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Quantitative Techniques
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Decision-making is an integral part of the process of management. It is the competence as a decision-maker
that distinguishes a good manager from others. While traditionally, decision-making has been considered
as an art, the growing complexity and competitive nature of environment in which managements have to
operate now necessitate that decisions be made more systematically using as much quantitative information
as possible, along with a consideration of qualitative factors. This book is about the use of quantitative
techniques in managerial decision-making.
Broadly speaking, decision-making involves choosing a course of action from various available alternatives.
The job of a manager, in the process of selecting from among available alternatives, is facilitated in a large
measure by the application of appropriate quantitative techniques when, and to an extent, a problem
is quantified. This approach to decision-making is known by several names like operations research,
management science, quantitative analysis, so on.
The contents of this chapter will help a manager to understand questions like the following:
€ What is operations research and how has it evolved? What are its characteristic features?
€ What is the methodology used in operations research? In this context, what is problem formulation, model
building, acquisition of input data, solution and interpretation of the results obtained, model validation and
implementation of the solution?
€ What are different types of models, and what is the use of mathematical models in operations research?
€ What are different classifications of solutions–feasible and infeasible; optimal and non-optimal; and
unique and multiple optimal solutions?
€ What is sensitivity analysis?
€ How quantitative analysis is an integral part of the modern computer-based information systems and
how are quantitative tools used in each of the subsystems?
Thus, this introductory chapter gives some details about the decision-making process and an idea about the
nature and methodology of the quantitative analysis. Finally, a plan of the book is presented which contains
a brief account of the contents of each of the chapters to follow.
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Learning Objectives
After reading this chapter, you should be able to:
LO 1
LO 2
LO 3
LO 4
LO 5
LO 6
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LO 1 Identify the elements
of a decision and various
decision-making situations
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LO 2 Know the role of
quantitative analysis in
decision-making
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We live in a world of shortages since the supply of our resources is limited. This is all the more true at the
micro level. Thus, there is always a problem as to how to allocate the given resources in the best possible
manner. Linear programming is a technique which provides the answer in a wide variety of cases.
Some situations where a manager can use linear programming include the following:
€ How to allocate the advertising budget among various alternate advertising media which have different
degrees of effectiveness in reaching audiences and involve different costs?
€ In case of a multi-product firm, what product-mix will yield the maximum profit, when different products
are known to have different profitability coefficients and different resource requirements?
€ How should the given funds be allocated between different investment opportunities that yield varying
returns and involve different degrees of risk?
€ How should a dietician decide about the foods that contain varying proportions of ingredients like
carbohydrates, vitamins, proteins, etc. to be given to the patients so that their nutrition requirements are
met with at the minimum cost?
€ How should the land on an agricultural farm be allocated between different crops which involve different
costs on account of labour, manure, seeds, etc. and have different yields, resulting in unequal profitability
of the agricultural products produced?
€ How should the HR manager of a hospital decide about the employment of nurses that involves lowest
cost and yet meets the requirements at different times of the 24-hour day?
The next few chapters are devoted to a detailed account of linear programming. It must be kept in mind
that the most important thing is to develop the skill and ability to translate a given real-life situation into a
linear programming format, keeping in mind its assumptions and limitations. This chapter illustrates this
with examples. Later in the chapter, the graphic solution to some such problems is provided.
The chapter heavily uses the inequalities of ‘less than’ and ‘greater than’ types and equations. You should
be conversant with two-dimensional graphs and their use. Plotting of equations and inequalities on a graph
and the ability to determine the space on the graph over which they are satisfied, called the feasible area,
holds the key to successful graphic solution to the problems.
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Learning Objectives
After reading this chapter you should be able to:
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Once a linear programming problem is formulated, the next issue is to solve it for optimal values of
the decision variables, which leads to maximisation (say, of profit) or minimisation (may be, of cost).
The graphic method is indeed available but, as we know, it has its limitations. When only two items are
produced by a firm, two investment opportunities are available to an investor, two advertising media
are available to an advertising manager—in other words when only two decision variables are involved
— then we can think of a graphic solution to an LPP. In case of more than two variables, however, the
manager has to turn to the Simplex method. The Simplex method is a very powerful tool for solving linear
programming problems. The Simplex and its variants can handle any complexities in the LPPs irrespective
of the number and nature of the decision variables and constraints.
The application of Simplex method for solving a linear programming problem helps a manager to know
answers to questions like the following:
€ What mix of the products involved would yield the maximum profit?
€ What maximum profit can be obtained from the optimal product-mix?
€ Whether all resources would be fully utilised by the optimal mix, or are there any unutilised resources
and to what extent?
€ If the problem is of the minimisation type, then what mix of the products/items would minimise the
cost and what is the cost involved?
€ Does the problem have an alternate solution which is as good as the optimal solution obtained? If yes,
what is that solution?
€ Is any of the given restrictions of no consequence? To illustrate, a particular raw material may be
available in plenty in a given situation.
€ Does the problem have no solution that can meet all the requirements?
While the solution to relatively small or moderately large problems may be found while working
manually with the method, large-scale problems can be solved with the help of computers where software
are available. Real life problems obviously require a computer for solution.
Knowledge of simple algebraic manipulations and a good hand at arithmetic calculations are the prerequisites
for applying Simplex algorithm. A command over arithmetic operations on fractional values is necessary
and desirable for Simplex calculations. Unfortunately, an ordinary electronic calculator is not of much help
in this case.
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Learning Objectives
After reading this chapter, you should be able to:
LO 1
LO 2
LO 3
LO 4
LO 5
LO 6
LO 7
LO 8
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' method and conditions for
its application
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a standardised form
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ŶŽŶͲŶĞŐĂƟǀĞ͕ƚŚĞƐŽůƵƟŽŶŝƐƐĂŝĚƚŽďĞbasic feasible solution͘
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D & ' D LO 3 Describe the steps
for obtaining solution to an
LPP using Simplex method
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TABLE 3.1
Simplex Tableau
Coefficient values from
constraint equations
Basic variables with their
coefficients in the objective
function
Basis
x1
x2
S1
S2
bi
S1
2
3
1
0
60
S2
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per unit
0
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values
0
cj
4
3
0
1
40
35
0
0
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0
0
60
96
Dj = cj – zj
40
35
0
0
96
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TABLE 3.2
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TABLE 3.4
Simplex Tableau 3: Optimal Solution
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TABLE 3.12
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Simplex Tableau 3: Optimal Solution
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START
Add necessary slack, surplus and artificial variables to convert inequalities into equations
Write the initial tableau and obtain the solution corresponding to the identity
Calculate Dj = cj – zj
Is the
problem
maximisation or
minimisation?
Maximisation
Minimisation
Select the most negative Dj.
Designate the column as
‘key column'
Select the largest Dj. Designate
the column as ‘key column’
Divide the coefficients in the key column into RHS elements, bi’s
Select the row with smallest non-negative quotient and call it ‘key row’
Designate the intersection of key row and key column as ‘key element'
Divide all elements of the key row by key element to get replacement
row, for the revised simplex tableau
Solve for each remaining row in the matrix
Element in Corresponding
the key
element in
New Row i = Old Row i column
the replacement
and row i
row
Calculate Dj = cj – zj
Maximisation
No
Are
all Dj 0 or
- ve?
Yes
Is the
problem
maximisation or
minimisation?
The solution is optimal
Minimisation
Yes
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all Dj 0 or
+ ve?
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TABLE 3.17
Simplex Tableau 1: Non-optimal Solution
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TABLE 3.18
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+
,
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,
+
,
,'
+
+
?
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0+
+M@
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0+M@
2???
1?
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?
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B?
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TABLE 3.19
Simplex Tableau 3: Optimal Solution
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,
,
?
?
+
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0+M+=?
1?
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3KDVH,, &/ '& 2 ' " +,& ' &2,?2,+
TABLE 3.20
Simplex Tableau 4: Non-optimal Solution
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?
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1?
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TABLE 3.21
¨
Simplex Tableau 5: Optimal Solution
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% ' LO 7 Summarise multiple
optimal solutions, infeasibility, unboundedness and
degeneracy
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TABLE 3.22
&2,,02,@
Simplex Tableau 1: Non-optimal Solution
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TABLE 3.23
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+=?
2
≠
Simplex Tableau 2: Non-optimal Solution
$ +
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2
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?
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1
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TABLE 3.24
Simplex Tableau 3: Optimal Solution
$
+
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2
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TABLE 3.25
Simplex Tableau 4: Optimal Solution
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TABLE 3.26
$
Simplex Tableau 1: Non-optimal Solution
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TABLE 3.27
Simplex Tableau 2: Non-optimal Solution
$
+
?
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2M,I
+
0+M,
?
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TABLE 3.28
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0
?
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Simplex Tableau 3: Final, Non-optimal Solution
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TABLE 3.29
$
Simplex Tableau 1: Non-optimal Solution
+
,
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2
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≠
TABLE 3.30
$ +
?
Simplex Tableau 2: Non-optimal Solution
+
,
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,
2
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M'
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2M,I
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TABLE 3.31
Simplex Tableau 3: Final, Non-optimal Solution
$ +
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,
2
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,
?
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+, + ,+, ≥?
Simplex Tableau 1: Non-optimal Solution
+
,
+
+
,
,
M'
+
0
,
@
0+
?
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@
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0
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+=
+?42
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0
0
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≠
TABLE 3.33
$ Simplex Tableau 2: Non-optimal Solution
+
,
+
,
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,
M'
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≠
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TABLE 3.34
$
Simplex Tableau 3: Non-optimal Solution
+
,
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?
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TABLE 3.35
$
,
0=M3
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>M2
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≠
+@ ¨
Simplex Tableau 4: Unbounded Solution
+
,
+
,
+
,
,'
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+
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x2
8
3x 1
+ x2
7
=8
A
6
Degenerate, optimal solution
6x 1
+3
5
x2
8
=1
4
4x
1 +
3
2
5x
2
=
30
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1
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& &223
TABLE 3.36
Simplex Tableau 1: Non-optimal Solution
$ +
,
+
,
2
M'
+
?
3
2
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3
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2
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1
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&22>
TABLE 3.37
$
Simplex Tableau 2: Optimal Solution
+
,
+
,
2
,
+
+M2
?
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3
,
2?
,
?
+
?
0+M2
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TABLE 3.38
$
+
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Simplex Tableau 3: Non-optimal Solution
+
,
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+
+
,
?
2
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?
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?
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3
+=M,
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03
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& $ D . ,1¥?42?¥3<+1?& &22B
TABLE 3.39
$
Simplex Tableau 4: Optimal Solution
+
,
+
,
2
+
,1
+
?
=M+1
?
0+M3
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x2
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+
,
<=+4,,
6
$
"
?
?
+
2
?
1
3
?
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+=
4
Non-degenerate
optimal solution
8 A
B
4x
1
3x
0
x2
1 +
FEASIBLE
REGION
2
+2
x2
=1
6
=9
1
+
=
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C
3
2
1
3x
x2
x1
4
–2
& $ " " ' " % & . ' ' " +<2,<?
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solution
–4
–6
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2+0,4 2 <B
+, + , 2 ≥?
& &2@?
TABLE 3.40
$
+
Simplex Tableau 1: Non-optimal Solution
+
?
@
,
+
?
?
+3
@
,
?
2
+
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B
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/ " # , # 2
:KHQ6LV(OLPLQDWHG
( ,
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TABLE 3.41
Simplex Tableau 2: Non-optimal Solution
$
+
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,M2I
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0@M2
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+
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TABLE 3.42
$
Simplex Tableau 3: Optimal Solution
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,
2
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2
,
,
?
+
2M,
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3
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=
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0+M,
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TABLE 3.43
$
Simplex Tableau 4: Non-optimal Solution
+
,
+
,
2
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≠
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& &2@@
TABLE 3.44
$ +
Simplex Tableau 5: Non-optimal Solution
+
,
+
,
2
?
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?
+
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# <0+M,
TABLE 3.45
$
Simplex Tableau 6: Optimal Solution
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,
+
,
2
2
?
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+
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& &2@= &2@,
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infeasibility͘
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ƉƌŽďůĞŵŝƐunbounded͘
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/ #;/6' & LO 8 Solve linear
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Start
Obtain initial solution to
the problem
Any
basic variable
with solution value
=0?
No
Yes
Solution is non-degenerate
Solution is degenerate
Is the
problem
maximisation or
minimisation
?
Maximisation
Minimisation
Are
all D j ≥ 0?
Are
all D j £ 0?
No
Yes
Any
artificial
variable in
the basis
?
Yes
No
All
key column
elements £ 0?
No
Yes
Any
non-basic
variable
D j = 0?
Obtain revised
solution
Infeasibility
No
Yes
Unbounded solution
Unique optimal
solution
End
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TABLE 3.49
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TABLE 3.50
$ Simplex Tableau 3: Non-optimal Solution
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TABLE 3.51
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TABLE 3.53
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TABLE 3.54
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TABLE 3.55 Simplex Tableau 1: Non-optimal Solution
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TABLE 3.56
Simplex Tableau 2: Non-optimal Solution
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TABLE 3.57
Simplex Tableau 3: Non-optimal Solution
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TABLE 3.66
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The application of simplex method allows not only the solution to linear programming problems but also
provides a fund of information which can be usefully employed by the manager in decision-making. The
numbers generated in the course of applying simplex provide inputs to obtain answers to many ‘what-if’
questions.
Further, every linear programming problem has a mirror image problem which is also a linear programming
problem. Using a set of rules, the mirror image problem, called the dual, can be obtained for a given linear
programming problem, known as primal problem in this context. This ‘primal-dual’ relationship is very
important. Since the two problems are connected to each other, there is obviously a relationship to be
expected between their solutions; there indeed is. This chapter explores and explains this, and does more.
