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UGC NET Computer Science Dec 2022 Shift II (Paper II)

An OS follows round-robin scheduling with time quantum of $4$ ms. Assuming that the 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 CPU is _____.

Solution

In Round Robin scheduling, each process gets CPU for one time quantum.

Given:

Time quantum $=4$ ms

Number of processes $=20$

Maximum waiting time occurs for the last process in the ready queue.

Before the last process gets CPU, the other $19$ processes will execute.

So,

Maximum waiting time $=19 \times 4$

$=76$ ms

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