

and (2, 2) the sums of the payoffs are not equal to zero.

11. Payoff: it is the outcome of the game. Payoff (gain or game) matrix is the

table showing the amounts received by the player named at the left-hand-side

after all possible plays of the game. The payment is made by player named at

the top of the table.

Let player A have in courses of action and player B have n courses of action.

Then the game can be described by a pair of matrices which can be constructed

as described below.

(a) Row designations for each matrix are the courses of action available to player A.

(b) Column designations for each matrix are the courses of action available to

player B.

(c) The cell entries are the payments to A for one matrix and to B for the other

matrix. The cell entry a„ is the payment to A in A's payoff matrix when A chooses

the course of action i and B chooses the course of action j.

(d) In a two-person zero-sum game, the cell entries in B's payoff matrix will be

the negative of the corresponding cell entries in A's payoff matrix. A is called

maximizing player as he would try to maximize the gains, while B is called

minimizing player as he would try to minimize his losses.

5.7 Some definitions

1. Strategy : A strategy of a player has been defined as an alternative course

of action available to him in advance by which player decides the course of

action that he should adopt. Strategy may be to types:

(a) Pure Strategy: If the players select the same strategy each time, then it is

referred to as pure strategy. In this case each player knows exactly what

101

the other is going to do and the objective of the players is to maximize

gains or minimize losses.

(b) Mixed Strategy: When the players use a combination of strategies and

each player always kept guessing as to which course of action is to be

selected by other player at a particular occasion then this is known as

mixed-strategy. Thus, there is a probabilistic situation and objective of the

player is to maximize expected gains or to minimize losses.

Mathematically, a mixed strategy, for a player with two or more possible courses

of action is denoted br- the set S of m non-negative real numbers (probabilities)

whose sum is unity. If xt (j = 1, 2, ..., n) is the probability of choosing course of

action j, then we have

S = (x1 x2, ..., xn)

Subject to the constraints

x1 + x2 + ….. xn = 1

xj >0; j=1,2,...,n

Optimal Strategy: The course of action or a complete plan that leaves a player in

the most preferred position regardless of the actions of his competitors is called

optimal strategy. Here by most preferred position we mean any deviation from

the optimal strategy or plan would result in decreased payoff.

Payoff: A quantitative measure (e.g. money, percent of market share or utility) of

satisfaction, a player gets at the end of game is called the payoff or outcome.

Value of the Game: It refers to the expected outcome per play when players

follow their optimal strategy.

5.8 TWO-PERSON ZERO-SUM GAME

There are two types of two-person zero sum games. In one, the most

preferred position is achieved adopting a single strategy and therefore the game

is known as the pure strategy game. The second me requires the adoption by

102

both players a combination of different strategies in order to achieve almost

preferred position and is therefore referred to as the mixed strategy game.

Payoff Matrix

A two person zero-sum game is conveniently represented by a matrix as shown

in Table 10.1. The which shows the outcome of the game as the players select

their particular strategies, is as the payoff matrix. It is important to assume that

each player knows not only his own list of possible courses of action but also of

his opponent.

Let player A has M courses of action (Al, A2, ..., A,,) and player B has n courses

of action (B1, B2 …… Bn). The numbers n and m need not be equal. The total

number of possible outcome is therefore ^i. These outcomes are shown in Table

.

By convention, the rows denote player A's strategies and the columns denote

player B's strategies. The element aij (i = 1, 2, ..., m; j =1, 2, ..., n) represents the

payments to player A by player B for any combination of strategies. Each round

of the game consists of a simultaneous choice of strategy Ai by player A and

strategy BI by player B. The payoff is then equal to aq. Of course the payment

make to B must be -a 1J . The above pay off matrix is a profit matrix for the

player A and a loss matrix for the player B. The positive elements represent profit

to the player A while negative elements represent loss to him and vice-versa for

player B.

103

Illustrative Example Let us consider the labour union and management

collective bargaining situation. The union's bargaining position is summarized as

follows :

U1 = 15% wage increase

U2 = 10 days of sick leave with pay.

The position adopted by management is as

follows : Ml -- 10% wage increase

M2 = 5 days of sick leave with pay.

The payoff matrix from union's point of view is:

This is a two-person zero-sum game with two alternative choices of strategies

available to union as well as to management, since the gains of one is taken

exactly equal to losses for the other. In this collective bargaining situation, if

union selects strategy Ul and management selects strategy Ml, then union wins

Rs. 50, i.e. increase in wage.

104

In this game, union's objective is to adopt a strategy which enable them to

gain as much as possible, while management's objective is to adopt a strategy

which enable to lose as little as possible.

5.9 PURE STRATEGIES: GAMES WITH SADDLE POINTS

How to select the optimal strategy for each player without knowledge of the

competitor's strategy is the basic problem of playing a game? Since the payoff

matrix is usually expressed in terms of

the payoff to player A (whose strategies are repressed by the rows), the criterion

calls for A is to select the strategy (pure or mixed) that maximizes his minimum

gains. For this reason, player A is called the maximized. Player B in turn, will act

so as to minimize his maximum losses and is called the minimizer.

The minimum value in each row represents the least gain (payoff)

guaranteed to player A, if he plays his particular strategy. These are indicated in

the matrix by row-minima. Player A will then select the strategy that maximizes

his minimum gains. Player A's selection is called the maximin strategy (or

principle) and his corresponding gain is called the maximin value of the game.

Player B, on the other hand, likes to minimize his losses. The maximum

value in each column represents the maximum losses to player B, if he plays his

particular strategy. These are indicated in the matrix by column maxima. Player B

will then select the strategy that minimizes his maximum losses. Player B's

selection is called the minimax strategy (or principle) and his corresponding loss

is called the minimax value of the game.

If the maximin value equals the minimax value, then the game is said to have

a saddle or equilibrium point and the corresponding strategies are called optimal

strategies. The amount of payoff at an equilibrium point is known as the value of

the game. If may be noted that if player A adopts mutimax criterion, then player B

has to adopt maximin criterion as it is a two-person zero-sum game. A game may

have more than one saddle points. A game with no saddle point is solved by

employing mixed strategies.

105

Example Two companies A and Bare competing for their competitive

product. To improve its market share, company A decides to launch the following

strategies :

A1 = Home delivery

services A2 = Mail order

services

A3 = Free gift for customer

As a countermove, the company B decides to use media advertising to promote

its product :

B1 = Radio

B2 = Magazine

B3 = Newspaper

Past experience and recent studies reveal that the payoff matrix to company A

for any combination of strategies is,

What is the optimal strategy for both the companies and the value of the game ?

Solution Using maximin principle, Company A selects that strategy among Al, AZ

and A3 which can maximize its minimum gains.

The strategy to be chosen will be determined based on the values of row minima,

i.e.,

106

Company A, will chose strategy AZ which yields the maximum payoff of 1, i.e.,

Similarly Company B will adopt that strategy among its strategies Bl, BZ and

B3 wluch can minimize its maximum losses. For this, Company B has to use the

minimax principle, i.e.,

Thus Company B will choose strategy Bl which leads to a minimum loss of 1.

Since the value of maximin coincides with the value of the minimax, an

equilibrium or saddle point is determined in this game. It is apparent that a saddle

point is that point which is both maximum of the row minima and the minimum of

the column maxima. The amount of payoff at an equilibrium point is also known

as value of the game. Hence the optimal pure strategy for both the companies

are : Company A must select strategy AZ and Company B must select strategy

Bt. The value of the game is 1, which indicates that Company A will gain 1 unit

and Company B will lose 1 unit.

107

5.10 MIXED STRATEGIES: GAMES WITHOUT SADDLE POINTS

Pure strategies are available as optimal strategies only for those games which

have a saddle point. For games which do not have a saddle point can be solved

by applying the concept of mixed strategies. A mixed strategy game can be

solved by (i) algebraic method, (ii) analytical or calculus method, (iii) graphical

method, and (iv) linear programming method.

Example Two breakfast food manufacturing firms A and Bare competing for

an increased market share. To improve its market share, both the firms decide to

launch the following strategies:

Al, BI = Give coupons

A2 B3 = Decrease price

A3, B3 = Maintain present strategy

A4, B4 -- Increase advertising

The payoff matrix, shown in the following table describe the increase in market

share for firm A and decrease in market share for firm B.

Determine the optimal strategies for each firm and the value of the game.

Solution First, we apply the maximin (minimax) principle to analyses the game.

(minimax) game result is shown in Table

108

In Table , it may be noted that there is no saddle (equilibrium) point. Thus, firm A

might adopt strategy A4 (Increase Advertising) in order to maximize its minimum

gains and firm B might adopt strategy B4 (Increase Advertising). However, firm B

quickly realised that if firm A selected strategy A4, it could reduce its losses to 10

by adopting to strategy B3 (Maintain Present Strategy). Firm B, would have

initially avoided strategy B3 due to the possibility of firm A selecting strategy A1

(Give Coupons) yielding a loss to firm B of 25 as compare to minimum loss of 5

for strategy B4'

Since in the game, one player makes the first move which is then followed by

the other player and then back to the first player and so on. Thus, the game will

never end and a point of equilibrium can not be reached. The moves adopted by

each firm are shown by the arrows in the Table

It is apparent from Table that for firm A, only strategies A1 and A4 are

relevant and strategies B3 and B1 for firm B. This can be further verified by

reducing the payoff matrix with the rule of dominance. The application of the rule

of dominance is shown in next table .

109

In Table , we see that strategy AZ is dominated by A1 and strategy A3 is

dominated by Ay. Similarly strategy Bl and BZ are dominated by strategy B3 and

By respectively. Hence we can drop out these dominated strategies by enclosing

these in dotted lines as shown in Table The reduced payoff matrix is therefore

shown in Table 10.6.

The Algebraic Method

We shall illustrate this method by solving the game shown in Table 10.6. Since

the payoff matrix has no saddle point, it is desirable for each firm A and B to play

a combination of strategies with certain probabilities.

For Firm A. Let firm A selects strategy A1 with a probability of p and therefore

selects strategy A4 with a probability of (1-p). Suppose that firm B, selects

strategy B3. Then the expected gain to firm A for this game is given by

110

25p + 10(1 - p) = 15p + 10

On the other hand, if firm B selects strategy B4, then firm A's expected gain is

5p + 15(1 - p)= -10p + 15

Now, in order for firm A to be indifferent to which strategy, firm B selects, the

optimal plan for firm A requires that its expected gain to be equal for each of firm

B's possible strategies. Thus equating two equations of expected gains, we get

15y + 10 = -10p + 15

or 25p = 5

or p=1/5=0.2

and q=1-p=1-0.2=0.8

Hence firm A would select strategy At with probability of 0.2 and strategy A4

with a probability of 0.8.

For Firm B. Let firm B selects strategy B3 and B4 with a probability of q and (1 - q)

respectively.

The expected loss to firm B when firm A adopts strategies A1 and A4

respectively are

25q+5(1-q)=20q+5

and 10q + 15(1 - q) = -5q + 15

By equating expected losses of firm B, regardless of what firm A would

choose, we get

20q+5=-5q+15

or 25q = 10

or q = 10/25 = 0.4

and p=1-q=1-0.4=0.6

Hence firm B would select strategy B3 and Bq with a probability of 0.4 and 0.6

respectively. The value of the game is determined by substituting the value of p

or q in any of the expected value and is determined as 13, i.e.

Expected gain to Firm A:

(i) 25x0.2+10 x0.8=13

(ii) 5 x0.2+15x0.8=13

111

Expected loss to Firm B:

(i) 25x0.4+ 5x0.6=13

(ii) 10 x0.4+15x0.6=13

From this expected loss to one firm and gain to another firm, we observe that by

using mixed strategies both firms have improved their market share as compared

to the maximin (minimax) values as shown in Table 10.4. Firm A has increased

its expected gain from 10 to 13 and firm B has decreased its expected loss from

15 to 13.

The Analytical Method

The method is almost similar to the previous method expect instead of equating

the two expected values, the expected value for a given player is maximized. To

illustrate this method let us take the same Example 10.2, discussed in the

previous method.

Suppose firm A selects strategy A1 with a probability p and obviously selects

A4 with a probability (1-p). Similarly, let firm B select strategy B3 with a probability

q, then strategy By with a probability (1 - q).

Firm A's expectation is given by

E(p, q) = 25pq + 10(1 - p)q + 5p(1 - q) + 15(1 - p)(1 - q)

If the expectation is to be maximized, then

The value of the game can be obtained by substituting the value of p, 1 - p, q and

1 - q in the expression of expected value E(p, q). The value of the game is found

to be 13 as before.

The Graphical Method

112

Since the optimal strategies for both the firms (or players) assign non-zero

probabilities to the same number of pure-strategies, thus it is obvious that if one

firm (or player) has only two strategies the other will also use two strategies.

Hence, graphical method is helpful in finding out which of the wo strategies can

be used.

Graphical method is useful if the nature of the game is of the form (2 x n) or (m

x 2). The graphical method consists of two graphs: (i) the payoffs (gains)

available to firm (or player) A versus :!~s strategies options and (ii) the payoffs

(losses) faced by firm (or player) B versus his strategies options.

Consider the following (2 x n) payoff matrix:

it is assumed that the game does not have a saddle point. Player A has two

strategies A1 and A2. He may select A2 with a probability p1 and A2 with a

probability p2 such that p1 + p2 = 1(pl, P? 0). The objective is to determine the

optimal values of p1 and p2. Thus, for each of the pure strategies, B1, B2, ..., Bn

available to player B, the expected payoff for player A would be as follows:

According to the maximin criterion for mixed strategy games. Player A should

select the value of pl and pz so as to maximize his minimum expected payoffs.

This may be done by plotting the straight lines :

113

The lower boundary of these lines will give the minimum expected payoff and the

highest point on this lower boundary will then give the maximum expected payoff

and hence the value of p1 and P2.

We now determine only two strategies of player B corresponding to those

lines which pass through the maximin point. This helps in reducing the size of the

game to 2 x 2 size, which can be solved by any method described earlier.

The (m x 2) games are also treated in the same manner except that minimax

point is the lowest point on the upper boundary of the straight lines.

Example Solve the following game graphically and find the value of the

game.

Solution: The game does not have a saddle point as shown in Table 10.10

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Player A's expected payoff corresponding to Player B's pure strategies is given

below. Table 10.9'V

These four expected payoff lines can be plotted on the graph to solve the game.

The Graph for Player A

Draw two parallel lines one unit apart and mark a scale on each. These two lines

represent the two strategies available to player A.

Player A determines the expected payoff for each alternative strategy

available to him. If player B selects strategy Bl, player A will gain 70 by selecting

strategy A1 and 10 by selecting strategy AT The value 70 is plotted along the

vertical axis under strategy A1 and the value 10 is plotted along the vertical axis

under strategy Az. A straight line joining the two points is then drawn. This line

represents the maximum possible payoff to player A. Proceeding in the same

manner, we draw another three lines.

We assume that player B will always select the alternative strategies yielding

the worst result to player A. Thus, the payoffs (gains) to A represented by the

lower boundary (shown by thick line in the figure) for any probabilistic value of A1

and AZ between 0 and 1. According to the maximin criterion, player A will always

select a combination of strategies A1 and AZ such that he maximizes his

minimum gains. In this case the optimum solution occurs at the intersection of

the two payoff lines.

The point of optimum solution (i.e. highest or maximin point on the lower

boundary) occurs at the intersection of two lines :

E2 = 25p1 + 60p2 = 25p1 + 60(1 - p1)

115

E3 = 45p1 + 30p2 -- 45p1 + 30(1 - p1)

Figure clearly indicates that Player A's expected payoff depends on which

strategy Player B selects. At the point where the two lines Ez and E3 intersect, the

payoff is the same for the player A no matter which counter strategy player B

uses. We find this unique payoff by setting Ez equals E3 and solving for pl, i.e.

25p1 + 60(1 - pl) = 45p1 + 30(1 - pl)

Therefore Pi = 3/5

and pz = 1 - 3/5 = 2/5

Then, substituting the value of pl in the equation for Ez (or

E3), we have V = 25(3/5) +

60(2/5) = 39

This is the optimal value of the game, when p1= 3/5 and pz= 2/5, for player A.

116

Guided by the minimax principle, player B should also select a pair of

probabilities q2 and q3 for his strategies BZ and B3 such that he will minimize the

maximum expected losses. Thus, if the player A selects strategy Ai, player B's

expected loss is

L2 = 25q2 + 45q2

Similarly, if player A selects strategy A, player B's expected loss is

To solve for qz, equate the two equations, i.e.

