Draw the Gantt Charts for FCFS,SJF,Preemptive priority and RR(Quantum=2)

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written 6.9 years ago by

gp4803 • 200

modified 2.9 years ago by

binitamayekar ★ 6.6k

Subject: Operating System

Topic: Process Management And Memory Management

Difficulty: Medium

Consider the following set of processes,with length of CPU bursts given in Millisecond as follows:

Process	Burst Time	Arrival Time	Priority
P1	8	0	3
P2	1	1	1
P3	3	2	2
P4	2	3	3
P5	6	4	4

1.Draw the Gantt Charts for FCFS,SJF,Preemptive priority and RR(Quantum=2)

2.What is the turn around time of each process for above algorithm?

3.What is the waiting time of each process for each of the above algorithm?

4.Which algorithm results in minimum average waiting time.

operating systems

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1 Answer

2.6kviews

written 2.9 years ago by

binitamayekar ★ 6.6k

Turn Around Time = Process Completion Time – Process Arrival Time

Waiting time = Turn Around time – Burst time

First Come First Serve (FCFS) Scheduling

FCFS 2

Process	Arrival Time	Burst Time	Completion Time	Turn Around Time	Waiting Time
P1	0	8	8	8 – 0 = 8	8 – 8 = 0
P2	1	1	9	9 – 1 = 8	8 – 1 = 7
P3	2	3	12	12 – 2 = 10	10 – 3 =7
P4	3	2	14	14 – 3 = 11	11 – 2 = 9
P5	4	6	20	20 – 4 = 16	16 – 6 = 10

Average Turn Around Time = (8 + 8 + 10 + 11 + 16)/5 = 10.6 ms

Average Waiting time = (0 + 7 + 7 + 9 +10)/5 = 6.6 ms

SJF 2

Process	Arrival Time	Burst Time	Completion Time	Turn Around Time	Waiting Time
P1	0	8	20	20 – 0 = 20	20 – 8 = 12
P2	1	1	2	2 – 1 = 1	1 – 1 = 0
P3	2	3	5	5 – 2 = 3	3 – 3 = 0
P4	3	2	7	7 – 3 = 4	4 – 2 = 2
P5	4	6	13	13 – 4 = 9	9 – 6 = 3

Average Turn Around Time = (20 + 1 + 3 + 4 + 9)/5 = 7.4 ms

Average Waiting time = (12 + 0 + 0 + 2 + 3)/5 = 3.6 ms

PPB

Process	Arrival Time	Burst Time	Priority	Completion Time	Turn Around Time	Waiting Time
P1	0	8	3	12	12 – 0 = 12	12 – 8 = 4
P2	1	1	1	2	2 – 1 = 1	1 – 1 = 0
P3	2	3	2	5	5 – 2 = 3	3 – 3 = 0
P4	3	2	3	14	14 – 3 = 11	11 – 2 = 9
P5	4	6	4	20	20 – 4 = 16	16 – 6 = 10

Average Turn Around Time = (12 + 1 + 3 + 11 + 16)/5 = 8.6 ms

Average Waiting time = (4 + 0 + 0 + 9 + 10)/5 = 4.6 ms

Quantum = 2 ms

Process	Arrival Time	Burst Time	Completion Time	Turn Around Time	Waiting Time
P1	0	8	20	20 – 0 = 20	20 – 8 = 12
P2	1	1	3	3 – 1 =2	2 – 1 = 1
P3	2	3	12	12 – 2 = 10	10 – 3 = 7
P4	3	2	7	7 – 3 = 4	4 – 2 = 2
P5	4	6	18	18 – 4 = 14	14 – 6 = 8

Average Turn Around Time = (20 + 2 + 10 + 4 + 14)/5 = 10 ms

Average Waiting time = (12 + 1 + 7 + 2 + 8)/5 = 6 ms

Shortest Job First (SJF) Results in minimum Average Waiting time from all other schedulings

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