Consider the following processes, with the arrival time and the length of the CPU burst given in milliseconds. The scheduling algorithm used is preemptive shortest remaining-time first. The average turn around time of these processes is ________ milliseconds.

We apply the Preemptive Shortest Remaining Time First (SRTF) scheduling algorithm:

### Gantt Chart Execution:

• At $t = 0$: $P_1$ arrives with remaining time $10$. $P_1$ starts executing.

• At $t = 3$: $P_2$ arrives with burst time $6$. Remaining time of $P_1 = 10 - 3 = 7$. Since remaining time of $P_2$ ($6$) < $P_1$ ($7$), $P_1$ is preempted and $P_2$ executes.

• At $t = 7$: $P_3$ arrives with burst time $1$. Remaining time of $P_2 = 6 - (7-3) = 2$. Since $P_3$ ($1$) < $P_2$ ($2$), $P_2$ is preempted and $P_3$ executes.

• At $t = 8$: $P_3$ finishes execution. $P_4$ arrives with burst time $3$.

  • Available processes: $P_2$ (remaining $2$), $P_4$ (remaining $3$), $P_1$ (remaining $7$).

  • Shortest remaining is $P_2$ ($2$). $P_2$ executes from $t = 8$ to $t = 10$ and finishes.

• At $t = 10$: Available processes: $P_4$ ($3$), $P_1$ ($7$).

  • $P_4$ executes from $t = 10$ to $t = 13$ and finishes.

• At $t = 13$: Available process: $P_1$ ($7$).

  • $P_1$ executes from $t = 13$ to $t = 20$ and finishes.

### Completion and Turnaround Times:

• $P_1$: Completion Time (CT) = $20$, Turnaround Time (TAT = CT - AT) = $20 - 0 = 20$

• $P_2$: Completion Time (CT) = $10$, Turnaround Time (TAT) = $10 - 3 = 7$

• $P_3$: Completion Time (CT) = $8$, Turnaround Time (TAT) = $8 - 7 = 1$

• $P_4$: Completion Time (CT) = $13$, Turnaround Time (TAT) = $13 - 8 = 5$

### Average Turnaround Time: