Question

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 Operation 1 (hours) Operation 2 (hours)
A 4 8
B 10 5
C 3 11
D 14 14
E 6 1
F 7 12

A) Using? Johnson's rule for? 2-machine scheduling, the sequence? is:

b) For the schedule developed using? Johnson's rule, the total length of time taken to complete the six jobs? (including the 2nd? operation) =

Answer 1

Answer:

A) Job sequence-   C.A.F.D.B.E

b) Total length of time taken to complete the six jobs (including the 2nd operation) = 54 hrs

Explanation

Johnson’s Rule

1. all jobs and times for each work center are listed first
2. The job with the shortest activity time is chosen. If that time is in the first work center, the job is scheduled in the beginning side of the sequence. If it is in the second work center, the job is scheduled in the bending side of the sequence
3. the   scheduled is eliminated from the list
4. now,   steps 2 and 3 are repeated for remaining jobs working toward the center of the sequence

Johnson’s Rule
Job	Operation 1 (hours)	Operation 2 (hours)	Job Sequence
A	4	8	2
B	10	5	5
C	3	11	1
D	14	14	4
E	6	1	6
F	7	12	3

.
Job with the shortest activity time, E . The shortest activity is in work center 2. So schedule E last in the sequence. _._._._._.E(->)

Remaining activities A,B,C,D,F

Now the Job with the shortest activity time, C   . The shortest activity is in work center 1. So allot C first(beginning side) in the sequence. C._._._._.E

Remaining activities A,B,D,F

Now the Job with the shortest activity time, A   . The shortest activity is in work center 1. So allot A beginning side of the sequence. C.A._._._.E

Remaining activities B,D,F

Now the Job with the shortest activity time, B   . The shortest activity is in work center 2. So allot B in the end side of sequence. C.A._._.B.E

Remaining activities D,F

Now the Job with the shortest activity time, F   . The shortest activity is in work center 1. So allot F in the beginning side of sequence. C.A.F._.B.E

Now the Job, D   is filled in the remaining position. C.A.F.D.B.E

Job sequence-   C.A.F.D.B.E

Time

We process jobs in the sequence given above. In work center 1 the jobs are processed in sequence, c takes 3 hours, the a takes 4 hrs and so on

After a job is processed in work center (operation center 1), it moves immediately to work center 2.

So C leaves work center 1 and enters work center 2 at the beginning of 4 the hour. After 11 hours of operation it exits work center 2. Now A is ready (from work center 1) and enters the work center 2 . A takes 8 hrs at work center 2. The other jobs are done the same way.

Finally, the six jobs are completed when E is finished processing at work center 2

Total length of time taken to complete the six jobs (including the 2nd operation) = 54 hrs