Transportation Problem: minimum cell cost Method

With the minimum cell cost method, the basic logic is to allocate to the cells with the lowest costs.

Notice that all of the remaining cells in column A have now been eliminated, because all of the wheat was demanded at destination A, Chicago, has now been supplied by source 3, Des Moines.

The next allocation is made to the cell that has the minimum cost and also is feasible. This is cell 3B which has a cost of \$5. The most that can be allocated is 75 tons (275 tons minus the 200 tons already supplied). This allocation is shown in Table B-4.

The third allocation is made to cell 1B, which has the minimum cost of \$8. (Notice that cells with lower costs, such as 1A and 2A, are not considered because they were pre- viously ruled out as infeasible.) The amount allocated is 25 tons. The fourth allocation of 125 tons is made to cell 1C, and the last allocation of 175 tons is made to cell 2C. These allocations, which complete the initial minimum cell cost solution, are shown in Table B-5.

The total cost of this initial solution is \$4,550