Consider the following three processes with their arrival times and CPU burst…

Key idea: use preemptive shortest-job-first (shortest remaining time first): at every arrival choose the process with the smallest remaining CPU time.

• Time 0–1: P1 runs (remaining 7 ms).

• Time 1–5: P2 arrives at 1 and runs (4 ms), completing at time 5.

• Time 5–12: P1 resumes (7 ms), completing at time 12.

• Time 12–21: P3 runs (9 ms), completing at time 21.

Completion and waiting times:

1. P1: arrival 0, burst 8, completion 12 → turnaround = 12 − 0 = 12 ms → waiting = 12 − 8 = 4 ms.

2. P2: arrival 1, burst 4, completion 5 → turnaround = 5 − 1 = 4 ms → waiting = 4 − 4 = 0 ms.

3. P3: arrival 2, burst 9, completion 21 → turnaround = 21 − 2 = 19 ms → waiting = 19 − 9 = 10 ms.

Average waiting time = (4 + 0 + 10) / 3 = 14 / 3 ≈ 4.67 ms. Therefore the numeric answer 4.66 (approx) is correct.