GATE | GATE 2017 MOCK II | Question 51
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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Last Updated :
28 Jun, 2021
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