AMITY UNIVERSITY

OPERATION RESEARCH MBA INTERNATIONAL BUSINESS

SEMESTER II

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

Table of Contents Preface .......................................................................................................................................................... 5

Unit 1 :Introduction to operation research .................................................................................................. 6

Development of operation research : ....................................................................................................... 6

Definitions of OR ....................................................................................................................................... 7

Characteristic of OR : ................................................................................................................................ 8

NATURE OF OPERATIONS RESEARCH ........................................................................................................ 8

Necessity/signifiance of operations research in industry ........................................................................ 9

SCOPE OF OPERATIONS RESEARCH ......................................................................................................... 11

APPLICATIONS OF VARIOUS OR TECHNIQUES ........................................................................................ 16

MODELS IN OR ........................................................................................................................................ 17

LASSIFICATION SCHEMES OF MODELS .................................................................................................... 17

CHARACTERISTICS OF A GOOD MODEL .................................................................................................. 23

ADVANTAGES OF A MODEL .................................................................................................................... 23

LIMITATIONS OF A MODEL...................................................................................................................... 24

CONSTRUCTING THE MODEL .................................................................................................................. 24

Selecting Components of the System ..................................................................................................... 25

APPROXIMATIONS (SIMPLIFICATIONS) IN OR MODELS .......................................................................... 27

Omitting Certain Variables ...................................................................................................................... 27

TYPES OF MATHEMATICAL MODELS ....................................................................................................... 30

Unit –2 LINEAR PROGRAMMING................................................................................................................ 31

INTRODUCTION ....................................................................................................................................... 31

BASIC TERMINOLOGY, REQUIREMENTS, ASSUMPTIONS, ADVANTAGES AND LIMITATIONS ................. 32

Basic Requirements................................................................................................................................. 32

Basic Assumptions .................................................................................................................................. 33

Advantages of Linear Programming ........................................................................................................ 34

Limitations of Linear Programming......................................................................................................... 35

APPLICATION AREAS OF LINEAR PROGRAMMING .................................................................................. 36

FORMULATION OF LINEAR PROGRAMMING MODELS ........................................................................... 39

GRAPHICAL METHOD OF SOLUTION ....................................................................................................... 42

Simplex method : EXAMPLE .................................................................................................................... 45

ARTIFICIAL VARIABLES TECHNIQUES....................................................................................................... 48

The Big M-Method .................................................................................................................................. 48

Unit -3 ........................................................................................................................................................ 53

The Transportation Model ...................................................................................................................... 53

ASSUMPTIONS IN THE TRANSPORTATION MODEL ................................................................................. 54

DEFINITION OF THE TRANSPORTATION MODEL ..................................................................................... 54

MATRIX TERMINOLOGY .......................................................................................................................... 57

FORMULATION AND SOLUTION OF TRANSPORTATION MODELS .......................................................... 57

UNIT 4......................................................................................................................................................... 89

Assignment Problem ............................................................................................................................... 89

MATHEMATICAL REPRESENTATION OF THE ASSIGNMENT MODEL ....................................................... 90

COMPARISON WITH THE TRANSPORTATION MODEL............................................................................. 90

SOLUTION OF THE ASSIGNMENT MODELS ............................................................................................. 91

THE HUNGARIAN METHOD FOR SOLUTION OF THE ASSIGNMENT PROBLEMS ..................................... 95

FORMULATION AND SOLUTION OF THE ASSIGNMENT MODELS ........................................................... 99

UNIT : 5 ..................................................................................................................................................... 106

THE THEORY OF GAMES ........................................................................................................................ 106

CHARACTERISTICS OF GAMES ............................................................................................................... 107

GAME MODELS ..................................................................................................................................... 108

DEFINITIONS .......................................................................................................................................... 108

Some definitions ................................................................................................................................... 112

TWO-PERSON ZERO-SUM GAME .......................................................................................................... 113

PURE STRATEGIES: GAMES WITH SADDLE POINTS ............................................................................... 115

MIXED STRATEGIES: GAMES WITHOUT SADDLE POINTS ...................................................................... 119

UNIT : 6 ..................................................................................................................................................... 133

Network Techniques ............................................................................................................................. 133

NETWORK LOGIC (NETWORK OR ARROW DIAGRAM) .......................................................................... 134

MERITS AND DEMERITS OF AON DIAGRAMS .................................................................................... 138

CRITICAL - PATH METHOD..................................................................................................................... 139

PROGRAMME EVALUATION AND REVIEW TECHNIQUE (PERT) ............................................................ 154

OBJECTIVES OF NETWORK ANALYSIS .................................................................................................... 156

ADVANTAGES OF NETWORK TECHNIQUES ........................................................................................... 157

LIMITATIONS OF NETWORKS ................................................................................................................ 158

DIFFICULTIES IN USING NETWORK METHODS ...................................................................................... 158

COMMENTS ON THE ASSUMPTIONS OF PERT/CPM ............................................................................. 159

APPLICATIONS OF NETWORK TECHNIQUES .......................................................................................... 159

DISTINCTION BETWEEN PERT AND CPM ............................................................................................... 161

