In this method we allocate maximum possible in the lowest cost cell of the first row. The idea is to exhaust either the capacity of the first source or the demand at destination center is satisfied or both. Continue the process for the other reduced transportation costs until all the supply and demand conditions are satisfied.
In the above problem we first allot in cell AX of first row as it has the lowest cost of Rs. 2000. So we allocate minimum out of (1000, 2200) i.e., 1000. This exhausts the supply capacity of plant A and thus the first row is crossed off. The next allocation is in cell BX as the minimum cost in row 2 is in this cell. We allocate minimum of (1500, 1300) i.e., 1300 in this cell. This exhausts the demand requirements, of destination center X and so column 1 is crossed off.
Distribution centers


X 
Y 
Supply 

A 
Rs.2000
1000 
Rs.=5380 
1000 
Plants 
B 
Rs.2500
1300 
Rs. 2700
200 
1500 

C 
Rs.2550  Rs.1700
1200 
1200 

Demand 
2300 
1400 
3700 
Now we proceed to row No.3 in which the minimum cost Rs. 1700 is in cell CY. Here we allot minimum out of 1400 and 1200. Since the demand of distribution centers is 1400 and we have allotted only 1200 we allot 200 in cell BY. Now column Y is satisfied and we cross out column Y. Also since in row two the complete supply of 1500 is satisfied of (1300 + 200 = 1500) row two is also satisfied and can be crossed out. Similarly, row three is also satisfied and can be crossed out,
Hence Z = Rs. (2000 × 1000 + 2500 × 1300 + 2700 200 + 1700 × 1200 = Rs. 78,30.000.
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