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The initial basic feasible solution to the following transportation problem using Vogel's approximation method is

D1D2D3D4Supply
S1121430
S2332150
S3425920
Demand20403010

  • A.\(x_{11} = 20, x_{13} = 10, x_{21} = 20, x_{23} = 20, x_{24} = 10, x_{32} = 10\), Total cost = 180
  • B.\(x_{11} = 20, x_{12} = 20, x_{13} = 10, x_{22} = 20, x_{23} = 20, x_{24} = 10\), Total cost = 180
  • C.\(x_{11} = 20, x_{13} = 10, x_{22} = 20, x_{23} = 20, x_{24} = 10, x_{32} = 10\), Total cost = 180
  • D.None of the above

Answer: D

Vogel's approximation method first finds the penalty of each row and column (the difference of the two smallest costs). The largest penalty, 3, is in column D4 (costs 4, 1, 9), so the cheapest cell there, S2-D4, takes 10 units. Next the largest penalty is 2, and column D1 is chosen, which gives x11 = 20. Then row S3 has the largest penalty and takes 20 units in D2, x32 = 20, followed by x13 = 10, x22 = 20 and x23 = 20. The cost is

\[ 20 + 10 + 10 + 40 + 60 + 40 = 180 \]

The solution has x32 = 20, while options (A) to (C) show x32 = 10, which leaves 10 units of S3 unassigned, so none of them lists the solution correctly. The keyed answer is none of the above.

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