Discussion
Three processes P, Q and R enter the ready queue simultaneously with the length of the cpu burst given in milliseconds as 20, 10 and 15 respectively. The wailing time for process R in case of shortest job first scheduling is :
Answer: C
The waiting time of R is 10 milliseconds.
Shortest job first runs the process with the shortest burst first. All three arrive together, so the order is Q (10), then R (15), then P (20):
\[ \begin{aligned} Q &: 0 \text{ to } 10 \\ R &: 10 \text{ to } 25 \\ P &: 25 \text{ to } 45 \end{aligned} \]
R waits while Q runs, so its waiting time (printed 'wailing time') is 10 ms. 25 is R's completion time, 15 its own burst, and 20 the burst of P.
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