An operating system uses shortest remaining time first scheduling algorithm for pre-emptive scheduling of processes. Consider the following set of processes with their arrival times and CPU burst times (in milliseconds): The average waiting time (in milliseconds) of the processes is ________.

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

1. Gantt Chart Execution:

   • At $t = 0$: $P_1$ arrives (Burst $= 12$). $P_1$ executes until $t = 2$.

   • At $t = 2$: $P_2$ arrives (Burst $= 4$). Remaining time for $P_1 = 10$. Since $4 < 10$, $P_2$ preempts $P_1$.

   • At $t = 3$: $P_3$ arrives (Burst $= 6$). Remaining time for $P_2 = 3$. Since $3 < 6$, $P_2$ continues.

   • At $t = 6$: $P_2$ completes. Ready queue: $P_1$ (rem $= 10$), $P_3$ (rem $= 6$). $P_3$ starts executing.

   • At $t = 8$: $P_4$ arrives (Burst $= 5$). Remaining time for $P_3 = 4$. Since $4 < 5$, $P_3$ continues.

   • At $t = 12$: $P_3$ completes. Ready queue: $P_4$ (rem $= 5$), $P_1$ (rem $= 10$). $P_4$ executes.

   • At $t = 17$: $P_4$ completes. $P_1$ resumes and executes until $t = 27$.

2. Completion Times ($CT$):

   • $CT(P_1) = 27$

   • $CT(P_2) = 6$

   • $CT(P_3) = 12$

   • $CT(P_4) = 17$

3. Turnaround Time ($TAT = CT - AT$):

   • $TAT(P_1) = 27 - 0 = 27$

   • $TAT(P_2) = 6 - 2 = 4$

   • $TAT(P_3) = 12 - 3 = 9$

   • $TAT(P_4) = 17 - 8 = 9$

4. Waiting Time ($WT = TAT - BT$):

   • $WT(P_1) = 27 - 12 = 15$

   • $WT(P_2) = 4 - 4 = 0$

   • $WT(P_3) = 9 - 6 = 3$

   • $WT(P_4) = 9 - 5 = 4$

5. Average Waiting Time:

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