Apart from a mathematical connection between primal and dual, it explains the economic significance of
the dual and helps a manager to answer questions like the following:
€ Is it advisable to produce all the items that can be produced in a given situation? If some item is left out
in the optimal product-mix, then under what condition will it be advisable to produce it?
€ What is the marginal profitability of each of the resources of the firm? In turn, this means by how much
the profit will increase if more quantity of a particular resource is added or how much reduction in profit
will result if its availability is reduced?
€ To what extent will additional quantities of a resource (or its reduction) cause an increase (or decrease)
at a uniform rate?
€ Will the optimal product-mix need to change it the profitability of the various products changes? Obviously,
when some small changes (increases or decreases) occur in the profitability of the products it would not
cause changes in the quantities of the products being produced, but then what are the limits of these
price changes?
€ Is it advisable to introduce a new product given the amounts of resources required for its production
and its profitability?
€ How would the product-mix change, if at all, its technological changes cause the resource requirements
of a product to change?
This chapter involves handling of inequalities; arithmetical operations on fractional values; plotting of
equalities and inequalities on graphs, and their understanding. Further, a knowledge of the concepts of
matrices and their transpose is needed.
ഩ‹ഩYƵĂŶƟƚĂƟǀĞdĞĐŚŶŝƋƵĞƐŝŶDĂŶĂŐĞŵĞŶƚ
Learning Objectives
After reading this chapter, you should be able to:
LO 1
LO 2
LO 3
LO 4
LO 5
LO 6
LO 7
LO 8
<ŶŽǁƚŚĞĐŽŶĐĞƉƚŽĨĚƵĂůŝƚLJ
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džĞĐƵƚĞϭϬϬƉĞƌĐĞŶƚƌƵůĞŝŶƐĞŶƐŝƟǀŝƚLJĂŶĂůLJƐŝƐǁŝƚŚŵƵůƟƉůĞƉĂƌĂŵĞƚĞƌĐŚĂŶŐĞƐ
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LO 1 Know the concept
of duality
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+ , $ + , + , + , . + , . + , - "## * LO 2 Identify the
conditions necessary for
obtaining dual to an LPP
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:ULWLQJWKH'XDO
= & > ? > .
LO 3 Construct the dual
to a given LPP
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ƚŚĞĚƵĂůƚŽĂŐŝǀĞŶƉƌŝŵĂůƉƌŽďůĞŵĂƌĞ;ŝͿĂůůƚŚĞǀĂƌŝĂďůĞƐĂƌĞŶŽŶͲŶĞŐĂƟǀĞĂŶĚ;ŝŝͿĂůůƚŚĞĐŽŶƐƚƌĂŝŶƚƐƐŚŽƵůĚ
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11
28
10
9
24
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TABLE 4.1
[ 2
Simplex Tableau: Optimal Solution
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TABLE 4.2
Simplex Tableau 1: Non-optimal Solution
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TABLE 4.3
≠
Simplex Tableau 2: Non-optimal Solution
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TABLE 4.4
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TABLE 4.6
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TABLE 4.7
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TABLE 4.8
Simplex Tableau: Optimal Solution
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TABLE 4.9
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TABLE 4.12
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TABLE 4.13
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TABLE 4.17
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TABLE 4.20
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TABLE 4.25
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The distribution of goods produced by a factory from various warehouses (or sources) to different
markets (or other destinations) where they are required causes problems to almost every business. The
transportation method is developed to deal with the transportation of goods from different sources to
different destinations, given the relevant data like available quantities at various sources, demand at
each of the destinations, the cost of shipping along each route and non-availability of certain routes, if
any. It may be interesting to note that while such problems can also be handled as linear programming
problems; the transportation method provides an efficient means to solve them.
The method allows the manager to seek answers to the questions like the following:
€ What is the optimal way of shipping goods from various sources (warehouses) to different markets so
as to minimise the total cost involved in the shipping?
€ How to handle a situation when some routes are not available or when some units have to be necessarily
transported from a particular source to a particular market?
€ How would the optimal shipping schedule change if some routes become cheaper/costlier?
€ If it were possible to increase supply, which of the sources should be preferred?
€ Instead of allowing shipping of goods only from listed sources to different destinations, if it were possible
to ship goods from a particular source to another source or destination and then from there to a further
destination, how much cost can be saved? This is what is called a transhipment problem.
€ If an item can be produced at different locations at varying costs and sold in different markets at different
prices, then what production and shipping plan will yield maximum profit?
€ How can the production be scheduled, given the cost of production of an item and the cost of carrying
stock, so as to meet the requirements in different periods?
The use of transportation method is not limited to solution of the transportation problems alone. Problems
of scheduling production, controlling inventory and management of funds over different time periods
illustrate some other areas which lend themselves to handling by the transportation method.
This chapter basically requires elementary arithmetic calculations. However, familiarity with summation
notation, a basic knowledge of inequalities, matrices and their transpose is also necessary. While working
through this chapter, master the skill of drawing of a closed path.
^ƉĞĐŝĂůůLJ^ƚƌƵĐƚƵƌĞĚ>ŝŶĞĂƌWƌŽŐƌĂŵŵĞƐ/͗dƌĂŶƐƉŽƌƚĂƟŽŶĂŶĚdƌĂŶƐŚŝƉŵĞŶƚWƌŽďůĞŵƐഩ‹ഩ
Learning Objectives
After reading this chapter, you should be able to:
LO 1
LO 2
LO 3
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LO 9 /ůůƵƐƚƌĂƚĞƐĞŶƐŝƟǀŝƚLJĂŶĂůLJƐŝƐĨŽƌƚƌĂŶƐƉŽƌƚĂƟŽŶƉƌŽďůĞŵƐ
LO 10 /ŵƉůĞŵĞŶƚƉƌŽĚƵĐƟŽŶƐĐŚĞĚƵůŝŶŐĂŶĚŝŶǀĞŶƚŽƌLJĐŽŶƚƌŽů
ƚŚƌŽƵŐŚƚƌĂŶƐƉŽƌƚĂƟŽŶŵŽĚĞů
LO 11 ^ŽůǀĞƚƌĂŶƐŚŝƉŵĞŶƚƉƌŽďůĞŵƐ
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TABLE 5.1
Transportation Tableau
Destination ( j)
Origin (i)
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TABLE 5.3
To
From
Initial Feasible Solution: NWC Method
Q
P
180
A
12
R
Supply
13
500
170
150
10
S
12
180
B
7
11
8
120
300
14
200
C
6
16
11
7
200
Demand
180
150
350
320
1,000
Total cost = 12 ¥ 180 + 10 ¥ 150 + 12 ¥ 170 + 8 ¥ 180 + 14 ¥ 120 + 7 ¥ 200 = ` 10,220
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TABLE 5.4
To
From
A
Obtaining Initial Feasible Solution: LC Method
P
Q
12
10
R
150
Supply
S
50
12
13
300
500
300
B
7
11
8
14
C
6
16
11
7
200
20
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180
150
350
50
320
300
1,000
180
300
20
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TABLE 5.5
Initial Feasible Solution: Least Cost Method
To
P
Q
A
12
10
12
B
7
11
C
6
Demand
180
From
R
50
150
Supply
S
300
13
500
8
14
300
16
11
7
200
150
350
320
1,000
300
180
20
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TABLE 5.6
Obtaining Initial Feasible Solution: VAM
To
Q
From
P
R
A
12
B
7
11
8
14
C
6
16
11
Demand
180
150
I
1
II
III
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Supply
S
I
II
III
500
2
2
2
300
120
1
1
3
7
200
1
–
–
350
230
320
120
1,000
1
3
6
5
1
4
1
–
1
4
1
10
150
230
12
180
13
120
120
200
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TABLE 5.7
Initial Feasible Solution: VAM
To
P
Q
A
12
10
B
7
11
8
14
C
6
16
11
7
200
Demand
180
150
350
320
1,000
From
R
230
150
12
180
Supply
S
13
120
500
120
300
200
Total cost = 10 ¥ 150 + 12 ¥ 230 + 13 ¥ 120 + 7 ¥ 180 + 8 ¥ 120 + 7 ¥ 200 = ` 9,440
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START
For each row and column,
find the difference between
the two least cost cells that
have not been allocated
Select the largest of the
differences of rows and
columns, and, in case of a tie,
choose the one corresponding
to which largest number of
units can be assigned or the
cost value is the lowest
Assign the largest quantity
permissible by the rim requirements to the cell in that row/
column with the smallest cost
Eliminate the row/column
that has been satisfied
No
Are
all rim
conditions
satisfied?
Yes
STOP
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TABLE 5.8
To
From
Initial Feasible Solution: Testing for Optimality
Q
P
180
A
12
R
Supply
13
500
170
150
10
S
–
12
+
180
B
7
11
C
6
16
Demand
180
150
+
8
120
–
300
14
200
11
7
200
350
320
1,000
Total cost = 12 ¥ 180 + 10 ¥ 150 + 12 ¥ 170 + 8 ¥ 180 + 14 ¥ 120 + 7 ¥ 200 = ` 10,220
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TABLE 5.9
To
From
Improved Solution: Non-Optimal
Q
P
180
R
50
150
Supply
S
120
13
500
8
14
300
16
11
7
200
150
350
320
1,000
A
12
10
12
B
7
11
C
6
Demand
180
300
200
Total cost = 12 ¥ 180 + 10 ¥ 150 + 12 ¥ 50 + 13 ¥ 120 + 8 ¥ 300 + 7 ¥ 200 = ` 9,620
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TABLE 5.10
To
Improved Solution: Optimal
P
Q
A
12
10
12
B
7
11
C
6
Demand
180
From
R
230
150
Supply
S
120
13
500
8
14
300
16
11
7
200
150
350
320
1,000
180
120
200
Total cost = 10 ¥ 150 + 12 ¥ 230 + 13 ¥ 120 + 7 ¥ 180 + 8 ¥ 120 + 7 ¥ 200 = ` 9,440
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TABLE 5.11
Initial Feasible Solution (LCM): Testing for Optimality
To
P
Q
A
12
10
B
7
11
8
14
C
6
16
11
7
200
Demand
180
150
350
320
1,000
From
R
Supply
S
50
150
13
12
300
500
300
300
20
180
Total cost = 10 ¥ 150 + 12 ¥ 50 + 13 ¥ 300 + 8 ¥ 300 + 6 ¥ 180 + 7 ¥ 20 = ` 9,620
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TABLE 5.12
To
From
Improved Solution: Optimal
Q
P
R
230
150
A
10
12
12
180
Supply
S
120
13
500
300
120
B
7
11
8
14
C
6
16
11
7
200
Demand
180
150
350
320
1,000
200
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TABLE 5.13
Initial Solution: Testing for Optimality
To
Q
P
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R
150
180
A
12
10
B
7
11
–
12
170
+5
+
S
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ui
13
500
0
300
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180
+1
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8
120
–
14
200
C
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6
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16
– 10
11
7
200
1,000
Demand
180
150
350
320
vj
12
10
12
18
Total cost = 12 ¥ 180 + 10 ¥ 150 + 12 ¥ 170 + 8 ¥ 180 + 14 ¥ 120 + 7 ¥ 200 = ` 10,220
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TABLE 5.14
Improved Solution: Non-optimal
To
Q
P
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B
+
+1
12
10
7
11
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ui
