Question 24

We are given four processes:

• P1: Arrival = 0, Burst = 7
• P2: Arrival = 3, Burst = 3
• P3: Arrival = 5, Burst = 5
• P4: Arrival = 6, Burst = 2

Trace the scheduling step by step:

• At $t = 0$: Only P1 is available (remaining burst = 7). P1 runs from $t = 0$ to $t = 3$.
• At $t = 3$: P1 has remaining burst $7 - 3 = 4$. P2 arrives with burst 3. Since $3 < 4$, P2 preempts P1. P2 runs.
• At $t = 5$: P2 has remaining burst $3 - 2 = 1$. P3 arrives with burst 5. Remaining times: P2 = 1, P1 = 4, P3 = 5. P2 continues running.
• At $t = 6$: P2 completes (completion time = 6). P4 arrives with burst 2. Remaining times: P4 = 2, P1 = 4, P3 = 5. P4 has the shortest remaining time, so P4 runs from $t = 6$ to $t = 8$.
• At $t = 8$: P4 completes (completion time = 8). Remaining processes: P1 (rem = 4), P3 (rem = 5). P1 runs from $t = 8$ to $t = 12$.
• At $t = 12$: P1 completes (completion time = 12). Only P3 remains (rem = 5). P3 runs from $t = 12$ to $t = 17$.
• At $t = 17$: P3 completes.

Calculate turnaround and waiting times (Waiting time = Turnaround time - Burst time):

• P1: Turnaround = $12 - 0 = 12$, Waiting = $12 - 7 = 5$
• P2: Turnaround = $6 - 3 = 3$, Waiting = $3 - 3 = 0$
• P3: Turnaround = $17 - 5 = 12$, Waiting = $12 - 5 = 7$
• P4: Turnaround = $8 - 6 = 2$, Waiting = $2 - 2 = 0$

Average waiting time across all 4 processes: