If the time-slice used in the round-robin scheduling policy is more than the…

Answer: If the time-slice (quantum) is larger than the maximum CPU burst of any process, round-robin degenerates to first come first serve.

Why:

• Round-robin normally preempts a running process when its time-slice expires and moves the process to the back of the ready queue.

• If the quantum is larger than every process's CPU burst, no process will ever be preempted for exceeding the quantum; each process runs to completion when scheduled.

• Because the scheduler dispatches the next ready process from the queue and no preemption rearranges the queue, processes complete in the order they arrived — the defining behavior of first come, first serve.

• This is not shortest-job-first because no selection by burst length occurs, and it is not priority scheduling because priorities are not involved in changing the order.

Short example: Three processes arrive in order with CPU bursts 2, 4, and 3, and the quantum is 5. Each process will finish during its first CPU allocation in arrival order (2, then 4, then 3), so the execution order matches first-come-first-serve.