25q2 + 45(1 - q2) = 60q2 + 30(1 – q2) q2 = 3/10 1-q2 = 1-3/10=7/10

Therefore and Substituting the value of qz and q3 in the equation for L2 (or L3), we

have V = 25(3/10) + 45(7/10) = 39

This is the optimal value of the game, when q1 = 0, q2 = 3/10, q3 = 7/10 and qy =

0, for player B.

Linear Programming Method

The major advantage of using linear programming technique is to solve mixed-

strategy games or larger dimensions than games of (2 x 2) size. However, in

order to explain the procedure, we shall use linear programming method to solve

the game shown in Table 10.6.

Let us take the following notations :

V = value of the game

p1, p2 = probabilities of selecting strategies A1 and A4

respectively.

q1, q2 = probabilities of selecting strategies B3 and B4

respectively.

117

Firm A's objective is to maximize its expected gains which can be achieved

by maximizing the value of the game (V), i.e., it might gain more than V if Firm B

adopts a poor strategy. Hence, the expected gain for Firm A will be as follows :

25p1 + 10p2 > V (for if Firm B adopts strategy B3)

5p1 + 15p2 > V (for if Firm B adopts strategy By)

pl + p2 = 1 (sum of probabilities)

and p1, p2 > 0

Dividing each inequality and equality by V, we get

25p1 / V + 10p2/ V >_ 1

5p11 V + 15P2/ V ?

p1/V + p2/V = 1/V

In order to simplify, we define new variables,

p1/V = xl and P2/ V = x2

The objective of Firm A is to maximize the value of V, which is equivalent to

minimizing 1/V (as V becomes larger the value of 1/V becomes smaller). The

resulting linear programming problem can now be given as

Maximize Z = V or Minimize Z = 1/ V = xl + xz subject to the constraints

25x1 + 10xz ? 1

5x1 + 15x2 ? 1 (10.1) and xl, xz ? 0

Firm B's objective is to minimize its expected losses which can be reduced

by minimizing the value of the game (V) i.e., it might lose less than V if Firm A

adopts a poor strategy. Hence the expected loss for Firm B will be as follows :

25q1 + 5q2 <_ V (for if Firm A adopts strategy Al)

10q1 + 15q2 <_ V (for if Firm A adopts strategy Ay)

ql + q2 = 1 (sum of probabilities)

and ql, q2 > 0

Dividing each inequality and equality by V, we have

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25q1/V + 5q2/V < 1

10q1/ V + 15q2 / V < 1

q1/V + q2lV = 1/V

In order to simplify, we define new variables

q1lV = y1 and q2lV = y2

Since, minimizing of V is equivalent to maximizing 1/ V and yl + y2 = 1/ V, the

resulting linear programming problem can now be given as

Minimize Z = V or Maximize Z = I/ V = y1 + y2

subject to the constraints

25y1 + 5y2 < 1

10y1 + 15y2 < 1 (10.2)

and y1 y2 > 0

It may be noted that (10.1) is the dual of (10.2). Therefore, solution of the

dual problem can be obtained from the optimal simplex table of primal.

To solve the dual of the linear programming problem, introduce slack

variables to convert the two inequalities to equalities. The problem becomes

Maximize = y1 + y2 + Os1 + Os2

subject to the constraints

25y1 + 5y2 + s1 = 1

10y1 + 15y2 + s2 = 1

y1 y2 > 0

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UNIT : 6 Network Techniques

6.1 Introduction : A network (also called network diagram or network technique)

is a symbolic representation of the essential characteristics of a project. PERT

and CPM are the two most widely applied techniques.

(a) Programme Evaluation and Review Technique (PERT)

It uses event oriented network in which successive events are joined by arrows.

It is preferred for projects that are non-repetitive and in which time for various

activities cannot be precisely pre-determined. There is no significant past

experience to guide; they are once-through projects. Launching a new product

in the market by a company, research and development of a new war weapon,

launching of satellite, sending space craft to Mars are PERT projects. Three

time estimates - the optimistic time estimate, pessimistic time estimate and the

most likely time estimate are associated with each and every activity to take

into account the uncertainty in their times.

(b) Critical Path Method (CPM)

It uses activity oriented network which consists of a number of well recognised

jobs, tasks or activities. Each activity is represented by arrow and the activities

are joined together by events. CPM is generally used for simple, repetitive

types of projects for which the activity times and costs are certainly arid

precisely known. Projects like construction of a building, road, bridge,

physical verification of store, yearly closing of accounts by a company can be

handled by CPM. Thus it is deterministic rather than probabilistic model.

6.2 NETWORK LOGIC (NETWORK OR ARROW DIAGRAM)

Some of the terms commonly used in networks are defined below.

Activity

It is physically identifiable part of a project which requires time and resources

for its execution. An activity is represented by an arrow, the tail of which

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represents the start and the head, the finish of the activity. The length, shape

and direction of the arrow has no relation to the size of the activity.

Event

The beginning and end points of an activity are called events or nodes. Event is

a point in time and does not consume any resources. It is represented by a

circle. The head event, called the jth event, has always a number higher than

the tail event, called the ith event i.e., j > i. For example

Activity

Event Event

Fig. 6.1

Making the pattern of impeller' is an activity. `Start making the pattern of

impeller' is an event. `Pattern making completed' is an event.

Path

An unbroken chain of activity arrows connecting the initial event to some other

event is called a path.

Network

It is the graphical representation of logically and sequentially connected arrows

and nodes representing activities and events of a project. Networks are also

called arrow diagrams.

Network Construction

Firstly the project is split into activities. Start and finish events of the project are

then decided. After deciding the precedence order, the activities are put in a

logical sequence by using the graphical notations. While constructing the

network, in order to ensure that the activities fall in a logical sequence, following

questions are checked:

Network Analysis in Project Planning (PERT and CPM)

(i) What activities must be completed before a particular activity starts ?

(ii) What activities follow this ?

i j

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(iii) What activities must be performed concurrently with this ?

Activities which must be completed before a particular activity starts are called

the predecessor activities and those which must follow a particular activity are

called successor activities.

While drawing the network following points should be kept in mind:

1. Each activity is represented by one and only one arrow. But in some

situations where an activity is further subdivided into segments, each

segment will be represented by a separate arrow.

2. Time flows from left to right. Arrows pointing in opposite direction are to be

avoided.

3. Arrows should be kept straight and not curved.

4. Angles between the arrows should be as large as possible.

5. Arrows should not cross each other. Where crossing cannot be avoided,

bridging should be done as shown in Fig. 14.6.

6. Each activity must have a tail and a head event. No two or more activities

may have the same tail and head events.

7. An event is not complete until all the activities flowing into it are completed.

8. No subsequent activity can begin until its tail event is completed.

9. In a network diagram there should be only one initial event and one end

event.

Dummy

An activity which only determines the dependency of one activity on the other,

but does not consume any time is called a dummy activity. Dummies are

usually represented by dotted line at rows.

To illustrate the use of dummy, refer to Fig. 14.2 (a) and assume that the start

of activity C depends upon the completion of activities A and B and that the

start of activity E depends only on the completion of activity B. For this

situation, figure 14.2 (a) is a faulty representation. This is corrected by

introducing a dummy activity D as shown in Fig. 14.2 (b).

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A dummy activity is introduced in the network for two basic reasons:

1. To maintain the precise logic of the precedence of activities. Such a dummy

is called `logical dummy'. It is shown in Fig. 14.2 (b).

2. To comply with the rule that no two or more activities can have the same tail

and head events. Such a dummy is called `grammatical dummy'. In Fig. 14.2

(c), both activities A and B have the same tail event 10 and same head event

20, which is incorrect since no two activities can have the same pair of tail and

head events. Such activities are called duplicate activities. This difficulty is

resolved by the introduction of a dummy activity in any of the four ways

represented in Fig. 14.2 (d), (e), (n or (g).

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Looping (Cycling)

Sometimes due to faulty network sequence a condition illustrated in figure 14.3,

arises. Here the activities D, E and F form a loop (cycle). Activity D cannot start

until F is completed, which, in turn, depends upon the completion of E. But E is

dependent upon the completion of D. Thus the network cannot proceed. This

situation can be avoided by checking the precedence relationship of the

activities and by numbering them in a logical sequence.

Fig. 14.3

6.3 MERITS AND DEMERITS OF AON DIAGRAMS

The greatest merit of AON diagram is its simplicity. It is easier to draw,

interpret, review and revise. Absence of dummies makes it more readily

understood by non-technical users. However, in spite of simplicity and

other merits of AON diagrams, arrow diagrams continue to enjoy

popularity among the users of network techniques. Perhaps the main

reasons of popularity are their early development and suitability to PERT.

Arrow diagrams became well established before AON diagrams came into

existence. PERT emphasises more on events, which form the nodes of

arrow diagrams and due to this reason arrow diagrams became the basis

of PERT analysis. Secondly, activities on arrows suggest the flow of work

or the progress of the project, while activities on nodes make the network

appear static. Thirdly, the users of arrow diagrams argue that

identification of the activity in numeric form, that is by the tail and head

event numbers (t, j) makes it more suitable for computer programming.

The numeric job description, supplies all the connected dependency

relations. In ease of AON diagram, for every activity, its predecessor as

well as successor activities have to be supplied, which makes the

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computer programming more involved as well as requiring more storage

space in computer memory. However, the merits and demerits of both

diagrams discussed above are only the subjective opinions of its users.

Computer programs for both the systems are available and are being

used. From the application point of view, arrow diagram seems to enjoy

better popularity, basically because of its better suitability to PERT, which

is very popular technique of project management. On the other hand AON

diagram is being increasingly used in construction industry, where CPM is

used for planning, scheduling and controlling the project. Thus it may be

concluded that while arrow diagrams are more common with probabilistic

networks, AON diagrams are more popular with deterministic networks.

6.4 CRITICAL - PATH METHOD

Measure or Activity

Each task or activity takes sometime for its completion. This time duration

depends upon the nature of the activity. Some activities are rarely performed and

no data exists for their time durations. Their time consumption involves a

considerable degree of uncertainty. Such activities are called `variable activities'

and stochastic modelling techniques are applied in their time estimation. Under

this category fall the activities which demand creative ability, such as, research,

design and development work and the activities which are performed under

uncertain environments, such as, construction work during rainy season.

Oil the other hand, there are activities for which the associated time duration can

be accurately estimated. Such activities are said to be deterministic in nature or

deterministic These activities are usually repetitive in nature. Also it is presumed

that (i) skilled persoils experienced in method study are available to do the job (ii)

sufficient additional resources are available to allow uninterrupted activity. Above

all, it is the assumption of confidence that a(i will go well. Figures 14.20 (a) and

(b) show frequency distribution curves for the tkvo types of activities. In Figure

14.20 (a), the dispersion of the curve is more and hence more is the uncertainty.

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In Fig. 14.20 (b), for deterministic type activity, the dispersion is less and the

system fends to be more deterministic.

The projects which comprise of the variable type activities associated with

probabilistic time estimates, employ PERT version of the networks and the

projects comprising of deterministic type of activities are handled by CPM version

of networks.

This is the main difference between the two techniques. The other

difference between the two is that PERT is event-oriented while the CPM is

activity-oriented. In Fig. 14.20 (a) and (b),

Time Units

Any convenient time unit can be used, but it must be consistent throughout

the network. Depending upon the project length and level of detail, time unit

may be working days, shifts or weeks. Full time units are usually used, for

instance activity estimated at 3 days and 6 hours will be assigned 4 days.

Critical Path Analysis

The Critical path of a network gives the shortest time in which the whole

project can be completed. It is the chain of activities with the longest time

durations. These activities are called critical activities. They are critical in

the sense that delay in any of them results in the delay of the completion of

the project. There may be more than one critical path in a network and it is

possible for the critical path to run through a dummy. The critical path

analysis consists of the following steps:

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1. Calculate the time schedule for each activity : It involves the

determination of the time by which an activity must begin and the time

before which it must be completed. The time schedule data for each activity

include the calculation of the earliest start, the earliest finish, the latest

start, the latest finish times and the float.

2. Calculate the time schedule for the completion of the entire project : It

involves the calculation of project completion time.

3. Identify the critical activities and find the critical path : Critical activities

are the ones which must be started and completed on schedule or else the

project may get delayed. The path containing these activities is the critical

path and is the longest path in terms of duration.

Consider the network shown in Fig. 14.21 which consists of the following

activities:

The earliest start time l(E) for an activity represents the time at which an

activity can begin at the earliest. It assumes that all the preceding

activities start and finish at their earliest times. For instance earliest start

times of activities 1-2 and 1-3 are zero each or the earliest occurrence

time of event 1 is zero. Earliest start times of activities 2-3 and 2-5 or the

earliest occurrence time of event 2 is obtained by adding activity time t i2 to

earliest occurrence time of event 1 i.e., it is 0+15=15.

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Next consider event 3. It can be reached directly from event 1 or via event

2, the times for the two sequences being 15 and 15 + 3 = 18. Since event

3 can occur only when all the preceding activities and events have taken

place, its earliest occurrence time or the earliest start times of activities

emanating from even 3 is 18, the higher of the two values 15 and 18. This

is represented by putting E = 18 around its node in the network. Likewise,

the earliest occurrence time of each event can be determined by

proceeding progressively from left to right i.e., following the forward pass

method according to the following rule:

If only one activity converges on an event, its earliest start time E is given

by E of the tail event of the activity plus activity duration. If more than one

activity converges on it, E's via all the paths would be computed and the

highest value chosen and put around the node.

The E's calculated for the problem at hand are shown in the network

diagram.

The latest finish time (L) for an activity represents the latest by which an

activity must be completed in order that the project may not be delayed

beyond its targeted completion time. This is calculated by proceeding

progressively from the end event to the start event. The L for the last event

is assumed to be equal to its E and the L's for the other events are

computed by the following rule (using backward pass method):

If only one activity emanates from an event, compute L by subtracting

activity duration from L of its head event. If more than one activity

emanates from an event, compute L's via all the paths and choose the

smallest and put it around the event at hand.

The L's calculated for the problem at hand are shown in the network

diagram.

Next, the earliest finish time (Tp,p) and the latest start time (T LS) for an

activity are computed:

TEF = E + t ij,

T L S -L- tij

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where tij is the duration for activity i - j. Float (also called total float) for an

activity is then calculated:

F=L-TEF or F=TLS-E.

Float is, thus, the positive difference between the finish times or the

positive difference between the start times. The following analysis table is

then compiled:

TABLE

Start time Finish time Total

Activity (i -j) Duration (D) Earliest Latest Earliest Latest Float

(1) (2) (3) (4) (5) (6) (7)

1-2 15 0 0 15 15 0

1-3 15 0 3 15 18 3

2-3 3 15 15 18 18 0

2-5 5 15 32 20 37 17

3-4 8 18 18 26 26 0

3-6 12 18 28 30 40 10

4-5 1 26 36

36

27 37 10

4-6 14 26 26 40 40 0

5-6 3 27 37 30 40 10

6-7 14 40 40 54 54 0

Columns 1 and 2 contain the activities and their durations is weeks. Under

column 3 are noted the E's for the tail events and under column 6 are noted the

L's of the head events. Other columns are then computed as follows:

column 4 = column 6-column 2,

column 5 = column 3 + column 2,

and column 7 = column 6-column 5,

= column 4 - column 3.

As an example, consider activity 1-2. Its tail event 1 has E = 0. Put 0 against

this activity under column 3. Its head event 2 has L = 15. Put I S against it

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under column 6. Under column 4 note the value in column 6 minus activity

duration i. e., 15 - 15 = 0. Under column 5 note the value in column 3 plus

activity duration i.e., 0 + I S = 15. Compute total float by subtracting column S

from 6 or column 3 from 4. Total float for activity 1-2 is 0. Similarly, calculate

total float for other activities. Critical path is the path containing activities with

zero float. These activities demand above normal attention with no freedom of

action. For the problem at hand it is 1-2-3-4-6-7 and is shown by double arrows

in Fig. 14.21. The project duration is 54 weeks. Sometimes, there may be more

than one critical path i.e., two or more paths with the same maximum

completion time. Noncritical activities have positive float (slack or leeway) so

that we may slacken while executing them and concentrate on the critical

activities. While delay in any critical activity will delay the project completion,

this may not be so with the non-critical activities.

The Three Floats

Total float : It is the difference between the maximum time available to perform

the activity and the activity duration. The maximum time available for any

activity is from the earliest start time to the latest completion time. Thus for an

activity i - j having duration tii,

Maximum time available = L-E.

Total Float = L - E - tii,

=(L- tii,) -E or L - (E+ tii,)

TLS – E or L- TEF.

