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47 votes
47 votes

Consider the $3$ processes, $P1, P2$ and $P3$ shown in the table. $$\small \begin{array}{|c|c|c|} \hline \textbf{Process} & \textbf{Arrival Time} & \textbf{Time Units Required} \\\hline \text{P1} & 0 & 5\\\hline \text{P2} & 1 & 7 \\\hline \text{P3} & 3 & 4 \\\hline  \end{array}$$The completion order of the $3$ processes under the policies FCFS and RR$2$ (round robin scheduling with CPU quantum of $2$ time units) are

  1. FCFS: $P1, P2, P3$  RR2: $P1, P2, P3 $
  2. FCFS: $P1, P3, P2$  RR2: $P1, P3, P2$
  3. FCFS: $P1, P2, P3$  RR2: $P1, P3, P2$
  4. FCFS: $P1, P3, P2$  RR2: $P1, P2, P3$
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6 Answers

Best answer
54 votes
54 votes

FCFS First Come First Server


RR2

In Round Robin We are using the concept called Ready Queue. 
 

Note
 at $t=2$ ,

  • $P1$ finishes and sent to Ready Queue
  • $P2$ arrives and schedules $P2$

This is the Ready Queue


At $t=3$

  • $P3$ arrives at ready queue

At $t =4$

  • $P1$ is scheduled as it is the first process to arrive at Ready Queue


Option (C) is correct

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18 votes
18 votes
FCFS :- First come first serve.

Here arrival times of all processes are different, So for completion time just order them according of their arrival time. We get P1,P2,P3. So this eliminates option B & D.

Round Robin - Here when you run round robin algorithm on this 3 processes. completion sequence is P1, P3, P2.As Burst Time for P2 is big, P3 completes before P2.

 

So answer is (C)

Referemce :-

https://en.wikipedia.org/wiki/Round-robin_scheduling

https://en.wikibooks.org/wiki/Operating_System_Design/Scheduling_Processes/FCFS
3 votes
3 votes
option C  is correct  here FCFS is completion order               P1   P2   P3

                                                                                       0    5      12     16

 and RR scheduling    P1   P2   P1   P3   P2   P1   P3    P2

                             0     2      4     6      8     10   11     13     16

  ready queue -  P1,P2,P1,P3 ,P2,P1,P3,P2

           completion order P1, P3 ,P2  hence optionC is correct
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