transaction

Question

S:R1(x),w1(x),R2(x),R1(y),w2(x),C2,A1

why it is violating Ti->Tj condition?

Comments

i dont think it is violating any such condition...if there was a loop between T1 and T2 then the condition would have been violated....but we are getting a serialisable schedule from T1->T2...

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please visit this link

https://www.youtube.com/watch?v=Wa_z_8q0I8Y

lecture time 13.40...question no 5..

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ohh ohkk...u wre asking about recoverability..

then yes it is violating..

see T2 is reading x from T1..and it is getting commited before T1..so it is not recoverable....if u r reading from a transition then let that gets commited first then u commit..but it is not happening here...i hope u got it..

why actually this is not right...because see T1 is getting aborted after some time.....so whatever T2 has read from it earlier before getting commited is a fault now.....so if u r reading from something let that commit first else ur schedule will not be recoverable..

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firstly m sorry for not completing my question...

yeah got it.. thank,..but in the line  "whatever T2 has read from it earlier before getting commited is a fault now.....so if u r reading from something let that commit first else ur schedule will not be recoverable.."i m stuck here..

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means like if u r taking some food from a stranger(reading from someone)..then what will u do to be safe....let him eat first(let that transition commit first)...then u can be sure and can eat(now u commit)..else u may die(not recoverable).....got it

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yeah got it..thnx bro

Answer 1

2nd transaction reads the value that is updated by the first transaction and commit of second transaction is done first.

if there is any failure ocurs first tranaction is rollback but 2nd transaction we cannot because it is already commited.

so it is irrecoverable schdeule.

so commit of 2nd transaction is delayed until commit or rollback of first transaction