Question

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Consider following four processes arriving in the order of P1, P2, P3 and P4. Process Arrival...

Consider following four processes arriving in the order of P1, P2, P3 and P4.

Process Arrival Time CPU burst Time

P1 0.0 6

P2 0.0 8

P3 0.0 7

P4 0.0 3

What is the average waiting time for RR scheduling (time quantum = 5)?

What is the average waiting time for SJF scheduling?

What is the average waiting time for FCFS scheduling?

Solutions

Expert Solution

Question 1) What is the average waiting time for RR scheduling (time quantum = 5)?

P1 completed at 19

P2 completed at 22

P3 completed at 24

P4 completed at 18

Turnaround Time = Complete Time - Arrival Time

19 - 0.0 = 19

22 - 0.0 = 22

24 - 0.0 = 24

18 - 0.0 = 18

Waiting Time = Turnaround Time - CPU burst Time

19 - 6 = 13

22 - 8 = 14

24 - 7 = 17

18 - 3 = 15

Average waiting time for RR scheduling (time quantum = 5) = (13 +14 + 17 + 15) / 4

= 14.75

Average waiting time for RR scheduling = 14.75

Question 2)

What is the average waiting time for SJF scheduling?

P1 completed at 9

P2 completed at 24

P3 completed at 16

P4 completed at 3

Turnaround Time = Complete Time - Arrival Time

9 - 0.0 = 9

24 - 0.0 = 24

16 - 0.0 = 16

3 - 0.0 = 3

Waiting Time = Turnaround Time - CPU burst Time

9 - 6 = 3

24 - 8 = 16

16 - 7 = 9

3 - 3 = 0

Average waiting time for RR scheduling (time quantum = 5) = (13 +14 + 17 + 15) / 4

AAverage waiting time for SJF scheduling = (3 + 16 + 9 + 0) / 4 =

= 28 / 4

Average waiting time for SJF scheduling = 7

Question 3

What is the average waiting time for FCFS scheduling?

P1 completed at 6

P2 completed at 14

P3 completed at 21

P4 completed at 24

Turnaround Time = Complete Time - Arrival Time

6 - 0.0 = 6

14 - 0.0 = 14

21 - 0.0 = 21

24 - 0.0 = 24

Waiting Time = Turnaround Time - CPU burst Time

6 - 6 = 0

14 - 8 = 6

21 - 7 = 14

24 - 3 = 21

Average waiting time for FCFS scheduling = (0 + 6 + 14 + 21) / 4

= 41 / 4

Average waiting time for FCFS scheduling = 10.25

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venereology answered 1 year ago

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