Question 42

We trace execution cycle by cycle according to SRTF:

• $P_1$: arrival $0$, burst $12$
• $P_2$: arrival $2$, burst $4$
• $P_3$: arrival $3$, burst $6$
• $P_4$: arrival $8$, burst $5$

1. Gantt Chart Construction:

   • Time $0$ to $2$: Only $P_1$ is present. $P_1$ runs for $2$ ms (remaining: $12 - 2 = 10$).
   • Time $2$: $P_2$ arrives with burst $4$. Since remaining time of $P_2$ ($4$) < remaining time of $P_1$ ($10$), $P_1$ is preempted and $P_2$ runs.
   • Time $3$: $P_3$ arrives with burst $6$. $P_2$ has remaining $3 < 6$, so $P_2$ continues.
   • Time $6$: $P_2$ finishes (total run $4$ ms). Ready queue: $P_3$ (burst $6$), $P_1$ (remaining $10$). $P_3$ has shorter remaining time ($6 < 10$), so $P_3$ starts running.
   • Time $8$: $P_4$ arrives with burst $5$. At this instant, $P_3$ has run from $6$ to $8$ ($2$ ms), so $P_3$ has remaining time $6 - 2 = 4$ ms. Since remaining time of $P_3$ ($4$) < burst of $P_4$ ($5$), $P_3$ is NOT preempted and continues running.
   • Time $12$: $P_3$ finishes (total run $6$ ms). Ready queue: $P_4$ (remaining $5$), $P_1$ (remaining $10$). $P_4$ runs ($5 < 10$).
   • Time $17$: $P_4$ finishes (total run $5$ ms). $P_1$ resumes with remaining $10$ ms.
   • Time $27$: $P_1$ finishes.

   Gantt chart: P1 (0..2) | P2 (2..6) | P3 (6..12) | P4 (12..17) | P1 (17..27)

2. Completion and Waiting Times:

   • $P_1$: Completion = $27$. Turnaround = $27 - 0 = 27$. Waiting time = $27 - 12 = 15$.
   • $P_2$: Completion = $6$. Turnaround = $6 - 2 = 4$. Waiting time = $4 - 4 = 0$.
   • $P_3$: Completion = $12$. Turnaround = $12 - 3 = 9$. Waiting time = $9 - 6 = 3$.
   • $P_4$: Completion = $17$. Turnaround = $17 - 8 = 9$. Waiting time = $9 - 5 = 4$.

3. Average Waiting Time:

   $$\text{Average WT} = \frac{15 + 0 + 3 + 4}{4} = \frac{22}{4} = 5.5\text{ ms}.$$