GATE CS 2015 Set 3 Q.39 (CS Q.29): Transactions, Serializability and Concurrency Control

Consider the following partial Schedule $S$ involving two transactions $T1$ and $T2$ . Only the read and the write operations have been shown. The read operation on data item $P$ is denoted by $\text{read}(P)$ and the write operation on data item $P$ is denoted by $\text{write}(P)$ .

Time instance	T1	T2
1	read(A)
2	write(A)
3		read(C)
4		write(C)
5		read(B)
6		write(B)
7		read(A)
8		commit
9	read(B)

Suppose that the transaction $T1$ fails immediately after time instance 9. Which one of the following statements is correct?

$T2$ must be aborted and then both $T1$ and $T2$ must be re-started to ensure transaction atomicity

Schedule $S$ is non-recoverable and cannot ensure transaction atomicity

Only $T2$ must be aborted and then re-started to ensure transaction atomicity

Schedule $S$ is recoverable and can ensure atomicity and nothing else needs to be done

Official answer and solution

Official answer: B
From: Official GATE 2015 answer key

Solution

Inspect the schedule for uncommitted reads and the commit ordering of participating transactions.

Examine the operations in Schedule $S$ :

Time 2: $T1$ writes to data item $A$ (write(A)).
Time 7: $T2$ reads data item $A$ (read(A)), which was modified by $T1$ .
This is a dirty read because $T1$ has not yet committed.
Time 8: $T2$ commits (commit).
Time 9: $T1$ reads data item $B$ .
After time 9: $T1$ fails.

A schedule is recoverable if and only if, whenever a transaction $T_j$ reads a value written by $T_i$ , $T_i$ commits before $T_j$ commits.

Here, $T2$ reads $A$ written by $T1$ and commits at time 8, while $T1$ has not committed and subsequently fails after time 9.
Because $T2$ has already committed, its effects are durable (ACID Durability) and it cannot be rolled back or aborted.
Consequently, transaction atomicity for $T1$ cannot be maintained without violating $T2$ 's durability.

Therefore, Schedule $S$ is non-recoverable and cannot ensure transaction atomicity (Option B).

Final answer

Common trap

Believing T2 can simply be aborted, forgetting that once a transaction commits (time 8), DBMS durability rules forbid aborting or rolling it back.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 3
Question: 39 of 65, printed as CS Q.29
Section: Computer Science and Information Technology

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