Operating System MCQ - Processes

6 Consider a set of 5 processes whose arrival time. CPU time needed and the priority are given below

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

smaller the number, higher the priority.
If the CPU scheduling policy is SJF with pre-emption, the average waiting time will be

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

7 Consider a set of 5 processes whose arrival time. CPU time needed and the priority are given below

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

smaller the number, higher the priority.
If the CPU scheduling policy is priority scheduling without pre-emption, the average waiting time will be

30 + 0 + 3 + 3 + 18 divided by 5, i.e. 10.8 ms.

8 Consider a set of 5 processes whose arrival time. CPU time needed and the priority are given below

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

smaller the number, higher the priority.
If the CPU scheduling policy is priority sche duling with pre-emption, the average waiting time will be

Here the process which will start at the initial millisecond will be P2 as it has more priority that P1.

ms	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)

So, waiting time---- P1 -30 P2 -3 P3 -0 P4 -5 P5 -0
Average---- (30+3+0+5+0)/5= 7.6 ms

9 Cascading termination refers to termination of all child processes before the parent terminates