DocumentOR

2007

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Production Engineering Department

Shantilal Shah Engineering College

OPERATIONS RESEARCH

PRODUCTION ENGINEERING DEPT. 2

CERTIFICATE

This is to certify that Mr./Miss

_______________________________________________________________

Student of semester - VIII Production

Engineering,Roll No. _______ of S.S.Engineering College has

Satisfactorily accomplished his/her term work by

submitting this file of ______________________________________

on Date: ___________

Examinor Head of the Dept

OPERATIONS RESEARCH


INDEX

Sr.

No.

Name of Experiment Page

From To

Date of

Start

Date of

Complet

-ion

Sign

1. OPERATION RESEARCH

2. STUDY OF LINEAR

PROGRAMMING METHODS,

MODELS FORMULATION &

ITS ROLE IN OPERATION

RESEARCH

3. STUDY & IMPORTANCE OF

SIMPLEX METHOD IN LINEAR

PROGRAMMING

4. STUDY OF DUAL PROBLEM,

ITS INTERPRETATION &

SENSITIVITY ANALYSIS

5. SPECIAL CASE OF SIMPLEX

METHOD

6. PROJECT EVALUATION AND

REVIEW TECHNIQUES &

CRITICAL PATH METHOD

(PERT/CPM)

7. GAMES THEORY

8. INVENTORY MANAGEMENT

9. QUEUING THEORY

10. REPLACEMENT THEORY

11. STUDY ABOUT SEQUENCING

PROBLEM.

OPERATIONS RESEARCH


EXPERIMENT.NO: 1

OPERATION RESEARCH

HISTORICAL DEVELOPMENT

It is generally agreed that Operation Research (OR) came into existence as a discipline

during World War II. However, a particular model and technique of Operation Research

can be traced back much earlier. The term Operation Research was coined as a result of

research on military operations during this war. Since the war involved strategic and tactical

problems, which were so complicated, that to expecting adequate solution from individuals

Operation Research specialists in a single discipline was unrealistic. Therefore, groups of

individuals who collectively and physical science were formed as special units within the

armed forced to deal with strategic and tactical problems of various military operations.

Such groups were first in England and the United States. One of the groups in England

came to be known as Blackett‟s Circus. This group under the leadership of Prof.

P.M.S.Blackett was attached to the Radar Operation Research unit and assigned the

problem of analyzing the co-ordination of Radar equipment at gun sites. Following the

success of this group, such mixed team approach was adopted in other allied nations.

A key person in post war development of Operation Research was George B Dantzig. In

1947, he developed linear programming and its solution methods known as simplex

method. Besides linear programming, many others were well developed before the end of

1950‟s.

During the 1950‟s, there was substantial progress in the applications of Operation Research

techniques for civilian activities along with a great interest in the professional development

and education in Operation Research. Many college and universities introduce Operation

Research in their curricula. They were generally schools of engineering, public

administration, business management, applied mathematics, economics, computer services,

etc. Today, however service organization such as banks, hospitals, libraries, airlines,

railways, etc. recognize the usefulness for in improving their efficiency. In 1948, an

Operation Research was formed in England which later changed the name to Operation

Research Society of America (ORSA) was found in 1952 and its journal, Operation

Research, first published in 1953. In the same year, the Institute of Management Science

(TIMS) was founded as an International Society to identify, extend and unify scientific

knowledge pertaining to management. Its journals, Management, Science, first appeared in

1954.

In India, Operation Research came into existence in 1949 when an Operation Research unit

was established at Regional Research Laboratory, Hyderabad. At the same time, Prof. R S

OPERATIONS RESEARCH


Verma, also ser up an Operation Research team at Defence Science Laboratory to solve

problems of store, purchase and planning. In 1933, Prof. P C Mahalanobis established an

Operation Research team in the Indian statistical institute, Calcutta to solve to solve

problems related to national planning and survey. The Operation Research Society of India

was founded in 1957 and started publishing its journal OPSEARCH. In same year, India

along with Japan became in London. The other member of IFORS was USA, UK,

FRANCE and what was when WEST GERMANY.

Because of OR‟s multi disciplinary character and application in varied fields, it has a good

future provided people devoted to OR study can help meet the needs of society. Some of the

problems the area of hospital management, energy conservation, environmental pollution

etc. which have been solved by OR specialists is an indication that OR can also contribute

towards the improvements in the pattern of social life and areas of global need. However, in

order to make the future of OR more bright, it specialists have to make good use of the

avenues open to them.

SIGNIFICANCE OF OPERATIONS RESEARCH

As the term implies OR involves research in military operations. This indicates the

approach as well as the area of its application. The OR approach is particularly useful in

balancing conflicting objectives (goals OR interests) where there are many alternatives

courses of action available to the decision makers.

In view of any problem situation involving the whole system, the decision maker, whatever

his specialization will need help and it is in the attempt to provide this assistance that OR

has been developed. OR attempts to resolve the conflicts of interest among various sections

of the organization and seeks the optimal solution which may not be acceptable to one

department but is in the interest of the organization as a whole. Further, OR is concerned

with providing the decision maker with decision aids (or rules) derived from:

A total system orientation.

Scientific methods of investigation.

Models of reality generally based on quantitative measurement technique.

Thus, successful application of Operation Research technique for solving a problem must

involve:

Constructing mathematical, economic and statistical model of the problem under

study to treat situations of complexity and uncertainly. This helps to view the problem

in it‟s entirely.

Analyzing the relationships among different variables and parameters associated with

the problem so as to determine consequences of decision alternative.

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Suggesting suitable measures of desirability in order to evaluate the relative merit of

decision alternatives.

OPERATION RESEARCH APPROACH

The future of Operation Research approach to any decision and control problems can be

considered under the following methodologies:

1. INTERDISCIPLINARY APPROACH: Interdisciplinary teamwork is essential

because while attempting to solve a complex management problem one person may not

have the complete knowledge of all aspect (such as economic, social, political,

psychological, engineering etc.) of it. This means we should not expect a desirable

solution to managerial problems. Therefore, a team of individuals specializing in that

field in order to arrive at an appropriate and desirable solution of the problem. However,

there are certain problem situations, which may be analyzed even by one individual.

2. METHODOLOGICAL APPROACH: “Operation Research is the application of

scientific methods, techniques and tools to problem involving the operation of system so

as to provide those in control of operations with optimum solutions to the problems”.

The scientific method consist of observing and defining the problems; formulating and

testing the hypothesis; and analyzing the results of the test. The data so obtained are

then used to decide whether the hypothesis is accepted, then the results should be

implemented. Otherwise, an alternative hypothesis has to be formulated.

3. WHOLISTIC APPROACH: Arriving at a decision, an operation research team

examines the relative importance of all confliction and multiple objectives and validity

of the claims of various departments of the organization from the perspective of the

whole organization.

4. OBJECTIVE APPROACH: An Operation Research approach seeks to obtain an

optimal solution to the problem under analysis. For this a measure of desirability is

defined based on the objectives of the organization. A measure of desirability so defined

is then used to compare alternative course of action with respect to their out come.

SCIENTIFIC METHOD IN OPERATION RESEARCH

The most important feature of Operation Research is the use of the scientific method and

building of decision models. There are three phases of the scientific method on which

Operation Research practice is based.

1. JUDGMENT PHASE: This phase includes

a) Identification of the real-life problem.

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b) Selection of an appropriate objective and the values of various variables related to

these objectives.

c) Application of the appropriate scale of measurement, i.e. deciding the measures of

effectiveness.

d) Formulation of an appropriate model of the problem, abstracting the essential

information‟s so that a solution at the decision maker‟s goals can be sought.

2. RESEARCH PHASE: This phase is the largest and longest between other two phases.

However, the remaining two also equally important as they provide the basis for a

scientific method. This phase utilize:

a) Observations and data collection for a better understanding of the problem.

b) Formulation and experimentation to test the hypothesis on the basis of additional

data.

c) Analysis of the available information and verification of the hypothesis using pre-

established measures of desirability.

d) Predictions of various results form the hypothesis.

e) Generalization of the result and consideration of alternative method.

3. ACTION PHASE: This phase consists of making recommendation for implementing

the decision by an individual who is the position to implement results. This man must

be aware of the environment in which the problem occurred objective, assumption and

omissions of the model of the problem.

MODELS OF OPERATION RESEARCH

PHYSICAL MODELS

These models provide a physical appearance of the real object under study either reduced in

size/scaled up. Physical models are useful only in design problems because they are easy to

observe, build and describe. Problem such as portfolio selection, media selection,

production scheduling etc. cannot be analyzed with a physical model. Physical model are

classified into the following two categories:

1. Iconic Model: Iconic models retain some of the physical properties and

characteristics of the system they represent. An iconic model is either in an idealized

formula scaled version of the system. In other words, such models represent the

system as it is by scaling it up Operation Research down.

The iconic models are simple to conceive, specific and concrete. An iconic model is

used to describe the characteristics of the system rather than explanatory. This

means such models are used to represent a static event and characteristics, which are

not used in determining/predicting effects due to certain changes in actual system.

OPERATIONS RESEARCH


For example, color of an atom does not play any vital role in the scientific study of

its structure.

2. Analogue Model: These models represent a system by the set properties different

from that of the original system does not resemble physically. After the problem is

solved, the solution is reinterpreted in terms of the original system.

For example, the organizational chart represents the state of formal relationships

existing between members of the organization. Map in different colors may

represent water, desert and other geographical features. These models are less

specific and concrete but easier to manipulate and more general than iconic models.

SYMBOLIC MODELS:

These models use letters, numbers and other symbols to represent the properties of the

system. These models are also used to represent relationship, which can be represented in a

physical form. Symbolic models can be classified into two categories:

1. Verbal Models: These models describe a situation in written Operation Research

spoken language. Written sentences, books, etc. are examples of a verbal model.

2. Mathematical Models: These models involve the use of mathematical symbols,

letters, numbers, and mathematical operators (+, -, *, /) to represent relationship

among various variable of the system to describe its properties Operation Research

behavior.

The solution to such models is then obtained by applying suitable mathematical

techniques. Symbolic models are precise and abstract and can be manipulated by

using laws of mathematics.

BASIC OPERATIONS RESEARCH MODELS

There is no unique set of problems, which can be solved by using Operation Research

models Operation Research techniques. Several Operation Research models/techniques can

be grouped into some basic categories as given below:

1. ALLOCATION MODEL: Allocation model are used to allocate resources to activate

in such a way that some measure of effectiveness is optimized. Mathematical

programming is the broad term for the Operation Research technique solves allocation

problems.

When the solution values Operation Research decision variables for the problem are

restricted to being integer values / just zero-one values, the problem is classified as an

integer programming problem Operation Research a zero-one programming problem

respectively.

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The problem having multiple, conflicting and incommensurable objective functions

subject to linear constraints is called a goal-programming problem. If the decision

variables in the linear programming problem depend on chance, then such a problem is

called a stochastic programming problem.

2. INVENTORY MODELS: Inventory models deal with the problem of determination

of how much to order at a point in time and when to place an order. The main objective

is to minimize the sum three conflicting inventory costs: the cost of holding Operation

Research carrying extra inventory, the cost of storage Operation Research delay in the

delivery of items when it is needed, a cost of ordering or set-up. These are also useful in

dealing with quantity discounts and selective inventory control.

3. WAITING LINE MODELS (QUEUING): These models have been developed to

establish a trade-off between costs of providing service and the waiting time of a

customer in the queuing system. Constructing a models entails describing the

components of the system: arrival process, queue structural and service process and

solving for the measure of performance – average length of waiting time, average time

spent by the customer in the line, traffic intensity, etc. of the waiting system.

4. COMPETITIVE MODELS (GAME THEORY): These models are used to

characterize the behaviors of two/more opponents who compete for the achievement of

conflicting goals. These models are classified according to several factors such as

number of competitors, sum of loss and gain and type of strategy which would yield

him the best / the worst outcomes.

5. NETWORK MODELS: These models are applied to the management of large-scale

projects. PERT/CPM technique help in identifying potential trouble spots in project

through the identification of the critical path. These techniques improve project co-

ordination and enable the efficient use of resources. Network methods are also used to

determine time-close trade off, resource allocation and updating of activity times.

6. SEQUENCING MODELS: These models are used whenever there is program in

determining the sequence in which a number of tasks can be performed by a number of

service facilities such as hospital, plant etc. in such a way that some measure of

performance, for example, total time to process all the jobs on all the machines, is

optimized.

7. REPLACEMENT MODELS: These models are use when one must decide the

optimal time to replace equipment for one reason Operation Research the other-for

instance, in the case of equipment whose efficiency deteriorates with time Operation

Research fails immediately and completely. For example, in case of an automobile the

user has his own measure of effectiveness. So there will not be single optimal answer

for everyone even if each automobile gives exactly the same service.

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8. DYNAMIC PROGRAMMING MODELS: Dynamic programming may be

considered as an out growth of mathematical programming and involves the

optimization of multistage decision processes. The method starts by dividing a given

problem into stages Operation Research sub-problems and then solves that sub-problem

sequentially until the solution to the original problem is obtained.

9. MARKOV-CHAIN MODELS: These models are used for analyzing a system that

changes over a period of time among various possible outcomes/states. The models

while dealing with such systems describe transitions in terms of transition probabilities

of various states. These models have been used to test brand-loyalty and brand-

switching tendencies of consumers, where each system state is considered to be a

particular brand purchase. These have also been used in reliability analysis, where the

states of the system are the various levels of performance of the equipment being

monitored.

10. SIMULATION MODELS: These models are used to develop a method to evaluate

the merit of alternative courses of action by experimenting with a mathematical model

of the problems where various variables are random. That is, these provide a means of

or generating representative samples of the measures of performance variable. Thus,

repetition of the process by using the simulation model provides an indication of the

merit of alternative course of action with respect to the decision variable.

GENERAL METHODS

In general, the following three models are used solving Operation Research models. In all

these methods, values of decision variables are obtained that optimize the given objectives

function.

1. ANALYTICAL METHOD (DEDUCTIVE): In this method, classical optimization

techniques such as calculus, finite difference and graphs are used for solving and

Operation Research model. In this case, we have a general solution specified by symbol

and we can obtain the optimal solution in a non-iterative manner. For example, in

inventory models, in order to calculate economic order quantity the analytical method

requires that the first derivative of the mathematical expression

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

Where, TC = Total variable inventory cost;

Cp = Ordering cost per;

Q = size of an order;

D = annual demand;

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Ch = carrying cost per time period.

Be taken and equated to zero as the first step toward identifying optimum value of Q* =

sqrt (2DCp/Ch). This is because of the concept of maximum and minimum for

optimality. Here, it may be noted that the validity of the conclusion depends only on the

validity of the mathematical approach that is being used.

2. NUMERICAL METHODS (ITERATIVE): When analytical methods fail to obtain

the solution of a particular problem due to its complexity in terms of constraints/number

of variables, then numerical methods is used to get the solution. In this method instead

of solving the problem directly, a general is applied to obtain a specific numerical

solution.

