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If we have only one process in ready queue with burst time "m", then how many context switching will happen using round robing scheduling with time quantum q ,where q<m.Assume that dispatching the process first time is not counted as  a context switch.
in Operating System by Boss (25.6k points) | 940 views
+1
m/q if we are considering last context switch otherwise m/q-1.
+1
For one process does context switch happen? Saving and loading same process :o
0
Short term scheduler schedules process from ready queue to CPU and dispatcher changes the context which is on the disk so I don`t think they will ever come to know that it's the same process under round robin scheduling ! what do you think?
0
Theocratically i think you are right. I thought the same way.But i am not able to convince myself:(
+1
what will happen if there are n processess is it $(m/q)^{n}$????.correct me if i am wrong.
0
but don't you think that once the process starts executing then the ready queue will be empty as there is only one process in the system?

2 Answers

+4 votes
Best answer
It clearly says " .Assume that dispatching the process first time is not counted as  a context switch."

so number of context switch = 0

as there is only one single process.
by Veteran (74.4k points)
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+3 votes
It will not context switch since only one process is there . So ans is 0.
by Boss (10.6k points)
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