complete book

Introduction of Quantitative Technique

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INTRODUCTION OF QUANTITATIVE TECHNIQUE

INTRODUCTION

People have been using mathematical tools to help solve problems for thousands of year;

however, the formal study and application of quantitative techniques to practical decision

making is largely a product of the 20th century. The technique we study in quantitative

analysis have been applied successfully to an increasingly wide variety of complex

problem in business, government, healthcare, education, and many other areas.

To get aware of the mathematics of how a particular quantitative technique works; one

must also be familiar with the limitations, assumption and specific applicability of the

technique. The successful use of quantitative techniques usually results in a solution

that is timely, accurate, flexible, economical, reliable, an easy to understand and use.

WHAT IS QUANTITATIVE TECHNIQUE?

Quantitative technique is the scientific approach to managerial decision making. The

approach starts with data. Like raw material for a factory, these data are manipulated

or processed into information that is valuable to people making decision. This

processing and manipulating of raw data into meaningful information is the heart of

quantitative technique. e.g., we can use quantitative technique to determine how

much our investment will be worth in the future when deposited at a bank at a given

interest rate for a certain number of years. Quantitative technique can also be used in

computing financial ratio for the balance sheets for several companies whose stock we

are considering.

DEFINITION

Different mathematician have given various definition of operation research.

―Operation research is a scientific method of producing executive departments with a

quantitative basis for decision regarding the operation under their control‖

-P.M. MORSE & G.E.KIMBALL

―Operation research is a act of giving bad answers to problems which otherwise have

worse answers‖

-T.S. SAATY

―Operation research is a scientific research an approach to problem solving for

executive management‖

-H.M.WAGNER


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―Operation research has been described as a method, an approach, set of techniques a

team activity, a combination of many disciplines, an extension of particular

disciplines (mathematics engineering economics), a new discipline, vocation, even a

religion. It is perhaps some of all this things.‖

-S.L.COOK

THE QUANTITATIVE TECHNIQUE APPROACH

CHARACTERISTICS OF QUANTITATIVE TECHNIQUE

1. Flexibility:-Models should be capable of adjustment with new formulas

without having any significant changes in its frame.

2. Less no of variable: - The number of variable in a model should not be very

large but at the time no variable selected to any important factors should be

left ones.

3. Less time: - A model should not take much time in construction.

ADVANTAGES OF QUANTITATIVE TECHNIQUES

1. It describes the problem much more concisely.

2. It provides systematic and logical approach to the problem.

Acquiring Input Data

Testing the Solution

Analyzing the Results

Implementing the Results

Defining the problem

Developing a Model

Developing a Solution


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3. It indicates the scope and limitation of the problem.

4. It enables the use of high power mathematical tools to analyze the problems.

5. It helps in finding avenues for new research and improvement in a system.

TECHNIQUES

A brief description of some of the commonly used techniques is given below. Details

are covered in relevant chapters of this book. Some of the techniques which is

covered in this book are:-

LINEAR PROGRAMMING: It is an allocation technique where the objective

function is linear and the constraints are modeled as linear inequalities. Example-

Graphical Representation, Simplex Method etc.

TRANSPORTATION MODEL: A special case of linear programming which

matches sources of supply to destinations on cost or distance considerations. For

example; movement of raw materials from different sources to manufacturing

plants at different locations based on availability of raw materials at various

sources, the requirements at different plants and the cost of transportation involved.

ASSIGNMENT MODEL: A special case of transportation model where the

aim is to assign a number of ‗origins‘ to the same number of destinations at a

minimum total cost. For example; assigning of man/machines to same number

of jobs /tasks.

REPLACEMENT MODEL: These models deal with the formulation of

appropriate replacement policies when some items have to be replaced due to

obsolescence or wear and tear. Replacement models also deal with equipments

and items which fail completely and instantaneously.

QUEUING THEORY: It studies the random arrivals at servicing or processing

facility of limited capacity. These models attempt to predict the behavior of

waiting lines, i.e.; the time spend waiting for a service. The technique is

descriptive and describes behavior that can be expected given certain

parameters. It is not prescriptive in nature and does not offer an optimal

solution. The models deal with the tradeoffs between cost of providing service

and the value of time spent in waiting for a service.

DECISION THEORY: Decision situations can be classified into deterministic

or certainty, probabilistic or risk and uncertainty. Decision making under certainty

can be dealt with by various optimization techniques. Decision theory deals

largely with decision making under risk where the probabilities of certain

conditions occurring (such as demand for an item) are predicted and various

options assessed based on these probabilistic values. In situations of uncertainty

there can be no specific approach. A set of decision rules can be applied and


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insight gained into the decision maker‘s style of functioning. This is particularly

applicable to studying a competitor‘s style of decision making and then predicting

how he would react to a certain condition so as to gain advantage for oneself.

GAME THEORY: This deals with decision making under conditions of

competition. Its assumptions currently restrict its usage.

APPLICATION & SCOPE OF QUANTITATIVE TECHNIQUES

Some of the industrial, government, business problems which can be analyzed by or

approach have been arranged by functional areas as follows: -----

Finance and accounting:

Dividend policies, investment and portfolio management, auditing, balance

sheet and cash flow analysis

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

planning

Claim and complaint procedure, and public accounting

Establishing costs for by-products and developing standards costs

Marketing:

Selection of product-mix, marketing and export planning

Best time to launch a new product

Sales effort allocation and assignment

Predicting customer loyalty

Advertising, media planning, selection and effective packing alternatives

Purchasing, procurement and exploration:

Optimal buying and recording under price quantity discount

Bidding policies

Transportation planning

Vendor analysis

Replacement policies

Production management:

Facilities planning

Location and size of warehouse or new plant, distribution centers and retail

outlets

Logistics, layout and engineering design

Transportation, planning and scheduling


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Manufacturing

Aggregate production planning, assembly line, blending, purchasing and

inventory control

Employment, training, layoffs and quality control

Allocating R&D budgets most effectively

Maintenance and project scheduling

Maintenance policies and preventive maintenance

Maintenance crew size and scheduling

Project scheduling and allocation of resources

Personal management:

Manpower planning, wage/salary administration

Negotiation in a bargaining situation

Designing organization structures more effectively

Skills and wages balancing

Scheduling of training programmes to maximize skill development and

retention

Techniques and general management:

Decision support systems and MIS; forecasting

Making quality control more effective

Project management and strategic planning

Government:

Economic planning, natural resources, social planning and energy

Urban and housing problems

Militar, police, pollution, control, etc.

Limitation of Quantitative Techniques

1. All the problems cannot be converted into numerical values. So it is not solving

by Quantitative technique.

2. Not understandable to everyone who are not aware of the technique of

Quantitative technique.

3. The some of the technique of the Quantitative technique are very complex in

solving

4. Mathematical model: as most of the operation research techniques are

mathematical in nature. It is therefore necessary to put the problem in the term


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of mathematical model. In many situations it is not possible to represent the

problem in mathematical form and hence operation research techniques cannot

be applied.

5. Expensive: As application of operation research for the problem solving

requires service of specialist and the use of computer. Therefore it is expensive

to use operation research technique for solving day to day problems.

6. Imperfection of solution: By operation research technique we cannot obtain the

perfect answer to our problems. But only the quality of the solution is improved

from worse to bad answer.

7. The techniques used must be applicable to the problem and must reflect the

purpose and scope of the problem.

Assignment Problems

7

ASSIGNMENT PROBLEMS

INTRODUCTION

In the world of trade Business Organizations are confronting the conflicting need for

optimal utilization of their limited resources among competing activities. The course

of action chosen will invariably lead to optimal or nearly optimal results.

The assignment problem is a special case of transportation problem in which the

objective is to assign a number of origins to the equal number of destinations at the

minimum cost (or maximum profit). Assignment problem is one of the special cases

of the transportation problem. It involves assignment of people to projects, jobs to

machines, workers to jobs and teachers to classes etc., while minimizing the total

assignment costs. One of the important characteristics of assignment problem is that

only one job (or worker) is assigned to one machine (or project). Hence the number of

sources are equal the number of destinations and each requirement and capacity value

is exactly one unit.

Although assignment problem can be solved using either the techniques of Linear

Programming or the transportation method, the assignment method is much faster and

efficient. This method was developed by D. Konig, a Hungarian mathematician and is

therefore known as the Hungarian method of assignment problem. In order to use this

method, one needs to know only the cost of making all the possible assignments. Each

assignment problem has a matrix (table) associated with it. Normally, the objects (or

people) one wishes to assign are expressed in rows, whereas the columns represent the

tasks (or things) assigned to them. The number in the table would then be the costs

associated with each particular assignment. It may be noted that the assignment

problem is a variation of transportation problem with two characteristics.(i)the cost

matrix is a square matrix, and (ii)the optimum solution for the problem would be such

that there would be only one assignment in a row or column of the cost matrix .

Application Areas of Assignment Problem

Though assignment problem finds applicability in various diverse business situations,

we discuss some of its main application areas:

(i) In assigning machines to factory orders.

(ii) In assigning sales/marketing people to sales territories.

(iii) In assigning contracts to bidders by systematic bid-evaluation.

(iv) In assigning teachers to classes.

(v) In assigning accountants to accounts of the clients.

Assignment Problems

8

Format of Assignment Problems

In assigning police vehicles to patrolling area.

Job

Persons j1 j2 ------ jn

I1 X11 X12 ----- X1n

I2 X21 X22 ----- X2n

--- ----- ---- ----- -----

In ----- ---- ----- Xnn

Cij is the cost of performing jth job by ith worker. Xij is the number ith individual

assigned to jth job.

Total cost = X11 * C11 + X12 * C12 + ----- + Xnn * Cnn.

Solution Methods

There are various ways to solve assignment problems. Certainly it can be formulated

as a linear program, and the simplex method can be used to solve it. In addition, since

it can be formulated as a network problem, the network simplex method may solve it

quickly.

However, sometimes the simplex method is inefficient for assignment problems

(particularly problems with a high degree of degeneracy). The Hungarian Method

used with a good deal of success on these problems and is summarized as follows.

Step 1. Determine the cost table from the given problem.

(i) If the no. of sources is equal to no. of destinations, go to step 3.

(ii) If the no. of sources is not equal to the no. of destination, go to step 2.

Step 2. Add a dummy source or dummy destination, so that the cost table becomes a

square matrix. The cost entries of the dummy source/destinations are always zero.

Step 3. Locate the smallest element in each row of the given cost matrix and then

subtract the same from each element of the row.

Step 4. In the reduced matrix obtained in the step 3, locate the smallest element of

each column and then subtract the same from each element of that column. Each

column and row now have at least one zero.

Step 5. In the modified matrix obtained in the step 4, search for the optimal

assignment as follows:

Assignment Problems

9

(a) Examine the rows successively until a row with a single zero is found. In

rectangle this row (�) and cross off (X) all other zeros in its column. Continue

in this manner until all the rows have been taken care of.

(b) Repeat the procedure for each column of the reduced matrix.

(c) If a row and/or column has two or more zeros and one cannot be chosen by

inspection then assign arbitrary any one of these zeros and cross off all other

zeros.

Step 6. If the number of assignment (�) is equal to n (the order of the cost matrix), an

optimum solution is reached.

If the number of assignment is less than n(the order of the matrix), go to the next step.

Step7. Draw the minimum number of horizontal and/or vertical lines to cover all the

zeros of the reduced matrix.

Step 8. Develop the new revised cost matrix as follows:

(a) Find the smallest element of the reduced matrix not covered by any of the lines.

(b) Subtract this element from all uncovered elements and add the same to all the

elements laying at the intersection of any two lines.

Step 9. Go to step 6 and repeat the procedure until an optimum solution is attained.

Assignment Problems

10

See diagrammatic Representation of Hungarian Approach

Assignment Problems

11

MINIMIZATION PROBLEM (BALANCED)

Example 1: A Company has 5 machines and 5 jobs. The relevant cost matrix is given

below:

Find the assignment that minimizes the total cost:

Machines Jobs

J1 J2 J3 J4 J5

M1 10 4 5 3 11

M2 13 11 9 12 10

M3 12 3 10 1 9

M4 9 1 11 4 8

M5 8 6 7 3 10 ,

Solutions:

Step 1: First we have to check that the given matrix is square matrix or not. Here the

given matrix is square matrix.

Step 2: Subtract least entry of each row from all the entries of that row. The first

reduced cost matrix will be as given below:

J1 J2 J3 J4 J5

M1 7 1 2 0 8

M2 4 2 0 3 1

M3 11 2 9 0 8

M4 8 0 10 3 7

M5 5 3 4 0 7

Then in the above matrix subtract least entry of each column from all entries of that

column. The second reduced cost matrix or the total opportunity cost matrix will be as

follows:

J1 J2 J3 J4 J5

M1 3 1 2 0 7

M2 0 2 0 3 0

M3 7 2 9 0 7

M4 4 0 10 3 6

M5 1 3 4 0 6

Assignment Problems

12

Step3: For testing the optimality we draw minimum number of straight lines to cover

all the zeros. Since three lines cover all the zeros, which is not equal to the matrix size

(number of row= number of column≠ number of draw line).

Step4: Select the smallest entry from all the entries which are not covered by a

straight line (here it is 1). Subtract this smallest entry from all the uncovered entries

and add it to all those entries which are at the intersection of two lines and other

covered entries remain unchanged. The revised reduced cost matrix is given below:

J1 J2 J3 J4 J5

M1 2 0 1 0 6

M2 0 2 0 4 0

M3 6 1 8 0 6

M4 4 0 10 4 6

M5 0 2 3 0 5

Again we see that only four straight lines are required to cover all the zeros of the

revised cost matrix, therefore optimal assignment cannot be made at this stage.

Repeating the step 4 we get the following revised cost matrix:

J1 J2 J3 J4 J5

M1 1 0 0 0 5

M2 0 3 0 5 0

M3 5 1 7 0 5

M4 3 0 9 4 5

M5 0 3 3 1 5

Since 5 straight lines (equal to the number of rows and columns) are required to cover

all the zeros, optimal assignment can be made at this stage. It is given below:

Assignment Problems

13

J1 J2 J3 J4 J5

M1 1 0 0 5

M2 0 3 0 5

M3 5 1 7 5

M4 3 9 4 5

M5 3 3 1 5

Step5: Select one row containing exactly one zero and surrounded it by

Here we select row no. 3. We can also select 4 or 5.

Cut all the zeros of that column (If has).

Machine1 – Job3 Cost Rs 5





Total 25

Q.1) Unbalanced Minimization:

A B C D E F

I 17 25 11 08 16 31

II 23 13 44 16 19 17

III 32 19 31 28 12 25

IV 26 24 27 21 29 07

V 28 21 19 45 23 43

Ans.

(STEP-I) The number of Column are less than number of Row, therefore we add one

Dummy in Column with relative cost zero. The assignment problem is given below:

0

0

0

0

0

Assignment Problems

14

Balancing:

A B C D E F

I 17 25 11 08 16 31

II 23 13 44 16 19 17

III 32 19 31 28 12 25

IV 26 24 27 21 29 07

V 28 21 19 45 23 43

Dummy 0 0 0 0 0 0

(STEP-II) Subtract least entry of each Row from all the entries of that Row. The

matrix will be as given below:

Row minimization:

A B C D E F

I 09 17 03 00 08 23

II 10 00 31 03 07 04

III 20 07 19 16 00 13

IV 19 17 20 14 22 00

V 09 02 0 26 04 24

Dummy 00 00 00 00 00 00

(STEP-III) Subtract least entry of each Column from all the entries of that Column.

The matrix will be as given below:

Column minimization:

A B C D E F

I 09 17 03 00 08 23

II 10 00 31 03 07 04

III 20 07 19 16 00 13

IV 19 17 20 14 22 00

V 09 02 0 26 04 24

Dummy 00 00 00 00 00 00

Assignment Problems

15

(STEP-IV)Draw minimum line & covers maximum zero:

A B C D E F

I 09 17 03 00 08 23

II 10 00 31 03 07 04

III 20 07 19 16 00 13

IV 19 17 20 14 22 00

V 09 02 00 26 04 24

Dummy 00 00 00 00 00 00

(STEP-V) Make box to assign the particular jobs:

A B C D E F

I 09 17 03 08 23

II 10 31 03 07 04

III 20 07 19 16 13

IV 19 17 20 14 22

V 09 02 26 04 24

Dummy 00 00 00 00 00 00

I=D=08

II=B=13

III=E=12

IV=F=07

V=C=19

Dummy=A=00

Total= 59(Ans.)

00

00

00

00

00

Assignment Problems

16

Example 2: Company XYZ has 5 jobs. 5 people applied for those jobs. The company

needs to find the minimum salary taken for those jobs. Expected salary of those jobs

of person A, B, C, D, E is given below as matrix form:

Person Jobs (in thousand)

I II III IV V

A 4 3 1 5 2

B 7 4 2 10 5

C 8 7 4 6 4

D 3 5 8 7 9

E 5 6 3 8 10

Condition

a) A cannot get job 5.

b) Job 3 get only C.

c) E cannot get any job.

Step 2: Since A cannot get job no. 5, therefore we have to put ∞ at job 5 of A.

C get job no. 3, therefore we have to remove row of C and job no. 3 column from the

matrix.

E cannot get any job; therefore we have to remove all the row of E. It is shown in

table below:

Person Jobs

I II IV V

A 4 3 5 ∞

B 7 4 10 5

D 3 5 7 9

Step2: Here the matrix is unbalanced so we have to take dummy person as a row.

New matrix as follows:

Person Jobs

I II IV V

A 4 3 5 ∞

B 7 4 10 5

D 3 5 7 9

Dm 0 0 0 0

Assignment Problems

17

Row Minimization:

Person Jobs

I II IV V

A 1 0 2 ∞

B 3 0 6 1

D 0 2 4 6

Dm 0 0 0 0

Column Minimization:

Person

Jobs

I II IV V

A 1 0 2 ∞

B 3 0 6 1

D 0 2 4 6

Dm 0 0 0 0

Step3: Select the smallest entry from all the entries which are not covered by a

straight line (here it is 1). Subtract this smallest entry from all the uncovered entries

and add it to all those entries which are at the intersection of two lines and other

covered entries remain unchanged. The matrix is given below:

Person

Jobs

I II IV V

A 0 0 1 ∞

B 2 0 5 0

D 0 3 4 6

Dm 0 0 0 0

Here no. of line = matrix size, therefore optimal assignment can be made at this stage.

Person

Jobs

I II IV V

A 0 0 1 ∞

B 2 0 5 0

D 0 3 4 6

Dm 0 0 0 0

Assignment Problems

18

The solution is:

A – Job2 = 3

B – Job5 = 5

C – Job3 = 4

D – Job1 = 3

Dm – Job2 =0

Total = 15

MAXIMIZATION

Maximization Case –

Question: Four sales men are to be assigning to four sales territories (1 to each).

Estimates of the sales revenues in thousand of rupees for each sales man are as under:

Sales territories

Salesman T1 T2 T3 T4

S1 25 38 43 20

S2 45 12 19 4

S3 43 16 29 24

S4 9 40 45 44

You are required:

To obtain the optimal assignment pattern that maximizes our sales revenue.

Solution:

Step 1- As given matrix gives revenues which are to be maximized. In order to use

minimization technique obtain relative loss matrix by subtracting all the revenues

from maximum revenue i.e. 45

Relative loss matrix

20 7 2 25

0 33 26 41

2 29 16 21

36 5 0 1

Assignment Problems

19

Step 2- We have to check that the given matrix is square or not. Here the matrix is

square.

Step 3- Subtract least entry of each row from all entries of that row to obtain row

reduce matrix. It is given below:

18 5 0 23

0 33 26 41

0 27 14 19

36 5 0 1

Step 4- In the above matrix subtract least entry of each column from all the entries of

that column to obtain total reduce matrix. Try to cover all the zeros of this matrix by

using minimum number of lines. It is shown below-

18 0 0 22

0 28 26 40

0 22 14 18

36 0 0 0

Since the number of covering lines is 3 which is less than size of the matrix optimal

solution cannot be obtained.

Subtract minimum uncovered element (14 in this case) from all the uncovered entries

and add it to all those entries which are at the intersection of two lines. Draw

minimum number of lines to cover all the zeros of this new matrix.

32 0 0 22

0 14 12 26

0 8 0 4

50 0 0 0

Since number of covering lines is 4, which is equal to size of the matrix. Optimal

assignment can be made. It is given in the following table-

Assignment Problems

20

32 0 0 22

0 14 12 26

0 8 0 4

50 0 0 0

Thus optimal assignment is as follows:

Salesman Sales Territory Sales revenue

S1 T2 38000

S2 T1 45000

S3 T3 29000

S4 T4 44000

1,56,000

MAXIMIZATION UNBALANCED

Ques.1. There is 5 jobs for product selling and only 4 executives applied for it. You

have to find maximum selling out of them and assign the job.