Thus the total float of an activity is the difference of its latest start and earliest

start time or the difference of its latest finish and earliest finish times. Total float

represents the maximum time within which an activity can be delayed without

affecting the project completion time.

Free Float : It is that portion of the total float within which an activity can be

manipulate= without affecting the floats of subsequent activities. It is computed

by subtracting the head eve--. slack from the total float. The head event slack is

(L - E) of the event.

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Free float of activity

i - j = T. F. - (L - E) of event j.

Thus free float is the time by which completion of an activity can be delayed

without delaying its immediate successor activities.

Independent Float : It is that portion of the total float within which an activity can

be delayed for start without affecting the floats of preceding activities. It is

computed by subtracting the - event slack from the free float. If the result is

negative, it is taken as zero.

Independent float of activity

i - j = F. F. - (L - E) of tail event i.

Apart from the above three floats, there is another float, namely the interfering

float for the activities.

Interfering Float : Utilization of the float of an activity can affect the floats of n!

subsequent activities in the network. Thus, interfering float can be defined as

that part of the total float which causes a reduction in the floats of the

succeeding activities. In other words it can be defined as the difference

between the latest finish time of the activity under consideration and the

earliest start time of the following activity, or zero, whichever is larger. Thus,

interfering float refers to that portion of the activity float which cannot be

consumed without adversely affecting the floats of the subsequent activities.

It is numerically equal to the difference between the total float and the free

float of the activity. It is also equal to the head erc;u slack of the activity.

Thus interfering float of an a:aisity = T.F. - F.F. -- (L - E) of the head event

of the activity. Suberitical Activity : Activity having next higher float than the

critical activity is called the subcritical activity and demands normal attention but

allows some freedom of action. The path connecting such activities is named as

the .subcritical path. A network may have more than one subcritical path.

Supercritical Activity : An activity having negative float is called supercritical

activity. Such an activity demands very special attention and action. It results

131

when activity duration is more than the time available. Such negative float,

though possible, indicates an abnormal situation requiring a decision as to how to

compress the activity. It can be done by employing more resources so as to

make the total float zero or positive. Compression of the network, however,

involves an extra cost.

Slack : It is the time by which occurrence of an event can be delayed. It is

denoted by S and is the difference between the latest occurrence time and

earliest occurrence time of the event.

i. e., S = L - E of the event.

In the above discussion, the term float has been used in connection with the

activities and slack for the events. However, the two terms are being used

interchangeably i.e., slack for the activities and float for the events by some of

the writers.

It is numerically equal to the difference between the total float and the

free float of the activity. It is also equal to the head erent slack of the

activity.

Thus interfering float of an activity= T.F. -F.F. _ (L- E) of the head event of the

activity. Subcritical Activity : Activity having next higher float than the critical

activity is called the subcritic2} activity and demands normal attention but allows

some freedom of action. The path connecting such activities is named as the

subcritical path. A network may have more than one subcritical path.

Supercritical Activity : An activity having negative float is called supercritical

activity. Such an activity demands very special attention and action. It results

when activity duration is more than the time available. Such negative float,

though possible, indicates an abnormal situation requiring a decision as to how to

compress the activity. It can be done by employing more resources so as to

make the total float zero or positive. Compression of the network, however,

involves an extra cost.

132

Slack : It is the time by which occurrence of an event can be delayed. It is

denoted by S and is the difference between the latest occurrence time and

earliest occurrence time of the event. i.e., S = L - E of the event.

In the above discussion, the term float has been used in connection with the

activities and slack for the events. However, the two terms are being used

interchangeably i.e., slack for the activities and float for the events by some of

the writers.

EXAMPLE 14.12-1

Tasks A, B, C,..., H, I constitute a project. The precedence relationships

are A<D;A<E;B<F;D<F,W<G;C<H;F<I:G<I.

Draw a network to represent the project and find the minimum time of

completion of the project when time, in days, of each task is as follows:

Task A B C D E F G H I

Time 8 10 8 10 16 17 18 14 9

Also identify the critical path.

Solution

The given precedence order reveals that there are no predecessors to activities

A, B, and C and hence they all start from the initial node. Similarly, there are no

successor activities to activities E, H and I and hence, they all merge into the end

node of the project. The network obtained is shown in Fig. 14.22 (a).

The nodes of the network have been numbered by using the Fulkerson's rule.

The activity descriptions and times are written along the activity arrows. To

determine the minimum project completion time, let event 1 occur at zero time.

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The earliest occurrence time (E) and the latest occurrence time (L) of each

event is then computed.

The E and L values for each event have been written along the nodes in Fig

14.22 (b).

The critical path is now determined by any of the following methods:

Method 1. The network analysis table is compiled as below.

TABLE

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Activity Duration Start

Earliest

time

Latest

Finish

Earliest

time

Latest

Total

float

1-2 8 0 0 8 8 0

1-3 8 0 9. 8 17 9

1-4 10 0 8 10 18 8

2-4 10 8 8 18 18 0

2-6 16 8 28 24 44 20

3-5 18 8 17 26 35 9

3-6 14 8 30 22 44 22

4-5 17 18 18 35 35 0

5-6 9 35 35 44 44 0

Activities 1-2, 2-4, 4-5 and 5-6 having zero float are the critical activities and

1-2-4-5-6 is the critical path.

Method 2. For identifying the critical path, the following conditions are

(i) E = L for the tail event.

(ii) E = L for the head event.

(iii) Ej- Ef = LJ - L; = tij.

Activities 1-2, 2-4, 4-5 and 5-6 satisfy these conditions. Other activities do

not fulfil all the three conditions. The critical path is, therefore, 1-2-4-5-fi.

Method 3. The various paths and their duration are:

Path Duration (days)

1-2-6 24

I-2-4-5-6 44

1-4-5-6 36

1-3-5-6 35

1-3-6 22

Path 1-2-4-5-6, the longest in time involving 44 days, is the critical path.

EXAMPLE A project schedule has the following characteristics:

Activity Time (weeks) Activity Times (weeks)

1-2 4 5-6 4

1-3 1 5-7 8

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2-4 1 6-8 1

3-4 1 7-8 2

3-5 6 8-10 5

4-9 5 9-10 7

(i) Construct the network.

(ii) Compute E and L for each event, and

(iii) Find the critical path.

Solutions

The given data results in a network shown in Fig. 14.23. The figures along

the arrows represent the activity times.

The earliest occurrence time (E) and the latest occurrence time (L) of

each event are now computed by employing forward and backward pass

calculations.

In forward pass computations,

136

E values are represented in Fig.

In backward pass communications

137

L values are also represented in Fig. 14.23. Network analysis table is

given below.

TABLE 14.3

Activity Duration

(weeks)

Start

Earliest

time

Latest

Finish

Earliest

time

Latest

Total

float

1-2 4 0 5 4 9 5

1-3 1 0 0 1 1 0

2-4 1 4 9 5 10 5

3-4 1 1 9 2 10 8

3-5 6 1 1 7 7 0

4-9 5 5 10 10 15 5

5-6 4 7 12 11 16 5

5-7 8 7 7 15 15 0

6 - 8 1 11 16 12 17 5

7 - 8 2 15 15 17 17 0

8 - 10 5 17 17 22 22 0

9 - 10 7 10 15 17 22 5

Path 1-3-5-7-8-10 with project duration of 22 weeks is the critical path.

138

6.5 PROGRAMME EVALUATION AND REVIEW TECHNIQUE (PERT)

Time Estimates

The CPM system of networks omits the probabilistic considerations and is based on

a Single Time Estimate of the average time required to execute the activity.

[n PERT analysis, time duration of each activity is no longer a single time estimate,

but is a random variable characterised by some probability distribution - usually a 0-

distribution- To estimate the parameters of the (3-distribution (the mean and

variance), the PERT system is based on Three Time Estimates of the

performance time of an activity. They are

(i) The Optimistic Time Estimate: The shortest possible time required for the

completion of an activity, if all goes extremely well. No provisions are made for

delays or setbacks while estimating this time.

(ii) The Pessimistic Time Estimate : The maximum possible time the activity

will take if everything goes bad. However, major catastrophes such as

earthquakes, floods, storms and labour troubles are not taken into account while

estimating this time.

(iii) The Most Likely Time Estimate: The time an activity will take if executed

under normal conditions. It is the medal value.

For determining the single time estimates used in CPM, some historical data may

be available, but the best way of predicting the three time estimates is by

intelligent guessing. The experienced person who may be an engineer, foreman

or worker having sufficient technical competence is asked to guess the various

time estimates. For estimation the activity should be taken randomly, so that the

guess of the assessor is not prejudiced by the predecessor and the successor

activities.

Frequency Distribution Curve for PERT

We have three time estimates for a PERT activity, the optimistic (to), pessimistic

(tp) and the most likely time (tm). In the range from optimistic to pessimistic, there

can be a number of time estimates for the activity. If a frequency distribution

curve for the activity times is plotted,

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it will look like the one shown in figure 14.30. It is assumed to be a (3-distribution

curve with a unimodal point occurring at t,„ and its end points occurring at to and

tP. The most likely time need not be the midpoint of to and tp and hence the

frequency distribution curve may be skewed to the left, skewed to the right or

symmetric.

Though the curve is not fully described by the mean (µ) and the standard

deviation (б), yet in PERT the following relations are approximated for µ and б:

Expected time or average time of an activity is taken equal to mean. This is the

time that the activity is expected to consume while executed. Thus

The expected time is then used as the activity duration and the critical path is

obtained by the analytical method explained earlier.

The variance or standard deviation is used to find the probability of completing

the whole project by a given date. The underlying procedure is as follow:

Compute the variance of all the activity durations of the critical path. Add them

up and take the square root to find the standard deviation of the total project

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duration and denote it by a. Now. while a (3-distribution curve approximately

represents the activity-time frequency distribution, the project expected time

follows approximately a normal distribution curve. The standard normal

distribution curve has an area equal to unity and a standard deviation of one

and is symmetrical about the mean value as shown in Fig. 8.8. ± 3 a give the

limits of the total possible duration with 99 per cent confidence i.e., 99 per cent

of the area under normal distribution curve is within ± 3 c from the mean. In

other words, to find the probability of completing the project in time T, w1-

calculate the standard normal variate,

where Z is the number of standard deviations the scheduled time or target date

lies away from the mean or expected date.

The probability is then read from the standard normal probability distribution

table (tab.. - C.2 at the end of the book) for the value of Z calculated above.

6.6 OBJECTIVES OF NETWORK ANALYSIS

Some of the main objectives of network analysis are:

(i) to complete the project within the stipulated period.

(ii) optimum utilization of the available resources.

(iii) minimization of cost and time required for the completion of the

project.

(iv) minimization of idle resources and investments in inventory.

(v) to identify the bottlenecks, if any, and to focus attention on critical

activities.

(iv) to reduce the set-up and changeover costs.

6.8 ADVANTAGES OF NETWORK TECHNIQUES

1. They are most valuable and powerful for planning, scheduling and control of

operations in large and complex projects.

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2. They are useful tools to evaluate the level of performance by comparing actual

performance against the planned targets.

3. They help to determine the interdependence of various activities for proper

integration and coordination of various operations.

4. They help to evaluate the time-cast trade off and determine the optimum

schedule. 5. These techniques are simple and can be easily oriented towards

computers.

6. The networks clearly designate the responsibilities of different supervisors.

Supervisor of an activity knows his time schedule precisely and also the

supervisors of other activities with whom he has to coordinate. .

7. These techniques help the management in achieving the objective with

minimum of time and least cost and also in predicting the probable project

duration and the associated cost.

8. Applications of PERT and CPM have resulted in saving of time which

directly results in saving of cost. Saving in time or early completion of the

project results in earlier return of revenue and introduction of the product or

process ahead of the competitors, resulting in increased profits.

9. They help to foresee well ahead of actual execution the difficulties and

problems that are likely to crop up during the execution of the project.

10. They also help to minimize the delays and hold-ups during execution.

Corrective action can also be taken well in time.

11. Application of network techniques has resulted in better managerial control,

improved utilisation of resources, improved communication and progress

reporting and better decision-making.

6.9 LIMITATIONS OF NETWORKS

1. Construction of networks for complex projects is complicated and time

consuming due to trial and error approach.

2. Estimation of reliable and accurate duration of various activities is a difficult

exercise.

3. With too many resource constraints the analysis becomes very difficult.

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4. Time-cost trade off procedure, in many situations, is complicated.

6.10 DIFFICULTIES IN USING NETWORK METHODS

Following are some of the problems faced in the managerial use of network

methods :

1. Difficulty in securing the realistic time estimates. [n the case of new and

non-repetitive type projects, the time estimates are often mere guesses.

2. The natural tendency to oppose changes results in the difficulty of

persuading the management to accept these techniques.

3. The planning and implementation of networks require personnel trained in

the network methodology. Managements are reluctant to spare the existing staff

to learn these techniques or to recruit trained personnel.

4. Developing a clear logical network is also troublesome. This depends upon

the data input and thus the plan can be no better than the personnel who

provides the data.

5. Determination of the level of network detail is another troublesome area. The

level of detail varies from planner to planner and depends upon the judgement

and experience.

6.11 COMMENTS ON THE ASSUMPTIONS OF PERT/CPM

1. In PERT analysis p-distribution curve is assumed for expected times of all

the activities. However, actually p-distribution curve may not be applicable to

each and every activity.

2. The formulae for the expected duration and standard deviation are simplified.

In certain cases the errors, due to these simplifications, may even be of the

order of 33%.

3. PERT analysis assumes independence of activities. Limitation of resources

may invalidate the independence of activities.

4. It is not always possible to sort out completely identifiable activities and their

start and finish times.

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5. Time estimates have an element of subjectiveness in them. The whole

analysis, being based on them is, therefore, weak. The analysis can, at best, be

as good as the time estimates.

6. The CPM model has an assumption that the duration of an activity is linearly

and inversely related to the cost of resources consumed for the activity. Cost-

time trade-off relationships are difficult to obtain in many cases either because

data are not available or b-,cause their estimation is too complex and

expensive. A great deal of effort and expertise is required to estimate them.

6.12 APPLICATIONS OF NETWORK TECHNIQUES

Networks provide a comprehensive study of the entire project in terms of

precedence and succession of various activities as well as resources available

to perform them to evolve some better and quicker plan to complete the project.

They can be used for complicated large scale projects involving financial and

administrative problems.

The list containing PERT and CPM applications is very large and the

applications are expanding to many new areas. Following are a few typical

areas in which these techniques are widely accepted:

1. Construction lndastry : It is one of the largest areas in which the network

techniques of project management have found application. These techniques

are used in the construction of buildings, roads, highways, bridges, dams and

irrigation projects.

2. Manufacturing : The design, development and testing of new machines,

installing machines and plant layouts are a few examples of how it can be

applied to the manufacturing function of a firm. It has been used in

manufacturing of ships, aeroplanes, etc.

3. Maintenance planning : R and U has been the most extensive area where

PERT has been used for development of new products, processes and

systems. It has been used in missile development, space programmes,

strategic and tactical military operations. etc.

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4. Administration : Networks have been used by the administration for

streamlining paperwork system, for making major administrative system

revisions, for long range planning and developing staffing plans, etc.

5. Marketing : Networks have been used for advertising programmes, for

development and launching of new products and for planning then- distribution.

6. Inventory planning : Installation of production and inventory control,

acquisition of spare parts, etc. have been greatly helped by network techniques.

7. Other areas of application are preparation of budget and auditing,

installation of computers and large machinery, best traffic flow patterns,

orglnisation of big conferences and public works, advertising and sales

promotion strategies, etc.

6.13 DISTINCTION BETWEEN PERT AND CPM

The PERT and CPVI techniques are similar in terms of their basic structure,

rationale and mode of analysis. However, there are certain distinctions between

PERT and CPM networks which are described below:

1. CPM is activity oriented i.e, CPM network is built on the basis of

activities. Also results of various calculations are considered in terms of

activities of the project. On the other hand, PERT is event oriented. Here,

emphasis is on the completion of a task rather than the activities required to be

performed to reach a particular event or task.

2. CPM is a deterministic model. It does not take into account the

uncertainties involved in the estimation of time for the execution of an activity.

Each activity is assigned a single time based on past experience. PERT,

however, is a probabilistic model. It uses three estimates of the activity time

namely, optimistic, pessimistic and most likely, with a view to take into account

uncertainty in time. Thus expected duration of each activity is probabilistic and

expected duration indicates that there is 50% chance of completing the activity

within that time.

3. CPM places dual emphasis on project time as well as cost and finds the

trade-off between project time and project cost. By employing additional

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resources, it helps to manipulate project duration within certain limits so that

project duration can be reduced at optimum cost. On the other hand, PERT is

primarily concerned with time only. It helps to schedule and coordinate various

activities so that project can be completed in scheduled time.

