| Job | 1 | 2 | 3 | 4 | 5 |
| Machine A | 5 | 1 | 9 | 3 | 10 |
| Machine B | 2 | 6 | 7 | 8 | 4 |
Determine a sequence for the five jobs that will minimize the elapsed time.
Solution:
In the given problem, the jobs must be processed through the machines A and B in the order AB, i.e., the first machine to be processed is A and then B. Establish a sequence table containing 5 job cells to sequence the jobs. Now, select the least processing time considering both the machines. The least time is taken by job 2 on machine A (i.e., 1). Place job 2 in the first cell from left to right of the Table no. 2 (since it occurs in row 1).
Delete the allotted Job 2. Now select the least processing time available in the existing Table no. 3.
The least processing time in the reduced Table no. 4 is 2 for Job l, on machine B. Place this job 1 from right to left in the last cell (since it occurs in row 2).
The new reduced table after deleting job 1 is given (Table 5)
The least time is 3 for Job 4, on machine A. Place this job left to right (since it occurs in row 1) next to job 2 as shown in Table no. 6.
The reduced table after deleting job 4 is given in Table 7
The least processing time is 4 for Job 5. Place the job right to left (since it occurs in row 2) next to job 1 as shown in Table no. 8.
Finally we have only one job left, that is Job 3. Place the job in the empty cell as shown in Table no. 9.
The optimal sequence is thus obtained and the flow of Job 1 to Job 5 through the machine A and the machine B will be
2 → 4 → 3 → 5 → 1
The minimum elapsed time and the idle time for machine A and machine B is shown in Table no. 10.
In Table 10, the minimum elapsed time, (from Job 2 to Job 1) is 30 hours. The idle time on machine A is 2 hours and on machine B is 3 hours.