13
500
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300
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200
1,000
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150
350
320
vj
12
10
12
13
Total cost = 12 ¥ 180 + 10 ¥ 150 + 12 ¥ 50 + 13 ¥ 120 + 8 ¥ 300 + 7 ¥ 200 = ` 10,220
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TABLE 5.15
Improved Solution: Optimal
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10
12
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C
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180
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13
500
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300
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16
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7
200
1,000
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180
150
350
320
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11
10
12
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Total cost = 10 ´ 150 + 12 ´ 230 + 13 ´ 120 + 7 ´ 180 + 8 ´ 120 + 7 ´ 200 = ` 9,440
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7KHJLYHQWUDQVSRUWDWLRQSUREOHPLVXQEDODQFHGVLQFHDJJUHJDWHGHPDQG XQLWVZKLOH
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TABLE 5.16
Transportation Problem: Unbalanced, with Prohibited Routes
-2
* 'HPDQG
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TABLE 5.17
Initial Basic Feasible Solution by VAM
B
A
Iteration
Market
Warehouse
Supply
C
I
II
III
IV
V
180
5
5
–
–
–
100
60
5
5
5
3
3
M
160
4
4
4
4
–
9
120
80
2
2
2
4
4
0
0
100
0
–
–
–
–
660
1
M
2
14
3
9
4
11
5
0
Demand
240
140
200
40
220
40
I
9
5
6
II
2
2
1
III
2
2
3
IV
2
2
-
V
3
4
-
60
12
7
11
6
180
5
80
100
7
160
40
40
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UHSURGXFHGLQ7DEOHDQGWHVWHGIRURSWLPDOLW\6LQFHDOOD £WKHVROXWLRQLVVHHQWREHRSWLPDO1RWH
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LWLVERXQGWREHH[WUHPHO\QHJDWLYHVLQFHLWZRXOGLQYROYH0- 7KHVROXWLRQJLYHQLQ7DEOHLVQRWXQLTXHEHFDXVHLQFHOOD 7RREWDLQDQDOWHUQDWLYHRSWLPDO
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TABLE 5.18
Initial Feasible Solution: Optimal
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LO 7 Deal with degenerate
solutions and maximisation
problems
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TABLE 5.19
Alternate Optimal Solution
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PHWKRGDQGWKHPRGL¿HGGLVWULEXWLRQPHWKRG 02', DUHLQRSHUDWLYHLQVXFKDFDVH7KHIRUPHUFDQQRWEH
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PRUHLIWKHQHHGEH HPSW\FHOODQGWUHDWWKHFHOODVDQRFFXSLHGFHOO7KLVDPRXQWLVUHSUHVHQWHGE\D*UHHN
OHWWHUe HSVLORQ DQGLVWDNHQWREHVXFKDQLQVLJQL¿FDQWYDOXHWKDWZRXOGQRWDIIHFWWKHWRWDOFRVW7KXVLWLV
ELJHQRXJKWRFDXVHWKHSDUWLFXODUURXWHWRZKLFKLWLVDVVLJQHGWREHFRQVLGHUHGDVDEDVLFYDULDEOHEXWQRW
ODUJHHQRXJKWRFDXVHDFKDQJHLQWKHWRWDOFRVWDQGRWKHUQRQ]HURDPRXQWV$OWKRXJKeLVWKHRUHWLFDO\QRQ
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WKHFHOOFKRVHQIRULQVHUWLQJHSVLORQPXVWEHDQLQGHSHQGHQWFHOORULJLQDWLQJIURPZKLFKDFORVHGORRSFDQQRW
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:KLOHDQXPEHURILQGHSHQGHQWFHOOVPD\EHDYDLODEOHLQDJLYHQVLWXDWLRQLWPD\EHDJRRGLGHDWRFKRRVHWKH
RQHZLWKWKHVPDOOHVWXQLWFRVWYDOXH)URPDSUDFWLFDOVWDQGSRLQWLWLVQRWQHFHVVDU\WRHYDOXDWHHDFKDQGHYHU\
HPSW\FHOOIRULQGHSHQGHQFHWRLQVHUWHSVLORQ,QVWHDGDFHOOZLWKPLQLPXPFRVWPD\EHH[DPLQHGZKHWKHULW
LVLQGHSHQGHQWRUQRW,ILWLVIRXQGWREHLQGHSHQGHQWWKHQSODFHHSVLORQLQLWDQGLIQRWWKHQPRYHWRDQRWKHUFHOO
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3
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3
3
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TABLE 5.20
Initial Feasible Solution: Non-optimal
To Æ
M1
M2
M3
M4
P1
30
6 –
4
9
+1
P2
20
From
Ø
–4
–11
M5
Supply
ui
0
40
0
40
2
0
50
2
0
90
1
220
10
–2
20
P3
–12
1
7
–4
6
–11
11
–
3
50
+
1
0
1
–11
14
20
+
–
0
e
30
60
P4
7
1
Demand
90
30
50
30
20
vj
6
0
–2
1
–2
–13
12
–4
6
–1
`
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TABLE 5.21
Revised Solution: Optimal
To Æ
From
Ø
M1
P1
6
P2
20
M2
M3
M4
4
9
1
11
3
M5
Supply
ui
0
40
0
0
40
2
0
50
1
0
90
1
220
10
30
–4
–10
–2
20
20
– 12
–4
6
–10
e
50
P3
7
P4
7
1
–12 12
–4 6
Demand
90
30
50
30
20
vj
6
0
–1
1
–2
0
0
1
–12
14
–1
30
60
–1
Total cost = 6 ¥ 30 + 1 ¥ 10 + 3 ¥ 20 + 0 ¥ 50 + 7 ¥ 60 + 1 ¥ 30 = ` 700
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VLJQRQWKHFORVHGORRS7KHRSWLPDOLW\WHVW02',VXJJHVWVWKDWWKHVROXWLRQJLYHQLQ7DEOHLVDQRSWLPDO
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LIQHFHVVDU\ RIWKHUHFHQWO\YDFDWHGFHOOVZLWKWKHPLQLPXPFRVW$QGWKHQZHSURFHHGZLWKWKHSUREOHPLQ
WKHXVXDOPDQQHU
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0DUNHW
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0
0
0
0
3
3
3
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TABLE 5.22
Initial basic feasible solution: Non-optimal
Market
Supply
ui
8
150
0
5
125
–1
3
Plant
M1
M3
M2
65
P1
85
7
5
10
–
–3
+
110
P2
M4
–4
15
6
8
4
–
+
–2
–2
65
P3
+1
+
9
120
12
– 10
7
185
460
–4
Demand
110
80
150
120
vj
7
5
7
4
Total cost = 65 ¥ 5 + 85 ¥ 7 + 110 ¥ 6 + 15 ¥ 4 + 65 ¥ 10 + 120 ¥ 7 = ` 3,130
TABLE 5.23
Revised Solution: Optimal
Market
Supply
ui
8
150
0
5
125
0
3
Plant
M1
M2
M3
10
5
7
M4
150
P1
–4
–4
–1
45
P2
80
6
8
4
–1
–1
e
65
P3
9
–5
120
12
10
7
185
460
Demand
110
80
150
120
vj
6
4
7
4
Total cost = 150 ¥ 7 + 45 ¥ 6 + 80 ¥ 4 + 65 ¥ 9 + 120 ¥ 7 = ` 3,065
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TABLE 5.24
Opportunity Loss Matrix: Initial Solution
Market
Warehouse
A
B
13
7
17
+ 18
C
Availability
ui
200
0
500
11
+ 5
300
15
1,000
D
200
X
– 17
Y
– 10
180
22
19
– 20
100
–4
– 7
15
–5
20
14
0
100
Z
11
Demand
180
320
100
400
vj
–4
7
–1
–4
–
+6
400
Total profit = 18 ¥ 200 + 7 ¥ 100 + 18 ¥ 400 + 14 ¥ 180 + 3 ¥ 20 + 11 ¥ 100 = ` 15,180
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TABLE 5.25
Revised Solution: Non-optimal
Market
Warehouse
A
B
13
7
17
18
C
Availability
ui
200
0
500
11
9
D
200
X
– 11
19
– 14
–4
0
380
120
Y
–4
Z
11
Demand
180
vj
2
180
+ 1+
– 7
15
100
–6
20
14
+ 5
300
320
100
400
1,000
7
5
–4
22
–
Total profit = 18 ¥ 200 + 7 ¥ 120 + 18 ¥ 380 + 14 ¥ 180 + 11 ¥ 100 + 20 ¥ 20 = ` 15,300
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DEDVLFYDULDEOH*LYHQLQ7DEOHWKLVLVWHVWHGWREHRSWLPDO
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TABLE 5.26
Revised Solution: Optimal
Market
Warehouse
A
B
13
7
17
18
C
Availability
ui
200
0
500
11
9
D
200
X
Y
– 11
–4
–15
19
0
–4
280
100
120
15
7
14
5
300
1,000
120
180
Z
11
Demand
180
320
100
400
vj
2
7
4
–4
–6
22
–1
Total profit = 18 ¥ 200 + 7 ¥ 120 + 10 ¥ 100 + 18 ¥ 280 + 14 ¥ 180 + 20 ¥ 120 = ` 15,400
START
Write the problem in tabular form
Is it
balanced?
No
Balance the table using dummy
row/column
Yes
Is it a
maximisation
problem?
Convert it into a minimisation
problem: subtract each element
of the profit matrix from its
highest value
Yes
No
Find an initial basic feasible
solution using NWC rule, VAM etc.
Is it
degenerate
Yes
Eliminate degeneracy by assigning
e to requisite number of cells
No
Generate an improved solution
No
Is it
optimal?
Yes
The problem is solved
STOP
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TABLE 5.27
P
Q
12
10
12
–1
11
8
16
11
7
C
6
ui
13
500
0
14
300
–4
7
200
–6
1,000
230
120
120
180
B
Supply
S
R
150
A
Optimal Solution
To
From
. 7. 8* –5
–5
200
–1
– 12
–5
Demand
180
150
350
320
vj
11
10
12
13
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TABLE 5.28
Transportation Tableau: Sensitivity Analysis
To
From
P
Q
12
10
12
11
8
16
11
153
A
–1
7
C
6
ui
13
503
0
14
300
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7
200
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1,003
230
120
120
180
B
Supply
S
R
–5
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200
–1
–12
–5
Demand
180
153
350
320
vj
11
10
12
13
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TABLE 5.29
Transportation Tableau: Sensitivity Analysis
To
From
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Q
12
10
150
A
–1
+
B
C
–1
–
7
190
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11
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6
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500
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300
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110
110
210
11
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210
1,010
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240
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16
Supply
S
R
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190
150
350
320
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12
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ഩ‹ഩYƵĂŶƟƚĂƟǀĞdĞĐŚŶŝƋƵĞƐŝŶDĂŶĂŐĞŵĞŶƚ
7KHWUDQVKLSPHQWSUREOHPFDQEHGHSLFWHGLQWDEXODUIRUPDVVKRZQLQ7DEOH
TABLE 5.33
#
Transhipment Problem
>
±
±
" ± ± " + ± S S # Ø # Æ
"=
'
=
"
"
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7KH¿UVW URZVUHSUHVHQWWKH FRQVWUDLQWVJLYHQLQ ZKLOHWKHUHPDLQLQJ URZVVKRZWKHFRQVWUDLQWV
JLYHQLQ 7KHFRQVWUDLQWVLQ DQG DUHUHSUHVHQWHGE\WKH¿UVW FROXPQVDQGWKHUHPDLQLQJ FROXPQV
UHVSHFWLYHO\$OOWKHYDOXHVDUHSODFHGRQWKHGLDJRQDOIURPOHIWWRSWRULJKWERWWRP(DFKRIWKHPEHDUVQHJDWLYH
VLJQZKLFKPXVWEHFRQVLGHUHGFDUHIXOO\ZKHQDLVLQYROYHGLQWKHUHDGMXVWPHQW GXULQJWKHVROXWLRQSURFHVV ƚƌĂŶƐƉŽƌƚĂƟŽŶƉƌŽďůĞŵŝƐĐĂůůĞĚĂƐĂtranshipment problemǁŚĞŶƐŚŝƉŵĞŶƚƐŽĨŐŽŽĚƐĂƌĞĂůƐŽĂůůŽǁĞĚĨƌŽŵ
ŽŶĞƐŽƵƌĐĞƚŽĂŶŽƚŚĞƌĂŶĚĨƌŽŵŽŶĞĚĞƐƟŶĂƟŽŶƚŽĂŶŽƚŚĞƌƐŽƚŚĂƚĂƚƌĂŶƐƉŽƌƚĂƟŽŶƉƌŽďůĞŵǁŝƚŚmͲŽƌŝŐŝŶƐ
ĂŶĚnͲĚĞƐƟŶĂƟŽŶƐďĞĐŽŵĞƐĂƚƌĂŶƐŚŝƉŵĞŶƚƉƌŽďůĞŵǁŝƚŚmнnƐŽƵƌĐĞƐĂŶĚĂŶĞƋƵĂůŶƵŵďĞƌŽĨĚĞƐƟŶĂƟŽŶƐ͘
tŝƚŚƐŽŵĞŵŝŶŽƌŵŽĚŝĮĐĂƟŽŶƐ͕ŝƚŝƐƐŽůǀĞĚĂƐĂƚƌĂŶƐƉŽƌƚĂƟŽŶƉƌŽďůĞŵ͘
([DPSOH
5HFRQVLGHU([DPSOH1RZDVVXPHWKDWLWLVSRVVLEOHIRUWKHLWHPLQTXHVWLRQWREHVKLSSHG
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3
4
5
6
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4
&
5
6
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7KHJLYHQSUREOHPLVZULWWHQDVDWUDQVKLSPHQWSUREOHPLQ7DEOH:HVWDUWZLWKWKHRSWLPDOVROXWLRQREWDLQHG
SUHYLRXVO\$SSOLFDWLRQRIWKH02',PHWKRGIRUGHWHUPLQLQJWKHRSWLPDOLW\RIWKHVROXWLRQVXJJHVWVWKDWWKH
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KDVWKHWRWDOWUDQVKLSPHQWFRVWHTXDOWRCZKLFKLVVPDOOHUWKDQWKHWRWDOFRVWRICREWDLQHGHDUOLHU
$QRWKHULPSURYHPHQWOHDGVWRRSWLPDOVROXWLRQWRWKHSUREOHP7KLVLVJLYHQLQ7DEOH
TABLE 5.34
Initial Feasible Solution: Non-optimal
B
A
–0
C
2
P
8
R
Q
12
10
150
– 12
Supply
ui
500
0
300
–4
7
200
–6
8
0
– 11
2
0
– 10
10
0
– 12
– 13
S
230
13
120
A
0
B
5
– 0
10
9
0
7
6
0
16
8
0
5
4
0
4
8
0
0
1,000
–9
+2
+
–2
–0
–1
7
7
180
–5
11
+8
16
11
–5
120
–5
14
200
–0
C
– 16
– 11
6
–1
– 12
–5
–0
P
– 23
Q
– 20
R
– 24
12
10
12
– 14
– 17
– 16
11
8
– 11
– 20
– 17
11
–7
–6
–7
–6
6
–6
–0
–1
5
–6
3
+1
–0
–9
–0
S
– 26
13
– 23
14
– 14
7
– 11
9
–7
–9
Demand
0
0
0
180
150
350
320
vj
0
4
6
11
10
12
13
Total cost = 10 ¥ 150 + 12 ¥ 230 + 13 ¥ 120 + 7 ¥ 180 + 8 ¥ 120 + 7 ¥ 200 = ` 9,440
)URP7DEOHWKHRSWLPDOVROXWLRQLV
6HQGIURPSODQW
6HQGIURPSODQW
6HQGIURPSODQW
6HQGIURPZDUHKRXVH%
XQLWVWRSODQWDQGXQLWVWRZDUHKRXVH%
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XQLWVWRZDUHKRXVH'
