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Example : Solve the following transportation problem to maximize profit and give criteria for optimality.
Profits (Rs) / Unit
Destination Origin 1 2 3 4 Supply A B C
Demand 40 20 60 30
Solution: We first convert the elements of profit matrix by multiplying by -1 and then we adopt the method of minimization. This is done in table 44
Table 44
Destination Origin 1 2 3 4 Supply A B C
Demand 40 20 60 30
In the above problem total supply is not equal to total demand. (supply demand) and hence it is an unbalanced transportation model. To balance, we introduce a dummy column 5. to find the initial feasible solution by VAM. We put the elements in the dummy column 90 as in table 45
Table 45
Destination Origin 1 2 3 4 Supply (Penalty A B C
Demand 40 20 60 30 Penalty (4) (3) (2) (3)
Allot in the row B having maximum penalty and in the cell (B, I) with least cost.
The origin B is deleted for further analysis as it is exhausted. We have the reduced matrix given in table 46.
Table 46
Destination Origin 1 2 3 4 5 Supply Penalty A -40 -25 100 (7) C -38 20
Demand 10 20 60 30 50 Penalty (2) (13) (6) (3) (0)
Allot in the column (2), which has a maximum penalty (13) and in the cell (C, 2) with least cost. Column 2 is deleted as the supply is satisfied and we have the reduced matrix as in table 46
Table 47
Origin Destination Supply Penalty 1 3 4 5 A C
10
Demand 10 60 30 0 50 (8) Penalty (2) (6) (3) (0)
Allot (C, 1) with 10 units and delete column 1. Then we have table 48
Table 48
Origin Destination Supply Penalty 3 4 5 A
C
Demand 60 30 50 Penalty (6) (3) (0)
Allot 30 to cell (A, 4) and delete column 4. Then we have table 49
Table 49
Origin Destination Supply Penalty 3 5 A
C
40 -28 (^0 40) (28) Demand 60 50 Penalty (6) (0)
Table 54
Origin Destination 1 2 3 4 5 A B C
( ui + vj ) for unalloted cells
From table 55 see that the cell (A, 1) has a negative value indicating that there is scope for optimality. This is a potential box to be alloted. Therefore we go back to the allotment matrix in table 51. Allotting to cell (A, 1) we have to remove from cell (A, 3), add to cell (C, 3) and remove from cell (C, 1) so that the loop is complete indicating that reallotment is possible. Now we have to decide the amount to be alloted and this is decided with the idea that in the process of reallotment no cell can have negative allotment. Hence we have the new allotment as per the following table 56
Table 55
Origin Destination 1 2 3 4 5 A B C
Cell Evaluation
Table 56
Origin Destination 1 2 3 4 5 A B C
Reallotment
Table 57
Origin Destination 1 2 3 4 5 ui A -40 - -22 -33 0 0 B -44 - - - - - C -38 -28 - - - vj -40 -32 -22 -33 0
Evaluation of ui and vj
Table 58
Origin Destination 1 2 3 4 5 A B C
Cost of unalloted cells
Table 59
Origin Destination 1 2 3 4 5 A B C
( ui + vj ) for unalloted cells
Table 60
Origin Destination 1 2 3 4 5 A B C
Cell Evaluation
The matrix of cell evaluation of cell evaluation in table 60 shows a negative entry in the cell (B, 3) indicating that this is a potential cell for allotment so that we get a better result. Hence we make a reallotment as shown by dotted lines in table 61. Table 61
Origin Destination 1 2 3 4 5 A B C
Reallotment We can conduct once again the optimality test. One more iteration would reveal that the allotments in the table 61 are optimal. This iteration is left to the reader. Hence the profit
(20 40) (30 33) (20 44) (50 0) (10 30) (20 38) (50 20) Rs .5130.
