GATE IT 2008 | Question: 55

Question

If the time-slice used in the round-robin scheduling policy is more than the maximum time required to execute any process, then the policy will

• A. degenerate to shortest job first
• B. degenerate to priority scheduling
• C. degenerate to first come first serve
• D. none of the above

Answer 1

Answer is (C).

RR follows FCFS with time slice if time slice is larger than the max time required to execute any process  then it is simply converged into  fcfs as every process will finish in first cycle itself

Comments

why not SJF?

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RR follows FCFS with time slice if time slice is large then it is simply fcfs as every process will finish in first cycle itself .SJF has nothing to do here

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because when the round robin pick up those process who are at the head of the ready queue which is not the case with the SJF which may pickup the later one if it has short burst time.

Answer 2

	AT	BT
p0	0	4
p1	2	3
p2	3	1

LET T=5

PO    P1     P2

0      4    7       8

Same result with FCFS

Using SJF

PO    P2     P1

0      4    5      8

The reason being that RR uses a queue.And when the tq is pretty higher than the max burst of  all processes then all processes finish in the same order as they arrive in time less than tq.There  is no process which executes a part gets pre empted and added to queue end awaiting future turn.

Answer 3

C option

In Round Robin Scheduling, scheduler picks up the jobs based on Arrival time and execute it for the Time quantum if execution time is greater than the time quantum. So now when time quantum is greater than the Execution time of the job it is as good as First Come First Serve Policy.

Comments

Why not priority scheduling pls gives me ans

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Soni Bisen Is there any priority in Round Robin Scheduling

Answer 4

You can check it taking an example also. I'll explain why it happens so.

Actually we can say Round Robin is nothing but FCFS with preemption. Actually the TIME QUANTUM used in the algorithm, introduces preemption and hence makes it better than FCFS in terms of response time.

Now, if my time quantum is as large as the largest execution  in the set of proccesses then, even the largest execution time process will not get preempted. In a nutshell, I can say that we removed preemption from the algorithm, hence it is just left with FCFS, and hence it degenerates into FCFS.