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CRQDFFRXQWRIWKHSRVVLELOLW\RIWUDQVKLSPHQW
0
vj
13
12
10
0
– 26
– 22
– 20
Demand
S
R
Q
2
0
14
8
– 14
– 15
– 20
–9
6
0
7
11
16
–0
– 13
–6
–9
–3
–3
9
180
9
5
8
0
6
7
12
P
–0
180
–7
–4
–7
– 12
–3
10
150
4
4
0
6
16
11
10
Q
–0
150
– 11
–3
–4
–7
–2
10
350
8
0
3
5
11
8
12
R
–0
350
–7
+1
–4
–3
13
320
0
–0
1,000
0
– 13
– 10
0
2
0
– 10
0
8
10
–9
200
7
–6
–2
300
14
200
0
ui
500
120
Supply
13
S
Total cost = 2 ¥ 230 + 10 ¥ 150 + 13 ¥ 120 + 7 ¥ 180 + 8 ¥ 350 + 7 ¥ 200 = ` 8,980
– 25
– 16
– 19
11
6
– 21
– 14
7
P
12
0
– 13
9
– 16
–3
7
10
C
–7
– 230
0
5
8
B
–2
2
230
C
0
–0
B
A
A
Revised Solution: Non-optimal
+
+
TABLE 5.35
ഩ‹ഩYƵĂŶƟƚĂƟǀĞdĞĐŚŶŝƋƵĞƐŝŶDĂŶĂŐĞŵĞŶƚ
0
vj
13
12
10
12
0
– 25
– 22
– 20
– 21
– 15
10
Demand
S
R
Q
P
C
5
B
–7
0
A
–0
2
0
14
8
– 14
– 16
5
0
7
11
– 12
–6
–9
9
180
9
5
8
– 21
16
11
0
–2
6
6
– 10
–0
7
7
7
–3
12
–0
180
–6
–4
–7
– 11
–3
10
270
– 120
150
4
4
0
6
16
11
10
Q
– 10
–3
–4
–6
–2
10
350
8
0
3
5
11
8
12
R
–0
350
–8
–5
–4
–1
12
320
0
10
2
8
7
14
13
S
Total cost = 2 ¥ 230 + 10 ¥ 270 + 7 ¥ 180 + 8 ¥ 350 + 7 ¥ 200 + 2 ¥ 120 = ` 8,860
– 24
– 16
– 19
– 14
– 12
–4
–3
8
P
0
– 230
230
C
To
9
0
2
B
Revised Solution: Optimal
A
TABLE 5.36
–0
120
200
1,000
0
0
0
0
200
300
500
Supply
– 12
– 10
– 10
–9
–5
–2
0
ui
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7REHJLQZLWKOHWXVODEHOWKHSODQWV$%DQG&DVDQGUHVSHFWLYHO\DQGWKHZDUHKRXVHVDW345DQG
6UHVSHFWLYHO\DVDQG
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)LJXUH7KHTXDQWLWLHVWREHVKLSSHGDUHVKRZQLQFROXPQ$ZKLOHYDULRXVURXWHVDQGWKHLUQRGHVDUH
LQGLFDWHGLQFROXPQV%&'DQG(7KHXQLWVFRVWVLQYROYHGLQVHQGLQJWKHLWHPRQYDULRXVURXWHVDUHVKRZQ
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–
–
–
–
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5(9,(:,//8675$7,216
([DPSOH
6ROYHWKHIROORZLQJSUREOHPXVLQJWKHWUDQVSRUWDWLRQPHWKRGREWDLQLQJWKHLQLWLDOIHDVLEOH
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TABLE 5.37
From
To
1
2
– 24
– 22
Initial Feasible Solution: Non-optimal
1
2
3
80
69
103
3
47
– 59
4
12
12
+
100
72
6
+
64
10
16
4
86
– 87
Supply
ui
61
12
0
40
16
– 31
94
20
– 40
25
8
– 57
19
8
– 52
64
10
65
–20
16
3
5
4
– 71
+
103
15
– 24
6
87
36
2
57
19
– 11
8
– 63
e
20
Demand
16
14
18
6
10
vj
56
72
103
76
71
– 23
– 21
94
+
27
5
72
– 11
– 70
Total cost = 103 ¥ 12 + 72 ¥ 6 + 40 ¥ 10 + 16 ¥ 16 + 36 ¥ 4 + 15 ¥ 6 + 19 ¥ 2 + 20 ¥ 8 = ` 2,756
ഩ‹ഩYƵĂŶƟƚĂƟǀĞdĞĐŚŶŝƋƵĞƐŝŶDĂŶĂŐĞŵĞŶƚ
TABLE 5.38
To
From
1
2
– 36
– 34
Improved Solution: Non-optimal
1
2
3
80
69
103
3
47
– 59
10
64
+
72
Supply
ui
+ 61
12
0
40
16
– 31
94
20
– 28
25
8
– 57
– 52
2
10
100
5
4
8
8
65
– 32
16
4
3
16
4
86
15
27
20
Demand
16
14
vj
44
72
– 59
103
– 12
87
36
– 51
8
– 99
– 11
57
– 12
19
– 11
2
6
5
– 35
– 21
72
94
19
8
18
6
10
64
103
64
71
– 82
Total cost = 103 ¥ 10 + 64 ¥ 2 + 72 ¥ 8 + 40 ¥ 8 + 16 ¥ 16 + 36 ¥ 4 + 15 ¥ 8 + 20 ¥ 6 + 19 ¥ 2 = ` 2,732
TABLE 5.39
From
To
1
– 36
Improved Solution: Optimal
1
2
3
80
69
103
100
72
103
87
–7
2
64
Supply
ui
61
12
0
40
16
– 31
94
20
– 28
25
8
– 47
– 42
5
4
8
2
16
2
– 34
47
– 69
– 32
65
16
– 10
4
3
16
4
86
15
27
20
Demand
16
14
vj
44
62
– 69
– 12
36
– 61
8
– 89
–1
57
–2
19
– 11
2
6
5
– 25
– 11
72
94
19
8
18
6
10
64
103
64
61
– 72
Total cost = 103 ¥ 2 + 64 ¥ 2 + 61 ¥ 8 + 72 ¥ 16 + 16 ¥ 16 + 36 ¥ 4 + 15 ¥ 8 + 20 ¥ 6 + 19 ¥ 2 = ` 2,652
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TABLE 5.40
Initial Solution–VAM
Buyer
Plant
A
–2
D
E
8
4
B
9
C
6
T
–1
4
600
7
150
–1
–2
5
F
100
–4
100
10
9
–2
100
8
3
3
Demand
750
200
500
vj
6
4
6
400
Supply
ui
100
0
800
3
150
0
400
–3
1,450
Total cost = 4 ¥ 100 + 9 ¥ 600 + 7 ¥ 100 + 9 ¥ 100 + 6 ¥ 150 = ` 8,300 and Penalty = 3 ¥ 400 = ` 1,200
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TABLE 5.41
Initial Solution-VAM
Destination
Source
P
A
6 –
B
13
–8
C
9+
D
12
–5
200
Q
R
11
9
e
–5
–5
150
9
8
250
14
–2
–3
+
+1
10
10
–2
Supply
ui
8
200
0
15
300
–1
500
3
100
1
S
300
–7
–
12
12
10
Demand
350
250
300
200
vj
6
6
9
9
100
100
1,100
Total cost = 6 ¥ 200 + 8 ¥ 300 + 9 ¥ 150 + 9 ¥ 250 + 12 ¥ 100 + 10 ¥ 100 = ` 9,400
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TABLE 5.42
Revised Solution–Optimal
Destination
Source
P
A
6
B
13
–8
C
9
D
12
–4
Q
R
11
9
10
8
Supply
ui
200
0
15
300
–1
12
500
3
100
2
S
e
100
–5
–5
250
9
–2
10
250
–2
–1
14
8
300
–8
–1
12
10
Demand
350
250
300
200
vj
6
6
9
8
100
100
1,100
Total cost = 6 ¥ 100 + 8 ¥ 100 + 8 ¥ 300 + 9 ¥ 250 + 9 ¥ 250 + 10 ¥ 100 = ` 9,300
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TABLE 5.43
Revised Solution–2
Destination
Source
P
Q
R
A
M
11
9
B
13
10
8
S
e
–9
–6
200
C
9
D
12
–5
–6
9
–3
10
250
–1
–1
8
300
–8
50
15
14
12
12
10
Demand
200
250
300
200
vj
5
5
9
8
50
100
Supply
ui
50
0
300
–1
500
4
100
2
950
Total cost = 6 ´ 150 + 8 ´ 50 + 8 ´ 300 + 9 ´ 200 + 9 ´ 250 + 12 ´ 50 + 10 ¥ 100 = ` 9,350
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2£
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2£
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TABLE 5.44
Initial Feasible Solution - Optimal
Noida
Supply
ui
1,350
0
750
– 50
M
1,200
– 300
350
900
– 200
0
750
– 650
Greater Noida
Gurugram
500
250
450
200
1,350
Dehradun
M
Aligarh
600
Bhopal
350
0
–50
450
–50
150
600
1,200
300
Kanpur
450
Dummy
0
Demand (tons)
1,800
1,500
1,650
vj
650
450
250
250
750
– 200
0
– 300
– 400
300
F 7KHGXDORIWKHJLYHQSUREOHPLV
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1 £-
1 £
1 £
1 £
1 £
1 £
1 £
1 £
1 £
1 £
1 £-
1 £
1 £
1 £
1 £
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TABLE 5.45
Advertising Campaign Schedule
- =?0=B
=C0>D
>E0?D
79
5DGLR
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0LQ([SRVXUHV LQPLOOLRQV 7
-
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SODFLQJDQeLQWKHFHOOUHSUHVHQWHGE\ODVWFROXPQRIWKHVHFRQGURZ
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DVVKRZQ
7KHLPSURYHGVROXWLRQLVJLYHQLQ7DEOH ,WLVWHVWHGDQGIRXQGWREHRSWLPDO,WPD\EHQRWHG
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TABLE 5.46
Initial Solution: Non-optimal
Age-groups
Media
13–18
19–25
12
7
26–35
25
TV
–1
5
– 10
36 & over
10 10
1
Dummy
Max.
Exp.
ui
+ 0
40
0
30
–1
–1
e
30
10
Magazines
14
–4
–3
–6
9
–3
12
12
+ 9
–1
10
10
0
10
12
–
Radio
0
20
90
–3
Min. Exp
30
25
15
10
10
vj
11
7
10
10
1
Dummy
10
36 & over
10 10
12
10
0
12
TABLE 5.47
Improved Solution: Optimal
Age-groups
Media
13–18
19–25
12
7
26–35
25
TV
–2
–1
Max.
Exp.
ui
40
0
30
0
0
20
0
90
5
0
e
30
Radio
10
Magazines
14
–2
9
–3
0
15
–4
–5
12
9
5
–2
Min. Exp
30
25
15
10
10
vj
10
7
9
10
0
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LWHQWLUHO\PHHWVWKHUHTXLUHPHQWRIDWOHDVWPVXFKH[SRVXUHV1RZWRHQVXUHPH[SRVXUHV
WKURXJK79LQWKHDJHJURXS±ZHSODFHDLQWKHFHOO DQGGUDZDFORVHGSDWKDQG
PDNHDGMXVWPHQWVLQWKHFHOOVZKLFKOLHRQLW7KHFHOOVLQFOXGH DQG 7KHUH
VXOWLQJVROXWLRQLVSURYLGHGLQ7DEOH
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TABLE 5.48
Revised Solution
Age-groups
Media
19–25
Dummy
Max.
Exp.
7
10
36 & over
10
10
10
9
12
10
0
Magazines
14
12
9
12
0
20
Min. Exp
30
25
15
10
10
90
13–18
25
4
TV
12
Radio
26–35
1
0
40
26
4
30
5
15
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7KURXJK79
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JURXS±PSHRSOHWRUHDFKLQWKHDJHJURXSDQGDERYH
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TABLE 5.49
Initial Solution: VAM
Distribution centres
Factory
Z
W
X
Y
10
8
5
Supply
ui
7,000
0
8,000
10
10
7,000
A
– 14
–9
6,000
B
–1
7
9
1,000
–
4
–6
+ 8
14
– 8
10,000
25,000
4,000
6,000
1,000
15
C
6
Demand
6,000
6,000
8,000
5,000
vj
–4
–1
5
–2
–1
10
1 +
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TABLE 5.50
Improved Solution: Optimal
Distribution centres
Factory
W
X
Y
10
8
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7,000
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8,000
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2,000
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15
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6,000
3,000
10
14
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10,000
6,000
6,000
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6
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TABLE 5.51
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Distribution centres
Factory
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7,000
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8,000
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6,000
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9
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10
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TABLE 5.52
Revised Problem II: Initial Solution
Distribution centres
Factory
W
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Y
10
8
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7,000
0
8,000
10
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9
15
8
10
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8
9,500
24,500
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6,000
C
6
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6,000
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5,000
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TABLE 5.53
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TABLE 5.54
Initial Solution—Non-optimal
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TABLE 5.55
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TABLE 1
Cost of Transportation, Capacity and Overhead Cost Information
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TABLE 2
Transportation Cost (per tyre) from Various Regional Warehouses
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In every workplace, there are jobs to be done and there are people available to do them. But everyone is not
equally efficient at every job. Someone may be more efficient on one and less efficient on the other job, while it
might be otherwise for someone else. The relative efficiency is reflected in terms of the time taken for, or the cost
associated with, performance of different jobs by different people. An obvious problem for a manager to handle
is to assign jobs to various workers in a manner that they can be done in the most efficient way. Interestingly,
such problems can be formulated as linear programming problems or as transportation problems and solved
as such, but a method, called Hungarian Assignment Method provides an easy route. It allows a manager
to obtain answers to questions like the following:
€ How to assign the given jobs to some workers on a one-to-one basis when completion time of performance
is given for each combination and it is desired that the jobs are completed in the least time or at the least
cost. Further, in the assignment pattern so obtained, is only one option available or there are other equally
attractive ones also?
€ How to deal with situations when the number of jobs do not match with the number of job performers,
when some job (s) cannot be performed by, or is not be given to, a particular performer, or when a certain
job has to be given to a particular individual?
