Galvin

Question

The Following processes are being scheduled using a Premptive RR scheduling Alogrithm. Each Process is Assigned a numeric Proirity, with higher number indicating higher priority. In addition it also has ideal task which consumes no CPU Resources and is identified by Pidle. This task has Priority 0 and is scheduled whenever the system has no other process to run. The length of time Quantum = 10 units.If a process is prempted by a higher priority process,the prempted process is placed at the end of queue.

Thread	Priority	Burst	Arrival
P1	40	20	0
P2	30	25	25
P3	30	25	30
P4	35	15	60
P5	5	10	100
P6	10	10	105

A. Turn Around Time of Each Process ?

B. Waiting time of Each Process ?

C. CPU Utilization Rate ?

Comments

I am getting Stuck in between. My main doubt is here when the time Quanta of a higher priority process will complete then which to schedule Next. The lower priority process in head of queue of the Higher priority process again. Because i follow first one then the chance of again coming the higher priority will be long, so may be it should be Second thing i.e. The higher priority process will keep on executing, but then How come it is behaving as RR because it will be simply as Priority Scheduling Algorithm

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If a process is preempted by a higher priority process, the preempted process is placed at the end of the queue.

What I understood from the question is, this line tells us everything about the implementation.
RR comes into play only when we have to take a decision between the processes of the same priority like $P_2$  and $P_3$.  

Answer 1

RR scheduling is simply FCFS with preemption added. The preemption occurs on the basis of quantum. No matter what the priority the processes will be preempted once the time quantum is over and put at the end of a queue. The process at the start of the queue is brought up for execution next.