The numerical methods starts with a solution obtained by trail and error and as set of

rules for improving it towards optimally. The solution so obtained is then replaced by

the improved solution and the process of getting improved solution is repeated until

such improvement is not possible/ the cost of further calculation cannot be justified.

3. MONTE CARLO METHOD: This method is based upon the idea of experimenting

on a mathematical model by inserting into the model specific values of decision

variables at different points of time and under different conditions and then observing

their effect on the criterion chosen for variables. In this method, random samples of

specified random variables are drawn to know what is happening to the system for a

selected period of time under different conditions. The random samples form a

probability distribution that represents the real life system and from this probability

distribution, the value of the desired random variable can be estimated.

APPLICATION AND SCOPE OF OPERATON RESEARCH

Some of the industrial/ government/ business problems which can be analyzed by Operation

Research approach have been arranged functional area wise as follows: -

1. FINANCE AND ACCOUNTING:

Divided policies, investment and portfolio management, auditing balance sheet,

cash flow analysis.

Break-even analysis, capital budgeting, cost allocation and control, financial

planning.

Claim and complaint procedure, public accounting.

2. MARKETING:

Selection of product – mix, marketing planning, exports planning

Sales effort allocation and assignment.

Advertising and media planning.

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3. PURCHASING, PROCUREMENT AND EXPLORATION:

Optimal buying and recording under price quantity discount.

Bidding policies.

Transportation planning.

Vendor analysis.

Replacement policies.

4. PRODUCTION MANAGEMENT:

Facilities Planning

Manufacturing

Maintenance and project scheduling

5. PERSONNEL MANAGEMENT:

Manpower planning, wage/salary administration

Skills and wages balancing

Scheduling of training programmers

6. TECHNIQUES AND GENERAL MANAGEMENT:

Decision support system and MIS; forecasting

Organizational design and control

Project Management, strategic planning

7. GOVERNMENT:

Economic planning, natural resources, social planning, energy

Urban and housing problem

Military, police, pollution control etc.

OPERATIONS RESEARCH


EXPERIMENT.NO: 2

STUDY OF LINEAR PROGRAMMING METHODS, MODELS FORMULATION & ITS ROLE IN OPERATION RESEARCH

GUIDELINES ON LINEAR PROGRAMING MODEL

FORMULATION

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

allocation is based on its applicability to many diversified decision problems. The effective

use and application require, as a first step, the mathematical formulation of the LP model

when the problem in words.

STEPS OF LP MODEL FORMULATION

Step 1: - Define Decision Variable:

a) Express each constraint in words. For this first see, whether the constraint is of form,

(at least than). Operation Research of the for (no longer than) Operation Research

= (Exactly equal to)

b) Then express the objective function in words.

c) Step 1(a) and 1(b) should then allow you to verbally identify the decision variable.

Step 2: - Formulate the Constraints:

Formulate all the constraints imposed buy the resource availability and express them as

linear equality Operation Research inequality in terms of the decision variables defined in

step 1.

Step 3: - Formulate the Objective Function:

Define the objective function. That is, determine whether the objective function is to be

maximized /minimized. Then express it as a linear function of decision variables multiplied

by their profit/cost contributions.

INTRODUCTION

An optimal as well as feasible solution an LP problem is obtained by choosing among

several values of decision variables X1, X2,…., Xn the one set of values that satisfy the

given set of constraints simultaneously and also provides the optimal values of the given

objective function.

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GRAPHICAL SOLUTION METHODS OF LP PROBLEMS

While obtaining the optimal solution to the LP problem with the graphical method, the

statement of the following theorems of linear programming is used.

1. The collection of all feasible solutions to an LP problem constitutes a convex set whose

extreme points correspond to the basic feasible solutions.

2. There are finite numbers of basic feasible solutions within the feasible solution space.

3. If the optimal solution occurs at feasible solutions on the system Ax = b, x>0, is a

convex polyhedron, then at least one of the extreme points gives an optimal solution.

4. If the optimal solution occurs at more than one extreme point, then the value of

objective function will be the same for all convex combinations of these extreme points.

EXTREME POINT ENUMERATION APPROACH

This solution method for an LP problem divided into five steps.

Step 1:- State the given problem in the mathematical form as illustrated previously.

Step 2:- Graph the constraints, by temporarily ignoring inequality sign and decide about the

area of feasible solutions according to the inequality sign of the constraints.

Indicate the area of feasible solutions by a shaded area, which forms a convex

polyhedron.

Step 3:- Determine the co-ordinate so the extreme points of the feasible solution space.

Step 4:- Evaluate the values of the objective function at each extreme point.

Step 5:- Determine the extreme point to obtain the optimum value of the objective function.

SOME SPECIAL CASES IN LINEAR PROGRAMMING

1. ALTERNATIVE OPTIMAL SOLUTION: So far we have seen that the optimal

solution of any linear programming problem occurs at extreme point of the feasible

region and the solution is unique, i.e. no other solution yields the same value of the

objective function. However, in certain cases a given LP problems may have more than

one optimal solution yielding the same objective function value.

There are two conditions that should satisfy in order that an alternative optimal solution

exists.

The given objective function is parallel to a constraint that forms the boundary of

the feasible solutions region. In other words, the slope of the objective function is

same as that of the constraint forming the objective function is same as that of the

constraint forming the boundary of the feasible solutions region.

The constraint should form a boundary on the feasible region in the direction of

optimal movement of the objective function. In other words, the constraint should

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be an active constraint. Graphically an active constraint is one that passes through

one of the extreme points of the feasible solution space.

2. AN UNBOUNDED SOLUTION: When the value of decision variables in linear

programming is permitted to increase infinitely without violation the feasibility

condition, then the solutions said to be unbounded. The objective function value can be

increased infinitely. However, an unbounded feasible region may yield some definite

value of the objective function.

3. AN INFEASIBLE SOLUTION: If it is not possible to find a feasible solution that

satisfies all the constraint, then LP problem, I said to have an infeasible solution /

alternatively inconsistency. Infeasibility depends mainly on the constraints and has

nothing to do with the objective function.

LINEAR PROGRAMMING WITH THE HELP OF SIMPLEX

METHOD

STANDARD FORM OF LP

The use of the simplex method to solve an LP problem requires that the problem be

converted into its standard form. The standard form of the LP problem should have the

following characteristics:

All the constraints should be expressed as equations by adding slack operation

research surplus and artificial variables.

The right hand side of each constraint should be made non-negative, if kit is not;

multiplying both sides of the resulting constraint by –1 should do this.

The objective function should be of the maximization is expressed as:

For your ready reference the standard for of the LP problem is expressed as,

Optimize (Max/Min) Z = C1X1 + C2X2 +….+ CnXn + 0S1 +….+ 0Sm

Subject to the linear constraints

A11X1 + A12X2 + …+ A1nXn + S1 = B1

A21X1 + A22X2 + … + A2nXn +S2 = B2

|

|

Am1X1 + Am2X2 + … + AmnXn +Sm = Bm

And X1, X2,…, Xn, S1, S2,…,Sm 0

In matrix notation the standard form is expressed as:

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Optimize Z = CX + 0S

Subject to the linear constraints

AX + S = B and X, S 0

Where C = (C1,C2,…,Cn) is the row vector;

X = (X1,X2,…,Xn), B = (B1,B2,…,Bm)

S = (S1,S2,…,Sn) are columns vectors and

A11 A12… A1n

A21 A22… A2n

… … …

Am1 Am2 Amn is the (m x n) matrix of coefficients of variables X1,

X2,…, Xn in the constraints.

There types of additional a variables, namely

1. Slack variable (S)

2. Surplus variables (-S)

3. Artificial variables (A)

TWO PHASE METHOD

In the first phase of the method the sum of the artificial variables is minimized subject to

the given constraints to get a basic feasible solution of the LP problem. The second phase

minimizes the original objective function starting with the basic feasible solution obtained

at the first phase. Since the solution of the LP problem is completed in two phases, this is

called the two-phase method.

STEPS OF ALGORITHM

PHASE 1:

Step 1:

a) If all the constraints in the give LP problem are (<) type, then phase II can be directly

use to solve the problem. Otherwise, a sufficient number of artificial variables are

added to get a basis matrix (identity matrix).

b) If the given LP problem is of minimization, then convert it to the maximization type by

the usual method.

Step 2: Solve the following LP problem by assigning a coefficient of –1 to each artificial

variable and zero to all other variable and zero to all other variables in the objective

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function and with the basic feasible solution X1 = X2 = … = Xn = 0 and Ai = b1; i =

1,2,…,m

Maximization Z* = (-1) A1

Subject to the constraints

Z* = aij xj = Aj = bi; i = 1,2,…,m; and xj , Aj > 0

Apply the simplex algorithm to solve this LP problem. The following three cases may arise

at optimally:

Max Z* = 0 and at least one artificial variable is present in the basis with position

value. Then no feasible solution exists for the original LP problem.

Max Z* = 0 and no artificial variable is present in the basis. Then the basis consists of

only decision variables (Xjs) and hence we may move to phase II to obtain an optimal

basic feasible solution to the original problem.

Max Z* = 0 and at least one artificial variable is present in the basis at zero value.

The feasible solution to the above LP problem is also a feasible solution to the

original LP problem. Now orders to arrive at the basic feasible solution we may

proceed directly to phase II operation research else eliminate the artificial basic

variables and then proceed to phase II.

THE BIG – M METHOD

The big – M method is another method of removing artificial variables from the basis. In

this method, we assign coefficients to artificial variables, under sizable from the objective

function point of view. If objective function Z is to be minimized, then a very large positive

price is assigned to each artificial variable. Similarly, if Z is to be maximized then a very

large negative price is assigned to each of these variables. The penalty will be designated by

= M for a maximization problem and +M for a minimization problem where M>0. The big

– M method for solving an LP problem can be summarized in the following steps:

Step 1: Express the LP problem in the standard form by adding surplus variables and

artificial variables. Assign a zero coefficient to surplus variables and a very large

positive number +M to artificial variables in the objective function.

Step 2: The initial basic feasible solution is obtained by assigning zero value to original

variables.

Step 3: Calculate the values of Cj – Zj in last row of the simplex table and examine these

values:

If all Cj – Zj 0 then the current basic feasible solution is optimal.

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If for a column, k, Ck – Zk is most negative and all entries in this column are negative,

then the problem has an unbounded optimal solution.

If one or more Cj – Zj < 0 then select the variables to enter into the basic with the

largest negative Cj – Zj value. That is, Ck – Zk = Min(Cj – Zj:Cj – Zj < 0)

The column to enter is called key or pivot column.

Step 4: Determine the key row and key element in the same manner as discussed in the

simplex algorithm of the maximization case.

LINEAR PROGRAMMING FORMULATION OF DUAL PROBLEM

The term „dual‟ in a general sense implies two/double. The concept of duality is very useful

in mathematics, physics, and statistics and engineering.

In context of linear programming duality implies that each linear programming problem can

be analyzed in two different ways but having equality solutions. Each LP problem stated in

its original form has associated with another linear-programming problem, which is unique

based on the same data. In general it is immaterial which of the two problems is called

primal / dual since the dual is primal.

SYMMETRICAL FORM

Suppose the primal LP problem is given in the form,

Maximize Zx = C1X1 + C2X2 +…+ CnXn


A11X1 + A12X2 +…+ A1nXn + S1 = B1

A21X1 + A22X2 +…+ A2nXn + S2 = B2

|

Am1X1 + Am2X2 +…+ AmnXn +Sm = Bm

And X1 ,X2,… ,Xn 0

Then corresponding dual LP problem is defined as:

Minimize Zy = B1Y1 + B2Y2 +…+ BmYm


A11Y1 + A12Y2 +…+A1nYn + S1 = B1

A21Y1 + A22Y2 +…+ A2nYn +S2 = B2

|

Am1Y1 + Am2Y2 +…+ AmnYn + Sm = Bm

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And Y1, Y2,…,Yn 0

Therefore above pair of LP problem can be expressed in the general LP model form as:

PRIMAL DUAL

n

Max Zx = CjXj

j=1

Subject to

n

AjiXj Bi; i = 1,2,…,m

j=1

And Xj 0; j = 1,2,..,m

n

Min Zy = BiYi

j=1

Subject to

n

AjiYi Cj; j = 1,2,…,n

j=1

And Yi 0; i = 1,2,..,m

A summary of the general relationship between primal and dual LP problem is given in

table below:

IF PRIMAL THEN DUAL

Objective is to maximize Objective is to minimize

Variable Xj Constraint j

Constraint i Variable Yi

Variables Xj unrestricted in sing Constraint j is = type

Constraint i is = type sign Variable Yi is restricted in sign

type constraints type constraints

0 variables inequality constraints

inequality constraints 0 variable

ECONOMIC INTERPRETATION

In the primal LP model we may define each parameter as follows:

Z = return

Xj = units of variable j

Cj = return per unit of variable Xj

Aij = requirement of resource I by per unit of variable j

Bi = units of resource i

The new parameters introduced in the dual problem are Zy and Yi. Since Zx = Zy so Zy is

also in terms of “return”. The interpretation associated with the dual variables Yi, i =

1,2,…,m is discussed below:

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Rewriting the primal LP problem in terms of new terminology

PRIMAL:

n

Maximize Zx = { return / units of variable Xj}(units of variable Xj)

j=1


{ units of resource I / units of variable Xj}(units of variable) unit of resource i;

i = 1,2,…,m and Xj 0

DUAL:

n

Minimize Zy = (units of resource i) Yi , subject to the constraints

j=1

n

{ units of resource i/ units of variable Xj} Yi {return / units of variables Xj}

j=1

i = 1,2,…,n and Yi 0

TRANSPORTATION MODEL

One important application of linear programming has been in the area of physical

distribution of resources, from one place to another to meet a specific set of requirements. It

is easy to express a transportation problem mathematically in terms of an LP model, which

can be solved by simplex method. Since transportation problem involves a large number of

variables and constraints it takes a very long time to solve it by the simplex ways, involving

transportation algorithm, namely the stepping stone method, MODI method have been

involved for this purpose.

The transportation algorithm discussed in this chapter is applied to minimize the total cost

of transporting a homogeneous commodity from one plane to another. However, it can also

be applied to the minimization of some total value / utility, for example, financial resources

are distributed in such a way that the profitable return is maximized.

TRANSPORTATION PROBLEM

The transportation problem applied to situation in which a single product is to be

transported form several sources to several sinks. In general, let there be m sources, S1,

S2,…,Sm having Ai units of supplies or capacity respectively table transported among n

destination D1,D2,…,Dn with Bj units of requirement respectively. Let Cij be the cost of

shipping one unit of commodity from sources i to destination j for each route. If Xij

represents the units shipped per route from source i to destination j, the problem is to

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determine the transportation schedule so as to minimize the total transportation cost

satisfying supply and demand conditions. Mathematically the problem can be stated as:

m n

Minimize (total cost) = Cij Xij

i=1 j=1


n

Xij = Ai, i = 1,2,…,m

j=1

m

Xij = Bj, j = 1,2,…,n

i=1

And Xij 0 for all i and j.

