In the row minima method, the first row that is the lowest cost cell is exhausted. Our aim will be to allocate the maximum either at the first source or demand at the destinations or to satisfy both. This process must be continued for all the other reduced transportation costs until and unless the supply and demand are satisfied.
In this problem, we begin with the lowest cost i.e. Rs. 2000 present at the first cell AX of the first row. So we allocate the lowest out of 1000 and 2200, so it is 1000. This deletes the first row and exhausts the supply capacity of factory A. now comes the next allocation in the next row, row 2, and cell BX. Similar as before we choose the minimum value between 1500 and 1300. So it is 1300. Hence meets the demand requirements of center X and deleting the column 1.
X Y Supply
A Rs.2000 Rs.5380
Factories B Rs. 2500 Rs.2700
C Rs.2550 Rs.1700
Demand 2300 1400 3700
Similarly we move to row 3 where the minimum cost is Rs. 1700 in the cell CY. Then we allocate the minimum out from 1400 and 1200. As now we see that the demand of the distribution centers is 1400 and we could allocate only 1200 so we add more 200 in the cell BY. This satisfies the column Y and hence crossed. The similar thing we do with row 2. In row 2 the supply was satisfied by 1300+200=1500 and similarly crossed out. Same with row3.
Therefore Z = Rs. (2000 × 1000 + 2500 × 1300 + 2700 200 + 1700 × 1200 = Rs. 78, 30.000.