Classical

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

 
 

Option: A

Explanation :


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

 
 

Option: D

Explanation :


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

 
 

Option: A

Explanation :


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

 
 

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.


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

 
 

Option: C

Explanation :

8 + 0 + 3 + 15 + 8 divided by 5, i.e. 6.8 ms.




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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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