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EXPERIMENT.NO: 3

STUDY & IMPORTANCE OF SIMPLEX METHOD IN LINEAR PROGRAMMING

GRAPHICAL SOLUTION

INTODUCTION

An optimal as well as feasible solution to an LP problem is obtained by choosing among

several values of decision variables x1, x2, …., xn. The one set of values that satisfies the

given set of constraints simultaneously and also provide the optimal values of a given

objective function.

For LP problems that have only two variables it is possible that the center set of feasible

solution can be displayed graphically by plotting linear constraints to locate a best optimal

solution. A technique used to identify the optimal solution is called the graphical solution

technique for an LP problem with two variables.

DEFINITION

Basic Solution:

In a system of m equations and n unknowns where n>m,

AX= b and XT belongs to Rn

When A is a (m x n) matrix of rank m.

Let B be any x sub matrix formed by linearly independent columns of A, then the solution

is obtained by setting n-m variables, not associated with the columns of B, equal to zeroes

and solving the resultant system is called a Basic Solution to the given system of equations.

The n-m variables whose values did not appear in the solution are called non-basic

variables and the remaining m variables are called basic variables.

Solution:

Solution values of decision variables xj(j=1,2,…n) which satisfy the constraint of general

LP model is called the solution of the LP model.

Feasible Solution:

Solution values of decision variables xj (j=1,2,…n) which satisfy the constraints and non-

negativity condition of general LP model are said to constitute the feasible solution to that

LP model.

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Basic Feasible Solution:

A feasible solution to an LP problem, which is also the basic solution, is called the Basic

Feasible Solution. That is, all basic variables assume non-negative values. Basic Feasible

Solution are of two types:

Degenerate: A basic feasible solution is called Degenerate if at least one basic

variable possesses zero value.

Non-Degenerate: A basic feasible solution is called Non-degenerate if all m basic

variables are positive and non-zero.

Optimum Basic Feasible Solution:

A basic feasible solution which optimizes the objective function of the given LP models

called an Optimum Basic Feasible Solution.

Unbounded Solution:

A solution which can be increased operation research decreased the value of objective

function of LP problem indefinitely is called Unbounded Solution.

This solution method for an LP problem is divided into five steps:

Step1: State the given problem in the mathematical form.

Step2: Graph the constraints, by temporarily ignoring the inequality sign and decide about

the area of feasible solution according to the inequality sign of constraints. Indicate

the area of feasible solutions by a shaded area which forms a convex polyhedron.

Step3: Determine the coordinate so the extreme points of the feasible solution space.

Step4: Evaluate the value of the objective function at each extreme point.

Step5: Determine the extreme points to obtain the optimum value of the objective function.

SIMPLEX METOD:

STANDARD FORM OF LP:

The use of the simplex method to solve an LP problem requires that the problem be

converted into its standard form. The standard of the LP problem should have the following

characteristic.

1. All the constraints should be expressed as equations by adding slack, operation

research surplus and artificial variables.

2. The right hand side of each constraint should be made non-negative, if it is not, this

should be done by multiplying both sides of the resulting constraints by-1.

3. The onjective function should be of the maximization is expressed as:

Optimize (max/min) Z = C1X1 + C2X2 +…..+ CnXn + 0S1 +….+ 0Sm

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Subject to linear constraints,

A11X1 + A12X2 +…..+ A1nXn + S1 = B1

A21X1 + A22X2 +…..+ A2nXn +S2 = B2

|

Am1X1 + Am2X2 +…..+ AmnXn + Sm = Bm

And X1,X2,……..,Xn,S1,S2 0

In matrix notation the standard form is expressed as:

Optimize Z = CX+0S

Subject to linear constraints

AX+S = b and X,S 0

Where C=[C1,C2,…..Cn]is the row vector.

X=[X1,X2,….Xn], B=[B1,B2,…..Bm]T

S= [S1, S2, …. Sn] are column vectors.

A11 A12 A1n

A21 A22 A2n

| | |

Am1 Am2 Amn

Is the matrix of coefficients of variables X1, X2, …, Xn in the constraints. Three types of

additional variables namely

1. Slack variables [S]

2. Surplus variables [-S]

3. artificial variables [A]

STEPS IN SIMPLEX METHOD:

Step1: Formulation of LPP

Step2: Convert constraints into equality form.

Step3: Construct the starting simplex table.

Step4: Test optimality by analysis.

Step5: Find “incoming” and “outgoing” variables and rewrite the table as per given set of

rules.

Step6: Repeat step 4 onwards again till optimum basic feasible solution is obtained.

THE BIG - M METHOD

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

method, we assign coefficients to artificial variables, undesirable from the objective

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function point of view, if objective function Z is to be minimized, then a very large positive

price is assigned to each artificial variable. Similarly, if Z is to be maximized, then a very

large negative price is assigned to each of this variable. The penalty will be designated by

=M for maximization problem and +M for a minimization problem, where M>0. The big M

method for solving and LP problem can be summarized in the following steps:

1. Express the LP problem in the standard form by adding surplus variables and artificial

variables. Assign a zero coefficients to surplus variables and a very large positive

number +M and-M to artificial variables in the objective function.

2. The initial basic feasible solution is obtained by assigning zero value to original

variables.

3. Calculate the values of Cj-Zj in fast row of the simplex table and examine these values.

a) [a] If all Cj-Zj>0, then the current basic feasible solution optimal.

b) [b] If for a column, k, Ck-Zk is most negative and all entries in this column are

negative then the problem has an Unbounded Optimal Solution.

c) [c] If one / more Cj-Zj<0 then select the variable to enter into the basic with the

largest negative Cj-Zj value. That is,

d) Ck-Zk= Min{Cj-Zj; Cj-Zj<0}

e) The column to be entered is called the key / pivot column.

4. Determine the key row and key element in the same manner as discussed in the

simplex procedure of the maximization case.

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EXPERIMENT.NO: 4

STUDY OF DUAL PROBLEM, ITS INTERPRETATION & SENSITIVITY ANALYSIS

DEFINATION

The dual is an auxiliary LP problem defined directly and systematically from the original of

primal LP model. In most LP treatments, the dual is defined for various forms of the primal

depending on the types of the constraints, the signs of the variables and the sense of

optimization.

The general standard for the primal is defined as

n

Maximize or Minimize Z = CjXj

j =1

Subject to

n

AijXj = Bi, i = 1,2,……,m

j=1

Xj 0, j = 1,2,….,n

CONSTRUCTION OF DUAL PROBLEM

Following steps are as under:

If the primal problem has its objective function of maximization its all constraints must

have type of inequality. Like that, the primal problem having its objective function to

be minimized should have all its constraints type of inequality.

The nature of optimization is changed, i.e. if objective of primal problem is to be

maximized, the objective of dual problem is to be minimized and vice versa.

Right hand constants of primal problem become objective function coefficients of dual

problem. This demands that for each constraint in primal problem there is a separate

dual variable. And vice versa objective function coefficients of primal problem become

right hand constants of dual problem giving a constraint of each variable in the primal

problem. For example, if primal problem ha m number of constraints and n number of

variables then its dual problems has n number of constraints and m number of variables,

i.e. if primal is of order m x n the dual is of order n x m.

Inequalities of constraints are reversed.

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Row coefficients of the matrix of primal problem become column coefficients of the

matrix of dual problem in the same order i.e. 1st row of primal become 1

st column of

dual.

Primal as well as dual variables are non-negative.

INTERPRETATION AND PROPERTIES OF DUAL

Following are a few important interpretation and properties of dual problem:

The objective function value of any feasible solution to the primal will be less than or

equal to the objective function value of every feasible solution to the dual i.e. ZZmin.

If either the primal or the dual has an unbounded solution, the solution to the other

problem is infeasible.

If both primal and dual problems have feasible solution then both also have optimum

solution and Zmax = Zmin.

If the primal has a feasible solution but the dual has no feasible solutions, then the

primal does not have a finite optimum solution. Likewise, if the dual has a optimum

solution.

The above conclusions are derived as the statements of dual theorems.

Some of the important of Prima-Dual are:

The index row coefficient under the slack variables in primal optimal solution is the

optimal values of the dual variables.

The index row coefficient under variable in the primal optimal gives the difference

between the left and right hand sides of the associated optimal dual corresponding dual

constraints.

DUAL SIMPLEX METHOD

In simplex method we start with initial feasible solution and through stages of iteration

along simplex algorithm we improve the solution till optimal solution is one, in terms of

algorithm for maximization in which index row coefficients are non-negative i.e. dual

variables are feasible. Now, from our understanding of Primal-Dual relations, we know that

index row coefficients are the values of dual variables and hence it suggests dual feasible

solution. There are situations where a primal solution may be infeasible, but corresponding

variables indicate that dual is feasible. Thrust solution is primal infeasible but optimum. In

a dual simplex method, initial solution is infeasible but optimum, and through iteration, it

reaches feasibility at which stage it also reaches true optimum.

To summaries the procedure:

The solution associated with a basis is optimal if all index row coefficients are 0.

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The basic variable having the largest negative value is the departing variable and its row

is the key row.

SENSITIVITY ANALYSIS

Management‟s interest in optimization is not confined to obtaining optimum solutions for

the given set of data. In fact, management for its decision-making in changing situations has

interest in analyzing the present optimal solutions for projecting its future picture. Such an

investigative analysis is called sensitivity analysis. As this analysis is based on the optimal

solution presently obtained, it is called post-optimally analysis.

Sensitivity analysis will seek answers to the following questions: (these questions are put in

relation to a product-mix problem for profit maximization with the given set of constraints

due to resources. However, the analysis will draw general conclusion.)

…. if objective coefficient of a particular basic variable in optimal solution changes, does

the present solution remain optimal? If objective coefficient of a particular non-basic

variable in optimal solution changes, does the present solution remains optimal?

…. when will happen to the present optimal solution if a certain resources is augmented or

curtailed?

…. what will happen to the present optimal solution, if a product is added or dropped?

CHANGE IN THE OBJECTIVE FUNCTION COEFICIENT

Variation in the objective function coefficient may belong to a basic variable or to a non-

basic variable. We shall take the two cases separately. Consider a change in the objective

function coefficient of the non-basic variable in the optimal solution. Any change in the

objective function coefficient of the non-basic variable will affect only its index row

coefficient and not others. Here X3 is non-basic variable. Any decrease in its coefficient

will not affect the present solution.

Now consider a change in the objective coefficient of the basic variable in the optimal

solution. Here, it affects the index now coefficients of all the variable. Hence, as soon as the

index row coefficient of basic variable becomes positive, it leaves the solution, and that of

non-basic variable becomes negative, it qualifies for entry into the solution. In either case,

the present optimal solution changes.

CHANGES IN RIGHT HAND SIDE CONSTANTS

The right hand side constants denote present level of availability of resources. When this is

increased or decreased, it will have the effect on objective function. Besides, it may also

change the basic variables in the solution.

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EXPERIMENT.NO: 5

SPECIAL CASE OF SIMPLEX METHOD

Special applications of simplex methods are:

11)) Transportation technique.

22)) Assignment model.

1. TRANSPORTATION TECHNIQUE

This is a special case of linear programming. The aim of this technique is that within a

given time period, a single homogenous specified commodity to be transported form „m‟

number of sink on destination available units so that, total cost of transportation is

minimum. Here size and location of plant are decided. The technique helps to determining

optimal solution.

Consider „m‟ number of source and „n‟ number of destination, also consider ith

source of

availability of Ai number of units of a product and n number of destinations as jth

destination requires Bj number of units for the same product. Let the cost associated with

transportation is Cij and decision variable is Xij. Let mathematical statement of problem,

m n

Minimize Z = Cij Xij ………………… (1)

i=1 j=1

equation (1) is called objective function.

m

Subject to Xij = Bj

i=1

n

& Xij = Ai

j=1

These are called constraints of the function.

For a balanced problem, suppose there are 4 sources and 3 destinations, minimize,

Z = C11X11 + C12X12 + C13X13 + C21X21 + C22X22 + C23X23 + C31X31 + C32X32 + C33X33 +

%S C41X41 + C42X42 + C43X43

Subject to X11 + X12 + X13 = A1

X21 + X22 + X23 = A2

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X31 + X32 + X33 = A3

X41 + X42 + X43 = A4

X11 + X12 + X13 + X14 = B1

X21 + X22 + X23 + X24 = B2

X31 + X32 + X33 + X34 = B3

For balance problem,

A1 + A2 + A3 + A4 = B1 + B2 + B3

It can be represented in matrix as under:

Origin Destination Availability

1 2 3

1 X11 C11 X12 C12 X13 C13 A1

2 X21 C21 X22 C22 X23 C23 A2

3 X31 C31 X32 C32 X33 C33 A1

4 X41 C41 X42 C42 X43 C43 A4

Requirement B1 B2 B3

TECHNIQUES OF FEASIBLE SOLUTION

11)) North-West Corner Method: It may be recalled that feasible solution is the one,

which satisfies the given set of constraints. Here first, the north-west corner is

selected and the availability of with source utilized and the row is eliminated. Again

after Ath

row is utilized next new north-west corner is selected and its availability

and requirement is satisfied, and this is done until all the requirement is satisfied, and

as the allocation done, cost of respective allocation is known and total cost of

transportation is calculated.

22)) Vogel‟s Approximation Method: This method is an improved algorithm to obtain

better feasible solution. The process is to evaluate penalty number for each row and

column. It is given, as the difference between the lowest cost coefficients is each

row and column. The penalty number physically amounts to cost penalty associated

with by not accepting that row or column for assigning an allocation to its lowest

cost cell. Select largest of these penalty numbers and determine the row or column

to which an allocation has to be make. Thus row or column selected against largest

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penalty cost will be assigned more possible units and its lowest cost. The process

will be repeated for each allocation to be made after eliminating row or column.

33)) Inspection Method: If size of problem is small, inspection method is used. Here

dummy row or column will be allocated. Select lowest cost cell and assign more

feasible units to it. Repeat this until Ai < Bj or Ai > Bj and until availability and

requirement become equal.

DIFFERENT METHODS FOR OPTIMAL SOLUTION

Before any other method is used for optimally test a check has to be made over number of

allocations assigned in the solution; which must be equal to (m + n) having non-degenerate

solution.

11)) Stepping Stone Method: Any empty cell in the feasible solution gives non-basic

variables. Each vacant on unfilled cell has Xij = 0. We never know if any one or

more of them may become basic variables there by reducing the transportation cost

or improving solution further. Each empty cell must be evaluated. Opportunity cost

of an empty cell is the net increase or decrease in the total transportation cost if a unit

is assigned to it. As sum conditions are to be honored, one unit from the present

allocation in the row as well as column of this empty cell has to be deducted and

some unit has to be added to the present allocation in the column as well as row at

which deduction took place and whole orthogonal path is completed.

