Q T whole book

1. Introduction of Quantitative Technique
1
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

2. Introduction of Quantitative Technique
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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

3. Introduction of Quantitative Technique
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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

4. Introduction of Quantitative Technique
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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

5. Introduction of Quantitative Technique
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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

6. Introduction of Quantitative Technique
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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.

7. 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.

8. 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:

9. 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.

10. Assignment Problems
10
See diagrammatic Representation of Hungarian Approach

11. 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

12. 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:

13. 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
Machine2 – Job5 Cost Rs 10
Machine3 – Job4 Cost Rs 1
Machine4 – Job2 Cost Rs 1
Machine5 – Job1 Cost Rs 8
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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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-

20. 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

21. 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

22. 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

23. 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.

24. 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 ∞

25. 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

26. 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

27. 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

28. 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

29. 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

30. 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

31. 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

32. 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. Transportation Problem
33
STEPS OR PROCEDURE OF SOLVING TRANSPORTATION PROBLEMS
Make Empty Cells(4)
YES
NO
NO
Calculate Value ofRi and Cj (5)
Xij =TCij – (Ri + Cj)
YES
Is Xij <0 ? Select Minimum Value ofXij
Construct Loop
Revise solution and check for rim
value condition
Find aBasic 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)
OptimalSolution

34. Transportation Problem
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. Transportation Problem
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:
P1 P2 P3 Capacity(tons)
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
Requirement 700 500 800

36. Transportation Problem
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
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:

37. Transportation Problem
37
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 700 500 800
Determine optimal allocation of requirements.
Solution:
P1 P2 P3 Capacity(tons)
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
Requirement 700 500 800
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. Transportation Problem
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.
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 700 500 800
Determine optimal allocation of requirements.

39. Transportation Problem
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
2000COLUMN
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
entries. After that we select the row C3, because it has largest difference. In C3, find
the least value is 9 after that assign the maximum possible number 200 to the cell so
that it exhausted the Row C3 completely.

40. Transportation Problem
40
After that we will find remaining rows and columns, the difference for two lowest
entries. After that we select the row C2, because it has largest difference. In C7, find
the least value is 4 after that assign the maximum possible number 400 to the cell so
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)
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 700 500 800 2000
Determine optimal allocation of requirements.

41. Transportation Problem
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

42. Transportation Problem
42
R4+C3 = TC R4+8 = 11 R4 =3
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

43. Transportation Problem
43
TC-(R3+C2) = 18-(5+12) = 1
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

44. Transportation Problem
44
W.H.
CAPACITY
3 3 4 5 15
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
CapacitiesC1 C2 C3 C4 C5
F1 4 2 3 2 6 8
F2 5 4 5 2 1 12

45. Transportation Problem
45
Solutions:
Factory
Customer
CapacitiesC1 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
F3 6 5 4 7 7 14
Required 4 4 6 8 12
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. Transportation Problem
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. Transportation Problem
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
Step4: 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. Transportation Problem
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. Transportation Problem
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. Transportation Problem
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. Transportation Problem
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
Price1 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. Transportation Problem
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. Transportation Problem
53
Case Study:

54. Game Theory
54
GAME THEORY
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.

55. Game Theory
55
 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.
 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.

56. Game Theory
56
NO
YES
NO
YESYES
Law of dominance
START
Find the saddle point
If saddle is
there
If Matrix
is 2*2
If matrix is
m*2 or 2*n
NO
YES

57. Game Theory
57
FLOW CHART REPRESENTING FLOW OF GAME THEORY
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.

58. Game Theory
58
 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
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
m*n
2*2 Arithmetic Method
m*n LPP (Linear
Programming Problem)
m*2
Arithmetic
method/graphical
method
2*n
Arithmetic
method/Graphical
method (lower envelop
& upper point)

59. Game Theory
59
 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.
 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.

60. Game Theory
60
 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.

61. Game Theory
61
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.

62. Game Theory
62
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:

63. Game Theory
63
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

64. Game Theory
64
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

65. Game Theory
65
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

66. Game Theory
66
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

67. Game Theory
67
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 15H1
H2

68. Game Theory
68
18 18
17 17 M3
16 16
15 15
M2
14 14
13 13
12 12
11 11
10 10
9 9
8 8V=11.19
7 7
6 6
5 5
4
4
M1
3 3
2 2
1 1

69. Game Theory
69
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

70. Game Theory
70
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

71. Replacement Theory
71
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.

72. Replacement Theory
72
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.

73. Replacement Theory
73
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)
Operatingcostof
newequipment
Cost of operating
Replacement Replacement
Time
(b)

74. Replacement Theory
74
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

75. Replacement Theory
75
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.

76. Replacement Theory
76
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.

77. Replacement Theory
77
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, 5th 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.

78. Replacement Theory
78
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:

79. Replacement Theory
79
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

80. Replacement Theory
80
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.

81. Replacement Theory
81
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)

82. Replacement Theory
82
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?

83. Queuing Theory
83
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

84. Queuing Theory
84
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
facilivy 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 FastLow
High
Total cost of the service
service
Cost
Low
optical service level

85. Queuing Theory
85
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.

86. Queuing Theory
86
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).

87. Queuing Theory
87
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.
(ii i) 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.

88. Queuing Theory
88
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.

89. Queuing Theory
89
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)

90. Queuing Theory
90
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-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

91. Queuing Theory
91
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 possibl
e 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