Given CPU time slice of \(2ms\) and following list of processes.…
2019
Given CPU time slice of \(2ms\) and following list of processes.
\(\begin{array}{ccc} \text{Process} & \text{Burst time} & \text{Arrival time} \\ & \text{(ms)} \\ p_1 & 3 & 0 \\ p_2 & 4 & 2 \\ p_3 & 5 & 5 \end{array}\)
Find average turnaround time and average waiting time using round robin CPU scheduling?
Answer: B. 5.66,1.66 — Solution: Use Round Robin scheduling with time quantum = 2 ms. Gantt chart and remaining times: 0–2: p1 (remaining 1); 2–4: p2 (remaining 2); 4–5: p1…
- A.
4,0
- B.
5.66,1.66
- C.
5.66,0
- D.
7,2
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Correct answer: B
Solution:
Use Round Robin scheduling with time quantum = 2 ms.
Gantt chart and remaining times:
0–2: p1 (remaining 1); 2–4: p2 (remaining 2); 4–5: p1 (completes at 5); 5–7: p2 (completes at 7); 7–9: p3 (remaining 3); 9–11: p3 (remaining 1); 11–12: p3 (completes at 12).
p1: completion time = 5 ms; turnaround = completion − arrival = 5 − 0 = 5 ms; waiting = turnaround − burst = 5 − 3 = 2 ms.
p2: completion time = 7 ms; turnaround = 7 − 2 = 5 ms; waiting = 5 − 4 = 1 ms.
p3: completion time = 12 ms; turnaround = 12 − 5 = 7 ms; waiting = 7 − 5 = 2 ms.
Average turnaround time = (5 + 5 + 7) / 3 = 17 / 3 ≈ 5.67 ms.
Average waiting time = (2 + 1 + 2) / 3 = 5 / 3 ≈ 1.67 ms.
Answer: average turnaround ≈ 5.67 ms, average waiting ≈ 1.67 ms.
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