SSC IMD SA Mock 04 – Question 147

Which of the following is a conflict serializable schedule?
\(S_1: R_1(A); R_1(B); R_2(A); R_2(B); W_2(B); W_1(A)\)
\(S_2: R_1(A); R_2(A); R_2(B); W_2(B); R_1(B); W_1(A)\)

• S1 is conflict serializable but S2 is not.
• S1 is not conflict serializable but S2 is conflict serializable.
• Both are not conflict serializable.
• Both are conflict serializable.

To test conflict serializability, we inspect the precedence graph for cycles:
– **For schedule \(S_1\)**:
– Item \(B\): \(R_1(B)\) comes before \(W_2(B)\), creating directed edge \(T_1 \to T_2\).
– Item \(A\): \(R_2(A)\) comes before \(W_1(A)\), creating directed edge \(T_2 \to T_1\).
– Because there is a cycle (\(T_1 \leftrightarrow T_2\)), \(S_1\) is **NOT conflict serializable**.
– **For schedule \(S_2\)**:
– Item \(A\): \(R_2(A)\) comes before \(W_1(A)\), creating edge \(T_2 \to T_1\).
– Item \(B\): \(W_2(B)\) comes before \(R_1(B)\), creating edge \(T_2 \to T_1\).
– The graph has no cycle (only \(T_2 \to T_1\)), so \(S_2\) is **conflict serializable** (equivalent to serial schedule \(T_2 \to T_1\)).