#Solved | Operating System | Round Robin

Question

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Option D) has been given as a correct option in the MSQ question but why the E's (arrival time = 4ms) turn was not placed after D?

Comments

At the end of 3, A again added to queue where E has not in the queue till that time. E added at t=4. So, A get the second chance before E get the first chance.

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Sir @Shaik Masthan, did not understand the meaning of,

At the end of 3, A again added to queue

Time quantum is of 3ms, till D, (3ms * 4) 12ms would be used and E arrived at 4ms so E would be already waiting in the ready queue. 

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At t=0, A is added to Queue.

Only one process is in queue. That is scheduled.

At t=1, B added to ready queue. As A time quantum not completed, B will still wait in the ready queue.

At t=2, C will be added to the ready queue.

At t=3, A time quantum expire and D also arrived.

So both A and D added to the queue in no particular order. B will scheduled for running.

At t=4, E will be added to the ready queue.

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Always manage a ready queue for Round Robin Questions.

Answer 1

AT- arrival time

BT - burst time

CT- completion time

TAT- turn around time  , TAT=CT-AT

WT-waiting time , WT = TAT-BT

RT- response time , RT=  first time when process got cpu - arrival time

key point to remember: try to solve round robin question by taking 1 interval time gap that will be benificial for you but how??

simple see always first put whichever process comes into queue first then  take out a process from front of queue then,

make a process run till time quantum when the process is running if some process has arrived u should put that process into the queue based on arrival time.

when a process completes its execution if it has still some burst time left then again put this process  in the rear/ back of the queue

then take out a process from front of the queue and make it run for the given time quantum.

Answer 2

At t = 3, process A's time quantum expires, causing it to be preempted and enqueued since it has not yet completed execution. At this moment, the ready queue contains: B, C, D, A.

Next, B is scheduled (now ready queue contains: C, D, A.) and runs until t = 6. Meanwhile, at t = 4, process E arrives and is added to the queue. As a result, the updated queue at t = 6 becomes C, D, A, E.

Consequently, after D finishes its time quantum, A gets scheduled before E.

Comments

I think A and C are also correct.

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Doubt was only regarding option D.