A B C D E

I 26 33 14 53 27

II 37 17 22 08 11

III 55 13 24 41 12

IV 42 38 32 27 49

Sol. Here we have to find maximum profit so we have to reduce all the values from

the maximum value. For maximization to minimization. So the new table is

A B C D E

I 29 22 41 02 28

II 18 38 33 47 44

III 00 42 31 14 43

Assignment Problems

21

IV 13 17 23 28 06

This is not a square matrix so we have to balance it with the help of dummy row.

A B C D E

I 29 22 41 02 28

II 18 38 33 47 44

III 00 42 31 14 43

IV 13 17 23 28 06

Dm 00 00 00 00 00

Row minimization

A B C D E

I 27 20 39 00 26

II 00 20 15 29 26

III 00 42 31 14 43

IV 07 11 17 22 00

Dm 00 00 00 00 00

Colum minimization

A B C D E

I 27 20 39 00 26

II 00 20 15 29 26

III 00 42 31 14 43

IV 07 11 17 22 00

Dm 00 00 00 00 00

Assignment Problems

22

Draw minimum line for covers maximum zeros

A B C D E

I 27 20 39 00 26

II 00 20 15 29 26

III 00 42 31 14 43

IV 07 11 17 22 00

Dm 00 00 00 00 00

Numbers of lines is not equal to numbers of row or Colum.

Then we have to find least number(11) of free numbers and subtract it from free

numbers and add with there when lines are intersects and don‘t do any things where

lines are pass. And again draw the minimum lines which covers maximum zeros.

Free numbers are those which are not in lines.

A B C D E

I 27 09 28 00 26

II 00 09 04 29 26

III 00 31 20 14 43

IV 07 00 06 22 00

Dm 11 00 00 11 11

Number of lines is not equals to row or column then we have to apply the same step.

A B C D E

I 27 05 24 00 22

II 00 05 00 29 22

III 00 27 16 14 39

IV 11 00 06 26 00

Dm 15 00 00 15 11

Assignment Problems

23

Candidate I assign the job D, II assign the job C, III Assign the job A,

IV assign the job E and Dm will be assign the job B.

So the minimum salaries which is given to employees is

First option is

I*D= 53

II*C= 37

III*A= 55

IV*E= 49

Dm*B= 0

194

So maximum sell is 194 Ans.

Assignment Problems

24

MULTIPLE CONDITIONED QUESTIONS

If there is any unbalanced, multiple conditioned question, first we will minimize it (if

question is maximization) then we will balance the question. After that fulfill the

conditions and if needed balance the question again. Then after follow the general

procedure.

Q.1) Profit earned by different sales man in different territories are as follows:-

TERRITORY

SALES MAN

T1 T2 T3 T4 T5 T6

A 31 16 14 13 15 30

B 25 19 18 17 19 26

C 38 17 22 21 23 22

D 15 22 26 25 27 18

E 14 23 30 29 31 14

CONDITIONS

1. C has to be assigned job T5

2. E should not get T2 & T6

3. D should not get any job

SOLUTION

Minimization and applying conditions.

STEP-I

TERRITORY

SALES MAN

T1 T2 T3 T4 T5 T6

A 31 16 14 13 15 30

B 25 19 18 17 19 26

C 38 17 22 21 23 22

D 15 22 26 25 27 18

E 14 ∞ 30 29 31 ∞

Assignment Problems

25

STEP-II

Balancing the problem

TERRITORY

SALES

MAN

T1 T2 T3 T4 T6

A 7 22 24 25 8

B 13 19 20 21 12

E 24 ∞ 8 9 ∞

D1 0 0 0 0 0

D2 0 0 0 0 0

STEP-III

Row minimization

TERRITORY

SALES MAN

T1 T2 T3 T4 T6

A 0 15 17 18 1

B 1 7 8 9 0

E 16 ∞ 0 1 ∞

D1 0 0 0 0 0

D2 0 0 0 0 0

STEP-IV

Column minimization

TERRITORY

SALES MAN

T1 T2 T3 T4 T6

A 0 15 17 18 1

B 1 7 8 9 0

E 16 ∞ 0 1 ∞

D1 0 0 0 0 0

D2 0 0 0 0 0

Assignment Problems

26

STEP-V

Assigning the job.

SALES MAN TERRITORIES SALES PROFIT

A T1 31

B T6 26

C T5 23

D NO JOB 00

E T3 30

D1 T2/T4 00

EXERCISES

MINIMIZATION PROBLEMS

1.

A company has 5 machines and 5 jobs. The revelant cost is given below

Find the assignment that minimizes total cost.

Machines JOBS

J1 J2 J3 J4 J5

M1 10 4 5 3 11

M2 13 11 9 12 10

M3 12 3 10 1 9

M4 9 1 11 4 8

M5 8 6 7 3 10

Ans.

Machine 1 JOB 3 COST RS.3

Machine 2 JOB 5 COST RS.1O

Machine 3 JOB 4 COST RS.1

Machine 4 JOB 2 COST RS. 1

Machine 5 JOB 1 COST RS. 8

TOTAL. 25

Assignment Problems

27

2.

A car hire company has one car at each of the five depots a,b,c,d,e. A customer

requires car at each town vit. A,B,C,D and E. Distance (in kms) between deposits

(origins) and towns (destination) are given in the following distance matrix. How

should the cars be assigned to customers so as to minimize the total distance travelled.

A B C D E

A 160 130 175 190 200

B 135 120 130 160 175

C 140 110 155 170 185

D 50 50 80 80 110

E 55 35 70 80 105

Ans. A * e =200 km

B * c =130 km

C * b =110 km

D * a =50 km

E * d =80 km

Total = 570 km

MAXIMIZATION PROBLEMS

1.

A marketing manager has 5 sales man & 5 sales area, considering the capabilities of

the salesmen nature of areas, the maraketing manager estimates that sales per month

(in thousands of rupees) for salesman in each area would be as follow:

Salesman AREAS

A1 A2 A3 A4 A5

S1 42 48 50 38 50

S2 50 34 38 31 36

S3 51 37 43 40 47

S4 32 48 51 46 46

S5 39 43 50 45 49

Assignment Problems

28

Final Optimal Assignment

Ans.

TOTAL COST 241000

S1=A2 S1=A2

S2=A1 S2=A5

S3=A5 OR S3=A1

S4=A3 S4=A4

S5=A4 S5=A5

<

2.

Quantity of clothes of different brands sold in different cities per mnth are as follows

Wrangler Pantaloon Pvogue Lee Levis

Delhi 25 22 55 42 48

Gurgaon 78 41 40 46 41

Noida 18 33 52 50 37

Jaipur 32 18 30 37 20

Chandigarh 40 20 42 48 51

Find the showroom for different brands which should be opened in different cities.

Ans = 254

Unbalance Assignment problem (Minimization)

Q. 1) A company is face with the problem of assigning six different machines to

Five different job. The cost are estimated as follows (in hundred of Rs.)

Solve the problem assuming that the objective is to minimize the total cost.

A1 A2 A3 A4 A5

M1 05 10 02 12 02

M2 04 10 03 14 06

M3 06 13 04 16 06

M4 07 14 04 18 09

M5 08 14 06 18 12

M6 12 18 10 20 12

Assignment Problems

29

Ans.

Machine Cost Machine Cost

M1 A5 02 M1 A5 02

M2 A1 04 M2 A4 14

M3 A4 16 M3 A1 06

M4 A3 04 M4 A3 04

M5 A2 14 M5 A2 14

M6 A6 00 M6 A6 00

Total=40 Total=40

Q. 2) Solve the following unbalance Assignment problem of the minimizing total

time for doing all jobs.

Jobs

J1 J2 J3 J4 J5

O1 16 12 15 12 16

O2 12 15 18 17 17

O3 17 18 16 19 18

O4 16 12 13 14 15

O5 19 13 18 19 17

O6 14 17 14 16 18

Ans:

Operation Job Time

O1 J4 12

O2 J1 12

O3 J6 00

O4 J5 15

O5 J2 13

O6 J3 14

Total=66

Assignment Problems

30

Unbalance Assignment problem(Maximization)

Q.1) Four different Airplanes are to be assigned to handle three cargo consignment

with a view to maximize profit . The profit matrix is a follow in thousand of

Rupees

Cargo Consignment

Airplane I II III

W 08 11 12

X 09 10 10

Y 10 10 10

Z 12 08 09

Ans:

W to III profit 12,000 W to III profit 12,000

X to II profit 10,000 X to Dummy profit 0

Y to Dummy profit 00 Y to II profit 10,000

Z to I profit 12,000 Z to I profit 12,000

Total=34,000 Total=34,000

Q.2) Solve the following assignment problem. The data give him the table refer to

production in central unit.

Machine

A B C D

I 10 05 07 08

II 11 04 09 10

III 08 04 09 07

IV 07 05 06 04

V 08 09 07 05

Assignment Problems

31

Ans.

Operation Machine No. of Units

1 A 10

2 D 10

3 C 9

4 Dummy 0

5 B 9

Unit=38

CONDITION

EXERCISE- Solve the following assignment problem in which 5 jobs and 7 peoples

are given and some conditions are considered when jobs assign, the conditions are-

Job 3 has to be assigned to B.

Job 1 and 4 can not be assigned to G.

D has to get job 5

1 2 3 4 5

A 36 33 25 30 26

B 42 25 35 26 45

C 54 50 52 35 28

D 28 35 40 28 36

E 20 28 32 40 52

F 26 30 48 25 27

G 24 28 22 26 30

ANSWER- 128

Transportation Problem

32

TRANSPORTATION PROBLEMS

“INTRODUCTION”

The Transportation model deals with situations where some commodity or product is

distributed from multiple sources to multiple destinations. The model may be used to

find the minimum transportation cost or the maximum profit, depending upon the

amounts shipped from each source to each destination.

The problem solution will be the optimum distribution scheme, showing exactly how

much of the commodity should be transported via each possible route.

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

cost of transporting a homogeneous commodity from supply origins to demand

destinations. However, it can also be applied to the maximization of some total value

or utility, for example, financial resources are distributed in such a way that the

profitable return is maximized.

ADVANTAGES

Proper utilization of resources take place without much of the losses.

The selections and allocations of resources to their destinations become more

accurate.

This process helps in cost minimization and profit maximization, which major

objective of organizations.

It helps in planning and decision making.

Some Important Terms or Definitions

1) Feasible solution-: Set of Non Negative values

xij=1,2,3,4….m,j=1,2,3,4……….n. which satisfy the following condition is

called feasible solution.

2) Basic feasible solution-: a feasible solution with an allocation of (m+n-1)

number of variables . xij,i=1,2,3…..m, j=1,2,3,4…….n. is called a basic

feasible solution.

3) Optimum Solution-: A basic feasible solution of transportation problem

which minimizes the total transportation cost or maximizes total revenue.

4) Rim requirement-: The quantity required or available are called rim

requirement.

5) Balanced transportation problem-: If in any transportation problem total

number of units available is equal to the total number of units required, then it

is called balanced transportation problem.


33

STEPS OR PROCEDURE OF SOLVING TRANSPORTATION PROBLEMS

Make Empty Cells (4)

YES

NO

NO

Calculate Value of Ri and Cj (5)

Xij = TCij – (Ri + Cj)

YES

Is Xij < 0 ? Select Minimum Value of Xij

Construct Loop

Revise solution and check for rim

value condition

Find a Basic feasible Solution (3)

f

YES

Balance the Problem (2)

Is R + C -1= Filled

Cell?

?

NO Is Problem

Balanced?

?

NO

START

YES Is Problem of

maximization?

Convert to minimization (1)

Optimal Solution


34

Notes:

1) Convert to minimization-: If the problem is maximization then convert it

into minimization by subtracting the whole value of table from the largest.

2) Balance the problem-: If the problem is unbalanced then balance the problem

by using the dummy and give the extra capacity to the dummy. Now the

problem will be balanced.

3) Find a Basic Feasible Solution-: After balancing the problem find a basic

feasible solution that which one is right for problem among following

methods .north west corner method, least cost method, Vogel‘s.

4) Make empty cell-: If R+C-1= filled cell does not satisfy the table then make a

empty cell on any unfilled cell.

5) Calculate value of R and C-: By using formula x= TC-(R+C).We will

calculate the value of R and C.

THE TRANSPORTATION METHOD

There are several methods available to obtain an initial basic feasible solution. But the

general steps are discussed below:

Step 1. The solution algorithm to a transportation problem may be summarized into

the following steps:

The formulation of the transportation problem is similar to the LP problem

formulation. Here the objective function is the total transportation cost and the

constrains are the supply and demand available at each source and destination,

respectively.

Step 2. Obtain an initial basic feasible solution.

North-West Corner Method (NWCM)

It is a simple and efficient method to obtain an initial solution. This method does not

take into account the cost of transportation on any route of transportation. The method

can be summarized as follows;

Step1 Start with the check at the upper left (north-west) corner of the transportation

matrix and allocate as much as possible equal to the minimum of the rim value for the

first row and first column.


35

Step 2

(a) If allocation made in Step 1 is equal to the supply available at first source (a1, in

first row), then move vertically down to the cell (1, 2) in the second row and

first column and apply Step 1 again, for next allocation.

(b) If allocation made in Step 1 is equal to the demand of the first destination (b1 is

first column), then move horizontally to the cell (1, 2) in the first row and

second column and apply Step 1 again for next allocation.

(c) If a1 = b1, allocate w11 = a1 or b1 and move diagonally to the cell (2, 2).

Step 3. Continue the procedure step by step till an allocation is made in the south-east

corner cell of the transportation table.

Example-: A construction company wants cement at three of its project sites P1, P2

and P3. It procures cement from four plants C1, C2, C3 and C4. Transportation costs

per ton, capacities and requirements are as follows:

P1 P2 P3 Capacity(tons)

C1 5 8 12 300

C2 7 6 10 600

C3 13 4 9 700

C4 10 13 11 400

Requirement 700 400 800

Determine optimal allocation of requirements.

Solution:


C1 5

(300) 8 12 300

C2 7

(400)

6

(200) 10 600

C3 13 4

(300)

9

(400) 700

C4 10 13 11

(400) 400



36

The cell (C1, P1) is the North-West corner cell in the given transportation table, the

rim values for row C1 and column P1 are compared. The smaller of two, i.e. 300 is

assign as the first allocation. This means that 300 units of commodity are to be

transported from plant C1 to project site P1. However, this allocation leaves a supply

of 700 – 300 = 400 unit of commodity at C1.

Move vertically and allocated as much as possible to cell (C2, P1). The rim value for

column P1 is 400 and for row C2 is 600. The smaller of the two, i.e. 400, is placed in

the cell.

Proceeding to column P2, the rim value for column P2 is 500 and for row C2 is 200.

The smaller of the two, i.e. 200, is placed in the cell.

Proceeding to column P3, the rim value for column P3 is 800 and for row C3 is 400.

The smaller of the two, i.e. 400, is placed in the cell.

Now, the rim value for column P3 is 400 which is balanced by the rim value of row

C4 that is 400.

The total transportation cost of the initial solution derived by the North-West corner

method is obtain by multiplying the quantity in the occupied sales with the

corresponding unit cost and adding. Thus the total transportation cost of the solution is-:

Total cost = 300 x 5 + 400 x 7 + 200 x 6 + 300 x 4 + 400 x 9 + 400 x 11

= Rs. 14,700

Least Cost Method

Since the objective is to minimize the total transportation cost, we must try to

transport as much as possible through those routes where the unit transportation cost

is lowest.

Step 1 – Select the cell with the lowest unit cost in the entire transportation table and

allocate as much as possible to this cell and eliminate that row and column in which

either supply or demand is exhausted.

Step 2 – Repeat the procedure until the entire available supply at various sources and

demand at various destinations is satisfied





37


C1 5 8 12 300

C2 7 6 10 600

C3 13 4 9 700

C4 10 13 11 400



Solution:


C1 5

(300) 8 12 300

C2 7

(400) 6

10

(200) 600

C3 13 4

(500)

9

(200) 700

C4 10 13 11

(400) 400


In the above Solution we have,

The cell (C3, P2) contain the lowest transportation cost in the given transportation

table, the rim values for row C3 and column P2 are compared. The smaller of two, i.e.

500 is assign as the first allocation. This means that 500 units of commodity are to be

transported from plant C3 to project site P2. However, this allocation leaves a supply

of 700 – 500 = 200 unit of commodity at C3.

Now, search for the least cost in table without considering P2 because the rim value of

P2 is zero. The lowest transportation cost is at cell (C1, P1) the rim values for row C1

and column P1 is compared. The smaller of two, i.e. 300 is assigned. However, this

allocation leaves a supply of 700 – 300 = 400 unit of commodity at P1.

Now, search for the least cost in table without considering column P2 and row C1

because the rim value of P2 and C1 is zero. The lowest transportation cost is at cell

(C2, P1) the rim values for row C2 and column P1 is compared. The smaller of two,

i.e. 400 is assigned. However, this allocation leaves a supply of 600 – 400 = 200 unit

of commodity at c2.


38

Now, the rim value for row C2, C3, and C4 are respectively 200, 200 and400 which is

balanced by the rim value of column P3 that is 800.

The total transportation cost of the initial solution derived by the ―Least Cost Method‖

is obtained by multiplying the quantity in the occupied sales with the corresponding

unit cost and adding. Thus the total transportation cost of the solution is-:

Total Cost = (C1, P1) + (C2, P1) + (C2, P3) + (C3, P2) + (C4, P3)

= 300 X 5 + 400 X 7 + 200 X 10 + 500 X 4 + 400 X 11

= 1500 + 2800 + 2000 + 2000 + 4400

= Rs. 12,700

VOGELS APPROXIMATION METHOD

Step 1: Construct the transportation tableau as described earlier.

Step 2: For each row and column, the difference between the two lowest cost entries

.If the lowest cost entries are tied, the difference is Zero.

Step 3: Select the row or column that has the largest difference .In the event of a tie

selection is arbitrary.

Step 4: In the row or column, identified in step 3, select the cell that has lowest cost

in tree.

Step 5: Assign maximum possible number of unit to the cell selected in step 4(the

smaller of two between demand and the availability).This will completely exhaust a

row or a column. Omit the exhausted row and column.

Step 6: Reapply step 2 to step 5. Iteratively using the remaining row and columns

until the total demand is met and supply exhausted.





C1 5 8 12 300

C2 7 6 10 600

C3 13 4 9 700

C4 10 13 11 400




39

Solution:

P1 P2 P3 CAPACITY ROW

DIFFERENCE

C1

5

300 8 12 300 3 7 - -

C2

7

400 6

10

200 600 200 1 3 3 3

C3

13

4

500

9

200 700 200 5 4 4 -

C4

10 13

11

400 400 1 1 1 1

REQUIREMENT 700

400 500

800

600

2000

COLUMN

DIFFERENCE

2 2 1

2 - 1

3 - 1

3 - 1

Total cost = C1×P1+C2×P1+C2×P3+C3×P2+C3×P3+ C4×P3

= 300×5+400×7+200×10+500×4+200×9+400×11

= 1500+2800+2000+2000+1800+4400

= Rs. 14,500

Firstly we will find, for each row (C1, C2, C3, and C4) and each column (P1, P2, and

P3), the difference for two lowest entries. After that we select the row C3, because it

has largest difference. In C3, find the least value is 4 after that assign the maximum

possible number 500 to the cell so that it exhausted the column P2 completely.

After that we will find remaining rows and columns the difference for two lowest

entries. After that we select the row C1, because it has largest difference. In C1, find

the least value is 5 after that assign the maximum possible number 300 to the cell so

that it exhausted the row C1 completely.

After that we will find remaining rows and columns the difference for two lowest



that it exhausted the Row C3 completely.


40

After that we will find remaining rows and columns, the difference for two lowest



that it exhausted the column P1 completely.

By continuous doing this process we will assign the all unit to the cell and will find

the minimum transportation cost.

Optimality Test

Once initial solution has been found the next step is to test that solution for

optimality. Modified Distribution Method (MODI method) is generally used for

testing the optimality of the existing solution.

MODIFIED DISTRIBUTION METHOD (MODI Method)

The following steps involved in the MODI method:-

Step 1- R+C−1= Filled cell (In this step we will subtract 1 from the sum of rows and

columns, after that we will compare with filled cells. which will be equal to the filled

cell.)

Step 2- Assume any row or column equal to zero.

