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Question
Find the sequence that minimizes total elapsed time to complete the following jobs in the order XY. Find the total elasped time and idle times for each machine.
Jobs | A | B | C | D | E |
Machine X | 10 | 2 | 18 | 6 | 20 |
Machine Y | 4 | 12 | 14 | 16 | 8 |
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 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 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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