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A. Process Burst Time Arrival Time Completion Waiting

This document contains 3 examples of non-preemptive shortest job first (SJF) scheduling. In each example, it lists the processes, their burst times, arrival times, completion times, and waiting times. It then calculates the average waiting time (AWT) for each example. The first example has an AWT of 2.67, the second is 2.75, and the third is the highest at 4.75.

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Joseph Escovidal
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67 views1 page

A. Process Burst Time Arrival Time Completion Waiting

This document contains 3 examples of non-preemptive shortest job first (SJF) scheduling. In each example, it lists the processes, their burst times, arrival times, completion times, and waiting times. It then calculates the average waiting time (AWT) for each example. The first example has an AWT of 2.67, the second is 2.75, and the third is the highest at 4.75.

Uploaded by

Joseph Escovidal
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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NON-PREEMPTIVE SJF

A.
Process Burst Time Arrival Time Completion Waiting
P1 4 0 4 0
P2 2 1 6 3
P3 3 1 9 5

P1 P2, P3
P1 P2 P3
0 1 2 3 4 5 6 7
8 9

Waiting Time:
P1 = 0 – 0 =0
P2 = 4 – 1 = 3
P3 = 6 – 1 = 5

AWT = 0 + 3 + 5 = 8/3 = 2.67

B.
Process Burst Time Arrival Time Completion Waiting
P1 4 0 7 3
P2 1 0 1 0
P3 6 0 13 7
P4 2 0 3 1

P1, P2
P3, P4
P2 P4 P1 P3
0 1 2 3 4 5 6 7 8 9 10 11 12 13

Waiting Time:
P1 = 3 – 0 = 3
P2 = 0 – 0 = 0
P3 = 7 – 0 = 7
P4 = 1 – 0 = 1

AWT = 3 + 0 + 7 +1 = 11/4 = 2.75

C.
Process Burst Time Arrival Time Completion Waiting
P1 3 3 9
P2 5 2 18
P3 4 4 13
P4 6 1 6

P4 P2 P1 P3
P4 P1 P3 P2
0 1 2 3 4 6 9 13 18

WAITING TIME:
P1 = 6 – 3 = 3
P2 = 13 – 2 = 11
P3 = 9 – 4 = 5
P4 = 1 – 1 = 0

AWT = 3 + 11 + 5 + 0 = 19/4 = 4.75

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