€ How should the salesmen of a company be assigned to different sales zones so that the total expected
sales are maximised?
€ Given the order of preferences that different managers have for different rooms in a hotel on one of its
floors, what pattern of assignment of rooms to the managers will satisfy their requirements the most?
€ How to schedule the flights or the bus routes between two cities so that the layover times for the crew
can be minimised?
In fact, the assignment method works for any problems in which one-to-one matching is called for in the
light of the given payoffs, where the total pay-off is sought to be minimised or maximised.
Only ordinary arithmetical skills are required for working through this chapter. Also needed for some part
of it is familiarity with the summation notation, algebraic inequalities, matrices and their transpose.
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Learning Objectives
After reading this chapter, you should be able to:
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PDWWHUZKRDPRQJWKHPKDVEHHQDVVLJQHGWKHMRE7KXVLQDQDVVLJQPHQWSUREOHPWKHTXHVWLRQLVKRZWKH
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ZRUNHUVLQSHUIRUPLQJWKHVHMREVDVJLYHQLQ7DEOH
TABLE 6.1
Time Taken by Workers on Various Jobs (in minutes)
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:HVKDOOQRZFRQVLGHUDQDO\WLFDOO\WKHVROXWLRQWRWKLVSUREOHPRIWKHVXSHUYLVRU
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LQYROYLQJWKHPLQLPXPFRVW RUPD[LPXPSUR¿WLIWKHSUREOHPLVRIWKH
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LO 2 Illustrate how
to solve an assignment
problem by complete
enumeration and
transportation method
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LVWKHRSWLPDORQHDVLWLQYROYHVWKHOHDVWWRWDOWLPHHTXDOWRPLQXWHV
TABLE 6.2
Worker–Job Assignments
O
,WPD\EHQRWHGWKDWLQJHQHUDOIRUDQZRUNHUMRESUREOHPWKHUHDUH QXPEHURIZD\VLQZKLFKWKHMREV
FDQEHDVVLJQHGWRWKHZRUNHUV7KXVWKHOLVWLQJDQGHYDOXDWLRQRIDOOWKHSRVVLEOHDVVLJQPHQWVLVDVLPSOH
PDWWHUZKHQLVDVPDOOQXPEHU%XWZKHQ LVDQHYHQPRGHUDWHO\ODUJHQXPEHUWKLVPHWKRGLVQRWYHU\
SUDFWLFDO)RUH[DPSOHIRUD¿YHSHUVRQDQG¿YHMRESUREOHPDWRWDORI DVVLJQPHQWVZLOOQHHGWREH
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DVVLJQPHQWVZRUNVRXWWREH7KHUHIRUHWKHPHWKRGLVQRWVXLWDEOHIRUUHDOZRUOGVLWXDWLRQV
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DVVLJQHGWRDZRUNHUZRXOGEHDQGRWKHUYDOXHVLQWKDW WK URZDQGWKH WK FROXPQVKRXOGHTXDOWR
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WRQRWHKHUH7KHVROXWLRQREWDLQHGE\DSSO\LQJWKLVPHWKRGZRXOGEHVHYHUHO\GHJHQHUDWH7KLVLVEHFDXVH
WKHRSWLPDOLW\WHVWLQWKHWUDQVSRUWDWLRQPHWKRGUHTXLUHVWKDWWKHUHPXVWEH ± ±ZKHQ EDVLFYDULDEOHV)RUDQDVVLJQPHQWSUREOHPRIWKHRUGHU ¥ KRZHYHUWKHUHZRXOGEHRQO\EDVLFYDULDEOHV
LQWKHVROXWLRQEHFDXVHKHUHDVVLJQPHQWVDUHUHTXLUHGWREHPDGH7KXVDODUJHQXPEHURIHSVLORQVPD\EH
UHTXLUHGWREHLQWURGXFHGLQWKHVROXWLRQLQRUGHUWRSURFHHGZLWKWKLVPHWKRG
7KXVDQDVVLJQPHQWSUREOHPLVDYDULDWLRQRIWKHWUDQVSRUWDWLRQSUREOHPZLWKWZRFKDUDFWHULVWLFV)LUVWWKH
SD\RIIPDWUL[IRUWKHSUREOHPZRXOGEHDVTXDUHPDWUL[DQGVHFRQGWKHRSWLPDOVROXWLRQWRWKHSUREOHPZRXOG
DOZD\VEHVXFKWKDWWKHUHZRXOGEHRQO\RQHDVVLJQPHQWLQDJLYHQURZRUFROXPQRIWKHSD\RIIPDWUL[
7KHLQIRUPDWLRQJLYHQLQ([DPSOHLVGHSLFWHGLQ7DEOHDVDWUDQVSRUWDWLRQPRGHO7KLVLVD¥DVVLJQPHQW
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TABLE 6.3
The Assignment Problem as a Transportation Problem
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TABLE 6.4
Initial Solution: VAM
Job
Supply
Worker
B
A
C
D
51
67
1
55
1
e
e
1
1
45
40
2
57
42
63
3
49
52
48
64
1
4
41
45
60
55
1
Demand
1
1
1
1
4
1
e
1
1
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VROXWLRQLVIRXQGWREHQRQRSWLPDO$FFRUGLQJO\DFORVHGSDWKLVGUDZQVWDUWLQJZLWKWKHFHOOZKLFK
KDVWKHODUJHVW DYDOXHHTXDOWR
TABLE 6.5
Initial Solution: Non-optimal
Job
Supply
ui
67
1
0
55
1
2
64
1
4
55
1
–4
4
Worker
B
A
e
1
45
2
57
– 10
+
–
C
D
e
40
–7
–
51
1
42
– 17
+
14
63
e
1
1
3
49
4
41
Demand
1
1
1
1
vj
45
40
44
67
–8
52
48
7
1
–9
45
– 20
60
8
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TABLE 6.6
Revised Solution: Optimal
Job
Worker
B
A
e
C
Supply
ui
1
0
55
1
2
64
1
4
55
1
–4
4
D
1
1
45
40
2
57
42
–7
51
– 14
67
1
– 10
– 17
63
e
1
3
49
4
41
Demand
1
1
1
1
vj
45
40
44
53
52
–8
48
–7
1
45
–9
– 20
60
–6
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LO 3 Determine how to
solve assignment problem
as an LPP and using HAM
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,QJHQHUDOOHW
Ï1 if ith person is assigned the jth job
xij = Ì
Ó0 if ith person is nott assigned the jth job
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n
n
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i = j =
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j =
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i =
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ŶĂƐƐŝŐŶŵĞŶƚƉƌŽďůĞŵĐĂŶďĞƐŽůǀĞĚŝŶŵĂŶLJǁĂLJƐ͘/ƚĐĂŶďĞƐŽůǀĞĚďLJĐŽŵƉůĞƚĞĞŶƵŵĞƌĂƟŽŶŽĨĂůůƚŚĞƉŽƐƐŝďůĞ
ĂƐƐŝŐŶŵĞŶƚƉĂƩĞƌŶƐďƵƚƚŚŝƐŵĞƚŚŽĚŝƐǀĞƌLJƟŵĞͲĐŽŶƐƵŵŝŶŐ͘ŶĂƐƐŝŐŶŵĞŶƚƉƌŽďůĞŵĐĂŶĂůƐŽďĞĨŽƌŵƵůĂƚĞĚĂŶĚ
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" ' ( ) " * ) " +
" " +
START
Write the problem
in tabular form
Is it
a balanced
problem?
No
Add dummy
row(s)/column(s)
Yes
Is it
a maximisation
problem?
Yes
Convert it into a minimisation
problem, either (i) by changing
the signs of the elements of
the table or by subtracting all
the values from the largest
value.
No
Obtain reduced cost tables.
(i) Subtract from all entries in each
row the least value in the row.
(ii) From this table, subtract from
all entries in each column the
least value in the column.
Can all
zeros be covered by less
than n
lines?
No
Yes
Improve the solution. For this:
(i) Select the minimum of the
uncovered (by lines) cell values.
(ii) Subtract this value from all
uncovered cell values.
(iii) Add this value to the cells
lying on the intersection of
any pair of lines.
(iv) Leave the cell values covered
by only one line undisturbed.
Make assignments on
one-to-one match
basis considering zeros
in rows/columns.
STOP
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TABLE 6.7
!
Time Taken (in minutes) by four Workers
6;
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8
;;
8
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6
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TABLE 6.8
Reduced Cost Table 1
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6WHS * + #
" ? TABLE 6.9
Reduced Cost Table 2
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6WHS 2 +
5 +
" ? " !< TABLE 6.10
Reduced Cost Table 3
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TABLE 6.11
Assignment of Jobs
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Reduced Cost Table 1
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TABLE 6.14
Reduced Cost Table 3
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2 Cost Table
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TABLE 6.16
Reduced Cost Table 1
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TABLE 6.17
Reduced Cost Table 2
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dŚĞ,ƵŶŐĂƌŝĂŶƐƐŝŐŶŵĞŶƚDĞƚŚŽĚŝŶǀŽůǀĞƐŵĂŬŝŶŐĂƐƐŝŐŶŵĞŶƚƐŽŶĂŽŶĞͲƚŽͲŽŶĞďĂƐŝƐ͘,ĞŶĐĞ͕ĂŶĂƐƐŝŐŶŵĞŶƚ
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TABLE 6.18
6E Balancing the Assignment Problem
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TABLE 6.19
Reduced Cost Table 1
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TABLE 6.20
Reduced Cost Table 2
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TABLE 6.21
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TABLE 6.22
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TABLE 6.23
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Reduced Cost Table 2
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TABLE 6.24
Reduced Cost Table 3
District
Salesman
D1
D2
D3
D4
D5
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12
0
0
7
0
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0
10
8
10
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0
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TABLE 6.25
Obtaining Optimal Values of Dual Variables
Job
Worker
B
A
1
45
2
57
3
49
4
41
vj
45
e
1
40
e
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WRRQHW\SLVWDQGWKHW\SLVWLVSDLGIRUDIXOOKRXUHYHQZKHQKHZRUNVIRUDIUDFWLRQRIDQKRXU)LQGWKHOHDVW
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W\SLVWV7KLVLVVKRZQLQ7DEOH7KHHOHPHQWVRIWKHPDWUL[DUHREWDLQHGDVIROORZV7RLOOXVWUDWHLIW\SLVW
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7KLVUHVXOWVLQDFRVWRICIRUWKLVFRPELQDWLRQ
TABLE 6.26
Total Cost Matrix
2
B
&
1
6XEWUDFWLQJ WKH PLQLPXP HOHPHQW RI HDFK URZ IURP DOO LWV HOHPHQWV ZH REWDLQ 5&7 DV VKRZQ LQ
7DEOH
TABLE 6.27
Reduced Cost Table 1
2
B
&
1
1RZVXEWUDFWLQJWKHPLQLPXPHOHPHQWRIHDFKFROXPQIURPDOOWKHHOHPHQWVZHJHW5&7JLYHQLQ7DEOH
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TABLE 6.28
Reduced Cost Table 2
2
B
&
1
+HUHWKHPLQLPXPQXPEHURIOLQHVWRFRYHUDOO]HURVLVHTXDOWRZKLFKLVVPDOOHUWKDQWKHRUGHU RIWKHJLYHQ
PDWUL[$FFRUGLQJO\WKHUHYLVHGWDEOHLVSUHSDUHGE\FRQVLGHULQJWKHOHDVWXQFRYHUHGYDOXHHTXDOWRDQG
DGMXVWLQJLWZLWKXQFRYHUHGFHOOYDOXHVDQGWKRVHO\LQJDWWKHLQWHUVHFWLRQRIOLQHV7DEOHFRQWDLQV5&7
TABLE 6.29
Reduced Cost Table 3
2
B
&
1
,Q7DEOHDVZHOOIRXUOLQHVFDQFRYHUDOO]HURV$FFRUGLQJO\5&7LVREWDLQHGGUDZQXSDVWKHUHYLVHG
WDEOH7KLVLVJLYHQLQ7DEOH
TABLE 6.30
Reduced Cost Table 4
Job
Typist
P
Q
R
S
T
A
2
1
2
3
0
B
4
1
0
7
0
C
0
0
2
0
2
D
0
1
0
1
0
E
3
0
3
0
3
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,QWKLVFDVHWKHPLQLPXPQXPEHURIOLQHVWRFRYHUDOO]HURVHTXDOV ZKLFKPDWFKHVZLWKWKHRUGHURIWKH
PDWUL[$FFRUGLQJO\DVVLJQPHQWVKDYHEHHQPDGHDVGHVFULEHGEHORZ