PROBABILITY STATEMENTS Or PROJECT DURATION ............................................................................ 162

Unit :7 ........................................................................................................................................................ 168

Inventory Theory ................................................................................................................................... 168

PRINCIPAL CATEGORIES OF INVENTORIES AND THEIR FUNCTIONS ..................................................... 170

REASONS FOR CARRYM INVENTORIES .................................................................................................. 173

STRUCTURE OF INVENTORY MANAGEMENT SYSTEM .......................................................................... 175

FACTORS INVOLVED IN INVENTORY ANAIYSIS ...................................................................................... 178

THE BASIC DETERMINISTIC INVENTORY MODELS ................................................................................. 193

THE EOO MODELS WITH PRICE (OR QUANTITY) DISCOUNTS ............................................................... 202

UNIT : 8 ..................................................................................................................................................... 205

Queuing Theory .................................................................................................................................... 205

Historical Development. ....................................................................................................................... 208

Queuing Process/System. ..................................................................................................................... 208

Classification of Queues and their problems. ..................................................................................... 214

The various queuing problems are related with ................................................................................... 214

SINGLE-CHANNEL QUEUING THEORY ................................................................................................... 220

Assumptions and Limitations of Queuing Model.................................................................................. 231

MULTI-CHANNEL QUEUING THEORY MODEL VI : ................................................................................. 234

End Chapter Quizzes ................................................................................................................................. 236

Chapter 1 ............................................................................................................................................... 236

Chapter 2 ............................................................................................................................................... 239

Chapter 3— ........................................................................................................................................... 242

Chapter 4 ............................................................................................................................................... 245

Chapter 5— ........................................................................................................................................... 248

Chapter 6— ........................................................................................................................................... 251

Chapter -7 ............................................................................................................................................. 254

Reference .................................................................................................................................................. 257

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.

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.

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.

(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.

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.

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 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.

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 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.

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,

(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.

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, 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 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.

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.

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.

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

(c) Distribution policy.

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.

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:

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.

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.

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. 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-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.

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

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.

CHARACTERISTICS OF A GOOD MODEL

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

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.

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.

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.

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),

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

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. 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

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.

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

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- 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 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

1.17 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.

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 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.

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.

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

Unit –2 LINEAR PROGRAMMING

INTRODUCTION

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 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.

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.

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.

(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.

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 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.

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.

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.

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.

APPLICATION AREAS OF LINEAR PROGRAMMING

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 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 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.

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.

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 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

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

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

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

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 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.

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.

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

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.

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.

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. 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

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.

Unit -3

The Transportation Model

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.

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

warehouses, warehouses to wholesalers, wholesalers to retailers and retailers to

customers.

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.

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 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

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:

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 non-

negative 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.-. .

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.

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. -

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:

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

(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 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.

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:

(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.

(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, 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.

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 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 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.

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. 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.

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 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.

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 terminates in the

chosen cell. All other corners of the path lie in the cells for which al locations 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

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 u i 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 v j (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.

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.

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:

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.

(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

= 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.

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.

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.

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.

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

UNIT 4

Assignment Problem

DEFINITION OF THE ASSIGNMENT MODEL

An assignment problem concerns as to what happens to the effectiveness funct ion

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 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.

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.

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 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:

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 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.

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

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.

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:

(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.

Flow chart of these steps is shown in Fig. 4.1 below.

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,

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.

(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.

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.

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

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.

UNIT : 5

THE THEORY OF GAMES

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.

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.

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.

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.

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.

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 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,

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.

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 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.

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 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.

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.

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.

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.

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.,

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.

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

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 .

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

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

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

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 :

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

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)

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.

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.

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

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

UNIT : 6

Network Techniques

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 l ike

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.

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 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

i j

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 ?

(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).

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).

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

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 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.

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. 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:

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.

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

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 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.

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 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.

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

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. 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

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

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,

E values are represented in Fig.

In backward pass communications

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.

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,

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 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.

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.

ADVANTAGES OF NETWORK TECHNIQUES

1. They are most valuable and powerful for planning, scheduling and control of

operations in large and complex projects.

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.

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.

4. Time-cost trade off procedure, in many situations, is complicated.

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.

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.

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.

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 manu-

facturing 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.

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.

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 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.

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:

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

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 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.

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.

Unit :7

Inventory Theory

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 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.

PRINCIPAL CATEGORIES OF INVENTORIES AND THEIR FUNCTIONS

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 (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

(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, 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.

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.

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.

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.

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 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 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.

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.

(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 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.

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 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.

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.

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

=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

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.

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

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.

where t j is the time interval during which items are drawn from inventory and t z 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,

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

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, tt, 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.

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

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.

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

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

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

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

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.

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

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.

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

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 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

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 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.

(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.

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 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.

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:

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.

(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:

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 :

(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 b e

possible to identify the order of arrivals:

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.

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.

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.

(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 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

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 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 studied 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

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.

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 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.

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.

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

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 :

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.

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.

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.

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:

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

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

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.

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.

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

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).

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,

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:

End Chapter Quizzes

Chapter 1

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.

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

Chapter 2

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 + 6y2

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.

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

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.

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

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

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

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

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

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

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

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

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

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