# Operating System - Process Scheduling

1:

Dijkstra's banking algorithm in an operating system, solves the problem of

 A. deadlock avoidance B. deadlock recovery C. mutual exclusion D. context switching Answer Report Discuss Option: A Explanation : Click on Discuss to view users comments. Write your comments here:
2:

In Round Robin CPU scheduling, as the time quantum is increased, the average turn around time

 A. increases B. decreases C. remains constant D. varies irregularly Answer Report Discuss Option: D Explanation : Click on Discuss to view users comments. Write your comments here:
3:

Pre-emptive scheduling is the strategy of temporarily suspending a running process

 A. before the CPU time slice expires B. to allow starving processes to run C. when it requests I/O D. none of these Answer Report Discuss Option: A Explanation : Click on Discuss to view users comments. Write your comments here:
4:

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 FCFS, the average waiting time will be

 A. 12.8 ms B. 8 ms C. 6 ms D. none of these Answer Report Discuss 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  So answer is 'A'. 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. Click on Discuss to view users comments. Write your comments here:
5:

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

 A. 16 ms B. 12.8 C. 6.8 ms D. none of these Answer Report Discuss Option: C Explanation : 8 + 0 + 3 + 15 + 8 divided by 5, i.e. 6.8 ms. Click on Discuss to view users comments. manoj kumar said: (3:32am on Wednesday 23rd August 2017) please explain the above question so that we clearly understand this . Write your comments here:

## Suggest an improvement

Syllabus covered in this section is-

• Main functions of operating systems
• Multiprogramming, multiprocessing and multitasking
• Memory Management- Virtual memory, paging, fragmentation
• Concurrent Processing: Mutual exclusion, Critical regions, locks and unlock.
• Scheduling: CPU scheduling. I/O scheduling, Resource scheduling
• Deadlock and scheduling algorithms.
• Banker's algorithm for deadlock handling.

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