Given processes with arrival times and burst times — P1(0,5), P2(1,3), P3(2,8), P4(3,6) — what is the average waiting time using Shortest Job First (SJ

Using SRTF: at t=0, only P1 runs (remaining=5). At t=1, P2 arrives (burst 3 < P1's remaining 4) → P2 preempts, runs. At t=2, P3 arrives (remaining 8; P2 has 2 left) → P2 continues. At t=3, P4 arrives (remaining 6; P2 has 1 left) → P2 continues and finishes at t=4 (completion=4). Next shortest among P1(4 remaining), P3(8), P4(6) → P1 runs from t=4 to t=8 (completion=8). Next shortest among P3(8) and P4(6) → P4 runs from t=8 to t=14 (completion=14). Finally P3 runs from t=14 to t=22 (completion=22). Waiting time = Completion − Arrival − Burst: P1 = 8−0−5 = 3, P2 = 4−1−3 = 0, P4 = 14−3−6 = 5, P3 = 22−2−8 = 12. Average = (3+0+5+12) / 4 = 20/4 = 5.0.