Question

Question 3. Scheduling

ArtWork is a graphic design company. All jobs it receives must go through sketching (Process X) and then colouring (Process Y) in sequence. Each of the processes is operated by a dedicated designer.

Job	Process X (Hour)	Process Y (Hour)
A	11	6
B	8	3
C	7	10
D	4	3
E	1	2
F	5	9
G	2	4
H	12	4
I	6	1
J	4	9

1. The above 10 jobs have just arrived. Use Johnson’s rule to determine the optimal sequence to schedule the jobs. If there is more than one optimal sequence, show them all.

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2. Draw a Gantt chart for your schedule. If there is more than one optimal sequence, show anyone of them.

3. Calculate the average flow time.

4. If you are allowed to split the last job in the sequence (pay attention, not necessarily the
   1. last job in the table) into two equal halves. For example, a job spends 6 hours in X and 4 hours in Y can be split into two jobs of 3 hours in X and 2 hours in Y. Without changing the sequence, do you think this arrangement is quicker? Why or why not?
5. The company has received four special jobs. These jobs have to go through sketching (Process X), colouring (Process Y) and then printing and coating (Process Z) in sequence. Each of the processes is operated by a dedicated designer. What is the optimal sequence to schedule the jobs? Show your steps. [hint: self-study is required]

Job	Process X (hour)	Process Y (hour)	Process Z (hour)
P	6	1	5
Q	2	2	6
R	5	1	4
S	3	3	8

Answer 1

NOTE: Due to paucity of time, we can solve only 1st four questions i.e. only for section a of the above question.

i) Using Johnsons method, of assigning jobs with lowest time in process X first and jobs with lowest processing time in process Y last in the schedule, we get the following sequences:

Sequence 1	E	G	J	F	C	A	H	B	D	I
Sequence 2	E	G	D	J	F	C	A	H	B	I
Sequence 3	E	G	D	J	F	C	A	B	H	I
Sequence 4	E	G	D	J	F	C	A	H	B	I
Sequence 5	E	G	J	D	F	C	A	B	H	I

ii) The Gant chart will be shown for the first sequence among the above 5 combinations i.e. for sequence: E-G-J-F-C-A-H-B-D-I

iii) The total process time for each of the two processes are calculated as shown in the below table:

	Process X (Hour)	Process Y (Hour)
Sequence	Tin	Tpro	Tout	Tin	Tpro	Tout
E	0	1	1	1	2	3
G	1	2	3	3	4	7
J	3	4	7	7	9	16
F	7	5	12	16	9	25
C	12	7	19	25	10	35
A	19	11	30	35	6	41
H	30	12	42	42	4	46
B	42	8	50	50	3	53
D	50	4	54	54	3	57
I	54	6	60	60	1	61
	Total	60		Total	51
	Idle	1		Idle	10

Thus the average flow time for both the processes = (flow time X + flow time Y )/2 = (60 + 51)/2 = 55.5 Hours

Thus the average flow time of the entire process for the given sequence is 55.5 Hours.

iv) Now if we split the last job of the sequence i.e. Job I into 2 activities I1 and I2 with process times of 3 Hours and 0.5 Hours in processes X & Y each respectively. Job table is as below:

Job	Process X (Hour)	Process Y (Hour)
A	11	6
B	8	3
C	7	10
D	4	3
E	1	2
F	5	9
G	2	4
H	12	4
I1	3	0.5
I2	3	0.5
J	4	9

We also follow the same sequence E-G-J-F-C-A-H-B-D-I determined by Johnson's rule, we get the following calculations of process flow time as depicted in the below table:

	Process X (Hour)	Process Y (Hour)
Sequence	Tin	Tpro	Tout	Tin	Tpro	Tout
E	0	1	1	1	2	3
G	1	2	3	3	4	7
J	3	4	7	7	9	16
F	7	5	12	16	9	25
C	12	7	19	25	10	35
A	19	11	30	35	6	41
H	30	12	42	42	4	46
B	42	8	50	50	3	53
D	50	4	54	54	3	57
I1	54	3	57	57	0.5	57.5
I2	57	3	60	60	0.5	60.5
	Total	59		Total	49
	Idle	1.5		Idle	11.5

Thus we see that process flow times of Processes X and Y have reduced to 59 Hours and 49 Hours respectively (but with an increase in idle time too)

Thus the average flow time for both the processes = (flow time X + flow time Y )/2 = (59 + 49)/2 = 54 Hours

Thus the average flow time of the entire process for the given sequence is 54 Hours.

Thus there is a decrease in average flow time by 55.5-54 = 1.5 Hours. Thus we can safely conclude that splitting the last job into 2 sub jobs have made the arrangement more quicker by 1.5 Hours.