Timestamp Ordering Protocol

Question

Assume basic timestamp ordering protocol and that time starts from $1$, each operation takes unit amount of time and start of transaction $T_i$ is denoted as $S_i$. The table of timestamp is given below:

Find $rts(a), wts(a), rts(b)$ and $wts(b)$ at the end

1.   1, 5, 2, 5
2.   1, 7, 3, 3
3.   3, 7, 3, 7
4.   1, 7, 3, 7

Comments

is it Option D

Youngest transaction who executed the given read/write will be its TS

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Yes D is correct.. I am not getting..Can u please explain  ?

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did you find a solution??

Answer 1

Here,

T1 starts at TS =1

T2 starts at TS = 3

T3 starts at TS =7.

While giving the the TS for any read or write always look for youngest.

RTS(a) = a is first read by T1 hence RTS(a) =1. (Read(a) is never done anywhere again hence it is youngest)

WTS(a) = a is first written by T2 hence WTS(a) = 3. But again written by T3 which has higher TS (youngest) Hence final TS of W(a) = 7

RTS(b) = b is first read by T2 hence RTS(b) =3. (Read(b) is never done anywhere again hence it is youngest)

WTS(b) = b is first written by T2 hence WTS(b) = 3. But again written by T3 which has higher TS (youngest) Hence final TS of W(b) = 7

Hence answer is 1,7,3,7

Comments

Well explained..Thank you so much :) ... I understood it but want to learn more about it.. Can u please give me some video reference or any other reference related to it ?

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what is the significance of serial no 1-9? As there is timestamp is 2 for r1(a) then why taking 1, similarly, so on.  i m not getting this. please explain!

Answer 2

Given information will be represented as in figure above. Now TS to a data item is assigned as the TS of latest transaction which accessed it. So RTS(a)= TS(S1)=1

WTS(a) = TS(S3) = 7, RTS(b) = TS(S2) = 3, WTS(b) = TS(S3)= 7 and thus option D is right answer.

"We can do it directly from given table also, then no need to make above table"

Comments

Thank you so much..  Now I got it.