Q58,Paper 3,J 13. Consider the following processes with time slice of 4 milliseconds (I/O requests are ignored) :

Process	A	B	C	D
Arrival time	0	1	2	3
Cpu cycle	8	4	9	5

The average turn around time of these  processes will be

(A) 19.25 milliseconds

(B) 18.25 milliseconds

(C) 19.5 milliseconds

(D) 18.5 milliseconds.

Answer B.

Explanation.

Round robin scheduling. Time slice 4 ms

Gantt chart is.

A(4)-----------b(8)---------c(12)------d(16)-------a(20)-----c(24)---d(25)-c(26)

turnaround time =completion time- submission time.

A turnaround= 20-0.

B turnaround=8-1=7.

C turnaround=26-2.

D turnaround =25-3=22.

Average turnaround time=(20+7+24+22)/4=18.25ms