Ex6.2

The round robin scheduling algo is not supposed to put preempted proceses at the tail of the queue. I cant quite understand to the answer of ex 6.2 unit 6 in the study guide. Can anyone help?

This is how I understand it. Lets say we have 3 processes busy running, say P1, P2 and P3 (in that order, using RR scheduling). Now, just after process P2 has completed, P4 arrives. Even though P4 could theoretically execute now, it does not. What happens is, P3 and P1 will first want to execute, as they are already in the queue (in this scenario P1 is now last in the queue). This means that P4 wil be placed into the queue between P1 and P2, changing the execution order to P1, P4, P2 and P3. (basically when a new process enters, it gets put into the queue just before the process that was executed last).
This is the way I see it.
Hope it helps.

Thanks.