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Assign the programmers U, V, W to the programmes I, II, III in such way that the total computer time is minimum. The estimated computer time required by the programmers to execute the programmes is given in the table.

Programmes / ProgrammersUVW
I12010080
II8090110
III110140120

The optimal assignment is :

  • A.\(I \to W\), \(\mathrm{II} \to V\), \(\mathrm{III} \to U\)
  • B.\(I \to V\), \(\mathrm{II} \to U\), \(\mathrm{III} \to W\)
  • C.\(I \to U\), \(\mathrm{II} \to W\), \(\mathrm{III} \to V\)
  • D.Optimal assignment cannot be done

Answer: A

The optimal assignment is I → W, II → V, III → U, with total time 280.

With three programmers there are only six assignments, so they can be compared directly:

Assignment (I, II, III)Total
U, V, W120 + 90 + 120 = 330
U, W, V120 + 110 + 140 = 370
V, U, W100 + 80 + 120 = 300
V, W, U100 + 110 + 110 = 320
W, U, V80 + 80 + 140 = 300
W, V, U80 + 90 + 110 = 280

The Hungarian method gives the same result. Option B costs 300 and option C 370, and an optimal assignment always exists for a square cost table, so D is wrong.

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