سوال ۹۷ کنکور ارشد مهندسی کامپیوتر ۱۴۰۳

In Round Robin each process gets the CPU for one quantum $q$, then a context switch of $s$ ms hands the CPU to the next process. Here $n = 20$, $q = 10$ ms, $s = 2$ ms.

When a given process finishes its quantum, it must wait while all the other $n-1 = 19$ processes each take their turn before it runs again. Each turn of another process occupies one quantum plus the context switch that follows it:

$q + s = 10 + 2 = 12 \text{ ms per other process.}$

The maximum wait until this process's next quantum begins is therefore

$(n-1)\,(q+s) = 19 \times 12 = 228 \text{ ms.}$

The $19$ slots account for the $19$ other quanta ($19 \times 10 = 190$ ms) plus the $19$ context switches that occur between them and the final switch back to this process ($19 \times 2 = 38$ ms), giving $190 + 38 = 228$ ms.

So the answer is 228 ms (option 3).

Pitfall: Use $n-1=19$, not $n=20$ — the waiting process does not wait for itself. And do not forget to add the context-switch overhead $s$ to each of the other turns; ignoring it gives $190$ (the trap option 2).