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 $read(P)$ and the write operation on data item $P$ is denoted by $write(P)$. Suppose that the transaction $T1$ fails immediately after time instance 9. Which one of the following statements is correct?

In the given schedule $S$:

• At time instance 2, transaction $T1$ performs $\text{write}(A)$.

• At time instance 7, transaction $T2$ performs $\text{read}(A)$, reading the uncommitted value written by $T1$.

• At time instance 8, transaction $T2$ performs $\text{commit}$.

A schedule is defined as recoverable if, whenever transaction $T_j$ reads a value written by transaction $T_i$, the commit operation of $T_i$ appears before the commit operation of $T_j$.

Here, $T2$ reads from uncommitted transaction $T1$ and commits at time 8 while $T1$ has not yet committed. When $T1$ fails after time instance 9, it must be aborted and its updates rolled back to maintain atomicity. However, because $T2$ has already committed (durability), its actions cannot be undone (cannot be rolled back).

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