F(x) = x2 – x-1 What is the average rate of change of f over the interval -1 < x < 1?
2 answers:
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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.
