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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 SystemDefine serializability. Explain conflict and view serializability.20257mDATABASE MANAGEMENT SYSTEMSerializability of concurrent transaction is ensured by - (i) locking - (ii) time stamping - (iii) both (a) and (b) - (iv) none of these20232m
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