In the previous section, we used matrix minimum method (Least cost method) to solve a transportation problem. In this section, we provide another example. Let's concentrate on the following example:
Consider the transportation problem presented in the following table:
Factory | Warehouse | Supply | ||
---|---|---|---|---|
W1 | W2 | W3 | ||
F1 | 16 | 20 | 12 | 200 |
F2 | 14 | 8 | 18 | 160 |
F3 | 26 | 24 | 16 | 90 |
Demand | 180 | 120 | 150 | 450 |
Solution.
We observe that F2W2 = 8, which is the minimum transportation cost and allocate 120 units to it. The demand for the second column is satisfied.
Table 1
Factory | Warehouse | Supply | ||
---|---|---|---|---|
W1 | W2 | W3 | ||
F1 | 16 | 20 | 12 | 200 |
F2 | 14 | 18 | ||
F3 | 26 | 24 | 16 | 90 |
Demand | 180 | 150 | 450 |
The resulting feasible solution is shown in the following table.
Final Table
Factory | Warehouse | Supply | ||
---|---|---|---|---|
W1 | W2 | W3 | ||
F1 | 20 | |||
F2 | 18 | |||
F3 | 24 | 16 | ||
Demand | 450 |
Number of basic variables = m + n –1 = 3 + 3 – 1 = 5.
The total transportation cost associated with this solution is calculated
as given below:
50 X 16 + 150 X 12 + 40 X 14 + 120 X 8 + 90 X 26 = 6460.