&
B
2
1
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URRPKDVLWVRZQDGYDQWDJHVDQGGLVDGYDQWDJHV
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DOOGLIIHUHQWVL]HVDQGVKDSHV(DFKRIWKH¿YHPDQDJHUVZHUHDVNHGWRUDQNWKHLUURRPSUHIHUHQFHVDPRQJVWWKH
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0
0
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&$1RYHPEHU
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TABLE 6.31
&
Assignment Problem
#
#
#
#
#
#
#
#
#
#
#
#
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HDFKRIWKHURZV7KLVLVJLYHQLQ7DEOH
TABLE 6.32
&
Reduced Cost Table 1
#
#
#
#
#
#
#
#
#
#
#
#
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TABLE 6.33
Assignment of Rooms to Managers
Manager
Room
M1
M2
M3
M4
M5
301
M
3
1
M
0
302
0
0
4
0
1
303
1
M
0
3
M
304
1
0
1
1
1
305
M
1
2
0
M
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#
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#
#
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E )RUFHUWDLQUHDVRQVLIVDOHVUHSUHVHQWDWLYH%FDQQRWEHDVVLJQHGWRVDOHVWHUULWRU\,,,ZLOOWKHRSWLPDO
DVVLJQPHQWVFKHGXOHFKDQJH",IVRZKDWZLOOEHWKDWVFKHGXOHDQGZKDWZLOOEHWKHHIIHFWRQWKHWRWDO
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7KHJLYHQSUREOHPEHLQJRIWKHPD[LPLVDWLRQW\SHQHHGVWREHFRQYHUWHG¿UVWLQWRPLQLPLVDWLRQW\SH7KLV
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TABLE 6.34
Opportunity Loss Matrix
C
CC
CCC
CD
6XEWUDFWLQJ PLQLPXP YDOXH LQ HDFK URZ IURP HDFK RI WKH URZ HOHPHQWV ZH GHULYH 5&7 VKRZQ LQ
7DEOH
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TABLE 6.35
Reduced Cost Table 1
C
CC
CCC
CD
1RZVXEWUDFWLQJPLQLPXPYDOXHLQHDFKFROXPQIURPHDFKRIWKHFROXPQHOHPHQWVRI5&7ZHJHW5&7
DVVKRZQLQ7DEOH
TABLE 6.36
Reduced Cost Table 2
C
CC
CCC
CD
6LQFHWKHPLQLPXPQXPEHURIOLQHVWRFRYHUDOO]HURV LVVPDOOHUWKDQWKHPDWUL[RUGHU 5&7LVGHULYHG
DVDUHYLVHGWDEOH7KLVLVJLYHQLQ7DEOH
TABLE 6.37
Reduced Cost Table 3
I
II
III
IV
A
10
55
30
0
B
0
35
0
30
C
10
0
0
10
D
0
0
10
0
:LWKWKHQXPEHURIOLQHVWRFRYHUDOO]HURVEHLQJHTXDOWRWKHRUGHURIWKHJLYHQPDWUL[DVVLJQPHQWVFDQEH
PDGHDVVKRZQLQ7DEOH+RZHYHUWKHSUREOHPKDVDQDOWHUQDWLYHRSWLPDOVROXWLRQDVZHOO%RWKRIWKHVH
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,IVDOHVPDQ FDQQRWEHDVVLJQHGWRWHUULWRU\,,, DOWHUQDWLYH WKHQDOWHUQDWLYHDERYHPD\EHDGRSWHG
ZLWKRXWDGYHUVHHIIHFWRQVDOHLQFUHDVH
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RIWKHVHIRXUSURGXFWV7KH¿UPUHFRUGVKRXUVDGD\DQGDOORZVPLQXWHVIRUOXQFK7KHSURFHVVLQJWLPH
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KRXUVDQGPLQXWHVWKDWLVPLQXWHV7KHQXPEHURIXQLWVRIGLIIHUHQWSURGXFWVZKLFKFRXOGEHSURGXFHG
E\WKHIRXURSHUDWRUVFDQEHFDOFXODWHGE\GLYLGLQJE\WKHJLYHQSURFHVVLQJWLPHV:LWKWKHSUR¿WSHUXQLW
RIHDFKSURGXFWEHLQJJLYHQZHPD\FDOFXODWHWKHSUR¿WUHVXOWLQJIURPHDFKSRVVLEOHDVVLJQPHQW7KHSUR¿W
PDWUL[LVJLYHQLQ7DEOH7KHYDOXHVLQWKLVPDWUL[DUHGHULYHGDVIROORZV)RUH[DPSOHRSHUDWRUFDQ
SURGXFH XQLWVRISURGXFW ZKLFKDWDSUR¿WUDWHRICSHUXQLWLPSOLHVDWRWDOSUR¿WRIC
TABLE 6.38
( Profit Matrix
2 %
7RVROYHWKHSUREOHPLWLV¿UVWFRQYHUWHGLQWRDPLQLPLVDWLRQSUREOHPE\REWDLQLQJRSSRUWXQLW\ORVVPDWUL[
E\VXEWUDFWLQJHDFKYDOXHIURP±WKHKLJKHVWSUR¿WYDOXHLQWKHWDEOH,WLVJLYHQLQ7DEOH
TABLE 6.39
( Opportunity Loss Matrix
2 %
ഩ‹ഩYƵĂŶƟƚĂƟǀĞdĞĐŚŶŝƋƵĞƐŝŶDĂŶĂŐĞŵĞŶƚ
8VLQJWKH+XQJDULDQPHWKRGZH¿UVWREWDLQ]HURVLQHYHU\URZDQGFROXPQDVJLYHQLQ7DEOHVDQG
TABLE 6.40
( Reduced Cost Table 1
2 %
TABLE 6.41
( Reduced Cost Table 2
2 %
,WLVREVHUYHGWKDWDOO]HURVDUHFRYHUHGE\WKUHHOLQHVZKLFKLVRQHOHVVWKDQWKHRUGHURIWKHPDWUL[7KH
LPSURYHGPDWUL[LVREWDLQHGDVVKRZQLQ7DEOHZLWKDGMXVWPHQWZLWKWKHOHDVWXQFRYHUHGYDOXHRI
TABLE 6.42
Reduced Cost Table 3
Product
Operator
A
B
C
D
1
0
0
21
0
2
30
0
146
150
3
0
24
0
4
4
0
0
21
120
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DVVLJQPHQWVFDQEHPDGHDVVKRZQLQ7DEOH7KXVWKHRSWLPDODVVLJQPHQWLV
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( 2 %
2 ?
7RWDO
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RIÀLJKWVWKDWPLQLPLVHVOD\RYHUWLPHDZD\IURPKRPH)RUDQ\JLYHQSDLULQJWKHFUHZZLOOEHEDVHGDWWKHFLW\
WKDWUHVXOWVLQWKHVPDOOHVWOD\RYHU
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DP
DP
DP
DP
DP
DP
DP
DP
SP
SP
SP
SP
SP
SP
SP
SP
0&RP'HOKL
7REHJLQZLWKZH¿UVWDVVXPHWKDWDOOWKHFUHZLVEDVHGDW1HZ'HOKL8VLQJWKLVDVVXPSWLRQZHFDQREWDLQ
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ZKLFKLVDPWKHQLWFDQGRVRRQO\DIWHUKRXUVVLQFHDPLQLPXPOD\RYHUWLPHRI¿YHKRXUVLVUHTXLUHG
6LPLODUO\OD\RYHUWLPHVIRURWKHUÀLJKWFRPELQDWLRQVFDQEHREWDLQHGDVVKRZQLQWKHWDEOH
TABLE 6.43
Layover Time—Crew at New Delhi
E$
-6,
-6-
-6.
-6/
7KHOD\RYHUWLPHVIRUYDULRXVÀLJKWFRPELQDWLRQVZKHQFUHZLVDVVXPHGWREHEDVHGDW0XPEDLDUHVLPLODUO\
FDOFXODWHGDQGVKRZQLQ7DEOH
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TABLE 6.44
Layover Time—Crew at Mumbai
E$
-6,
-6-
-6.
-6/
1RZVLQFHWKHFUHZFDQEHEDVHGDWHLWKHURIWKHVWDWLRQVPLQLPXPOD\RYHUWLPHVFDQEHREWDLQHGIRUGLIIHUHQW
ÀLJKWFRPELQDWLRQVE\VHOHFWLQJWKHFRUUHVSRQGLQJORZHUYDOXHRXWRIWKHDERYHWZRWDEOHV)RULQVWDQFHLQ
FRPELQLQJ±ZHVHOHFWZKLFKLVORZHURIWKHWZRYDOXHVDQG 7DEOHVDQG 7KH
YDOXHVDUHVKRZQLQ7DEOH
TABLE 6.45
Layover Time—Crew at New Delhi/Mumbai
E$
-6,
-6-
-6.
-6/
ZKHQFUHZLVEDVHGDW0XPEDL
7RFRQWLQXHZHFDQREWDLQWKHRSWLPDOSDLULQJRIÀLJKWVVRDVWRPLQLPLVHWKHWRWDOOD\RYHUWLPHXVLQJWKH
+XQJDULDQ$VVLJQPHQW0HWKRG7DEOHJLYHVWKH5&7O
TABLE 6.46
Reduced Cost Table 1
Flight
201
202
203
204
101
14
13
0
3
102
13
12
7
0
103
0
1
6
1
104
1
0
7
12
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RIOLQHVFRYHULQJDOO]HURVLVHTXDOWRWKHRUGHURIWKHJLYHQPDWUL[ZHFDQREWDLQWKHRSWLPDODVVLJQPHQWDV
VKRZQLQWKHWDEOH
7KHRSWLPDOSDLULQJRIÀLJKWVLVJLYHQRQQH[WSDJH
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E$ E$ > ' @ % 1HZ'HOKL
1HZ'HOKL
1HZ'HOKL
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WHOHYLVLRQSURPRWLRQDOFDPSDLJQ,WKDVGHFLGHGWRVFKHGXOHDVHULHVRIRQHPLQXWHFRPPHUFLDOVGXULQJSHDN
DXGLHQFHYLHZLQJKRXUVRI±SP7RUHDFKWKHZLGHVWSRVVLEOHDXGLHQFHWKHFRPSDQ\ZDQWVWRVFKHGXOH
RQHFRPPHUFLDORQHDFKRIWKHQHWZRUNVDQGWRKDYHRQO\RQHFRPPHUFLDODSSHDUGXULQJHDFKRIWKHIRXU
RQHKRXUWLPHEORFNV7KHH[SRVXUHUDWLQJVIRUHDFKKRXUZKLFKUHSUHVHQWWKHQXPEHURIYLHZHUVSHUC
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KRXUV
$
%
&
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E +RZZRXOGWKHVFKHGXOHFKDQJHLILWLVGHFLGHGQRWWRXVHQHWZRUN$EHWZHHQDQGSP"
0&RP'HOKL
D 7RVROYHWKLVSUREOHPZH¿UVWPXOWLSO\HDFKYDOXHLQWKHPDWUL[E\WRH[SUHVVH[SRVXUHUDWLQJVSHU
C ODNK ,W VLPSOL¿HV WKH FDOFXODWLRQ ZRUN VRPHZKDW )XUWKHU EHLQJ D PD[LPLVDWLRQ SUREOHP ZH
VXEWUDFWHDFKYDOXHIURPWKHODUJHVWYDOXHWRJHWWKHRSSRUWXQLW\ORVVPDWUL[7KHUHVXOWRIWKHVHVWHSV
LVJLYHQLQ7DEOH
TABLE 6.47
Opportunity Loss Matrix
Network
Viewing
hours
A
B
C
D
1–2 p.m.
0
90
158
176
2–3 p.m.
82
116
100
165
3–4 p.m.
79
86
172
194
4–5 p.m.
156
57
103
143
ഩ‹ഩYƵĂŶƟƚĂƟǀĞdĞĐŚŶŝƋƵĞƐŝŶDĂŶĂŐĞŵĞŶƚ
1RZZHDSSO\+$0WRVROYHWKHSUREOHP5RZUHGXFWLRQVDQGFROXPQUHGXFWLRQVDUHVKRZQLQ7DEOHV
DQGUHVSHFWLYHO\
TABLE 6.48
Reduced Cost Table 1
Network
Viewing hours
A
B
C
D
1–2 p.m.
0
90
158
176
2–3 p.m.
0
34
18
83
3–4 p.m.
0
7
93
115
4–5 p.m.
99
0
46
86
TABLE 6.49
Reduced Cost Table 2
Network
Viewing hours
A
B
C
D
1–2 p.m.
0
90
140
93
2–3 p.m.
0
34
0
0
3–4 p.m.
0
7
75
32
4–5 p.m.
99
0
28
3
6LQFH WKH QXPEHU RI OLQHV FRYHULQJ DOO ]HURV LV ZH LPSURYH WKH VROXWLRQ DV VKRZQ LQ 7DEOHV DQG
TABLE 6.50
Viewing hours
Reduced Cost Table 3
Network
A
B
C
D
1–2 p.m.
0
90
137
90
2–3 p.m.
3
37
0
0
3–4 p.m.
0
7
72
29
4–5 p.m.
99
0
25
0
^ƉĞĐŝĂůůLJ^ƚƌƵĐƚƵƌĞĚ>ŝŶĞĂƌWƌŽŐƌĂŵŵĞƐ//͗ƐƐŝŐŶŵĞŶƚWƌŽďůĞŵഩ‹ഩ
TABLE 6.51
Reduced Cost Table 4
Network
Viewing hours
A
B
C
D
1–2 p.m.
0
83
130
83
2–3 p.m.
10
37
0
0
3–4 p.m.
0
0
65
22
4–5 p.m.
106
0
25
0
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TABLE 6.52
Reduced Cost Table 5
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Viewing hours
A
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1–2 p.m.
M
0
68
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2–3 p.m.
M
16
0
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0
7
93
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4–5 p.m.
99
0
46
21
TABLE 6.53
Viewing hours
Reduced Cost Table 6
Network
A
B
C
D
1–2 p.m.
M
0
47
0
2–3 p.m.
M
37
0
0
3–4 p.m.
0
28
93
50
4–5 p.m.
78
0
25
0
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TABLE 6.54
Total Cost Matrix
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TABLE 6.55
Reduced Cost Table 1
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TABLE 6.56
Reduced Cost Table 2
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TABLE 6.57
Total Cost Matrix
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TABLE 6.58
Reduced Cost Table 1
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TABLE 6.59
Reduced Cost Table 2
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TABLE 6.60
Reduced Cost Table 3
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In continuation with the linear programming problems that seek to maximise or minimise the objective
function, we now move on to the situations that are closer to reality. Formulation and solution of such
problems as linear programming problems are based on two simplifying assumptions that there is only a
single objective to achieve (such as maximisation of profit) and that the variables involved are continuous so
that, for example, solution to a profit maximising problem of a multi-product firm may call for producing in
numbers involving fractional values as well, like 18.7, 32.6, so on. It obviously sounds impractical to produce
fractional unit, give fractional number of advertisements in a media-mix problem, engage employees in
fractional numbers and so forth; and the firm may have multiple objectives to achieve. The present chapter
addresses both these issues.
