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Consider 4 processes sharing the CPU in round robin fashion. If context switch time is 1 sec , what must be the time quantum q such that the number of context switches are reduced , but at the same time each process is guaranteed to get the turn at the CPU for every 10 secs?


Ans 2 or 1.5 .

Is here any formula used like (n-1)q+ns

explain in detail
in Operating System 518 views
logically if we thought like this -

P1  |  P2  |  P3  |  P4  |  P1        (vertical lines denotes Context switches)

then before again P1 will be in running state, there will be 4 Context switches, hence 10-4 = 6 units of time is remaining,

To make full use of it, each process  P2, P3 AND P4 should run for 2 units of time. (i.e time quantum = 2)
if every one taking 2 unit of time

means 1 unit for process execution, and 1 unit for context switch

then 2 *4=8 unit utilized

how 10 unit utilized?
I'm saying between P1 and again P1, we have 10 units, there will be 4 context switches , so 10-4 = 6 units will be there.

Now in remaining 6 units if we gave process P2, P3, and P4 equally 2 units for running (this will be TQ) then overall 10 units are utilized.
ok, u mean time between P1 ended and P1 started


U r not taking P1 time

IMO, the question lacks clarity, its slightly ambiguous.

 each process is guaranteed to get the turn at the CPU for every 10 secs?

Suppose process A executed at time 0ms, and now the current time is 10ms.

Then the question asks which of the following exactly? :-

  1. Before this very moment (i.e.10ms) the process A should have already been executed? OR
  2. At exactly this moment (i.e.10ms), Process A should be guaranteed to get started?

See this question :-

1 Answer

2 votes
Best answer
$4$ processes given , Round Robin scheduling always use Circular queue data strucure.

So ordering will be  $P_1\rightarrow P_2 \rightarrow P_3 \ \rightarrow P_4\rightarrow P_1...$ till they exhaust their Burst time.

Now in question its mentioned that in b/w two scheduling on CPU , time should be $10\ sec.$

So considering that scenario would be $Q_1\ |CS \ |Q_2\ |CS \ |Q_3\ |CS\ |Q_4\ |CS\ |Q_1..$

$E_{qn}=3 \times Q \ + 4 \times CS=10$ sec

Given = $CS  \ = \ 1$sec

$\therefore Q=\color{RED}{2sec}$

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yes, I was taking P1 time at first

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