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Solution of example by using the
Johnson's Rule rules:
Given
table:
| Job |
Work Centre 1
(Drill) |
Work Centre 2
(Lathe) |
| A |
5 |
2 |
| B |
3 |
6 |
| C |
8 |
4 |
| D |
10 |
7 |
| E |
7 |
12 |
Set the sequence that will
minimise the total processing time for the five jobs:
Step 1 : The job with the shortest processing time is A, in work centre 2(with a
time of 2 hours).
Because it is at the second centre, schedule A
last.
Step 2
: Job B
has the next shortest time(3).
Because that time is the first work centre, we should schedule it first and
eliminate it from the consideration.
Step 3
: The next shortest time is Job C(4)
on the second machine.
Therefore, place it at last.
Step 4 : There is a tie(at 7) for the shortest
remaining job.
We can place E, which was on
the first work centre, first.
Then D is placed in the last
sequencing position.
The sequencial times are:
| Work Centre 1 |
3 |
7 |
10 |
8 |
5 |
| Work Centre 2 |
6 |
12 |
7 |
4 |
2 |
Figure9
The Time-phased flow of this job sequence is illustrated graphically as below :
 Simulation on Johnson's Rule
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