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GATE | GATE 2017 MOCK II | Question 51

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  • Last Updated : 28 Jun, 2021

On a system using round-robin scheduling let ‘s’ represent the time needed to perform a process switch, ‘q’ represents the round-robin time quantum, and ‘r’ represents the average time a process runs before blocking on I/O. The CPU efficiency when s = q < r is:

(A) r/(r+s)

(B) q/(q+s)

(C) 1/2
(D) 0

Answer: (C)

Explanation: ‘r’ is the average time a process runs before I/O block and ‘s’ is the time needed for switch.
‘q’ represents the round-robin time quantum.
The efficiency of the CPU when s < r and q < r.
= r/(r/q)*s+r
= q/(q+s)

Here s = q.
Hence efficiency is ½.

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