Step 3- Filled cell – R+C= TC (In this step we find the value of row and column)

Step 4 – Unfilled cell – Find the sign of unfilled cell with the help of TC−(R+C)





C1 5 8 12 300

C2 7 6 10 600

C3 13 4 9 700

C4 10 13 11 400

Requirement 700 500 800 2000



41

Solution:

C1=5 C2=3 C3=8

P1 P2 P3 CAPACITY ROW

DIFFERENCE

R1=0 C1

5

300 8 12 300 3 7 - -

R2=2 C2

7

400 6

10

200 600 200 1 3 3 3

R3=1 C3

13

4

500

9

200 700 200 5 4 4 -

R4=3 C4

10 13

11

400 400 1 1 1 1

Requirement 700

400 500

800

600

2000

Column

Difference

2 2 1

2 - 1

3 - 1

3 - 1

Total cost = C1×P1+C2×P1+C2×P3+C3×P2+C3×P3+C4×P3

= 300×5+400×7+200×10+500×4+200×9+400×11

= 1500+2800+2000+2000+1800+4400

= Rs. 14,500

Step 1- R+C–1 = Filled cell

4+3–1 = 6 (BALANCED MATRIX)

Step 2- For filled cell

Let R1= 0

R1+C1 = TC 0+C1 = 5 C1 = 5

R2+C1 = TC R2+5 = 7 R2 = 2

R2+C3 = TC 2+C3 = 10 C3 =8

R3+C3 = TC R3+8 = 9 R3 = 1

R3+C2 =TC 1+C2 = 4 C2 = 3

R4+C3 = TC R4+8 = 11 R4 =3


42

Step 3- For unfilled cell (Find sign)

TC–(R1+C2) = 8–(0+3) = 5

TC–(R1+C3) = 12-(0+8) = 4

TC–(R2+C2) = 6–(2+3) = 1

TC–(R3+C1) = 13–(1+5) = 7

TC–(R4+C1) = 10–(3+5) = 2

TC–(R4+C2) = 13–(3+3) = 7

There is no negative sign so that the total cost will be same.

Question - 2

C1

=13

C2 =

12

C3 =

10 C4

W1 W2 W3 W4 PRODUCTION

CAPACITY

ROW

DIFFERENCE

R1 = 0 F1 13

(2) 11 15 40 2 2 - - -

R2 = 2 F2 17 14

(3)

12

(3) 13 6 3 1 1 - 4

R3 = 5 F3 18

(1) 18

15

(1)

12

(5) 7 2 3 3 3 6

W.H.

CAPACITY 3 1 3 4 1 5

COLUMN

DIFFERENCE

4 3 3 1

1 4 3 1

1 - 3 1

1 - 3 -

Total Cost = F1W1+F2W2+F2W3+F3W1+F3W3+F3W4

= 26+42+36+18+15+60

= Rs. 197

Step 1- First we will find filled cell

TC-(R1+C4) = 40-(0+7) = 33

TC-(R2+C1) = 17-(2+13) = 2

TC-(R2+C4) = 13-(2+7) = 4

TC-(R3+C2) = 18-(5+12) = 1


43

If there will be negative sign than we will make house in L shape or Square shape. In

this process, we will assign the sign alternatively. Then we will subtract the whole

corner vales from the least positive value which will also at any corner of house.

R+C−1 = 3+4−1 = 6

Step 2 -After that we will find the value of row and column by assuming any row is

equals to zero.

For filled cell (let R1 = 0)

R1+C1 = TC = 0+C1 = 13 = C1 = 13

R3+C1 = TC = R3+13 = 18 = R3 = 5

R3+C3 = TC = 5+C3 = 15 = C3 = 10

R3+C4 = TC = 5+C4 = 12 = C4 = 7

R2+C3 = TC = R2+10 = 12 = R2 = 2

R2+C2 =TC = 2+C2 =14 = C2 = 12

Step 3- After that we will find the sign.

For Unfilled Cell

TC-(R1+C2) = 11-(0+12) = -1

TC-(R1+C3) = 15-(0+10) = 5

C1 = 15 C2 =14 C3 =12 C4=9

W1 W2 W3 W4 PRO.

CAPACITY

R1=-2 F1

13 +

(2)

11 -

(H) 15 40 2

R2=0 F2 17

14

+

(3)

12

-

(3)

13 6

R3=3 F3

18

-

(1)

18

15

+

(1)

12

(5) 7

W.H.

CAPACITY 3 3 4 5 15


44

W1 W2 W3 W4 PRO.

CAPACITY

F1 13

(1)

11

(1) 15 40 2

F2 17 14

(2)

12

(4) 13 6

F3 18

(2) 18

15

12

(5) 7

W.H.

CAPACITY 3 3 4 5 15

Total Cost = F1W1+F1W2+F2W2+F2W3+F3W1+F3W4

= 13+11+28+48+60+36

= Rs. 196

Special Situations

There is certain special situation or case which generally happened in practical life,

some of them are discussed below:

Maximizing Transportation Problem:

The main objective of an organization is to cost minimization and profit

maximization. The problem of maximizing can be solved by following process:

A maximization transportation problem can be converted into the usual minimizing

problem by subtracting all the contributions from the highest contribution involved in

the problem.

Example:

Priyanka Steel Co. has three factories and five customers, profit (` per unit) are given.

Find a schedule where co. gets maximum profit.

Factory

Customer

Capacities C1 C2 C3 C4 C5

F1 4 2 3 2 6 8

F2 5 4 5 2 1 12

F3 6 5 4 7 7 14

Required 4 4 6 8 12


45

Solutions:

Factory

Customer

Capacities C1 C2 C3 C4 C5

F1 3 5 4 5 1 8

F2 2 3 2 5 6 12

F3 1 2 3 0 0 14

Required 4 4 6 8 12

In above problem the highest contribution or per unit profit is 7 and now we subtract

all the contribution no from 7,

Now applying the VAM method in the above table.

Total Cost according to VAM and after applying MODI method is as follows:

= 8 X 1 + 2X 2 +3X 4 + 2 X 6 + 1 X 2 + 0 X 8 + 0 X 4

=8 + 4 +12 +12 +2 + 0 + 0

= Rs. 38

Factory

Customer Capacities

P1 P2 P3 P4

C1 C2 C3 C4 C5

C1=2 C2=3 C3=2 C4=1 C5=1

F1 R1=0 3 5 4 5 1 (8) 8 2 2 X X

F2 R2=0 2

(2)

3

(4)

2

(6)

5

6 12 0 O 1 1

F3 R3=-

1

1

(2) 2 3

0

(8)

0

(4) 14 0 1 1 1

Required 4 4 6 8 12

P1 1 1 1 5 1

P2 1 1 1 x 1

P3 1 1 1 x 6

P4 1 1 1 x x


46

Unbalanced Transportation problem

Transportation problems that have the supply and demand equal is a balanced

transportation problem. In other words requirements for the rows must equal the

requirements for the columns. An Unbalanced transportation problem is that in

which the supply and demand are unequal. There are 2 possibilities that make the

problem unbalanced which are

(i) Aggregate supply exceeds the aggregate demand or

(ii) Aggregate demand exceeds the aggregate supply.

Such problems are called unbalanced problems. It is necessary to balance them before

they are solved.

The following table gives the cost of transportation, the availabilities and

requirements of an organization:

Ware House

Demand Points Available Capacity

A B C D

W1 10 20 20 15 50

W2 15 40 15 35 100

W3 25 30 40 50 150

Requirements 25 115 60 30

Solution:

In the above problem total demand is less than total supply by (300-230) 70 units.

Step 1: So, we will create a new demand point and name this as ―DUMMY‖.

Step 2: Assume Transportation cost as to be ―Zero‖ and write in DUMMY column in

place of Transportation cost.

Step 3: Write the difference of total demand and supply in the required column cell.

Aggregate supply exceeds the

aggregate demand

Aggregate demand exceeds the

aggregate supply

2 Possibilities That Make the Problem Unbalanced


47

A B C D DUMMY CAPACITY

W1 10 20 20 15 0 50

W2 15 40 15 35 0 100

W3 25 30 40 50 0 150

25 115 60 30 70 300

Step 4: Apply method of basic feasible solution. (By Vogel’s Approximation Method).

A B C D DUMMY CAPACITY P1 P2 P3 P4 P5

W1 10

(20) 20 20

15

(30) 0 50 10 5 10 - -

W2 15

(5)

40

(35)

15

(60) 35 0

100

5 0 0 0 25

W3 25 30

(80)

40

50

0

(70)

150 30 5 5 5 5

25 115 60 30 70 300

P1 5 10 5 20 0

P2 5 10 5 20 -

P3 5 10 5 - -

P4 10 10 25 - -

P5 10 10 - - -

Total Cost =10×20+40×35+15×30+15×5+40×35+15×60+30×80

=200+1400+450+75+1400+900+2400

= Rs. 6825

Step 5: Now, we will test this solution for optimality.

C1= 15 C2= 30 C3= 15 C4= 25 C5= 0

A B C D DUMMY CAPACITY

R1= -

10 W1

10

20

(20) 20

15

(30) 0 50

R2= 0 W2 15

(25) 40

15

(60) 35

0

(15)

100

R3= 0 W3 25 30

(95)

40

50

0

(55)

150

25 115 60 30 70 300

The Optimal Cost for above sum

= 20×20+15×30+15×25+15×60+30×95

= 400+450+375+900+2850 = Rs. 4975


48

Prohibited Routes

At times there are transportation problems in which one of the sources is unable to

ship to one or more of the destinations. When this occurs, the problem is said to have

an unacceptable or prohibited route. In a minimization problem, such a prohibited

route is assigned a very high cost to prevent this route from ever being used in the

optimal solution. After this high cost is placed in the transportation table, the problem

is solved using the techniques previously discussed. In a maximization problem, the

very high cost used in minimization problems is given a negative sign, turning it into

a very bad profit.

Question: Solve the following transportation problem with the restriction that nothing

can be transported from factory F2 to warehouse WE3 :

FACTORY W1 W2 W3 W4

NO OF

UNITS

AVAILABLE

F1 40 60 60 10 15

F2 70 30 50 20 4

F3 30 90 20 40 11

REQUIREMENT 10 5 10 5 30

Transportation root is restricted in the cell F2-W3. So we have to replace the

particular cell with any maximum value. Here we are taking maximum value as M

and solve the problem as usual method.

Now, apply VAM

W1 W2 W3 W4 Availability P1 P2 P3 P4

F1

40

(9)

60

(1)

60

10

(5) 15/10 30 30 20 20

F2

70

30

(4) M 20 4/0 810 10 40 40

F3

30

(1) 90

20

(10) 40 11/1/0 10 10 60

Requirement 10/9/0 5/1/0 10/0 5/0

P1 10 30 40 10

P2 10 30 10

P3 10 30

P4 30 30


49

Now test for optimality.

C1 = 40 C2 =60 C3= 30 C4 = 10

W1 W2 W3 W4 Availability

R1 = 0 F1 40

(9)

60

(1)

60

10

(5) 15

R2 = -30 F2 70

30

(4) M 20 4

R3 = -10 F3 30

(1) 90

20

(10) 40 11

Requirement 10 5 10 5

Total Optimal Cost = 9 x 40 + 1 x 60 + 5 X 10 + 4 X 30 + 1 X 30 + 10 X 20

= 360 + 60 + 50 + 120 + 30 + 200

= Rs. 820

Some Necessary Allocation

In our today‘s business it is often happens that company has to send their product to

pre determined or fixed customers or warehouse or distribution centre, in spite of their

new or any other customer‘s .so, in this type of problem, transportation cost is pre –

allocated between two particular parties.

The following examples illustrate the details of the problem and put more light on the

concept.

For example:

A firm producing a single product has three plants and four distributors. The three

plants will produce 60, 80 and 100 units respectively during the next period. The firm

has made a commitment to sell 90 units to distributor a,40 units to distributor B, 60

units to distributor C and 50 units to D. if the management wants to transport at least

40 units from plant P1 to distributor A .find the optimal transportation schedule. The

net distribution cost to a distributor is given in the following table:

Plant

Distributor

A B C D

P1 8 7 5 2

P2 5 2 1 3

P3 6 4 3 5


50

Solution:

Since the firm wants to apply at least 40 units from plant P1 to distributor C, we

subtract 40 units from the number of units available at plant P1and from the

requirement of distributor A i.e. now the number of units available at plant P1 will be

only 20 (60-40) and demand of distributor A will be only50 (90-40) units.

Plant

Distributor

A B C D

No. of units

available

P1 8 7 5 2 20

P2 5 2 1 3 80

P3 6 4 3 5 100

Demand 50 40 60 50 200

Now, apply Vogel‘s approximation method, the following table comes:

C1= 6 C2= 4 C3= 3 C4= 5

A B C D

R1= -3 P1 8 7 5 2

(20) 20 3 - -

R2= -2 P2 5 2

(20)

1

(60) 3 80 1 1 1

R3= 0 P3 6

(50)

4

(20) 3

5

(30) 100 1 1 1

Demand

50 40 60 50 200

1 2 2 1

1 2 2 2

1 2 - 2

Total Cost = 20 X 2 + 20 X 2 + 60 X 1 + 50 X 6 + 20 X 4 + 30 X 5

= 40 + 40 +60 + 350 + 80 + 150

= Rs. 720

Unsolved Problems

Q.1 A manufacturing firm must produce a product in sufficient quantity to meet

contractual sales in next four months. The production capacity and unit cost of

production vary from month to month. The product produced in any month

may be held for sale in later months but at an estimated storage cost of Re.1

per unit per month. No storage cost is incurred for goods sold in the same


51

month in which they are produced. There is no opening inventory and none is

desired at the end of fourth month. The necessary details are given in the

following table:

Month Contractual

Sales

Max.

Production

Unit cost of

Production

1 20 40 14

2 30 50 16

3 50 30 15

4 40 50 17

How much should the firm produce each month to minimize total cost?

Ans.

Cost Rs. 2210

Q.2 A company has four manufacturing plants and five warehouses. Each plant

manufactures the same product which is sold to different warehouse at

different prices. The details are given below.

Plants 1 2 3 4

Manufacturing cost 12 10 8 7

Raw material cost 8 7 7 5

Capacity 100 200 120 80

Warehouses Transportation Cost

Demand Sales

Price 1 2 3 4

A 4 7 4 3 80 30

B 8 9 7 8 120 32

C 2 7 6 10 150 28

D 10 7 5 8 70 34

E 2 5 8 9 90 30

Formulate the above as a transportation problem to maximize profit and obtain the

optimal transportation schedule.

Ans: Rs. 4580


52

Q.3 The following table gives all the necessary information on the available supply

to each warehouse, the requirement of each customer and unit transportation

cost from warehouse to each customer:

Warehouse Customer Available

8 9 6 3 18

20

18

6 11 5 10

3 8 7 9

Required 15 16 12 13 56

Find optimal transportation schedule.

Ans. Rs. 301


53

Case Study:

Game Theory

54

INTRODUCTION

In practical life, it is required to take decisions in a competing situation when there are

two or more opposite parties with conflicting interest and the outcome is controlled

by the decisions of all the parties are concerned.

In all the above problems where the competitive situations are involved one act in a

rational manner and tries to resolve the conflict of interest in his favor.

In these situations GAME THEORY was developed in the twentieth century.

However, the mathematical treatment of the game theory was given in 1944 by John-

Von-Newmann through ―THEORY OF THE GAMES AND ECONOMIC

BEHAVIOR”.

DEFINITION

It may be defined as ―a body of knowledge that deals with making decisions when

2 or more intelligent and rational opponents are involved under conditions of

conflict and competition‖. The approach is to seek to determine a rival‘s most

profitable counter –strategy to one‘s own best moves and to formulate the appropriate

defensive measures.

ASSUMPTIONS OF GAME THEORY

The number of competitors in the competition is known.

The participants simultaneously choose their respective course of action.

The profit and loss in the competition is fixed and determined in advance.

It is assumed that each competitor behaves rationally and intelligently.

It is assumed that players attempts to maximize gains and minimize losses.

The decision of the game can be positive, negative or zero to each player.

It is assumed that players have the knowledge of all the information relating to

the game they play.

KEY CONCEPTS AND TERMS

Game – A game represents a conflict between two parties / countries / persons

/ and is played with certain predetermined rules.

Dominance – One of the strategies of either player may be inferior to atleast

one of the remaining ones. The superior strategies are said to dominate the

inferior ones. A procedure that is used to reduce the size of the game.

Game Theory

55

Mixed strategy – In a game without saddle point, the optimal policy is to use

mixed strategies i.e. to use some optimal combination of available strategies.

Saddle point – A saddle point of a pay off matrix is that position in the matrix

where the maximum of the row minima is equal to the minimum of the

column maxima. The pay off at the saddle point is called the value of the game

and the corresponding strategies are the pure strategies.

Strategy – The number of competitive actions that are available for a player

are called strategies to that player.

Value of the game – It is the expected pay off of the play when all players of

the game follow their optimal strategies. The game is called fair if the value of

game is zero and unfair if it is non zero.

Game Theory

56

FLOW CHART REPRESENTING FLOW OF GAME THEORY

NO

YES

NO

YES YES

Law of dominance

START

Find the saddle point

If saddle is

there

Value of game

Strategies

Arithmetic OR

Graphical method LPP Arithmetic method

If Matrix

is 2*2

If matrix is

m*2 or 2*n

STOP

NO

YES

YES YES

Game Theory

57

FINDING SADDLE POINT

See the highest number in column draw on that number

And See the lowest number in row draw on that number.

If and on a particular number. That is Saddle point.

The saddle point is the value of game.

PRINCIPLE OF DOMINANCE

If no saddle point:

When the elements of a row are less than or equal to the elements of another

row then that lesser row is cut.

When the elements of a row are less than or equal to the average elements of

another two rows, then that lesser row is cut.

When the elements of a column are greater than or equal to the elements of

another row then that greater column is cut.

When the elements of a Column are Greater than or equal to the average

elements of another Column, then that Greater column is cut.

After the law of dominance we find the following situations

m*n

2*2 Arithmetic Method

m*n LPP (Linear Programming Problem)

m*2 Arithmetic

method/graphical method

2*nArithmetic

method/Graphical method (lower envelop

& upper point)

Game Theory

58

After the law of dominance if m*n matrix become 2*2 matrix then we find the

solution as follows;

2*2

Find the difference of second row and put it in front of first row

Find the difference of first row and put it in front of second row

Find the difference of second column and put it in below of first column

Find the difference of first column and put it in below of Second column

Calculate the total of the difference of the columns or rows that should be equal

For calculating the value of game we multiply difference of columns with

corresponding any of the column, and divided by total of the difference

For calculating the strategies of players we divided the value putted in front of

the player strategies

GRAPHICAL METHOD

2*n

Steps to solve:

For 2*n game, first draw to vertical parallel lines.

Let line first be the first row and line second be the second row.

Now plot scale on these lines.

Then plot the element of first row and first column on first line and the element

of second row and first column on second line. Similarly second column third

column so on.

We find the lower envelop and upper point and intersections lines of upper

point.

Then with the help of two parallel lines strategies and upper point intersections

lines we make 2*2 matrix we find the value of game and strategies.

m*2

Steps to solve:

For m*2 game, first draw to vertical parallel lines.

Let line first be the first column and line second be the second column.

Now plot scale on these lines.

Game Theory

59

Then plot the element of first row and first column on first line and the element

of first row and second column on second line. Similarly second column third

column so on.

We find the upper envelop and lower point and intersections lines of lower point.

Then with the help of two parallel lines strategies and lower point intersections

lines we make 2*2 matrix we find the value of game and strategies.

ARITHMETIC METHOD

Steps to solve

In the game matrix of m*2 and 2*n, make various combinations of 2*2 matrix.

Now out of these combinations we calculate value of game, the maximum value

of game in case of m*2 and minimum value of game in case of 2*n, we take and

their corresponding strategies.

APPLICATIONS OF GAME THEORY

Used by poker and chess player to win their games.

Army generals make use of this technique to plan war strategies.

Used to analyze a board range of activities, including dating and mating

strategies, legal and political negotiations, and economic behavior.

The game theory technique is designed to evaluate situations where individuals

and organizations can have conflicting objectives.

In wage negotiations between unions and firms

SIGNIFICANCE

IT IS BASED ON TWO BASIC ASSUMPTIONS-

A least in two person zero sum game this theory outlines a scientific quantitative

technique which can be fruitfully used by players to arrive at an optimum

strategy, given firms objective

Game theory gives insight into several less known aspects which arise in

situation of conflicting interest.

Game Theory

60

LIMITATIONS

The assumptions that the player makes is unrealistic as he can only make a

guess of his own and his rival‘s strategy.

It becomes complex and difficult as the number of players increases in the game

The assumption made does not seem practical.

Mixed strategies are also not very helpful.

Example: Suppose there are two players A and B with different strategies.

A

a1 a2 a3 a4

b1 20 15 12 35

b2 25 14 08 10

B b3 40 02 10 05

b4 - 5 04 11 00

Now make a square on the largest number column wise and encircle the lowest

number row wise.

A

a1 a2 a3 a4

b1 20 15 12 35

b2 25 14 08 10

B b3 40 02 10 05

b4 - 5 04 11 00

Thus,

Saddle point is at no. 12

Value of game= 12

B(1,0,0,0) and A(0,0,1,0) Ans-

Now let's take another example:

Suppose there are two competitors Reliance and Airtel with different strategies

against each other.

Game Theory

61

Reliance

R1 R2 R3

A1 10 09 10

Airtel A2 15 10 16

A3 20 12 13

Now make square around largest number column wise and encircle the smallest

number rpw wise.

