Consider the following CPU processes with arrival times (in milliseconds) and length of CPU bursts (in milliseconds) as given below : If the pre-emptive shortest remaining time first scheduling algorithm is used to schedule the processes, then the average waiting time across all processes is ________ milliseconds.

We schedule the processes using the Preemptive Shortest Remaining Time First (SRTF) algorithm:

• At time $t = 0$: Process $P_1$ arrives with remaining burst time $7$. $P_1$ executes.

• At time $t = 3$: $P_2$ arrives with burst time $3$. Remaining burst time of $P_1$ is $7 - 3 = 4$. Since $\text{Remaining}(P_2) = 3 < 4$, $P_1$ is preempted and $P_2$ starts execution.

• At time $t = 5$: $P_3$ arrives with burst time $5$. Remaining burst time of $P_2$ is $3 - 2 = 1 < 5$. So $P_2$ continues.

• At time $t = 6$: $P_4$ arrives with burst time $2$. $P_2$ finishes execution at $t = 6$.

• At time $t = 6$: Ready queue has $P_4$ (remaining: $2$), $P_1$ (remaining: $4$), $P_3$ (remaining: $5$). The shortest is $P_4$, so $P_4$ runs from $t = 6$ to $t = 8$ and finishes at $t = 8$.

• At time $t = 8$: Ready queue has $P_1$ (remaining: $4$) and $P_3$ (remaining: $5$). $P_1$ runs from $t = 8$ to $t = 12$ and finishes at $t = 12$.

• At time $t = 12$: $P_3$ runs from $t = 12$ to $t = 17$ and finishes at $t = 17$.

Gantt Chart:

Turnaround Time (TAT = Completion Time - Arrival Time) and Waiting Time (WT = TAT - Burst Time):

• For $P_1$: $\text{CT} = 12$, $\text{TAT} = 12 - 0 = 12$, $\text{WT} = 12 - 7 = 5$

• For $P_2$: $\text{CT} = 6$, $\text{TAT} = 6 - 3 = 3$, $\text{WT} = 3 - 3 = 0$

• For $P_3$: $\text{CT} = 17$, $\text{TAT} = 17 - 5 = 12$, $\text{WT} = 12 - 5 = 7$

• For $P_4$: $\text{CT} = 8$, $\text{TAT} = 8 - 6 = 2$, $\text{WT} = 2 - 2 = 0$