GATE CS 2025 Set 2 Q16 — CPU Scheduling

To find the average turnaround time, we trace the execution of the processes using the Shortest Remaining Time First (SRTF) scheduling algorithm:

• At $t = 0$: Only $P_1$ (burst time = 10) is available. $P_1$ starts executing.
• At $t = 1$: $P_2$ (burst time = 13) arrives. The remaining time for $P_1$ is 9. Since $9 < 13$, $P_1$ continues execution.
• At $t = 2$: $P_3$ (burst time = 6) arrives. The remaining time for $P_1$ is 8. Since $P_3$ has the shortest remaining time ($6 < 8$), $P_1$ is preempted, and $P_3$ starts executing.
• At $t = 8$: $P_3$ completes execution. $P_4$ (burst time = 9) arrives. The remaining times of the active processes are:
  • $P_1$: 8
  • $P_2$: 13
  • $P_4$: 9
  $P_1$ has the shortest remaining time (8) and executes from $t = 8$ to $t = 16$, completing at $t = 16$.
• At $t = 16$: The remaining processes are $P_2$ (13) and $P_4$ (9). $P_4$ executes from $t = 16$ to $t = 25$, completing at $t = 25$.
• At $t = 25$: Only $P_2$ (13) remains. It executes from $t = 25$ to $t = 38$, completing at $t = 38$.

Now, we calculate the Turnaround Time ($\text{TAT} = \text{Completion Time} - \text{Arrival Time}$) for each process:

The average turnaround time is calculated as: