Johnson's Rule in Sequencing Problems
Last Updated :
01 Sep, 2022
The sequencing problem deals with determining an optimum sequence of performing a number of jobs by a finite number of service facilities (machine) according to some pre-assigned order so as to optimize the output. The objective is to determine the optimal order of performing the jobs in such a way that the total elapsed time will be minimum.
Consider there are jobs 1,2,3,.......,n to be processed through m machines. (Machine A, Machine B, Machine C, ......, Machine n). The objective is to find a feasible solution, such that the total elapsed time is minimum. We can solve this problem using Johnson's method. This method provides solutions to n job 2 machines, n job 3 machines, and 2 jobs m machines.
Johnson's Algorithm:
Johnson's rule in sequencing problems is as follows:
- Find the smallest processing time on Machine 1 and Machine 2.
- a) If the smallest value is in Machine 1 process that job first.
b) If the smallest value is in the Machine 2 process that job last. - a) if the minimum time for both Machine 1 and Machine 2 is equal, then perform the job of Machine 1 first and then the job of Machine 2.
b) if there is a tie for a minimum time among jobs in Machine 1, select the job corresponding to the minimum of Machine 2 and process it first.
c) if there is a tie for a minimum time among jobs in Machine 2, select the job corresponding to the minimum of Machine 1 and process it last. - And then calculate the total elapsed time( the time interval between starting the first job and completing the last job) and Idle time( the time during which the machine remains idle during the total elapsed time.
In this article, we will understand this problem through n job 2 machines. We have to find the sequence of jobs to be executed in different machines to minimize the total time.
Example:
| Jobs | Machine 1 | Machine 2 |
|---|
| J1 | 9 | 7 |
|---|
| J2 | 5 | 4 |
|---|
| J3 | 10 | 9 |
|---|
| J4 | 1 | 5 |
|---|
| J5 | 3 | 2 |
|---|
Solution: The smallest time is 1 for Machine 1, process it first.
Next, the smallest time is 2 for Machine 2, process it last
The next smallest time is 3 of job 5 but Job 5 is already being processed by Machine 2. Discard it.
Next, the smallest time is 4 for Machine 2, process it last just before J5.
The next smallest time is 5 for job 2 but job 2 is already being processed by Machine 2. Discard it.
Next, the smallest value is 7 for Machine 2, process it last just before J2.
Next, the smallest value is for Machine 1 & Machine 2, J1 of Machine 1 is already being processed on Machine 2 . So, take J3 of Machine 2 and process it last just before J3.
So, the final sequence of processing the jobs is J4, J3, J1, J2, and J5.
| Job | Machine 1 | Machine 2 |
|---|
| In Time | Out Time | In Time | Out Time |
|---|
| J4 | 0 | 0+1=1 | 1 | 1+5=6 |
|---|
| J3 | 1 | 1+10=11 | 6 | 6+9=15 |
|---|
| J1 | 11 | 11+9=20 | 15 | 15+7=22 |
|---|
| J2 | 20 | 20+5=25 | 22 | 22+4=26 |
|---|
| J5 | 25 | 25+3=28 | 26 | 26+2=28 |
|---|
Total Elapsed Time = 28
Idle time of Machine 1 = 0
Idle time of Machine 2 = 1 (Machine 2 has to wait for Machine 1 for 1 unit during the execution of Job 4.
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