Given a schedule S for transactions T1 and T2 with set of read and write operations, S: R1(X) R2(X) R2(Y) W2(Y) R1(Y) W1(X). Identify, whether given schedule is equivalent to serial schedule or not?

To determine if a given schedule is equivalent to a serial schedule, we need to check if it preserves the order of conflicting operations in the transactions. Conflicting operations are pairs of operations that access the same data item and at least one of them is a write operation.

Let's analyze the given schedule S: R1(X) R2(X) R2(Y) W2(Y) R1(Y) W1(X).

The conflicting operations are:

• R1(X) and R2(X) because they both read the same data item X.
• R1(X) and W1(X) because they both access the same data item X and one of them is a write operation.
• R2(X) and W1(X) because they both access the same data item X and one of them is a write operation.
• R2(Y) and W2(Y) because they both access the same data item Y and one of them is a write operation.
• R1(Y) and W2(Y) because they both access the same data item Y and one of them is a write operation.

Now, let's check if the order of these conflicting operations is preserved in the given schedule S.

• The order of R1(X) and R2(X) is preserved because R1(X) comes before R2(X) in the schedule S.
• The order of R1(X) and W1(X) is preserved because R1(X) comes before W1(X) in the schedule S.
• The order of R2(X) and W1(X) is not preserved because R2(X) comes before W1(X) in the schedule S, but in the transaction T2, the write operation W1(X) comes before the read operation R2(X).
• The order of R2(Y) and W2(Y) is preserved because R2(Y) comes before W2(Y) in the schedule S.
• The order of R1(Y) and W2(Y) is not preserved because R1(Y) comes after W2(Y) in the schedule S, but in the transaction T1, the read operation R1(Y) comes before the write operation W2(Y).

Therefore, the given schedule S is not equivalent to a serial schedule because it does not preserve the order of all conflicting operations in the transactions.