#Operations Research
#Job Sequencing
#Johnson's Rule
#Processing 'n' jobs on 3 machines
Steps for solution:
Step - 1:
To check the conditions for converting the data with only 2 machines.
Johnson's Rule can be applied only in the case of processing with two (2) machines. So, where there are 3 machines we need to convert the data in such a way that as if there are only two machines (of course, virtual) to process the job. But, we can convert the data with 3 actual machines into a data with 2 virtual machines only if ANY ONE OR BOTH of the following two conditions are satisfied:
(i) Minimum processing time on Machine-1 should be "greater than or equal to" the maximum processing time on Machine-2
And / Or
(ii) Minimum processing time on Machine-3 should be "greater than or equal to" the maximum processing time on Machine-2
Only if any one or all the two of the above conditions are satisfied, we can convert the data into the data having two virtual machines and apply Johnson's Rule.
Step - 2:
Let there be only two (virtual) machines, say X and Y, such that
(i) the time on X is the total of the time on the original Machine-1 and Machine-2 (i.e. Time on X = Time on (M-1) + Time on (M-2)
and
(ii) the time on Y is the total of the time on the original Machine-2 and Machine-3 (i.e. Time on Y = Time on (M-2) + Time on (M-3)
Now, we can apply the Jonson's Rule to this new data to decide the optimal job sequencing (i.e. preference order) in the same manner as in the case of the data originally having two machines.
Step - 3
Arrange the data in a tabular form showing job-wise processing
time on each of the two machines.
Step - 4:
Find out the least processing time in the whole table
Step - 5
(i) If the least processing time is on Machine-X, then process that job first (place it in the beginning of the 'Job Sequencing Schedule')
(ii) If the least processing time is on Machine-Y, then process that job last (place it at the end of the 'Job Sequencing Schedule')
(iii) If there is a tie in the least processing time on Machine-X, then the job with the least total time will be processed first and then the other job(s) after that job in the same manner.
(iv) If there is a tie in the least processing time on Machine-Y, then the job with the least total time will be processed last and the other job(s) before that job in the same manner.
Step - 4
To prepare Schedule/Statement of Minimum Total Time Elapsed and Idle Time of machines.
PLEASE TRY TO UNDERSTAND HOW TO PREPARE THIS STATEMENT AND HOW TO FIND OUT THE IDLE TIME OF EACH MACHINE. I HAVE DISCUSSED DIFFERENT WAYS OF FINDING IDLE TIME IN DIFFERENT CASES... :)
Case: (Processing time in hours per unit)
Job: A B C D E
Machine - I: 8 10 6 7 11
Machine - II: 5 6 2 3 4
Machine - III: 4 9 8 6 5
Determine the optimal job sequencing order and also find out the minimum total time elapsed and idle time of each machine.
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