Discussion
Consider the following assignment problem
| Jobs / Machines | A | B | C | D |
|---|---|---|---|---|
| I | 10 | 25 | 15 | 20 |
| II | 15 | 30 | 5 | 15 |
| III | 35 | 20 | 12 | 24 |
| IV | 17 | 25 | 24 | 20 |
The solution to this problem is given by :
Answer: A
The optimal assignment is I → A, II → C, III → B, IV → D.
Its total cost is the lowest possible:
\[ 10 + 5 + 20 + 20 = 55 \]
The Hungarian method (or checking all 24 assignments) confirms that no other assignment costs less; the next best cost 62. The other options cost more: I → A, II → B, III → C, IV → D costs 10 + 30 + 12 + 20 = 72; I → B, II → A, III → C, IV → D costs 25 + 15 + 12 + 20 = 72; and I → B, II → C, III → A, IV → D costs 25 + 5 + 35 + 20 = 85.
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