22)) Modi Method: This method is also called modified method. In this method, we

generate a column under beading of Ui and Vj of respectively right of the solution

and row of at the bottom of the solution. Here number of variables „m‟ and „n‟ at Ui

and Vj are determined. Select any one of the Ui variables and assign any arbitrary

value (normally zero) to it and determine rest of them the relationship.

Ui + Vj = Cij

After all the values Ui and Vj are determined, opportunity costs of empty cells are

evaluated using the relationship.

Dij = Cij – (Ui + Vj)

2. ASSIGNMENT MODEL

Consider that there are „n‟ persons having varying degrees of proficiency to accomplish „n‟

different tasks in an organization. These „n‟ tasks are to be assigned to each one of „n‟

persons in such a manner that the effective performance of the organization as a whole is

maximized. This is called assignment problem. The objective function to be optimized

may be effective performance (more) total cost of finishing „n‟ jobs or total time of

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completing „n‟ jobs or any other suitable criteria for objective function depending upon the

assignment problem.

MATHEMATICAL STATEMENT OF ASSIGNMENT MODEL

With „n‟ tasks and „n‟ agents, let Cij be the cost of its agent, performing jth

task.

n n

Min Cij Xij

i=1 j=1

n

Sub to Xij = 1

j=1

The statement appears similar to that of transportation techniques and Xij takes value either

0 or 1.

Xij = 1 (if task is assigned to agent 1)

Xij = 1 (if task is not assigned to agent 1)

TECHNIQUE FOR ASSIGNMENT MODEL

Hungarian method follows steps as under:

Add as many dummy rows on dummy column to make matrix square. This

corresponds to balancing of transportation problem. In a problem of assignment, each

of the origin has availability of a unit and each of destinations has requirement of a

unit.

Take the first row and subtract its smallest cost element from all the elements. Repeat

this for all the rows.

Similar do for column. In this there will be at least one zero element in each row and

column.

Starting with the row that has only one zero give assignment there and cross mark

there.

If there are more than one zero, start with column and find the solution.

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EXPERIMENT.NO: 6

PROJECT EVALUATION AND REVIEW TECHNIQUES & CRITICAL PATH METHOD (PERT/CPM)

In real world time can‟t be known with total certainty. Particularly there are several

projects involving research, innovation, and design, development where there is no previous

experience or record of handling such activities. For such activities forecasting of time

estimates will involve uncertainty. This necessitates probabilistic approach and is popularly

known as PERT. This is its one and only feature which distinguishes it from CPM.

Time t for an activity is assumed top have its probability density p(t) following Beta

distribution in PERT; thus it is

P(t) = k (t-to)tp-t)

where to = the optimistic time,

tp = the pessimistic time,

= exponents values of which are decided on the basis of the mode and mean ab

of the distribution

Optimistic time means estimated duration of activity where all factors concerning the

activity operate favorably. Although it is easier said than obtained, optimistic time

presumes that everything goes all right for the activity right from the word „get-set-go‟.

Pessimistic time means estimated duration of activity where in all factors concerning the

activity operate unfavorably. Both of these estimates are qualified guesses and are

additionally consider the concept of most likely time, tm that is the mode of the distribution.

Figure 1 shows beta distribution followed by activity times, where TE is the average time in

which activity is expected to be

completed. The value of TE

depends on how close the values

of to and tp are relative to tm.

Expected time of an activity

TE = ( to + 4tm + tp

) / 6

Since the actual time of an activity

is likely to vary form mean value

(TE), we need the variance of the

activity time distribution. Being

unimodal distribution, it spread

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may be approximated to six times the standard deviation value. Thus,

6= tp - to

Having three time estimates for all the activities in a project, we can compute average time

and variance for each activity. Considering average time as actual time, the critical path is

found. Project time will then be the sum of all the activity times in the critical path. But

the activity time is random variable, sum of all such times is also a random variable, and we

must give average time of the project and its variance. The variance of the project duration

is the sum of all the variances of the activities in the critical path.

PERT & CPM

Attempt is often made to bring out difference between PERT and CPM, besides that of

time-estimates being probabilistic in PERT and deterministic in CPM. In fact, all other

aspects are common. Historically PERT came up from different problem area. This was in

relation to plan and accelerate development of the Polaris ballistic missile.

When activity durations are uncertain, attention is principally focused on vents of activity.

This is because events are identified. For example, to develop a process for synthetic

cryogenic rubber. Here terminal point-event is known, though activity duration is not

known with certainty. It is in this context of more familiarity of events involved and less

certainty of activities. The PERT is called event-oriented. In fact, the difference is not

organic but that of focus.

EXPECTED DURATION OF CPM-PERT

By calculating expected activity time for each activity from three times estimates, PERT

network is prepared and critical path is found on that. As each activity time has standard

deviation and variance, the project duration obtained is also expected duration having

standard deviation and variance. The critical path length is the summation of activities on

the path. As the variance of a sum of independent activities is equal to the sum of their

residual variance, the variance of critical path duration can be calculated,

VT = variance of critical path = variances of critical activities

VT = Vt1 + Vt2 + Vt3

VT = variance of any activity in PERT = ()2

Where is the standard deviation of each activity time

=(tp – to) / 6

ST = standard deviation of the project = VT

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Variance VT standard deviation ST of the project gives the measure of uncertainty of the

expected duration of the project TE.

CRASHING

Being more practical on scheduling problems, know if the activities involved in the project

can be reduced in duration by providing extra resources. These added resources will

increase the cost of completing the activity, but other advantage out weigh this increased

cost, the activity should be expedited, or crashed. The duration of the project will be

reduced, only if duration of one or more critical activities is reduced. If there are activities

using similar resources and if these activities have slack, they can be carried out at their

normal or most efficient pace, with allocation of less resource, so that part of these can be

diverted to the activity on the critical path without additional cost. In general, this is not

possible for all activities and thus we use the concept of criticality to apply extra resources

to project activities selectively, so that maximum reduction of project time is achieved with

the least additional cost. The increased cost of an activity and is stated in terms of

additional expenditure required for reduction in activity duration of one unit time period.

The total cost of a project is not merely the sum of all direct costs of all the activities in the

project, but other indirect expenses, which depend on the duration of project, must also be

included.

Thus, project schedules influence two types of costs – direct cost of all activities in the

project, which increases when activities are expedited and indirect cost which decreases

when the project duration is reduced.

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Figure 2-a gives the relation of direct cost of the activity versus activity duration, presuming

linear relationship. On the other hand, relation of indirect costs with respect to project

duration assumed linear is given in figure 2-b.

If activity „A‟ has the direct cost defined by the slope A any reduction of the duration of

the activity by time TA will involve additional cost of (TA) *A. As against that it will

reduce the indirect cost of the project by (TA) * . Thus, crashing of activity A is

beneficial only if

(TA) * (TA) *A

i.e., A

It, therefore, suggests that it is possible to have an optimum schedule – the lower – cost

schedule, which strikes a balance between direct costs and indirect costs.

For simplicity, time cost relationships is approximated as linear one. For better

understanding, activity time-cost relationship is further elaborated. Initially, the time

estimate is so set that the resources required are used in the most efficient way to

accomplish the task at minimum cost. These are known as normal time and normal cost.

Suppose there is an activity, which requires three men for its completion in normal time. In

order to expedite it, every time one more man may be added and reduction in time and

increase in cost may be noted and plotted on graph. Every time the slope will be steeper.

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Finally it reaches to the point, where increase in number of men does not reduce the time

further, thus this is the shortest time in which this activity can be completed. This is called

crash time and cost against it as crash cost. By joining the crash and normal points, the line

is accepted as entire relationship. Beyond normal time, if the activity is delayed, the cost

may remain constant for small duration but will increase if the activity is excessively

delayed. Figure 3 shows the activity time cost relationship.

There is no reason to believe that every contractor should have some normal and crash

points for a selected activity. Thee tow points will depend on experience of the ken,

available equipments, teamwork, administrative ability that differ from organization to

organization.

The CPM model finds the least cost

schedule after verifying the possible

crashing of the project. To begin

with, a preliminary schedule is

generated with each activity at its

normal point. This is the maximum

length schedule. Then activities on

critical part are marked as critical

activities. For each of these critical

activities, direct cost versus reduction

in activity duration relation

established. At the same, relation

between project duration versus

indirect cost of project is also known.

If for one or more activities increase in

direct cost is less than the saving in the

indirect cost, then a less expensive schedule can be obtained. While considering critical

activities, the activity that has the highest potential for affecting reduction in duration are

the lowest additional cost is chosen first. The improvements are made in steps. New

schedules are generated as long as activities can be crashed with the net reduction in total

project costs.

NETWORK MODELS

Example: Routine problem on telecom networks.

At a given time some links are available, from the available ones, select the

cheapest/fastest/non reliable/shortest path from a given origin to a destination.

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REPRESENTATION

Label the modes 1,2,…..,m.

Label the links 1,2,…..,m.

DISCRETE FORMULATION

For a minimally connected network, n = m – 1.

Each link may change attributes [cost, capacity, etc.].

Each node may have same attributes [demand, capacity, etc.].

Put 1 for end mode.

Put –1 for beginning mode.

A useful concept is to define flows on links of a network.

Let Xj be the row on link j, if Cj is the cost of flow on j.

OBJECTIVE

Xj = 1 j leaving

Cj = 1 j entering

General minimum cost flow problem (also called optimum distribution problem)

Min CjXj

Subject to CijXj = Bi for each i j = arcs, i = nodes.

Is Xj Cij

The shortest path problem is a special case of this with

Bi = 1 for the origin

Bi = -1 for the destination

Bi = 0 for all other nodes

Cij = cost / time for traveling link j, ij = 0, ij = 1.

The bound constraints are treated separately. This Cp has integer solution (if the data is

integer). This is because of the constraint matrix.

OPERATIONS RESEARCH


EXPERIMENT.NO: 7

GAMES THEORY

INTRODUCTION

In business and economics literature, the term „games‟ refers to the general situation of

conflict and competition in which two or more competitions are engages din decision-

making activities in anticipation of certain outcomes over time. The competitors are called

players. A player may be an individual, a group of individuals, or an organization. A few

examples of the competitive and conflict situations where techniques of theory of games

may be used to resolve them are; two or more candidates contesting and election with the

objective of winning with more votes; advertising and other marketing campaigns between

competing business firm; contractors filling bid to win business contract etc theoretically

theory of games provides mathematical models that can be quite useful in explaining

interactive decision-making concept. But as a practical tool, theory of games is limited in

scope. However, it is one type of decision-making technique where on competitor‟s choice

of course of actions does determined after taking in to account all possible alternative

courses of action of another competitor engaged in the competition.

The models in the theory of games can be classified depending upon the following factors:

1. NUMBER OF PLAYERS: If a game involving only two players, then it is called a

two – person game. However, if the numbers of players are more than two, the game is

referred to as n-person game.

2. SUM OF GAINS & LOSSES: If in a game the gains to one player are exactly equal to

the losses to another player, so that sum of the gains and losses equal zero, then the

game is said to be a zero-sum game of it said to be non-zero game.

3. STRATEGY: The strategy for a player is the list of all possible actions that the he will

take for every payoff that might arise. Here it is on necessary that players have definite

information about each other‟s strategies.

The particular strategy by which a player optimizes his gains or losses without knowing the

competitor‟s strategies is called optima strategy. The expected out come per play when

players follow their optimal strategy is called the value of game. Generally, players in a

game employ two types of strategies as given below:

1. PURE STRATEGY: It is decision rule, which is always used by the player to select

the particular course of action. Thus, each player knows in advances of all strategies

out of which he always selects only one particular strategy irrespective of the strategy

OPERATIONS RESEARCH


others may choose and the objective of the players is to maximize gains or minimize

losses.

2. MIXED STRATEGY: When both the players are guessing as to which course of

action is to be selected on a particular occasion with some fixed probability, it is mixed

strategy game.

TWO – PERSONS – ZERO – SUM – GAME

A game with only two players, say player A ands B is called Two – persons - Zero – Sum –

Game, if say player A‟s gain is equal to the loss of player B, so that total sum is zero. The

payoffs in terms of gains or losses, when players select their particular strategies, can be

represented in the form of a matrix, called the payoff matrix.

ASSUMPTIONS OF THE GAME

1. Each player has available to him a finite number of possible courses of action. The

list may not be same for each player.

2. Player A attempts to maximize gains and Player B minimize losses.

3. The decisions of both players are made simultaneously and announce simultaneously

so that neither player has an adventives resulting from direct knowledge of the other

player‟s decision.

4. Both the players know not only possible to themselves by also of each other

PURE STRATEGIES

GAMES WITH SADDLE POINT

The selection of an optimal strategy by each player without the knowledge of the

competitor‟s strategy is the basis problem of playing games. Since the payoff for either

player provides all the essential information, therefore only one player‟s payoff table to

required evaluating the decisions. By convention, the payoff table for the player whose

strategies are represented by (say player A) rows is constructed. Now the objective of the

study 1 to know how these players select their respective strategies so that they may

optimize their payoff. Such a decision making criterion is referred to as the minimax –

maximin principle. Such principle I pure strategies game always leads to the best possible

selection of a strategies game always leads to the best possible selection of a strategy for

both players.

RULES TO DETERMING THE SADDLE POINT

The readers are advised to follow the following steps in order to determine the saddle point

in the payoff matrix.

OPERATIONS RESEARCH


1. Select the minimum element in each row of the payoff matrix and write them under

row minima heading. Then select a largest element among these elements and

enclose it in a rectangle.

2. Select the maximum elements in each column of the payoff matrix and write them

under column maxima heading. Then select a lowest element among these elements

and enclose it in a circle.

3. Find out the elements that are same in the circle as well as rectangle and mark the

position of such elements in the matrix. This element represents the value of the

games and it is called the saddle point.

MIXED STRATEGIES

There are certain payoffs where pure strategies do not lead to saddle point and hence it is

not possible to decide optimal strategies. There is no stable solution of the payoff matrix.

The shifting from one strategy to another in case of unstable solution results in selection of

the strategies by both players on random basis. The two players have to decide their plans

for playing the game and assign probability values X and Y by which they will play their

respective strategies. These plans are referred as mixed strategies.

ARITHMETIC MODEL

This method enables us to decide odd elements of each strategy in mixed strategies. From

odd elements, we can calculate the optimal frequency of each strategy and also the value of

the game.

Consider the following payoff matrix:

B1 B2

A1

A2

a11

a21

a12

a22

1 = a11 – a12

2 = a21 – a22

3 = a11 – a21 4 = a12 – a22

Odd elements of a given row are obtained by the numerical difference of elements in

another row. Similarly, oddments of a given column are obtained by the numerical

differences of elements in another column.

