Consider a single processor system with four processes A, B, C, and D,…

SRTF (Preemptive Shortest Remaining Time First): simulate timeline and compute waiting times.

• t=0–2: A runs (A remaining becomes 8 at t=2).

• t=2–4: B runs (B remaining becomes 4 at t=4).

• t=4–7: C (burst 3) runs and completes at t=7 (waiting time for C = completion − arrival − burst = 7−4−3 = 0).

• t=7–11: B resumes and completes at t=11 (waiting for B = 11−2−6 = 3).

• t=11–18: D runs and completes at t=18 (waiting for D = 18−6−7 = 5).

• t=18–26: A resumes and completes at t=26 (waiting for A = 26−0−10 = 16).

Per-process waiting times (SRTF):

• A = 16

• B = 3

• C = 0

• D = 5

Average waiting time (SRTF) = (16 + 3 + 0 + 5) / 4 = 6.

NP-SJF (Non-preemptive Shortest Job First): simulate timeline and compute waiting times.

• t=0–10: A runs to completion (A waiting = 0).

• t=10–13: Among arrived jobs B, C, D, C has shortest burst (3) so C runs and completes at t=13 (waiting for C = 13−4−3 = 6).

• t=13–19: Next shortest is B (6), so B runs and completes at t=19 (waiting for B = 19−2−6 = 11).

• t=19–26: D runs and completes at t=26 (waiting for D = 26−6−7 = 13).

Per-process waiting times (NP-SJF):

• A = 0

• B = 11

• C = 6

• D = 13

Average waiting time (NP-SJF) = (0 + 11 + 6 + 13) / 4 = 7.5.

Conclusion: The correct averages are SRTF = 6 and NP-SJF = 7.5.