Consider the database transactions T1 and T2, and data items X and Y. Which of the schedule(s) is/are conflict serializable?

To determine whether a schedule is conflict serializable, we construct the precedence (serializability) graph for each schedule:

1. Option (A):

   $$\text{R}_1(X), \text{W}_2(X), \text{W}_1(Y), \text{W}_2(Y), \text{R}_1(X), \text{W}_1(X), \text{COMMIT}(T2), \text{COMMIT}(T1)$$

   • $\text{R}_1(X)$ occurs before $\text{W}_2(X) \implies T_1 \to T_2$

   • $\text{W}_2(X)$ occurs before $\text{R}_1(X)$ and $\text{W}_1(X) \implies T_2 \to T_1$

   • There is a cycle ($T_1 \to T_2 \to T_1$), so the schedule is not conflict serializable.

2. Option (B):

   $$\text{W}_2(X), \text{R}_1(X), \text{W}_2(Y), \text{W}_1(Y), \text{R}_1(X), \text{COMMIT}(T2), \text{W}_1(X), \text{COMMIT}(T1)$$

   • $\text{W}_2(X)$ occurs before $\text{R}_1(X)$ and $\text{W}_1(X) \implies T_2 \to T_1$

   • $\text{W}_2(Y)$ occurs before $\text{W}_1(Y) \implies T_2 \to T_1$

   • There is no conflicting pair where $T_1$ precedes $T_2$. The graph has only the edge $T_2 \to T_1$, which is acyclic. Thus, this schedule is conflict serializable.

3. Option (C):

   $$\text{R}_1(X), \text{W}_1(Y), \text{W}_2(X), \text{W}_2(Y), \text{R}_1(X), \text{W}_1(X), \text{COMMIT}(T1), \text{COMMIT}(T2)$$

   • $\text{R}_1(X)$ occurs before $\text{W}_2(X) \implies T_1 \to T_2$

   • $\text{W}_2(X)$ occurs before $\text{R}_1(X)$ and $\text{W}_1(X) \implies T_2 \to T_1$

   • There is a cycle ($T_1 \to T_2 \to T_1$), so the schedule is not conflict serializable.

4. Option (D):

   $$\text{W}_2(X), \text{R}_1(X), \text{W}_1(Y), \text{W}_2(Y), \text{R}_1(X), \text{COMMIT}(T2), \text{W}_1(X), \text{COMMIT}(T1)$$

   • $\text{W}_2(X)$ occurs before $\text{R}_1(X)$ and $\text{W}_1(X) \implies T_2 \to T_1$

   • $\text{W}_1(Y)$ occurs before $\text{W}_2(Y) \implies T_1 \to T_2$

   • There is a cycle ($T_1 \to T_2 \to T_1$), so the schedule is not conflict serializable.

Hence, only option (B) is conflict serializable.