UGCNET CSE December 2022: 88

Question

 

An $\text{OS}$ follows round-robin scheduling with time quantum of $4 \mathrm{~ms}$. Assuming that the $\mathrm{CPU}$ is free now and there are $20$ processes waiting in the ready queue, the maximum amount of time that a process waits before getting into the $\mathrm{CPU}$ is

• A. $80 \mathrm{~ms}$
• B. $76 \mathrm{~ms}$
• C. $84 \mathrm{~ms}$
• D. None of the above

(Option $1[39649]) 1$
(Option $2[39650]) 2$
(Option $3[39651]) 3$
(Option $4[39652]) 4$

Answer Given by Candidate: $2$

Answer 1

The $20^{th}$ process will get started after quantum expiration of $19^{th}$ process.

Thus $20^{th}$ process will be delayed by $19*4 = 76ms$