Question

Let $R(z)$ and $W(z)$ denote read and write operations on a data element $z$ by a transaction $T_i$, respectively. Consider the schedule $S$ with four transactions.
\[ S: R_1(x) R_2(x) R_3(y) W_1(y) W_2(x) W_3(y) R_4(y) \] Which one of the following serial schedules is conflict equivalent to $S$?

• A. $T_1 \rightarrow T_3 \rightarrow T_4 \rightarrow T_2$
• B. $T_1 \rightarrow T_4 \rightarrow T_3 \rightarrow T_2$
• C. $T_4 \rightarrow T_1 \rightarrow T_3 \rightarrow T_2$
• D. $T_3 \rightarrow T_1 \rightarrow T_4 \rightarrow T_2$

Hint:

A conflict-equivalent schedule maintains the same order of dependent operations, ensuring the same execution semantics.

Answer 1

The Correct Option is A

To determine the conflict equivalent schedule, we analyze the dependencies between the operations on the data elements: - $R_1(x)$ and $R_2(x)$ are read operations on the same variable, so their order does not matter. Similarly, the operations $W_2(x)$ and $R_4(y)$ do not conflict. - $R_3(y)$, $W_1(y)$, and $W_3(y)$ form a chain of dependent operations, so the order of these operations is important. By examining the schedule and the dependencies: - $T_1$ should appear before $T_2$ because $W_2(x)$ and $R_4(y)$ create dependencies. - $T_3$ can appear before $T_2$, but after $T_1$, as $W_1(y)$ depends on $R_3(y)$. - $T_4$ can be placed at the end because $R_4(y)$ is the last operation involving $y$. The order $T_1 \rightarrow T_3 \rightarrow T_4 \rightarrow T_2$ satisfies all the dependencies and is thus the conflict equivalent schedule. Final Answer: (A)