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+2 votes
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in Databases by Active (2k points) | 155 views
0
c recoverable.
0
Ans is D which is Also right but I didn't understand why not recoverable?
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there is not write - > read pair so no posibility of dirty read. yes d is also correct.
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Yes, Both are looking right but which would be best?

2 Answers

+1 vote
Answer is D .

Draw precedence graph there is a cycle so it is not serializable .

There is no dirty read here so it is better to go for D
by Active (1.4k points)
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cycle in precedence graph only indicates that schedule is not conflict serializable. However the schedule can still be view serializable.
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answer is C as polygraph contains cycle. T1 does initial read on X so T1 should come in the order. again, T1 does final write on Y so T1 must be last in the order making a cycle

the schedule is recoverable as if T1 aborts after writing on Y, T2 and T3 will rollback too. there s cascading rollback. cascading rollback ensures consistency in database and recoverability
0 votes
Ans is D.

1) cycle present . T1 to T2(R1x->W2x), and T2 to T1(W2y->W1y) so not conflict serializable

2)Since T2 writes Y first followed by T3 and T1 hence the schedule to be recoverable T2 must commit first followed by T3 and T1. but here T1 commits first then T2 and T3. So schedule not recoverable. Here as T1 commits before T2 , suppose before T2 commits,T2 fails, then as per atomicity property of DBMS T2 will rollback but since T1 committed already and T1 can not rollback (durability property), but T1 has modified a common data object i.e, y ,but as T2 rollbacks all changes on the data objects are undone and hence updates of T1 is lost and unrecoverable.

3)here is blind write T2 to T3 and T2 to T1 and T3 to T1. So schedule may be view serializable. so checking initial read , done by T1. No updated read in between, final read is done by T1 too. now same transaction initial and final read is not possible in serial sequence. so schedule is not view serializable.

Hence Schedule is not serializable.
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