In the column minima method, we begin with the first column and allocate gradually moving towards the lowest cost cell of the column. This system is continued until the first destination centre is satisfied or the capacity of the second is exhausted, or both happens. So there are three cases which are as follows:
This process is continued until all the reduced transportation table conditions are satisfied. The matrix below explains the whole row minima method.
X Y Supply
A Rs. 2000 Rs.5380
Factories B Rs. 2500 Rs. 2700
1300 200 1500
C Rs. 2500 Rs. 1700
Demand 2300 1400 3700
So let’s now start solving in this mentioned method. So here we see the lowest cost cell in the column AX. We allocate the minimum amount which is 1000 out of 2300, 1000. With this, the capacity of the factory A is exhausted and the row one is crossed out. The next allocation is done with the cell BX. Similarly, the taking the minimum amount of the first column is 2500; we allocate with the minimum 1300 in this cell. This again satisfies the distribution centre X and hence the first column is also crossed.
Now moving to the second column with the minimum cost cell CY. We allocate 1200 in this cell among 1400 and 1200. Now considering the next one BY where we can allot, and all the conditions are satisfied.
Transportation cost associated with this solution is
Z = Rs. (2000 x 1000 + 2500 x 1300 + 1700 x 1200 + 2700 x 200)
= Rs. 78.34, 000
Which is same as a result obtained by row minima method.
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