# #OS #Galvin 9th edition RR Cpu scheduling alogrithm.

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### For example, with five processes and a time quantum of 20 milliseconds, each process will get up to 20 milliseconds every 100 milliseconds.

Can someone please explain this in a simple way, I'm not getting how each process will get 20 milliseconds every 100 ms?

Brother, they are talking about round robbin algorithm, where time quantum ='q' sec.

if you have n processes in ready queue with each have large burst time, the ready queue look like as : P1,P2,...Pn

now, if you run P1....how much time it will run? it should be 'q' sec, but it is not completed

now analyze, next time P1 should run when after running of P2,P3,...Pn ====> Each process must wait no longer than (n − 1) × q time units its next time quantum.

what is the meaning of each process gets 1/n of the CPU time in chunks of at most q time units. ?

therefore in the time of n * q sec, each process can get only $\frac{1}{n}$ * q sec only.

for you example, 5 processes... P1,P2,P3,P4,P5 with tq=20 msec and assume every process requires more than 20 msec

P1
0->20 ( after 20 sec P1 looses its turn )

P1 P2
0->20 20->40

P1 P2 P3
0->20 20->40 40->60

P1 P2 P3 P4
0->20 20->40 40->60 60->80

P1 P2 P3 P4 P5
0->20 20->40 40->60 60->80 80->100

Ready Queue:- P1,P2,P3,P4,P5 ==> ready queue is now look like as initial ==> P1 got its turn again after 80 msec => (n-1)*q

P1 P2 P3 P4 P5
0->20 20->40 40->60 60->80 80->100

total of 100 msec, each process get only 20 msec ==> therefore in the time of n * q sec, each process can get only $\frac{1}{n}$ * q sec only.

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@ Shaik Masthan  Thank you so much brother for explaining it very simple and for all the effort and time you put in, Thanks!

Look every process will get q quantum of time for excn on its turn. So as in exmple u took there time qauntum is 20ms.

So every process gets 20 ms to execute on its turn all so after first process complete excn 2nd process will excte similarly for 20ms...like this  5th process will complete its  excn at 100 ms...now after 100ms your 1st process gets chance to execute again so in this way a process needs to wait (n-1)*q to get its turn back to execute ...here n=5 ,q=20 so one prcoess would wait for 80 ms max not more than this.(if process burst time is more than time quantum)

And it can be said in this way too each process will get 20 ms in 100 ms ...as after 100ms process 1 will again get chance to execute.
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@ Shubham Shukla 6  Got it brother! Thanks for explaining ir nicely.

## Related questions

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Let's say we have four processes as (P1, P2, P3, P4) and have Arrival time as (0,1,2,3) respectively, but all of them has same CPU Burst time as (10ms) then does SRTF will simply be behaving like FCFS?
Consider a system having $10$ IO bound jobs and $1$ CPU bound job.. If IO bound job issue an IO request once for every ms of CPU computation and that each IO request takes $10$ ms. If context switch overhead is $0.1$ ms.Using round -robin scheduling with a time quantum of $10$ ms. , the CPU efficiency is __________________