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 C_{ij} but also the costs that just exceeds C_{ij}. The following steps are involved in
this method.
Step I Write down the cost matrix as shown below.
Distribution centers 
Supply 


A 
X_{11} 

Rs.2000 1000 

X_{12} 

Rs.5380 
1000 (3380) 

Plants 
B 
X_{21} 

Rs.2500 

X_{22} 

Rs.2700 
1500 (200) 


C 
X_{23} 

Rs.2550 

X_{32} 

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 X_{21} 
Rs.2700 X_{22} 
1500 (200) 
C 
Rs.2550 X_{23} 
Rs.1700 X_{32}=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 X_{21}=1300 
Rs.2700 X_{22}=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 X_{11} 1000 
Rs.5380 X_{12}

Rs.2500 X_{21} 1300 
Rs.2700 X_{22} 2000 
Rs.2550 X_{23}

Rs.1700 X_{32} 1200 
The cost of transportation associated with this solution is
Z = Rs. (2000 × 1000 + 2500 × 1300 + 1700 × 1200 + 2700 × 200)
= Rs, 78,30,000.
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