Consider the following set of processes that need to be scheduled on a single CPU. All the times are given in milliseconds. Using the *shortest remaining time first* scheduling algorithm, the average process turnaround time (in msec) is _________.

We schedule the processes using the preemptive Shortest Remaining Time First (SRTF) policy:

• At $t = 0$: Process A arrives (Burst = $6$). CPU is allocated to A.

• At $t = 3$: Process B arrives (Burst = $2$). A's remaining time is $6 - 3 = 3$. Since B has a smaller remaining time ($2 < 3$), B preempts A.

• At $t = 5$: Process B completes. Process C arrives (Burst = $4$). Ready queue contains:

  • A with remaining time $3$

  • C with remaining time $4$

  CPU is given to A (remaining time $3$).

• At $t = 7$: Process D arrives (Burst = $6$). A has remaining time $1$. A continues running.

• At $t = 8$: Process A completes. Ready queue contains:

  • C with remaining time $4$

  • D with remaining time $6$

  CPU is allocated to C.

• At $t = 10$: Process E arrives (Burst = $3$). C has remaining time $2$. Since $2 < 3$, C continues.

• At $t = 12$: Process C completes. Ready queue contains:

  • E with remaining time $3$

  • D with remaining time $6$

  CPU is allocated to E.

• At $t = 15$: Process E completes. CPU is allocated to D.

• At $t = 21$: Process D completes.

### Completion and Turnaround Times:

• A:$\text{CT} = 8 \implies \text{TAT} = 8 - 0 = 8$

• B:$\text{CT} = 5 \implies \text{TAT} = 5 - 3 = 2$

• C:$\text{CT} = 12 \implies \text{TAT} = 12 - 5 = 7$

• D:$\text{CT} = 21 \implies \text{TAT} = 21 - 7 = 14$

• E:$\text{CT} = 15 \implies \text{TAT} = 15 - 10 = 5$

### Average Turnaround Time: