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प्रश्न
A company has three jobs on hand. Each of these must be processed through two departments, in the order AB where
Department A: Press shop and
Department B: Finishing
The table below gives the number of days required by each job in each department
Job | I | II | III |
Department A | 8 | 6 | 5 |
Department B | 8 | 3 | 4 |
Find the sequence in which the three jobs should be processed so as to take minimum time to finish all the three jobs. Also find idle time for both the departments.
उत्तर
Job | I | II | III |
Department A | 8 | 6 | 5 |
Department B | 8 | 3 | 4 |
Observe that Min(A, B) = 3, corresponds to job II on depatment B.
∴ Job II is placed last in sequence
II |
Then the problem reduces to
Job | I | III |
Department A | 8 | 5 |
Department B | 8 | 4 |
Now, Min (A, B) = 4, corresponds to job III on department B.
∴ Job III is placed second in sequence and job I on remaining.
∴ Optimal sequence is
I | III | II |
Total elapsed time.
Job | Dept. A | Dept. B | ||
In | Out | In | Out | |
I (8, 8) | 0 | 8 | 8 | 16 |
III (5, 4) | 8 | 13 | 16 | 20 |
II (6, 3) | 13 | 19 | 20 | 23 |
∴ Total elapsed time = 23 days
Idle time for Department A = 23 – 19 = 4 days
Idle time for Department B = 8 days.
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संबंधित प्रश्न
There are five jobs, each of which must go through two machines in the order XY. Processing times (in hours) are given below. Determine the sequence for the jobs that will minimize the total elapsed time. Also find the total elapsed time and idle time for each machine.
Job | A | B | C | D | E |
Machine X | 10 | 2 | 18 | 6 | 20 |
Machine Y | 4 | 12 | 14 | 16 | 8 |
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Machine A | 6 | 7 | 5 | 11 | 6 | 7 | 12 |
Machine B | 4 | 3 | 2 | 5 | 1 | 5 | 3 |
Machine C | 3 | 8 | 7 | 4 | 9 | 8 | 7 |
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