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Problem 9.1. Given the following jobs that are to be processed on a single processor: rocesS Burst Time in ms Arri Time in ms 0 2 6 16 20 12 3 6 10 a. Show the schedule using the scheduling algorithms shortest p rocess next, a non- preemptive algorithm, and shortest remaining time, a preemptive algorithm. Use a GANTT chart to show the schedules in a manner similar to the FCFS schedule shown below: (1 unit 2 ms) 16 b. Compute the turnaround time and relative delay for each algorithm Solution: Solution for p9.1. goes here

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Answer #1

Answer:- a)

2. 2. Is 22 3マ40 4422

b.) Turnaround time:- Turnaround time is the time difference between completion time and arrival time.

Turn Around Time = Completion Time - Arrival Time

TurnAround time for non preemptive algorithm

Turnaround time for process p1= 4-0=4

Turnaround time for process p2=16-2=14

Turnaround time for process p3=30-6=24

Turnaround time for process p4 22-16=6

Turnaround time for process p5= 40-20=20

Total average turn around time = (4+14+24+6+20)/5=68/5=13..23

Waiting time :- Waiting time is the time difference between turn around time and burst time.

Waiting Time = Turn Around Time - Burst Time

Waiting time for non-preemptive algorithm

Wating time for p1= 4-4=0

Wating time for p2=14-12=2

Wating time for p3=24-8=14

Wating time for p4=6-6=0

Wating time for p5=10

Turn around time for preemptive algorithm

Turnaround time for process p1= 4-0=4

Turnaround time for process p2=30-2=28

Turnaround time for process p3=14-6=8

Turnaround time for process p4 22--16=6

Turnaround time for process p5= 40-20=20

Total average turnAround time=13.2

Waiting time for preemptive algorithm

Wating time for p1= 4-4=0

Wating time for p2=16

Wating time for p3=0

Wating time for p4=0

Wating time for p5=10

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