To ensure that the solution involves only integer values, we use integer programming, and the other extension
of linear programming covered in this chapter is goal programming, which postulates setting up of multiple
goals (may be conflicting in nature) instead of a single objective that calls for maximisation or minimisation.
The contents of this chapter help a manager to obtain answers to the questions like the following:
€ What is the optimal mix of the products in terms of the exact number of completed units to be produced
for each one?
€ What combination of the investment proposals is best to undertake?
€ How to obtain optimal assignment of workers to jobs by using a different approach and how to determine
the optimal route for a travelling salesman who wants to cover a number of cities in a tour?
€ How best can the multiple objectives be achieved by assigning penalties for not satisfying each of the
objectives and weaving these penalties in the objective function?
This chapter requires higher skills in formulating the problem—both integer and goal programming—in
the first instance. It will be evident from the variety of IPPs given and discussed. Further, in the context of
goal programming, difference between goals and constraints should be understood clearly for the correct
formulation of the problems. For solution to both, the integer as well as goal programming problems, while
a knowledge of the Simplex method is essential, much more needs to be understood. This is in keeping with
the requirements for each of the two types of problems.
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Learning Objectives
After reading this chapter, you should be able to:
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ŝŶƚĞŐĞƌĂŶĚĂŶŽŶͲŶĞŐĂƟǀĞĨƌĂĐƟŽŶ͘dŚŝƐŝŶǀŽůǀĞƐĐŚĂŶŐŝŶŐƚŚĞĐŽĞĸĐŝĞŶƚƐŽĨƚŚĞůŝŶĞĂƌĐŽŶƐƚƌĂŝŶƚƐ;ŝŶĐůƵĚŝŶŐ
ƚŚĞZ,^ĂƐǁĞůůͿĂƐĨŽůůŽǁƐ͗
Original constraints
Gomory’s cut
;ĂͿ/ŶƚĞŐĞƌ;ƉŽƐŝƟǀĞŽƌŶĞŐĂƟǀĞͿ
;ďͿWŽƐŝƟǀĞĨƌĂĐƟŽŶ;&ŽƌĞdžĂŵƉůĞ͕ϮͬϯͿ
;ĐͿEĞŐĂƟǀĞĨƌĂĐƟŽŶ;&ŽƌĞdžĂŵƉůĞ͕оϮͬϯͿ
ĞƌŽ
hŶĐŚĂŶŐĞĚ;ϮͬϯͿ
WŽƐŝƟǀĞĐŽŵƉůĞŵĞŶƚ;ϭͬϯͿ
/ŶƚŚĞƐĞĐŽŶĚƐƚĞƉ͕ĐŚĂŶŐĞƚŚĞ͚с͛ƐŝŐŶƚŽ͚ш͛͘
0 &
# :< #( 7 3
#+73# .(J+#(8+J<#+≥(J+# %&& %
& %&&! & % džƚĞŶƐŝŽŶƐŽĨ>ŝŶĞĂƌWƌŽŐƌĂŵŵŝŶŐ͗/ŶƚĞŐĞƌWƌŽŐƌĂŵŵŝŶŐĂŶĚ'ŽĂůWƌŽŐƌĂŵŵŝŶŐഩ‹ഩ
%&&% & Z
%&&$ x2
3x1 + 4x2 = 18 is cutting plane
IPP feasible point
6
3x1 + 3x2 = 15
5
'ŽŵŽƌLJ͛ƐĐƵƚŚĂƐƚǁŝŶƉƌŽƉĞƌƟĞƐ͗;ŝͿŝƚŝƐŶŽƚƐĂƟƐĮĞĚ
ďLJ ƚŚĞ ŽƉƟŵĂů ƐŽůƵƟŽŶ ƚŽ ƚŚĞ >W ƌĞůĂdžĂƟŽŶ͕ ĂŶĚ ;ŝŝͿ
ǁŚŝůĞŝƚĐƵƚƐƚŚĞŽƉƟŵĂůƐŽůƵƟŽŶƚŽƚŚĞ>WƌĞůĂdžĂƟŽŶ͕
ŝƚ ĚŽĞƐ ŶŽƚ ĐƵƚ ĂŶLJ ĨĞĂƐŝďůĞ ƐŽůƵƟŽŶƐ ƚŽ ƚŚĞ ŝŶƚĞŐĞƌ
ƉƌŽŐƌĂŵŵŝŶŐƉƌŽďůĞŵ͘
(J+#(8+J<#+≥(J+ :(
4
Optimal solution of
LP relaxation x1 = 3/2, x2 = 7/2
3
3x1 + 4x2 = 18
2
2x1 + 4x2 = 17
1
x1
0
1
2
3
4
5
6
7
8
9
10
% %&&
)LJXUH *UDSKLF 5HSUHVHQWDWLRQ RI &
,33³&XWWLQJ3ODQH
(7<J++7:J+ #
#$ %&&+(8=+8#(7(: <(8<+8#+7()##(7(:4+(4=+#+7()4<(4<+/ (J+#(8+J<#+≥(J+ <(8=+£(9
# # := TABLE 7.4
""
# :< Revised Simplex Tableau 1
(
+
#(
#+
#<
$ +
<33
3
(
(J+
4(J< 3
:J+
4+(J+
(
+33
(
3
4(J+
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<J+
AJ=
#<
3
3
3
4 (J+
4+J<[
(
4(J+
<J=
+33
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3
3 3
<J+
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3
3 4(J+
3
3
4)3
4(33J< 3
/ D ≠
% . @ # D #< . ! K4 )3∏ 4 (J+ 7 (33L K4(33J<∏4+J<7)3L0 #+ " ഩ‹ഩYƵĂŶƟƚĂƟǀĞdĞĐŚŶŝƋƵĞƐŝŶDĂŶĂŐĞŵĞŶƚ
" " % %
! 1 . 4 +(J+AJ=<J=" " #<##< # :) " 0 # :) D 0 $ 3(8(+8<J=#(83#+4(J+#<7()J=
% 3(8(+8<J=#(83#+4#<8(J+#<7<8<J=
#
<J=#(8(J+#<7<J=8<4+8#<
0 <J=#(8(J+#< ≥<J=
4<J=#(4(J+#< £4 <J=
4 <J=#(4(J+#<8#= 74<J=
#=" TABLE 7.5
""
Revised Simplex Tableau 2
(
+
# (
#+
# <
+
<33
3
(
<J=
3
4 (J+
()J=
(
+33
(
3
4 (
3
(
(
#+
3
3
3
<J=
(
4 <J+
<J=
+33
<33
3
3
3
/ (
()J=
3
<J=
3
D 3
3
4+)
3
4)3
/
# :;
džƚĞŶƐŝŽŶƐŽĨ>ŝŶĞĂƌWƌŽŐƌĂŵŵŝŶŐ͗/ŶƚĞŐĞƌWƌŽŐƌĂŵŵŝŶŐĂŶĚ'ŽĂůWƌŽŐƌĂŵŵŝŶŐഩ‹ഩ
TABLE 7.6
""
Revised Simplex Tableau 3
(
+
# (
# +
# <
# =
$ +
<33
3
(
<J=
3
4 (J+
3
()J=
)
(
+33
(
3
4 (
3
(
3
(
4(
#+
3
3
3
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(
4 <J+
3
<J=
(
#=
3
3
3
4 <J=
3
4(J+
(
4<J=
( ¨
+33
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3
3
3
3
/ (
()J=
3
<J=
3
4 <J=
3
3
4 +)
3
4 )3
3
≠
D $ '
#('4+)∏4 <J=7 (33J<
#<'4)3∏4(J+7 (33
0 #( #= " # # ::
TABLE 7.7
Revised Simplex Tableau 4: Optimal Solution
(
+
# (
# +
# <
# =
+
<33
3
(
3
3
4 (
(
<
(
+33
(
3
3
3
)J<
4 =J<
+
#+
3
3
3
3
(
4 (
4 +
3
#(
3
3
3
(
3
+J<
4 =J<
(
+33
<33
3
3
3
3
/ +
<
(
3
3
3
D 3
3
3
3
4 (33J<
4(33J<
""
0 %&&% (7++7<7(<33
% % & %&& J$ %&&/(+ %&&
#
'
# . $ G ഩ‹ഩYƵĂŶƟƚĂƟǀĞdĞĐŚŶŝƋƵĞƐŝŶDĂŶĂŐĞŵĞŶƚ
" 0 "<+(8+£(); <+(8(3+£();
0 " X-/ X-/ (J+
%UDQFKDQG%RXQG0HWKRG
# % % " # "G $ \ G
B # " - 34( dŚĞďƌĂŶĐŚĂŶĚďŽƵŶĚŵĞƚŚŽĚŝƐƵƐĞĨƵůůLJĞŵƉůŽLJĞĚŝŶƉƌŽďůĞŵƐǁŚĞƌĞƚŚĞƌĞĂƌĞĂĮŶŝƚĞŶƵŵďĞƌŽĨƐŽůƵƟŽŶƐ͘LJ
ĂƉƉůLJŝŶŐƐŽŵĞƌƵůĞƐ͕ƚŚĞƐĞƐŽůƵƟŽŶƐĂƌĞĚŝǀŝĚĞĚŝŶƚŽƚǁŽƉĂƌƚƐരͶരŽŶĞƚŚĂƚŵŽƐƚƉƌŽďĂďůLJĐŽŶƚĂŝŶƐƚŚĞŽƉƟŵĂů
ƐŽůƵƟŽŶĂŶĚ͕ƚŚĞƌĞĨŽƌĞ͕ƐŚŽƵůĚďĞĞdžĂŵŝŶĞĚĨƵƌƚŚĞƌ͖ĂŶĚƚŚĞƐĞĐŽŶĚƉĂƌƚƚŚĂƚǁŽƵůĚŶŽƚĐŽŶƚĂŝŶƚŚĞŽƉƟŵĂů
ƐŽůƵƟŽŶĂŶĚ͕ƚŚƵƐ͕ďĞůĞŌŽƵƚŽĨĨƵƌƚŚĞƌĐŽŶƐŝĚĞƌĂƟŽŶ͘dŚĞƉƌŽĐĞƐƐŝƐĐŽŶƟŶƵĞĚƵŶƟůŽƉƟŵĂůƐŽůƵƟŽŶŝƐŽďƚĂŝŶĞĚ͘
6ROXWLRQWR$VVLJQPHQW3UREOHPV
! $ 0
- $ C " . <;+9933 ]# $ #
([DPSOH
7KHFRVW LQC WRSHUIRUPGLIIHUHQWMREVE\GLIIHUHQWZRUNHUVLVJLYHQDVIROORZV
!
"
/
2EWDLQWKHRSWLPDODVVLJQPHQWRIMREVWRZRUNHUVXVLQJEUDQFKDQGERXQGPHWKRG
džƚĞŶƐŝŽŶƐŽĨ>ŝŶĞĂƌWƌŽŐƌĂŵŵŝŶŐ͗/ŶƚĞŐĞƌWƌŽŐƌĂŵŵŝŶŐĂŶĚ'ŽĂůWƌŽŐƌĂŵŵŝŶŐഩ‹ഩ
- "> # 6WHS, # =]7+= $ # "# ^ # ." % +38(98
(+8+)7:) ( + < = % . % $ 6WHS,, % " ! (
" # $ ( CA3D $ ($ "
. +9(++) / ( $ # # :9
TABLE 7.8
Assignment of Job 1 to Each Worker
"" & ' ! ( " (4+4<4=4
A38+98(+8+)7())
(4+4<4=4
:+8(98(+8+)7(+:
(4+4<4=4
)<8(98=98+)7(==
% (4+4<4=4
+38(98(+8)37(33
% % $ # :+ #:) " % %% (33 "
/()) ( " (+: ( " (
" ())\ (== (
" ( # Job 1 to A
Job 1 to B
155 Feasible
127 Feasible
75
Job 1 to C
Job 1 to D
144 Infeasible
100 Infeasible
)LJXUH 7UHH'LDJUDP$VVLJQPHQWRI -REWR(DFK:RUNHU
ഩ‹ഩYƵĂŶƟƚĂƟǀĞdĞĐŚŶŝƋƵĞƐŝŶDĂŶĂŐĞŵĞŶƚ
6WHS,,, % %% (
"0 (33 J + $
( # # :A
TABLE 7.9
Assignment of Job 1 to Worker D and Job 2 to Each Worker
"" & ' ! ( " (4+4<4=4
+38(98(+8)37(33
% (4+4<4=4
+38+98(+8)37((3
(4+4<4=4
+38A+8(+8)37(:=
% # @ (G " +G " <G "
=4 " ((3# :<-
(33 Job 1 to A
Job 1 to B
75
Job 1 to C
155 Feasible
127 Feasible
144 Infeasible
Job 2 to A
Job 1 to D
Infeasible
100
Job 2 to B
Job 2 to C
100 Infeasible
110 Feasible
174 Infeasible
)LJXUH 7UHH'LDJUDP$VVLJQPHQWRI -REWR'-REWR(DFK:RUNHU
# ( " + " (:=
( + ((3 6WHS,9 # ( + (33 $ ! (+ "
' < = < = C +38(989)8:97+3(+38(98(+8
937(<3 #
:=
džƚĞŶƐŝŽŶƐŽĨ>ŝŶĞĂƌWƌŽŐƌĂŵŵŝŶŐ͗/ŶƚĞŐĞƌWƌŽŐƌĂŵŵŝŶŐĂŶĚ'ŽĂůWƌŽŐƌĂŵŵŝŶŐഩ‹ഩ
Job 1 to A
Job 1 to B
75
Job 1 to C
Job 1 to D
155 Feasible
127 Feasible
Job 2 to A
144
Job 2 to B
100
Job 2 to C
100
Job 3 to B
Job 4 to C
Job 3 to C
Job 4 to B
110 Feasible
201 Feasible
130 Feasible
174 Infeasible
)LJXUH %UDQFKDQG%RXQG7UHH2SWLPDO$VVLJQPHQW
% ((3 " %%%# %%% ((3# ) * "
=
)3
+
+9
<
(+
(
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# ((3
# $ - 0 6ROXWLRQWR7UDYHOOLQJ6DOHVPDQ3UREOHPV
# '
LO 3 Explain solution
0 - to travelling salesman
problems
% $ . / " ! #
@- ! # ([DPSOH 0U,\HULVDVDOHVPDQZLWK'HOLWH0DQXIDFWXULQJ&RPSDQ\+HZDQWVWRYLVLWVL[FLWLHVVD\
DQGVWDUWLQJZLWKFLW\ZKHUHKHLVVWDWLRQHG7KHGLVWDQFHVEHWZHHQYDULRXVFLWLHVDUHJLYHQ
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0U,\HUZDQWVWRGHYHORSDWRXUWKURXJKWKH¿YHRWKHUFLWLHVDQGUHWXUQWRKLVKRPHFLW\LQVXFKDZD\WKDW
KHKDVWRWUDYHOWKHPLQLPXPGLVWDQFH
,WPD\EHQRWHGWKDWWKHPDWUL[RIWKHGLVWDQFHVEHWZHHQGLIIHUHQWFLWLHVLVQRWDV\PPHWULFDORQH)RULQVWDQFH
WKHGLVWDQFHIURPFLW\WRLVQRWJLYHQWREHWKHVDPHDVWKHGLVWDQFHIURPFLW\WR6XFKVLWXDWLRQVPD\
ഩ‹ഩYƵĂŶƟƚĂƟǀĞdĞĐŚŶŝƋƵĞƐŝŶDĂŶĂŐĞŵĞŶƚ
GHYHORSEHFDXVHRIVHYHUDOUHDVRQV,QWKHRQHZD\WUDI¿FV\VWHPIRUH[DPSOHWKHGLVWDQFHWREHFRYHUHGWR
UHDFKDSRLQWPLJKWEHGLIIHUHQWIURPWKHGLVWDQFHWREHFRYHUHGZKLOHUHWXUQLQJ
TABLE 7.10
Inter-city Distances (in km)
(
+ ,
-
.