4. CPM is primarily used for projects which are repetitive in nature and

comparatively small in size. PERT is generally used for projects where time

required to complete the activities is not known a priori. Thus PERT is used for

large, one time research and development type of projects.

6.14 PROBABILITY STATEMENTS Or PROJECT DURATION

As shown previously that a standard deviation and variance for each activity 'time

could be calculated, for the PERT time estimate. The PERT technique makes

use of the central limit theorem of statistics, which states that the s!!m of n

independent activity distributions mill tend to be normally distributed with mean

equal to file sum of activity and variance equal to the sum of activity variance.

This is based upon the assumption that the summation of an expected activity

times and variance along the critical path aggregates to form a normal

distribution of project duration.

Table 15.5 provides the time estimates of the critical path activities of the

previously considered,:

Table 15.5 Time Estimates of Critical Path Activities

The variance and standard deviation are:

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Probability of completing the project on or before a specified time

On the basis of normal distribution, the probability of obtaining any specified date

or directed date(td) can be easily derived. For example, the probability of

completing the project on or before 20 days with standard deviation of 0.66 days

(shown in Table 15.5) can be obtained from the Z-transformation as shown

below:

From the table of standard normal distribution values, the value of Z = -1.51

corresponds 0.0655. Thus the probability of completing the project on time with a

directed date of 20 daps - 6.55%. This is a poor expectation and calls for

replanning.

PERT Algorithm

The various steps involved in developing PERT network for analysing any project

are summarize, below:

Step 1: Prepare a list of all activities involved in the project.

Step 2: Draw a PERT diagram as per the rules discussed earlier for drawing a

network diagram.

Step 3: Number all events in the ascending order from left to right.

Step 4: Estimate the expected performance time of each activity, i.e.,

(i) optimistic (shortest) time (a)

(ii) pessimistic (longest) time

(b) (iii) most likely (model) time

Step 5: Upon the assumption of Beta probability distribution associated with

each activity time, the expected value of the activity time can be approximated by

the linear combination of the three time estimates calculated in step 4:

Estimated average activity time = tc = a+4m+b / 6

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Step 6: Using the expected value of the activity time, determine earliest event

time and latest event time.

Step 7: Determine the slack associated with each activity or event. Determine

the critical path by connecting all these events where slack is zero (i.e., E1 = L).

Step 8: Determine the variance (Vte) of each activity's time by using the following

formula:

Estimated standard deviation of activity time ((Yte) = b – a / 6

Therefore variance (Vte (dte)2 = [(b –a)/ 6]

Step 9: Calculate the probability of completing the project on or before a

specified completion date by using the Z-transformation (standard normal

equation) as shown below:

where Z = number of standard deviations the desired date lies from the mean or

expected time.

Step 10: Establish time-cost trade-off, if the project is running behind the

schedule, resource allocation may have to be performed if resources are limited.

Float of an Activity

The float or slack analysis is useful not only for behind-schedule projects but also

for those that are on or ahead of schedule, since it helps to provide a basis for

effective resource allocation and cost reduction. For example, float analysis of

several projects may alert management to the possibility that by shifting attention

from a program that is ahead of schedule to one that is in trouble, both may be

completed on time without the need of costly overtime.

The value of slack can be either positive, negative or zero depending upon the

relationship between E-values and L-values. Positive slack indicates that the

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project is ahead of the schedule, she greater the slack, the greater the margin of

safety will be, i.e., resources may be reallocated or activity can be delayed by the

time equal to the difference between E and L values, (i.e., L - E). The zero slack,

the project is on schedule, but this situation is usually considered a potential:

problem by the project-managers. If any difficulty arises, the project is likely to fall

behind schedule The negative slack indicates that the project is behind schedule,

i.e., resources are not adequate. Thus a corrective action must be planned for it

immediately, i.e., there is a need to crash the time of critical activities to reduce

the amount of negative float. However, this will increase total coe: of the project.

The basic difference between slack and float times is that slack is used for

events only whereas float is sued for activities.

1. Total Float (or slack): The amount of time by which the completion of an

activity can be delaye4 beyond the earliest expected completion time

without affecting the project duration times is called total float. If

Ei = earliest expected completion time of tail event

= earliest starting time for an activity (i, j)

and Ll = latest allowable completion time of head event

= latest finish time for an activity (i, j)

then the total available time (T) for performing the activity (i, j) is given by

T=Lj –Ei

The necessary time for activity is the required duration time d ii. Thus the activity

time can be increased by an amount given by

F = T-dij = (Lj –Ei)-dij,

or = LSii - ESij = LFij - EFij

This value of F is defined as total float of an activity. If more than this time is

given to the activity it may result in change of the critical path and may increase

the overall project duration.

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2. Free Float (slack): In calculating the total float only a particular activity has

been considered with respect to tail and head event times. But it may be

necessary to find out how much an activity can float (move or flexibility) without

affecting the flexibility of movement of the immediate succeeding activity. The

amount of constrained float is called free float. The terra free indicates that the

delaying the activity will not delay successor activity The free float is given by

Free float = (Ej – Ei) - dij

or Min. (ES for all immediate successors) - EF

3. Independent Float: In some cases, the float of an activity affects neither the

predecessor nor the successor activities. The float is then called independent

float and is given by

Independent float = (Ei - Ld - da

= LFij (for preceding activity)

= ESij (for successor activity)

In other words, independent float provides a measure of variation W starting time

of an activity without affecting, preceding and succeeding activities.

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Unit :7 Inventory Theory

7.1 INTRODUCTION

An inventory may be defined as a stock of idle resources of any kind having a

economic value. These could be in the form of physical resources such as raw

materials, semi-finished goods used in the production process, finished products

ready for delivery to consumers; human resource such as unused labour or

financial resource such as working capital, etc. Service industries are often

unable to inventory their final-products, although they must manage their raw

materials anc supplies inventories just as any other organization. For example,

an airline company may have the seating capacity to provide a transportation

service but lacks the demand necessary to create the end product (final

inventory) of a transportation service. Table 7.1 gives an idea of the types of

inventories maintained by various organizations.

For many organizations inventories represent amounts of tied-up capital

usually 25 to 60 per cent of total assets, depending upon the type of the

organization and the industry. Insufficient inventories hamper production and fail

to generate adequate sales, whereas excessive inventories adversely affect the

firm's cash flow and liquidity position. Moreover one cannot rely purely on

intuitive methods of establishing optimum order quantities and setting up

optimum inventor stock level. Hence, all this calls for a scientific inventory

management so that someone can set policies, establish guidelines for inventory

levels and ensure that appropriate control systems are functioning well.

While inventories must be held to facilitate production activities, it must be

noted that larger inventories do not necessarily lead to high volume of output,

where lack of inventories might hamper production. Inventories cost money to

acquire as well as hold them. The cost of acquisition is reduced as larger

quantities are purchased each time, and the decrease in inventories reduce: the

inventory carrying costs. Thus, the problem is to balance between the advantage

of having inventories (or losses that may be expected from not having adequate

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inventories) and costs c: carrying them to arrive at an optimal level of inventories

to minimize the total inventory cost

Table 7.1

System

Factory Raw materials, parts, semifinished goods, finished goods, etc.

Bank Cash reserves tellers

Hospital Number of beds, specialized personnel, stocks of drugs, etc.

Airline Company Aircraft seat miles per route, parts for engine repairs,

stewardness, mechanics, etc.

Putting it in another words we can state that basic objective of inventory control is

to released capital for more productive use. Inventories should be adequate to

achieve maximum product and sales. At the same time it should not be so

excessive as to restrict the ability of organization to earn high rate of return. The

problem of keeping inventory at the optimal level are of two fold

(i) To forecast the demand precisely at various points of time, and

(ii) To take steps to keep inventory at an optimal level, i.e. to find more

economical method for its management.

Inventory Control Models

Impact on Profitability: The inventory problems considered as a national scale

reveals interesting points. An estimate of the capital locked-up in inventories in

India is Rs. 1000 crores. Even as small saving (sa) 20%), resulting from careful

analysis may represent an impressive saving (Rs. 200 crores) without any

discremental effects on the service level provided by the system. This saving of

Rs. 200 crores indicates how effective inventory control on a national scale would

lead to industrial expansion and more employment opportunities.

7.2 PRINCIPAL CATEGORIES OF INVENTORIES AND THEIR FUNCTIONS

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Since inventories normally represent a sizable investment in a system, therefore

a basic question can be raised with respect to the causes for the existence of

inventories as well as the functions that they perform. In industries, in general,

inventories can be classified in four categories:

1. Process (or pipeline) inventories. This consists of materials actually being

worked on, or moving between work centres or being in transit to distribution

centres and customers, Therefore in order to satisfy demand un-interrupted,

it is necessary to hold extra stock at various points to handle demand while

replenishments are in transit from proceeding stage. The amount of pipe-line

inventories depend on the time required for transportation and the rate of

use.

2. Lot-size (or cycle) inventories. ln many cases, production and material

procurement takes place in batches. This may be because of the following

reasons :

a) Economics of scale. If the average cost of producing, purchasing,

or moving inventory decreases as the lot size increases, then it is

better to start with larger quantities at a time. For example, when a

fixed setup or an administrative cost is incurred whenever an item

has to be produced or ordered from outside vendor. A larger order

quantity results in a reduced fixed cost per unit of item.

b) Technological requirements. The design of the process may

impose certain batch size. For example, in a chemical reactor

processing by thankfuls might be necessary in order to achieve

desired reaction parameters.

3. Seasonal inventories. When the demand of the items vary with time, it

may become economical to build inventory during periods low demand to

ease the strain of peak demand periods upon the production facilities

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(assuming that item(s) are perishable). The extent to which this policy

should be used is determined by balancing the cost of carrying seasonal

inventories against the cost of changing the production rate and not of

meeting demand entirely.

4. Substep (or buffer) stocks. Inventories maybe carried because of

uncertainties of future requirements. Further requirements are estimated

by forecasting, but forecasts are always accompanied by errors. I planning

is done disregarding the possible forecasting errors, shortages may

incurred when materialized requirements exceed the forecast. To prevent

the losses normally associated with shortages, safety stocks have to be

build in the form of extra inventories above the level that would result from

planning on the basis of the demand forecast alone.

Safety stocks can offer protection not only against demand uncertainties but also

whenever the quantities delivered vary from what is ordered, or when

procurement lead times show a probabilistic behaviour, safety stocks are

effective tools W hedging against supply uncertainties.

5. Decoupling inventories. These are those types of inventories used to

reduce the interdependence of various stages of the production system.

The decoupling inventories may be classified into four groups :

(a) Raw materials and component parts. In any organization, bulk of

stores in terms of volume and value will generally be raw materials,

since bulk of raw materials only are processed to finished products

in the same form or in the processed form. Thus raw materials are

used to decuple the producer from suppliers. In other words, raw

material and component parts are useful to provide for

(i) Economic hulk purchasing

(ii) To control production of finished goods

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(iii) To act as a buffer stock against delay in shipment

(iv) Seasonal fluctuations.

(b) Work-in-process inventory. Work-in-process inventory is the semi-

finished stock accumulate in between two operations. It might

merely because of the changes in production cycle time. This is

due to unbalanced loading of machines, holdups, during

manufacturing shortage of tools due to high consumption and

deterioration of machines capability due to long use. The size of the

work in-process inventory is dependent on the production cycle

time, the percentage of machine utilization, and the make or buy

policies of the company. It is used to decouple successive

production stages. In other words, the working process inventory

serves the following purposes:

(i) Enable economical lot production

(ii) Cater to the variety of products

(iii) Replacement for wastages

(iv) Maintain uniform production even though sales may vary.

(c) Finished goods inventory. The finished goads inventory is

maintained to ensure a free flowing supply to the customers, to

allow stabilization of the level of production and to promote sales,

i.e. it is used to decouple the consumer from producer. The size of

the inventory depends on the demand and the ability of the

marketing department to push the product, the company's ability to

stick to the delivery schedule of the customers and the shelf-life of

the product and the warehousing capacity.

(d) Spare parts inventory. A company manufacturing equipment or

appliances derives about 15 to 20 per cent of its sales from the sale

of its spare parts. After sales-service to customer makes it,

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essential for these companies to maintain a stock of spares so that

they are made available when need arises. Similarly a buyer in an

organization is constantly looking to procuring suitable spares to

utilize the equipment fully. Hence, It is essential to have spare parts

inventory and the size of inventory depends on the average life of

the components.

7.3 REASONS FOR CARRYM INVENTORIES

Some of the important reasons for carrying inventories are listed below:

1. Smooth Production. The demand for an item fluctuates widely due to a

number of factors such as seasonality and production schedule. Thus, a

manufacturing firm carries stock of race materials, semi-finished goods and

finished goods because of the following reasons:

(i) To ensure continuous production by ensuring that inputs are always

available and economic production run can be made.

(ii) To facilitate intermittent production of several products on the lime

facility.

(iii) To decouple successive stages in processing a product so that

downtime in one stage does not stop the entire process.

(iv) To help level production activities, stabilize employment, and

improve labour relations by storing human and machine effort.

2. Customer Satisfaction. Inventory of goods are carried in order to :

(i) Ensure an adequate and prompt supply of items to the customer to

avoid the reputation for constantly being out of stock and avoid the

shortage at a minimum cost. It may lose a significant number of

customers permanently.

(ii) Provide service to customers with varying demand and in various

locations by maintaining a adequate supply to meet their immediate

and seasonal needs.

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3. Delayed Deliveries. Inventory of goods are carried in order to provide a

means of hedging against future price and delivery uncertainties, such as

strikes, price increase, high rate of usage, and inflation.

4. Financial Gain.

(i) It makes use of available capital (and/or storage space) in a most

effective way and avoids and unnecessary expenditure on high

inventories, etc.

(ii) It reduces the risk of loss due to the changes in prices of items

stocked at the time of making the stock.

(iii) It provides a means of obtaining economic lot size and gaining

quantity discounts on bulk purchases.

(iv) It eliminates the possibility of duplicating and frequency of ordering

in order to minimize accumulation and build up of surplus stock.

(v) It minimizes the losses due to deterioration, obsolescence, damage

and pilferage.

7.4 STRUCTURE OF INVENTORY MANAGEMENT SYSTEM

An inventory system can be defined as a coordinated set of rules and procedures

that allow for routine decisions such as :

(i) When it is necessary to place an order (or set up production) to replenish

inventory?

(ii) How much is to be ordered (or produced) for each replenishment?

The objective of a well-designed procedure should be the minimization of the

costs incurred the inventory system, achieving at the same time the customer

service level specified by the company policies.

In order to arrive at the optimum inventory policy, let us first look at the various

input and output factors of an inventory system as shown in Fig. 12.1.

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Fig. Input-output Factors of an Inventory System

Regardless of items held in stock, an inventory management system can be

viewed as be---E structured of following sub-systems:

1. Accounting for inventories: It concerns with the careful record keeping of

inventory Control namely, delivery lead times source of acquisition, ordering

restrictions, dates of items received or issued; cost of each item, auditing,

control, parties of each transaction and existing stocks.

The periodic valuation of inventories is an essential accounting function that

includes both the verification of records by thorough inspection of all

quantities (i.e. stock on hand and or order; customer order status).

For various practical or financial reasons, accounting methods such as first-in

first- out (FIFO) may be employed, assuming that the oldest inventory (first

in) is the first to be used (first out).

In order to achieve more realistic financial results, many firms are adopting

last-in-first-out (LIFO) accounting method. Here it is assumed that the value

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of an item leaving inventory in its replacement cost and eliminates the

otherwise taxable, 'phantom profits' of a firm during inflationary period.

2. Decision Rules: These are concerned with the management of inventories

involving two fundamental functions namely: (a) Planning (what to store;

where are the best resource for procurement) and (b) Control (when to order

and how much to order). The questions of what and from where are

important from inventory planning aspects, but they are beyond the scope

and this book.

The how much is to order (or produce) for each replenishment is largely a

function of costs and ultimately based upon the concept of economic order

quantity. The when it is necessary to place an order (or set-up production)

question is a function of the organization's forecast scheduled requirements.

The answers to these questions may be given in one of the two ways:

(a) As the level of inventory item drops to a particular level (Reorder

point or level), replenishment may take place at fixed time interval.

The size of an order can vary in order to bring inventory to a

desired level;

(b) Management may order a fixed amount every time when the level

of inventory item drops to a certain reorder level at variable cycle

time.

