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VOGEL’S APPROXIMATION METHOD (VAM)

This method usually provides a better initial (starting) solution than the method described already. In fact, VAM generally, yields an optimum or very close to optimum starting solution. This method takes into account not only the lest cost Cij but also the costs that just exceeds Cij. The following steps are involved in
this method.

Step I Write down the cost matrix as shown below.

Distribution centers

Supply

 

A

X11

Rs.2000

1000

X12

Rs.5380

1000 (3380)

Plants

B

X21

Rs.2500

X22

Rs.2700

1500 (200)

C

X23

Rs.2550

X32

Rs.1700

1200 (850)

Demand

2300

1400

 

(500)

(1000)

 

Find out the difference between the smallest and second smallest cost elements in each column and write it below the column in brackets i.e., in column X the difference is 500 and in the second column it is 1000

Find out the difference between the smallest and second smallest cost elements in each row and write it on the right side of each row in brackets i.e. ill row A 3380, in row B 200 and in row C 850.

It may be noted that the ‘difference’, which is indicated under columns or rows actually indicates the unit penalty incurred by failing to make an allocation to the least cost cell in the row or column.

Step II

Select the row or column with the maximum difference and allocate as much as possible (keeping the restrictions of supply and demand in mind) to the least cost cell in the row or column selected. In case of a tie, take up anyone. Now, in this example, since 3380 is the greatest difference, we choose row A and allocate 1000 to least cost cell i.e., AX.

Step III.

Cross out the row or column which satisfies the condition by allocation just made. So row A is crossed out. The matrix without row A is as shown below

X

Y

B

Rs.2500

X21

Rs.2700

X22

1500 (200)

C

Rs.2550

X23

Rs.1700

X32=1200

1200 (850)

 

Repeat step I to II till all the allocations have been made. Now column Y shows maximum difference, so we allocate to the least cost cell in Y column i.e., CY an amount of 1200 but this does not satisfy column Y completely.

Also, row C shows maximum difference (850) out of the two rows. We allocate 1200 to cell CY which is the least cost cell in row C. Since this allocation completely satisfies row C, we cross row C and the shrunken matrix is shown below.

X

Y

B

Rs.2500

X21=1300

Rs.2700

X22=200

 

Since cell BX has the least cost, maximum possible allocation of 1300 is made here. In cell BY, we allocate 200.

All the above allocations made can now be shown in one single matrix as below.

Rs.2000

X11

1000

Rs.5380

X12

Rs.2500

X21

1300

Rs.2700

X22

2000

Rs.2550

X23

Rs.1700

X32

1200

The cost of transportation associated with this solution is

Z = Rs. (2000 × 1000 + 2500 × 1300 + 1700 × 1200 + 2700 × 200)

= Rs, 78,30,000.