Test: Transactions & Concurrency Control- 1 - Computer Science Engineering (CSE) MCQ

2 Phase Locking (2PL) is a concurrency control method that guarantees serializability. The protocol utilizes locks, applied by a transaction to data, which may block (interpreted as signals to stop) other transactions from accessing the same data during the transaction’s life. 2PL may be lead to deadlocks that result from the mutual blocking of two or more transactions. See the following situation, neither T3 nor T4 can make progress.

Timestamp-based concurrency control algorithm is a non-lock concurrency control method. In Timestamp based method, deadlock cannot occur as no transaction ever waits.

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.

CONCEPT

A schedule is called conflict serializable if it can be transformed into a serial schedule by swapping non-conflicting operations.

For a schedule to be conflict serializable, there should be no cycle present in its precedence graph.

EXPLANATION: 

Option_1 – Not conflict serializable

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.

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.

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

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.

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.

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.

if transaction fails, atomicity requires effect of transaction to be undone. Durability states that once transaction commits, its change cannot be undone (without running another, compensating, transaction). Recoverable schedule: A schedules exactly where, for every set of transaction Ti and Tj. If Tj reads a data items previously written by Ti, then the commit operation of Ti precedes the commit operation of Tj. Aborting involves undoing the operations and redoing them since by the time stamp it is aborted.
Option (A): T2 must be aborted and then both T1 and T2 must be re-started to ensure transaction atomicity. It is incorrect because it says abort transaction T2 and then redo all the operations. But there is no gaurantee that it will succeed this time as again T1 may be fail.
Option(B): Schedule S is non-recoverable and cannot ensure transaction atomicity. Correct, it is by definition an irrecoverable schedule so now even if we start to undo the actions one by one(after t1 fails) in order to ensure transaction atomicity. Still we cannot undo a commited transaction. hence this schedule is irrecoverable by definition and also not atomic since it leaves the database in an inconsistent state. Simply dirty read so nonrecoverable.
Option (C): Only T2 must be aborted and then re-started to ensure transaction atomicity. It is incorrect because it says abort only transaction T2 and then redo all the T2 operations. But this is dirty read problem as it is reading the data item A which is written by T1 and T1 is not committed. Again it will be the dirty read problem. So incorrect.
Option (D): Schedule S is recoverable and can ensure transaction atomicity and nothing else needs to be done. Incorrect, it is clearly saying that schedule s is recoverable but it is irrecoverable because T2 read the data item A which is written by T1 and T1 failed and rollback, at the rollback T1 start undo all operations and modified the value of A with previous value but T2 is already committed so T2 can't change the read value of A which was earlier taken from T1.

• Page level locking locks whole page i.e all rows therefore highly restrictive
• Table locking is mainly used for concurrency control with DDL operations
• A row share table lock is the least restrictive, and has the highest degree of concurrency for a table.It indicates the transaction has locked rows in the table and intends to update them.

Therefore Row level provides highest level of concurrency.Hence Answer is C

D refers to Durability.

The above scenario is Conservative 2PL(or Static 2PL). In Conservative 2PL protocol, a transaction has to lock all the items it access before the transaction begins execution. It is used to avoid deadlocks. Also, 2PL is  conflict serializable, therefore it guarantees serializability. Therefore option A Advantages of Conservative 2PL :

• No possibility of deadlock.
• Ensure serializability.

Drawbacks of Conservative 2PL :

• Less throughput and resource utilisation because it holds the resources before the transaction begins execution.
• Starvation is possible since no restriction on unlock operation.
• 2pl is a deadlock free protocol but it is difficult to use in practice.

Cycle in precedence graph tells that schedule is not conflict serializable. DFS and BFS traversal of graph are possible even if graph contains cycle. And hence DFS and BFS are also possible for non serializable graphs. But Topological sort of any cyclic graph is not possible. Thus topological sort guarantees graph to be serializable. Option D is not valid because in a transaction with more indices might have to come before lower one. Also two non- conflicting schedule can occur simultaneously.

As we can see in figure,

• T2 overwrites a value that T1 writes
• T1 aborts: its “remembered” values are restored.
• Cascading Abort could have arised if - > Abort of T1 required abort of T2 but as T2 is already aborted , its not a cascade abort. Therefore, Option C

Option A - is not true because the given schedule is recoverable Option B - is not true as it is recoverable and avoid cascading aborts; Option D - is not true because T2 is also doing abort operation after T1 does, so NOT strict.

All the conflicting pairs like (3WA, 1WA) are in the same order in both S1 and S2.

Basic two phase locking protocol ensures only conflict serializability and strict two phase locking protocol ensures recoverability as well. So statement A is correct. Checkpoints are inserted to minimize the task of undo-redo in recoverability. So, statement B is not correct. Hence correct option B is correct choice.