Moreover, because of the uncertain demand and incorrect forecast,

safety stocks decision rules are needed in order to guarantee some

desired level of customer service.

3. Operating constraints. These constraints bring the decision rules together

with optimal inventory policies. Various items being controlled depending on

their inherent characteristics, limited warehouse space, limited budget

available for inventory; degree of management attention towards various

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items in the inventory, and ser•: ice levels that can be achieved by using

some appropriate stock policy.

4. Systems measure of performance. This is directly related to the total (or

incremental) minimum inventory cost necessary to satisfy forecasted

demand. In the following sections, various types of costs involved in the

analysis of inventory policy will be discussed.

7.5 FACTORS INVOLVED IN INVENTORY ANAIYSIS

Inventory models can be classified largely according to the following factors:

Inventory Related Costs (Economic Parameters)

Four categories of inventory costs are associated with keeping inventories of

items. These are (i) purchase (or production) costs, (ii) ordering (or set-up) costs,

(iii) carrying (or holding) costs, and (iv) shortage (or stock out) costs.

(i) Purchase (or Production) Costs: The cost of purchasing (or producing)

a unit of an item is known as purchase (or production) cost. The

purchase price will become important when quantity discounts or price

breaks can be secured for purchases above a certain quantity or when

economies of scale suggest that the per unit production cost can be

reduced by a larger production runs.

(ii) Ordering or Set-up) Costs: If any item is purchased, an ordering cost is

incurred each time an order is placed. These costs include the

following factors administrative (paper work, telephone calls, postage),

transportation of items ordered, receiving and inspection of goods,

processing payments, etc. If a firm produces its own inventory instead

of purchasing the same from an outside source, then production set-up

costs are analysis to ordering costs.

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(iii) Carrying (or Holding) Costs: The costs associated with holding

inventories in stock are known as holding costs. These are dependent

on the level of inventory held in the stock and the time for which an

item is held in stock.

It consists of all those costs that are incurred due to cost of money

invested in inventory, storage cost, insurance, depreciation, taxes, etc.

This cost may also be expressed as a percentage of average rupee -

value of inventory held rather than some specified rupee carrying cost

her unit held. The variables for the carrying cost portion are as follows:

I = Average amount of inventory held per unit time as a percentage of

average rupee value of inventory.

P = Price (or value) of holding one unit per unit time.

Therefore the total carrying cost may now be expressed as :

Carrying cost = I x P.

(iv) Shortage (or Stock out) Costs: The penalty costs for running out of stock

(i.e., when an item can not be supplied on the customer's demand) are

known as shortage costs.

These costs include the loss of potential profit through sales of items

demanded and loss of goodwill, in terms of permanent loss of customer

and its associated lost profit in future sales.

On the other hand, when customers can wait, the unfilled demand can be

fulfilled when item demanded becomes available. Costs incurred in this

case are costs for extra paper work, possible expediting of orders, etc. But

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if customer is unwilling to wait the profit on the sales as well as the

goodwill is lost.

To minimize shortage penalty costs, additional inventory, called safety

stock or buffer stock is generally carried.

The optimal inventory policy is usually based on, and is determined from,

the above discussed four categories of costs and their relationship to

different inventory levels. Therefore, for any inventory situation, total

inventory cost can be determined from the following relation:

Total inventory cost = Purchase costs of inventory items +

Ordering costs + Carrying costs +

Shortage costs.

Minimizing just one of these three costs, viz. ordering costs, carrying costs

and shortage costs of inventory is easy but of little value. For example, to

minimize carrying cost, a firm can simply stop carrying any inventory. This

action, however, can be expected to create unreasonable shortage of

items or very high ordering costs. The actual process for minimizing total

inventory costs entails two basic decisions - how much to order and when

to order. In fact these are the two decision variables that inventory models

use in optimizing an inventory system.

(v) Salvage Costs (or Selling Price): When the demand for an item is affected

by its quantity in stock, the decision depends upon the underlying criterion

and includes the revenue from sale of the item. Salvage costs are

generally combined with the storage costs and not considered

independently.

162

The inventory model showing the relationship of inventory costs with order

quantity and inventory level over time can be illustrated graphically as

shown W Fig. 12.12.

Other characteristics besides costs, which play an important role in the

formulation, solution and performing sensitivity analysis on the inventory

models are as follows:

Fig. Economic Order Quantity Graph

Demand

Customer's demand, that is, size of demand, rate of demand and the patte-rn of

demand for a given item is extremely important in the determination of an optimal

inventory policy

The size of demand is the number of items required per period. It may not be the

number of items sold, as some demand may remain unfilled due to shortage or

delay. It can be either deterministic or probabilistic. In deterministic case the

demand during each time period is known with certainty. This can be fixed

(static) or can vary (dynamic) from time to time. But in the probabilistic case the

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demand for a period of time is not known with certainty but its pattern can be

described by a known probability distribution.

The rate of demand is the size of demand over a particular unit of time. It can be

variable or constant, deterministic or probabilistic, depending on the size of

demand.

The pattern of demand is the manner in which items are drawn from inventory.

Some items may be drawn at the beginning of time period (instantaneously), or

at its end or at a uniform rate during the period. These patterns certainly affect

the total carrying cost o' inventory.

Order Cycle

The time period between placement of two successive orders is referred to as

am order cycle. The order may be placed on the basis of following two types of

inventory review systems:

(a) Continuous Review: The record of the inventory level is checked

continuously until a specified point (called reorder point) is reached where

a new order is placed. This is often referred to as the two-bin system. This

divides the inventory into two parts and places it physically, or on paper, in

two bins. Items are drawn from only one bin, and when it is empty, a new

order is placed. Demand is then satisfied from the second bin until the

order is received. Upon receipt of the order, enough items are placed in

the second bin to make up the earlier total. The remaining items are

placed in the first bin. This procedure is then repeated.

(b) Periodic Review: In this system the inventory levels are reviewed at equal

time intervals and orders are placed at such intervals. The quantity

ordered each time depends on the available inventory level at the time of

review.

164

Time Horizon

The period over which the inventory level will be controlled is referred to as time

Horizon. This can be finite or infinite depending on the nature of demand.

Lead Time

The time between ordering a replenishment of an item and actually receiving the

item into inventory is referred to as lead tone. The lead time can be either

deterministic, constant or variable, or probabilistic. If the lead time is zero, then

we have the special case of instantaneous delivery, i.e. no need for placing an

order in advance. If the lead time exists (i.e., it is not zero) and also demand

known, then it is required to place an order in advance by an amount of time

equal to the lead time.

Stock Replenishment

The rate at which items are added to inventory is one of the important

parameters in inventory models. The actual replenishment of items (or stock)

may occur at a uniform rate or instantaneous over time. Usually the uniform

replacement occurs in the case when the item is manufactured within the factory

while instantaneous replacement occurs when the items (or stock) are purchased

from outside sources. ,

Reorder Level

The level between maximum and minimum stock, at which purchasing (or

manufacturing) activities must start for replenishment.

Reorder Quantity

This is the quantity of replacement order. In certain cases it is the economic

order quantity.

165

12.4 CLASSIFICATION OF FIXED ORDER QUANTITY INVENTORY

MODELS

Model 1. (b) (Demand Rate Non-uniform, Replenishment Rate Infinite)

In this model all assumptions are same as in model 1 (cr) with the exception

that instead of uniform demand rate R, we are given some total demand D, to

be satisfied during some long time period T. Thus demand rates are different in

different order cycles.

Let q be the fixed quantity ordered each time the order is placed.

D Number of orders. N = -. q

If tt is the time interval between orders t and 2, r, the time interval between

orders 2 and 3

and so on, the total time T will be

166

=t1+t2+...+tn . . .(12.8)

This model is illustrated schematically in figure 12.3.

Fig. Inventory situation for different rates of demand in different

cycles. Holding costs for time period T will be

167

From equations (12.10) and (12.11) we find that results for this model can be

obtained if the uniform demand rate R in model I (a) is replaced by average

demand rate D/T.

Model 1 (c) (Demand Rate Uniform, Replenishment or Production

Rate Finite)

In the classical EOQ model the replenishment rate was assumed to be infinite;

the entire quantity ordered was delivered in a single lot. T ha is possible only for

bought-out items and is simply unthinkable for made-in items. Such items are

produced by the production department of the organisation at a constant rate

and are also supplied to the customers at a constant rate. When the production

starts, a fixed number of units are supposed to be added to inventory each day

till the production run is completed; simultaneously, the items will be demanded

at a constant rate, as stipulated earlier. Obviously. U-..e rate at which they are

produced has to be higher than the consumption rate, for only then can there

be the built-up of inventory.

It is assumed that run sizes are constant and that a new run will be started

whenever inventory is zero. Let

R = number of items required per unit time,

K = number of items produced per unit time,

Ci = cost of holding per item per unit time,

C3 = cost of setting up a production run,

q = number of items produced per run, q = Rt,

t = interval between runs.

168

Figure shows the variation of inventory with time.

Fig.Inventory situation with finite rate of production.

Here, each production run of length t consists of two parts t i and r,, where

(i) ti is the time during which the stock is building up at a constant rate of K - R

units pe: unit time,

(ii) t2 is the time during which there is no production (or supply or

replenishment) anc inventory is decreasing at a constant dentand rate R per

unit time.

Model 2 (a) (Demand Rate Uniform, Replenishment Rate Infinite,

Shortages Allowed)

In the earlier models the shortages and hence back ordering was not permitted.

Hence the models involved a trade-off between carrying cost and ordering cost.

However, in actual practice shortages may take place and hence shortage cost

also needs to be considered. Shortages may

also be allowed to derive certain advantages. One advantage of allowing

shortages is to increase the cycle time, and hence spreading the ordering (or

setup) cost over a long period, thereby reducing the total ordering cost over the

planning period. Another advantage is decreased net stock in inventory, resulting

in reduced inventory carrying cost.

This model is just the extension of model 1 (a), allowing shortages. Let

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R = number of items required per unit time i.e., demand rate,

C1 = cost of holding the item per unit time,

C2 = shortage cost per item per unit time,

C3 = ordering cost/order,

q = number of items ordered in one order, q=Rt,

t = interval between orders,

Im = number of items that form inventory at the beginning of time interval t.

Lead time is assumed to be zero. Figure 12.5 shows the variation of inventory

with time.

It is assumed that when shortages occur and customers are not served

immediately, they leave their orders with the supplier and these back orders are

filled as soon as the stock is received, such as point D in the Fig. Out of the total

quantity q received, all shortages equal to an amount S are first taken care and

the remaining quantity Z„, = q - s forms the inventory for the next cycle.

Inventory Level

Fig. Inventory situation for model 2 (a).

Here, the total time period T is divided into n equal time intervals, each of

value t. The time interval t is further divided into two parts t1 and t2.

i. e., t = t1 + t2.

170

where t j is the time interval during which items are drawn from inventory

and tz is the interval during which the items are not filled. Using the

relationship of similar triangles,

9

Now total inventory during time t= area of 40AB = 1 Z [m . ti.

.. Inventory holding cost during time t = 2 Ci Im . ti.

Similarly, total shortage during time t = _area of 48CD = 2 (q- Im) t,

.'. Shortage cost during time t = Z Cz (q- Im) tz, and ordering cost during

time t = C3.

Total cost during time t= 2 C1 Im t, + Z Cz (q- Im) tz -F C3 or total average

cost per unit time,

171

Total average cost per unit time C (Im, q) being a function of two variables

Im and q, has to be partially differentiated w.r.t. im and q separately and

then put equal to zero.

= 0, which gives

The minimum average cost per unit time from equation (12.27) is given by

From equation (12.28) we observe that unless C, is zero, optimum order level

Im is less than the demand q during the time interval t. Therefore, it is

advantageous to plan for shortages.

Model 2 (c) (Demand Rate Uniform, Production Rate Finite, Shortages

Allowed)

This model has the same assumptions as in model 2(a) except that production

rate is finite. Figure 12.6 shows the variation of inventory with time.

Referring to Figure 12.6, we find that inventory is zero in the beginning. It

increases at constant rate (K - R) for time t, until it reaches a level Im. There is

no replenishment during time tz, inventory decreases at constant rate R till it

becomes zero. Shortage starts piling up at constant rate R during time t3 until

this backlog reaches a level s. Lastly, production starts and backlog is filled at a

constant rate K - R during time 14 till the backlog becomes zero. This completes

one cycle; the total time taken during this cycle is

172

Inventory situation for model 2 (c).

t=t1 +t2 + t3 + t4.

This cycle repeats itself over and over again. Now holding cost during 1916

interval t

Now C is a function of six variables I„„ s, t i, tz, ty and 4 but we can derive

relationships which determine the values of Im, t t, ty t3 and tq in terms of only two

variables q and s. An inventory policy is given when we know how much to

produce i.e., q and when to start production, which can be found if s is known.

The manufacturing rate multiplied by the manufacturing time gives the

manufactured quantity.

7.6 THE BASIC DETERMINISTIC INVENTORY MODELS

In this section, we shall consider four different types of inventory models, starting

from the basic economic order quantity model. Other three models will simply

reflect one or more changes in the basic assumptions of the initial model.

The notations used in the development of models are as follows:

Q = Number of units (or quantity) ordered (supplied) per order (units /order)

D = Demand W units (usage) of inventory per year (units/year)

N = Number of orders placed per year TC = Total inventory cost (Rs./year)

Co = Ordering cost per order placed

173

C = Purchase or manufacturing price per unit inventory

Ch = Carrying or holding cost per unit per period of time the inventory is held

(Rs. per unit)

Cs = Shortage cost, per unit of inventory or expressed as a percentage of

average rupee value of inventory (%)

R = Reorder point units

L = Lead time (weeks or months)

t= Reorder cycle (the elapsed time between placement of two successive

orders measured as a fractional part of the standard time period)

rp = Replenishment (or production) rate at which lot size Q is added to

inventory.

In the development of these models, we are assuming a standard time horizon of

one year (which is more common in actual practice) but depending on the

requirement of a particular situation this period could be different, say a month, a

week or even a day.

Model I : Economic Order Quantity Model with Uniform Demand

As inventory is used to cover actual demand and there is a need for

replenishment, it is important to decide on how much to order at a time, it is

desirable to order in quantities that will balance he costs of holding too much

stock against that of ordering in small quantities too frequently. This order

quantity is called an Economic Order Quantity.

The objective of the study of this model is to determine an optimum order

quantity (EOQ) such fiat the total inventory cost is minimized. We illustrate this

model after making the following assumptions:

1. Demand

D = demand rate is constant and known throughout the reorder

cycle time.

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2. Replenishment

r = replenishment (or production) rate is instantaneous, i.e. the entire

order quantity Q is received at one time as soon as the order is

released

L = lead time is constant and zero

C = purchase price or cost per unit is constant, i.e. quantity discounts are

not allowed

3. Costs

Ch, Co = unit costs of carrying inventory and ordering are known

and constant

Cs = shortage is not allowed

4. Decision variables

Q = order quantity (replenishment size)

The behaviour of inventory at hand with respect to time is illustrated graphically

in figure. The downward sloping line in the figure shows that the level of inventory

is reducing a t a uniform and known rate over time. Since the demand is uniform

and known exactly and supply is instantaneous, the reorder point is that when

the inventory level falls to zero.

Fluctuation W inventory levels for a given item over time reflects 'the repetitive

cycles of depletion and replenishment as shown in the Fig. 12.3. Since actual

amount invested in inventory varies constantly, this can be simplified by using the

concept of average inventory.

With a constant rate of demand, average inventory is simply the arithmetic mean

of maximum and minimum levels of inventory, i.e.

Maximum level + Minimum Level Average inventory = 2

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Fig. EOO Model with Uniform Demand;

Now find the optimum level of Q so that total inventory cost is minimized.

The inventory costs are determined as follows:

1. Ordering cost = Total annual demand/Quantity ordered each time x Co

= D/Q x Co

2. Carrying cost = Average units in inventory x Carrying cost per unit

= Q/2 x Ch

The total variable inventory cost then, is the sum of these two costs:

TC = D/Q Co + Q/2 Ch

A generalized graph of the ordering cost, carrying cost and total cost is shown in

Fig. 12.2. From this graph, we can see that the ordering cost, which varies with

the inverse of the order quantity generates a downward sloping hyperbola, and

carrying cost, which varies directly with the order quantity, can be represented by

an upward sloping line. The total cost is minimum at a point where ordering costs

equal carrying costs. Thus, economic order quantity occurs at a point where:

Ordering cost = Carrying cost

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Thus optimal Q* (EOQ) is derived to be

This expression is widely known as the wilson lot size formula. Characteristics

of Model I

1. Optimum number of orders placed per year

2. Optimum length of time between orders (duration of supply)

3. Minimum total yearly inventory cost

Remarks

177

1. If the carrying cost is given as a percentage of average value of inventory

held, then total annual carrying cost may be expressed as

Ch= I x P

The total annual inventory cost then becomes

Thus optimal Q* (EOQ) will be

2. The general formula for optimal order quantity Q* can also be obtained by

using calculus (concept of maxima & minima). The method is illustrated below:

Differentiating TC with respect to Q, we have

Equating d(TC)/dQ equal to zero so as to find out the optimum value of Q,

because the value of second derivative of TC with respect to Q is positive, we get

Model II: Economic Order Quantity with Different Rates of Demands in

Different Cycles

178

In this model all those assumptions which were used to derive the economic

order quantity in model I are valid except that the demand rate is different in'

different cycle. The total demand D is specified as demand during time horizon T.

