Question

The following processes are being scheduled using a preemptive, priority-based, round-robin scheduling algorithm. Each process is assigned a numerical priority,with a higher number indicating a higher relative priority. The scheduler will execute the highest-priority process. For processes with the same priority, a round-robin scheduler will be used with a time quantum of 10 units. If a process is preempted by a higher-priority process, the preempted process is placed at the end of the queue.

Process            Burst Time Arrival Time Priority

P1 15 0 8

P2 20 0 3

P3 20 20 4

P4 20 25 4

P5 5 45 5

P6 15 55 5

What is the average waiting time of these processes?

Answer 1

Solution

We know that:

Turnaround time(TAT) = Completion time(CT) - Arrival time(AT)

Waiting time(WT) = Turnaround time(TAT) - Burst time

The Gantt chart is shown below:

Process	Arrival Time	Burst Time	Completion Time	Turnaround Time	Waiting time
P1	0	15	15	15 - 0 = 15	15 - 15 = 0
P2	0	20	95	95 - 0 = 95	95 - 20 =75
P3	20	20	55	55 - 20 = 35	35 - 20 = 15
P4	25	20	80	80 - 25 = 55	55 - 20 = 35
P5	45	5	50	50 - 45 = 5	5 - 5 = 0
P6	55	15	70	70 - 55 = 15	15 - 15 = 0

Average waiting time = (0 + 75 + 15 + 35 + 0 + 0)/6 = 125/6 = 20.833