Consider the following 3 processes with the length of the CPU burst time given in milliseconds:

Process	Arrival Time	Burst Time
P1	0	8
P2	1	4
P3	2	9

What is the average waiting time for these processes if they are scheduled using preemptive shortest job first scheduling algorithm?

Preemptive SJF is also called Shortest Remaining Time First.

At time $0$, only $P1$ is available, so $P1$ runs.

At time $1$, $P2$ arrives. Remaining time of $P1=7$ and burst time of $P2=4$, so $P2$ runs.

At time $2$, $P3$ arrives, but $P2$ has remaining time $3$, which is shortest. So $P2$ continues.

Gantt chart:

$0\to1:P1,\ 1\to5:P2,\ 5\to12:P1,\ 12\to21:P3$

Completion times:

$CT(P1)=12,\ CT(P2)=5,\ CT(P3)=21$

Waiting time formula:

$WT=CT-AT-BT$

$WT(P1)=12-0-8=4$

$WT(P2)=5-1-4=0$

$WT(P3)=21-2-9=10$

Average waiting time:

$\frac{4+0+10}{3}=\frac{14}{3}=4.66$