consider the following snapshot of the process to be executed .Draw the Gantt chart and determine the average waiting time and average turn around time for FCFS, SJF(pre-emptive),SJF(non pre-emptive)

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written 6.9 years ago by

devikaraniroy • 640

modified 3.1 years ago by

apsareena • 670

PROCESS	ARRIVAL TIME	BURST TIME
P1	0	7
P2	1	4
P3	3	3
P4	5	1
P5	7	5

Subject: Operating System

Topic: PROCESS MANAGEMENT

Difficulty: Hard

operating systems

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1 Answer

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written 3.1 years ago by

apsareena • 670

First Come First Serve

Gantt chart:

   A     B      C      D    E
0     7     11     14    15    20

Job	Arrival Time	Burst Time	Finish Time	Turnaround time	Waiting time
A	0	7	7	7	0
B	1	4	11	10	6
C	3	3	14	11	8
D	5	1	15	10	9
E	7	5	20	13	8
			Average	51 / 5 = 10.2	31 / 5 = 6.2

Shortest Job First( Non-preemptive)

Gantt chart:

   A    D    C    B    E
0    7    8    11    15    20

Job	Arrival Time	Burst Time	Finish Time	Turnaround Time	Waiting Time
A	0	7	7	7	0
B	1	4	15	14	10
C	3	3	11	8	5
D	5	1	8	3	2
E	7	5	20	13	8
			Average	45 / 5 = 9	25 / 5 = 5

Shortest job first (preemptive):

Gantt chart:

  A    B    D    C    E    A
0    1    5    6    9    14    20

Job	Arrival Time	Burst Time	Finish Time	Turnaround Time	Waiting Time
A	0	7	20	20	13
B	1	4	5	4	0
C	3	3	9	6	3
D	5	1	6	1	0
E	7	5	14	7	2
Average				38 / 5 = 7.6	18 / 5 = 3.6

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