Solution :

Reliance

R1 R2 R3

A1 10 09 10

Airtel A2 15 10 16

A3 20 12 13

So, Value of game= 12

Reliance(0,1,0)

Airtel (0,0,1) Ans-

1. Problem without saddle point: When there is no saddle point then we require to

apply the law of dominance.

Steps to solve:

1. Apply law of dominance- In this we remove the highest column and remove the

lowest row compared to other columns and rows.

2. If we get a 2/2 matrix then difference of the column will be multiplied with the

columns and difference of the rows will be multiplied by rows.

Case of Henry and Dave

Suppose there are two brokers accused of fraudulent trading activities: Dave and

Henry. Both Dave and Henry are being interrogated separately and do not know what

the other is saying. Both brokers want to minimize the amount of time spent in jail

and here lies the dilemma. The sentences vary as follows:

Game Theory

62

Henry

A1 A3

B1 02 05 2/5

B2 05 03 3/5

2/5 3/5 5

V= 2x2+5x3 = 19

5 5

Strategies

Henry (2/5,3/5)

Dave (2/5,3/5) Ans-

1) If Dave pleads not guilty and Henry confesses, Henry will receive the minimum

sentence of one year, and Dave will have to stay in jail for the maximum

sentence of five years.

2) If nobody makes any implications they will both receive a sentence of two

years.

3) If both decide to plead guilty and implicate their partner, they will both receive a

sentence of three years.

Dave

Game Theory

63

4) If Henry pleads not guilty and Dave confesses, Dave will receive the minimum

sentence of one year, and Henry will have to stay in jail for the maximum five

years.

Obviously, pleading guilty is the most attractive should the other plead not guilty

since the sentence is only one year. However, if the other party also chooses to plead

guilty, both will have to serve three years. On the other hand, if both parties plead not

guilty, they'd have to serve two years in jail. Consequently, the risk of pleading not

guilty is a five-year sentence, should the other choose to confess.

Example: There are two player A and B with their a different strategies:

A

A1 A2 A3

B1 30 20 15

B B2 25 18 10

B3 19 15 25

SOLUTION :-

A

A1 A2 A3

B1 30 20 15

B B2 25 18 10

B3 19 15 25

B1 B2 B3

A1 30 20 15

A2 25 20 15

A3 19 15 25

Game Theory

64

B2 B3

A1 20 15 10/15

A3 15 25 5/15

10/15 5/15 15

V = 20X10 + 15X5 = 275

15 15

V = 15X10 + 25X5 = 275

15 15

B ( 0, 10/15, 5/15)

A ( 10/15, 0, 5/15) Ans-

EXAMPLE 2: There are two competitors Ramesh and Suresh having different

strategies.

Suresh

S1 S2 S3

R1 10 5 -2

Ramesh R2 13 12 15

R3 16 14 10

Suresh

S1 S2 S3

R1 10 5 -2

Ramesh R2 13 12 15

R3 16 14 10

Game Theory

65

S2 S3

R1 12 15 4/7

R3 14 10 3/7

5/7 2/7 7

V = 12X4+ 14x3 = 90 = 12.85

7 7

V = 12X5+ 15X2 = 90 = 12.85

7 7

Strategies:

S(0,5/7,2/7)

R(4/7,0,3/7) ANS-

Example 2: The two car companies Maruti and Hyundai are competing with each

other having different strategies.

Maruti

M1 M2 M3 M4

Hundai H1 18 13 5 10

H2 7 14 15 11

H3 6 3 14 16

Maruti

M1 M2 M3 M4

Hundai H1 18 13 5 10

H2 7 14 15 11

H3 6 3 14 16

Game Theory

66

Maruti

M1 M2 M3

Hundai H1 18 13 5

H2 7 14 15

Maruti

M1 M2

Hundai H1 18 13 7

H2 7 14 5

1 11 12

V =18x7 +7x5 = 161 = 13.4

12 12

Maruti

M1 M3

Hundai H1 18 5 8/21

H2 7 15 13/21

10/21 11/21 21

V =18x8 +7x13 = 235 = 11.19

21 21

In this case we will take the minimum value of the game that is 11.19

Strategies:

Maruti ( 10/21,0,11/21,0)

Hyundai ( 8/21,13/21,0) Ans-

Graphical Representation of the above problem:

Maruti

M1 M2 M3

Hundai H1 18 13 5

H2 7 14 15

H1 H2

Game Theory

67

18

18

17

17 M3

16

16

15

15

M2

14

14

13

13

12

12

11

11

10

10

9

9

8

8

V=11.19

7

7

6

6

5

5

4

4 M1

3

3

2

2

1

1

Game Theory

68

Graphic Method at a glance:

Row × Column Value of game Envelope Intersecting point

2 × m minimum Lower Highest

m × 2 Maximum Upper Lowest

Practice Problems:

1). There are two players Rafael Nadal and Roger Federer. They both have

different strategies. Solve the game problem and find the Value of the game.

Rafa

R1 R2 R3

F1 20 28 14

Fedex F2 16 15 11

F3 24 19 12

Ans. Value of game=14

2). There are two players Lisa and Mona with different strategies. Find the value of

game.

Lisa

L1 L2 L3 L4

M1 12 20 15 28

Mona M2 15 16 11 15

M3 16 25 17 26

Ans: Value of game=16

3). Two competitors X and Y are competing with each other.Both have different

strategies against each other. Find value of game.

Player x

X1 X2 X3

Y1 16 25 13

Player y Y2 12 20 7

Y3 8 10 40

Ans: Value of the Game= 15.31

Game Theory

69

4). There are two players A and B having different strategies against each other.

Find the Value of game.

Player B

B1 B2 B3

A1 1 7 2

Player A A2 6 2 7

A3 5 1 6

Ans: Value of game= 4

<

5). There are two players Gillette and Vi John. Both have different strategies to

acquire better market share. Find the Value of Game. Show graphical

representation.

Gillette

G1 G2 G3

V1 07 16 12

Vi John V2 09 17 19

V3 14 16 08

V4 12 15 11

Ans: Value of Game=12.125

6). Times of India and The Hindu are competing with each other having different

strategies to win the market. Find the value of game and give the graphical

representation.

TOI

T1 T2 T3 T4 T5

H1 03 06 03 07 09

Hindu H2 07 09 02 02 06

H3 09 10 07 04 15

Ans: Value of game= 5.28

Replacement Theory

70

REPLACEMENT THEORY

The aim of the replacement theory was to show that quantitative analysis could help

decision making within an organisation.

INTRODUCTION

Replacement theory is concerned with the problem of machinery, men or equipment

is one which arises in every organisation. In any organisation, sooner or later all

equipment needs to be replaced. Suppose, an organisation purchase an equipment and

after few years or within a month it become less effective or useless due to either

sudden or gradual deterioration in their efficiency, failure or breakdown and needs

substantial expenditure on its maintenance. The problem, in such situation is, to

determine the best policy to be adopted with respect to replacement of the equipment.

The replacement theory provides answer to the following three types of situation in

terms of optimal replacement period.

a. Items such as machines, vehicles, tyres, etc. whose efficiency deteriorates with

age due to constant use and which need increased operating and maintenance

cost. In such cases deterioration level is predicted and is represented by (a)

increased maintenance/operation cost, (b) its waste or scrap value and damage to

them and safety risk.

b. It such as light bulbs and tubes, radio and television parts, etc. which do not give

any indication of deterioration with time but fail all of sudden and become

completely useless. Such cases requires an anticipation of failure involving

probabilities if failure. The optimum replacement policy is formulated to

balance the wasted life of items replaced before against the costs incurred when

items fail in service.

c. In respect of replacement of employee in an organisation gradually reduces due

to retirement, death, retrenchment and other reasons.

DEFINITION: “Replacement theory deals with the analysis of assets/equipment

which deteriorates with time and fix the optimal time of their replacement so that

the total cost is the minimum.”

Advantage of Replacement Theory:

1. It helps in decision making within an organisation.

2. It deals with the analysis of men, machines, equipment, etc which deteriorates

with time and fix the optimal time to solve the issue.

3. By fixing the optimal time of the organisation with the help of replacement, it

reduces the cost of maintenance and increased the production.

Replacement Theory

71

4. It helps management to taking decision when the old item has deteriorate and

requires expensive maintenance.

5. It helps the management to taking the decision whether to replace now the

expensive item which has deteriorate, or if not, when to reconsider replacement

of the item in question.

Limitation of Replacement Theory:

1. Future costs and resale prices of an equipment are predictable is quite

Unrealistic.

2. Replacement theories in general are likely to be random variables calling for

probabilistic approach to the analysis.

3. The historical costs in respect of an equipment requiring replacement are

appropriate estimates of the costs for the equipment by which it is sought to be

replaced may not hold.

4. Maintenance costs do not always follow a smooth pattern over a time. Very

often, they are incurred in discrete lumps caused by such things as, for example,

installing a new gearbox in van or reconditioning the body of the truck. In such

cases, the best policy might be to be decide whether to scrap the vehicles (or

other equipment) or undertake the major repairs to work and use it further.

5. In respect of the group replacement decision in case of items that fail suddenly,

a practical objection is that all the items are replaced when group replacement

take place whether they are working or not even if they are the ones that were

installed only recently the ones that were.

6. The replacement could be carried out only at certain fixed time at the year ends

or weekends.

7. In reality, the replacement can be, and is, done at any point during a given time.

There are two types of items that lend themselves to replacement application.

a. Items that deteriorate with age.

b. Items that fail completely.

When consideration items that deteriorates with age, chief elements of concern are the

factors of:

a. Increased operating cost,

b. Idle time, and

c. Increased repair cost.

Replacement Theory

72

Replacement models for the above situation should determine:

Timing – When the equipment should be replaced.

Selection – The type of replacement.

Timing is complicated by decreasing salvage value, increasing Operating and

maintenance cost and technological innovation. In essence, the problem also becomes

that of estimating future cost, value and developments.

There are given below graph of two possible approaches to the problem. First,

equipment may be replaced when its performance declines to the point where it is no

longer acceptable, the out put may be too low, quality too poor, breakdowns too

frequent, etc. A drawback of this approach is that its response is too late as the

equipment is already unsatisfactory. A better alternative would be to analyse costs and

keep the equipment operating for the specific time which minimize to total costs.

Performance of new equipment

Minimal acceptable performance

Replacement Replacement

Time

Performance (a)

Operating cost of

new equipment

Cost of operating

Replacement Replacement

Time

(b)

Replacement Theory

73

Performance varying with age: (a) Performance deteriorating with time until

replacement when a minimal acceptable is reached; (b) Increasing operating cost over

time minimized by replacement policy.

Types of Failure

There are two types of failure are discussed below;

Gradual Failure: It is a progressive in nature. That is, as the life of an item increases,

its operational efficiency also deteriorates resulting in

a. Increased the running cost.

b. Decreased in its productivity

c. Decreased in the resale or salvage value

Mechanical items like pistons, rings, bearing tyres etc. fall under this category.

Sudden Failure: This type of failure occurs in terms after some period of giving

desired service rather than deterioration while in service. The period of giving desired

services is not constant but follows some frequency distribution which may be

progressive, retrogressive or random in nature.

Types of replacement

1. Replacement of equipment which deteriorates with time

2. Replacements of items that fail completely

3. Staffing problems

1. Replacement of equipment which deteriorates with time

Replacement decisions of such items are based on the economic life cycle cost

concept or average monthly/ annual, in economic life cycle of, maintenance and

operating cost in general increase with time and a stage come when these costs

become so large that it is better to replace the item with new one. Some times in a

Failure

Gradual failure

Sudden failure

Replacement Theory

74

system there are a number of alternatives in which we have to make comparison

between the choices on the basis of cost involvement.

2. Replacement of items that fail completely.

The second type of replacement problem is concerned with items that either work or

fail completely. It is sometimes more economical to replace the group as a whole even

if some of the items are functioning satisfactorily then to replace each item as it fails.

Failures items have to be replace as they occur but it may be profitable at some stage.

It is of two types:-

a) Individual Replacement

b) Group Replacement

3. Staff replacement

The staff of an organization calls for replacement because people leave the

organization due to inefficiency, resignation, retirement or death of employee from

time to time. Therefore to maintain suitable strength of employee in system there is a

need to formulate some recruitment policy. For this we assume that life distribution of

the staff in a system is known to us.

Steps for solving the problem;

Step 1: Find the present value of the maintenance cost for each of the years, by

multiplying the cost value by an appropriate present value factor (based on the given

rate of discount and the time).

Step 2: Accumulate present values obtained in step 1 up to each of the years 1, 2, 3...

Add the cost of the equipment to each of these values.

Step 3: Accumulate present value factors up to each of the year 1, 2, 3.…

Step 4: Divide the cost plus cumulative maintenance costs for each year, obtained

instep 2, by the corresponding cumulative present value factors in step 3. This gives

the annualized costs for the various years. It may be noted that the annualized cost is

also the weighted average of costs- the price value (in short PV) factors serve as

weight here and the average is calculated as in the usual way, by dividing the

summation of the products of the costs and their respective weights by the summation

of the weights.

KEY CONCEPTS:

Year of Service- It is the year from purchasing year of the equipment till the year we

are taking the service from it.

Replacement Theory

75

Running cost- It is the day to day cost incurred in operating equipment or performing

unit.

Cumulative running cost- It is the cost which become greater by stages or increasing

by successive addition of running cost.

Depreciation cost- It is the cost when there is a decrease in price or value of

equipment due to its obsolescence or use.

Total cost- It is the sum of cumulative cost and depreciation cost.

Average cost- It is the total cost for all units produced divided by the year of service.

Illustration: 1

A machine owner finds from his past records that costs pre year of maintaining a

machine whose purchase price in ` 6,000 in as follows:

Year 1 2 3 4 5 6 7 8

Maintenance

Cost 1000 1200 1400 1800 2300 2800 3400 4000

Resale Price 3000 1500 1500 375 200 200 200 2000

Determine at what age machine should be replaced.

Solution: Purchase Price (C) = ` 6,000

Determination of optimal replacement period

Year Maintenance

cost

Cumulative

Maintenance cost

Resale

Price

(S)

Capital

Loss

C –S

Total

Cost

(T)

Average

Cost

T/Year

1

2

3

4

5

6

7

8

1,000

1,200

1,400

1,800

2,300

2,800

3,400

4,000

1,000

2,200

3,600

5,400

7,700

10,500

13,900

17,900

3,000

1,500

1,500

375

200

200

200

2,000

3,000

4,500

5,250

5,625

5,800

5,800

5,800

4,000

4,000

6,700

8,050

11,025

13,500

16,300

19,700

23,700

4,000

3,350

2,950

2,756.25

2,700

2,716.6

2,814.29

2,962.50

The above table shows that the lowest average cost per year is at the end of 5th year,

which is Rs. 2700.

Replacement Theory

76

Decision: Machine should be replaced after 5 yrs.

Illustration: 2

The data of the operating cost per year and resale prices of equipment A whose

purchase price is RS. 10000 are given here.

Year 1 2 3 4 5 6 7

Operating

Cost 1500 1900 2300 2900 3600 45000 55000

Resale

Value 5000 2500 1250 600 400 400 400

A: What is the optimum period for replacement?

B: When equipment A is 2 year old, equipment B , which is a new model for the

same usage, is available the optimum period for replacement is 4 year with an

AVG cost of RS.3600 should we change equipment A with that of B? If so

when?

Solution:

Year Maintenance

cost

Cumulative

Maintenance

cost

Capital

Loss

C –S

Total

Cost

(T)

Average

Cost

T/Year

1 1500 1500 5000 6500 6500.0

2 1900 3400 7500 10900 5450.0

3 2300 5700 8750 14450 4816.7

4 2900 8600 9400 18000 4500.0

5 3600 12200 9600 21800 4360.0*

6 4500 16700 9600 26300 4383.3

7 5500 22200 9600 31800 4542.9

Since the AVG. cost corresponding to the 5 yearly is the least, the optimal period of

replacement = 5 years.

(B) As the minimum average cost for equipment B is smaller than that for equipment.

A, it is product to change the equipment. To decide the time of change, we would

determine the cost of keeping the equipments in its 3rd

, 4th, 5

th year of life and

compare each of these values with RS.3600 (the AVG. cost of equipment B). The

equipment A shall be held as long as the marginal cost of holding it would be smaller

than the minimum AVG. Cost for equipment B. The calculations are given below.

Replacement Theory

77

YEAR OPERATING

COST DEPRECIATION TOTAL COST

3 2300 1250(2500-1250) 3550

4 2900 650(1250-600) 3550

5 3600 200(600-400) 3800

Illustration: 2

An airline requires 200 assistant hostesses, 300 hostesses, and 50 supervisors, Women

are recruited at the age of 21, and if still in service retire at 60. Given the following

life table, determine:

a) How many women should be recruited in each year?

b) At what age should promotion take place?

Airline Hostesses Life Record

Age

No in services

21

1,000

22

600

23

480

24

384

25

307

26

261

27

228

28

206

Age

No in services

29

190

30

181

31

173

32

167

33

161

34

155

35

150

36

146

Age

No in services

37

141

38

136

39

131

40

125

41

119

42

113

43

106

44

99

Age

No in services

45

93

46

87

47

80

48

73

49

66

50

59

51

53

52

46

Age

No in services

53

39

54

33

55

27

56

22

57

18

58

14

59

11

-

-

Solution:

If 1,000 women had been recruited each year for the past 39 years, then the total

number of them recruited at the age of 21 and those serving up to the age of 59 is

6,480. Total numbers of women recruited in the airline are: 200+300+50=550.

(a) Number of women to be recruited every year in order to maintain a strength of

550 hostesses

550 × (1,000/6,480) = 85 approx.

(b) If the assistant hostesses are promoted at the age of

X, then up to age (x - 1), 200 assistant hostesses will be required. Among 550 women,

200 are assistant hostesses. Therefore, out of strength of 1,000 there will be:

Replacement Theory

78

200 × (1,000/550) = 364 assistant hostesses.

But from the life table given in the question, this number is available up to the age of

24 years. Thus, the promotion of assistant hostesses is due in the 25th year.

Since out of 550 recruitments only 300 hostesses are needed, if 1,000 girls are

recruited, then only 1000 × (300/500) = 545 (approx). will be hostesses.

Hence, total number of hostesses and assistant hostesses in recruitment of 100 will be:

545 + 364 = 909.

This means, only 1,000 – 909 = 91 supervisors are required. But from life table this

number is available up to the age of 46 years. Thus promotion of hostesses to

supervisors will be due in 47th year.

ILLUSTRATION 3: A truck owner finds from his past records that maintained cost

per year of a truck whose purchase price is Rs. 80000 are given below:

Year 1 2 3 4 5 6 7 8

Maintenance

cost 10000 13000 17000 22000 29000 38000 48000 60000

Resale price 40000 20000 12000 6000 5000 4000 4000 4000

Find the optimal replacement period.

SOLUTION:

Using the given data the minimum average annual cost of the equipment is computed

in the table:

Year Running cost

M

Total

running cost

∑Mi

Depreciation

C-S

Total cost

C-S+∑Mi

Average

annual cost

1 10000 10000 40000 50000 50000

2 13000 23000 60000 83000 41500

3 17000 40000 68000 108000 36000

4 22000 62000 74000 166000 34000

5 29000 91000 75000 205000 33200

6 31000 129000 76000

136000 34167

7 48000 177000 76000 253000 36143

8 60000 237000 76000 313000 39125

Replacement Theory

79

Clearly the minimum average cost is in fifth year and is Rs. 33200. Hence the

equipments must be replaced after 5 years

ILLUSTRATION 4: A plant manager is considering replacement policy for a new

machine. He estimates the following:

Year

Cost in Rupees

Replacement Cost at

the beginning of the

year

Salvage Value at the

end of the Year (S) Operating Cost

1 30,000 18,000 7,500

2 33,000 15,000 9,000

3 37,500 12,000 12,000

4 42,000 7,500 15,000

5 48,000 3,000 19,500

6 57,000 0 24,000

SOLUTION:

Table for different replacement costs will be as under:

Year Maintenance

Cost

Cumulative

Maintenance

Cost ∑Mi

Replacement

Cost at the

beginning of

the year R

Capital

Loss R-S

(Salvage

Value)

Total Cost

∑Mi + R-S

Average

Cost

∑Mi-R-

S/i

1 7,500 7,500 30,000 12,000 19,500 19,500

2 9,000 16,500 33,000 18,000 34,500 17,250

3 12,000 28,500 37,500 25,500 54,000 18,000

4 15,000 43,500 42,000 34,500 78,000 19,500

5 19,500 63,000 48,000 45,000 1,08,000 21,600

6 24,000 87,000 57,000 57,000 1,44,000 24,000

Since the average cost for 2nd

year (17250) is minimum the optimal replacement

period is two year.

Replacement Theory

80

Numerical Question:

1. The following table gives the running costs per year and resale values of certain

equipment whose purchase price is Rs6500. At what year is the replacement are

optimally.