Thus the inventory costs are as follows:

Ordering cost = D/ Q C0

Carrying costs = Q/2 ChT

The total inventory cost is the sum of above two costs:

TC = D/C C0 + Q/2 ChT

For calculating Q*, equating ordering costs and carrying costs we get

D/ Q C0 = Q/2 ChT

Thus optimal Q* (EOQ) is derived to be

The minimum total yearly inventory cost is

Limitations of EOQ Model

Several types of assumptions made in the derivation of EOQ model formula have

been a subject of persistent controversy. Following are the limitations of the EOQ

model :

1. Demand is assumed to be known with certainty and uniform. However, in

actual practice the demand is neither known with certainty nor uniform.

Therefore, when the fluctuations are more, then the model loses its validity.

2. Ordering is not linearly related to number of orders. As the number of orders

increases, the ordering cost rises in stepped manner.

3. Ordering cost may not be independent of the order quantity.

4. Instantaneous supply of inventory when inventory level touches zero is not

possible.

5. The formula is not applicable when inventory cost is meaningless as in the

case of departmental stores from main warehouse to sub-stores.

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6. It is very laborious to calculate inventory carrying cost for B and C class of

items.

Example 12.1 Novelty Ltd. carries a wide assortment of items for its

customers. One item, Gaylook, is very popular. Desirous of keeping its inventory

under control, a decision is taken to order only the optimal economic quantity, for

this item, each time. You have the following information. Make your

recommendations:

Annual demand : 1,60,000 units

Price per unit : Rs. 20

Carrying cost : Re. 1 per unit or 5 per cent per rupee of

inventory value

Cost per order : Rs. 50

Determine the optimal economic quantity by developing the following table:

Size of order

No. of orders 1 10 20 40 80 100

Average inventory

Carrying costs

Order costs

Total costs

Solution Total cost associated with the six different order sizes is calculated as

shown in Table 12.2.

Orders

per Year Lost Size

Average .

Inventory.

Carrying

Cost (Re. 1)

Ordering Cost

(Rs. 50 per order)

Total Cost

per Year

1 1,60,000 80,000 80,000 50 80,050

10 16,000 8,000 8,000 500 8,500 20 5,000 4,000 4,000 1,000 5,000 40 4,000 2,000 2,000 2,000 4,000 SO 2,000 1,000 1,000 4,000 5,000 100 1,600 800

.

800 5,000 5,800

180

Since the total cost per year is minimum when number of orders in a year are 40,

therefore the economic order quantity is 4,000 units.

181

7.7 THE EOO MODELS WITH PRICE (OR QUANTITY) DISCOUNTS

In the previous EOQ models, the price per unit of the item held in the inventory

was constant regardless of the amount ordered, i.e., this cost was independent of

the order size Q. However, there are many situations in which the order size Q

should be influenced by the fact that a lower per unit price may be offered (price

discount) by the suppliers to encourage large orders from their customers. In

such cases it is desirable to ensure whether the savings in purchase cost due to

price discount(s) for large orders combined with a decrease in ordering costs are

sufficient to balance the additional carrying costs due to increased average

inventory.

Such discounts help (i) suppliers in moving more inventory forward in the

distribution channel and lowering carrying costs if buyers purchase in large

quantities and (ii) buyers in trading off lowered purchasing and ordering cost with

higher carrying costs.

If we were to ignore the price discount factor then Q* = 2DCo /Ch . However if

price discount is available then total cost per unit of the inventory system and

items would be

where C1 represents the per unit cost of the items stocked at price break points

and ordinarily C1 > C2 > C3, . . .. From there it is apparent that we do not wish to

order less than Q* because both costs (cost of inventory system and cost of

items) would be higher. However, if we were to order more than Q, it is possible

that the increase in the inventory system cost would be more than compensated

for by the savings in the cost of items. If, it was desirable to increase the order

size to obtain the price-break then the increase should be enough to get the

lower price; anything more would incur unnecessary high inventory costs.

In summary the method of determining optimal order quantity is as follows :

1. Determine the EOQ (Q*) on the basis of the non-discounted original price

2. Determine optimal cost at this EOQ (Q*) point

182

3. Calculate total cost for quantity discount EOQ (D*) points

4. Compare the total cost at EOQ (Q*) with non-discounted price with that for

quantity discount EOQ (D*) at higher volumes. Select either EOQ (Q*) or

EOQ (D*) with the lowest cost.

Quantity Discounts

Example 12.11 A factory requires 1500 units of an item per month, each costing

Rs. 27. The cost per order is Rs. 150 and the inventory carrying charges work

out to 20% of the average inventory. Find out the economic order quantity and

the number of orders per year.

Would you accept a 2% price discount on a minimum supply quantity of 1200

units? Compare the total costs in both the cases.

Solution : From the data of the problem, we have

Annual demand (D) = 1500 x 12 = 18,000 units

Purchase cost (C) = Rs. 27 per item

Ordering cost (Ca) = Rs. 150 per order

Carrying cost (Ch) = C x I = 27 x 0.20 = Rs. 5.40

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UNIT : 8 Queuing Theory

WAITING LINE OR QUEUING THEORY

8.1 Introduction. In everyday life there is u flow of customers to., avail some service

facility at some service station

n: The rate of flow depends on the nature of the service and the servicing capacity of the

station. In many situations there is a congestion of items arriving for service because an

item cannot be serviced immediately on arrival and each new arrival has to wait for some

time before it is attended. This situation occurs where the total number of customers

requiring service exceeds the number of facilities. A group of customers/items waiting at

some place to receive attention/service. including those receiving the service, is known as

queue.

Similarly softie service facility waits for arrival of customers when the total capacity of

system is more than the number of customers. Thus, in the absence of a perfect balance

between the service facility and the customers, waiting is required either by the service

facility or- by the customer. The imbalance between the customers and service facility;

known as congestion, cannot be eliminated completely but efforts/techniques can be

evolved to reduce the magnitude of congestion or wating time of a new arrival in the

system.

A resonably long waiting line may result in loss of customers to the organisation.

The method of reducing congestion by the expansion of servicing counter may result in

an increase in adle time of the service station and may become uneconomical for the

organisation. Thus both the situations namely of unreasonably long queue or expansion

of servicing counters are uneconomical to individual or managers of the system. The

arrival pattern and servicing time of the units in the system are influenced by a number of

factors and can never be controlled or assessed in advance. The waiting line phenomenon

is the direct result of randomness in the operation of service facilities. The customers

184

arrival or his service time are not known in advance, for otherwise the operation of the

facility cannot be scheduled so as to eliminate waiting completely. Hence the problem of

waiting line is quite complicated and requires careful study. For this, some sort of

equilibrium between the costs associated with waiting and costs of preventing waiting is

to be evolved by determining the optimal service time and arrival rate of the units in the

system.

In this chapter we shall try to develop a mathematical theory to study the problem. Here

we shall try to find expected waiting time for a particular arrival and the time it shall take

for servicing in a given situation (by calculating probability of customer's entering the

system and the probability of their servicing times). The theory can help us in making

predictions about the behaviour of . the system in terms of its service, estimation of

possible congestion and methods to eliminate them. The following are some of the

instances where we generally come across waiting line problem.

(i) Check-out stations and 'personnel determination. It can determine the desired

number of check-out stations and personnel needed in super markets and

departmental stores to ensure smooth and economic operations at different times

of the day.

(ii) Aeroplane departure analysis using queuing theory enables airlines to schedule

the departures of their planes with respect to the schedules of competitor airlines.

(iii) Machine breakdown and repairs analysis determines the number - of repair

personnel required to handle the breakdown with minimum overall costs. Here the

broken machine is considered as customer needing service of a repairman.

(iv) The theory can also be applied to the staffing; of clerical operations. Letters

arriving at typists desk, the letters are the customers and the typist the server:

(v) Queuing theory has significant impact on the design of inventory and production

control systems.

(vi) In dock yards the dock costs and demurrage costs can be quite large and one

should find optimum number of docks which minimises these two costs.

185

(vii) In many cases the workers are assigned different volume of tasks e.g. one worker

may be required to operate one machine where ,as other operates two machines at

a time. The basic salary of the two workers may be same but bonus as incentive

may be given on the basis of extra production. It is observed that variation in

down time due to repairs of machines the worker operating with two machines

would have to operate at a greater level of efficiency to earn the same amount of

bonus. Queuing theory can be applied to decide waage incentive schemes.

(viii) Queuing situations commonly experienced are

(a) In banks; cafeterias. (b) Jobs waiting for processing by a computer. (c)

Employees waiting for promotion (d) Cars waiting for traffic lights to turn green

(e) Doctor's office, hospitals (f) Barber's shop (g) Telephone booths, or calls

arriving at a telephone switch board (h) Booking offices, Post offices etc. (i)

Students depositing fees at various counters, book stores, libraries (j)

Automobile's repair shop, petrol pumps etc.

It is evident from the above illustrations, that if service capacity is not sufficient [o cope

with the demand then waiting line is generated in the system and unless some remedial

measure is introduced, it is likely to grow with time. Similarly, if the service facility is

more than the demand rate then the idle time for the service facility is likely to increase.

Both these situations are not in the interest of the customer as well as the organisation.

Naturally, the organisation providing the service facility would like to find that what

should be the level of waiting time of an arrival in the system at which it is worthwhile to

install extra/additional capacity in the system. Alternately, the management may allow

the arrivals to wait for extra time and risk the losses than to face the situation of idle

capacity. All these situations need exhaustive analysis and study by some analytical tools.

8.2 Historical Development. The telephone industry is responsible for queuing theory.

Theoretical research into the properties of queues first of all started in the problem of

telephone calls by a Sweedisn engineer Mr. A.K. Erlang in 1903. The theory was further

developed by Molllns in 1927 and then by Thornton D. Fry. A systematic approach to the

problem was made by Mr. D.G. Kendall in 1951 by using -model terminology and since

186

then significant work has been done in this direction. Now queueing theory has been

applied to wide variety of operations. The basic feature of all these operations is that the

sequence of units amviing at the service stations are eventually discharged after service.

8.3 Queuing Process/System. Basically, a queuing process is centred around a service

system (facility).

All queuing situations involve the arrival of customers (input) at a service facility, where

some time may be spent in waiting and then receiving he desired service. The customers

arriving for service may or may not enter he system. Thus, the input pattern in the system

depends on the nature of he system as well as the behaviour of the customer. The

combination of these two determines the arrival pattern. After the service is completed

the customer leaves the service system (output). The departure pattern mainly fepends on

the service discipline. There can he many types of qneung systems depending on the

nature of inputs, service mechanism and customer characteristics.

Components of a Queuing System:

187

Input implies the mode of arrival of customers at the service facility. The number of

customers emanate from finite, or infinite sources. Typically customers arrive at the

system randomly singly or in batches. The input process is characterised by the nature of

the a.-rivals, capacity of the system and the behaviour of the customers.

(A) The size of customers arriving for servicing depends on the nature of the

population which can be finite or infinite. From practical view point a population

is considered to be finite, if the probability of an arrival is greatly changed when

one member of the population is already receiving service.

The periods between the arrival of individual customers may be constant or

scattered in some fashion. Most queuing models assume that the same inter-

arrival time distribution applies for all customers thoughout the period of study.

The most convenient way is to designate some random variables corresponding to

the times between arrivals. In general the arrivals follow a Poisson' distribution

when the total number of arrivals during any given time interval is independent of

the number of arrivals that have already occured prior to the beginning of time

Interval.

188

(B) In many systems the capacity of the space where the arrivals have to wait before

taken into service is limited. In such cases when the length of waiting line crosses

a certain limit, no further units/arrivals are permitted to enter the system till some

waiting space becomes vacant. Such queue systems are known as systems with

finite capacity and considerably affects the arrival pattern of the system e.g. a

doctor may give appointment to fixed number of patients each day.

(C) The human behaviour and the facilities of servicing in any system are important

factors for the development of queuing problem. The behaviour of the customer,

mainly his impatience may affect the nature of the system. Customer's behaviour

can be classified in following categories.

(i) Balking. A customer may not like to join the queue seing it very long and

he may not like to wait.

(ii) Reneging. He may leave the queue due to impatience after joining i[.

(iii) Collusion. Several customers may collaborate and only one of them may

stand in the queue.

(iv) Jockeying. If there are number of queue; then one may leave one queue to

join another.

2. Service Facility/Mechanism. This means the arrangement of server's facility to

serve the arriving customer. Service time in waiting line problems is also a

statistical variable and can be studied either as the number of services completed

in a given period of time or alternately the completion time of a service. Service

mechanism of any system is mainly determined by:

189

A : Service Facility Design : The facilities at the service station can be divided in two

main categories (i) Single channel and (ii) Multi-channel facilities.

(i) Single channel Queues : There may be only one counter for servicing and

as such only one unit can be served at a time. The next unit can be taken

into service when the servicing operations on the previous unit are

completed. The single channel queue can be divided in two types :

(a) Single phase (b) Multi-Phase.

In a single phase queue, the whole service operations are completed in one

stage fig (6-2A)

(a) Single channel single phase queue

(b) Single channel multi-phase queue : Here the unit taken for service

has to pass through many stages before the unit goes out of the servicing

channel. All the phases of service are arranged in a ordered sequence (see

Fig. 6.2 B).

Single channel-k-Phases arranged in series.

(ii) Multi channel. Due to rush of customers, management may decide to provide a

number of service counters so that queue length may not become unreasonably large and

the organisation may not loose customers'

due to long queue. But loo many counters may result in long idle time of counters due to

shortage of customers.

(a) Multi channel queue discipline with single phase :

190

(b) A mixed arrangement of servicing facilities arranged in parallal and series can be

termed as multi-channel multi-phase queue discipline. Here the servicing of any

unit taken into service is completed into a number of stages arranged in series.

See Fig. (6-2 D) e.g. several ticket counters in a cinema may send all customers to

one ticket collector and vice-versa.

B. Queue/Service Discipline : Queue Discipline identifies the order in which

arrivals in the system are taken into service. The Queue discipline does not always

take into account the order of arrivals. Various methods are available to solve

queuing problems under different queuing disciplines but most of these introduces

complications in the analysis. The most common discipline is First In First Out

(FIFO) or First come First Served (FCFS) discipline. Here the customers are

serviced strictly in the order of their joining the system e.g. queues at booking

stations.

The Last Come First Served (LCFS) or Last in First Out (LIFO) System is one

where the item arriving last are first to go into service e.g. in big stores the items

arriving last are issued first. Similarly in elevators passenger entering last may stand

near the gate and thus may leave first.

Service In Random Order (SIRO) rule implies that arrivals are taken into service

randomly, irrespective of the order of their arrivals in the system. Here the server

chooses one of the customers to offer service at random. e.g. in a government office

processing of papers often takes place in an indiscriminate order. These disciplines

are useful in allocation of an item whose demand is high and supply is low viz

allotting the shares to applicants by a company. Sometimes SIRO is the only

alternative to assign service as it may not be possible to identify the order of

arrivals:

191

Priority Disciplines are those where any arrival is chosen for service ahead of some

other customers already in queue. In the case of 'Pre-emptive' priority the preference

to any arriving unit is so high that the unit already in service ,is renowned/displaced

to take it into service: A non-pre-emptive rule of priority is one where an arrival

with low priority is given preference for service than a high priority item.

8.4 Classification of Queues and their problems.

The mathematical description of a Queue can be formulated by means of a model

expressed as A/B/S : (d/,f ) where

A : Arrival pattern of the units, given by the probability distribution of inter-

arrival time of units.

B : The probability distribution of servicing time of individuals being actually

served.

S: The number of servicing channels in the system.

D: Capacity of the system i.e. the maximum number of units the system can

accommodate at any time.

F: The manner/order in which the arriving units are taken into service i.e.

FIFO/LIFO/SIRO/Priority.

