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I\'m having a lot of trouble understanding how to do this. Any help would be gre

ID: 3722186 • Letter: I

Question

I'm having a lot of trouble understanding how to do this. Any help would be great! Would prefer something thorough - really trying to understand how the table is built, how to compute waiting and turnaround time, and what the differences between the 2 are.

Thanks!!

Assume the following processes P1,P2, P3, P4 and P5 arrive at 1, 2, 4, 5, 5 respectively The CPU burst and the priority assigned to each process are: 45 5 15 20 3 P1: P2: P3: P4: P5: 25 2 4 For FCFS, RR, SJF and RR scheduling, determine a) the turnaround time for every process, b) waiting time for every process and the average waiting time, c) throughput for the system. Use a time quantum of 10 time units, and negligible context time.

Explanation / Answer

FCFS

0--1-------------------46-------------------51--------------------66-----------------86-------------------111

TurnAround Time

P1 = 46

P2 = 51

P3 = 66

P4 = 86

P5 =111

Waiting Time

P1 = 0

P2 = 51-2 = 49

P3 = 66-4 = 62

P4 = 86-5 = 81

P5 = 111-5 = 106

Average Waiting time = (0+49+62+81+106)/5 = 59.6

SJF

0-1-----------------2-----------------7----------------22----------------42---------------67--------------111

TurnAround Time

P1 = 111

P2 = 7

P3 = 22

P4 = 42

P5 =67

Waiting Time

P1 = 67-1 = 66

P2 = 0

P3 = 7-4 = 3

P4 = 22-5 = 17

P5 = 42-5 = 37

Average Waiting time = (66+0+3+17+37)/5 = 123/5 = 24.6

Priority

0--1-------------4--------------5--------------25------------39--------------81-----------106-----------111

TurnAround Time

P1 = 81

P2 = 111

P3 =39

P4 = 25

P5 =106

Waiting Time

P1 =39-1= 38

P2 = 106-5 = 101

P3 = 25-4 = 21

P4 = 5-5 = 0

P5 = 81-5 = 76

Average Waiting time = (38+101+21+0+76)/5 = 47.2

RR(quantum = 10)

0-1----11------16-----26-----36-----46------56------61-----71------81-----91----96-----106------111

TurnAround Time

P1 = 111

P2 = 16

P3 = 61

P4 = 71

P5 =96

Waiting Time

P1 = 106-1= 105

P2 = 11-2 = 9

P3 = 16-4 = 12

P4 = 26-5 = 21

P5 = 91-5 = 86

Average Waiting time = (105+9+12+21+86)/5 = 46.6

c) Throughtput = 5 process/111 unit time = 0.045 processes per unit time

P1 P2 P3 P4 P5
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