Optimal frequency of the strategy A1, X1 = 1 / (1 + 2)

Optimal frequency of the strategy A2, X2 = 2 / (1 + 2)

Optimal frequency of the strategy B1, Y1 = 3 / (3 + 4)

Optimal frequency of the strategy B2, Y2 = 4 / (3 + 4)

If total of oddments of rows is equal to the total of oddments of columns, the value of the

game, V will be

Vb1 = a11X1 + a21X2 ….. value if B plays 1

OPERATIONS RESEARCH


Vb2 = a12X1 + a22X2 ….. value if B plays 2

Va1 = a11Y1 + a12Y2 ….. value if A plays 1

Va2 = a21Y1 + a22Y2 ….. value if A plays 2

METHOD OF SUB – GAMES

This is applicable to any m x 2 or 2 x n game. The whole game is split into several 2 x 2

sub-games, which are then solved by pure strategy, arithmetic method or algebraic method

as applicable. From amongst these sub-games, a sub-game which gives maximum game

value m x 2 game (where row player has more than 2 strategies) or minimum game value in

2 x n games (where column player has more than 2 strategies) is selected as the solution.

METHOD OF MATRICES

Previously we discussed arithmetic method as applicable 2 x 2 payoff matrix. In fact, it can

also be extended 3 x 3 matrix, as under, which is self – explanatory.

B1 B2 B3

A1 a11 a12 a13

A2 a21 a22 a23

A3 a31 a32 a33

Differences of column 1 & 2 Differences of column 2 & 3 Oddments

1 = a11 – a12 4 = a12 – a13 1

2 = a21 – a22 5 = a22 – a23 2

3 = a31 – a32 6 = a32 – a33 3

Differences of row 1 & 2 Differences of row 2 & 3 Oddments

7 = a11 – a21 10 = a21 – a31 1

8 = a12 – a22 11 = a22 – a32 2

9 = a13 – a23 12 = a23 – a33 3

If sum of row oddments is equal to sum of column oddments, optimum frequencies are:

X1 = 1 / (1 + 2 + 3); X2 = 2 / (1 + 2 + 3); X3 = 1 / (1 + 2 + 3).

Y1 = 1 / (1 + 2 + 3); Y2 = 2 / (1 + 2 + 3); Y3 = 3 / (1 + 2 + 3).

If sum of row oddments is not equal to sum of column oddments, the methods of matrices

fail.

OPERATIONS RESEARCH


LINEAR PROGRAMMING METHOD FOR GAME PROBLEM

As remarked in graphical method for 2 x n or m x 2 games, the characteristic of linear

relations can be used for solving the games by simplex method of linear programming.

This method is applicable for more general m x n games between two players. It also

comes handy when oddments of rows are not equal to oddments of columns. The game

problem can be solved by simplex algorithm and the value of the game obtained for new

matrix will be reduced by the constant added to get the value of the game for original

matrix. Frequency values of strategies in both matrices remain same.

OPERATIONS RESEARCH


EXPERIMENT.NO: 8

INVENTORY MANAGEMENT

INTRODUCTION Inventory is material in stock and idles at a given point of time that will be used to satisfy

some further demand of that material. Inventory decreases as demand process operates and

increases as replenishment process operates. Demand may be random function and perhaps

normally not under control of decision-maker, while replenishment in monitored by the

decision-maker. The replenishment is the input process to the inventory system and gives

the rate at which material is added to the inventory system. The source of replenishment

may be either production or purchase of the material. The rate at which material is

withdrawn from the inventory is the output process of the inventory. The output is always

in response to demand. The output rate equals to demand rate unless the policy decision

provides otherwise or in event of inventory level decreasing to zero level, when the

situation is described as stock out or out-of-stock condition.

The mathematical models of inventory system under different operating conditions involve

mainly two-decision variable i.e. how much to order and when to order, when source of

replenishment is purchase of how much to produce and when to produce, when production

is the source or replenishment. With the associated cost of material and its storing,

inventory system operating cost. It is therefore necessary to appreciate what benefits are

derived from having inventory and then derive the concept of optimum inventory. The

following benefits of inventory control, which resemble with its function, are self-

explanatory:

1. Inventory enables in decoupling production system from demand fluctuation. If demand

increases inventory of finished and semi-finished goods take care of it. It demands

decreases the inventory of finished and semi-finished goods could be built up.

2. Inventory enables to derive advantages of planning optimum lot size in production as a

trade-off against setting time.

3. Inventory enables the sequencing of different components for optimum machine loading

and set-up costs.

4. Inventory enables to have uniform labor force resulting labor force resulting in good

labors relation. In production of the product with seasonal demand.

5. Inventory enables to sustain customer goodwill, as it is possible to honors the orders by

committed dates.

6. In the absence of adequate, inventory a situation my come where “rush-order” has to be

made for which, in general, very high price as to be paid.

OPERATIONS RESEARCH


All these advantages justify the policy of having large inventory. However, there are

considerations, which warrant such a policy. These considerations are as under:

1. Inventory means locked up capital, which could be used for alternative purposed.

2. Locked up capital involves interest.

3. Inventory being physical goods in likely to deteriorate on storage.

4. It involves the risk of fire ad theft. Alternatively, if involves insurance cost. This

suggests that although inventory, because it has to be only adequate. Thus, excess

inventories are undesirable. This calls for controlling the inventories in most profitable

way.

INVENTORY MODELS

By inventory model, we mean to represent the inventory state, operating condition and costs

involved through mathematical relationships, so as to fine Economic Order Quantity when

source of procurement is purchase or to fine Economic Batch Quantity when source of

procurement is manufacturing.

Model 1:

The model is Deterministic. Single item model i.e. it ignores all uncertainties and deals only

with one item. It is characterized by: (1) demand is constant, at uniform rate and known

with certainty (2) lead time and other system parameters such as costs are constant,

independent of replenishment quantity and are known with certainty (3) replenishment is

infinite and (4) shortage of stock is not permitted. A production system needs R units of a

purchased material over a schedule period of time T. the rate of usage is uniform i.e. R/T. if

all R units are purchased at the beginning of the schedule period T, the inventory position is

as shown in fig, which is the graphical presentation of inventory level v/s time.

OPERATIONS RESEARCH


T

Ouantity

in s

tock

Time, t

As shortage is not permitted, the replenishment has to commence at b. As replenishment is

instantaneous, it is shown by the vertical line and the next cycle commences.

OPERATIONS RESEARCH


Thus, for the schedule period T there is average inventory of R/2 units, which involves

holding cost. There is one order cost. Therefore, total associated cost for order R units:

TCR (R) = C1 (R/2) T + C2 …..(1)

Instead, if q units are ordered at the beginning and the order of the same quantity is

repeated R/q times, the position will be shown in following figure.

tc

Ouantity

in s

tock

tc tc

Time, t

The total associated cost in schedule period T will then be R/q times cost associated with

each order.

TAC (q) = [C1 * (q/2) * tc] R/q

TAC (q) = C1 * (R/2) * tc + C2 (R/q)…………(2)

Comparing equations (1) and (2), it can be seen that by not ordering the full quantity in the

beginning the cost-component due to holding decreases, as tc < T, and cost-component due

to re-ordering increases R/q times. So, there has to be certain value of q at which TAC(q) is

min. This value of q is optimum order-size of Economic Ordering Quantity designated as

qo. Number of purchase-cycle in schedule time T = R/q = T/ tc.

Therefore equation (2) can be rewritten as:

TAC (q) = C1 * (R/2) * (Tq/R) + C2 (R/q)

TAC (q) = C1 * (T/2) * q + C2 (R/q)…………..(3)

From above equation TAC(q) depends only on q. Two individuals cost component

(C1T/2)*q and C2R/q, as well as TAC(q) are shown in the following figure.

To determine qo we differentiate TAC(q) in equation (3) with respect to q, for maximum

and minimum TAC(q). Thus,

d(TAC)/d(q) = C1T/2 – C2R/q2 = 0

So, qo = 2C2R / (C1/T)

OPERATIONS RESEARCH


C R/q

q

Cost

2

C Tq/21

TAC (q)

Second differentiation of TAC is Positive. Hence, the value of q is for minimum TAC, and

corresponding value of q is optimum.

So, qo = 2C2R / (C1/T).…………(4)

This is called optimum lot size, also known as economic lot size. R/T is the rate of demand,

in units per unit time.

Designating it as r = R/T equation (4) becomes,

qo = 2C2r/C1

If i is the interest rate on capital per unit time T and if holding cost essentially consists of

purchases-investment. Then C1 = i * P where P is price per unit quantity.

So, qo = 2C2R / (iPT)

By substituting the value of qo of equation (4) in equation (3) we get minimum total

associated cost during scheduled period T

TAC (qo) = [C1C2RT/2] + [C1C2RT/2] = 2C1C2RT

Minimum total associated cost per unit time TAC (q) = 2C1C2r

This shows that for optimum value of qo when TAC minimum the cost-component due to

holding of inventory is equal to the cost component due to re-ordering. Thus in for point P

lies on the ordinate from the point of intersection of the graphs of individual cost-

components.

Now, tcc optimum cycle times is given by,

Ttcc = Tqo/R = T/R * [2C2R/C1T] = 2C2T/(C1R) = 2C2/(C1r)

OPERATIONS RESEARCH


MODEL 4

Having considered deterministic demand models earlier, we may now consider the cases

where demand is probabilistic, which is quite realistic in every day world.

CHARACTERISTIC:

Demand is in discrete quantities; demand is probabilistic and is met instantaneously.

Set-up cost is zero. Therefore, relevant cost elements are holding cost and shortage

cost.

Let r = number of discrete units required with probability, pr,

l = stock level in discrete units,

C1 = holding cost per unit quantity per unit time,

C3 = shortage cost per unit quantity per unit time,

When l > r the inventory cost is due to holding.

When r > l the inventory cost is due to shortage.

1. As demand is probabilistic the probable units in inventory is (l - r) pr and its cost is

C1 (l - r) pr. The total inventory cost due to holding is,

r=1

C1 (l - r)pr

r=0

As the demand exceeds and becomes l + 1 and above the cost is due to shortage. The

probable units in shortage is (r - l) pr and its cost is C3(r - l) pr. Thus, if the demand is to

vary from zero to infinity,

r=1 ∞

TAC (l) = C1(l - r)pr + C3(r - l)pr .....(5)

r=0 r=l+1

Now, optimum stock level lo can be considered as one in which one more unit well as one

less unit will increase the cost. That is at lo.

TAC (lo + 1) > TAC (lo) and

TAC (lo - 1) > TAC (lo)

We may re-write the equation (5) for l+1 stock level

l+2 ∞

TAC (l + 1) = C1(l + 1 - r)pr + C3(r - l - 1)pr .....(6)

0 l+2

l+1 1

OPERATIONS RESEARCH


Now C1(l + 1 - r)pr = C1(l + 1 - r)pr + C1(l + 1 - l - 1)pl+1

0 0

1

= C1 (l + 1 - r)pr

0

l+1 1 1

So, C1(l + 1 - r)pr = C1(l - r)pr + C1(l - r)pr .....(7)

0 0 0

∞

Similarly for C3(r - l - 1) pr

l+2

∞

= C3(r - l - 1) pr - C3(l + 1 - l - 1) pl+1

r=l+1

∞

= C3(r - l – 1 )pr

l+1

∞ ∞ ∞

So, C3 (r - l - 1)pr = C3 (r - l)pr - C3pr .....(8)

l+2 l+1 l+1

Substituting equations (7) and (8) in equation (6),

l l ∞ ∞

TAC (l + 1) = C1(l - r)pr + C1pr + C3(r - l)pr - C3pr

0 0 l+1 l+1

So,

l ∞ l ∞

TAC (l + 1) = C1(l - r)pr + C3(r - l)pr + C1(l - r)pr - C3pr .....(9)

0 l+1 0 l+1

First two terms in equation (9) make TAC (lo) as in equation (5), therefore,

1 ∞

TAC (l + 1) = TAC (l) + C1pr - C3pr

0 l+1

∞ l ∞

OPERATIONS RESEARCH


Now, pr = 1 = pr + pr

0 0 l+1

l l

Therefore, TAC (l + 1) = TAC (l) + C1pr - C3 [1 - pr] 0 0

l

So, TAC (l + 1) = TAC (l) + (C1 + C3) pr - C3

0

For l to be optimum, TAC (lo + 1) > TAC (lo)

lo

i.e., (C1 + C3) Pr - C3 > 0

0

lo

So, Pr > C3

0 C1 + C3

So, P (r / lo) > C3 .....(10)

C1 + C3

P (r / lo) means summation of P (r) for the value of r from zero to lo. If stock level is l - 1,

l-1 ∞

TAC (l - 1) = C1 (l - 1 - r) pr + C3 (r - l + 1) pr .....(11)

0 l

l-1

The first term C1 (l - 1 - r) pr can be rewritten as

0

l

C1 (l - 1 - r) pr + C1pr

0

l l

C1 (l - r) pr - C1 pr + C1pr .....(12)

0 0

∞

The second term C3 (r - l + 1) pr can be rewritten as

l

∞ ∞

OPERATIONS RESEARCH


C3 (r - l + 1) pr = C3 (r - l + 1) pr + C3pl

l l+1

∞ ∞

= C3 (r - l) pr + C3pr + C3pl

l+1 l+1

∞ l

= C3 (r - l) pr + C3 (1 - pr) + C3pl .....(13)

l+1 0

Substituting equation (12) and (13) in equation (11),

l l ∞ l

TAC (l - 1) = C1 (l - r) pr - C1pr + C3 (r - l)pr + C3 (1 - pr) + C3pl

0 0 l+1 0

l ∞ l

TAC (l - 1) = C1 (l - r) pr + C3 (r - l)pr - (C1 + C3) pr + C3 + C1pl + C3pl

0 l+1 0

So,

l-1

TAC (l - 1) = TAC(l) - (C1 + C3) pr + C3

0

For l to be optimum,

TAC (lo - 1) > TAC(lo)

l-1

So, - (C1 + C3) pr + C3 > 0

0

l-1

So, C3 > pr

C1 + C3 0

So, C3 > p(r / lo - 1) .....(14)

C1 + C3

Combining equations (10) and (14), for minimum TAC (lo) so the optimum stock level

should be such that,

p (r / lo - 1) < C3 < p (r / lo)

C1 + C3

OPERATIONS RESEARCH


p (r / lo - 1) means summation of p (r) for the value of r from zero to (lo - 1).

OPERATIONS RESEARCH


EXPERIMENT.NO: 9

QUEUING THEORY

INTRODUTION

Most of the basic work on queues or waiting line theory was carried out by A. K. Erlang in

early nineteenth century. He experienced congestion of telephone traffic during his work at

Copenhagen Telephone Company and came out with most of the work for the mathematical

models in queuing theory.

Service-seekers and serves make a system, which needs consideration. Service may be

anything that occupies a man, facility or location for some period of time so that potential

service-seekers known as arrivals at any time and require the use of server. If at any time, in

a queue may drive out a potential customer. On the other hand, if more serves are provided

that reasonably necessary so that no queue or very short queue is formed, it costs more

money. Therefore, a trade-off becomes necessary in designing service facility system. This

needs an insight into the growth of queue in a given system.

INPUT PROCESS:-

In order to describe the input processes, a random variable such as time between successive

arrival events is designated. It is assumed that random variable is independent and

identically distributed. Thus knowing the probability distribution, the input process can be

described. Often the arrival of customers is completely random in nature and it is assumed

that the number of customers arriving in any time interval „t‟ follows through Poisson

distribution, with parameter t. Here, is the average number of arrivals per unit time.

