Process Burst Priority P1 8 4 P2 6 1 P3 1 2 P4 9 2 P5 3 3 Us
Process
Burst
Priority
P1
8
4
P2
6
1
P3
1
2
P4
9
2
P5
3
3
Using the above data:
Find the Avg Wait time , Avg Turnaround, Avg Response Time using:
FCFS
SJF
Non preemptive priority
Preemptive Priority
Round Robin(Quantum of 1)
| Process | Burst | Priority |
| P1 | 8 | 4 |
| P2 | 6 | 1 |
| P3 | 1 | 2 |
| P4 | 9 | 2 |
| P5 | 3 | 3 |
Solution
FCFS
P1--- > P2--- >P3--- >P4--- >P5
0--->8-->14-->15--->24-->27
Average Waiting Time
=(0+8+14+15+24)/5
=61/5
=12.2 milliseconds
Turnaround Time = Burst Time + Waiting Time
Process Turnaround Time
P1 8+0=8
P2 6+8=14
P3 1+14=15
P4 9+15=24
P5 3+24=27
Average Turnaround Time
=(8+14+15+24+27)/5
=88/5
17.6 milliseconds

