Answer : option D
Round Robin QUEUE :-- P,Q,R,S,P,Q,R,S,P,Q,R,S…..
Given that processes arrived in the order P,Q,R and S.
- Switching to ready processes after termination of currently executing process also a context switch.
- There is no context switch from S to P. ( Therefore P should be complete with in 1 TQ )
- There are exactly two context switches from Q to R ( Therefore Q and R both should requires more than 1 TQ )
- There is exactly one context switch from R to Q ( Therefore Q require more time than R )
- Exactly one context swich from R to S. ( Therefore S should be complete with in 1 TQ otherwise another CS from R to S required.)
- Exactly one conext switch from S to Q
Accumulating all the above points
QUEUE :--- P,Q,R,S,Q,R,Q
Given that time quantum = 4.
$0<P_{BT}\leq4$
$0<S_{BT}\leq4$
$8<Q_{BT}$
$4<R_{BT}\leq8$
in Option D, Q’s BT=7. Hence it is not possible.