8.5 The various queuing problems are related with

(i) Queue length. Number of persons in the-system at any time. We may be

interested in studying the distribution of queue length.

(ii) Waiting time. It is the time upto which an unit has to wait before it is taken

into service after arriving at the servicing station. This is studied with the

help of waiting time distribution.

The waiting time depends on

(a) the number of units already there in the system,

(b) the number of servicing stations in the system,

(c) the schedule in which units are selected for service.

192

(iii) Servicing time. It is the time taken for servicing a particular arrival.

(iv) Average length of tine. The number of customers in the queue per unit of

time.

(v) Average idle -time. The average time for which the system remains idle.

Notations.

X : The inter-arrivai time between two successive customers (arrivals).

Y : The service time required for any customer.

w : The wailing time for any customer before it is taken into service.

V : Time spent by a customer in the system.

n : Number of customers in the system i.e. in the waiting line at any

time, including the number of customers being Serviced.

Pn (t) : Probability that n Customers arrive in the system in time

0n (t) : Probability that n units are serviced in time t.

U (T) : Probability distribution of inter -arrival time P (t < T).

V (T) ; Probability distribution of servicing time P (t < T).

F (N) : Probability distribution of queue length a[ any time.

P (N < n)

En : Denotes same state of the system at a time when there are n units in

the system.

λn : average number of customers arriving per unit of time when there

are already n 'units in the system.

λ : average number of customers arriving 'per unit of time.

μn : average number of customers being served per unit of time when

there are already n units in the system.

μ : average number of Customers being served per unit of time.

(λ/ μ) = p : is known as traffic intensity.

Steady, Transient and Explosive states in a Queue system. The study of waiting

line depends on the distribution of arrivals and the distribution of customer's service

times. It is observed that under fixed conditions of customer arrivals and servicing

193

facility a queue length is a function of time. As such a queue system can be considered

some sort of random experiment and the various events of the experiment can be taken

to be the various changes occurring in the system at any time.

In the case of arrivals in a queue system three states of nature are possible, namely, the

steady state, transient state, and the explosive state. A short description of these states is

given below :

(i) Steady State. If the average rate of arrival is less than the average rate of

service, and both are constant, the system will eventually settle down into steady state

and becomes independent of the initial state of the queue. Then the probability of

finding a particular length of queue at any time will be same. The size of queue will

fluctuate in the steady state but the statistical behaviour of the queue remains steady.

A necessary condition for the steady state to be reached is that the elapsed time since

the start of the operation becomes sufficiently large i.e. ( t æ ) , but this condition is

not sufficient as the existence of steady state also depends upon the behaviour of the

system e.g. if rate of arrival is greater than the rate of service then a steady state cannot

be reached. Here we assume that the system acquires a steady state as t æ i.e. the

number of arrivals during a certain interval becomes independent of time

Hence in the steady state system, the probability distributions of arrivals, waiting time,

and servicing time does not depend on time.

(ii) Transient State. A system is said to be in "Transient state'. when its operating

characteristics are dependent on time. Usually a system is in transient state during the

early stages of its operation when the behaviour of the system is dependent on the

initial state of queue. When, the probability distribution of arrivals, waiting time and

servicing time are dependent on time the system is said to be in Transient State.

(iii) Explosive State. Here u5e waiting line increases indefinitely with time e.g. arrivals

in a restaurant during rush hours.

In this chapter, it is assumed that the system has attained a steady

194

state.

Distribution of Arrivals and Service times in Queuing System. W- can see that

under fixed conditions of customer arrivals and servicing facility a queue length is a

function of time. A queue is some sort of random experiment and the probability

models for arrival and service times must be accurately ascertained before the

properties of the queue for a given system are studied. Most elementary Queue models

assume that the inputs and outputs follow special type of process known as Birth and

Death process. In this section we shall study some well known distributions of arrival

and service times.

Poisson Process. It is applied to a system where the changes are independent of time i.e.

the factors which affect the changes remain absolutely unchanged and the probability of

occurrence of any event at any time is independent of time. An infinite sequence of

independent events occurring at an instant of time form a Poison Process, if the

following conditions are satisfied :

(i) The total number of events in any time interval X does not depend on the events

which has occurred before the beginning of the period i.e. the number of arrivals in

non-overlapping interval are statistically independent and the process has independent

and identically distributed increments.

(ii) The probability of an event occurring in a small time interval At is X At + 0

(At)2 where A is some constant and 0 (At)

2 means some function of At of order

>_ 2, i.e. If (A1) I < K.as At---) 0 for any constant K {~t)

(iii) Two or more units cannot arrive or serviced at the same time i.e. the probability

of the occurrence of more than one event in the interval t and t + At is of the order 0

(At) which is negligible.

Distribution of Arrivals. Here we explain the concept by considering one

probability distribution for time between successive arrivals, known as exponential

distribution. The distribution of arrivals in a queuing system can be considered as a

195

Pure Birth Process. The term birth refers to the arrival of new calling units in the

system. Here the objective is to study the number of customers that enter the system i.e.

only arrivals are counted and no- departure takes place. Such process is known as pure

birth process, e.g. in an office, the computer operator waits until at least five records

accumulate before feeding the information to the computer.

Mathematically, our objective is to derive an expression for the probability Pn (t) of n

arrivals during the time interval (0, t), assuming that system started its operation at time

t = 0 i.e. number of arrivals in time interval (0, t) is taken to be a random variable

following a Poisson Distribution with parameter ?.t.

How to recognise a Poisson Distribution validity for a given -situation ?

In Queuing theory studies Poisson Probability distribution plays an important role.

One can ascertain the validity of Poisson distribution by analysing the arrivals and

departures using the following procedure :

I. If the queuing situation is already in existence then observe it for while

to identify the random/non-random pattern of successive arrival;. If the arrivals are

random, there is a good chance that the process may follow a Poisson distribution.

lI. Gather observations about the number of arrivals by recording the number

of customers arriving during appropriate time intervals. After collecting sufficient

amount of data compute its mean and variance. If these are approximately equal

then the distribution of arrival is Poisson.

The probability distribution of the arrival pattern can also be s tudied and identified

through analysis of past data. It is also known as the Poisson input i.e. when the

arrival pattern of customers in the system follows a Poisson Process.

Theorem. To show that under the three conditions of' a Poisson Process the number

of arrivals in a fixed time full the Poisson law i.e. if die probability of an arrival in

time interval t and t + then

196

Proof. Let us consider the consecutive time intervals (0, t,) and (t, t + t),

l-hcn in t:rc interval (t, t + At), n arrivals can uLkc place in following three

mutu~!Iv exclusive ways :

(i) there are n arrivals in the interval (t + t t) and no arrival in the interval (t + t t)

It is assumed that the number of arrivals in non-overlapping interval are statistically

independent i.e. total number of events in any time interval X does not depend on

the events which has occurred before the beginning of the period.

8.6 SINGLE-CHANNEL QUEUING THEORY

A single-channel queuing problem results from random interarrival time and

random service time at a single service station. The random arrival time can be

described mathematically by a probability distribution. The most common

distribution found in queuing problems is Poisson distribution. This is used in

single-channel queuing problems for random arrivals where the service time is

exponentially distributed. The sections ahead give the reader an insight into the

true nature of operations research the difficulties of developing OR models, the

need for logical assumptions and the utilization of higher mathematics.

Models for Arrival and Service Times

Generally, arrivals do not occur at fixed regular intervals of times but tend to be

clustered or scattered in some fashion. A Poisson distribution is a discrete

probability distribution which predicts the number of arrivals in a given time. The

Poisson distribution involves the probability _i occurrence of an arrival. Poisson

assumption is quite restrictive in some cases. It assumes that arrivals are

random and independent of all other operating conditions. The mean arrival rate

(i.e., -e number of arrivals per unit of time) )L is assumed to be constant over time

197

and is independent :f the number of units already serviced, queue length or any

other random property of the queue.

Since the mean arrival rate is constant over time, it follows that the probability of

an arrival between time t and t + dt is λ dt.

Thus probability of an arrival in time dt = λ dt. (10.1)

The following characteristics of Poisson distribution are written here without proof

Probability of n arrivals in time

Probability density function of inter-arrival time (time interval between two

consecutive arrivals)

,..(10.3)

Finally, Poisson distribution assumes that the time period dt is very small so that

(dt)2, (dt)3 etc. 0 and can be ignored.

Service time is the time required for completion of a service i.e., it is the time

interval between beginning of a service and its completion. The mean service

rate is the number of customers served per unit of time (assuming the service to

be continuous througout the entire time =it), while the average service time 1/μ is

the time required to serve one customer. Tile most common type of distribution

used for service times is exponential distribution. It involves the probability of

completion of a service. It should be noted that Poisson distribution cannot be

applied to servicing because of the possibility of the service facility remaining idle

for some time. Poisson distribution assumes fixed time interval of continuous

servicing, which can never be assured in all services.

Mean service rate p is also assumed to be constant over time and independent

of number of units already serviced, queue length or any other random property

of the system. Thus probability that a service is completed between t and t + dt,

provided that the service is continuous

= μdt.

198

Under the condition of continuous service, the following characteristics of

exponential distribution are written, without proof :

Probability of n complete services in time

Probability density function (p.d.f) of interservice time, i.e., time between two

consecutive services = ...(10.5) Probability that a customer shall be serviced in more than time t = e"~`. ...(10.6)

Model I. Single-Channel Poisson Arrivals with Exponential

Service, Infinite Population Model [(M/M/I) :

Let us consider a single-channel system with Poisson arrivals and

exponential service time distribution. Both the arrivals and service rates are

independent of the number of customers in the waiting line. Arrivals are

handled on `first come, first served' basis. Also the arrival rate 7,, is less

than the service rate p.

The following mathematical notation (symbols) will be used in connection

with queuing models:

n = number of customers in the system (waiting line + service facility) at time

λ = mean arrival rate (number of arrivals per unit of time).

μ =mean service rate per busy server (number of customers served per unit

of time).

λdt = probability that an arrival enters the system between t and t + dt time

interval i.e., within time interval dt.

1 - λdt = probability that no arrival enters the system within interval dt plus higher

order terms in dt.

μ = mean service rate per channel.

μ dt = probability of one service completion between t and t + dt time interval i. e.,

within time interval dt.

1 - μ.dt=probability of no service rendered during the interval dt plus higher order

terms in dt.

pn = steady state probability of exactly n customers in the system.

199

pn (t) = transient state probability of exactly n customers in the system at time t,

assuming the system started its operation at time zero.

pn+1 (t) = transient state probability of having n + 1 customers in the system at

time t.

pn-1 (t) = transient state probability of having n - 1 customers in the system at

time t.

pn (t + dt) = probability of having n customers in the system at time t + dt.

Lq = expected (average) number of customers in the queue.

Ls = expected number of customers in the system (waiting + being served).

Wq = expected waiting time per customer in the queue (expected time a

customer keeps waiting in line).

Ws = expected time a customer spends in the system. (in waiting + being served)

Ln = expected number of customers waiting in line excluding those times when

the line is empty i.e., expected number in non-empty queue (expected number of

customers in a queue that is formed from time to time).

Wn = expected time a customer waits in line if he has to wait at all i.e., expected

time in the queue for non-empty queue.

To determine the properties of the single-channel system, it is necessary to find

an expression for the probability of n customers in the system at time t i.e., pn (t),

for, if pn (t) is known, the expected number of customers in the system and hence

the other characteristics can be calculated. In place of finding an expression for

pn (t), we shall first find the expression for Pn (t + dt).

The probability of n units (customers) in the system at time t + dt can be

determined by summing up probabilities of all the ways this event could occur.

The event can occur in four mutually exclusive and exhaustive ways:

TABLE

Event No, of units No. of arrivals No. of

services

No. of units

at time t in time dt in time dt at time t + dt

1 n 0 0 n

2- n+l 0 1 n

200

3 n-I 1 0 n

4 n 1 1 n

Now we compute the probability of occurrence of each of the events,

remembering that the probability of a service or arrival is μdt or λdt and (dt)2 - 0.

:. Probability of event 1 = Probability of having n units at time t

x Probability of no arrivals

x Probability of no services

Note that other events are not possible because of the small value of dt that

causes (dt)2 to approach zero (as in event 4).

Since one and only one of the above events can happen, we can obtain pn(t +

dt) [where n > 0} by adding the probabilities of above four events.

Taking the limit when at ---> 0, we get the following differential equation which

gives the relationship between p„, p„-i, p„+ at any time t, mean arrival rate ),.

and mean service rate u :

201

After solving for pn(t + dt) where n > 0, it is necessary to solve for pn (t + dt)

where n = 0 i.e. to solve for po (t + dt). If n -- 0, only two mutually exclusive

and exhaustive events can occur as shown in table 10.2.

202

TABLE

Event No. of units No. of arrivals No. of services No. of units

at time t in time at in time at at time t + at

1 0 0 - 0

2 1 0 1 0

Note that if no units were in the system, the probability of no service would be

1. Probability of having no unit in the system at time t+dt is given by summing

up the probabilities of above two events.

When dt - 0, the differential equation which indicates the relationship between

probabilities pt, and p, at any time t, mean arrival rate λ and mean service rate

μ is

Equations (10.7) and (10.8) provide relationships involving the probability density

function pn(t) for all values of n but still we do not know the value of pn(t).

Assuming the steady state condition for the system, when the probability of

having n units (customers) in the system becomes independent of time, we get

Pn(t) =Pm d IPn(t)I = 0.

203

Therefore, for a steady state system the differential equations (10.7) and (10.8)

reduce to difference equations (10.9) and (10.10) :

Similarly, for n = 2, equation (10.9) gives

Equation (10.11) gives p. in terms of po, λ and μ. Finally, an expression for po in

terms of λ and μ must be obtained. The easiest way to do this is to recognize

that the probability that the channel is busy is the ratio of the arrival rate and

service rate, λ / μ . Thus po is 1 minus this ratio.

204

Having known the value of pn, we can find the various operating characteristics

of the system.

1. Expected number of units in the system (waiting + being served), L, is obtained

by using the definition of an expected value:

205

An Explanatory Note on the Queuing Formulae

1. Traffic intensity. The ratio λ / μ is called the traffic intensity or the

utilisation factor and it determines the degree to which the capacity of the

service station is utilised (expected fraction of time the service facility is

busy). For instance, if customers arrive at the rate of 9 per hour and the

service rate is 10 per hour, the utilisation of the service facility is 9/10 =

90%.

2. Average length o f the queue = λ / μ . λ / μ – λ

Consider that a statistician observes the queue at a service facility after

every one hour and that the length of the queue for, say, six observations is

as follows:

Observation No. Queue Service,fdciliy Length of queue

1 None None 0

2 ** * 2

3 None * 0

4 ***** * 5

5 ** * 2

6 *** * 3

Thus the average queue length = 2.

3. Average number of units in the system = λ / μ – λ

For the above situation, this average

(0+0)+(2+1)+(0+1)+(5+1)+(2+1)+(3+1) = 17

6 6

4. Average length of non-empty queue = λ / μ – λ

This will be calculated from observation no. 2, 4, 5 and 6, since during

observations 1 and 3 the queue was empty, though during observation 3

there was a unit being served. For the above situation, then this average

206

2+5+2+3 = 3

4

5. Average waiting time of an arrival = λ / μ – λ

At times when there are no units in the system, the arriving unit will not

have to wait. However, when there are units already in the system, the

arriving unit will have to wait. Let the waiting times of, say, 8 units observed

be

= 10, 8, 3, 0, 5, 9, 0, 6 minutes.

Then the average waiting time = 10+8+3+0+5+9+0+6 = 41

8 = 8 5.125 minutes.

6. Average waiting time of an arrival who waits = 1/μ – λ

In the above situation, ignoring the observations when the waiting time is

zero, this average

= 41/6 = 6.83 minutes.

7. Average time an arrival spends in the system = 1 / μ – λ.

Here, along with the waiting times, the service times must also be given.

Then this average

Total (waiting time + service time) minutes

8

8.7 Assumptions and Limitations of Queuing Model

The various results of section 10.9.2 have been derived under the following

simplifying assumptions :

1. The customers arrive for service at a single service facility at random

according to Poisson distribution with mean arrival rate )v or equivalently,

the inter-arrival times follow exponential distribution.

2. The service time has exponential distribution with mean service rate p..

3. The service discipline followed is first come, first served.

207

4. Customer behaviour is normal i.e., customers desiring servicejoin the

queue, wait for their turn and leave only after getting serviced; they do not

resort to balking, reneging or jockeying.

5. Service facility behaviour is normal. It serves the customers

continuously, without break, as long as there is queue. Also it serves only

one customer at a time.

