Test: Transactions & Concurrency Control- 3 - Computer Science Engineering (CSE) MCQ

T1 can complete before T2 and T3 as there is no conflict between Write(X) of T1 and the operations in T2 and T3 which occur before Write(X) of T1 in the above diagram.
T3 should can complete before T2 as the Read(Y) of T3 doesn’t conflict with Read(Y) of T2. Similarly, Write(X) of T3 doesn’t conflict with Read(Y) and Write(Y) operations of T2.
Another way to solve this question is to create a dependency graph and topologically sort the dependency graph. After topologically sorting, we can see the sequence T1, T3, T2.

To check for conflict-serializable, we need to make a precedence graph, if the graph contains a cycle, then it\'s not conflict serializable, else it is. Here, for the precedence graph there will be only two directed edges, one from T2 -> T3 ( Read- Write Conflict), and another from T2 -> T1( Read- Write Conflict), hence no cycle, so the schedule is conflict serializable. Now to check for Recoverable, we need to check for a dirty-read operation( Write by Transaction Ti, followed by Read by Transaction Tj but before Ti commits) between any pair of operations. If no dirty-read then recoverable schedule, if a dirty read is there then we need to check for commit operations. Here no dirty read operation ( as T3 and T1 commits before T4 reads the Write(X) of T3 and T1 , and T2 commits before T4 reads the Write(Y) of T2 ). Therefore the schedule is recoverable. Hence, Option C.

We must undo log record 6 to set B to 10000 and then redo log records 2 and 3 bcoz system fail before commit operation. So we need to undone active transactions(T2) and redo committed transactions (T1) Note: Here we are not using checkpoints. Checkpoint : Checkpoint is a mechanism where all the previous logs are removed from the system and stored permanently in a storage disk. Checkpoint declares a point before which the DBMS was in consistent state, and all the transactions were committed. Recovery: When a system with concurrent transactions crashes and recovers, it behaves in the following manner − =>The recovery system reads the logs backwards from the end to the last checkpoint. =>It maintains two lists, an undo-list and a redo-list. =>If the recovery system sees a log with and or just , it puts the transaction in the redo-list. =>If the recovery system sees a log with but no commit or abort log found, it puts the transaction in undo-list. All the transactions in the undo-list are then undone and their logs are removed. All the transactions in the redo-list and their previous logs are removed and then redone before saving their logs. So Answer is B redo log records 2 and 3 and undo log record 6

T1 and T2 have conflicting operations between them forming a cycle in the precedence graph. R(D2) of T2, and W(D2) of T1 ( Read-Write Conflict) R(D1) of T1, and W(D1) of T2 ( Read-Write Conflict) Hence in the precedence graph of the schedule there would be a cycle between T1 and T2 vertices. Therefore not a serializable schedule.

Consistency in database systems refers to the requirement that any given database transaction must only change affected data in allowed ways, that is sum of x and y must not change.

In option D, there is no interleaving of operations. The option D has first all operations of transaction 2, then 3 and finally 1 There can not be any conflict as it is a serial schedule with sequence 2 --> 3 -- > 1

For conflict serializability of a schedule( which gives same effect as a serial schedule ) we should check for conflict operations, which are Read-Write, Write-Read and Write-Write between each pair of transactions, and based on those conflicts we make a precedence graph, if the graph contains a cycle, it\'s not a conflict serializable schedule. To make a precedence graph: if Read(X) in Ti followed by Write(X) in Tj ( hence a conflict ), then we draw an edge from Ti to Tj ( Ti -> Tj) If we make a precedence graph for S1 and S2 , we would get directed edges for S1 as T2->T1, T2->T3, T3->T1, and for S2 as T2->T1, T2->T3, T3->T1, T1->T2. In S1 there is no cycle, but S2 has a cycle. Hence only S1 is conflict serializable. Note : The serial order for S1 is T2 -> T3 -> T1.

Since there is cycle found in precedence graph, it is not Conflict Serializable. Also, there is a blind write in T3 so we need to test further for view serializability using polygraph. From polygraph, we can derive that it is View serializable. Option (C) is Correct.

Option C is a normal operation. Option B is also fine as no write operation is involved. Option A can be recovered, but option D can\'t be. See this for an example.

Since T1 and T3 are not committed yet, they must be undone. The transaction T2 must be redone because it is after the latest checkpoint.