3.1k views

Consider the $3$ processes, $P1, P2$ and $P3$ shown in the table.

Process Arrival time Time Units Required
$P1$ $0$ $5$
$P2$ $1$ $7$
$P3$ $3$ $4$

The completion order of the $3$ processes under the policies FCFS and RR2 (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: $P1, P2, P3$
edited | 3.1k views

FCFS First Come First Server

 $P1$ $P2$ $P3$

$\mathbf{0}$           $\mathbf{5}$            $\mathbf{12}$         $\mathbf{16}$

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 $P2$

 -- -- -- $P1$

At $t=3$

• $P3$ arrives at ready queue
 -- -- $P3$ $P1$

At $t =4$

• $P1$ is scheduled as it is the first process to arrive at Ready Queue
 $p1$ $p2$ $p1$ $p3$ $p2$ $p1$ $p3$ $p2$

Option (C) is correct

edited by
+3
order of computation is actually this
+2
Is it the way we always follow with round robin ?
+1
@pc

At t = 1, P2 arrives, before P1 finishes its execution.
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.

Referemce :-

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

https://en.wikibooks.org/wiki/Operating_System_Design/Scheduling_Processes/FCFS
+3
I'm not understanding how P3 finishes after P1.

P1    P2    P3    P1    P2    P3    P1    P2

0       2      4      6      8      10    12     14      16

is the sequence I'm getting.
0
i am getting the sequance p3 p1 p2 for RR
+4
round robin takes the first process in ready queue,after time quantum 2 p1 is sent to ready queue and is joined by p3 at time quantum 3. When p2 is preempted in time quantum 4,p1 is executed as it is in ready queue before p3
+5
Someone explain how $RR2: P1,P3,P2$ ?

iam getting $RR2: P3,P1,P2$
–1

@mcjoshi It is P1 P3 P2 itself..!

+1

See the order of completion in round robin

+5

Order of computation is

 p1 p2 p1 p3 p2 p1 p3 p2

See my answer below for more explanation

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

completion order P1, P3 ,P2  hence optionC is correct
edited

edited

You can  appy this method to solve every question related to round robin process

–1 vote
Ans is c..

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