Stepping Stone Method:
Continue the above example containing the initial basic feasible solution obtained by least cost rule. In this there six empty cell :91,1) )1,2,) )1,3),(2,3) (,2,4)and (3,2). To start with let us assess the potential for improvement of the non basic variable( 1,1). A shipment of one unit of the item on this route would mean an increase of Rs. 3 but also will mean reduction of one unit from cell( 1,4) and there by a reduction or Rs( 3,1 ) and add one unit cell ( 3, 4). This has the effect of this operation would be an increase of Rs. 3 in the cost. It is shown as follows:
Increase in cost by increasing cell(1,1) by 1 unit = + 3
Decrease in cost by decreasing cell ( 1,4) by 1 unit = -2
Increase in cost by increasing cell ( 3,4,) by1 unit = + 3
Decrease in cost by decreasing cell ( 3,1,) by 1 unit = -2
An increase in the cost of Rs. 2 per unit can be affected by adopting the route AX. Implies that the opportunity cost of this route is - 2.
If we consider the empty cell ( 1,3) then the closed loop is formed by cells( 1,4) ( 3,3),and ( 3,4).
Now if one unit is transferred from cell (1,4) to cell (1,3) then adjustments have to be made in cells ( 3,4) and (3,3) so that demand restriction is not violated. In case of transferring one unit the new adjustment allocation for these cells will thus becomes the changes in cost by one unit of such transfer is 5-4-2+3= Rs2, which is positive. This is the per unit opportunity cost for cell ( 1,3).
In this way opportunity cost can be calculated for all the empty cells. If opportunity cost for all the cells is negative, then it implies that existing solution is optimal and there is no scope for improvement. But if opportunity cost for one or more unoccupied cells is positive then the most positive opportunity cell is selected for allocation and maximum possible allocation is made in this cell making necessary adjustment s in the allocation of the other occupied cells. Again the optimality is tested. The process till an optimal solution is obtained.