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round robin scheduling policy

operating system:

The pure round robin scheduling policy

a) responds poorly to short processes if the time slice is small

b)does not use any apriori information aabout the services times of processes

c)ensures  that the ready queue is always the same size

d)becomes equivalent to the SJF when time slice is made infinitely large

given ans is (b)

my doubt is, if we use round robin with SJF then we have to consider the service time of process then how (b) is possible? and if the time slice is small then it will respond poorly to all processes whether the process is small or big. then why (a) is not correct?

please clarify.

4Comments
Ranita Biswas @ranita
8 Dec 2014 12:34 pm

The correct answer is (b).

With the Round-Robin-scheduling policy, processes are dispatched in FIFO order but are given a limited amount of CPU time, called the time slice or quantum. If a process does not complete before the time slice expires, it is preempted, and the CPU is given to the next waiting process in the ready queue. The preempted process is placed at the back of the ready queue. Hence no a priori knowledge is used to determine the next process to schedule.

You are supposed to use general Round-Robin-scheduling in this case, not one with SJF.

jayendra @jayendra
8 Dec 2014 01:38 pm

ok, and why (a) is not correct? If time slice is small then it will respond poorly to all processes whether it is small or big.

Ranita Biswas @ranita
8 Dec 2014 03:15 pm

You are right that "If time slice is small then it will respond poorly to all processes whether it is small or big." As option (a) is specific to only short processes, you can not choose it as correct answer.

jayendra @jayendra
8 Dec 2014 01:35 pm

thank you