Schedule S is R2(Y),R1(X),R3(Z),R1(Y),W1(X),R2(Z),W2(Y),R3(X),W3(Z). P: S is conflict-serializable. Q: If T3 commits before T1 finishes, S is recoverable. Choose the truth values.
Conflicts on Y give T1→T2 because R1(Y) precedes W2(Y). On X, W1(X) precedes R3(X), giving T1→T3. On Z, R2(Z) precedes W3(Z), giving T2→T3. The graph is acyclic, with order T1,T2,T3, so P is true. T3 reads X from T1; committing T3 before T1 violates recoverability. Thus Q is false, B.
More Transactions and Concurrency Control questions
GATE 2026 CS (CS2) Q20– Consider concurrent execution of two transactions T 1 and T 2 in a DBMS, both of which…
GATE 2025 CS (CS1) Q15– A schedule of three database transactions T 1, T 2, and T 3 is shown. R i(A) and W i(A)…
GATE 2025 CS (CS2) Q27– An audit of a banking transactions system has found that on an earlier occasion, two joint…
GATE 2025 CS (CS2) Q53– Consider the database transactions T1 and T2, and data items X and Y. Which of the…
GATE 2024 CS (CS1) Q46– Consider the read-write schedule S over three transactions T 1,T 2,T 3: S: r 1(z); w 1(z);…
GATE 2024 CS (CS2) Q19– Once the DBMS informs the user that a transaction has been successfully completed, its…