Discussion
The assignment cost of assigning any one operator to any one machine is given in the following table :
| Machines | Operators | |||
|---|---|---|---|---|
| I | II | III | IV | |
| U | 10 | 5 | 13 | 15 |
| V | 3 | 9 | 18 | 3 |
| W | 10 | 7 | 3 | 2 |
| X | 5 | 11 | 9 | 7 |
The optimal assignment is :
Answer: B
The optimal assignment is U \(\to\) II, V \(\to\) IV, W \(\to\) III, X \(\to\) I, at a total cost of 16.
Hungarian method. Subtract each row minimum (5, 3, 2, 5):
U: 5, 0, 8, 10; V: 0, 6, 15, 0; W: 8, 5, 1, 0; X: 0, 6, 4, 2.
Column minima are 0, 0, 1, 0; subtracting 1 from column III gives U: 5, 0, 7, 10; V: 0, 6, 14, 0; W: 8, 5, 0, 0; X: 0, 6, 3, 2. A full set of independent zeros is U-II, V-IV, W-III, X-I, so this assignment is optimal, with cost \(5 + 3 + 3 + 5 = 16\) (the total reduction \(15 + 1\)).
The other options cost far more: U-II, V-III, W-I, X-IV costs \(5 + 18 + 10 + 7 = 40\); U-III, V-IV, W-II, X-I costs \(13 + 3 + 7 + 5 = 28\); U-IV, V-II, W-III, X-I costs \(15 + 9 + 3 + 5 = 32\).
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