SKEDSOFT

Operations Research

By drawing loops, let us try to avoid 17 hours cell and include a cell, which is having time element less than 17 hours. The basic feasible solution is having m n – 1 allocation.

Here also the maximum time of transport is 17 hours.

 

In this allocation highest time element is 11 hours. Let us try to reduce the same.

In this allocation also the maximum time element is 11 hours. Let us try to avoid this cell.

No more reduction of time is possible. Hence the solution is optimal and the time required for completing the transportation is 10 Hours. Tmax = 10 hours.