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Consider the following schedule S of transactions T1, T2, T3, T4: T1 T2 T3 T4   Reads(X)         Writes(X) Commit   Writes(X)  Commit         Writes(Y) Reads(Z) Commit           Reads(X) Reads(Y) Commit   Which one of the following statements is CORRECT?

Consider the following schedule S of transactions T1, T2, T3, T4: T1 T2 T3 T4   Reads(X)         Writes(X) Commit   Writes(X)  Commit         Writes(Y) Reads(Z) Commit           Reads(X) Reads(Y) Commit   Which one of the following statements is CORRECT? Correct Answer S is both conflict-serializable and recoverable

Concept:

  • Using precedence graph method we can find out Schedule is conflict serializable or not.

  • If precedence graph has any loop then that schedule is going to be not conflict-serializable.

  • If precedence graph doesn’t have any loop then that schedule is going to be conflict-serializable.

  • If one transaction ( T1 )is writing some value x and that value is read by some other transaction(T2) then T1 should commit first then only T2 should read that value and commit in order to become recoverable schedule.


Explanation:

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  • There is no loop in this graph so schedule is conflict serializable.
  • T2: T3: T1: T4 is the only serial schedule that is equivalent to given schedule. (Using Topological Sorting)
  • There are two write read dependency
  1. T1 write(X) and T4 read (X)
  2. T2 write (Y) and T4 read (Y)
  • In first case T1 is committed first then read by T4 then T4 is committed so this is recoverable.
  • In second case T2 is committed first then read by T4 then T4 is committed so this is also recoverable so schedule S is recoverable.


Hence option 3 is the correct answer.

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