GATE CS 2003 Q87 — Transactions and Concurrency Control

To determine if the given schedule is serializable, we construct its precedence (dependency) graph. A directed edge $T_i \to T_j$ exists if there is a conflict between an operation in $T_i$ and a subsequent operation in $T_j$ on the same data item (where at least one of the operations is a write).

1. Identify Conflicting Operations

• On Data Item $D2$:
  • $T2: W(D2)$ occurs before $T3: R(D2)$ $\implies T2 \to T3$
  • $T3: W(D2)$ occurs before $T1: R(D2)$ $\implies T3 \to T1$
• On Data Item $D1$:
  • $T1: W(D1)$ occurs before $T2: R(D1)$ $\implies T1 \to T2$
• On Data Item $D3$:
  • $T2: R(D3)$ occurs before $T3: W(D3)$ $\implies T2 \to T3$ (already captured)

2. Construct the Precedence Graph

Combining the dependency edges, we get:

The precedence graph contains a cycle: $T1 \to T2 \to T3 \to T1$.

Conclusion

Since the precedence graph contains a cycle, the schedule is not conflict serializable. Furthermore, because there are no blind writes (every write is preceded by a read on the same data item by the same transaction), conflict serializability is equivalent to view serializability. Thus, the schedule is not serializable at all.

Therefore, the correct option is D.