Consider the following four processes with the arrival time and length of CPU…
2017
Consider the following four processes with the arrival time and length of CPU burst given in milliseconds :
\(\begin{array}{ccc} \textbf{Process} & \textbf{Arrival Time} & \textbf{Burst Time} \\ P_1 & 0 & 8 \\ P_2 & 1 & 4 \\ P_3 & 2 & 9 \\ P_4 & 3 & 5 \end{array}\)
The average waiting time for preemptive SJF scheduling algorithm is ____________
- A.
6.5 ms
- B.
7.5 ms
- C.
6.75 ms
- D.
7.75 ms
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Correct answer: A
Answer: Average waiting time = 6.5 ms
Schedule (preemptive SJF / SRTF) timeline and completions:
t = 0 to 1: P1 runs (remaining P1 = 7)
t = 1 to 5: P2 runs and finishes at t = 5 (arrival 1, burst 4)
t = 5 to 10: P4 runs and finishes at t = 10 (arrival 3, burst 5)
t = 10 to 17: P1 resumes and finishes at t = 17 (arrival 0, total burst 8)
t = 17 to 26: P3 runs and finishes at t = 26 (arrival 2, burst 9)
Compute waiting times for each process:
P1: completion 17 − arrival 0 − burst 8 = 9 ms
P2: completion 5 − arrival 1 − burst 4 = 0 ms
P3: completion 26 − arrival 2 − burst 9 = 15 ms
P4: completion 10 − arrival 3 − burst 5 = 2 ms
Average waiting time = (9 + 0 + 15 + 2) / 4 = 26 / 4 = 6.5 ms.
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