Operating System - Processes
| Process Priority | Arrival Time (in ms) | CPU Time Needed (in ms) | Priority |
| P1 | 0 | 10 | 5 |
| P2 | 0 | 5 | 2 |
| P3 | 2 | 3 | 1 |
| P4 | 5 | 20 | 4 |
| P5 | 10 | 2 | 3 |
If the CPU scheduling policy is SJF with pre-emption, the average waiting time will be
- Option : C
- Explanation : Scheduling order will be
P2 , P3 , P1 , P5 , P1, P4
Waiting time of processes will be
P2 = 0
P3= 5-2=3
P1=10+2=12
P5=0
P4= 15
Average waiting time will be = (0+3+12+0+15)/5= 30/5=6ms
| Process Priority | Arrival Time (in ms) | CPU Time Needed (in ms) | Priority |
| P1 | 0 | 10 | 5 |
| P2 | 0 | 5 | 2 |
| P3 | 2 | 3 | 1 |
| P4 | 5 | 20 | 4 |
| P5 | 10 | 2 | 3 |
If the CPU scheduling policy is priority scheduling without pre-emption, the average waiting time will be
| Process Priority | Arrival Time (in ms) | CPU Time Needed (in ms) | Priority |
| P1 | 0 | 10 | 5 |
| P2 | 0 | 5 | 2 |
| P3 | 2 | 3 | 1 |
| P4 | 5 | 20 | 4 |
| P5 | 10 | 2 | 3 |
If the CPU scheduling policy is priority sche duling with pre-emption, the average waiting time will be
- Option : B
- Explanation : Here the process which will start at the initial millisecond will be P2 as it has more priority that P1.
So, waiting time---- P1 -30 P2 -3 P3 -0 P4 -5 P5 -0ms Process 0 to 2 P2 (P2 completed 2 ms here) 2 to 5 P3 (No wait for P3) 5 to 8 P2 (P2 had to wait 3 ms to get executed ) 8 to 10 P4 (P4 had to wait 3 ms to get started) 10 to 12 P5 (No wait for P5) 12 to 30 P4 (P4 had to wait 2 ms to complete its remaining) 30 to 40 P1 (Was waiting for 30 ms)
Average---- (30+3+0+5+0)/5= 7.6 ms
