Doubt 1 part 1 : What happens when TS(Ti) < W-timestamp(Qk)
Let Qk denote the version of Q whose write timestamp is the largest write timestamp less than or equal to TS(Ti).
So according to this condition, the situation "TS(Ti) < W-timestamp(Qk) " can't occur ever because it is already said that the W-timestamp(Qk)<=TS(Ti).
At the very beginning we maintain a version Q0 of dB and initialize the read and write TS with 0.
So any transaction Ti will have a version of Q whose W-timestamp(Qk) is lesser than TS(Ti) i.e. W-timestamp(Q0) < TS(Ti) => 0<TS(Ti)
Doubt 1 part 2: What happens when TS(Ti) > W-timestamp(Qk)
Let Qk denote the version of Q whose write timestamp is the largest write timestamp less than or equal to TS(Ti) i.e. W-timestamp(Qk)<=TS(Ti)
The primary condition is this only. Whenever we confront any operation this is the condition that we always check.
Doubt 2 : Multiversion protocol allows schedules which may not be conflict or even view serializable. So Yes you are right that this protocol allows some non conflict serializable schedules too.
Doubt 3: Multiversion allows more than view serializable schedules.
I had following doubts: is it worth to give so much importance to this topic fo GATE perspective? Can we safely skip it?
I am a student as well so i shouldn't comment on this :P But yes my teacher told that this is very less important for GATE.