# For a transportation problem with cost matrix as given below:

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For a transportation problem with cost matrix as given below: Find a solution by north west corner rule and find transportation cost.

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Here the availability; ai = requirement; bj = 112, thus the T.P. is balanced.

Therefore there exists a feasible solution.

Consider the north west corner cell at (1, 1) allocate as; minimum (a,, b,)

= min (100, 30) x11 = 30. Here a1 > b1 and replace b1 = 100 – 30 = 70. Next, Allocate at (1,2) as: Min (70, 50) = 50; X12 = 50.

Here again a1 > b1 and replace, b1 = 70 – 50 = 20 Next allocations meat (1,3) as; min (20, 32)

= 20 and replace (32 – 20) – 12 A; So, X13 = 20. Next allocation is mode at (2, 3) as:

Here b3 > a1

So, Min (12, 12) = 12 ; X23 = 12.

Thus, initial feasible solutions is:-

X11 = 30, X12 = 50, X13 = 20, X23 = 12

The transportation cost is: Z = ΣΣCij.Xij

= 3 × 30 + 7 × 50 + 4 × 20 + 2 × 12

= 90 + 350 + 80 + 24 = Rs. 544.

Note: In the problem m(rows) = 2, n(Columns) = 3

∴ m + n – 1 = 2 + 3 – 1 =4

The number of allocations = 4 is same as (m + n – 1).

The initial feasible solutions a basic feasible solution.