Question

The following set of seven jobs is to be processed through two work centers at George? Heinrich's printing company. The sequence is first? printing, then binding. Processing time at each of the work centers is shown in the following? table:

Job Printing (hours) Binding(hours)
T 12 4
U 8 11
V 5 14
W 6 2
X 9 10
Y 3 1
Z 7 13

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

Scheduled Order

Job

1.____

2. ____

3. ____

4.____

5.____

6.____

7.____

?c) For the schedule developed using? Johnson's rule, the total length of time taken to complete the seven printing and binding jobs? (including binding)? = ______ hours ?(enter your response as a whole? number).

?d) The idle time in the binding shop based on the sequence developed using ?Johnson's rule = ______ hours (enter your response as a whole? number).

?

e) By splitting Job V in? half, the binding? machine's idle time would be cut by ______ hour(s)?
??(enter your response rounded to one decimal? place).

Answer 1

As per Johnson’s Rule of sequencing; if there are two step operations for any job then select the one with the shortest processing time from both operations. If the shortest time is from the first operation then this job will schedule first and if it is from second operation then it will schedule last. This process we have to repeat again and again till any job left at any operation center.

Therefore apply this rule on the above operations

Job	Printing (Hours) (operation 1)	Binding (Hours) (operation 2)
T	12	4
U	8	11
V	5	14
W	6	2
X	9	10
Y	3	1
Z	7	13

Shortest processing time

Shortest processing time (time)	Job	Operation (1 Or 2)	Sequence
1	Y	Operation 2	Last
2	W	Operation 2	Second from last
4	T	Operation 2	Third from Last
5	V	Operation 1	First
7	Z	Operation 1	Second
8	U	Operation 1	Third
9	X	Operation 1	Fourth

Therefore the sequence is V-Z-U-X-T-W-Y

a-b.

Scheduled order	Job
1	V
2	Z
3	U
4	X
5	T
6	W
7	Y

c. Total length calculation:

Scheduled order	Job	Cumulative time Required at Printing (hours)	Cumulative time Required at Binding (hours)
1	V	5	5+14=19
2	Z	5+7=12	19+13=32
3	U	12+8=20	32+11=43
4	X	20+9=29	43+10=53
5	T	29+12=41	53+4=57
6	W	41+6=47	57+2=59
7	Y	47+3=50	59+1=60

Therefore total length of time taken to complete the seven printing and binding jobs is 60 hours. (The longest time at binding)

d. The idle time at binding shop = total length of time taken to complete the seven printing and binding – total time required to do only binding job

= 60 hours – (4+11+14+2+10+1+13) hours = 60 hours – 55 hours = 5 hours

Therefore the idle time at binding shop is 5 hours.

e. The idle time for binding machine is 5 hours which is basically the processing time of printing job at first job V; by splitting Job V in half, it will take only 2.5 hours

Therefore the idle time at binding shop will be 2.5 hours.