Question

Solve by Round Robin Scheduling, Quantum = 3 (preemptive). Calculate the average waiting time: Process Arrival Time Execution Time РО 5 P1 1 3 P2 N 8 P3 3 6

Answer 1

Round Robin Scheduling Algorithm is a preemptive scheduling algo that loads a process into the CPU only for a given time quantum and after the specified quantum is over whethe the process has completed or not it preempts the CPU and loads the next process in waiting queque. This in simple language can be called a schedduling algorithm which assigns each process a fixed time slot in a cyclic manner.

Now as we can see that according to the arrival time of the processses our waiting queue would we formed as :

P0 (5), P1 (3), P2(8), P3(6) where P0 is going to be first to be loaded on CPU and P3 last.

First of all lets draw the Gantt Chart that will clear most of the things:

PO (2) P1 (0) P2 (5) P3 (3) PO (0) P2 (3) P3 (0) P2 (0)

So in Gantt Chart above I have used () parenthesis with process name just to tell you how much execution time for a process is left after it has run for the specified time Quantum on the CPU.

Now lets calculate waiting time of each process:

P0: Its execution time was 5. It was allocated CPU in the starting as it come so it didnt wait for a unit. But after running for 3 units it was preempted and for execution of remaining 2 units it has to wait for 12-3 = 9 units of time as shown in the Gantt Chart above. So P0 waited for a total of 9 units.

P1: P1 was available at 1 but allocated CPU at 3 so it has to wait for 2 units. Now as its execution time was 3 units so it gets completed in the allocated time quantum and thus P1 has to wait for a total of 2 units.

P2: P2 was available at 2 but CPU was allocated to it at 6 so here it has to wait for 4 units. Further as its execution time was 8 so to complete the remaining it has to wait for additional 14 - 9 = 5 and 20 -17 = 3 so total waiting time for P2 was 4+5+3 = 12 units.

P3: P3 was allocated CPU at 9 whereas it was waiting from 3 so it has to wait for 9-3 = 6 units. Further as it has the execution time of 6 units it has to wait additionally for 17-12 = 5 units so in total P3 waited for 6+5 = 11 units of time.

Now the Average Waiting Time is : (9 + 2 + 12 + 11)/4 = 34/4 = 8.5 units

If any doubt left do comment it out in the comments section....