8/13/2019 CSC520 Chapter 4 Exercise Solutions

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Exercise

Suppose a system uses priority schedulingwhere a small integer means higher priority

Create a Gantt chart to illustrate the execution of

these processes What is the turnaround time for process P2 under

priority scheduling? What is the average waiting time for the

processes?

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Turnaround time for P2 160 Waiting time

P0 70 P1 0 P2 150 P3 160 P4 20

Average waiting time 400/5 = 80

P1 P4 P2P0 P30 20 70 150 160 180

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8/13/2019 CSC520 Chapter 4 Exercise Solutions

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Exercise

Suppose a system uses SJN scheduling wherea small integer means higher priority

Create a Gantt chart to illustrate the execution of

these processes What is the turnaround time for process P3 under

SJN scheduling? What is the average waiting time for the

processes?

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P3 Turnaround time 95 55 = 40

Waiting Time P0 0 P1 150 P2 85 P3 20 P4 65

Average waiting time 320/5 = 64

P0 P30 75 95

P2120

P1160

P4205

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Exercise

Based on the information given below, draw aGantt chart illustrating the execution of eachprocess using the following algorithms

FCFS SJF Priority Scheduling

Calculate the waiting time for Process B and Dfor each of the scheduling algorithm

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FCFS/SJF PB 9 FCFS/SJF PD 12

Priority PB 14 Priority PD 7

A

0 10

B

12

C

15

D

20

A

0 10

B

12

C

15

D

20

A

0 10

D

15

B

17

C

20

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Exercise

Based on the information given below, draw aGantt chart illustrating the execution of eachprocess using the priority scheduling

algorithm Calculate the average turnaround time for all

processes

Calculate the average waiting time for allprocesses

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P0 20 P1 45 P2 45 P3 50 P4 60 Average turnaround time 320/5 = 64 P0 0

P1 10 P2 25 P3 35 P4 35 Average waiting time 105/5 = 21

P0

0 20

P1

55

P2

75

P3

90

P4

115

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Exercise

Suppose a system uses RR (q =15) schedulingwhere a small integer means higher priority

Create a Gantt chart to illustrate the execution of these processes

What is the turnaround time for process P2 underRR scheduling?

What is the average waiting time for theprocesses?

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P2 turnaround time = 25 Waiting time

P0 25 + 35 + 20 + 15 + 5 = 100 P1 25 P2 15 P3 30 P4 10 + 20 + 15 + 15 = 60

Average waiting time 230/5 = 46

P0

0 15

P1

30

P2

40

P0

55

P1

60

P3

75

P4

90

P0

105

P3

110

P4

125

P0

140

P4

140 155

P0 P4

170 175 180

P0