Page 128 - DCOM303_DMGT504_OPERATION_RESEARCH

Unit 6: Transportation Problem – Optimality Tests

Solution: First, we find out an initial basic feasible solution by Matrix Minimum Method Notes
Factory D E F G Capacity
A 4 6 450 8 6 250 700
B 3 50 5 2 350 5 400

C 3 350 9 6 5 250 600
Requirement 400 450 350 500 1700
Here, m + n – 1 = 6. So the solution is not degenerate.
The cell AD (4) is empty so allocate one unit to it. Now draw a closed path from AD.

Factory D E F G Capacity
A 4 +1 6 450 8 6 249 700
B 3 50 5 2 350 5 400
C 3 349 9 6 5 251 600

Requirement 400 450 350 500 1700
The increase in the transportation cost per unit quantity of reallocation is + 4 – 6 + 5 – 3 = 0.
This indicates that every unit allocated to route AD will neither increase nor decrease the
transportation cost. Thus, such a reallocation is unnecessary.
Choose another unoccupied cell. The cell BE is empty so allocate one unit to it.
Factory D E F G Capacity
A 4 6 449 8 6 251 700

B 3 49 5 +1 2 350 5 400
C 3 351 9 6 5 249 600
Requirement 400 450 350 500 1700
The increase in the transportation cost per unit quantity of reallocation is + 5 – 6 + 6 – 5 + 3 –
3 = 0
This indicates that every unit allocated to route BE will neither increase nor decrease the
transportation cost. Thus, such a reallocation is unnecessary.
The allocations for other unoccupied cells are:
Unoccupied cells Increase in cost per unit of reallocation Remarks
CE + 9 – 6 + 6 – 5 = 4 Cost Increases

CF + 6 – 3 + 3 – 2 = 4 Cost Increases
AF + 8 – 6 + 5 – 3 + 3 – 2 = 5 Cost Increases
BG + 5 – 5 + 3 – 3 = 0 Neither increase nor decrease
Since all the values of unoccupied cells are greater than or equal to zero, the solution obtained
is optimum.
Minimum transportation cost is:
6 × 450 + 6 × 250 + 3 × 250 + 2 × 250 + 3 × 350 + 5 × 250 = ` 7350

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