Year 1 2 3 4 5 6 7 8

Running Cost 1400 1500 1700 2000 2400 2800 3300 3800

Resale Price 4000 3000 2200 1700 1300 1000 1000 1000

(Ans: After 5 years)

2. The cost of a machine is ` 10,000 and the other relevant information is given

below:

Period Running Cost Resale Price

1 1000 7000

2 1200 5000

3 1400 4750

4 1800 4375

5 2300 4200

6 2800 4200

7 3400 4200

8 4000 4200

Find the optimal replacement period for the machine.

(Ans: 5yrs optimal replacement cost per year ` 2700.)

3. The Cost of a Scooter is ` 30000 and other details are given below:

Year 1 2 3 4 5 6

Maintenance Cost 300 1000 1500 3000 4800 6000

Resale Value 20000 15300 12550 11000 6100 3400

Find the Optimal replacement period for the scooter.

(Ans: 4years)

Replacement Theory

81

4. A machine owner finds from his past records that costs pre year of maintaining a

machine whose purchase price in Rs. 6000 in as follows:

Year 1 2 3 4 5 6 7 8

Maintenance

cost 1000 1200 1400 1600 2300 2800 3400 4000

Resale price 3000 1500 1500 315 200 200 200 2000

Determine at what age machine should be replace?

5. A Fleet owner finds from his past records that the cost per year of running a

vehicle, whose purchase price is RS 50,000 is:

YEAR 1 2 3 4 5 6 7

RUNNING

COST (RS) 5000 6000 7000 9000 11500 16000 18000

RESALE

VALUE(RS) 30000 15000 7500 3750 2000 2000 2000

Thereafter, running cost increases by Rs 2000, but resale value remains constant

at Rs 2000. At what age is a replacement due?

6. The cost of a machine is Rs 6100 and it‘s scrap value is only Rs 100. The

maintenance cost are found from experience to be:

Year 1 2 3 4 5 6 7 8

Maintenance

cost(Rs) 100 250 400 600 900 1250 1600 2000

When should the machine be replaced?

Queuing Theory

82

QUEUING THEORY

INTRODUCTION

Queuing theory deals with problems that involve waiting (or queuing). It is quite

common that instances of queue occurs everyday in our daily life. Examples of

queues or long waiting lines might be

Waiting for service in bank and at reservation counter.

Waiting for a train or bus.

Waiting at barber saloon.

Waiting at doctors‘ clinic.

Whenever a customer arrives at a service facility, some of them usually have to wait

before they receive the desired service. This form a queue or waiting line and

customer feel discomfort either mentally or physically because of long waiting queue.

We infer that queues from because the service facilities are inadequate. If service

facilities are increased, then the question arise how much to increase? For example,

how many buses would be needed to avoid queues? How many reservation counters

would be needed to reduce the queue? Increase in number of buses and reservation

counters requires additional resources. At the same time, cost due to customer

dissatisfaction must also be considered.

Symbols and notations:

n = total number of customers in the system, both waiting and in service

µ = average number of customers being serviced per unit of time.

λ = average number of customers arriving per unit of time.

C = number of parallel service channels

Ls or E(n) = average number of customers in the system, both waiting in the service.

Lq or E(m) = average number of customers waiting in the queue

Ws or E(w) = average waiting time of a customer in the system both waiting and in

service

Wq or E(w) = average waiting time of a customer in the queue

Pn (t = probability that there are n customer in the queue

Queuing Theory

83

Queuing system

The customers arrive at service counter (single or in a group) and attended by one or

more servers. A customer served leaves the system after getting the service. In

general, a queuing system comprise with two components, the queue and the service

facility. The queue is where the customers are waiting to be served. The service

facility is customers being served and the individual service stations.

SERVICE SYSTEM

The service is provided by a service facility (or facilities). This may be a person (a

bank teller, a barber, a machine (elevator, gasoline pump), or a space (airport runway,

parking lot, hospital bed), to mention just a few. A service facility may include one

person or several people operating as a team.

There are two aspects of a service system—(a) the configuration of the service system

and (b) the speed of the service.

Configuration of the service system

The customers‘ entry into the service system depends upon the queue conditions. If at

the time of customers‘ arrival, the server is idle, then the customer is served

immediately. Otherwise the customer is asked to join the queue, which can have

several configurations. By configuration of the service system we mean how the

cost of waiting

cost of service

Level of Service Fast Low

High

Total cost of the service

service

Cost

Low optical service level

Queuing Theory

84

service facilities exist. Service systems are usually classified in terms of their number

of channels, or numbers of servers.

Single Server – Single Queue

The models that involve one queue – one service station facility are called single

server models where customer waits till the service point is ready to take him for

servicing. Students‘ arriving at a library counter is an example of a single server

facility.

Several (Parallel) Servers – Single Queue

In this type of model there is more than one server and each server provides the same

type of facility. The customers wait in a single queue until one of the service channels

is ready to take them in for servicing

Several Servers – Several Queues

This type of model consists of several servers where each of the servers has a

different queue. Different cash counters in an electricity office where the customers

can make payment in respect of their electricity bills provide an example of this type

of model.

Queuing Theory

85

Service facilities in a series

In this, a customer enters the first station and gets a portion of service and then moves

on to the next station, gets some service and then again moves on to the next station.

…. and so on, and finally leaves the system, having received the complete service. For

example, machining of a certain steel item may consist of cutting, turning, knurling,

drilling, grinding, and packaging operations, each of which is performed by a single

server in a series. Service Facility.

Characteristics of Queuing System

In designing a good queuing system, it is necessary to have a good.

Information about the model. The characteristic listed below would

Provide sufficient information.

1. The Arrival pattern.

2. The service mechanism.

3. The queue discipline.

4. The number of service channels.

5. Number of Service Stages

1. The Arrival pattern.

Arrivals can be measured as the arrival rate or the interarrival time (time between

arrivals).

Queuing Theory

86

Interarrival time =1/ arrival rate

These quantities may be deterministic or stochastic (given by a propbability

distribution). Arrivals may also come in batches of multiple customers, which is

called batch or bulk arrivals. The batch size may be either deterministic or stochastic.

(i) Balking: The customer may decide not to enter the queue upon Arrival, perhaps

because it is too long.

(ii) Reneging: The customer may decide to leave the queue after waiting a certain

time in it.

(iii) Jockeying: If there are multiple queues in parallel the customers may switch

between them.

(iv) Drop-o®s: Customers may be dropped from the queue for reasons outside of

their control. (This can be viewed as a generalisation of reneging.)

2. Service Pattern

As with arrival patterns, service patterns may be deterministic or stochastic. There

may also be batched services. The service rate may be state-dependent. (This is the

analoge of impatience with arrivals.)

Note that there is an important di®erence between arrivals and services. Services do

not occur when the queue is empty (i.e. in this case it is a no-op).

3. Queue Discipline

This is the manner by which customers are selected for service.

(i) First in First Out (FIFO).

(ii) Last in First Out (LIFO), also called

(iii) Service in Random Order (SIRO).

(iv) Priority Schemes. Priority schemes are either:

Preemptive: A customer of higher priority immediately displaces any customers of

lower priority already in service. The displaced customer's service may be either

resumed from where it was left o®, or started a new.

Non-Preemptive: Customers with higher priority wait current service completes,

before being served.

Queuing Theory

87

4. The number of service channels

5. Number of Service Stages

Customers are served by multiple servers in series.

In general, a multistage queue may be a complex network with feedback

Application of queuing theory:

Queing theory has been applied to a great variety of business situations. Here we

shall discuss a few problem s where the theory may be applied-

1) Waiting line theory can be applied to be determine the number of checkout

counters needed to secure smooth and economic operations of its stored at

various time during the day of a super market or a departmental store .

2) Waiting line theory can be used to analyze the delays at the toll booths of

bridges and tunnels.

Queuing Theory

88

3) Waiting line theory can be used to improve the customers service at restaurants,

cafeteria, gasoline service station, airline counters, hospitals etc,

4) Waiting line theory can be used to determine the proper determine the proper

number docks to be constructed in the building of terminal facilities for trucks

&ships.

5) Several manufacturing firms have attacked the problems of machine break down

& repairs by utilizing this theory. Waiting line theory can be used to determine

the number of personnal to be employed so that the cost of the production loss

from down time & the cost f repairman is minimized.

6) Queuing theory has been extended to study a wage incentive plan

Queuing theory (Limitations)

1) Most of the queuing models are quite complex & cannot be easily understood.

2) Many times form of theoretical distribution applicable to given queuing

situations is not known.

3) If the queuing discipline is not in‖ first in, first out‖, the study of queuing

problems become more difficult.

BASIC POINTS

Customer: (Arrival)

The arrival unit that requires some services to performed.

Queue: The number of Customer waiting to be served.

Arrival Rate (λ): The rate which customer arrive to the service station.

Service rate (µ): The rate at which the service unit can provide services to the

customer

If Utilization Ratio Or Traffic intensity i.e λ /µ

λ / µ > 1 Queue is growing without end.

λ / µ < 1 Length of Queue is go on diminishing.

λ /µ = 1 Queue length remain constant.

When Arrival Rate (λ) is less than Service rate (µ) the system is working.

i.e λ< µ (system work)

Queuing Theory

89

Formulas

µ=Service Rate

λ= Arrival Rate

1. Traffic Intensity (P)= λ /µ

2. Probability Of System Is Ideal (P0) =1-P

P0 = 1- λ /µ

3. Expected Waiting Time In The System (Ws) = 1/ (µ- λ)

4. Expected Waiting Time In Quie (Wq) = λ / µ(µ- λ)

5. Expected Number Of Customer In The System (Ls)= λ / µ(µ-λ)

Ls=Length Of System

6. Expected Number Of Customers In The Quie (Lq)= λ 2/ µ(µ- λ)

7. Expected Length Of Non-Empty Quie (Lneq)= µ/ (µ- λ)

8. What is the Probability Track That K Or More Than K Customers In The

System.

P >=K (P Is Greater Than Equal To K)

= (λ /µ)K

9. What Is The Probability That More Than K Customers Are In The System

(P>K)= (λ /µ)K+1

10. What Is The Probability That Atleast One Customer Is Standing In Quie.

P=K=(λ /µ)2

11. What Is The Probability That Atleast Two Customer In The System

P=K=(λ /µ)2

Solved Example

Question 1. People arrive at a cinema ticket booth in a poison distributed arrival rate

of 25per hour. Service rate is exponentially distributed with an average time of 2 per

min.

Calculate the mean number in the waiting line, the mean waiting time, the mean

number in the system , the mean time in the system and the utilization factor?

Solution:

Arrival rate λ=25/hr

Service rate µ= 2/min=30/hr

Queuing Theory

90

Length of Queue (Lq) = λ 2/ µ(µ- λ)

= 252/(30(30-25))

=4.17 person

Expected Waiting Time In Quie (Wq) = λ / µ(µ- λ)

=25/(30(30-25))

=1/6 hr= 10 min

Expected Waiting Time In The System (Ws) = 1/ (µ- λ)

=1/(30-25)

=1/5hr= 12 min

Utilization Ratio = λ /µ

=25/30

=0.8334 = 83.34%

Question 2. Assume that at a bank teller window the customer arrives at a average rate

of 20 per hour according to Poisson distribution .Assume also that the bank teller

spends an distributed customers who arrive from an infinite population are served on

a first come first services basis and there is no limit to possible queue length.

1. What is the value of utilization factor?

2. What is the expected waiting time in the system per customer?

3. What is the probability of zero customer in the system?

Solution:

Arrival rate λ=20 customer per hour

Service rate µ= 30 customer per hour

1. Utilization Ratio = λ /µ

= 20/30 = 2/3

2. Expected Waiting Time In The System (Ws) = 1/ (µ- λ)

=1/(30-20)

=1/10 hour = 6 min

3. Probability of zero customers in the system P0 = 1 – P

=1- 2/3 = 1/3

Queuing Theory

91

Question 3 : Abc company has one hob regrinding machine. The hobs needing

grinding are sent from company‘s tool crib to this machine which is operated one shift

per day of 8 hours duration. It takes on the average half an hour to regrind a hob. The

arrival of hobs is random with an average of 8 hobs per shift.

1. Calculate the present utilization of hob regrinding machine.

2. What is average time for the hob to be in the regrinding section?

3. The management is prepared to recruit another grinding operator when the

utilization of the machine increases to 80%. What should the arrival rate of hobs

then be?

Solution: Let us calculate arrival rate and service rate per shift of 8 hours.

Arrival rate λ=8 shift

Service rate µ=8x60/30=16 /shift

1. Percentage of the time the machine is busy

Pb =arrival rate/service rate=8/16=0.50=50%

2. Average time for the hob to be in the grinding section.

i.e., average time in the queue system=ws

ws = 1/( µ- λ)=1/16-8=1/8 shift=1/8x8=1 hour

3. Let λ‘

=arrival rate for which utilization of the machine will be 80%,

Therefore, Pb‘

= λ‘ / µ

i.e., λ‘ =

Pb‘

. µ=0.80x16=12.8 per shift.

Question: 4

(a) calculate expected number of persons in the system if average waiting time pf a

customer is 45 or more than 45 minutes .

b) if service rate is same.

c) if arrival rate is same.

Solution:-(a)expected no. of persons in a system(Ls ) =λ/μ- λ

=45/65-45

=9/4

=3/4=1/65- λ

λ =191/3

(b)Ws = 1/ μ –λ=1/65-45

=1/20 x60/1=3 mins.

Queuing Theory

92

(c)ws =1/ μ- λ=1/6-4

= 3/4=1/ μ-45

=3 μ-135=4

=3 μ=139

μ =46.33

Question 5: In a factory, the machines break down and require service according to a

poission Distribuation at the average of per day. What is the probability that exactly

six Machines.

Solution : Given λ = 4, n = 6, t = 2 p

P(n,t) = (6,4) when λ = 4

We know, p (n,t) = (λt)n e-λt/ n!

p(6,2) = (4×2)6 e-4×2/ 6!

=86 e-8/720

=0.1221

Question 6: On an average , 6 customer arrive in a coffee shop per hour. Determine

the probability that Exactly 3 customers will reach in a 30 minute period, assuming

that the arrivals follow poison Distribution.

Solution:

Given, λ = 6 customers / hour

t = 30 minutes = 0.5 hour

n = 2

we know, p(n,t) = (λt)n e-λt/n!

p(6,2) = (6×0.5)2 e-6×0.5/2!

= 0.22404

Question 7: In a bank with a single sever, there are two chairs for waiting customers.

On an average one customer arrives 12 minutes and each customer takes 6 minutes for

getting served. Make suitable assumption, find

(i) The probability that an arrival will get a chair to sit on,

(ii) The probability that an arrival will have to stand, and

(iii) Expected waiting time of a customer.

Queuing Theory

93

Solution following assumption are made for solving the given queuing problem:

1. The arrival rate is randomly distributed according to poission distribution.

2. The mean value of the arrival rate is λ .

3. The services time distribution approximated by an exponiential distribution and

a mean rate of services is μ.

4. The rate of services is greather than the rate of arrival (μ>λ)

5. The queue discipline id FIFO.

Arrival rate λ= 12min or 5 customer / hr

Services rate μ = 6 min or 10 customer/ hr

λ/μ = 5/10 = ½

there are two chairs including services one.

(i) The probality that an arrival get a chair to seat on is given by:

Pn (n<=2)= 1- Pn(n>2)

1-(λ/μ)3

1-(1/2)3 = 7/8

(II) The probability that an arrival will have to stand is given by

1-(P0+p1+P2)

= 1-(7/8)= 1/8

(III) Expected waiting time of a customer in the queue is given by

Wq =λ/μ(μ-λ)

=5/10(10-5) = 1/(2*5) hr = 6 min

Question 8: A television repairman finds that the time spent on his jobs has an

expontial distribution with a mean of 30 minutes. If he repairs sets in the order in which

they came in, and if the arrival of sets follow a passion distribution approximately with

an average rate of 10 per 8- hour day, what is the repairman‘s expected idle time each

day? How many jobs are ahead of the average set just brought in?

Solution from data of problem, we have

Λ= 10/8=5/4 set per hour; and μ=(1/30)60= 2set per hour

Queuing Theory

94

(i) Expected idle time of repairmen each day

Number of hour for a repairman remains busy in 8 hour day( traffic intensity) is

given by

(8) (λ/μ)=(8) (5/8)= 5 hour

Hence, the idle time for a repairman in an 8 hour day will be : (8-5) =3 hour

(ii) Expected (or average) number of TV set in the system

LS = λ/μ-λ = 5/4/2-(5/4)

=5/3

=2 (APPROX) T.V sets

Unsolved question

Question 1: Calculate expected number of person in the system. If average waiting

time of customer is 30 min or more than 30 min , then services provider starts another

windows.

Calculate Arrival rate if service rate is same.

Calculate service rate if arrival rate is same.

(answer: Ws=1/5 hr,

λ =13

µ = 2)

Question 2: At a certain petrol pump, Customer arrive according to a passion process

with a average time at 5 min between the arrivals. The service time is exponential

distribution with mean 2 mins on the basic of this information.

Find out:-

a. Traffic intensity

b. What would be the average quieting length?

c. What is the expected number of customer at petrol pump?

d. What is the expected number time one spend at petrol pump?

e. What would we expected waiting time?

f. What would be the proportion time the petrol pump is idle?

Queuing Theory

95

Answer

a. 0.4

b. 0.26

c. 0.66

d. 0.02

e. 0.05

f. 0.6

<

Question 3. The machines in production shop breakdown at an average of 2 per hour.

The non productive time of any machine costs rs.30 per hour. If the cost of repairman

is Rs.50 per hour.

Calculate:

a. Number of machines not working at any point of time.

b. Average time that a machine is waiting for the repairman.

c. Cost of non-productive time of the machine operator.

d. Expected cost of system per hour.

Answer. a:: 2 machines

b: 2/3 hours

c: Rs. 60

d: Rs.110

Question 4. In a bank cheques are cashed at a single ‗teller‘ counter. Customers

arrived at the counter in a Poisson manner at an average rate of 30 customers /hour.

The teller takes on an average, a minute and a half to cash cheque. The service time

has been shown to be exponentially distributed

a) Calculate the percentage of time the teller is busy.

b) Calculate the average time a person is expected to wait.

Answer

a) 3/4

b) 6 minutes

Queuing Theory

96

Question 5 : Telephone users arrive at a booth following a Poisson distribution with

an average time of 5 minutes between one arrival and the next. The time taken for a

telephone call is on a average 3 minutes and it follows an exponential distribution.

What is the probability that the booth is busy? How many more booths should be

established to reduce the waiting time less than or equal to half of the present waiting

time.

Answer a) 0.6

b) wq=3/40hrs.

Question 6: Assume that goods trains are coming in a yard @ 30 trains per day and

suppose that the inter arrival times follow an exponential distribution . the service

time for each train is assumed to be exponential with an average of 36 minutes if the

yard can admit 9 trains at a time(there being 10 lines one of which is reserved for

shunting purpose).calculate the probability that the yard is empty and find the average

queue length.

Answer

λ=1/48

μ=λ/16

p=0.75

Po=o.28

Lq=1.55

Question 7. At what average rate must a clerk at a supermarket in order to ensure a

probability of 0.90 so that the customer will not wait longer than 12 minutes ? It is

assumed that there is only one counter at which customers arrive in a Poisson fashion

at an average rate of 15/hour. The length of service by the clerk has an exponential

distribution.

Answer: 2.48 minutes /service

Question 8. The beta company ‗s quality control deptt. Is managed by a single clerk,

who takes an average 5 minutes in checking part of each of the machine coming for

inspection. The machine arrive once in every 10 min. on the average one hour of the

machine is valued at Rs 25 and cost for the clerk is at rs 5 per hour. What are the

average hourly queueing system cost associated with the quality control department.

Answer Rs 30 per hour

Decision Theory

97

DECISION THEORY

The success or failure of an individual or organization experiences, depends upon the

ability to make appropriate decision. For making appropriate decision it requires

certain course of action or strategies which should be feasible (possible) and viable

(exists) in nature. Decision theory provides an analytical and systematic approach to

depict the expected result of a situation when alternative managerial actions and

outcomes are compared.

Decision theory is the combination of descriptive and prescriptive business modeling

approach i.e., it is concerned with identifying the best decision to take, assuming an

ideal decision maker who is fully informed, able to compute with perfect accuracy,

and fully rational. The practical application of this prescriptive approach (how people

actually make decisions) is called decision analysis, and aimed at finding tools,

methodologies and software to help people make better decisions which can be

classified as a degree of knowledge. The knowledge of degree is divided into four

categories which are given below:-

Characteristics of problem formulation :

A. Decision alternatives: In this case, N numbers of alternatives are available with

the decision maker whenever the decision is made. These alternatives may

depend on the previous decisions made. These alternatives are also called

courses of action which are under control and known to decision maker.

B. States of nature: These are the future conditions ( also known as consequences,

events, or scenarios) which are not under the control of decision maker. A state

of nature can be inflation, a weather condition, a political development etc. it

usually is not determined by an action of an individual or an organization. But it

may identify through some technique such as scenario analysis. Ex-

stakeholders, long-time managers.

