2. Example-2
Find solution of Processing 7 Jobs Through 3 Machines Problem
| Job | 1 | 2 | 3 | 4 | 5 | 6 | 7 | | Machine-1 | 5 | 7 | 3 | 4 | 6 | 7 | 12 | | Machine-2 | 2 | 6 | 7 | 5 | 9 | 5 | 8 | | Machine-3 | 10 | 12 | 11 | 13 | 12 | 10 | 11 |
Solution:
| Job | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|
| Machine `M_1` | 5 | 7 | 3 | 4 | 6 | 7 | 12 |
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| Machine `M_2` | 2 | 6 | 7 | 5 | 9 | 5 | 8 |
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| Machine `M_3` | 10 | 12 | 11 | 13 | 12 | 10 | 11 |
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Since any of condition `min{T_(1j)} >= max{T_(ij)}` and/or `min{T_(mj)} >= max{T_(ij)}`, for j=2,3,...,m-1 is satisfied.
So given problem can be converted to 2-machine problem.
Machine-G
Machine-H
1. The smallest processing time is 7 hour for job 1 on Machine-G. So job 1 will be processed first.
2. The next smallest processing time is 9 hour for job 4 on Machine-G. So job 4 will be processed after job 1.
3. The next smallest processing time is 10 hour for job 3 on Machine-G. So job 3 will be processed after job 4.
4. The next smallest processing time is 12 hour for job 6 on Machine-G. So job 6 will be processed after job 3.
5. The next smallest processing time is 13 hour for job 2 on Machine-G. So job 2 will be processed after job 6.
6. The next smallest processing time is 15 hour for job 5 on Machine-G. So job 5 will be processed after job 2.
7. The next smallest processing time is 19 hour for job 7 on Machine-H. So job 7 will be processed last.
According to Johanson's algorithm, the optimal sequence is as below
| Job | `M_1`
In time | `M_1`
Out time | `M_2`
In time | `M_2`
Out time | `M_3`
In time | `M_3`
Out time | Idle time
`M_2` | Idle time
`M_3` | | 1 | 0 | 0 + 5 = 5 | 5 | 5 + 2 = 7 | 7 | 7 + 10 = 17 | 5 | 7 | | 4 | 5 | 5 + 4 = 9 | 9 | 9 + 5 = 14 | 17 | 17 + 13 = 30 | 2 | - | | 3 | 9 | 9 + 3 = 12 | 14 | 14 + 7 = 21 | 30 | 30 + 11 = 41 | - | - | | 6 | 12 | 12 + 7 = 19 | 21 | 21 + 5 = 26 | 41 | 41 + 10 = 51 | - | - | | 2 | 19 | 19 + 7 = 26 | 26 | 26 + 6 = 32 | 51 | 51 + 12 = 63 | - | - | | 5 | 26 | 26 + 6 = 32 | 32 | 32 + 9 = 41 | 63 | 63 + 12 = 75 | - | - | | 7 | 32 | 32 + 12 = 44 | 44 | 44 + 8 = 52 | 75 | 75 + 11 = 86 | 3 | - |
The total minimum elapsed time = 86
Idle time for Machine-1
`=86 - 44`
`=42`
Idle time for Machine-2
`=(5)+(9-7)+(44-41)+(86-52)`
`=5+2+3+34`
`=44`
Idle time for Machine-3
`=(7)+(86-86)`
`=7+0`
`=7`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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