Odd uid: consider the three processes with arrival and burst time asProcess Arrival time Burst timeP1 0 9P2 1 4P3 2 7The pre-emptive shortest job first scheduling is used. Scheduling is carried out at arrival orcompletion of processes. What is the average waiting time for the three processes?Also find TTAT.

To solve this problem, we need to follow the steps of the Preemptive Shortest Job First (SJF) scheduling algorithm. This algorithm selects the process with the smallest execution time first. If a new process arrives with less burst time, the current process is preempted.

Here are the steps:

1. At time t=0, process P1 is the only one in the queue, so it starts execution.

2. At time t=1, process P2 arrives. Since P2 has a shorter burst time than the remaining time of P1, P2 preempts P1.

3. P2 runs for its entire burst time of 4 units, so it finishes at time t=5.

4. At this point, both P1 and P3 are in the queue. P1 has a remaining burst time of 8 (since it was preempted after 1 unit of execution), and P3 has a burst time of 7. So, P3 is selected next.

5. P3 runs for its entire burst time and finishes at time t=12.

6. Now, only P1 is left in the queue, so it runs for its remaining burst time and finishes at time t=20.

Now, we can calculate the waiting time for each process:

• Waiting time for P1 = Finish time - Arrival time - Burst time = 20 - 0 - 9 = 11 units
• Waiting time for P2 = Finish time - Arrival time - Burst time = 5 - 1 - 4 = 0 units
• Waiting time for P3 = Finish time - Arrival time - Burst time = 12 - 2 - 7 = 3 units

So, the average waiting time = (11 + 0 + 3) / 3 = 4.67 units

The Turnaround Time (TTAT) for a process is the total time from its arrival to its completion. So:

• TTAT for P1 = Finish time - Arrival time = 20 - 0 = 20 units
• TTAT for P2 = Finish time - Arrival time = 5 - 1 = 4 units
• TTAT for P3 = Finish time - Arrival time = 12 - 2 = 10 units

So, the average TTAT = (20 + 4 + 10) / 3 = 11.33 units.