# Vogel’s Approximation Method (VAM)

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The Vogel’s approximation method or VAM gives a good start up to the solution than any other method. The solution of VAM provides us with an optimum result or very close to optimum result to the starting solution. This process involves not only lest cost Cij but also the cost that exceeds Cij. So the steps are as follows:

Step I – look at the cost matrix given below

Distribution centres                                                                                                             supply

A                        X11                    Rs. 2000           X12              Rs. 5380

1000                                                          1000 (3380)

Factories          B                         X21                  Rs. 2500             X22             Rs. 2700

1500 (200)

C                         X23                 Rs. 2550            X32 Rs.1700

1200 (850)

Demand                                            2300                                           1400

(500)                                       (1000)

Now the difference between the smallest and the second smallest cost elements have to be identified in each column and then note it down in the column within brackets, i.e., in column X there is a  difference of 500 and 100in the second column.

Now the difference between the smallest and the second smallest elements in the row are written at the right side of each bracket i.e. ill row A 3380, in row B 200 andRow C 850.

The difference is finally noted down because the elements written within the brackets below the columns or rows are the penalty incurred due to the failing in the least cost cell in the row or column.

Step II – the row or column with the maximum difference is allocated to the least cost cell is thus selected. If you find the same values, then you are free to choose anyone value. Things will be clear if you look at this example, here 3380 is the greatest difference, so row A is selectedas the least cost cell containing 1000 in the cell AX is thus allocated.

Step III– now the rows or columns that satisfy the conditions are crossed out. Here in this example row, A is crossed out. Below is the matrix without row A.

X                                    Y

B                                        Rs. 2500                           Rs. 2700

X21                                    X22                           1500(200)

C                                        Rs. 2550                           Rs.1700

X23                                 X32 = 1200                  1200 (850)

Step I and Step II is thus repeated till all the allocations are done. After row A, column Y shows the highest difference as CY has avalue of 1200. There the conditions are not fulfilled. Not only this, but row C also carries a huge difference (850) compared to the other rows. This is the reason we allocate 1200 to the cell CY which is the least cost cell in row C.

So here is the final form of the matrix given below:

X                                      Y

B                             Rs. 2500                               Rs. 2700

X21= 1300                              X22= 200

As the cell, BX has the least cost, so the maximum allocations of 1300 are made here. In the cell BY, 200 could be allocated.

Below is the list of all the allocations made in one matrix.

Rs. 200                                                  Rs.5380

X11                                                         X12

1000

Rs.2500                                                 Rs. 2700

X21                                                         X22

1300                                                       2000

Rs.2550                                                  Rs.1700

X23                                                         X32

1200

Therefore the cost of transportation is related to the following calculation:

Z = Rs. (2000 x 1000 + 2500 x 1300 + 1700 x 1200 + 2700 x 200) = Rs. 78, 30, 000

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