Decision Theory

98

C. Pay off: A numerical outcome resulting from each possible combination of

alternatives and states of nature is called payoff. The payoff values are always

conditional values because of unknown states of nature. The payoff is measured

within a specified period (e.g. after one year). This period is sometimes called

decision horizon. Payoff can be measured in terms of money market share, or

other measures.

D. Pay off table: A tabular arrangement of these conditional outcomes (profit or

loss values) is known as payoff matrix. To construct a payoff matrix, the

decision alternatives (courses of action or strategies) and states of nature are

represented in the tabular form as below:

States of

nature(events)

Decision alternative ( courses of action)

A1 A2 A3 …… AM

E1 A11 A12 A13 …… A1m

E2 A21 A22 A23 …… A2m

E3 A31 A32 A33 …… A3m

….. ….. …… ….. …… ……

EN An1 An2 An3 …… Amn

Steps in decision theory approach

Identify and define the problem.

Listing of all possible future events, called states of nature. Such events are not

under control of decision maker.

Identification of all the courses of action which are available to the decision-

maker.

Evaluating the alternatives such as, cost effectiveness, performance, quality,

output, profit.

Expressing the pay-offs resulting from each pair of course of action and state of

nature.

Choosing an appropriate course of action from the given list on the basis of

some criterion that result in the optimal pay-off.

The next step is to implement the decision.

Decision Theory

99

Types of decision making environments

Decisions are made based upon the data available about the occurrence of events as

well as the decision situation. There are four types of decision making environment

which are follows:

1. Decision–making under certainty: In this type the decision maker has the

perfect information about the consequences of every course of action or

alternatives with certainty. Definitely he selects an alternatives that gives the

maximum return (pay-off) for the given state of nature. For ex- one have a

choices either to purchase national saving certificate, Indira Vikas Patra or

deposit in national saving scheme. Obviously he will invest in one the scheme

which will give him the assured return. In these decision model only one

possible state of nature exists.

2. Decision-making under risk: in this type, the decision maker has less

information about the certainty of the consequence of every course of action

because he is not sure about the return. In these decision model more than one

state of nature exists for which he makes an assumption of the probability with

each state of nature which will occur. For ex- probability of getting head in the

toss of a coin is 50%.

3. Decision-making under uncertainty: In this type, the decision-maker is unable

to predict the probabilities of the various states of nature which will occur. Here

the possible states of nature are known but still there is a less information than

Define the

problem

Gather information

Search for alternatives

Evaluates alternatives

Select alternative for action

Implement decision

Decision Theory

100

the decision under risk. For ex- The probability that Mr. Y will be the captain of

the Indian cricket team for coming 10 years from now is not known.

4. Decision-making under conflict: In this type, the consequences of each act of

the decision maker are influenced by the acts of opponent. An example of this is

the situation of conflict involving two or more competitors marketing the same

product. The technique used to solve this category is the game theory.

Decision Making under risk: Decision under risk is a probabilistic decision

situation, in which more than one state of nature exists and the decision maker has

sufficient information to assign probability values to the likely occurrence of each of

these states. Knowing the probability distribution of the states of nature, the best

decision is to select the course of action which has the largest expected payoff value.

The expected (average) payoff of an alternative is the sum of all possible payoffs of

that alternative weighted by the probabilities of those payoffs occurring.

The most widely used criterion for evaluating various courses of action under risk :

1. Expected Monetary Value (EMV) or Expected utility.

2. Expected opportunity Loss (EOL).

3. Expected value of Perfect Information (EVPI)

Expected monetary value: The expected value (EMV) for a given course is the

weighted sum of possible payoffs for each alternative. It is obtained by summing the

payoffs for each course of action multiplied by the probabilities associated with each

state of nature. The expected (or mean) value is the long-run average value that result

if the decision were repeated a large number of times.

Steps for calculating EMV: The various steps involved in the calculation of EMV

are as follow:

1. Construct a payoff matrix listing all possible courses of action and states of

nature. Enter the conditional payoff values associated with each possible

combination of course of action and state of nature along with probabilities of

the occurrence of each state of nature.

2. Calculate the EMV for each course of action by multiplying the conditional

payoffs by the associated probabilities and add these weighted values for each

course of action.

3. Select the course of action that yields the optimal EMV.

Expected opportunity loss (EOL) : An alternative approach to maximizing expected

monetary value (EMV) is to minimize the expected opportunity loss (EOL) also

called expected value of regret. The EOL is defined as the difference between the

Decision Theory

101

highest profit (highest payoffs) for a state of nature and the actual profit obtained for

the particular course of action taken. In other words, EOL is the amount of payoff that

is lost by not selecting the course of action that has greatest payoff for the state of

nature that actually occur. The course of action due to which EOL is minimum is

recommended.

Since EOL is an alternative decision criterion for decision making under risk,

therefore the results will always be the same as those obtained by EMV criterion.

Thus only one of the two methods should be applied to reach to a decision. It is stated

as follows:

Steps for calculating EOL: The steps which are involved in the calculation of EOL

are as follows:

1. Prepare a conditional profit table for each course of action and state of nature

combination along with the associated probabilities.

2. For each state of nature calculate the conditional opportunity loss (COL) values

by subtracting each payoff from the maximum payoff for that outcome.

3. Calculate EOL for each course of action by multiplying the probability of each

state of nature with the COL value and then adding the values.

4. Select a course of action for which the EOL value is minimum.

Expected value of perfect information (EVPI): In these decisions making under

risk each state of nature as associated with the probability of its occurrence. Perfect

information about the future demand would remove uncertainty for the problem. With

these perfect information the decision maker would know in advance exactly about

the future demand and he will be able to select a course of action that yields the

desired payoff for whatever state of nature that actually occurs.

EVPI represents the maximum amount the decision maker has to pay to get to this

additional information about the occurrence of various events.

EVPI = (expected profit with perfect information)- (expected profit without perfect

information).

Examples:

Q) A shopkeeper buys apple for Rs 20/kg and sells them for Rs 30/kg. The past

records of the sales are as follows:

No of customers: 50 80 100 120 150

No of Days: 20 30 20 10 20

Decision Theory

102

Ans: Profit =Selling price – Cost price.

= 30-20

= 10

Probability Demand Supply

50 80 100 120 150

20/100= 0.20 50 500 -100 -500 -900 -1500

30/100= 0.30 80 500 800 400 0 -600

20/100= 0.20 100 500 800 1000 600 0

10/100= 0.10 120 500 800 1000 1200 600

20/100= 0.20 150 500 800 1000 1200 1500

Total : 2500 3100 2900 2100 0

Maximum value : 500 800 1000 1200 1500

Minimum value : 500 -100 -500 -900 -1500

Maxi max: 1500 out of all maximum values.

Maxi min: 500 out of all minimum values.

Laplace:

2500/5 3100/5 2900/5 2100/5 0/5

= 500 = 620 = 580 = 420 = 0

620 in 80 units.

step 1. Write the demand & probability in column and supply in row.

Step 2. Calculate the Expected Pay off table (probability × pay off)

Supply

Probability Demand 50 80 100 120 150

20/100= 0.20 50 100 -20 -100 -180 -300

30/100= 0.30 80 150 240 120 0 -180

20/100= 0.20 100 100 160 200 120 0

10/100= 0.10 120 50 80 100 120 60

20/100= 0.20 150 100 160 200 240 300 620

A

Decision Theory

103

Total 500 620 520 300 -120

Expected monetary value (EMV): 620 (the highest one among the total).

Expected profit on perfect information (EPPI): (100+240+200+120+300) = 960.

Expected value of perfect competition (EVPI): 960 – 620 = 340.

Expected opportunity loss table: Deduct the highest number from expected

payoff table from each row.

Probability Demand Supply

50 80 100 120 150

20/100= 0.20 50 0 120 200 280 400

30/100= 0.30 80 90 0 120 240 420

20/100= 0.20 100 100 40 0 80 200

10/100= 0.10 120 70 40 20 0 60

20/100= 0.20 150 200 140 100 60 0

Total 460 340 440 660 1080

Expected opportunity loss table will be 340. The minimum among all total.

Expected value of perfect information (EVPI)= Expected opportunity of loss

table (EOL).

Decision making under uncertainty:

In the absence of information about the probability of any state of nature occurring,

the decision-maker must arrive at a decision only on the actual conditional pay-offs

values, together with a policy. There are several different criteria of decision making

in these situation. The criteria are as follows:-

i. Optimism (Maximax or Minimin) criterion.

ii. Pessimism (Maximin or Minimax) criterion.

iii. Equal probabilities (Laplace) criterion.

iv. Coefficient of optimism (hurweiz) criterion.

v. Regret (salvage) criterion.

i. Optimism criterion: In this criterion the decision-maker always looks for the

maximum possible profit (Maximax) or lowest possible (Minimin).Therefore he

selects the alternatives that maximum of the maxima (or minimum of the

minima) pay-offs. The method are as follows:

340

Decision Theory

104

a) Find the maximum (or minimum) payoff values corresponding to each

alternative courses of action.

b) Select the alternative with the best anticipated payoff value i.e., maximum profit

and minimum profit.

Examples:

Strategies

States of nature S1 S2 S3

P1 2,00,000 5,00,000 3,00,000

P2 4,00,000 1,50,000 9,00,000

P3 0 4,50,000 7,00,000

Strategies


P1 2,00,000 5,00,000 3,00,000

P2 4,00,000 1,50,000 9,00,000

P3 0 4,50,000 7,00,000

Column maximum 4,00,000 5,00,000 9,00,000

.

Ans: The maximum of column maxima is 9,00,000.

Hence the company should adopt strategy S3.

ii. Pessimism criterion: in this criterion the decision-maker ensures that he should

not earn no less (or pay no more) than some specified amount. Thus, he selects

the alternative that represents the maximum of the minima payoff in case of

profits. The methods are as follows :

a) Find the minimum (or maximum in case of profits) payoff values in case of

loss (or cost) data corresponding to each alternative.

b) Select an alternative with the best anticipated payoff value (maximum for

profit and minimum for loss or cost).

Maximax

Decision Theory

105

Examples:

Strategies


P1 2,00,000 5,00,000 3,00,000

P2 4,00,000 1,50,000 9,00,000

P3 0 4,50,000 7,00,000

Solution:-

Strategies


P1 2,00,000 5,00,000 3,00,000

P2 4,00,000 1,50,000 9,00,000

P3 0 4,50,000 7,00,000

Column minimum 0 1,50,000 3,00,000

Ans: The maximum of the row is 3,00,000.

iii. Equal probabilities (Laplace) criterion : The probabilities of states of nature

are not known, so it is assumed that all states of nature will occur with equal

probability. As state of nature are mutually exclusive and collectively

exhaustive, so the probability of each of these must be 1/(number of states of

nature). The methods are as follows:-

a) Assign equal probability value to each state of nature by using the formula:

= 1/(number of states of nature).

b) Calculate the expected (or average) payoff for each alternative (course of

action) by adding all the payoffs and dividing by the number of possible

states of nature or by applying the formula:

= (probability of state of nature) *(payoff value for combination of

alternative, and state of nature)

c) Selected the best expected payoff value (maximum profit and minimum

cost).

Maximin

Decision Theory

106

Strategy Expected return

S1 2,00,000 + 4,00,000 + 0 = 6,00,000/3= 2,00,000

S2 5,00,000 + 1,50,000 + 4,50,000 = 11,00,000/3 = 3,66,666

S3 3,00,000 + 9,00,000 + 7,00,000 = 19,00,000/3 =6,33,333

Since the largest expected return is from strategy S3.. the executive must select

strategy S3.

iv. Coefficient of optimism (hurwicz) criterion: In this criterion a decision maker

should neither be completely optimistic nor of pessimistic. It should be a mixture

of both. Hurwicz who suggested this criterion, introduced the idea of coefficient

of optimism (denoted by α) to measure the degree of optimism. This coefficient

lies between 0 and 1 represents a complete pessimistic attitude about future and

1 a complete optimistic attitude about future. Thus if α is the coefficient of

optimistic, then (1-α) will represent the coefficient of pessimism.

Hurwicz approach suggests that the decision maker must select an alternative that

maximizes

H(criterion of realism) =α (maximum in column)+ (1-α) minimum in column.

The methods are as follows:

a) Decide the coefficient of optimism α and then coefficient of pessimism(1 – α)

b) For each alternative select the largest and the lowest payoff value and multiply

these with α and (1-α) values, respectively. Then calculate the weighted average,

H by using above formula.

c) Select an alternative with best anticipated weighted average payoff value.

Examples:

Let the degree of optimism being 0.7.

Strategies


P1 2,00,000 5,00,000 3,00,000

P2 4,00,000 1,50,000 9,00,000

P3 0 4,50,000 7,00,000

Decision Theory

107

Strategies

States of

nature

Maximum pay

off (i)

Minimum pay

off (ii) H =α(i) +(1-α) (ii)

P1 5,00,000 2,00,000 0.7×5,00,000

+0.3×2,00,000=4,10,000

P2 9,00,000 1,50,000 0.7×9,00,000

+0.3×1,50,000=6,75,000

P3 7,00,000 0 0.7×7,00,000+0.3×0=4,90,000

The maximum value of H = 6,75,000

v. Regret (savage) criterion: In this, criterion is also known as opportunity loss

decision criterion or minimax regret decision criterion because decision maker

feels regret after adopting a wrong course of action resulting in an opportunity

loss of payoff. Thus he always intends minimize this regret. The method are as

follows:

a) Find the best payoff corresponding to each state of nature.

b) Subtract all other entries (payoff values) in that row from this value.

c) For each course of action identify the worst or maximum regret table .

Record this number in a new row.

d) Select the course of action with the smallest anticipated opportunity- loss

value.

Examples:

Strategies


P1 2,00,000 5,00,000 3,00,000

P2 4,00,000 1,50,000 9,00,000

P3 0 4,50,000 7,00,000

Decision Theory

108

Solution:-

Strategies

States of

nature S1 S2 S3

P1 5,00,000-

2,00,000=3,00,000 5,00,000-5,00,000=0

5,00,000-

3,00,000=2,00,000

P2 9,00,000-

4,00,000=5,00,000

9,00,000-

1,50,000=7,50,000 9,00,000-9,00,000=0

P3 7,00,000-0=7,00,000 7,00,000-

4,50,000=2,50,000 7,00,000-7,00,000=0

Column

Maximum 7,00,000 7,50,000 2,00,000

Q 1) The following matrix gives the payoff of different strategies (alternatives)

S1,S2,S3, against conditions N1,N2,N3, and N4.

N1

Rs.

N2

Rs.

N3

Rs.

N4

Rs.

S1 4,000 -100 6000 18000

S2 20,000 5000 400 0

S3 20,000 15000 -2000 1000

Calculate the decision taken under the following approach :

i. Pessimistic (maximin). Ans: 0 with strategy S2.

ii. Optimistic (maximax) Ans: Rs 20,000 with strategy S2 and S3.

iii. Regret Ans: Rs 16,000 with strategy S1.

iv. Equal probability (Laplace) Ans: Rs 8500 with strategy S3.

v. Hurwicz criterion, the degree

Of optimism being 0.7 Ans: Rs 14,000 with strategy S2.

MiniMax Regret

Decision Theory

109

Q 2) A producer of boats has estimated the following distribution of demand for a

particular kind of a boat.

No. of demanded : 0 1 2 3 4 5 6

Probability : 0.14 0.27 0.27 0.18 0.09 0.04 0.01

Each boat costs him Rs 7000 and sells them for Rs 10,000 each. Any boats that are

left unsold at the end of the season must be disposed of for Rs 6000 each. How many

boats should be in stock so as to maximize his expected profit?

Ans: 3 boats to get a profit of Rs.4080

Linear Programming Problem

110

LINEAR PROGRAMMING

INTRODUCTION

In every sphere of Human endeavor, we come across with the problem of allocation

of limited resources in competitive activities. In business situations also we often face

problems of allocating of fixed set of resources among a set of competing demands

e.g. Situations where we have to decide a product mix within the manufacturing and

marketing constraints, preparation of a schedule for transporting goods from the

places where they are available to the places of demand, etc.. In business problems the

criteria for selecting among alternatives is minimizing total cost, maximization total

contribution, maximizing total labour utilization etc.

What is Linear Programming?

Linear Programming (LP) is a particular type of technique used for economic

allocation of ‗scarce‘ or ‗limited‘ resources, such as labour, material, machine, time,

warehouse space, capital, energy, etc. to several competing activities, such as

products, services, jobs, new equipment, projects, etc. on the basis of a given criterion

of optimality. The phrase ‗scarce resources‘ means resources that are limited in

availability during the planning period. We learn to optimum use of these resources

through linear planning. The criterion of optimality generally is either performance,

return on investment, profit, cost, utility, time, distance, etc.

Before discussing in detail the basic concepts and applications of linear programming,

let us be clear about the two words, ‗linear and programming‘.

The term ―linear‖ is used because of the fact that all the relations among the variables

are linear. On the other hand, the word ―programming‖ refers to modeling and solving

a problem mathematically that involves the economic allocation of limited resources

by choosing a particular course of action or strategy among various alternative

strategies to achieve the desired objective.

Historical Development

George Bernard Dantzig (November 8, 1914 – May 13, 2005) was an American

mathematician, and the Professor Emeritus of Transportation Sciences and Professor

of Operations Research and of Computer Science at Stanford. Dantzig is known for

his development of the simplex algorithm, an algorithm for solving linear

programming problems, and his work with linear programming, some years after it

was initially invented by Soviet economist and mathematician Leonid Kantorovich.

Dantzig is the subject of a tale, often thought to be fictional, of a student solving an

important unsolved problem after mistaking it for homework.


111

In 1946, as mathematical adviser to the U.S. Air Force Comptroller, he was

challenged by his Pentagon colleagues to see what he could do to mechanize the

planning process, "to more rapidly compute a time-staged deployment, training and

logistical supply program." In those pre-electronic computer days, mechanization

meant using analog devices or punched-card machines. "Program" at that time was a

military term that referred not to the instruction used by a computer to solve

problems, which were then called "codes," but rather to plans or proposed schedules

for training, logistical supply, or deployment of combat units. The somewhat

confusing name "linear programming," Dantzig explained in the book, is based on this

military definition of "program."

In 1963, Dantzig‘s Linear Programming and Extensions was published by Princeton

University Press. Rich in insight and coverage of significant topics, the book quickly

became ―the bible‖ of linear programming.

General Structure of Linear programming Model

The general structure of LP model consists of three basic components which are as

under:

Decision Variables (Activities):

First we evaluate various alternatives for arriving at the optimum value of objective

function. The evaluation of various alternatives is guided by the nature of objective

function and availability of resources. For this we use certain activities usually called

decision variables and denoted by X1, X2,………………………….Xn in an LP

model all decision variables are continuous, controllable and non negative. That is X1

≥0, X2≥0……………….. , Xn≥0.

Objective function:

The objective function of each LP problem is expressed in terms of decision variables

to optimize the criterion of optimality (also known as measure of performance ) such

as profit, cost, revenue, distance etc. In general form it is represented as Optimize

(maximize or minimize) Z = C1X1 + C2X2+…………………………… + CnXn.

Constraints:

There are certain limitations on the use of resources like labour, machine, raw

material, space, money etc. that limit of which an objective can be achieved. Such

constraints must be expressed as linear equalities or inequalities in terms of decision

variables. The solution of an LP model must satisfy these constraints.


112

Assumptions of Linear Programming

The following are the four basic assumptions which are necessary for all linear

programming models:

1. CERTAINTY:

In all LP models, it is assumed, that all model parameters such as availability of

resources, profit (or cost) contribution of a unit of decision variable and consumption

of resources by a unit of decision variable must be known and may constant. In some

cases, these may be either random variables represented by a known distribution

(general or may statistical) or may tend to change, then the given problem can be

solved by a stochastic LP model or parametric programming.

2. DIVISIBILITY (or CONTINUITY):

The solution values of decision variables and resources are assumed to have either

whole numbers (integers) or mixed numbers (integer and fractional). However, if only

integer variables are desired, e.g. machines, employees, etc. the integer programming

method may be applied to get the desired values.

3. ADDITIVELY:

The value of the objective function for the given values of decision variables must be

equal to the sum of the contributions (profit or cost) earned from each decision

variable and the total sum of resources used, must be equal to the sum of the resources

used by each decision variable. For example, the total profit earned by the sale of two

products A and B must be equal to the sum of the profits earned separately from A

and B. Similarly, the amount of a resource consumed by A and B must be equal to the

sum of resources used for A and B individually.

4. LINEARITY (or PROPORTIONALITY):

All relationships in the LP model (i.e. in both objective function and constraints) must

be linear. In other words, for any decision variable, the amount of particular resource

used and its contribution to the cost one in objective function must be proportional to

its amount. For example, if production of one unit of a product uses 5 hours of a

particular resource, then making 4 units of that product uses 4x5=20 hours of that

resource.

