Question 16.4: Using Johnson’s Rule to Sequence Jobs A group of six jobs is...
Using Johnson’s Rule to Sequence Jobs
A group of six jobs is to be processed through a two-machine flow shop. The first operation involves cleaning and the second involves painting. Determine a sequence that will minimize the total completion time for this group of jobs. Processing times are as follows:
| PROCESSING TIME (hours) | ||
| Job | Work Center 1 | Work Center 2 |
| A | 5 | 5 |
| B | 4 | 3 |
| C | 8 | 9 |
| D | 2 | 7 |
| E | 6 | 8 |
| F | 12 | 15 |
To employ Johnson’s rule, create a “blank” sequence first, such as:
a. Select the job with the shortest processing time. It is job D, with a time of two hours.
b. Since the time is at the first center, schedule job D first. Eliminate job D from further
consideration.
c. Job B has the next shortest time. Since it is at the second work center, schedule it last and eliminate job B from further consideration. We now have
d. The remaining jobs and their times are
| Job | 1 | 2 |
| A | 5 | 5 |
| C | 8 | 9 |
| E | 6 | 8 |
| F | 12 | 15 |
Note that there is a tie for the shortest remaining time; job A has the same time at each work center. It makes no difference, then, whether we place it toward the beginning or the end of the sequence. Suppose it is placed arbitrarily toward the end. We now have
e. The shortest remaining time is six hours for job E at work center 1. Thus, schedule that job toward the beginning of the sequence (after job D):
f. Job C has the shorter time of the remaining two jobs. Since it is for the first work center, place it third in the sequence. Finally, assign the remaining job (F) to the fourth position and the result is
g. Construct a Gantt chart to reveal flow time and idle time information. Be very careful not to schedule the beginning of work at center 2 before work at center 1 has been completed for any given job. Traditionally, it is assumed that center 1 must finish and pass the job to center 2, which can cause idle time in center 2’s schedule, such as in the case of job F as follows:
Thus, the group of jobs will take 51 hours to complete. The second work center will wait two hours for its first job and also wait two hours after finishing job C. Center 1 will be finished in 37 hours. Of course, idle periods at the beginning or end of the sequence could be used to do other jobs or for maintenance or setup/teardown activities.
