Let us consider the matrix of the above problem where we have already found out the feasible solution.
Distribution Centers
X 
Y 

A 
Rs.2000 X_{11} 1000 
Rs.5380 X_{12}


Plants 
B 
Rs.2500 X_{21} 1300 (100) 
Rs.2700 X_{11} 2000 (+100) 
C 
Rs.2550 X_{23} +10 0 
Rs.1700 X_{12} 1200 (100) 
Let us take up any arbitrary empty cell i.e., ex and allocate + 100 units to this cell. Now in order to take the restrictions of column X, we must allocate 100 to cell BX and to maintain the row B restriction we must allocate + 100 to cell BY. This will result in unbalance of column Y conditions and so we must allot 100 to cell CY.
Now let us work out the net change in the transportation cost by the changes we have made in allocations.
Evaluation of cell CX = Rs. (2550 × 100 – 2500 × 100 + 2700 × 100 – 1700 × 100)
= 255000 – 250000 + 270000 – 170000
= Rs, 105000
As the evaluation of the empty cell ex results in a positive value the total transportation cost cannot be reduced. The feasible solution is an optimal solution already.
We must carry out evaluation of all the empty cells to be sure that optimal solution has been arrived. The total number of empty cells are m× n – (m + n 1) = (m1) (n 1). Hence (m1) (n – 1) cells must be evaluated. In the present problem m = 3 and n = 2, so only two empty cells are there but in other problems, The number of empty cells could be much more and this procedure becomes very lengthy and cumbersome.
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