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CPU free percentage with I/O (round robin)

Short answer

With round-robin quantum 5 ms, CPU is idle during 0–2 ms and 29–33 ms, giving 6 ms free out of 33 ms total; CPU free = 18.18 %.

Three processes P1, P2 and P3 arrive at time zero. Their total execution time is 10 ms, 15 ms, and 20 ms respectively. They spent first 20% of their execution time in doing I/O, next 60% in CPU processing and the last 20% again doing I/O. For what percentage of time was the CPU free? Use Round robin algorithm with time quantum 5 ms.

The CPU is free whenever no process is ready to use it. Each process does I/O first, then CPU work, then I/O again; I/O of different processes is assumed to run in parallel (each has its own device), and only CPU bursts are scheduled by round robin with quantum 5 ms.

Phase times

ProcessTotalFirst I/O (20%)CPU (60%)Last I/O (20%)
P110262
P215393
P3204124

All three processes arrive at 0 and start with I/O, so they enter the CPU ready queue at 2, 3 and 4 ms respectively.

Schedule

|idle| P1 | P2 | P3 | P1 | P2 | P3 |P3|idle|
0    2    7    12   17   18   22   27 29   33
  • 0-2: CPU free (everyone is doing first I/O).
  • 2-7: P1 runs 5 ms (1 ms of its 6 ms left). P2 and P3 join the queue at 3 and 4.
  • 7-12: P2 runs 5 ms (4 left). 12-17: P3 runs 5 ms (7 left).
  • 17-18: P1 finishes its CPU work; it then does last I/O 18-20.
  • 18-22: P2 finishes its CPU work (4 ms); last I/O 22-25.
  • 22-27: P3 runs 5 ms (2 left); 27-29: P3 runs the last 2 ms (alone, no other process is ready).
  • 29-33: P3 does its last I/O; the CPU is free.

Everything ends at 33 ms (P1 at 20, P2 at 25, P3 at 33).

Calculation

Total CPU busy time = 6 + 9 + 12 = 27 ms

Total time = 33 ms

CPU free time = 33 - 27 = 6 ms (0-2 and 29-33)

CPU free percentage = 6 / 33 x 100 = 18.18 %

Answer: the CPU is free for about 18.18 % of the time (CPU utilisation 81.82 %).

✓ Answer checked by Claude Opus 5.5 (high effort)

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