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प्रश्न
Find the sequence of the following five jobs to be processed on three machines M1 M2,, M3 that will minimize the total elapsed time to complete the tasks. Each job is to be processed in the order M1 - M2 - M3:
| Jobs | 1 | 2 | 3 | 4 | 5 |
| Machine M1 | 5 | 11 | 5 | 7 | 6 |
| Machine M2 | 1 | 4 | 2 | 5 | 3 |
| Machine M3 | 1 | 5 | 2 | 3 | 4 |
योग
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उत्तर
Min(M1) = 5, Max(M2) = 5, Min(M3) = 1
Min (M1) ≥ Max (M2)
∴ Min(M3) ≥ Max (M2) is satisfied.
G = M1 + M3 and H = M2 + M3
| Jobs | 1 | 2 | 3 | 4 | 5 |
| MG | 6 | 15 | 7 | 12 | 9 |
| MH | 2 | 9 | 4 | 8 | 7 |
∴ The required sequence is
| 2 | 4 | 5 | 3 | 1 |
The table of total elapsed time is as follows:
| Job | Machine (M1) | Machine (M2) | Machine (M3) | |||
| In | Out | In | Out | In | Out | |
| 2 | 0 | 11 | 11 | 15 | 15 | 20 |
| 4 | 11 | 18 | 18 | 23 | 23 | 26 |
| 5 | 18 | 24 | 24 | 27 | 27 | 31 |
| 3 | 24 | 29 | 29 | 31 | 31 | 33 |
| 1 | 29 | 34 | 34 | 35 | 35 | 36 |
Total elapsed time = 36 units.
Idle time for M1 = 2 units
Idle time for M2 = 21 units
Idle time for M3 = 21 units
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Sequencing in Management Mathematics
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