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Database Management SystemConsider the three transactions T1, T2, and T3, and the schedules S1 and S2 given below. State whether each schedule is serializable or not. If a schedule is serializable, write down the equivalent serial schedule(s). T1: r1 (X); r1 (Z); w1 (X); T2: r2 (Z); r2 (Y); w2 (Z); w2 (Y); T3: r3 (X); r3 (Y); w3 (Y); S1: r1 (X); r2 (Z); r1 (Z); r3 (X); r3 (Y); w1 (X); w3 (Y); r2 (Y); w2 (Z); w2 (Y); S2: r1 (X); r2 (Z); r3 (X); r1 (Z); r2 (Y); r3 (Y); w1 (X); w2 (Z); w3 (Y); w2 (Y);20247mDatabase Management SystemConsider the transactions T1, T2 and T3 and schedules S1 and S2 given below: T1 : r1 (X); r1(Z); w1 (X); w1 (Z), T2 : r2 (Y); r2 (Z); w2 (Z) T3 : r3 (Y); r3 (X); w3 (Y) S1 : r1 (X); r3(Y); r3(X); r2 (Y); r2(Z); w3(Y); w2 (Z); r1 (Z); w1(X), w1 (Z). S2 : r1 (X); r3 (Y); r2(Y); r3(X); r1(Z); r2(Z); w3 (Y); w1 (X); w2(Z), w1 (Z). Determine which of the above schedules are conflict – serializable.20227mDatabase Management SystemExplain schedule and transaction. Define the concepts of recoverable, cascade less, and strict schedules, and compare them in terms of their recoverability.20247mDatabase Management SystemDefine serializability. Explain conflict and view serializability.20257m
PreviousFor a B-tree of order m, every non-root internal node must have at least: (i) m/2 children (ii) [m/2] children (iii) [m/2]-1 children (iv) m-1 childrenNextUnder strict two-phase locking: (i) Exclusive locks are held until commit/abort (ii) Deadlocks are impossible (iii) Locks can be released anytime (iv) Transactions become serial
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