Discussion
Consider the following assignment cost matrix :
| Machines / Jobs | J1 | J2 | J3 | J4 |
|---|---|---|---|---|
| M1 | 10 | 9 | 7 | 8 |
| M2 | 5 | 8 | 7 | 7 |
| M3 | 5 | 4 | 6 | 5 |
| M4 | 2 | 3 | 4 | 5 |
The optimal cost of assignment is :
Answer: C
The optimal cost is 20.
Hungarian method. Subtract each row's minimum (7, 5, 4, 2), then each column's minimum (0, 0, 0, 1). That removes 19 in total, so no assignment can cost less than 19.
In the reduced matrix, M2 and M4 both have their only zero in column J1, so a complete assignment of zeros is impossible. Covering the zeros and adjusting by the smallest uncovered value adds one more unit, giving 20, and then a full set of independent zeros exists:
\[ \begin{aligned} & M_{1} \to J_{3} = 7,\ M_{2} \to J_{1} = 5 \\ & M_{3} \to J_{4} = 5,\ M_{4} \to J_{2} = 3 \\ & \text{Total} = 7 + 5 + 5 + 3 = 20 \end{aligned} \]
Checking every one of the 24 assignments confirms that none costs less than 20, so 18 is impossible, and 24 and 26 are feasible but not optimal.
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