Consider the following interleaved schedule S involving transaction T₁, T₂ on data item X and Y.
S: R₁(X); W₁(X); R₂(X); W₂(X); R₂(Y); W₂(Y); R₁(Y); W₁(Y); C₂, C₁
Where R(X|Y); W(X|Y) represents read and write operations on items X and Y and C₁/₂ are commits of T₁ and T₂ respectively.
A. S is neither conflict-, nor view-serializable, non-serializable schedule.
B. S is serializable, but some data values of X and Y leads T₁ and T₂ into an inconsistent state, due to interleaving of transaction.
C. S is not serializable due to dirty read and cycle in the precedence graph.
D. S is serializable and consistent for all data items X and Y.
D is the correct answer. but why?
