Question

Using Johnson's rule for 2-machine scheduling, the sequence is: Scheduled Order Six jobs are to be processed through a two-step operation. The first operation involves sanding, and the second involves painting. Processing times are as follows: Job Job Operation 1 (hours) 10 Operation 2 (hours) H TO For the schedule developed using Johnson's rule, the total length of time taken to complete the six jobs (including the 2nd operation) =1 hours (enter your response as a whole number).
please explain every step

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 2^nd hour. Hence Job E operation – 1 will start at 2^nd hour and go till 8^th hour (as processing time is 6 hours).

Now for Job D: the Operation – 1 is over at 5^th hour, but Operation – 2 has been freed from Job E at 8^th hour. Hence the Job D will start at 8^th 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

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 2^nd hour. Hence Job E operation – 1 will start at 2^nd hour and go till 8^th hour (as processing time is 6 hours).

Now for Job D: the Operation – 1 is over at 5^th hour, but Operation – 2 has been freed from Job E at 8^th hour. Hence the Job D will start at 8^th 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.

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