Question

Consider the following set of processes, with the length of the CPU burst given in milliseconds:

Answer 1

Answer:

a) Drawing for grant charts that illustrates execution of the process is in the image attached

b) To find turn around time we use: completion time - arrival time

•Turn around time for FCFS scheduling algorithm will be:

P1 = 2-0= 2

P2= 3-0 = 3

P3 = 11-0 = 11

P4 = 15-0 = 15

P5 = 20-0 = 20

• Turn around time for SJF scheduling algorithm

P1 = 3-0= 3

P2 = 1-0 = 1

P3= 20-0 = 20

P4= 7-0 = 7

P5 = 12-0 = 12

• Turnaround time for non-preemptive algorithm

P1 = 15-0 = 15

P2 = 20-0 = 20

P3 = 8-0 = 8

P4 = 19-0 = 19

P5 = 13 - 0 = 13

• Turnaround time for RR

P1 = 2-0=2

P2= 3-0= 3

P3= 20 - 0 = 20

P4 = 19-0 = 19

P5 = 18-0 = 18

c) To find waiting time we use (turnaround time - burst time)

•Waiting time for FCFS

P1= 2-2 = 0

P2 = 3-1 = 2

P3 = 11-8 = 3

P4 = 15-4 = 11

P5 = 20-5= 15

• Waiting time for SJF

P1= 3-2 = 1

P2 = 1-1 = 0

P3 = 20-8 = 12

P4 = 7-4 = 3

P5 = 12-5 = 7

• Waiting time for non-preemptive

P1= 15-2 = 13

P2 = 20-1 = 19

P3 = 8-8 = 0

P4 = 19-4 = 15

P5 = 13 - 5 = 8

• Waiting time for RR

P1 = 2-2 = 0

P2 = 3-1 = 2

P3 = 20-8 =12

P4 = 13-4= 9

P5= 18-5=13

d) Average waiting time

•'For FCFS

= (0+2+3+11+15)/5

= 31/5

= 6.2milliseconds

•average waiting time For SJF

(1+0+12+3+7)/5

=23/5

= 4.6milliseconds

• Average waiting time For non-preemptive

(13+9+0+15+8)/5

=55/5

=11milliseconds

• average waiting time For RR

(0+2+12+9+13)/5

=7.2milliseconds

Since the SJF algorithm has 4.6milliseconds, it results in the minimum average waiting time.