There are some situations where customers arrive at fairly regular time intervals.

OPERATIONS RESEARCH


One variation of input process is the effect of status of queue on arrival. If a queue is

specified to be finite, a customer arriving at the queue after this is not permitted to join the

queue. Further, even if queue is not finite, a customer, seeing a long queue ahead of him

might not join the queue. This is called balking. It may also happen that customer having

joined the queue earlier might leave it with or without any intention of returning. This is

called reneging. It is pertinent to queue discipline.

So, probability of one arrival = 1 - (1-T) = T

Let us note that if the density of inter arrival is exponential, the density g(y) of the total

arrival time y for ant n consecutive arrivals is given by,

g(y) = (y)n-1

e-ny

/ (n-1)!; y 0

and probability of n arrivals in time interval, T = (T)n e - T/n!

Thus, assumption of exponential distribution of inter arrival times is equivalent to

assuming Poisson distribution for probability distribution of n arrivals in interval T.

Expectation, E (n/T) = T

Variance, 2 (n/T) = T

Instead of exponential distribution of inter arrival time, Erlang distribution of order n can be

used. This is,

f(t) = (n) (

nt) (n-1) e - nt/(n-1)!; t 0

Expectation, E(t) = 1 /

Variance, 2(t) = 1/n

2 (Erlang)

SERVICE TIMES

Assuming that one customer is served by one server at a time, and service time for one

customer is independent of that for other.

Let g(t) = probability density function of time t required to serve one customer. Then mean

time of service = 0∞ e

-t g(t) dt = 1 /

where is mean service rate or number of customers served per unit time. Let us assume

exponential (negative) distribution fro service time. Then,

g (t) = e-t

; t 0

probability that service is not completed in time interval T = e-t

for small values of T,

it is approximately.

OPERATIONS RESEARCH


Probability of the event that service is complete in time interval T = 1 - T and the

probability of the event that service is completed in time interval T = T

QUENUE DISCIPLINE

Queue discipline describes the policy of serving the service-seekers of a queue. Normally

first-in-first-out popularly known ad FIFO (first-come-first-served) is a common queue

discipline. It is generally followed in practice. Some manufacturing concerns consider the

problem of machine breakdowns and repairs as a queuing model. In a group of machines,

breakdown takes place at random time.

ITERARRIVAL TIMES:-

It is necessary to specify the probability distribution of inter arrival times. Considering f(t)

as the density function for the time interval t between any two consecutive arrivals, when t

0.

Meantime between two consecutive arrivals = 0∞ f(t) t dt

So, 1/ = 0∞ f(t) t dt

It arrival is assumed completely random; the density function of inter arrival time can be

described by negative exponential distribution.

i.e., f(t) = e-/t

, t0.

Now, let A be the event that there is no arrival in the interval (O, T). The probability that

there is no arrival in the interval (O, T) and no arrival in (O, T+T) is given by

P(AB) = T+T∞ e

-/t dt = e

-(T+T)

Now, P(B/A) = P(AB)/P(A) = e-(T+T)

/ e-/t

= e -t

Probability of no arrival in interval T = e -t

This depends only on T.

Expanding e -t

by Taylor‟s series;

Probability of no arrival in interval T = 1 + (-T)/1! + (-T)2/2! + (-T)

3!

= 1-(T) + 2(T)

2/2 -

3 (T)

3/6

OPERATIONS RESEARCH


Neglecting higher terms of (T) for very small values of T, approximately probability of

no arrival T = 1T.

In any small interval of time T, as it is presumed to be sufficiently small to permit only

one arrival at the most, only two mutually exclusive and collectively exhaustive events can

occur i.e. either there is one arrival or there is no arrival.

P [no arrival in any interval T ] + P [only one arrival in interval T] =1

KENDALL‟S NOTATIONS

Kendall‟s notations are to specify queuing models in standardized symbols, which mainly

relate to inter arrival time and service time distribution and number of serves in the system

as

Nature of input Nature of service Number of

Process time distribution servers

For symbols,

M designates Exponential distribution

En designates Erlang distribution of order n.

D designates Deterministic distribution

Thus, the queue model M/M/2 symbolizes Poisson input, exponential service time and two

servers.

It is believed that in order to specify a queuing model correctly one must specify the queue

discipline above. To cover this additional variations Kendall‟s rotations were extended by

three more terms in 1971 in a conference on Standardization of Notation in Queuing

Theory. Now it reads :

Input Service Number of Limit of number Number in Queue

Process process servers in system the source discipline

Last three terms are omitted is they are

(///∞/∞/ FIFO)

QUEUE M/M/1: BALANCE EQUATION METHOD

Let us consider a service system where there is only one server, input is Poisson and

se5vice time distribution is exponential. We want to obtain operating characteristics of the

queue. These characteristics will answer some of the following questions:

Will a queue be formed?

OPERATIONS RESEARCH


If so, will it be what is its probability of having n customers?

What will be the mean number of customers in the system?

What will be the queue length?

What will be the waiting time for a customer?

Let inter arrival time density function be e -t

and service time density function be e-t

.

Let at any time n be the number of customers in the system, including those waiting and

those being served. Let Pn(T) be the probability that there are n customers in system at time

T.

In that case, the probability of having n customer in the system at time T+T is the union of

four mutually exclusion and exhaustive events as:

One arrival and no departure in time interval T, T+T with n-1 customers at time T.

Neither arrival nor departure during time interval T, T+T with n customers at time T.

One departure and no arrival during time interval T, T+T with n+1 customers at time

T.

One arrival and one departure during time interval T, T+T with n u

Therefore, the balance equation can be set out as under:

Pn(T+T) = (Probability of 1 Arrive in T)(Probability to no departure inT)Pn-1(T) +

(Probability of no arrival in T) (Probability of no departure in T) Pn(T) +

(Probability of no arrival in T)(Probability to one departure in T)Pn+1(T) +

(Probability of 1 Arrive in T)(Probability to one departure in T) Pn(T)

Making use of relations,

Pn(T+T) = (T)(1-T) Pn-1(T) + (1-T)(1-T) Pn (T) + (1-T) T Pn+1(T)

OPERATIONS RESEARCH


Simplifying the terms on the right hand side,

[Pn(T+T) - Pn (T)] / T = Pn-1(T) - TPn-1(T) - Pn(T) - Pn(T) + 2T Pn(T) +

Pn+1(T) - T Pn+1(T)

lim

dPn / dT = T 0 {Pn(T+T) - Pn (T)} / T

dPn / dT = Pn-1(T) – (+)Pn(T) + Pn+1(T)

This expression is valid for n>0

For n=0, n-1 is meaningless, and departure has no significance. Therefore P0(T) is also

meaningless. Let this limiting value of probability distribution be n for equilibrium dPn /

dT must be zero, Pn becomes independent of T. Then equation becomes:

Pn-1(T) - (+)Pn + Pn+1 = 0, for n = 1,2,3……..

subscript (T) is dropped for brevity and in equation ,

- P0 + P1 = 0, for n = 0

P1 = {/}P0

Let / = . If >1, means rate of arrival is more than mean rate of service. As a

consequence, the queue continues to build up indefinitely and for this the stationary

probabilities do not exist. Only if 1, there exist stationary probabilities Pn.

Equation can be re-written as Pn+1 = {(+)Pn – Pn-1} /

Putting n=1, P2 = (+)P1 /

Putting = /, P0 = P0, P2 = 2 P0

Likewise P3 = 3P0 and it can be shown that Pn =

n P0

∞

Now, Pn = 1

n=0

P0 + P1 + P2 + P3 +…+ Pn = 1

P0 + P0 + 2P0 +

3P0 + … +

n P0 = 1.

P0 [1 + + 2

+3

+ … +n] = 1.

P0 = 1 / [1 + + 2

+3

+ … +n] = (1-).

And Pn = (1-)n,

for n = 0,1,2,…etc.

P1 / P0 = P2 / P1 = P3 / P2 =…= Pn+1 / Pn =

Hence, Pn is in geometric progression for n = 0,1,2, …

OPERATIONS RESEARCH


The mean number of customer in the system, ň, is the summation of products of

probabilities with corresponding values of n, n = 0, 1, 2, 3, …

n=∞

ň = n Pn = P1 + 2P2 + 3P3 +… = P0 + 22P0 +…

n=0

ň = P0 (1 + 2 +32 + ...) = P0 / (1-)

2

Putting Po = (1-);

ň = / (1-) = / (-)

P0 is the probability of having no customer in the system. Two events that the server is busy

and that he is not busy are collectively exhaustive events.

P0 [server is not busy] + P [server is busy] = 1.

P0 [server is busy] = 1- P0.

Thus, the traffic intensity is also the probability of server being busy.

The queue length of mean number of customer in the waiting line nq will be given by, the

summation of the product, pos probabilities and (n - 1) for n = 1 to infinite as one customer

is being served.

n=∞ n=∞ n=∞

nq = (n-1)Pn = nPn - Pn

n=1 n=1 n=1

n=∞ n=∞

nq = nPn - Pn - P0

n=1 n=1

nq = ň - [1 - P0]

But, ň = / (1 - ) and P0 = 1 -

nq = { / (1-)} - = 2

nq = / (-)

For a customer joining a queue two matters are important. Firstly, how long he will have to

be in the system? Secondly, how long he will to wait till server becomes available to serve

him?

In steady state queue, a customer who joins at average n customer in the system and after

moving along the queue eventually gets served leaves behind the same average n customer.

This means that his duration stay in the system Ts is such that provides for n arrivals at the

mean rate of . Therefore ň = Ts and it is called Little”s formula.

OPERATIONS RESEARCH


Here, Ts = Expected number of customer in the system / Mean rate of arrival

= ň / = / (1-) = 1 / (-)

Similarly,

Tq = Expected waiting time in the system - mean service time

= Ts – (1/) = {1/(-)} – {1/} = / (-).

OPERATIONS RESEARCH


EXPERIMENT.NO: 10

REPLACEMENT THEORY

INTRODUCTION

Any engineering equipment has to give production service. While doing that, it

essentially needs operation cost. Its operational economy is rated through amount of

production or service per unit cost. With usage and years, equipment undergoes wear

and tear, needs progressively more maintenance, goes down in production or service

and requires frequent repairs. This involves additional cost, which in general increases

with years. Therefore, at some stage one intuitively feels, the existing equipment has to

be replaced by new similar or technologically better equipment. This is “Replacement

Problem” and it seeks to answer „when to replace‟ and „with what?‟ The answers to

these questions can be had with “Replacement Theory”.

COST OF “KEEPING IT ON” AND “REPLACING”

If we do not replace the existing equipment, we incur „operation, maintenance and

repairs cost‟ which increases every year.

Let Cm(n) = annual cost „operation maintenance and repairs‟ in nth year

C = capital investment on the equipment which once incurred is charged as

88permanent cost

If existing equipment is replaced it is discarded either as scrap of as still productively

usable in some application fetching some price on desired either as scrap or as still

productively usable in some application fetching some price on disposal. This is called

salvage value.

The salvage value of the equipment naturally decreases with year the decrease being

very slow in initial years and then very steep, as it tends to be a scrap.

Let Sn = salvage value of the equipment at the end of nth year

Let us presume for simplicity that money value does not change with year. That is

interest and inflation is insignificant.

The cumulative total cost of „keeping it on‟ upto nth years

= Capital investment value at the end on nth year + cumulative cost of

88operation, maintenance and repairs for nth year

n

TCK(n) = C + Cm 0

OPERATIONS RESEARCH


If replacement is considered at the end of the nth year,

CCR(n) = cumulative cost if replacement at the end of nth year = TCK(n) - Sn

Average annual cost if replaced at the end of nth years

CCR(n) = [(C – Sn) + Cm] / n

Obviously when average annual cost begins to rise, the replacement is due. This is the

year of replacement n is given by when,

CCR(n – 1) > CCR(n) and

CCR(n + 1) > CCR(n)

Implicit in the above is an assumption that level of performance of the equipment as

been maintained the same over the years.

REPLACEMENT BY ALTERNATIVE EQUIPMENT

In the previous article, we decided the due age of replacement on the basis of average

annual related cost of keeping the equipment. It may happen that while existing

equipment is being used, similar but improved version of equipment with different

investment cost, operation cost pattern and resale value pattern is available in the

market. In such a situation, decision is to be made whether the existing equipment needs

to be replaced by new version and if so when such a replacement should be made.

Let A be the existing equipment. Its own minimum average annual cost of keeping is

given by,

n

(CCR)A = [(Ca-San) + Cma] / n

0

where, Ca = capital investment of A

San = resale value of A at the end of nth year

Cma = operation maintenance cost of A from year to yea

n = number of years after which the replacement of A is due.

Let B be the new equipment. Its own minimum average annual cost of keeping is given

by,

(CCR)B = [(Cb-Sbn1) + Cmb] / n1

where, Cb = capital investment of B

Sbn1 = resale value of B at the end of nth year

Cmb = operation maintenance cost of B from year to yea

n1 = number of years after which the replacement of B is due.

Obviously, if (CCR)B < (CCR)A replacement of A by B is apparently advisable. Next

question is when this replacement should be made? It is often suggested that in the year

K when,

OPERATIONS RESEARCH


[Ca – Sa(k+1) + SCma] [Ca + Sa(k) + SCma] > (CCR)B

when incremental cost of A next year over previous year is more than minimum of

annual average of B the replacement should be made. This is not correct because when

equipment B replaces equipment A there is (C – S) of A carried over as also cumulative

cost of maintenance of A till that year thus in composite arrangement of A replacing B

other cumulative costs of A till that year and that of B thereafter should be worked, and

on that basis the average annular cost be worked out, reckoning from the year of

installation of A.

CAPACITY CORRECTION IN ALTERNATIVE EQUIPMENT

When exiting equipment is replaced by new equipment the latter may not have in

general the same capacity specification. For example, existing lathe may have the

capacity for producing 10 units per hour, whereas new equipment may have capacity

for producing 12 units per hour. In such a situation comparison of costs has to be on the

basis of the same performance and the correction factor should be applied by an

appropriate multiplier.

MONEY VALUE CHANGING WITH TIME REPLACEMENT

The decisions regarding replacement of equipment are taking into account the money

values of related costs new purchase, salvage value, operation cost, maintenance cost,

repair cost from the year. So we have to work out the discounted value of money the

present worth of money against the future cost figures, and find out the numbers of

years, n at the end of which the equipment should be replaced.

If i is the net incremental rate per annum, after one year it becomes (1+i), after two

years (1+i)2, and after n years (1+i)

n. thus, the present valve of 1 Rupee of nth year is 1 /

(1+i)n.

If we define r = 1 / (1+i) as discount range, present worth of 1 Rupee of nth year is r

n. If

C is the capital investment of today, m1, m2,… are the maintained repair cost in 1, 2,

3,… year respectively.

