Question 39

Chronological Sequence of Operations:

1. $T_2: \text{Read}(X)$
2. $T_3: \text{Write}(X)$
3. $T_3: \text{Commit}$
4. $T_1: \text{Write}(X)$
5. $T_1: \text{Commit}$
6. $T_2: \text{Write}(Y)$
7. $T_2: \text{Read}(Z)$
8. $T_2: \text{Commit}$
9. $T_4: \text{Read}(X)$
10. $T_4: \text{Read}(Y)$
11. $T_4: \text{Commit}$

Conflict Serializability (Precedence Graph):

• Conflict pairs on $X$:
  • $R_2(X)$ before $W_3(X) \implies T_2 \to T_3$.
  • $R_2(X)$ before $W_1(X) \implies T_2 \to T_1$.
  • $W_3(X)$ before $W_1(X) \implies T_3 \to T_1$.
  • $W_1(X)$ before $R_4(X) \implies T_1 \to T_4$.
  • $W_3(X)$ before $R_4(X) \implies T_3 \to T_4$.
• Conflict pairs on $Y$:
  • $W_2(Y)$ before $R_4(Y) \implies T_2 \to T_4$.
• All directed edges: $T_2 \to T_3 \to T_1 \to T_4$. There are no cycles in the graph, so schedule $S$ is conflict-serializable (equivalent serial order $T_2, T_3, T_1, T_4$).

Recoverability Check:

• A schedule is recoverable if every transaction $T_j$ that reads data written by $T_i$ commits only after $T_i$ commits.
• $T_4$ reads $X$ written by $T_1$: $T_1$ commits at step 5, and $T_4$ commits at step 11 ($T_1$ commits before $T_4$).
• $T_4$ reads $Y$ written by $T_2$: $T_2$ commits at step 8, and $T_4$ commits at step 11 ($T_2$ commits before $T_4$).
• No other dirty reads exist. Hence, schedule $S$ is recoverable.

Conclusion:
$S$ is both conflict-serializable and recoverable, which corresponds to option (C).