Answer to Question #263370 in Databases | SQL | Oracle | MS Access for Ravi kumar pandey

Question

Consider the following schedule of transactions

R1(A) W1(A) R2(A) W2(B) W1(B) Commit1 Commit2

Which of the following properties are true for the above schedule and justify the reason for your answer

with explanation?

a) Conflict Serializable

b) View Serializable

c) Recoverable

Answer 1

SOLUTION TO THE ABOVE QUESTION

ANSWER

a) Conflict Serializable

ANSWER JUSTIFICATION.

This is because the given schedule can be converted to a serial schedule

by swapping non-conflicting pairs

Lets convert it to a serial schedule

S: R1(A), W1(A), R2(A), W2(B), W1(B)

we can get two transactions of schedule S as: 

T1: R1(A), W1(A), W1(B)

T2: R2(A), W2(B)

Lets swap the non-conflicting pair R2(A) and W1(B)

therefore our schedule becomes

S1: R1(A), W1(A), W1(B), W2(B), R2(A)

S1 is a serial schedule in which all operations of T1 are performed before starting any operation of T2. Since S has been transformed into a serial schedule S1 by swapping non-conflicting operations of S, S is conflict serializable.