/
0
1
(
4
+)
(9
<)
)3
<A
+
+(
4
+9
(;
<3
(<
<
++
+9
4
(=
(;
+3
=
<)
(+
(=
4
(+
(+
)
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<3
(;
(+
4
9
;
<A
()
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(+
:
4
# $ ! $ \ (+3 " # 6 % " (4+4<4=4)4;4(# +)8+98(=8(+898<A7(+;" D . (+;" # . # (+;" ! ' # #" " 2 5 2 ( ;# % # " # # :((:(+
TABLE 7.11
Reduced Cost Table 1
,
-
.
/
0
1
(
2
:
3
(:
<+
+(
+
9
2
()
<
(:
3
<
9
(=
2
3
+
;
=
+<
3
+
2
3
3
)
=+
++
9
=
2
3
;
<+
9
(<
)
3
2
džƚĞŶƐŝŽŶƐŽĨ>ŝŶĞĂƌWƌŽŐƌĂŵŵŝŶŐ͗/ŶƚĞŐĞƌWƌŽŐƌĂŵŵŝŶŐĂŶĚ'ŽĂůWƌŽŐƌĂŵŵŝŶŐഩ‹ഩ
TABLE 7.12
Reduced Cost Table 2
City
City
1
2
3
4
5
6
1
M
7
0
17
32
21
2
0
M
15
3
17
0
3
0
14
M
0
2
6
4
15
0
2
M
0
0
5
34
22
8
4
M
0
6
24
8
13
5
0
M
0 ' (4<+4(<4==4+)4;;4)
/ ' (4<4=4+4()4;4)
' ;)" ()"
%# :(+ @ .; - (4<+4(<4==4+)4;;4) 93" # # 93" \ % " "(4<4=4+4()4;4)
D (+; " 93 " ! 0 / ! " % " (4<4=4+4()4;4) ;)
()" /)4;4) " - G " )4; " ;4) # "
#
:)
Upper Bound = 126 km
ble
pta
e
c
ac Sub-tours:
Un
5– 6
1– 3– 4– 2– 1 (65 km)
LB
5–
6– 5
(15 km)
= 80
6– 5
Break 5– 6– 5
Un
acc
ept
abl
e
)LJXUH ,QLWLDO%UDQFKDQG%RXQG7UHH
ഩ‹ഩYƵĂŶƟƚĂƟǀĞdĞĐŚŶŝƋƵĞƐŝŶDĂŶĂŐĞŵĞŶƚ
7KHZRUNLQJRIWKHPHWKRGLVOLNHWKLV)LUVWVWDUWZLWKWKHRSWLPDOVROXWLRQREWDLQHGHDUOLHULQVHUW2LQWKH
FHOOFRUUHVSRQGLQJWRWKHURXWH±LQWKHWDEOHPDNLQJWKHURXWHLPSRVVLEOH1RZVROYHLWDVDQDVVLJQPHQW
SUREOHP,WLVVKRZQLQ7DEOHVDQG
TABLE 7.13
Reduced Cost Table 1
,
-
.
/
0
1
2
2
2
2
2
2
2
TABLE 7.14
Reduced Cost Table 2
City
City
1
2
3
4
5
6
1
M
7
0
17
32
21
2
0
M
15
3
17
0
3
0
14
M
0
2
6
4
15
0
2
M
0
0
5
30
22
4
0
M
M
6
24
8
13
5
0
M
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6XEWRXUV
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džƚĞŶƐŝŽŶƐŽĨ>ŝŶĞĂƌWƌŽŐƌĂŵŵŝŶŐ͗/ŶƚĞŐĞƌWƌŽŐƌĂŵŵŝŶŐĂŶĚ'ŽĂůWƌŽŐƌĂŵŵŝŶŐഩ‹ഩ
TABLE 7.15
Reduced Cost Table 1
,
-
.
/
0
1
2
2
2
2
2
2
2
TABLE 7.16
Reduced Cost Table 2
,
-
.
/
0
1
2
2
2
2
2
2
2
TABLE 7.17
Reduced Cost Table 3
City
City
1
2
3
4
5
6
1
M
7
0
19
32
23
2
0
M
13
3
15
0
3
0
12
M
0
0
6
4
17
0
2
M
0
0
5
34
20
6
4
M
0
6
19
1
6
0
M
M
$VVLJQPHQWV
7RXU
/HQJWK
±±±±±±
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FOHDUO\LQYROYHVDWRXU±±±±±±ZLWKDWRWDO
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= 126 km
Upper bound
Revised upper bound = 87 km
Revised lower bound = 84 km
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UHGXFHWKHXSSHUERXQGWR EHFDXVHPRYHPHQW
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le
LB
= 80
6
5–
84
ab
ept
Sub-tours:
1–3–1 (40 km)
2– 6–5–4–2 (44 km)
acc
Un
6–5
U
n
acc
1RZVLQFHDJDSH[LVWVEHWZHHQWKHORZHUDQGXSSHU
ept
abl
Tour:
e
ERXQGVZHVKDOODWWHPSWWR¿QGWKHSRVVLELOLW\RI
87 1–3–5–6– 4–2–1
QDUURZLQJLW)RUWKLVDJDLQZHEUHDNWKHVPDOOHVW
(87 km)
RIWKHVXEWRXUV²LQWKHSUHVHQWFDVHWKHVXEWRXULV
±±7KHUHZRXOGDJDLQEHWZREUDQFKHVRQHWKDW
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PDNHVWKLVURXWHLPSRVVLEOH7KHSUREOHPLVWKHQVROYHGLQWKHPDQQHURIDQDVVLJQPHQWSUREOHPDVVKRZQ
LQ7DEOHV±
TABLE 7.18
Reduced Cost Table 1
,
-
.
/
0
1
2
2
2
2
2
2
2
2
TABLE 7.19
Reduced Cost Table 2
,
-
.
/
0
1
2
2
2
2
2
2
2
2
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TABLE 7.20
Reduced Cost Table 3
City
City
1
2
3
4
5
6
1
M
0
M
10
25
14
2
0
M
13
3
17
0
3
0
14
M
0
2
6
4
15
0
0
M
0
0
5
30
18
2
0
M
M
6
24
8
1
5
0
M
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TABLE 7.21
Reduced Cost Table 1
,
-
.
/
0
1
2
2
2
2
2
2
2
2
TABLE 7.22
Reduced Cost Table 2
,
-
.
/
0
1
2
2
2
2
2
2
2
2
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TABLE 7.23
Reduced Cost Table 3
City
City
1
2
3
4
5
6
1
M
5
0
21
34
19
2
0
M
17
9
21
0
3
M
8
M
0
0
0
4
15
0
4
M
4
0
5
24
12
0
0
M
M
6
20
4
11
7
0
M
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Initial UB
= 126 km
Revised UB = 87 km
Initial LB
= 80 km
Revised LB = 84 km
Revised LB
le
tab
Initial
1– 3
84
e
bl
pta
LB
= 80
Un
ep
acc
93
Sub-tours:
1–3–4
(40)
2–6–5–4–2 (44)
Break 1–3–1
3–1
Un
e
acc
acc
n
ept
U
abl
6
Sub-tours:
5–
e
1–3–4–2–1 (65)
5–6–5
(15)
Break
5–6–5
6– 5
Un
acc
ept
abl
e
87
90
Tour: 1–2–6–5–4–3–1
(93 km)
Not to be
considered as
the current UB
is exceeded
Tour: 1–3–6–5–4–2–1
(90 km)
Revised UB
STAGE–I
STAGE–II
STAGE–III
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START
Obtain value of the tour given by visiting cities in
numerical order. This is initial upper bound.
Formulate an assignment model, taking inter-city distances
as costs and placing M in all diagonal positions making
assignment of a city to itself impossible.
Solve using HAM and write value of solution in node. It represents
the lower bound on the value of any solution obtainable by
branching from this node. [LB for the whole problem equals the
least of all the nodes for which branching is not completed.]
Set new UB equal to the value of the Yes
solution corresponding to new tour.
Does the optimal
solution yield a tour with a
value less than the current UB?
No
Is the lower bound
for the whole problem
equal to UB?
No
Yes
Solution corresponding to
UB is optimal
STOP
Branch from the node with the smallest value by breaking up the shortest sub-tour at that node. Take
a branch for each city-to-city trip in this sub-tour. Formulate a new assignment problem with one of
the routes unacceptable. Do it for each of the routes involved.
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TABLE 7.26
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Simplex Tableau 3: Non-optimal Solution
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TABLE 7.62
Reduced Cost Table 4: Sub-problem 2
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513
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TABLE 7.64
Reduced Cost Table 6: Sub-problem 4
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4 / / TABLE 7.67
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There are several situations in which the outcome of a particular course of action is dependent upon the
action taken by another party. Examples include a chess game being played by two players, a war between
two countries, competing firms operating in a market, and the management and workers’ union negotiations,
and so on. In such cases, the parties involved are the players who have conflicting interests so that a gain
to one party is a loss to the other, with each of them having a certain number of strategies to choose from,
and the outcome depends upon the particular pair of strategies chosen by the parties. All these ingredients
make it a game situation, hence the name Theory of Games. The theory helps to determine the best decision
for each of the parties involved.
An understanding of the theory helps the manager to answer questions like:
€ What are the consequences of the interplay of each combination of strategies of both the players?
€ Is there one (or more) strategy for each of the players that is clearly the best one to play in the given
situation? In the language of theory, does the game have a saddle point so that players play pure
strategies?
€ If there is no clear cut strategy for each player, then what combination of strategies should be played by
each one? Thus, when no saddle point exists, what is the optimal mix of strategies for the players and
what is the expected pay-off resulting from the game?
€ Is a particular strategy or a combination of strategies for a player superior to another strategy, so that the
dominated strategy can be eliminated? By this process, if the game reduces to a small size, then how to
solve it analytically?
€ If the game size cannot be reduced to a small size, then how to formulate the game as a linear
programming problem and solve accordingly? How to use the primal–dual relationship to obtain the
answers?
For this chapter, you should know the concept and calculation of expected value and an application
of simplex method to solve linear programming problems, together with the primal–dual connection.
The ideas and concepts to focus on in this chapter include developing a pay-off matrix using the given
information, obtaining a saddle point if it exists, the reduction of game size by the principle of dominance and
formulation of a game as an LPP (especially when some negative pay-offs are involved) and its solution by
simplex algorithm.
dŚĞŽƌLJŽĨ'ĂŵĞƐഩ‹ഩ
Learning Objectives
After reading this chapter, you should be able to:
LO 1
LO 2
LO 3
LO 4
LO 5
LO 6
LO 7
LO 8
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START
Write the pay-off
matrix
Is
there a pure
strategy solution?
Yes
Obtain the
Solution(s)
No
Is it a
2 x 2 game?
Yes
No
Use dominance rule to
reduce the matrix as far
as possible
Solve by using the
analytical formulamixed strategy
Yes
Is
this reduced to
a 2 x 2 game?
Alternatively
No
Use graphical method
to reduce the problem
to a 2 x 2 game
Yes
Is
this a 2 x n
or m x 2 game?
No
Formulate and solve as
a linear programming
problem
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of a game as an LPP
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TABLE 8.1
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TABLE 8.2
Simplex Tableau 2: Non-optimal Solution
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TABLE 8.3
Simplex Tableau 3: Non-optimal Solution
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TABLE 8.4
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