Equipment is used to provide service or give production. It is presumed that repairs and

maintenance cost are carried out from year to year to retain the same level of service or

production.

SALVAGE VALUE CONSIDERED

Previously change of worth of money was considered without salvage value of the

equipment to be replaced. If Si is the salvage value of the equipment, its discounted worth at

present is Si / (1 + k)i = Sir

i as the salvage value is reckoned at the end of the concerned

year.

OPERATIONS RESEARCH


Thus equation is modified as

Pn = C + (m1 + rm2 + r2m3 + … + r

n-1mn) - Snr

n

Its weighted annual average can be shown as

n n

= (C + miri-1 – Snrn) / ri

i=1 i=1

The replacement is due for that value of n for which is minimum.

GROUP OF REPLACEMENT POLICY:-

It may appear advantageous to replace and item as and when it fails, particularly that item

which does not deteriorate in its performance with lapse of time. However, the inherent risk

in this policy of replacement is that replacement, when needed may take some time may be

for producing item, putting a man to replace the item and time for replacing the item. This

will entail down time or non-working period of the system. If it is a production system, this

means loss of production. This implies cost. Hence, it is worth considering replacing an

item before it has failed. The concept is similar to „preventive maintenance‟ or „inventory

system‟ in which cost of shortage is very high.

Let us consider a system having N number of identical items. This constitutes a group. If all

items of group are replaced simultaneously at one time, the cost per item is Cg. Having

installed all N items as a group, the policy is to replace individual items as and when they

are in group. The cost of individual replacement on failure is Ci. Obviously Ci > Cg. The

decision variable is period „t‟. The information needed is failure data over a period of time.

This provides probability of failure. Let us consider following situation.

A large hospital complex has several operation theatres. Each operation table has special

light bulge attachments. The bulge is prone to failure. There are 200 bulbs installed in all.

Considering 500 hours as „period‟ the failure of similar bulb has been as under,

Out of 100 bulbs

9 failed by the end of first period,

20 failed by the end of second period,

33 failed by the end of third period,

51 failed by the end of fourth period,

77 failed by the end of fifth period,

90 failed by the end of sixth period, and

100 failed by the end of seventh period.

OPERATIONS RESEARCH


The management considers to make it a practice to replace all in a group at one time, then

replace individual bulb as and when it fails, and after „fixed interval‟ of time. It is given that

each bulb, when replaced in a group, costs Rs. 5, but when replaced individually on failure

costs Rs. 20.

Obviously, the criterion for optimization if the cost of the period. The period at the end of

which minimum „average cost per period‟ is obtained, that will be „the fixed interval‟ or

time for group management.

Age-wise failure Probability is as under:

0-1 period 0.09

1-2 period 0.20 – 0.09 = 0.11

2-3 period 0.33 – 0.20 = 0.13

3-4 period 0.51 – 0.33 = 0.18

4-5 period 0.77 – 0.51 = 0.26

5-6 period 0.90 – 0.77 = 0.13

6-7 period 1.00 – 0.90 = 0.10

Let nk be the number of replacement at the end of kth period.

OPERATIONS RESEARCH


PROBLEM

Two functionally identical machine tools P and Q are available in the market. There is no

market for machines even one-year old and when eventually disposed off as scrap it fetches

insignificant amount.

Price of machine tool P is Rs. 12,000. Its operation, maintenance and repair cost is Rs. 400

in the first year, progressively increasing by Rs. 100 in next two years, then by Rs. 200 in

next two years and finally by Rs. 300, 400, 600, 800 in subsequent years.

Price of machine tool Q is Rs. 12,000. Its operation, maintenance and repair cost is Rs. 200,

350, 550, 750, 1000, 1300, 1800, 2400, 3000 from year to year.

(a) If worth of money remains constant, which is a better choice, P or Q? And in that case,

what should be the replacement policy?

(b) If worth of money increases by 10% every year, which is a better choice, P or Q? And

in that case, what should be the replacement policy?

(c) If Q is offered for purchase, through interest-subsidized scheme by State Finance

Corporation, to the extent of 5%, what should be the replacement policy? State the

assumptions you make.

SOLUTION

It is assumed that running costs – operation, maintenance and repairs including downtime

costs – are increased at the beginning of each concerned year.

Table – 1

Machine Tool – P

C = 12000, K = 0, r = 1 / (1+K) = 1

Year

i

mi ri-1

miri-1

miri-1 r

i-1 C+miri-1

ri-1

1 400 1 400 400 1 12400

2 500 1 500 900 2 6450

3 600 1 600 1500 3 4500

4 800 1 800 2300 4 3575

5 1000 1 1000 3300 5 3060

6 1300 1 1300 4600 6 2767

7 1700 1 1700 6300 7 2615

8 2300 1 2300 8600 8 2575

9 3100 1 3100 11700 9 2634

OPERATIONS RESEARCH


(a) Table – 1 gives the weighted average of annual cost

i

C + miri-1

= i=1 i

ri-1

i=1

It can be seen with m8 = 2300, m9 = 3100, 8 = 2575

i.e., m8 < 8 < m9

Therefore, the minimum present worth of annual value for P is Rs. 2575

Corresponding value for Machine Tool Q, as given in Table – 2 is Rs. 2669.

Thus, machine tool P is a better choice and with this, the replacement policy is to affect the

replacement at the end of 8th

year.

Table – 2

Machine Tool – Q

C = 13000, K = 0, r= 1 / (1+K) = 1

Year

i

mi ri-1

miri-1

miri-1 r

i-1 C+miri-1

ri-1

1 200 1 200 200 1 13200

2 350 1 350 550 2 6775

3 550 1 550 1100 3 4700

4 750 1 750 1850 4 3713

5 1000 1 1000 2850 5 3170

6 1300 1 1300 4150 6 2859

7 1800 1 1800 5950 7 2708

8 2400 1 2400 8350 8 2669

9 3000 1 3000 11350 9 2706

OPERATIONS RESEARCH


(b) When change in money worth is taken into account, r = 1 / (1.10). The weighted average

of annual cost, is tabulated in Table – 3 for machine tool P, and in Table – 4 for

machine tool Q.

Table – 3

Machine Tool – P

C = 12000, K = 0.10, r= 1 / 1.10

Year

i

mi ri-1

miri-1

miri-1 r

i-1 C+miri-1

ri-1

1 400 1.000 400 400 1.000 13200

2 500 0.909 455 855 1.909 7081

3 600 0.826 496 1351 2.735 5109

4 800 0.751 601 1925 3.486 4168

5 1000 0.683 683 2635 4.169 3650

6 1300 0.621 807 3442 4.79 3568

7 1700 0.565 961 4403 5.355 3172

8 2300 0.513 1180 5583 5.868 3104

9 3100 0.466 1445 7028 6.334 3096

Table – 4

Machine Tool – Q

C = 13000, K = 0.10, r= 1 / (1.10)

Year

i

mi ri-1

miri-1

miri-1 r

i-1 C+miri-1

ri-1

1 200 1.000 200 200 1.000 13200

2 350 0.909 318 518 1.909 7081

3 550 0.826 454 972 2.735 5109

4 750 0.751 63 1535 3.486 4168

5 1000 0.683 683 2218 4.169 3650

6 1300 0.621 807 3025 4.790 3568

7 1800 0.565 961 3986 5.355 3172

8 2400 0.513 1231 5217 5.868 3104

9 3000 0.466 1398 6615 6.664 3096

OPERATIONS RESEARCH


For machine too P, least value of is Rs. 2996, whereas for machine tool Q this value

does not appear in the range of data given. However, the least value for Q as available is

Rs. 3096. Therefore, P is better choice with replacement at the end of 8th year.

(c) When interest for machine tool Q is subsidized, r = 1 / (1 + 0.05) and the value of are

tabulated in Table – 5. Comparing the least value of Rs. 2873 with the least value of D

for P in Table – 3 i.e. Rs. 296, it can be concluded that the machine tool Q is better

choice, with replacement at the end of 8th year.

Table – 5

Machine Tool – Q

C = 13000, K = 0.05, r= 1 / (1.05)

Year

i

mi ri-1

miri-1

miri-1 r

i-1 C+miri-1

ri-1

1 200 1.000 200 200 1.000 13200

2 350 0.952 333 533 1.925 6933

3 550 0.907 499 1032 2.859 4911

4 750 0.864 648 1680 3.723 3945

5 1000 0.823 823 2503 4.546 3411

6 1300 0.784 1019 3522 5.330 3101

7 1800 0.746 1268 4790 6.076 2929

8 2400 0.711 1706 6496 6.787 2873

9 3000 0.677 2031 8527 7.464 2884

OPERATIONS RESEARCH


EXPERIMENT.NO: 11

STUDY ABOUT SEQUENCING PROBLEM.

HEUSTERIC PROBLEM SOLVING. Problem solving methods discuss earlier posses well defined models. Analytical

techniques and algorithms to obtain optimum solution. As was stated in model formulation

phases in chapter of real life problems. In general defi. Clear cut crystallization and

trimming problem can be brought within the relation of models. The techniques to the tool

kit of O.R. will take care of the solution. However in many cases this may not be possible.

In such cases heustric approach in problem solving comes hardly.

It is now obvious that the ill-structured problems are those problems which are not

amenable to solution by analytical or algorithms solution techniques. The concept of ill-

structured problems is understood in contrast to well structured problem ( WSP), the

character being characterized by :

1. formulable as an acceptable model.

2. capable of attaining feasible solution.

3. defined criteria for feasibility and optimum solution.

4. availability of techniques for solving models.

SEQUENCING PROBLEM :- The problem of sequencing falls within the realm of heuristic methods. Sequencing

problems encompasses those scheduling problems in which objective function or

sequence for processing set of jobs or activity to be undertaken on given facility in a

defined technological order. Sequence is a very important function in production

planning and control. This chapter will concentrate machine to perform operation on the

job which have unique design futures and have resulted from a specific customer order.

In order to describe a specific case of job shop scheduling problem we want to study for

4 factors which are

1. the job arrival . if „n‟ jobs arrive at a time in a shop where „m‟ facilities on

which they are to be processed are idle and available . the problem is called

as static problem. The problem will be known as dynamic problem if jobs

arrive at random with some stock tic process.

2. number of machines „m‟ that are in the job shops.

3. the flow process for „n‟ jobs through „m, machines.

4. the measure of performance to be optimized which plays a decisive roll in

job shop schedule problem.

Gantt chart is a most simple and widely expected representation of scheduling are

shown along vertical axes and operations time along horizontal axis. Each operation

on various resources (facility over machines) is represented by a horizontal bar.

Since the chart breaks down resources allocation by time. It gives a display of max

span. Job waiting chart breaks down resources allocation by time it gives a display

OPERATIONS RESEARCH


of observation of this chart. And improvement to the schedule depend on analyst

critical observation.

The general problem of sequencing is „n‟ jobs are to be performed on 1,2,or

several or all machine. Each job having its own technological order on machine. The

sequence be so determine as to minimize total elapsed time for „n‟ jobs to be

completed.

Rules of heuristic methods in sequencing problems have so far been

establish for a few specific cases viz.

1. n jobs on 1 machine.

2. n jobs on 2 machine x & y in the same order.

3. n jobs on 3 machine x,y & z and subject to special restriction of data.

4. 2 jobs on m machines each jobs may have its technological order and

5. n jobs on m machines having in the same technological order.

„N‟ JOBS AND „1‟ MACHINE :-

The simplest case of scheduling is to decide sequence of n jobs on a single resource or

machine. In a basic single machine problem n jobs are available for processing at time

„0‟ (static problem). Their facility setup times are independent on their sequence and

can be added to their individual processing time for the power of scheduling . this

aggregate of setup times and processing times for each job is constant. And known with

certainty . the facility or machine is continuously available without interruption for

processing the jobs and will not be idle. So long a job is waiting . once processing of

jobs started will be completed without any break . the total number of distinct schedules

will be n! . since all the schedules will results in the same next pan different

performance are used to evaluate the schedules.

„N‟ JOBS AND „2‟ MACHINES:- For further detail description let us consider the illustrative problem –1 given in tutorial.

„N‟ JOBS AND „3‟ MACHINES :- let us consider n jobs on 3 machines . F1,F2 & F3 is same technological order. There is

no general optimum or satisfying solution obtain so far.

If a problem satisfied for n jobs and 2 machines as desired in art. The condition is

minimum time element on F1max. time element on F2 . and minimum time element on

F3 max. time element on F2.

In that case , 3 columns (. F1,F2 & F3) time elements are redesigneted for 2 machines

A & B elements as ,

Ai = ( F1) I + ( F2)I and Bi = ( F2) I + ( F3)I

„2‟ JOBS AND „M‟ MACHINES :- jobs J1 & J2 are to be made on machine F1, F2, F3, ….. Fm not necessarily in the same

technological order. Graphical methods provides a very, simple for obtaining solution.

OPERATIONS RESEARCH


ASSIGNMENTS

LP FORMULATION – ASSIGNMENT

1. A small manufacturer employs five skilled men and ten semi skilled men for making a

product in two qualities: a deluxe model and an ordinary model. The production of a

deluxe model requires 2-hours work by a skilled man and a 2-hour work by a semi

skilled man. The ordinary model requires 1-hour work by a skilled man and a 3-hours

work by a semi skilled man. According to worker union‟s rules, no man can work more

than 8-hours per day. The profit of the deluxe model is Rs.1000 per unit and that of the

ordinary model is Rs.800 per unit. Formulate a linear programming model for this

manufacturing situation to determine the production volume of each model such that the

total profit is maximized.

2. A firm manufactures three products A, B & C. Their profits per unit are Rs.300, Rs.200

and Rs.400, respectively. The firm has two machines and the required processing time

in minutes on each machine for each product is given in the following table:

Product

A B C

Machine 1 4 3 5

Machine 2 2 2 4

Machines 1 and 2 have 2000 and 2500 machine minute, respectively. The upper

limits for the production volumes of the product A, B & C are 100 units, 200 units and

50 units, respectively. But, the firm must produce a minimum of 50 units of the product

A. Develop a LP model for this manufacturing situation to determine the production

volume of each product such that the total profit is maximized. 3. The manager of an oil refinery has to decide on the optimal mix of two possible blending processes. The inputs and the outputs per

production run of the blending process are as follows:

Input Output

Process Crude A Crude B Gasoline G1 Gasoline G2

1 5 3 5 8

2 4 5 4 4

The maximum amounts of availability of crude A and crude B are 200 units and 150

units, respectively. Market requirements show that at least 100 units of Gasoline G1

and 80 units of Gasoline G2 must be produced. The profits per production run from

process 1 and process 2 are Rs.3 lacs and Rs. 4 lacs respectively. Formulate this

problem as a LP model to determine the number of production runs of each process

such that the total profit is maximized.

