Vogel Approximation Method

The Vogel approximation method (Unit cost penalty method) is an iterative procedure for computing a basic feasible solution of a transportation problem.

This method is preferred over the two methods discussed in the previous sections, because the initial basic feasible solution obtained by this method is either optimal or very close to the optimal solution.

This method is a little complex than the previously discussed methods. So go slowly and reread the explanation atleast twice.

Steps in Vogel Approximation Method (VAM)

The standard instructions are paraphrased below:

Identify the boxes having minimum and next to minimum transportation cost in each row and write the difference (penalty) along the side of the table against the corresponding row.
Identify the boxes having minimum and next to minimum transportation cost in each column and write the difference (penalty) against the corresponding column
Identify the maximum penalty. If it is along the side of the table, make maximum allotment to the box having minimum cost of transportation in that row. If it is below the table, make maximum allotment to the box having minimum cost of transportation in that column.
If the penalties corresponding to two or more rows or columns are equal, you are at liberty to break the tie arbitrarily.
Repeat the above steps until all restrictions are satisfied.

Example 1

Consider the transportation problem presented in the following table:

Destination
Origin	1	2	3	4	Supply
1	20	22	17	4	120
2	24	37	9	7	70
3	32	37	20	15	50
Demand	60	40	30	110	240

Solution.

Calculating penalty for table 1

17 - 4 = 13, 9 - 7 = 2, 20 - 15 = 5
24 - 20 = 4, 37 - 22 = 15, 17 - 9 = 8, 7 - 4 = 3

Table 1

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Destination
Origin	1	2	3	4	Supply	Penalty
1	20		17	4	~~120~~ 80	13
2	24	37	9	7	70	2
3	32	37	20	15	50	5
Demand	60	40	30	110	240
Penalty	4	15	8	3

The highest penalty occurs in the second column. The minimum cij in this column is c₁₂ (i.e., 22). So x₁₂ = 40 and the second column is eliminated. The new reduced matrix is shown below:

Now again calculate the penalty.

Table 2

Origin	1	3	4	Supply	Penalty
1	20	17		80	13
2	24	9	7	70	2
3	32	20	15	50	5
Demand	60	30	110
Penalty	4	8	3

The highest penalty occurs in the first row. The minimum c_ij in this row is c₁₄ (i.e., 4). So x₁₄ = 80 and the first row is eliminated. The new reduced matrix is shown below:

Table 3

Origin	1	3	4	Supply	Penalty
2	24		7	70	2
3	32	20	15	50	5
Demand	60	30	30
Penalty	8	11	8

The highest penalty occurs in the second column. The minimum cij in this column is c₂₃ (i.e., 9). So x₂₃ = 30 and the second column is eliminated. The reduced matrix is given in the following table.

Table 4

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