System Software and Operating System - Process Management

>>>>>>>>Process Management

 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

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

• Option : C
• Explanation : 30 + 0 + 3 + 3 + 18 divided by 5, i.e. 10.8 ms.

 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

• Option : B
• Explanation : 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

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• Process Management Questions can be used to give quizzes by any candidate who is preparing for UGC NET Computer Science
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• Process Management Questions for UGC NET Computer Science

Process Management Multiple choice questions

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