OPERATIONS RESEARCH


SOLVE EXAMPLES USING SIMPLEX METHOD

1. Minimize Z=2x+9y-4z

Subject to 2x+3y+4z 16

x+6y-4z16

x,y, z >0 [S1=8, y=8/3,Z=24]

2. Maximize z=4x+5y

Subject to 2x+3y8

3x+y4

x,y0

Use big M method. [S2=8, x=4, z=16]

3. Maximize Z=6x+10y+8z

Subject to 2x+3y8

2y+5z10

3x+2y+4z15

x,y,z0 [y=750/41, z=930/41, x=89/41, Z=212.78]

4. Maximize Z=40x+30y

Subject to 3x+2y300

x+y80

2x+y200

3x+4y300

x60

y60

x,y0 [S1=80, y=20, S3=60, S4=40, x=60, S6=40, Z=3000]

5. Minimize Z=20x1+10x2

Subject to x1+2x240

3x1+x2=30

4x1+3x260

x1,x 20

Use (i) Big „M‟ Method.

(ii) Two – Phase Method [x1=6, x2=12, S1=10, Z=240]

6. Solve the following problem:

Maximize Z= -x1-4x2-3x3

Subject to 2x1+x2+3x34

x1+2x2+2x33SS

x1, x2 0 and x3 is unrestricted in sign. [Problem is infeasible]

OPERATIONS RESEARCH


FORMULATE PROBLEM AND SOLVE USING SIMPLEX METHOD

1. Shri Rajendra Desai has booked orders for product A and product B at the sales

price of Rs. 6 and Rs. 8 per unit respectively, for M/S supreme Engineers with the

condition that materials cost will be borne by the company i.e. M/S Supreme

Engineers. Shri Desai negotiates the terms with the Kartik Engineering, a job order

shop, which can offer machines of type x for 60 hours per week and machines of

type y for 100 hours per week at hourly rates. Each machine can produce product A

and B in any combination.

Machine-hours of type x are available with the operators of the Kartik Engineering only

and the rates are Rs. 4 per machine-cum-operator-hour. Machines of type y are

available with two alternative offers viz. Desai can either hire the machines alone at the

rate of Rs. 1/machine hour as well as operator at the rate of Rs. 3 per hour or Desai can

hire his own operator for machines y at the rate of Rs. 3 per hour. The production rates

are given in the Exhibit A.

Production rate, units per hour Exhibit – A

Product Machines, Type x Machines, Type y

With operator of

Kartik Engg.

With Desai‟s

operator

A 10 6 8

B 20 12 18

Formulate L.P.P. to maximize profit of Shri Desai.

[ Maximize

Z = (6- 4/10) X1 + (8-4/20) X2 + (6-4/6) X3 + (8-4/12)X4 +(6-4/8)X5 + (8-4/18)X6

Subject to X1 /10 + X2 /20 60

X3 /6 + X4 /12 + X5 /8 + X6 /18 10

X1, X2, X3, X4, X5, X6, 0]

2. The State Industries Finance Corporation is to decide its investment of Rupees 20

lacs on two proposals of Private sector Industries X and Y> X undertakes to

guarantee annual return of 10 per cent, and Y of 15 per cent. As X is in the line of

basic consumer products and that too co-operative venture, the Government has laid

down that at least Rs.5 lacs should be invested in X. The corporation would like to

have investments so made as to procure the minimum of 12 per cent annual return.

For the reason best known to one of the Directors, following his rigid attitude in this

regard SIFC has decided not to have investment in Y and X more than in the ratio

1.6: 1.

Formulate this L.P. problem for maximum annual return. Solve it be either (1)

Graphical method or (2) Simplex method.

[maximize

Z = 0.1 X1 + 0.15 X2

Subject to X1 + X2 = 20

0.1 X1 + 0.15 X2 0.12 (X1 + X2 )

X1 5

OPERATIONS RESEARCH


X2 / X1 1.6 / 1

X1, X2 0

X1 = 765/91, X2 = 1055/91, S2 = 107/260, Z = 2.5797]

3. The following data is available for two products.

Product Production Alternatives Production Rate in units/hr.

Centre 1 Centre 2

1 1

2

0.4

-----------

---------

0.5

2 1 0.5 1.0

Production hrs. available/week 500 400

Production cost Rs./hr. 6 8

Material cost is same for both the products and is Rs. 5 per unit. Selling price Rs.

Per unit for product one is 25 and that of product two is 31. Solve the problem for

optimal product mix.

[X1 = 1250, X2 = 800, X3 =0, Z = 9450]

4. A plant can produce three products, A, B and C, Product A needs 4 hours of

department 2 and one hour of department 3. Product B needs two hours of

department 1 and two hours of department 2, while product C needs two hours of

department 1 and two hours of department 3. Their respective profit coefficients in

Rs. Per unit are 50, 40, and 55. Hours available per month with department 1, 2, and

3 are 1000, 1000, and 800 respectively.

A purchased part is used in assembly of product A and C. Only 4000 parts are

available for the month. Two such parts are used in each piece of product A and

three parts in each piece of product C. There is a sales commitment of 200 units of

product A. Find the optimal product mix.

DUALITY & SENSITIVITY ANALYSIS

Problem 1:

Construct the dual of the problem:

Maximize Z = 3x+5y

Subject to X- 2y 3

X + 3y 9

X – y 5

X 0, is unrestricted in sign

Solve the dual.

[DUAL IS MAXIMIZE Z = -3A-9B+5C

SUBJECT TO –A-B+C 3

-2A+3B+C -5

A,B,C 0.

C=11,A=8,S3=0,Z=31]

OPERATIONS RESEARCH


Problem: 2

Use dual simplex method to solve the following

Minimize Z = 2x + 3y

Subject to 3x + 4y 5

4x + 5y 7

x + 2y 4

x , y 0

[S1 = ¼, X= 7/4, S3 =9/4, Z = 7/2]

Problem: 3

The following tableau gives an optional solution to a standard linear programme:

Maximize Z = CX.

Subject to AX = B, X 0

Tableau 5 7 -4 0 0

X1 X2 X3 S1 S2

7 x2 x3 ¾ 1 -1/4 ¼ 0

0 s2 ½ ¼ 0 -3/4 -5/4 1

(zj – cj) 7/2 ¼ 0 9/4 7/4 0

S1 and S2 are the slack variables.

(a) How much can c1 be increased before the current basis is no longer optimal?

(b) How much can c2 be varied sp that the given basis ( x2 , S2) is still optimal?

(c) What are the shadow prices of two resources?

[(a) c1=21/4, x1=2/3, S2=1/3,Z=4 (b) 20/3 c216 (c) 7/4 and 0]

OPERATIONS RESEARCH


QUEING THEORY

1. Customer arrivals at a teller counter in a Commercial Bank are considered to be

following Poisson probability distribution with an average time of 10 minutes

between one arrival and the next. The time length of service that is rendered is

assumed to be distributed exponentially with mean three minutes

i) What is the probability that a person arriving at the teller will have to

wait?

ii) What is the average length of queue that forms from time to time?

iii) The Commercial Bank will install a second teller man when convince

that the arrival rate increase in order to justify the second booth. When

does this happen?

2. There is congestion on the platform of Ahmed Railway station. The trains arrive at

the rate of 30 trains per day. The waiting time for any train to flag-off is

exponentially distributed with an average of 36 minutes. Calculate the following:

i) The mean queue size.

ii) The probability that the queue size exceeds 10.

OPERATIONS RESEARCH


INVENTORY CONTROL ASSIGNMENT

1. The Bombay Shoe Company has found that it purchases a large amount of industrial

tapes for production of its shoes. Currently, it purchases Rs.10,00,000 a year of the

various sized tapes from the local manufactures. A proposal was made by its

supplier, was offer consist of 1.25% discount if BSC places an order quarterly BSC

has calculated the cost to purchase at Rs.22.5 per order. Rs.20 per tape & inventory

carrying costs at 18%. Should BSC accept discount offer from its supplier? If not,

what alternate offer should be made in term of a discount?

2. The Himavan manufacturing company wishes to determine the most economy

quantity for one of its products. Manufacturing cost amount to Rs.15 per unit. The

production is 5000 units per annum. Each lot now requires a set-up cost of Rs.25

and the inventory cost of 25% of the average inventory value. What is the most

economic lot size to manufacturers? What is the corresponding total yearly cost?

3. Demand for a certain part order by Shah brothers tends to be constant at the monthly

rate of 1000 units. The per unit carrying cost of this item is Rs.25 per year, & cost of

placing an order is Rs.75.

i) What is the optimal order size? How often should an order be placed?

ii) Show that the annual holding cost & ordering cost are equal when an

optimal order size is used. What is the total relevant cost?

iii) If the company wants to order only one every other week, by what

percentage would this increase the total relevant cost?

4. The Calcutta Tool Company can manufacture a certain type of tool at the rate of

1480 per run, the demand for this tool is quite steady at annual rate of 9000 units.

Unit cost of the tool is Rs.30, an set up cost per production run is Rs.500. The

annual cost per unit sort is Rs.20. The company has determined the optimal

production lot size to be 3000 units.

i) What is the annual inventory holding cost, expressed as percentage of

unit cost?

ii) What is the total relevant cost & the max. inventory level?

5. Let C1=50(Rs.), Cs=20C1, & the demand distribution is as under:

r 0 1 2 3 4 5 6

P(r) 0.9 0.05 0.02 0.01 0.01 0.01 0

What is the optimum quantity to be ordered?

5. Assume that demand during a certain time interval T is random with P( r ) being the

probability that the total demand is r during interval T. The demand rate is constant

during the interval T.

C1=10(Rs.), C2=20C1 and the demand distribution is as under:

r 0 1 2 3 4 5 6

P(r) 0.1 0.2 0.2 0.3 0.1 0.1 0

What should be the initial inventory level?

OPERATIONS RESEARCH


PROBLEMS ON GAMES THEORY

1. Linear programming method:

(i) (ii)

B B

1 2 3 1 2 3

1 1 7 2 A 1 5 7 9

A 2 6 2 7 2 4 8 11

3 6 1 6

(iii) (iv)

B B

1 2 3 1 2 3

A 1 -1 6 4 1 4 7 9

2 5 -3 -6 A 2 11 6 10

6 3 5 6 9

(v) (vi)

B B

1 2 3 1 2 3 4

1 -5 10 20 1 3 -1 1 2

A 2 5 -10 -10 A 2 -2 3 2 3

3 5 -20 -20 3 2 -2 -1 1

(vii) (viii)

B B

1 2 3 4 1 2 3 4

1 4 10 11 14 1 -1 -3 5 4

2 15 5 13 18 A 2 -3 4 -3 -2

A 3 4 9 6 10 3 -3 -2 4 3

4 17 12 10 7

5 14 18 17 8

6 11 14 15 9

OPERATIONS RESEARCH


TRANSPORTATION Problem: Find the initial basic feasible solution of the following transportation problem by

Vogel‟s approximation method:

Warehouses

W1 W2 W3 W4 Capacity

F1 10 30 50 10 7

F2 70 30 40 60 9

F3 40 8 70 20 18

Requirement 5 8 7 14 34

Problem: A company has received a contract to supply gravel for three new construction

projects located in towns A, B and C. Construction engineers have estimated the required

amounts of gravel which will be needed at these construction projects as shown below:

Project location Weekly requirement

(Truck loads)

A 72

B 102

C 41

The company has three gravel plants X, Y and Z located in three different towns.

The gravel required by the construction projects can be supplied by these three plants. The

amount of gravel which can be supplied by each plant is as follows:

Plant Amount available/week

(Truck loads)

X 76

Y 62

Z 77

The company has computed the delivery cost from each plant to each project site.

These costs (in rupees) are shown in the following table:

Cost per load

A B C

X 4 8 8

Plant Y 16 24 16

Z 8 16 35

OPERATIONS RESEARCH


(a) Schedule the shipment from each plant to each project in such a manner so as to

minimize the total transportation cost within the constraints imposed by plant

capacities and project requirements.

(b) Find the minimum cost.

(c) Is the solution unique? If it is not, find alternative schedule with the same minimum

cost.

ASSIGNMENT Problem: Solve the following assignment problem using Hungerian method. The matrix

entries are processing times in hours.

Operator

1 2 3 4 5

1 20 22 35 22 18

2 4 26 24 24 7

Job 3 23 14 17 19 19

4 17 15 16 18 15

5 16 19 21 19 25

Problem: Consider the problem of assigning four sales persons to four different sales

regions as shown below such that the total sales is maximized.

Sales region

1 2 3 4

1 5 11 8 9

2 5 7 9 7

Salesman 3 7 8 9 9

4 6 8 11 12

The cell entries represent annual sales figures in crores of rupees. Find the optimal

allocation of the sales persons to different regions.

Problem: The flight timings between two cities, X and Y are as given in the following two

tables. The minimum layover time of any crew in either of the cities is 3 hours. Determine

the base city for each crew so that the sum of the layover times of all the crews in non-base

cities is minimized.

Timing of Flights form City X to City Y

Flight number Departure time Arrival time

(from City X) (to City Y)

OPERATIONS RESEARCH


101 6 a.m. 8.00 a.m.

102 10 a.m. 12.00 noon

103 3 p.m. 5.00 p.m.

104 8 p.m. 10.00 p.m.

Timing of Flights from City Y to City X

Flight number Departure time Arrival time

(from City Y) (to City X)

201 5.30 a.m. 7.00 a.m.

202 9.00 a.m. 10.30 a.m.

203 4.00 p.m. 5.30 p.m.

204 10.00 p.m. 11.30 p.m.

OPERATIONS RESEARCH


NETWORK PROBLEMS:

Problem: Consider the details of a distance network as shown below:

Arc Distance Arc Distance

1-2 8 3-6 6

1-3 5 4-5 8

1-4 7 4-6 12

1-5 16 5-8 7

2-3 15 6-8 9

2-6 3 6-9 15

2-7 4 7-9 12

3-4 5 8-9 6

(a) Construct the distance network.

(b) Find the shortest path from Node 1 to Node 9 using the systematic method.

(c) Find the shortest path from Node 1 to Node 9 using Dijkstra‟s algorithm.

Problem: Consider the details of a distance network as shown below:

Arc Distance

1-2 3

1-3 8

1-4 10

2-3 4

2-4 7

3-4 2

3-5 8

3-6 6

(a) Construct the detail network.

(b) Apply Floyd‟s algorithm and obtain the final matrices, D5 and P

5.

(c) Find the shortest path and the corresponding distance for each of the following:

(i) from Node 1 to Node 5

(ii) from Node 2 to Node 5.

OPERATIONS RESEARCH


Problem: For the network shown in the Figure activity, resource requirements are given in

the table. Solve the problem for resource allocation. There are two cranes and 8 welders

available for the project.

Figure

Table

Activity A B C D E F

Duration

Days 10 10 8 14 3 9

Resources

Required -

2

Cranes

8

Welders

6

Welders -

1

Crane

1 2 4 6

3

5

A D E

A

C B

F

OPERATIONS RESEARCH


Problem: The critical path network analysis is given in Figure. The activity times are also

given for each activity. Determine the critical path and critical path time. Also determine

the floats for each activity.

Figure

1 2

4

3

6

8

5 7

2

A

12

12

8

9

3

4 5

7

10

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