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Q5: In DBMS, a graph precedence is used to test whether a schedule of concurrent transactions is conflict serializable or not

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Answer #1

1.

Using the given schedule, the following points are considered for the required precedence graph :

  • There is a read write conflict on data item A from transaction T2 to T3 and so there will be an edge from T2 to T3.
  • There is a read write, write read and write write conflict on data item B from transaction T1 to T2 and so there will be an edge from T1 to T2.

So, the required precedence graph is :

Т1 Т2 T3

2.

The schedule S is conflict serializable because there are no cycles formed in the precedence graph.

3.

In order to construct the serial schedule S ', topological sorting has to be performed on the precedence graph as follows :

  • The vertex with indegree 0 in the precedence graph is T1.
  • So, the first transaction in the serial order S ' is T1.
  • Remove the vertex T1 from the precedence graph.
  • Now the vertex with indegree 0 in the precedence graph is T2.
  • So, the next transaction in the serial order S ' is T2.
  • Remove the vertex T2 from the precedence graph.
  • The next transaction in the serial order S ' is T3 as there are no more vertices.

So, the serial order S ' is :

T1 ----> T2 ------> T3.

Hence, it is possible to construct a serial schedule S ' that is conflict equivalent to the schedule S.

4.

In the given schedule, there is a dirty read from transaction T2 to transaction T3 on data item A. In addition to this, the transaction T3 commits before the transaction T2.

For this reason, the given schedule is not recoverable.

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