RGPV 2020 CPU Scheduling Algorithm
Consider the following set of processes.
|
Process |
Burst Time |
Arrival Time |
|
P1 |
3 |
0 |
|
P2 |
5 |
1 |
|
P3 |
2 |
2 |
|
P4 |
5 |
3 |
|
P5 |
5 |
4 |
Develop a Gantt-chart and calculate the average waiting time using:
i) FCFS
ii) SJF
Solution:
i) FCFS
 |
| Gantt Chart for FCFS |
From above Gantt Chart waiting time for each process:
Waiting time = Turnaround time – Burst time
|
Process |
Waiting time |
|
P1 |
3-3=0 |
|
P2 |
7-5=2 |
|
P3 |
8-2=6 |
|
P4 |
12-5=7 |
|
P5 |
16-5=11 |
Average waiting time = Sum of waiting time / Number of processes
Average waiting time = (0+2+6+7+11)/5 = 5.2
ii) SJF
 |
| Gantt Chart SJF |
From above Gantt Chart waiting time for each process:
Waiting time = Turnaround time – Burst time
|
Process |
Waiting time |
|
P1 |
3-3=0 |
|
P2 |
9-5=4 |
|
P3 |
3-2=1 |
|
P4 |
12-5=7 |
|
P5 |
16-5=11 |
Average waiting time = Sum of waiting time / Number of processes
Average waiting time = (0+4+1+7+11)/5 = 4.6
iii) Round Robin (q = 1)
 |
| Gantt Chart RR |
From above Gantt Chart waiting time for each process:
Waiting time = Turnaround time – Burst time
|
Process |
Waiting time |
|
P1 |
11-3=8 |
|
P2 |
17-5=12 |
|
P3 |
6-2=4 |
|
P4 |
16-5=11 |
|
P5 |
16-5=11 |
Average waiting time = Sum of waiting time / Number of processes
Average waiting time = (8+12+4+11+11)/5 = 9.2