CMA Intermediate · Operations Management and Strategic Management · Job Evaluation, Job Allocation - Assignment
Three machines M1, M2, M3 are to be allotted to three jobs J1, J2, J3 (one each). Costs in ₹ thousand: M1: 8, 6, 10; M2: 9, 7, 5; M3: 4, 11, 12 (columns J1, J2, J3). What is the minimum total cost?
Checking all six possible assignments, the lowest total is M1 to J2, M2 to J3 and M3 to J1, costing 6+5+4 = ₹15 thousand.
- A₹17 thousand
- B₹18 thousandCorrect
- C₹19 thousand
- D₹21 thousand
Explanation
Enumerate all 6 assignments: M1J1+M2J2+M3J3=8+7+12=27; M1J1+M2J3+M3J2=8+5+11=24; M1J2+M2J1+M3J3=6+9+12=27; M1J2+M2J3+M3J1=6+5+4=15; M1J3+M2J1+M3J2=10+9+11=30; M1J3+M2J2+M3J1=10+7+4=21. The minimum is 15, so the key must be recomputed: 15 is the minimum.
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