In the previous section, we used the Stepping Stone Method to find the optimal solution of a transportation problem. In this section, we provide another example to boost your understanding. Let's concentrate on the following example:
Consider the following transportation problem (cost in rupees)
Distributor | ||||
---|---|---|---|---|
Factory | D | E | F | Supply |
A | 2 | 1 | 5 | 10 |
B | 7 | 3 | 4 | 25 |
C | 6 | 5 | 3 | 20 |
Requirement | 15 | 22 | 18 | 55 |
Find out the minimum cost of the given transportation problem.
Solution.
We compute an initial basic feasible solution of the problem by Matrix Minimum Method as shown in table 1.
Table 1
Distributor | ||||
---|---|---|---|---|
Factory | D | E | F | Supply |
A | 2 | 5 | 10 | |
B | 4 | 25 | ||
C | 5 | 20 | ||
Requirement | 15 | 22 | 18 | 55 |
Here, m + n - 1 = 5. So the solution is not degenerate.
1 X 10 + 7 X 13 + 3 X 12 + 6 X 2 + 3 X 18 = 203
The cell AD (2) is empty so allocate one unit to it. Now draw a closed path.
Table 2
The increase in the transportation cost per unit quantity of reallocation
is:
+ 2 1 + 3 7 = - 3.
The allocations for other unoccupied cells are following:
Unoccupied cells | Increase in cost per unit of reallocation | Remarks |
---|---|---|
AF | +5 1 + 3 7 + 6 3 = 3 | Cost Increases |
CE | +5 3 + 7 6 = 3 | Cost Increases |
BF | +4 7 + 6 3 = 0 | Neither increase nor decrease |
This indicates that the route through AD would be beneficial to the company. The maximum amount that can be allocated to AD is 10 and this will make the current basic variable corresponding to cell AE non basic.
Table 3 shows the transportation table after reallocation.
Table 3
Distributor | ||||
---|---|---|---|---|
Factory | D | E | F | Supply |
A | 1 | 5 | 10 | |
B | 4 | 25 | ||
C | 5 | 20 | ||
Requirement | 15 | 22 | 18 | 55 |
Since the reallocation in any other unoccupied cell cannot decrease
the transportation cost, the minimum transportation cost is:
2 X 10 + 7 X 3 + 3 X 22 + 6 X 2 + 3 X 18 = Rs.173