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The matrix showing the total profit earned by the company:
As all the opportunity cost of empty cells are positive (maximization problem), the solution is
optimal.
The allocations are:
Cell
Load
XB
300
300 26 = 7, 800
YA
400
400 29
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Initial Basic feasible solution by VAM:
As we have m +n 1 (= 11) allocations, the solution is feasible and all the opportunity costs of
empty cells are negative, the solution is optimal.
Allocations:
Cell
Load
Cost in Rs.
XA
300
30
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Tableau. II Cost of transportation and purchase
Market segments.
Final Allocation by MODI method.
Tableau. II Cost of transportation and purchase
Market segments.
Allocation:
From
To
Load
Bangalore
2
10,000
4,03,000
Bangalore
3
80,
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Let us multiply the matrix by 100 to avoid decimal numbers and get the basic feasible solution by VAM.
Table. Avail: Availability. Req: Requirement, Roc: Row opportunity cost, Coc: Column opportunity
cost.
Tableau. I Cost of transp
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By drawing loops, let us try to avoid 17 hours cell and include a cell, which is having time element
less than 17 hours. The basic feasible solution is having m + n 1 allocations.
Here also the maximum time of transport is 17 hours
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Revised programme: Cell O3D1 having positive opportunity cost is included in the revised
programme.
Revised Programme.
171
Linear Programming: II Transportation Model
As the opportunity costs of all empty cells are negative, the so
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Problem. 4.4. Solve the transportation problem given below:
Let us make initial assignment by using Northwest corner method. To modify the solution we
include the cell O1D3 in the programme, as it is having highest opportunity cost
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Solution by Northwest corner method:
Initial allocation show that the solution is not having (m+n1) allocations. Hence degeneracy occurs.
The smallest load is added to cell XB which does not make loop with other loaded cells.
167
L
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(2)
(3)
(4)
(5)
Linear Programming: II Transportation Model
165
4.6. DEGENERACY IN TRANSPORTATION PROBLEM
Earlier, it is mentioned that the basic feasible solution of a transportation problem must have (m + n 1)
basis variables or
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maximization
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(i) The given matrix is to be multiplied by 1, so that the problem becomes maximization problem.
Or ii) Subtract all the elements in the matrix from the highest element in the matrix. Then the problem
becomes maximizat
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2. By including the cell having zero as the opportunity cost, derive one more optimal solution,
let it be the matrix B.
3. The new matrix C is obtained by the formula: C = d A + (1 d) B, where 'd' is a positive
fraction less than 1
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Implied cost Actual cost
Action
ui + vj >
C ij
A better programme can be designed by including this cell
in the solution.
ui + vj =
C ij
Indifferent; however, an alternative programme with same
total cost can be written by includin
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Like this we have to write all allocations and calculate the cost and select the lowest one. If more
than one assignment has same lowest cost then the problem has alternate solutions.
3. Solution by Transportation method
Let us tak
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Subject to:
Xij = (Xij)2
i and j = 1 to n
n
J =1
n
X ij = 1 (d ) and
j
(Each machine to one job only)
And
Xij = 0 for all values of j and i.
X
i =1
ij
= 1 (bi)
Structural Constraints.
For i and j = 1 to n
(Each job to one machine o
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CHAPTER 5
Linear Programming : III
Assignment Model
5.1. INTRODUCTION
In earlier discussion in chapter 3 and 4, we have dealt with two types of linear programming problems,
i.e. Resource allocation method and Transportation model. We have seen that though
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Shipping cost in Paise per unit dispatch.
To
From
A
B
C
D
X
50
70
40
35
Y
45
75
40
55
Z
70
65
60
75
It is also possible to produce 25% higher than the capacity in each factory by working overtime
at 50% higher in direct costs.
(a)
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7. The total number of allocation in a basic feasible solution of transportation problem of
m n size is equal to:
(a) m n
(b) (m / n ) 1
(c) m + n +1
(d) m + n 1
( )
8. When the total allocations in a transportation model of m n si
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12. From three warehouses A, B, and C orders for certain commodities are to be supplied to
demand points 1, 2, 3, 4 and 5 monthly. The relevant information is given below:
Warehouses
Demand points (Transportation cost in Rs per uni
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Production cum transportation cost per unit in Rs.
As there are m + n 1 allocations and all the opportunity costs of empty cells are negative, the
solution is optimal.
The optimal allocations are:
Cell
XA
XB
XC
YA
YB
ZC
Z DC
Load
4
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Solution: We have to optimize production and shipment cost. Hence the transportation matrix elements
are the total of manufacturing cost plus transportation cost. For example, the manufacturing cost of
factory X is Rs. 10. Hence th
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(b) Basic variables
If unit cost of loaded cell i.e. basic variable is changed, it affects the opportunity costs of several cells.
Now let us take the same solution shown above for our discussion. In case the unit cost of transport
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(4)
W2
Y
A
B
C
D
Avail
ROC
10
X
X
X
X
160
10
160
W1
X
8
9
7
14
70
1
W2
0
8
7
11
12
350
7
Req.
230
80
100
70
100
580
COC
10
0
2
4
2
W2
A
B
C
D
(5)
Avail
ROC
W1
X
8
9
7
14
70
1
W2
0
8
7
11
12
350
7
420
70
Req.
70
80
100
70
100
COC
IN
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2. W2 has no capacity limitation. However, it deals partial direct distribution of 80 units. Therefore,
as a source its availability should be the difference between the total availability from all
factories i.e X, Y and Z less its
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As the opportunity costs of all empty cells are negative, the solution is optimal. The optimal
allocation is:
Cell
Route
Load
Cost in Rs.
XA
XW2A
50
50 18
XB
X  W2  B
100
YB
Y W2  B
YD
Y  W2 D
100
ZA
Z W1 A
ZC
Z  W1  C
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4.10. TRANSSHIPMENT PROBLEM
We may come across a certain situation, that a company (or companies) may be producing the product
to their capacity, but the demand arises to these products during certain period in the year or the
dema
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In the above matrix, two cells, MO M and JO J are having positive opportunity costs = 20.
Hence, they may be included in the revised programme. If we include them in the programme, the final
optimal solution will be as follows:
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