Question

Round Robin Schedulin 

Process	Burst Time in Ms
P1	4
P2	2
P3	2

Calculate the average waiting time and turnaround time using round robin scheduling, where the time quantum q = 2 ms 

According to the Round Robin algorithm, the arrival of processes is shown in the following Gantt chart. 

From the Gantt chart, the waiting time and the turnaround time are calculated as: 

Waiting time for P1= 

Waiting time for P2 = 

Waiting time for P2 = 

Therefore, the average waiting time = 

Turnaround time =

Answer 1

SOLUTION :

Hey there,

This question is simple and this method is one of the famous method(ROUND ROBIN) in time scheduling for CPU.

First let me explain you some of the terms.

All the processes to be performed are represented as Pi,where i=1,2,3....n for n processes.

ROUND ROBIN is a preemptive process scheduling algorithm.

QUANTUM : The maximum time for which a process is allowed to run while processing in the queue.

BURST TIME : It is the maximum time for a process that it can take to finish.generally a process may or may finish at once.It depends on the Quantum value.

TAT(TURN AROUND TIME) : It is the time difference between the completion time and arrival time according to the Gantt Chart . ex - if a process initially starts at 1 and finally completed at 10 (means deleted from the scheduling queue) then TAT of that process = 10-1=9.

WAITING TIME :The summation of time interval Between two intervals when a process is again comes for execution is called Waiting Time of that process. ex-process P1 ends at 18 hence it waits in the interval ( 2-6,8-12,14-16 ) so total WT=4+4+2=10 ms.

Now, Coming to the question ,

from given Gantt Chart,

Waiting Time for P1 : waits in the interval 2-6,8-12,14-16(note that here - is used for interval purpose not the negative sign).

WT( P1 )= (6-2)+(12-8)+(16-14)=4+4+2=10

Waiting Time for P1 = 10 ms.

Waiting Time for P2 : waits in the interval 0-2,4-8,10-14.

WT( P2 )= (2-0)+(4-8)+(10-14)=2+4+4=10

Waiting Time for P2 = 10 ms.

Waiting Time for P3 : waits in the interval 0-4,6-10,12-15.

WT( P3 )= (4-0)+(10-6)+(15-12)=4+4+3=11

Waiting Time for P3 = 11 ms.

SO AVERAGE WAITING TIME = (10+10+11) / 3 = 10.33

Now for TURN AROUND TIME we need to find arrival time(AT) and completion time(CT)

Process AT CT TAT

P1 0 18 18(18-0)

P2 2 14 12(14-2)

P3 4 16 12(16-4)

Thus above is the turn around time(TAT).

Answer 2

Draw the Round Robin scheduling algorithms by the following process given in table 2 and calculate the average waiting time when a quantum time = 20