Formulating a problem as an LP model

1) Once the problem has been described, the next step is to transform it into a

proper mathematical structure. General steps taken are as under:


113

2) Define the decision variables that are relevant to the problem and, ensure that

their units of measurement are explicitly stated.

3) Identify the contribution coefficient (the cj‘s) associated with each variable.

4) Formulate the objective function quantitatively and express it as a linear function

of decision variables.

5) Identify the physical rate of substitution coefficient(the aij‘s)

6) Identify the variable resources or requirement ,i.e. , the right-hand-side

coefficient(the bj‘s)

7) Formulate suitable mathematical constraints related to each respective resource

or requirement as linear equalities\inequalities in terms of decision variables.

8) Mention the non-negativity condition associated with the decision variables.

APPLICATION AREAS OF LINEAR PROGRAMMING:

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

business and industry and in various other fields. Here, we will discuss a few of the

broad application areas of linear programming.

MILITARY APPLICATIONS:

Military applications include the problem of selecting an air weapon system

against enemy so as to keep them pinned down and at the same time minimizing

the amount of aviation gasoline used. A variation of the transportation problem

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

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

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

the required level of protection at the lowest possible cost.

AGRICULTURAL APPLICATIONS:

The study of farm economics deals with inter-regional competition and optimum

allocation of crop production. Efficient production patterns can be specified by a

linear programming model under regional land resources and national demand

constraints. Linear programming can be applied in agricultural planning, e.g.

allocation of limited resources such as acreage, labour, water supply and

working capital, etc. in a way so as to maximize net revenue.


114

PRODUCTION MANAGEMENT:

(i) PRODUCT MIX:

A company can produce several different products, each of which requires the

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

quantity of each product to be produced knowing its marginal contribution and

amount of available resource used by it. The objective is to maximize the total

contribution, subject to all constraints.

(ii) PRODUCTION PLANNING:

This deals with the determination of minimum cost production plan over

planning period of an item with a fluctuating demand, considering the initial

number of units in inventory, production capacity, constraints on production,

manpower and all relevant cost factors. The objective is to minimize the total

operation costs.

(iii) ASSEMBLY-LINE BALANCING:

This problem is likely to arise when an item can be made by assembling

different components. The process of assembling requires some specified

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

(iv) BLENDING PROBLEMS:

These problems arise when a product can be made from a variety of available

raw materials, each of which has a particular composition and price. The

objective is to determine the minimum cost blend, subject to availability of the

raw materials, and minimum and maximum constraints on certain product

constituents.

(v) TRIM LOSS:

When an item is made to a standard size (e.g. glass, paper sheet), the problem

that arises is to determine which combination of requirements should be

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

FINANCIAL MANAGEMENT:

1) PORTFOLIO SELECTION :

This deals with the selection of specific investment activity among several other

activities. The objective is to find the allocation which minimizes the expected

return of minimizes risk under certain limitations.


115

2) PROFIT PLANNING:

This deals with the maximization of the profit margin from investment in plant

facilities and equipment, cash in hand and inventory.

MARKETING MANAGEMENT:

1) MEDIA SELECTION:

Linear programming technique helps in determining the advertising media mix

so as to maximize the effective exposure, subject to limitation of budget,

specified exposure rates to different market segments, specified minimum and

maximum number of advertisements in various media.

2) TRAVELLING SALESMAN PROBLEM:

The problem of salesman is to find the shortest route from a given city, visiting

each of the specified cities and then returning to the original point of departure,

provided no city shall be visited twice during the tour. Such type of problems

can be solved with the help of the modified assignment technique.

3) PHYSICAL DISTRIBUTION:

Linear programming determines the most economic and efficient manner of

locating manufacturing plants and distribution centers for physical distribution.

PERSONNEL MANAGEMENT:

1) STAFFING PROBLEM:

Linear programming is used to allocate optimum manpower to a particular job so

as to minimize the total overtime cost or total manpower.

2) DETERMINATION OF EQUITABLE SALARIES:

Linear programming technique has been used in determining equitable salaries

and sales incentives.

3) JOB EVALUATION AND SELECTION:

Selection of suitable person for a specified job and evaluation of job in

organizations has been done with the help of linear programming technique.

Other applications of linear programming lie in the area of administration,

education, fleet utilization, awarding contracts, hospital administration and

capital budgeting, etc.


116

Advantages and disadvantages of Linear Programming:

Advantages Drawbacks

1) Insights and perspective into problem

solutions. It helps in organization and

study of the information in the same way

that the scientific approach to the problem requires.

2) Consideration of all possible solutions to

the problem. Many management

problems are so complex that the difficulty is encountered in planning and

feasible solution. By using the LP

technique, the manager makes sure that he is considering the best (Optimal)

solution for solutions)

3) Better and more successful decisions.

With linear programming the executive builds into his planning a true reflection

of the limitations and restrictions under

which he must operate. When it becomes necessary to deviate from the best

programme, he can evaluate the cost or

penalty involved.

4) Better tools for adjusting to meet changing conditions once a basic plan is

arrived at through linear programming, it

can be reevaluated for changing

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

determined so as to adjust the remainder

of the plan for best results.

5) Highlighting of bottlenecks in the production process is the most significant

advantage of this technique

E.g.When bottleneck occurs, some

machines cannot meet demand while others remain idle for some of the time

6) It also helps the mangers to shave better

understanding about the phenomenon and

various activities of the organization for the organization construction of suitable

mathematical model visualizing the

relationship between variables if any ad making improvement over them.

1) In LP problem, fractional values are

permitted for the decision variables

However, many decision Problems

require that the solution for decision variables should be obtained in non-

fractional values. Rounding-of the values

obtained by linear programming techniques may not result into an optimal

Solution.

2) In LP problems, coefficients in the

objective function and the constraint equations must be completely known and

they should not change during the period

of study. In practical situation, it may not

be possible to state all coefficients with certainty.

3) It may not possible to solve those

problems using LP in which non-linearity arises because of joint

interactions between some of the

activities regarding the total measure of

effectiveness or total usage of some resources.

4) Does not take into consideration the

effect of time and uncertainty

5) Parameters appearing in the model are assutmed to be constants but in real-life

situations they are frequently either

known nor constants.

6) Many real-world problems are so complex in terms of the numbers of

variables ad relations constrained in

them, that they tax the capacity of even

the largest computer.


117

PROBLEM FORMULATION:

Q: manufacturer wants to manufacture tables and chairs. He will make the profit of

Rs.50 per table and Rs.35 per chair. The quantity of raw material which he has-

Raw Material Quantity Available (maximum)

Ply 40 meter

Steel 50 meter

Plastic 60 meter

Quantity used in one Table and one Chair

Raw Material Table Chair

Ply 2m 2m

Steel 5m 2m

Plastic 1m 4m

How many tables and chairs he should make to maximize his profit?

SOL:

PROBLEM FORMULATION:

Let assume x1 unit of table and x2 unit of chair made.

Then,

Max.Z= 50x1 + 35x2

Z= profit

Max.Z= maximum profit,

Subjective Equations:

2x1 + 2x2 ≤ 40 (for ply)

5x1 + 2x2 ≤ 50 (for steel)

x1 + 4x2 ≤ 60 (for plastic)

x1, x2 ≥ 0.

Graphical method:

The graphic solution procedure is one method of solving two variable linear

programming. The problems involve following steps:

1. Formulate the problem in terms of mathematical constraints and an objective

function.


118

2. Plot each of the constraints and convert each equality in the constraints equation

to equality.

3. Identify the feasible region i.e. the area which satisfy all the constraints

simultaneously

Some important definition:

1. Solution:

Values of decision variables xj(j=1,2,……..,n)which satisfy the constraints of

general L.P problem is called the solution to that L.P problem.

2. Feasible solution:

Any solutions that also satisfy the non-negative restriction of the general L.P

problem is called a feasible solution.

3. Basic solution:

For a set of m simultaneous equation in n unknowns (n>m), a solution obtained

by setting (n-m) of the variables equal to zero and solving the remaining m

equations in m unknowns is called a basic solution.

4. Basic feasible solution:

A feasible solution to a general L.P problem which is also basic is called a basic

feasible solution.

5. Optimal feasible solution:

Any basic feasible solutions which optimize (maximize or minimize) the

objective function of a general L.P problem is called an optimal feasible solution

to that L.P problem

6. Degenerate solution:

A basic solution to the system of the equations is called degenerate if one or

more of the basic variables become equal to zero.

ILLUSTRATION:

Q1) Max Z = 50x1 + 35 x2

Subject to:

2x1 + 2x2 ≤ 40

5x1 + 2x2 ≤ 50

x1 + 4x2 ≤ 60

Where x1 ,x2 ≥ 0


119


120

Soln:

x1 0 20

x2 20 0

2x1 + 2x2 ≤ 40

x1 0 10

x2 25 0

5x1 + 2x2 ≥ 50

x1 0 60

x2 15 0

x1 + 4x2 ≥ 60

So the coordinates for A

5x1 + 2x2 = 50

x1 + 4x2 = 60 …………….. (Multiply the equation By 5)

5x1 + 2x2 = 50 ……………….. (Subtract the equation)

5x1 + 20x2= -300

------------------

-18x2 = -250

x2 = 125/9

x1 + 4 x2 =60

x1 + 500/9 = 60

x1 =60 - 500/9 = 40/9

(x1 = 40/9)

(x2 = 125/9)

A = (40/9, 125/9)

B = (0,15)

C = (10,0)


121

Max Z =50 x1 + 35x2

(50 × 40/9) + (35 ×125/9)

2000/9 + 4375/9

6375/9--------------------------------------(1)

708.34

Max Z = (50 × 0) + (35 ×15)

525-------------------------------------------(2)

Max Z = (50 × 10) + (35 ×0)

500-------------------------------------------(3)

Max Z = (6375/9)---------------------(ANSWER)

Q2) Min Z = -x1 + 2 x2

Subject to :

-x1 + 3x2 ≤ 10

x1 + x2 ≤ 6

x1 - x2 ≤ 2

Where x1 ,x2 ≥ 0

Soln:

x1 0 -10

x2 10/3 0

-x1 + 3x2 ≤ 10

x1 0 6

x2 6 0

x1 + x2 ≤ 6

x1 0 2

x2 -2 0

x1 - x2 ≤ 2


122

So the coordinates for A

-x1 + 3x2 = 10

x1 + x2 = 6

-x1 + 3x2 = 10 ………………………………. (Addition of the equation)

x1 + x2 = 6

------------------

4x2 = 16

(x1 = 2)

(x2 = 4)

For the coordinates for B

x1 + x2 = 6 x1 + x2 = 6 (Addition of the equation)

x1 - x2 = 2 x1 - x2 = 2

------------------

2x1 = 8

(x1 = 4)

(x2 = 2)

A = (2,4)

B = (4,2)

C = (2,0)

D = (0, 10/3)

Min Z = -x1 +2 x2

-2 + 8 = 6--------------------------------------(A)

Min Z = -4 + 4 = 0-----------------------------------(B)

Min Z = -2 + 0 = -2-----------------------------------(C)

Min Z = 0 + 2 ×10/3 = 20/3------------------------(D)

Min Z = -2 (ANSWER)


123


124

Q3) Max Z = 6x1 - 4x2

Subject to :

2x1 + 4x2 ≤ 4

4x1 + 8x2 ≥ 16

Where x1, x2 ≥ 0

Soln:

x1 0 2

x2 1 0

2x1 + 4x2 ≤ 4

x1 0 4

x2 2 0

4x1 + 8x2 ≥ 16

Q4) Max Z = 3x1 + 2 x2

Subject to:

x1 - x2 ≥ 1

x1 + x2 ≥ 3

Where x1, x2 ≥ 0

Soln:

x1 0 1

x2 -1 0

x1 + x2 ≥ 1

x1 0 3

x2 3 0

x1 + x2 ≥ 3


125


126

According to the given condition x1 , x2 ≥ 0,

Hence 1st coordinate cannot exist. This is an UNBOUNDED SOLUTION.

x1 - x2 = 1

x1 + x2 = 3

--------------

2 x1 = 4

x1 = 2

(x1 = 2)

(x2 = 1)

Max Z = 3x1 + 2 x2

6 + 2

8-------------------------------------------(1)

Max Z = 3x1 + 2 x2

0 + 6

6-------------------------------------------(2)

It is an feasible solution.

Q5) Max Z = 7x1 + 3x2

Subject to :

x1 + 2x2 ≥ 3

x1 + x2 ≤ 4

0 ≤ x1 ≤ 5/2

0 ≤ x2 ≤ 3/2

Where x1 ,x2 ≥ 0

Soln:

x1 0 3

x2 3/2 0

x1 + 2x2 ≥ 3---------------------------(1)


127

x1 0 4

x2 4 0

x1 + x2 ≤ 4---------------------------(2)

x1 5/2 0

x2 0 0

0 ≤ x1 ≤ 5/2---------------------------(3)

x1 0 0

x2 0 3/2

0 ≤ x2 ≤ 3/2---------------------------(4)

For – A (1 and 4)

x1 + 2x2 = 3

0 + x2 = 3/2

A = (0, 3/2)

x1 + 3 = 3

(x1 = 0)

(x2 = 3/2)

For – B (2 and 4)

x1 + x2 = 4

0 + x2 = 3/2

B = (5/2, 3/2)

(x1 = 5/2)

(x2 = 3/2)

For – C (1 and 3)

x1 + 2x2 = 3

x1 + 0x2 = 5/2

A = (5/2, 1/4)

x2 = 3 – 5/2


128


129

=1/2

(x1 = 1/4)

(x2 = 5/2)

Max Z = 7x1 + 3x2

6 + 9/2

9/2-------------------------------------------(A)

Max Z = 7x1 + 3x2

(7 × 5/2) + (3 × 3/2)

22---------------------------------------------(B)

Max Z = 7x1 + 3x2

(7 × 5/2) + (3 × 1/4)

35/2 + 3/2

-----------------------------------------------(C)

MAX Z = 22 (ANSWER)

SIMPLEX METHOD

The simplex Algorithm is a systematic ad efficient algebraically procedure for finding

corner point solutions and taking them for optimality.

BASIC TERMS INVOLVED IN SIMPLEX PROCEDURE

Certain terms relevant for solving a linear programming problem through simplex

procedure are introduced below:

1. Standard Form:

A linear programme in which all of the constraints are written as equalities. The

optimal solution of the standard form of a linear programme is the same optimal

solution of the original formulation of the linear programme.

2. Slack Variable:

A Variable added to the left-hand side of a ‗less-than or equal to‘ constraint to

convert the constraint into an equality is called a slack variable in economic


130

terminology; the value of the negative variable can usually be interpreted as the

amount of unused resources.

3. Surplus variable:

A variable subtracted from the left hand side of the greater than or equal to

constraints to convert the constraints into equality is called a surplus variable.

4. Basic variable :

For a system of m simultaneous linear equations in n variables(n>m), a solution

obtained by setting (n-m)variables equal to zero and solving for the remaining m

variables is called a basic variable.

5. Basic feasible solution:

A basic feasible solution to a linear programming problem is a basic solution fro

which the variables solved for, are all greater than or equal to zero.

6. Optimal solution:

Any basic feasible solution which optimizes the objective function of a general

LP problem is called an optimal basic feasible solution to the general LP

problem. Tableau Form in which a linear programme must be written prior to

setting up the initial simplex tableau. When a linear programme is written in this

form.

7. Simplex Tableau:

A table used to keep track of the calculation made at each iteration when the

simplex solution method is employed.

8. Zj Row:

The numbers in this row under each variable represents the total contribution of

out going profit when one unit of a non basic variable is introduced in to this in

place of a basic variable.

9. Cj – Zj (or Net Evaluation of Index) Row:

The row containing the net net profit (for loss) that will result from introducing

one unit of the variable indicated in that column in the solution numbers in index

rows are also known as shadow prices).

10. Pivot (or Key) Column:

The column with the largest positive number in the et evaluation row of

maximization problem, or the largest negative number in the net evaluation row

in a maximization problem.


131

11. Pivot (or Key) Row:

The row corresponding to the variable that will leave the basis in order to make

room for the entering variable (as indicated by the new pivot column)

12. Pivot (Number Key:

The element at the intersection of the pivot row and pivot column.

13. Iteration:

An iteration of the simplex method consists of the sequence of steps performed

in moving form one basic feasible solution to another.

NUMERICALS

Q1. Max Z = 3X1+ 5X2+ 4x3

Subject to :

2X1 + 3X2 ≤ 8

2X2+ 5X3 ≤10

3X1+ 2X2 + 4X3≤ 15

X1, X2, X3 ≥0

Balance these equation

2X1 + 3X2 + S1= 8

2X2+ 5X3 + S2=10

3X1+ 2X2 + 4X3+ S3= 15

MaxZ = 3X1+ 5X2+ 4x3+0S1+0S2+ 0S3

Initial Solution

Cj 3 5 4 0 0 0

Qty X1 X2 X3 S1 S2 S3 Ratio

0 S1 8 2 3 0 1 0 0 8/3→

0 S2 10 0 2 5 0 1 0 5

0 S3 15 3 2 4 0 0 1 15/2

Zj 0 0 0 0 0 0 0

(Cj-Zj) 3 5 4 0 0 0 0

↑


132

Key Value = 3

Improved table

Cj → 3 5 4 0 0 0

Qty ↓ X1 X2 X3 S1 S2 S3 Ratio

5x2 8/3 2/3 1 0 1/3 0 0 -

0S2 14/3 -4/3 0 5 -2/3 1 0 14/15→

0S3 29/3 5/3 0 4 -2/3 1 9 29/12

Zj 40/3 10/3 5 0 5/3 0 0

(Cj-Zj) -1/3 0 4 -5/3 0 0 0

↑

R2(NEW):

R3(NEW):

10-8/3X2=14/3

15-8/3X2=29/3

0-2/3X2=-4/3

3-2/3X2=5/3

2-1X2=0

2-1X2=0

5-0X2=5

4-0X2=4

0-1/3X2=-2/3

0-1/3X2=-2/3

1-0X2=1

0-0X2=0

0-0X2=0

1-0X2=1

Improved table-1

Cj 3 5 4 0 0 0


x2 8/3 2/3 1 0 1/3 0 0 4

x3 14/15 -4/15 0 1 -2/15 1/5 0 -

S3 89/15 41/15 0 0 -2/15 -4/5 1 89/41

Zj 56/15 34/15 5 4 17/15 4/5 0

(Cj-Zj) 11/15 0 0 -17/15 -4/5 0


133

R1 (New): R2 (New):

8/3 – (0X 14/15) + 5/3 29/3 – (4X14/15) = - 89/15

8/3-(0X (-) 4/15) =2/3 5/3 – (4X -4/15) = 41/15

1- (0X0) = 1 0- (4X0) = 0

0- (0X1) = 0 4- (4x1) = 0

1/3- (0X -2/15) = 1/3 -2/3 – (4X 2/5) = -2/15

0-(0X1/5) = 0 0-(4X1/5) = -4/5

0- (0X0) = 0 1 – (4X0) = 1

Optimal Table

Cj 3 5 4 0 0 0


5 x2 50/41 0 1 0 15/41 8/41 -10/41 -

4 x3 62/41 0 0 1 -6/41 5/41 4/41 -

0 S3 89/41 1 0 0 -2/41 -22/41 -15/41 -

Zj 785/41 3 5 4 45/41 24/41 11/41 -

(Cj-Zj)

0 0 0 -45/41 -24/41 4/41

R1 (New) R2 (New)

14/15 – (89/41 X – 4/5) = 62/41 8/3X 89/4) = 50/41

-4/15 – (-4/5X0) = 0 2/3 – (2/3 X1) = 0

– (-4/5 X0) = 0 1 – (2/3 X0) = 1

1-(-4/5x0)=1 0-(2/3x0)=0

-2/5-(-4/5x -2/41)=-6/41 1/3-(2/3x -2/41)=15/41

1/5-(-4/5 x -12/41)= 5/41 0-(2/3x -12/41)=8/41

0-(-4/5 x -15/41)=4/41 0-(2/3 x -15/41)=-10//41

MaxZ = 3X1+ 5X2+ 4x3

=3x(89/41)+5x(50/41)+4x(62/41)

=765/41(ans.)


