Given a CPU time slice of 2 ms and the following processes, assume zero context-switch cost. When an arrival coincides with quantum expiry, enqueue the arriving process before requeuing the preempted process. Process Burst time (ms) Arrival time (ms) p 1 3 0 p 2 4 2 p 3 5 5 Find average turnaround time and average waiting time using round-robin CPU scheduling?
\(\frac{17}{3}\text{ ms},\ \frac{5}{3}\text{ ms}\)
The schedule is P1: 0–2, P2: 2–4, P1: 4–5, P2: 5–7, then P3: 7–12 (with quantum boundaries at 9 and 11). Completion times are 5, 7 and 12 ms. Turnaround times are 5, 5 and 7 ms; subtracting burst times gives waiting times 2, 1 and 2 ms. The averages are \(17/3\text{ ms}\) and \(5/3\text{ ms}\), approximately 5.67 ms and 1.67 ms.
Important Questions from Process Scheduling - Teaching
Consider the following 3 processes with the length of the CPU burst time given in milliseconds :
Process Arrival Time Burst Time P1 0 8 P2 1 4 P3 2 9 What is the average waiting time for these processes if they are scheduled using preemptive shortest job first scheduling algorithm?
Consider the following four processes with the arrival time and length of CPU burst given in milliseconds :
Process Arrival Time Burst Time P 1 0 8 P 2 1 4 P 3 2 9 P 4 3 5 The average waiting time for preemptive SJF scheduling algorithm is
Which of the following is not a CPU scheduling Algorithm?
Which of the following statement(s) is/are true about CPU scheduling algorithm?
I. Shortest remaining time first scheduling may cause starvation
II. Round Robin is better than First Come First Serve in terms of response time
Processes P1, P2, P3 and P4 have CPU burst times 5, 7, 6 and 4 ms, respectively. All arrive at time 0, and the initial ready-queue order is P1, P2, P3, P4. Using Round-Robin scheduling with a 4 ms time quantum and no context-switch overhead, what is P4’s total waiting time in milliseconds?
