A set of jobs A, B, C, D, E, F, G, H arrive at time t = 0 for processing on turning and grinding machines. Each job needs to be processed in sequence – first on the turning machine and second on the grinding machine, and the grinding must occur immediately after turning. The processing times of the jobs are given below. If the makespan is to be minimized, then the optimal sequence in which these jobs must be processed on the turning and grinding machines is

Job A B C D E F G H Turning (minutes) 2 4 8 9 7 6 5 10 Grinding (minutes) 6 1 3 7 9 5 2 4

Job	A	B	C	D	E	F	G	H
Turning (minutes)	2	4	8	9	7	6	5	10
Grinding (minutes)	6	1	3	7	9	5	2	4

If the makespan is to be minimized, then the optimal sequence in which these jobs must be processed on the turning and grinding machines is

The correct option is A-E-D-F-H-C-G-B.

Step-by-Step Explanation:

To minimize the makespan for n jobs on two sequential machines (Turning followed by Grinding), we use Johnson's Rule. The objective of Johnson's Rule is to sequence the jobs such that idle time on the machines is minimized, thereby minimizing the total elapsed time (makespan).

Johnson's Algorithm Procedure:
1. Find the minimum processing time among all remaining jobs on both machines.
2. If this minimum time occurs on the first machine (Turning), place that job in the first available position of the sequence (from left to right).
3. If this minimum time occurs on the second machine (Grinding), place that job in the last available position of the sequence (from right to left).
4. If there is a tie, we can choose arbitrarily, though specific tie-breaking choices lead to the exact sequence matching the options.
5. Remove the scheduled job from the list and repeat the process until all jobs are scheduled.

Let us apply Johnson's Rule step-by-step to our job table:

Iteration 1:
The minimum processing time in the entire table is 1 minute, which occurs for Job B on the Grinding machine (second machine).
Since this minimum time is on the second machine, we place Job B in the last available position of our 8-job sequence:
Current Sequence: [ _ , _ , _ , _ , _ , _ , _ , B ]

Iteration 2:
With Job B removed, the minimum processing time among the remaining jobs is 2 minutes. This occurs for two jobs:
- Job A: 2 minutes on the Turning machine (first machine).
- Job G: 2 minutes on the Grinding machine (second machine).

Following Johnson's Rule:
- Since Job A's minimum time is on the first machine, it is placed in the first available position from the left: 1st position.
- Since Job G's minimum time is on the second machine, it is placed in the last available position from the right: 7th position (just before B).

Current Sequence: [ A , _ , _ , _ , _ , _ , G , B ]

Iteration 3:
With jobs A, B, and G removed, the remaining jobs are C, D, E, F, H.
The minimum processing time among these is 3 minutes, which occurs for Job C on the Grinding machine (second machine).
Since this is on the second machine, we place Job C in the last available position (from the right, in the 6th position):
Current Sequence: [ A , _ , _ , _ , _ , C , G , B ]

Iteration 4:
With jobs A, B, C, G removed, the remaining jobs are D, E, F, H.
The minimum processing time among these is 4 minutes, which occurs for Job H on the Grinding machine (second machine).
Since this is on the second machine, we place Job H in the last available position (5th position):
Current Sequence: [ A , _ , _ , _ , H , C , G , B ]

Iteration 5:
The remaining jobs are D, E, F.
The minimum processing time among these is 5 minutes, which occurs for Job F on the Grinding machine (second machine).
Since this is on the second machine, we place Job F in the last available position (4th position):
Current Sequence: [ A , _ , _ , F , H , C , G , B ]

Iteration 6:
The remaining jobs are D and E. The times are:
- Job D: Turning = 9, Grinding = 7
- Job E: Turning = 7, Grinding = 9

The minimum processing time is 7 minutes, which occurs for:
- Job D on the Grinding machine (second machine).
- Job E on the Turning machine (first machine).

Applying Johnson's Rule:
- Since Job E's minimum time is on the first machine, it is placed in the first available position from the left: 2nd position.
- Since Job D's minimum time is on the second machine, it is placed in the last available position from the right: 3rd position.

Current Sequence: [ A , E , D , F , H , C , G , B ]

All jobs have been scheduled. The resulting optimal sequence to minimize the makespan is A-E-D-F-H-C-G-B.