GATE CS 2015 Set 3 Q29 — Transactions and Concurrency Control

To determine the correct statement, let us analyze the dependencies and recoverability of the given schedule $S$:

1. Identify the Write-Read Dependency (Dirty Read):
   • At time instance 2, transaction $T_1$ writes to data item $A$ (write(A)).
   • At time instance 7, transaction $T_2$ reads data item $A$ (read(A)).
   • Since $T_2$ reads a value of $A$ written by $T_1$ before $T_1$ has committed, $T_2$ has a dirty read dependency on $T_1$ ($T_1 \to T_2$).
2. Recoverability Condition:

   A schedule is recoverable if, for every pair of transactions $T_i$ and $T_j$ such that $T_j$ reads a data item previously written by $T_i$, the commit operation of $T_i$ appears before the commit operation of $T_j$.

   $$\text{Commit}(T_1) < \text{Commit}(T_2)$$
3. Checking the Schedule:
   • $T_2$ commits at time instance 8.
   • $T_1$ has not committed by time instance 8 (and fails immediately after time instance 9).
   • Since $T_2$ commits before $T_1$, the recoverability condition is violated. Thus, schedule $S$ is non-recoverable.
4. Impact of Failure:

   When $T_1$ fails after time instance 9, it must be aborted and rolled back. However, $T_2$ has already committed at time instance 8. By the Durability property of ACID, a committed transaction's effects cannot be undone. Since we cannot abort $T_2$ to undo the dirty read, transaction atomicity cannot be ensured.

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