6. The waiting space available for customers in the queue is infinite. 7. The

calling source (population) has infinite size.

8. The elapsed time since the start of the queue is sufficiently long so

that the system has attained a steady state or stable state.

9. The mean arrival rate 7v is less than the mean service rate p..

However, in most of the actual business situations the above assumptions

are hardly satisfied. The various limitations in a queuing model are :

1. The waiting space for the customers is usually limited.

2. The arrival rate may be state dependent. An arriving customer, on

seeing a long queue, may not joint it and go away without getting service.

3. The arrival process may not be stationary. There may be peak period and

slack period during which the arrival rate may be more or less than the average

arrival rate.

4. The population of customers may not be infinite and the queuing discipline

may not be first come, first served.

5. Services may not be rendered continuously. The service facility may

breakdown; also the service may be provided in batches rather than

individually.

6. The queuing system may not have reached the steady state. It may be,

instead, in transient state. It is commonly so when the queue just starts and the

elapsed time is not sufficient.

208

EXAMPLE

A self-service store employs one cashier at its counter. Nine customers arrive

on an average every S minutes while the cashier can serve 10 customers in 5

minutes. Assuming Poisson distribution for arrival rate and exponential

distribution for service time, find

1. Average number of customers in the system.

2. Average number of customers in the queue or average queue length. 3.

Average time a customer spends in the system.

4. Average time a customer waits before being served.

[P.T U. B.E., 2001; Karn. U. B.E. (Mech.) 1998, 95]

Solution

209

EXERCISES

1. In a machine shop there are two identical machines. Products arrive for

machining at the average rate of 4 products per hour. The average

machining time is 24 minutes per product. The production manager

complains of loss of production time and sugests the installation of a third

machine. What is the average waiting time per product under the present

circumstances 7 What is it likely to be if a third machine is installed '?

Assume Poisson pattern of arrival and exponentially distributed service

times.

2. In a machine shop there are 10 identical machines. These machines are

subjected to periodical breakdowns and require the service of a maintenance

department. Each machine breaks down in a Poisson pattern. The time

required to put a machine back into production line is exponentially

distributed. Formulate a queuing model to determine the following :

(i) Average down time of the machines (Ws).

(ii) Probability that all the machines are in working order (po).

8.9 MULTI-CHANNEL QUEUING THEORY MODEL VI :

Multi-channel queuing theory treats the condition in which there are several

service stations in parallel and each customer in the waiting line can be

served by more than one station. Each service facility is prepared to deliver

the same type of service. The new arrival selects one station without any

external pressure. When a waiting line is formed, a single line usually breaks

down into shorter lines in front of each service station. The arrival rate 1, and

service rate it are mean values from Poisson distribution and exponential

distribution respectively. Service discipline is first come, first served and

customers are taken from a single queue i.e., any empty channel is filled by

the next customer in line.

Let n = number of customers in the system,

210

pn = probability of n customers in the system,

c = number of parallel service channels (c > 1),

λ = arrival rate of customers,

μ = service rate of individual channel.

When n < c, there is no queue because all arrivals are being serviced, and

the rate of servicing will be n μ as only n channels are busy, each at the rate

of pt. When n = c, all channels will be working and when n > c, there will be

(n - c) customers in the queue and rate of service will be cu as all the c

channels are busy. There will be three cases in this system. To determine

the properties of multi-channel system, it is necessary to find an expression

for the probability of n customers in the system at time t i.e., p„(t).

Case I (When n = O) :

Let us first find po (t + dt). This event can occur only in two exclusive and

exhaustive ways:

211

End Chapter Quizzes

Chapter 1—Introduction

MULTIPLE CHOICE

1. The field of management science

a. concentrates on the use of quantitative methods to assist in decision making.

b. approaches decision making rationally, with techniques based on the scientific method.

c. is another name for decision science and for operations research.

d. each of the above is true.

ANS: D

2. Identification and definition of a problem

a. cannot be done until alternatives are proposed.

b. is the first step of decision making.

c. is the final step of problem solving.

d. requires consideration of multiple criteria.

ANS: B

3. Decision alternatives

a. should be identified before decision criteria are established.

b. are limited to quantitative solutions

c. are evaluated as a part of the problem definition stage.

d. are best generated by brain-storming.

ANS: A

4. Decision criteria

a. are the choices faced by the decision maker.

b. are the problems faced by the decision maker.

c. are the ways to evaluate the choices faced by the decision maker.

d. must be unique for a problem.

ANS: C

5. The quantitative analysis approach requires

a. the manager's prior experience with a similar problem.

b. a relatively uncomplicated problem.

c. mathematical expressions for the relationships.

d. each of the above is true.

ANS: C

6. A physical model that does not have the same physical appearance as the object being modeled is

a. an analog model.

b. an iconic model.

c. a mathematical model.

d. a qualitative model.

212

ANS: A

7. .Management science and operations research both involve

a. qualitative managerial skills.

b. quantitative approaches to decision making.

c. operational management skills.

d. scientific research as opposed to applications.

ANS: B

8. George Dantzig is important in the history of management science because he developed

a. the scientific management revolution.

b. World War II operations research teams.

c. the simplex method for linear programming.

d. powerful digital computers.

ANS: C

9. The first step in problem solving is

a. determination of the correct analytical solution procedure.

b. definition of decision variables.

c. the identification of a difference between the actual and desired state of affairs.

d. implementation.

ANS: C

10. Problem definition

a. includes specific objectives and operating constraints.

b. must occur prior to the quantitative analysis process.

c. must involve the analyst and the user of the results.

d. each of the above is true.

ANS: D

213

Chapter 2—An Introduction to Linear Programming

MULTIPLE CHOICE

1. The maximization or minimization of a quantity is the

a. goal of management science.

b. decision for decision analysis.

c. constraint of operations research.

d. objective of linear programming.

ANS: D

2. A solution that satisfies all the constraints of a linear programming problem except the nonnegativity

constraints is called

a. optimal.

b. feasible.

c. infeasible.

d. semi-feasible.

Ans : C

3. Decision variables

a. tell how much or how many of something to produce, invest, purchase, hire, etc.

b. represent the values of the constraints.

c. measure the objective function.

d. must exist for each constraint.

ANS: A

4. Which of the following is a valid objective function for a linear programming problem?

a. Max 5xy

b. Min 4x + 3y + (2/3)z

c. Max 5x2 + 6y

2

d. Min (x1 + x2)/x3

ANS: B

5. Which of the following statements is NOT true?

a. A feasible solution satisfies all constraints.

b. An optimal solution satisfies all constraints.

c. An infeasible solution violates all constraints.

d. A feasible solution point does not have to lie on the boundary of the feasible region.

ANS: C

6. Slack

a. is the difference between the left and right sides of a constraint.

b. is the amount by which the left side of a constraint is smaller than the right side.

c. is the amount by which the left side of a constraint is larger than the right side.

d. exists for each variable in a linear programming problem.

ANS: B PTS: 1 TOP: Slack variables

7. To find the optimal solution to a linear programming problem using the graphical method

a. find the feasible point that is the farthest away from the origin.

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b. find the feasible point that is at the highest location.

c. find the feasible point that is closest to the origin.

d. None of the alternatives is correct.

ANS: D PTS: 1 TOP: Extreme points

8. Which of the following special cases does not require reformulation of the problem in order to

obtain a solution?

a. alternate optimality

b. infeasibility

c. unboundedness

d. each case requires a reformulation.

ANS: A

9. The improvement in the value of the objective function per unit increase in a right-hand side is

the

a. sensitivity value.

b. dual price.

c. constraint coefficient.

d. slack value.

ANS: B

10. As long as the slope of the objective function stays between the slopes of the binding constraints

a. the value of the objective function won't change.

b. there will be alternative optimal solutions.

c. the values of the dual variables won't change.

d. there will be no slack in the solution.

ANS: C

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Chapter 3— MULTIPLE CHOICE

1.

1. The optimal solution is found in an assignment matrix when the minimum number of straight

lines needed to cover all the zeros equals

a. (the number of agents) 1.

b. (the number of agents).

c. (the number of agents) + 1.

d. (the number of agents) + (the number of tasks).

ANS: B

2. The stepping-stone method requires that one or more artificially occupied cells with a flow of

zero be created in the transportation tableau when the number of occupied cells is fewer than

a. m + n 2

b. m + n 1

c. m + n

d. m + n + 1

ANS: B

3. The per-unit change in the objective function associated with assigning flow to an unused arc in

the transportation simplex method is called the

a. net evaluation index.

b. degenerate value.

c. opportunity loss.

d. simplex multiplier.

ANS: A

4. The difference between the transportation and assignment problems is that

a. total supply must equal total demand in the transportation problem

b. the number of origins must equal the number of destinations in the transportation problem

c. each supply and demand value is 1 in the assignment problem

d. there are many differences between the transportation and assignment problems

ANS: C

5. Using the transportation simplex method, the optimal solution to the transportation

problem has been found when

a. there is a shipment in every cell.

b. more than one stepping-stone path is available.

c. there is a tie for outgoing cell.

d. the net evaluation index for each unoccupied cell is 0.

ANS: D

1 TOP: Transportation simplex method

6. To use the transportation simplex method, a transportation problem that is unbalanced requires

the use of

a. artificial variables.

b. one or more transshipment nodes.

c. a dummy origin or destination.

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d. matrix reduction.

ANS: C PTS: 1 TOP: Transportation simplex method

TRUE/FALSE

7. The transportation simplex method can be used to solve the assignment problem.

ANS: T

8. The transportation simplex method is limited to minimization problems.

ANS: F

9. When an assignment problem involves an unacceptable assignment, a dummy agent or task must

be introduced.

ANS: F

10. In assignment problems, dummy agents or tasks are created when the number of agents and tasks

is not equal.

ANS: T

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Chapter 4

MULTIPLE CHOICE

1. The options from which a decision maker chooses a course of action are

a. called the decision alternatives.

b. under the control of the decision maker.

c. not the same as the states of nature.

d. All of the alternatives are true.

ANS: D

2. A payoff

a. is always measured in profit.

b. is always measured in cost.

c. exists for each pair of decision alternative and state of nature.

d. exists for each state of nature.

ANS: C

3. Making a good decision

a. requires probabilities for all states of nature.

b. requires a clear understanding of decision alternatives, states of nature, and payoffs.

c. implies that a desirable outcome will occur.

d. All of the alternatives are true.

ANS: B

4. A decision tree

a. presents all decision alternatives first and follows them with all states of nature.

b. presents all states of nature first and follows them with all decision alternatives.

c. alternates the decision alternatives and states of nature.

d. arranges decision alternatives and states of nature in their natural chronological order.

ANS: D

5. For a maximization problem, the conservative approach is often referred to as the

a. minimax approach

b. maximin approach

c. maximax approach

d. minimin approach

ANS: B TRUE/FALSE

6. Sample information with an efficiency rating of 100% is perfect information.

ANS: T

7. Decision alternatives are structured so that several could occur simultaneously.

ANS: F

8 .Risk analysis helps the decision maker recognize the difference between the expected value of a

decision alternative and the payoff that may actually occur.

ANS: T

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9. The expected value of an alternative can never be negative.

ANS: F

10. A decision strategy is a sequence of decisions and chance outcomes, where the decisions

chosen depend on the yet to be determined outcomes of chance events.

ANS: T

219

Chapter 5— MULTIPLE CHOICE

1. The problem which deals with the distribution of goods from several sources to several

destinations is the

a. maximal flow problem

b. transportation problem

c. assignment problem

d. shortest-route problem

ANS: B

2. The parts of a network that represent the origins are

a. the capacities

b. the flows

c. the nodes

d. the arcs

ANS: C

3. The objective of the transportation problem is to

a. identify one origin that can satisfy total demand at the destinations and at the same time

minimize total shipping cost.

b. minimize the number of origins used to satisfy total demand at the destinations.

c. minimize the number of shipments necessary to satisfy total demand at the destinations.

d. minimize the cost of shipping products from several origins to several destinations.

ANS: D

4. PERT and CPM

a. are most valuable when a small number of activities must be scheduled.

b. have different features and are not applied to the same situation.

c. do not require a chronological relationship among activities.

d. have been combined to develop a procedure that uses the best of each.

ANS: D

5. The critical path

a. is any path that goes from the starting node to the completion node.

b. is a combination of all paths.

c. is the shortest path.

d. is the longest path.

ANS: D

TRUE/FALSE

6.. Whenever total supply is less than total demand in a transportation problem, the LP model does

not determine how the unsatisfied demand is handled.

ANS: T

7. Converting a transportation problem LP from cost minimization to profit maximization requires

only changing the objective function; the conversion does not affect the constraints.

ANS: T

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8. If a transportation problem has four origins and five destinations, the LP formulation of the

problem will have nine constraints.

ANS: T

9. The capacitated transportation problem includes constraints which reflect limited capacity on a

route.

ANS: T

10. The shortest-route problem is a special case of the transshipment problem.

ANS: T

221

Chapter 6—

MULTIPLE CHOICE

1. Inventory

a. is held against uncertain usage so that a supply of items is available if needed.

b. constitutes a small part of the cost of doing business.

c. is not something that can be managed effectively.

d. All of the alternatives are correct.

ANS: A

2. Inventory models in which the rate of demand is constant are called

a. fixed models.

b. deterministic models.

c. JIT models.

d. requirements models.

ANS: B

3. The EOQ model

a. determines only how frequently to order.

b. considers total cost.

c. minimizes both ordering and holding costs.

d. All of the alternatives are correct.

ANS: B

4.. For inventory systems with constant demand and a fixed lead time,

a. the reorder point = lead-time demand.

b. the reorder point > lead-time demand.

c. the reorder point < lead-time demand.

d. the reorder point is unrelated to lead-time demand.

ANS: A

5.. Safety stock

a. can be determined by the EOQ formula.

b. depends on the inventory position.

c. depends on the variability of demand during lead time.

d. is not needed if Q* is the actual order quantity.

ANS: C

TRUE/FALSE

6. To be considered as inventory, goods must be finished and waiting for delivery.

ANS: F PTS: 1 TOP: Introduction

7. When demand is independent, it is not related to demand for other components or items produced

by the firm.

ANS: T PTS: 1 TOP: Introduction

222

8. Constant demand is a key assumption of the EOQ model.

ANS: T PTS: 1 TOP: EOQ model

9. In the EOQ model, the average inventory per cycle over many cycles is Q/2.

ANS: T PTS: 1 TOP: EOQ model

10. The single-period inventory model is most applicable to items that are perishable or have seasonal

demand.

ANS: T PTS: 1 TOP: When-to-order decision

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Chapter -7

MULTIPLE CHOICE

1. Decision makers in queuing situations attempt to balance

a. operating characteristics against the arrival rate.

b. service levels against service cost.

c. the number of units in the system against the time in the system.

d. the service rate against the arrival rate.

ANS: B

2. Performance measures dealing with the number of units in line and the time spent waiting are

called

a. queuing facts.

b. performance queues.

c. system measures.

d. operating characteristics.

ANS: D

3. Operating characteristics formulas for the single-channel queue do NOT require

a. .

b. Poisson distribution of arrivals.

c. an exponential distribution of service times.

d. an FCFS queue discipline.

ANS: A

4. The total cost for a waiting line does NOT specifically depend on

a. the cost of waiting.

b. the cost of service.

c. the number of units in the system.

d. the cost of a lost customer.

ANS: D

5. The arrival rate in queuing formulas is expressed as

a. the mean time between arrivals.

b. the minimum number of arrivals per time period.

c. the mean number of arrivals per channel.

d. the mean number of arrivals per time period.

ANS: D

13. What queue discipline is assumed by the waiting line models presented in the textbook?

a. first-come first-served.

b. last-in first-out.

c. shortest processing time first.

d. No discipline is assumed.

ANS: A PTS: 1 TOP: Queue discipline

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TRUE/FALSE

6. A waiting line situation where every customer waits in the same line before being served by the

same server is called a single server waiting line.

ANS: F

7. Queue discipline refers to the assumption that a customer has the patience to remain in a slow

moving queue.

ANS: F

8. Before waiting lines can be analyzed economically, the arrivals' cost of waiting must be

estimated.

ANS: T

9. In a multiple channel system it is more efficient to have a separate waiting line for each channel.

ANS: F

10. If some maximum number of customers is allowed in a queuing system at one time, the system

has a finite calling population.

ANS: F

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Reference

1. Introduction to operation research : Hillier /Lieberman

2. Schaum’s outline of operation research : Richard Bronson

3. Operation research by B.S Goel and S.K. Mittal by Pragati Prakashan