Least cost Method

The Least Cost Method is another technique used to find an initial feasible solution to the transportation problem. This method starts by selecting the cell with the lowest cost in the cost matrix and allocating as much as possible to that cell, subject to the supply and demand constraints. Then, the cell with the next lowest cost is selected, and the allocation process continues until all supplies and demands are met.

Here’s an example to illustrate the Least Cost Method:

Suppose we have three factories (F1, F2, F3) with capacities of 120, 180, and 100 units, respectively, and four warehouses (W1, W2, W3, W4) with demands of 150, 120, 180, and 70 units, respectively. The transportation costs per unit from each factory to each warehouse are given in the following cost matrix:

To apply the Least Cost Method, we start by selecting the cell with the lowest cost, which is the cell in row F2 and column W3 with cost 1. We allocate as much as possible to this cell, subject to the supply and demand constraints. Since F2 has a capacity of 180 units and W3 has a demand of 180 units, we allocate all 180 units to this cell. The transportation table now looks as follows:

Next, we find the cell with the next lowest cost, which is the cell in row F1 and column W3 with cost 2. We allocate as much as possible to this cell, subject to the supply and demand constraints. Since F1 has a capacity of 120 units and W3 has a remaining demand of 180 – 180 = 0 units, we allocate 120 units to this cell. The transportation table now looks as follows:

To continue with the Least Cost Method, we select the cell with the next lowest cost, which is the cell in row F3 and column W1 with cost 2. We allocate as much as possible to this cell, subject to the supply and demand constraints. Since F3 has a capacity of 100 units and W1 has a demand of 150 units, we allocate 100 units to this cell. The transportation table now looks as follows:

We then select the cell in row F3 and column W4 with cost 5 and allocate the remaining 50 units from F3 to W4, which satisfies the remaining demand of 70 units. The final allocation table is:

We can verify that the total cost of this solution is (01) + (03) + (1202) + (04) + (03) + (1203) + (01) + (602) + (1002) + (105) = 780 units.