Consider the following set of processes (tuples look like process, burst size):
ID: 3600650 • Letter: C
Question
Consider the following set of processes (tuples look like process, burst size): (P0, 10), (P1, 30), (P2, 15), (P3, 5). Assume the processes arrive in this order as well. Compute the average waiting time and average turnaround time for the following algorithms, and answer the final question:
-First-Come First-Served
-Shortest-Job-First
-Priority Scheduling (Assume priority is calculated as 5*processnumber+bursttime)
-Round Robin Scheduling (use time quantum 10)
Which of these algorithms gives the best turnaround time for this list of processes?
Explanation / Answer
Hi,
Given the processes burst times are
P0-10
P1-30
P2-15
P3-5
Given, they arrive in same order, that is all of them arrive at t=0
lets define the parameters we want to calculate
Waiting time-Amount of time spent ready to run but not running
Turn around time- mean time from submission to completion of process
1. FCFS
In this there is no preemption and processes are executed as an when they come,
average wait time= (10+40+55)/4=26.25
average turn around time= (10+40+55+60)/4=41.25
2.Shortest Job first:
In this, the process having least burst time is executed first, therefore in this case the order of execution will be
P3,P0,P2,P1
average wait time= (5+15+30)/4=12.5
average turn around time= (5+15+30+60)/4=27.5
3.Priority Scheduling (Assume priority is calculated as 5*processnumber+bursttime)
Lets calculate the priority,
P0- 5*0+10=10
P1- 5*1+30=35
P2- 5*2+15=25
P3- 5*3+5=20
So the order of execution will be P1,P2,P3,P0
average wait time= (30+45+50)/4=31.25
average turn around time= (30+45+50+60)/4=46.25
4.Round Robin Scheduling (use time quantum 10)
In this every process is executed for 10 seconds of CPU time,
average wait time= (30+35+30)/4=23.75
average turn around time= (10+60+50+35)/4=38.75
As we can see, turn around time is least for SJF for this processes.
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