please explain every step Using Johnson’s rule for 2-machine scheduling, the sequence is: Scheduled Order Six…
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Question “please explain every step Using Johnson’s rule for 2-machine scheduling, the sequence is: Scheduled Order Six…”
Answer
Answer 1
Job | Operation 1 (hours) | Operation 2 (hours) |
A | 10 | 5 |
B | 7 | 4 |
C | 5 | 7 |
D | 3 | 8 |
E | 2 | 6 |
F | 4 | 3 |
According to Johnson’s rule,
We will create a sequence equal to number of Jobs given.
Here we have 6 jobs. Hence the sequence will have 6 boxes.
Now, we will check the lowest operation time among all tasks and both the operations. If the Task time belongs to Operation 1, we will enter the task (or Job) from left hand side. If the task time belongs to Operation 2, we will enter the task (or Job) from right hand side. Once we enter the Job into the sequence, we will strike the Job from the list in order to avoid counting it again.
Step – 1:
The lowest operation time is 2 hours. It belongs to Operation 1 and is for Job E. We will enter Job E into the left of the sequence.
E |
Now the lowest time is 3 hours. There is a tie. Operation 1 time 3 hours belongs to D and Operation 2 time 3 hours belongs to F. Hence D will be entered in Left and F in Right.
E | D | F |
Now the lowest time is 4 hours. It belongs to Operation 2 for Job B. hence B will be entered to right.
E | D | B | F |
Now the lowest time is 5 hours. 5 hours in Operation 1 belongs to Job C and in operation 2 belongs to A. hence C will be entered in left and A in right.
E | D | C | A | B | F |
Hence the Job sequence is:
Scheduled Order | Job |
1 | E |
2 | D |
3 | C |
4 | A |
5 | B |
6 | F |
Total time taken to complete all jobs:
We will make a table in order to show the time taken.
Operation 1 | Operation 2 | |||||
Job | Processing Time | IN | OUT | Processing Time | IN | OUT |
E | 2 | 0 | 2 | 6 | 2 | 8 |
D | 3 | 2 | 5 | 8 | 8 | 16 |
C | 5 | 5 | 10 | 7 | 16 | 23 |
A | 10 | 10 | 20 | 5 | 23 | 28 |
B | 7 | 20 | 27 | 4 | 28 | 32 |
F | 4 | 27 | 31 | 3 | 32 | 35 |
We are writing the Job as per sequence.
For Operation – 1, E is the first Job hence will start first. As soon as Operation – 1 is free from E, it will start D. Eventually as the Operation – 1 is free from B, it will start F.
The Out time is equal to the in time plus the processing time.
In time for next Job is equal to the Out time for the previous Job.
Now we will start filling IN and OUT for Operation – 2.
We will check the operation – 1 as well as operation availability in order to start a job for Operation – 2.
Job E is done with Operation – 1 by 2nd hour. Hence Job E operation – 1 will start at 2nd hour and go till 8th hour (as processing time is 6 hours).
Now for Job D: the Operation – 1 is over at 5th hour, but Operation – 2 has been freed from Job E at 8th hour. Hence the Job D will start at 8th hour and go till 16 hours.
Similarly, if operation – 2 is free when a job is done with Operation – 1, the job will start immediately with Operation – 2. But if Operation – 2 is not free, then the job will wait till Operation – 2 is free and then will start.
Hence for Operation – 2 the start time can be mention as: MAX(OUT time of previous task in operation – 2, Out time of same task in operation – 1).
Here, the total time taken = 35 hours.
Answer 2
Job | Operation 1 (hours). | Operation 2 (hours). |
A | 10 | 5 |
B | 7 | 4 |
C | 5 | 7 |
D | 3 | 8 |
E | 2 | 6 |
F | 4 | 3 |
Johnson’s rule states that
We will create a sequence that is equal to the number of Jobs we have.
We have six jobs. The sequence will therefore have six boxes.
We will now check which operation takes the least time to complete each task. If the Task time is Operation 1, then we will go to the Task (or Job) from the left side. If the Task time is Operation 1, we will enter the task or Job from the right side. To avoid counting it again, we will enter the Job in the sequence and then we will remove the Job from our list.
Step 1:
Operation time is approximately 2 hours. It is Operation 1 and it is for Job E.
E |
The lowest time now is 3 hours. There is a tie. Operation 1:3 hours belongs to D, and Operation 2:3 hours belongs to F. Therefore D will be entered in Left and Right.
E | D | F |
The current time is now 4 hours. It is Operation 2 for Job B. therefore B will be entered at right.
E | D | B | F |
The lowest time now is 5 hours. 5 hours in Operation 1 belong to Job C, and 5 hours in operation 2 to A. C will therefore be entered in the left and A in the right.
E | D | C | A | B | F |
The Job sequence is:
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5 | B |
6 | F |
Total time taken to complete all jobs:
To show how long it took, we will create a table.
Operation 1 | Operation 2 | |||||
Job | Processing Time | IN | OUT | Processing Time | IN | OUT |
E | 2 | 0 | 2 | 6 | 2 | 8 |
D | 3 | 2 | 5 | 8 | 8 | 16 |
C | 5 | 5 | 10 | 7 | 16 | 23 |
A | 10 | 10 | 20 | 5 | 23 | 28 |
B | 7 | 20 | 27 | 4 | 28 | 32 |
F | 4 | 27 | 31 | 3 | 32 | 35 |
We are creating the Job in a sequence.
Operation – 1 will begin with Operation – 1. E is the first job. Operation – 1 will begin D as soon as it is free of E. Eventually, Operation – 1 will start F as soon as it is free of B.
The Out time equals the in time plus processing time.
The In Time for the Next Job is the same as the Out Time for the Previous Job.
Now we will start filling IN and OUT for Operation –
2.
Before we can start an Operation – 2 job, we will verify the operation -1 and availability.
Job E will be completed with Operation – 1 by 2 hours. Therefore, Job E operation – 1 will begin at 2 nd hours and continue until 8 _th hours (processing time is 6hrs).
Now, for Job D: Operation – 1 ends at 5 th hours. Operation – 2 is now free from Job E at 8 th hours. The Job D will begin at 8 the hour, and last until 16 hours.
Similar to Operation -1, Operation 2 will immediately start if Operation 2 is free. If Operation – 2 cannot be paid, the job will be held until Operation – 2 becomes free.
Operation – 2 can be described as MAX(OUT time of previous task in operation – 2, Out time of
same task in operation – 1). for the start time
Here, the total time taken = 35 hours.
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