Find the initial basic feasible solution of the following transportation problem using Vogel's approximation method and find the cost.

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Adding 3 to all the costs,

Calculating the row and column penalties,

The maximum penalty, 4, occurs in the row $O_2$.

The minimum $c_{ij}$ in this row is $c_{23}$ = 0.

The maximum allocation in this cell is 8.

It satisfies demand of $D_3$ and adjusts the supply of $O_2$ from 9 to 1 (9 - 8 = 1).

Now, choosing arbitrarily, maximum penalty is in $D_1$ with minimum cost at $O_2$

Which is the initial basic feasible solution. Here the total cost is: 7×7+9×7+4×1+0×8+2×3+3×13+8×1=169.

As we added 3 to all the costs initially, thus subtracting it now, we get:

And the cost is: 4×7+6×7+1×1-3×8-1×3+0×13+5×1=49.

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