# 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 FCFS, the average waiting time will be

• Option : A
• Explanation : According to FCFS process solve are p1 p2 p3 p4 p5 so
for p1 waiting time =0 process time=10 then
for p2 waiting time = (process time of p1-arrival time of p2)=10-0=10 then
for p3 waiting time = (pr. time of (p1+p2)-arrival time of p3)=(10+5)-2=13 and
same for p4 waiting time=18-5=13
same for p5 waiting time=38-10=28
So total average waiting time=( 0+10+13+13+28)/5=12.8
0 + 10 + (15- 2) + (18 5) + (38- 10) divided by 5, i.e. 12.8 ms.
Note : Here we will not see priority, we only see who comes first. And here both p1 andp2 came simultaneously and so we take p1 first and it gives answer which matches in option.

 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, the average waiting time (without pre-emption) will be

• Option : C
• Explanation : 8 + 0 + 3 + 15 + 8 divided by 5, i.e. 6.8 ms.

## Description

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