CMA Final · Strategic Cost Management · Transportation
A firm wants to maximise profit in a transportation problem. The highest unit profit in the matrix is ₹30. To solve it using the minimisation method, each profit is converted to an opportunity-loss figure. What value replaces a cell whose unit profit is ₹22?
The cell becomes ₹8. In a maximisation transportation problem, every unit profit is subtracted from the largest profit in the matrix, here ₹30, to get opportunity losses. The minimisation method then applies, and 30 minus 22 equals 8.
- A₹52
- B₹22
- C₹8Correct
- D₹30
Explanation
For a maximisation problem, each cell is subtracted from the largest profit in the matrix. This gives 30 - 22 = 8. Adding the two figures (₹52) is a wrong operation, and leaving the profit as it is (₹22) would make the minimisation method pick the least profitable routes.
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