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 following table.

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

1. Determine the sequence that will minimize the total completion time for the jobs using Johnson’s rule.
2. Illustrate graphically the throughput time and idle time at the two operations by constructing a time-phased chart.

Answer 1

Answer-a:

Johnson's rule of scheduling jobs in two work centers:

The smallest time is located in Job E (2 hours). Since the time is in Operation 1, schedule this job first.

Eliminate Job E from further consideration.

E

The next smallest time is located in Job F (3 hours) and Job D (3 hours). Schedule Job F last and Job D first.

Eliminate Job D & F from further consideration.

E

D

F

The next smallest time is located in Job B (4 hours). Since the time is in Operation 2, schedule this job last.

Eliminate Job B from further consideration.

E

D

B

F

The next smallest time is located in Job A (5 hours) and Job C (5 hours). Schedule Job A last and Job C first.

E

D

C

A

B

F

So, the jobs must be processed in the order E → D → C → A → B → F, and must be processed in the same order on both operation centers to minimize the total completion time

Answer-b:

Total idle time = 34 -31 = 3