Matrix minimum (Least cost) method is a method for computing a basic feasible solution of a transportation problem, where the basic variables are chosen according to the unit cost of transportation. This method is very useful because it reduces the computation and the time required to determine the optimal solution.
The following steps summarize the approach.
Consider the transportation problem presented in the following table:
Factory | Retail Shop | Supply | |||
---|---|---|---|---|---|
1 | 2 | 3 | 4 | ||
1 | 3 | 5 | 7 | 6 | 50 |
2 | 2 | 5 | 8 | 2 | 75 |
3 | 3 | 6 | 9 | 2 | 25 |
Demand | 20 | 20 | 50 | 60 |
Solution.
We observe that c_{21} =2, which is the minimum transportation cost. So x_{21} = 20. The demand for the first column is satisfied. The allocation is shown in the following table.
Table 1
Factory | Retail Shop | Supply | |||
---|---|---|---|---|---|
1 | 2 | 3 | 4 | ||
1 | 3 | 5 | 7 | 6 | 50 |
2 | 5 | 8 | 2 | ||
3 | 3 | 6 | 9 | 2 | 25 |
Demand | 20 | 50 | 60 |
Now we observe that c_{24} =2, which is the minimum transportation cost, so x_{24} = 55. The supply for the second row is exhausted.
Table 2
Factory | Retail Shop | Supply | |||
---|---|---|---|---|---|
1 | 2 | 3 | 4 | ||
1 | 3 | 5 | 7 | 6 | 50 |
2 | 5 | 8 | |||
3 | 3 | 6 | 9 | 2 | 25 |
Demand | 20 | 50 |
Proceeding in this way, we observe that x_{34} = 5, x_{12} = 20, x_{13} = 30, x_{33} = 20. The resulting feasible solution is shown in the following table.
Final Table
Factory | Retail Shop | Supply | |||
---|---|---|---|---|---|
1 | 2 | 3 | 4 | ||
1 | 3 | 6 | |||
2 | 5 | 8 | |||
3 | 3 | 6 | |||
Demand |
Number of basic variables = m + n –1 = 3 + 4 – 1 = 6.
The total transportation cost associated with this solution is calculated
as given below:
20 X 2 + 20 X 5 + 30 X 7 + 55 X 2 + 20 X 9 + 5 X 2 = 650.