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Question
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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Solution
| Job | A | B | C | D | E |
| Machine X | 10 | 2 | 18 | 6 | 20 |
| Machine Y | 4 | 12 | 14 | 16 | 8 |
Observe that Min(X, Y) = 2, corresponding to job B on machine X.
∴ Job B is placed first in the sequence.
| B |
Then the problem reduces to
| Job | A | C | D | E |
| Machine X | 10 | 18 | 6 | 20 |
| Machine Y | 4 | 14 | 16 | 8 |
Now, Min(X, Y) = 4, corresponds to job A on machine Y.
∴ Job A is placed last in the sequence.
| B | A |
Then the problem reduces to
| Job | C | D | E |
| Machine X | 18 | 6 | 20 |
| Machine Y | 14 | 16 | 8 |
Now, Min(X, Y) = 6, corresponds to job D on machine X.
∴ Job D is placed second in sequence.
| B | D | A |
Then the problem reduces to
| Job | C | E |
| Machine X | 18 | 20 |
| Machine Y | 14 | 8 |
Now, Min(X, Y) = 8, corresponds to job E on machine Y.
∴ Job E is placed fourth and job C on remaining in sequence.
∴ The optimal sequence is
| B | D | C | E | A |
Total elapsed time,
| Job | Machine X | Machine Y | ||
| In | Out | In | Out | |
| B (2, 12) | 0 | 2 | 2 | 14 |
| D (6, 16) | 2 | 8 | 14 | 30 |
| C (18, 14) | 8 | 26 | 30 | 44 |
| E (20, 8) | 26 | 46 | 46 | 54 |
| A (10, 4) | 46 | 56 | 56 | 60 |
∴ Total elapsed time = 60 hrs.
Idle time for machine X = 60 – 56 = 4 hrs
Idle time for machine Y = 2 + 2 + 2 = 6 hrs.
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