Transportation Problem

Vogel Approximation Method

Example 2

Consider the transportation problem presented in the following table:

Solution.

Table 1

Destination
Origin	1	2	3	Supply	Penalty
1		7	4	5	2
2	3	3	1	8	2
3	5	4	7	7	1
4	1	6	2	14	1
Demand	7 2	9	18	34
Penalty	1	1	1

The highest penalty occurs in the first row. The minimum cij in this row is c₁₁ (i.e., 2). Hence, x₁₁ = 5 and the first row is eliminated.

Now again calculate the penalty. The following table shows the computation of penalty for various rows and columns.

Final table

Destination
Origin	1	2	3	Supply	Penalty
1		7	4	5	2	-	-	-	-	-
2	3			8	2	2	2	2	3	3
3	5		7	7	1	1	3	3	4	-
4		6		14	1	1	4	-	-	-
Demand	7	9	18	34
Penalty	1	1	1
	2	1	1
	-	1	1
	-	1	6
	-	1	-
	-	3	-

Initial basic feasible solution

5 X 2 + 2 X 3 + 6 X 1 + 7 X 4 + 2 X 1 + 12 X 2 = 76.

Now, you must take a break because you really deserve it. We will see you at the next section when you are ready again.

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