Find solution of Processing 5 Jobs Through 2 Machines Problem
| Job | 1 | 2 | 3 | 4 | 5 |
| Machine-1 | 5 | 1 | 9 | 3 | 10 |
| Machine-2 | 2 | 6 | 7 | 8 | 14 |
Solution:
| Job | 1 | 2 | 3 | 4 | 5 |
|---|
| Machine `M_1` | 5 | 1 | 9 | 3 | 10 |
|---|
| Machine `M_2` | 2 | 6 | 7 | 8 | 14 |
|---|
1. The smallest processing time is 1 hour for job 2 on Machine-1. So job 2 will be processed first.
2. The next smallest processing time is 2 hour for job 1 on Machine-2. So job 1 will be processed last.
3. The next smallest processing time is 3 hour for job 4 on Machine-1. So job 4 will be processed after job 2.
4. The next smallest processing time is 7 hour for job 3 on Machine-2. So job 3 will be processed before job 1.
5. The next smallest processing time is 10 hour for job 5 on Machine-1. So job 5 will be processed after job 4.
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 | Idle time
`M_2` |
| 2 | 0 | 0 + 1 = 1 | 1 | 1 + 6 = 7 | 1 |
| 4 | 1 | 1 + 3 = 4 | 7 | 7 + 8 = 15 | - |
| 5 | 4 | 4 + 10 = 14 | 15 | 15 + 14 = 29 | - |
| 3 | 14 | 14 + 9 = 23 | 29 | 29 + 7 = 36 | - |
| 1 | 23 | 23 + 5 = 28 | 36 | 36 + 2 = 38 | - |
The total minimum elapsed time = 38
Idle time for Machine-1
`=38 - 28`
`=10`
Idle time for Machine-2
`=(1)+(38-38)`
`=1+0`
`=1`
This material is intended as a summary. Use your textbook for detail explanation.
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