Round Robin

Question

In case of Round robin algorithm with time quantum = 2s, and process P1 starting at t=0 and have processes P2, P3 in queue

If a process P4  arrives at  t =2s, but at the same time t=2s, P1 which was executing is preempted  since the quantum expires.

Which process will be put ahead whether P1 ahead of P4 or P4 ahead of P1 ? 

I mean the waiting queue will have the order P2, P3, P4, P1 or P2, P3, P1, P4 because there is race between P4 and P1.

Comments

But for test what will we consider then?

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as per my knowledge, generally we go with which is having less pid, but normally it is implementation defined only

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Ohk..I knew lesser pids are given priority in case of FCFS time clash..

Answer 1

In case of ROUND ROBIN ...things are like ...

If any process is preempted and at the same time if a new process arrives then we should follow this .

"ADD FIRSTLY ALL INCOMING PROCESS IN QUEUE THEN AT LAST ADD THE ONE WHICH WAS PREEMPTED.."

Like P1 , P2  , P3 is there if P1 preempted at 2s and P4 also comes at  sec ...then in QUEUE ADD p4 First then P1 always ..

Comments

But why is it that we need to add all incoming requests firsts? Can't we follow FCFS method as we do in SRTN? ... any standard source where I can refer to for this particular doubt ?

Answer 2

P2, P3, P4, P1