Consider four processes P, Q, R, and S scheduled on a CPU as per round robin algorithm with a time quantum of 4 units. The processes arrive in the order P, Q, R, S, all at time t = 0. There is exactly one context switch from S to Q, exactly one context switch from R to Q, and exactly two context switches from Q to R. There is no context switch from S to P. Switching to a ready process after the termination of another process is also considered a context switch. Which one of the following is NOT possible as CPU burst time (in time units) of these processes?

A

P = 4, Q = 10, R = 6, S = 2

B

P = 2, Q = 9, R = 5, S = 1

C

P = 4, Q = 12, R = 5, S = 4

D

P = 3, Q = 7, R = 7, S = 3

Question 23 Explanation:

The Gantt chart can be assumed as :
The combinations of given context switches are possible for options A,B and C. However, we can observe that option D does not follow the context switch. Hence, D is the answer.

Correct Answer: D

Question 23 Explanation:

The Gantt chart can be assumed as :
The combinations of given context switches are possible for options A,B and C. However, we can observe that option D does not follow the context switch. Hence, D is the answer.