134

Q 2: Max Z=X1+2X2+3X3-X4

Sub to

X1+2X3+3X3=15

2X1+X2+5X3=20

X1+2X2+X3+X4=10

X1, X2, X3, X4≥0

Balancing the equation

X1+2X2+3X3+A1=15

2X2+X2+5X3+A2=20

X1+2X2+X3+X4=10

Max Z=X1+2X2+3X3-X4-MA1-MA2

Initial Table

Cj → 1 2 3 -1 -M -M

Qty ↓ X1 X2 X3 X4 A1 A2 Ratio

- MA1 15 1 2 3 0 1 0 5

- MA2 20 2 1 5 0 0 1 4

-X4 10 1 2 1 1 0 0 10

Zj (35M-10) -(3M+1) -(3M+2) -(8M+1) -1 -M -M

Cj-Zj

(3M+2) (3M+4) (8M+42) 0 0 0

Key Value = 5

Improved Table

Cj 1 2 3 -1 -M -M


- MA1 3 -1/5 7/5 0 0 1 -3/5 15/7 →

- MA2 4 2/5 1/5 1 0 0 1/5 20

-1X4 6 3/5 9/5 0 1 0 -1/5 30/9

Zj (6M-

3M)

(3/5+M/

5)

-

(7M/5+6/5)

3 -1 -M 4/5

+3M/5

Cj-Zj

-(M/5 -

2/5)

(7M/5

+16/5) 0 0 0

2M/5-

4/5)

↑

Key Value = 5


135

R1(New) R3 (New)

15-(3X4) =3 10-(1X4) =6

(3X2/5) = -1/5 1-(1X2/5) = 3/5

(3X1/5) =7/5 2- (1X1/5) = 9/5

(3X1) = 0 1- (1X1) =0

0-(3X0) = 0 1-(1X0) = 1

(3X0) =1 0-(1X0) = 0

(3X1/5) = -3/5 0-(1X1/5) = -1/5

Key Value = 7/5

Improved Table – 1

Cj 1 2 3 -1 -M -M


2X2 15/7 -1/7 1 0 0 5/7 -3/7 -

3X3 25/7 3/7 0 1 0 -1/7 2/7 25/3

-1X4 15/7 6/7 0 0 1 -9/7 4/7 15//6

Zj 90/7 1/7 2 3 -1 16/7 -4/7

Cj-Zj

6/7 0 0 0 -(M+16/7) 4/7-M

KEY VALUE=6/7

R2(NEW: R3(NEW)

4-(1/5x15/7)=25/7 6-(9/5X15/7)=15/7

2/5-(1/5x1/7)=3/7 3/5-(9/5X-1/7)=6/7

1/5-(1/5x1)=0 9/5-(9/5X1)=0

1-(1/5x0)=1 0-(9/5X0)=0

0-(1/5x0)=0 1-(9/5X 0)=1

0-(1/5x5/7) =-1/7 0-(9/5X5/7)=-9/7

1/5-(1/5x -3/7)=2/7 -1/5-(9/5X-3/7)=4/7


136

Improved Table - 2

Cj 1 2 3 -1 -M -M


2X2 15/6 0 1 0 1/6 1/2 1/3

3X3 15/6 0 0 1 -1/2 1/2 0

1X1 15/6 1 0 0 7/6 -3/2 2/3

Zj 15 1 2 3 0 1 4/3

Cj-Zj 0 0 0 -1 -(M+1) -(M+4/3)

R1(NEW): R2(NEW):

15/7-(-1/7X15/6)=15/6 25/7-(3/7X15/6)=15/6

-1/7-(-1/7X1)=0 3/7-(3/7X1)=0

1-(-1/7X0)=1 0-(3/7X0)=0

0-(-1/7X0)=0 1-(3/7X0)=1

0-(-1/7X7/6)=1/6 0-(3/7X7/6)=-1/2

5/7-(-1/7X-3/2)=1/2 -1/7-(3/7X-1/3)=1/2

-3/7-(-1/7X2/3)=-1/3 2/7-(3/7X2/3)=0

MAX Z=X1+2X2+3X3-X4

15/6+15/3+15/2-0

=15(ANS.)

Q.3 Max Z = X1 + 2X2+ 3X3 – X4

Subject to:

X1 + 2X2 + 3X3=15

2X1+ X2+ 5X3 = 20

X1+ 2X2+ X3 + X4 = 10

X1, X2, X3, X4 ≥0

Balance the equation

X1+ 2X2 + 3X3 + A1 = 15 ………………….. (1)

2X1+ X+ 5X3 + A2 = 20 ……………………. (2)


137

X1+ 2X2+ X3 + X4 = 10…………………….. (3)

Maxz = X1 + 2X2 + 3X3 – X4 – MA1 – MA2

INITIAL TABLE

Cj → 1 2 3 -1 -M -M


-MA1 15 1 2 3 0 1 0 5

-MA2 20 2 1 5 0 0 1 4→

-1X4 10 1 2 1 1 0 0 10

Zj = (35M-10), -(3M+1), -(3M+2), -(8M+1), -1, -M, -M

(Cj-Zj) = (3M+2), (3M+4), (8M+4), 0 0 0 0

Key Value = 5

Cj → 1 2 3 -1 -M -M


-MA1 3 -1/5 7/5 0 0 1 -3/5 15/7 →

3X3 4 2/5 1/5 1 0 0 1/5 20

-1X4 6 3/5 9/5 0 1 0 -1/5 30/9

Zj = (6-3M), (3/5 +M/5), -(7M/5 + 6/5), 3, -1, -M (4/5 + 3M/5)

(Cj-Zj) = (-M/5- 2/5, (7m/5 + 16/5, 0, 0,0, (2M/5 – 4/5)

R1 (New): R3 (New):

15-(3X4) = 3 10 – (1X4)=6

1-(3X2/55) = -1/5 1- (1X2/5) = 3/5

2-(3X1/5) = 7/5 2-(1X1/5) = 9/5

3-(3X1)= 0 1-(1X1) = 0

0-(3X0) = 0 1- (1X0) =1

1-(3X0) = 1 0-(1X0)=0

0-(3X1/5) = -3/5 0-(1X1/5) = -1/5

Key Value = 7/5


138

Cj → 1 2 3 -1 -M -M


2X2 15/7 -1/7 1 0 0 5/7 -3/7 -

3X3 25/7 3/7 0 1 0 -1/7 2/7 25/3

-1X4 15/7 6/7 0 0 1 -9/7 4/7 15/6

Zj = 90/7 1/7 2 3 -1 16/7 -4/7

(Cj-Zj) = 6/7 0 0 0, -(M+ 116/7), (4/7 – M)

R2 (New): R3 (New) :

4-(1/5 X 15/7) = 25/7 6- (9/5 X 15/7) = 15/7

2/5 – (1/5 X -1/7) = 3/7 3/5 – (9/5 Xx – 1/7) = 6/7

1/5 – (1/5 X1) = 0 9/5 – (9/5X0) = 0

1-(1/5X0) =0 1- (9/5X0) = 1

0-(1/5X0) = 0 1- (9/5 X0) = 1

0-(1/5 X 5/7) = -1/7 0-(9/5 X 5/7) = -9/7

1/5 – (1/5 X -3/7) = 2/7 -1/5 – (9/55X – 3/7) = 4/7

Key Value = 6/7

Cj → 1 2 3 -1 -M -M


2X2 15/6 0 1 0 1/6 ½ 1/3

3X3 15/6 0 0 1 -1/2 ½ 0

1X1 15/6 1 0 0 7/6 -3/2 2/3

Zj = 15 1 2 3 0 1 4/3

(Cj-Zj)= 0 0 0 0 -1 (M+1) -(M+4/3)

R1 (New): R2 (New):

15/7 – (-1/7 X 15/6) = 15/6 25/7 – (3/7 X 15/6) = 15/6

-1/7 – (-1/7 X1) = 0 3/7 – (3/7 X1) = 0

1- (- 1/7 X 0) = 0 0- (3/7 X0 ) = 0

0 – (-1/7 X 0) = 0 1- (3/7 X0) =1


139

0-(-1/7 X – 3/2) = 1/6 0- (3/7 X 7/6) = -1/2

5/7 – (-1/7 X-3/2) = ½ -1/7 – (3/7 X -3/2) = ½

-3/7- (1/7 X 2/3) = -1/3 2/7 – (3/7 X 22/3)=0

Max Z = X1 + 2X2 + 3X3 – X4

15/6 + 15/3 + 15/2 – 0

15 Ans

Q.4 MinZ = 5X1 + 3X2

Subject to

2X1 + 4X2 ≤ 12

2x1 + 2X2 = 10

5x1 + 2X2 ≥ 10

X1, X2 ≥0


2X1 + 4X2 + s1 = 12 …………………(1)

2x1 + 2X2 + A1 = 10………………..(2)

5x1 + 2X2 - s1 + A2 =10---------(3)

Z* = -Z

MaxZ* = -5X1 - 3x2 – 0s1 – 0s2 – MA1 – MA2

Cj → -5 -3 0 0 -M -M

Qty ↓ X1 X2 s1 s2 A1 A2 Ratio

0s1 12 2 4 1 0 0 0 6

-MA1 10 2 2 0 0 1 0 5

-MA2 10 5 2 0 -1 0 1 2 →

Zj = -20M, -7M -4M 0 M -M -M

(Cj-Zj) (7M-5) (4M-3) 0 -M 0 0

↑

Key Value = 5


140

Cj → -5 -3 0 0 -M -M


0s1 8 0 16/5 1 2/5 0 -2/5 5/2 →

-MA1 6 0 6/5 0 2/5 1 -2/5 5

-5x1 2 1 2/5 0 -1/5 0 1/5 5

Zj = -(6M + 10), -5 -(6M/2 + 2) 0 (1 – 2M/5) -M (2M/5 – 1)

(Cj-Zj) 0 (6M/5 - 1) 0 (2M/5-1) 0 (1 – 7M/5)

↑

Key Value = 16/ 5

R1 (New): R2 (New):

12 – (2 X 2) = 8 10 – (2 X 2) = 6

2 – (2 X 1) = 0 2 – (2X1) = 0

4 – (2 X 2/5) =16/5 2 – (2 X 2/5) =6/5

1 – (2 X 0) = 1 0 – (2 X 0) = 0

0 – (2 X-1/5) = 2/5 0 – (2 X-1/5) = 2/5

0 – (2 X 0) = 0 1 – (2 X 0) = 1

0 – (2 X-1/5) = -2/5 0 – (2 X-1/5) = 2/5

Cj → -5 -3 0 0 -M -M


-3x2 5/2 0 1 5/16 1/8 0 -1/8 8

-MA1 3 0 0 3/8 -7/20 1 -1/4 8 →

-5x1 1 1 0 -1/8 -1/4 0 1/4 -

Zj = -(25//2 + 3M), -5,-3, -(3M/8 +5/16),(3M/8 +5/16),(7/8 – 7M/20),-M,(7/8 – M/4)

(Cj-Zj) 0 0 (3M/8+5/16),(7/8 – 7M/20, 0,(7//8-5M/4)

↑

Key Value = 13/8

R1 (New): R2 (New):

6 – (6/5 X 5/2) = 3 2 – (2/5 X 5/2) = 1


141

0 – (6/5 X 0) = 0 1 – (2/5 X 0) = 1

6/5 –(6/5 X1 )=0 2/5 – (2/5 X1 ) =0

0 – (6/5 X 5/16) = 3/8 0 – (2/5 X 5/16) = -1/8

2/5– (6/5 X 1/8) = -7/20 -1/5 – (2/5 X 1/8) = -7/4

1 – (6/5 X 0) = 1 0 – (2/5 X 0) = 0

-2/5– (6/5 X -1/18) = -1/4 1/5 – (2/5 X-1/8) = 1/4

Cj → -5 -3 0 0 -M -M


-3x2 0 0 1 0 5/12 -5/6 1/12 0 →

0s1 8 0 0 1 -14/15 8/3 -2/3 -

-5x1 2 1 0 0 -11/30 1/3 1/6 12

Zj = -10, -5 -3 0 7/12 5/6 -13/12

(Cj-Zj) 0 0 0 -7/12 -(M + 5/6) (13/12 –M)

Key Value = 1/12

R1 (New): R2 (New):

5/2 – (5/16 X 8) = 0 1 – (-1/8 X 8) = 2

0 – (5/16 X 0) = 0 1 – (-1/8 X 0) = 1

1 – (5/16 X 0) =1 0 – (-1/8 X 0) =0

5/6 – (5/16 X 1) = 0 -1/8 – (-1/8 X1 ) = 0

1/8 – (5/16 X -14/15) = 5/12 -1/4 – (-1/8 X -14/15) = -11/30

0 – (5/16 X 8/3) = -5/6 0 – (-1/8 X 8/3) = 1/3

-1/8 – (5/16 X -2/3) = 1/12 1/4 – (-1/8 X -2/3) = 1/6

Cj → -5 -3 0 0 -M -M


-MA2 0 0 12 0 5 -10 1

0s2 8 0 8 1 36/15 -4 0

-5x1 2 1 -2 0 -6/5 2 0

Zj = -10, -5 (10 -12M) 0 (6- 5M),(10M -10),-M


142

(Cj-Zj) 0 (12M-13) 0 (5M -6) (10-11M),0

R1 (New): R2 (New):

8 – (-2/3 X 0) = 8 2 – (1/6 X 0) = 2

0 – (-2/3 X 0) = 0 1 – (1/6 X 0) = 1

0 – (-2/3 X 12) =8 0 – (1/6 X 12) = -2

1 – (-2/3 X 0) = 1 0 – (1/6 X 0) = 0

-14/15 – (-2/3 X 5) = 36/15 -11/30 – (1/6 X 5) = -6/5

8/3 – (-2/3 X -10) = -4 1/3 – (1/6 X -10) = 2

-2/3 – (-2/3 X 1) = 0 1/6 – (1/6 X 1) = 0

Since this is a continuous so its solution is UNBOUND.

Q.5 MinZ = X1 – 2X2 – 3X3

Subject to

-2X1 + X2 + 3X3 = 2

2x1 + 3X2 + 4 X3 =1

X1, X2, X3 ≥0


-2X1+ X2 + 3X3+ A1 = 2 …………………(1)

2x1 + 3X2 + 4 X3 +A2 =1………………..(2)

Z* = -Z

MaxZ* = -X1 + 2x2 + 3X3 – MA1 – MA2

Cj → -1 +2 3 -M -M

Qty ↓ X1 X2 X3 A1 A2 Ratio

-MA1 2 -2 1 3 1 0 2/3

-MA2 1 2 3 4 0 1 1/4 →

Zj = -3M, 0 -4M -7M -M -M

(Cj-Zj) -1 (2+4M) (3+7M) 0 0

↑

Key Value = 4


143

Cj → -1 +2 3 -M -M

Qty ↓ X1 X2 X3 A1 A2 Ratio

-MA1 5/4 -7/2 -5/4 0 0 -3/4 2/3

-MA2 1/4 1/2 3/4 1 1 1/4 1/4 →

Zj = ¾ - 5M/4), 3/2 +7M/2 0 -4M -7M -M -M

(Cj-Zj) = -(5/2+ 7M/2), -(1/4 + 5M/4), 0, 0, -(7M/4 + ¾)

R1(New) : 2-(3X1/4) = 5/4

-2- (3X1/2) = -7/2

1- (3X3/4) = -5/4

3- (3X1) = 0

(3X0) = 1

(3X ¼) = -3/4

Max Z* = 2X2 + 3X3 – X1

= 0+ ¾ -0

= ¾

(Optimal But not Feasible Solution)

DUALITY METHOD

Introduction

The term ‗dual‘ in a general sense implies two or double. The concept of duality is

very useful in mathematics, physics, statistics, engineering and managerial decision

making. For example, in two-person game theory, one competitor‘s problem is the

dual of the opponent‘s problem

In the context of linear programming duality implies that each linear programming

problem can be analyzed in to different ways but having equivalent solutions. Each

LP problem (Maximization and Minimization) stated in the original form has

associated with another linear programming problem (called dual linear programming

or in short dual), which is unique , based on the same data dual in general, it is

immaterial which of the two problems is called primal or dual, since the dual of the

dual is primer.


144

Thus the main focus of dual is to find for each resource its best marginal value of

shadow price.

Shadow price= change in optimal objective function value

Unit change in the availability of resource

Rules for constructing the dual from primal

The rules for constricting the dual from for the primal or primal or primal from the

dual when using the symmetrical form are:

1. If the objective function of the primal is to be maximized, the objective function

of the dual becomes minimized ition and vice versa.

2. For a maximization primal with all ≤ type constraints, their exist a minimization

dual problem with all ≥ type constraints and vice versa. Thus, the inequality sing

is reversed in all the constraints expect the non-negativity condition.

3. Each constants in the primal correspond to a dual variable in the dual an vice

versa. Thus given primal problem with M constraints and N variable, there exist

a dual problem with M variable and N constraints.

4. The right hand side constants b1, b2…….bm of the primal become the

coefficients of the dual variable y1, y2,……,Yn in the dual objective function Zy

Also the coefficients C1,C2,…Cn of the primal variable X1, X2, ….. xn in the

objective function become the right hand side constants in the dual.

5. The matrix of the coefficients of variables in the constraints of dual is the

transpose of the matrix of variables in the constraints of primal and vice versa.

That is coefficients of the primal variables x1,x2…..xn in the constraints of primal

LP problem are the coefficients of dual variables in first, second,…nth,

constraints for the dual problem respectively.

6. If a variable say J in the primal LP problem is unrestricted in sign, Jth

dual

constraint is=(equality) type and vice versa.

The primal dual relationship may also be remembered conveniently by using the

following table:

Dual variables Primal variables

X1 X2 ……..XN

Maximum zx

Y1 A11 A12………….A1n ≤ b1

Y2

.

A21 A22……….A22n

.

≤ b2

Ym Am1 Am2……...Amn ≤bm

Minimum Zy ≥ C1 ≥C2……… ≥Cn


145

The primal constraints should be read across the row and dual constraints should be

read across the columns

ILLUSTRATIONS:

1. Max Z = 30x1+20x2

Sub. To

2x1+7x2 ≤ 25

5x1+x2 ≤ 15 x1,x2 ≥0

Soln-

30 20 minZD

≤ ≤

2 7 25

5 1 15

minZD= 25w1+15w2

sub.to

7w1+w2 ≤ 20-----------------------(1)

2w1+5w2 ≤ 30---------------------(2)

W1,w2 ≥0

Balancing the eqn-

7w1+w2+s1=20

2w1+5w2+s2=30

NOW solve the problem using simplex method.

2. MaxZ= 30x1+20x2

Sub.to

2X1+7X2 ≤ 25

5x1+x2= 15 x1 , x2 ≥0

Ans.

2x1+7x2 ≤ 25-----------------------------------------------(1)


146

5x1+x2 ≥ 25 or -5x1-x2≤ 15--------------------------(2)

5x1+x2 ≤ 15------------------------------------------------(3)

30 20 minZD

≤ ≤

2 7 25

- 5 -1 15

5 1 15

minZD=25w1+15w2+15w3

2w1-5w2+5w3 ≤ 30

7w1-w2+w3 ≤ 20

Balancing these eqn

2w1-5w2+5w3 +s1= 30

7w1-w2+w3 +s2= 20

NOW Solve the problem using simples method.

3. MaxZ= 30x1+20x2

Sub.to

2X1+7X2 ≤ 25

5x1+x2= 15

x1 ≥0, x2 = URWS

IMAGINE X2= (Y1-Y2)

2X1+7 (Y1-Y2) ≤ 25---------------------(1)

5x1+ (Y1-Y2) = 15------------------------(2)

NOW Solve the problem using simples method.


147

Unsolved Questions:

Graphical method

Q.1 Maximize(Profit) Z=90X1+60X2

Sub to

5X1+8X2≤2000

X1≤175

X2≤225

7X1+4X2≤1400

X1,X2≥O

Sol: Profit Rs.19666X2/3, X1=800/9 & X2=1750/9

Q.2. Minimize(Cost) Z=200X1+400X2

Sub to

X1+X2≥200

X1+3X2≥400

X1+2X2≥350

X1,X2≥0

Sol: Cost Rs.60,000,X1=100&X2=100

Q.3 Maximize (Profit) Z=80X1+120X2

Sub to

X1+X2≤9

X1≥2

X2≥3

X1,X2≥0

Sol: Profit Rs.960, X1=3&X2=6

Simplex method

Q.1 Maximize (Profit) Z =3X1+2X2

Sub to

5X1+3X2≤16000

3X1+3X2≤14000


148

2X1+4X2≤12000

X1,X2 ≥0

Sol: Profit Rs. 10,000, X1=2000,&X2=2000


149

Q.2 Minimize (cost) Z=4X1+3X2

Sub to

200X1+100X2≥4000

X1+2X2≥50

40X1+40X2≥1400

X1,X2≥0

Sol: Cost Rs.110,X1=5,&X2=30

Q.3 Maximize (Profit) Z=2X1+3X2+4X3

Sub to

3X1+X2+6X3≤600

2X1+4X2+2X3≥480

2X1+3X2+3X3=540

X1,X2,X3≥0

Sol: Maximum Z =624,X1=0,X2=96 &X3=84

Bibliography

150

BIBLIOGRAPHY

Render, B., Ralph, M.S., Michal, E.H. Quantitative Analysis for Management

Edition 2008. Arora, M. N., Cost and Management Accounting, edition 2006.

Sharma, J.K. Quantitative Techniques for Managerial Decision, Edition 2009.

Khandelwal, R.S., Gupta, B.L., Agarwal, S., Ahmad, T. Quantitative Analysis

for Management, Edition 2009.

complete book

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