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Question
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.
Solution
| 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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