Consider the following table of arrival time and burst time for three processes P0, P1 and P2. The pre-emptive shortest job first scheduling algorithm is used. Scheduling is carried out only at arrival or completion of processes. What is the average waiting time for the three processes?

From the given process table:

• Process P0: Arrival Time = $0\text{ ms}$, Burst Time = $9\text{ ms}$

• Process P1: Arrival Time = $1\text{ ms}$, Burst Time = $4\text{ ms}$

• Process P2: Arrival Time = $2\text{ ms}$, Burst Time = $9\text{ ms}$

Gantt Chart Execution (Pre-emptive Shortest Job First / SRTF):

1. Time 0 to 1 ms: Only P0 is present. P0 executes for $1\text{ ms}$. Remaining burst of P0 = $8\text{ ms}$.

2. Time 1 ms: P1 arrives with remaining burst $4\text{ ms}$. Since $4 < 8$, P0 is pre-empted and P1 starts executing.

3. Time 2 ms: P2 arrives with burst $9\text{ ms}$. Remaining burst of P1 is $3\text{ ms}$, which is the shortest ($3 < 8 < 9$). P1 continues executing.

4. Time 5 ms: P1 completes execution.

5. Time 5 ms to 13 ms: Ready queue has P0 (remaining $8\text{ ms}$) and P2 (remaining $9\text{ ms}$). Since $8 < 9$, P0 executes to completion at time $5 + 8 = 13\text{ ms}$.

6. Time 13 ms to 22 ms: P2 executes to completion at time $13 + 9 = 22\text{ ms}$.

Calculation of Waiting Times:

• Completion Time ($CT$):

  • P0: $13\text{ ms}$

  • P1: $5\text{ ms}$

  • P2: $22\text{ ms}$

• Turnaround Time ($TAT = CT - AT$):

  • P0: $13 - 0 = 13\text{ ms}$

  • P1: $5 - 1 = 4\text{ ms}$

  • P2: $22 - 2 = 20\text{ ms}$

• Waiting Time ($WT = TAT - BT$):

  • P0: $13 - 9 = 4\text{ ms}$

  • P1: $4 - 4 = 0\text{ ms}$

  • P2: $20 - 9 = 11\text{ ms}$

Average Waiting Time:

Hence, the correct option is A.