Question

[Note: The order of jobs in a schedule is 1^st,2^nd,3^rd,4^th,5^th, for A,B,C,D,E, respectively.]

The jobs A,B,C,D,E, arrived in that order to be processed on two machines.

Job	A	B	C	D	E
Time on Machine 1	40	48	58	26	70
Time on Machine 2	37	60	56	48	72

Question 6. Which is the Johnson’s Rule schedule?

     (A) ‘DBECA’     (B) ‘ACEBD’     (C) ‘DCABE’     (D) ‘DABEC’   (E) none of the above

Question 7. Consider the FIFO schedule of (A,B,C,D,E).

How many statements are correct?   (A) 0   (B) 1   (C) 2   (D) 3   (E) 4

Statement 1. The total time of completion is 324

Statement 2. The flow time of job D is 252

Statement 3. The total queue time of job C is 90

Statement 4. The total idle time of machine 2 is 51

. . .

Answer 1

Queation 6)

1. The smallest processing time is 26 hour for job 4 on Machine-1. So job 4 will be processed first.
2. The next smallest processing time is 37 hour for job 1 on Machine-2. So job 1 will be processed last.
3. The next smallest processing time is 48 hour for job 2 on Machine-1. So job 2 will be processed after job 4.
4. The next smallest processing time is 56 hour for job 3 on Machine-2. So job 3 will be processed before job 1.
5. The next smallest processing time is 70 hour for job 5 on Machine-1. So job 5 will be processed after job 2.

According to Johanson's algorithm, the optimal sequence is as below

4

2

5

3

1

Job	M1 In time	M1 Out time	M2 In time	M2 Out time	Idle time M2
4	0	0 + 26 = 26	26	26 + 48 = 74	26
2	26	26 + 48 = 74	74	74 + 60 = 134	-
5	74	74 + 70 = 144	144	144 + 72 = 216	10
3	144	144 + 58 = 202	216	216 + 56 = 272	-
1	202	202 + 40 = 242	272	272 + 37 = 309	-

The total minimum elapsed time = 309

Idle time for Machine-1
= 309 - 242
= 67

Idle time for Machine-2
= 26 + 10 + (309 - 309)
= 36

Question 7)

Description	A	B	C	D	E
End time of Machine 1	40	88	146	172	242
End time of Machine 2	77	148	204	252	324
Waiting time of Machine 1	0	0	0	0	0
Waiting time of Machine 2	0	11	0	0	0

Number of Statement correct is 2

Answer (C)