| Job | I | II | III | IV | V |
| Lathe | 4 | 1 | 5 | 2 | 5 |
| Surface grinder | 3 | 2 | 4 | 3 | 6 |
Observe that Min (lathe, grinder) = 1, corresponds to job II on lathe.
∴ Job II is placed first in sequence.
Then the problem reduces to
| Job | I | III | IV | V |
| Lathe | 4 | 5 | 2 | 5 |
| Surface grinder | 3 | 4 | 3 | 6 |
Now, Min (lathe, grinder) = 2, corresponds to job IV on lathe.
∴ Job IV is placed after II in sequence.
Then the problem reduces to
| Job | I | III | V |
| Lathe | 4 | 5 | 5 |
| Surface grinder | 3 | 4 | 6 |
Now, Min (lathe, grinder) = 3, corresponds to job I on grinder.
∴ Job I is placed last in sequence.
Then the problem reduces to
| Job | III | V |
| Lathe | 5 | 5 |
| Surface grinder | 4 | 6 |
Now, Min (lathe, grinder) = 4, corresponds to job III on grinder.
∴ Job III is placed before I and V on remaining in sequence.
∴ Optimal sequence is
Total elapsed time
| Job | Lath | Surface Grinder |
| In | Out | In | Out |
| II (1, 2) | 0 | 1 | 1 | 3 |
| IV (2, 3) | 1 | 3 | 3 | 6 |
| V (5, 6) | 3 | 8 | 8 | 14 |
| III (5, 4) | 8 | 13 | 14 | 18 |
| I (4, 3) | 13 | 17 | 18 | 21 |
∴ Total elapsed time = 21 hrs
Idle time for lathe = 21 – 17 = 4 hrs
Idle time for surface grinder = 1 + 2 = 3 hrs.