The correct answer is option 1. Concept: Conflict-equivalent schedule: The two schedules S1 and S2 are said to be conflict equivalent if all the conflicts in both S1 and S2 must be the same. Test for Conflict-equivalent: Step 1: Construct a precedence graph where each transaction represents one vertice. And each conflicting operation represents one edge. i.e If from transaction Ti to Tj then draw an edge from vertices Ti to Tj. Step 2: If the precedence graph contains no cycles then the schedule is said to be conflict serializable. Step 3: The order of serializability is determined based on the topological sort of the directed graph. Explanation: Given schedule S with four transactions, T1 T2 T3 T4 R(x) R(X) R(X) R(Y) W(Y) W(X) W(Y) R(Y) [ alt="F1 Madhuri Engineering 28.03.2022 D33" src="//storage.googleapis.com/tb-img/production/22/03/F1_Madhuri_Engineering_28.03.2022_D33.png" style="width: 150px; height: 151px;"> The order of serializability is determined based on the topological sort of the directed graph. For a Directed Acyclic Graph (DAG), topological sorting is a linear ordering of vertices in which vertex u occurs before v in the ordering for any directed edge u v. If the graph is not a DAG, topological sorting is not feasible. [ alt="F1 Madhuri Engineering 28.03.2022 D34" src="//storage.googleapis.com/tb-img/production/22/03/F1_Madhuri_Engineering_28.03.2022_D34.png" style="width: 401px; height: 223px;"> Serial schedules are conflict equivalent to S: T1 → T3 → T4 → T2 Hence the correct answer is T1 → T3 → T4 → T2