Let RR3 means Round Robin Scheduling algorithm with time quantum of 3 units and FCFS means First Come First Serve [FCFS] scheduling algorithm. Now if 3 processes P1, P2, P3 has arrival time in order is (0, 2, 4) and execution time is (8, 5, 4) respectively then the completion order of P1, P2, P3 under RR3 and FCFS scheduling is :

Question

Accepted Answer

The correct option is (3):
RR3: P2, P1, P3 and FCFS: P1, P2, P3

To determine the correct answer, we must calculate the completion order for each scheduling algorithm separately.

The FCFS (First Come First Serve) scheduling algorithm is non-preemptive and executes processes in the strict order of their arrival time. Process P1 arrives first at time 0 and executes for its full burst time (8 units) until time 8. P2 arrives next at time 2, but it must wait until P1 completes. P2 then runs from time 8 to 13 (8 + 5). P3 arrives last at time 4 and waits until P2 completes at 13, running from 13 to 17 (13 + 4). Therefore, the completion sequence under FCFS is P1, P2, P3.

The RR3 (Round Robin with a time quantum of 3 units) scheduling algorithm is preemptive, sharing the CPU by giving each process a maximum of 3 time units. Following the Gantt chart calculation: P1 runs (0-3), P2 arrives; P2 runs (3-6), P3 arrives; P1 runs (6-9); P3 runs (9-12); P2 resumes (12-14) and *completes* (since its remaining burst time was 2); P1 resumes (14-16) and *completes* (remaining burst time 2); finally, P3 resumes (16-17) and *completes* (remaining burst time 1). The process completion times are P2 at 14, P1 at 16, and P3 at 17. Hence, the RR3 completion order is P2, P1, P3. Both results match Option 3.

The incorrect option is (1):
RR3: P2, P1, P3 and FCFS: P3, P2, P1.
This option incorrectly states the FCFS completion order. The core principle of FCFS is to execute jobs based on arrival time. Since P1 arrived at 0, P2 at 2, and P3 at 4, the execution order must be P1, P2, P3, regardless of their burst times. A sequence of P3, P2, P1 is logically impossible under FCFS as P3 would have to wait for both P1 and P2 to complete.

The incorrect option is (2):
RR3: P3, P2, P1 and FCFS: P1, P2, P3.
This option incorrectly states the RR3 completion order. While the FCFS order is correct, the Round Robin calculation shows P2 finishing first at time 14. This is because P2 had a smaller initial burst time (5) than P1 (8) and benefited from the preemption to complete relatively quickly once it entered the ready queue. The completion order of P3, P2, P1 for RR3 would mean P3 finished first, which is incorrect as P3 finished last at time 17.

The incorrect option is (4):
RR3: P1, P2, P3 and FCFS: P3, P2, P1.
This option is entirely incorrect because both scheduling orders are wrong. The FCFS order is wrong, as explained above, since P1, P2, P3 are scheduled by arrival time. The RR3 order is also wrong; the Round Robin algorithm with a time quantum of 3 leads to P2 completing first, followed by P1, and then P3, due to the interleaved execution and the relative remaining burst times.