Consider the following schedule $\mathbf{S}$ of transactions T1, T2, T3, T4: Which one of the following statements is CORRECT?

Let us analyze the given schedule $\mathbf{S}$ for conflict serializability and recoverability:

### 1. Conflict Serializability Construct the precedence graph by identifying conflicting operations (operations on the same data item by different transactions where at least one is a write):

• On data item $X$:

  • $R_2(X)$ occurs before $W_3(X) \implies T_2 \to T_3$

  • $R_2(X)$ occurs before $W_1(X) \implies T_2 \to T_1$

  • $W_3(X)$ occurs before $W_1(X) \implies T_3 \to T_1$

  • $W_1(X)$ occurs before $R_4(X) \implies T_1 \to T_4$

• On data item $Y$:

  • $W_2(Y)$ occurs before $R_4(Y) \implies T_2 \to T_4$

• On data item $Z$:

  • Only $R_2(Z)$ is present, so there are no conflicts.

The resulting precedence graph has edges:

Since there are no cycles in the precedence graph, the schedule is conflict-serializable (equivalent serial schedule: $T_2 \to T_3 \to T_1 \to T_4$).

### 2. Recoverability A schedule is recoverable if every transaction $T_j$ that reads a value written by transaction $T_i$ commits only after $T_i$ commits.

• $T_4$ reads $X$ written by $T_1$, and $T_1$ commits before $T_4$ commits.

• $T_4$ reads $Y$ written by $T_2$, and $T_2$ commits before $T_4$ commits.

• All other reads are from initial database values.

Since all write-read dependencies commit in the proper order, the schedule is recoverable.

Therefore, $\mathbf{S}$ is both conflict-serializable and recoverable.

Thus, (C) is the correct option.