Question

Given the operation times provided:

	JOB TIMES (minutes)
	A	B	C	D	E	F
Center 1	20	16	43	60	35	42
Center 2	27	30	51	12	28	24

a. Develop a job sequence that minimizes idle time at the two work centers.

The sequence is             (Click to select)  A-B-C-D-E-F  B-A-C-E-F-D  C-A-B-D-E-F  F-E-D-A-B-C  E-F-D-B-C-A  D-C-B-A-E-F  .

b. Determine idle time of center 2, assuming no other activities are involved.

Idle time             minutes

Answer 1

We will apply Johnson’s rule to solve the problem

Johnson’s Rule is:

1. From the list of unscheduled jobs, select the one with the shortest processing time in either work center.
2. If the shortest time is at the first work center, do the job first in the schedule otherwise do the job last in the schedule.
3. Remove the job assigned in Step 2 from the list of unscheduled jobs.
4. Repeat steps 1,2 and 3 filling in the schedule from the front and the back until all jobs have been scheduled.

The original table :

Job	Centre 1	Centre 2
A	20	27
B	16	30
C	43	51
D	60	12
E	35	28
F	42	24

Accordingly , order of jobs using Johnson’s rule :

B --- A --- C ----   E --- F ---- D

Time taken to finish all the jobs :

Centre 1: B ( 16) A ( 36) C (79) E (114) F (156) D ( 216)

Centre 2 :             B (46) A ( 73) C ( 130) E (158) F ( 182) D (228)

It will take 228 minutes to complete all the tasks.

Data in parenthesis () are the corresponding completion time of the activity.

Total job time of all activities at centre 2 = 172 minutes

But, completion time at centre 2 = 228 minutes

Therefore, idle time at centre 2 = 228 – 172 minutes = 56 minutes