Question

Consider concurrent execution of two transactions 𝑇1 and 𝑇2 in a DBMS, both of
which access a data object 𝐴. For these two transactions to not conflict on 𝐴, which
one of the following statements must be true?

• A. Both 𝑇1 and 𝑇2 only read 𝐴
• B. 𝑇1 reads 𝐴 and 𝑇2 writes 𝐴
• C. 𝑇1 writes 𝐴 and 𝑇2 reads 𝐴
• D. Both 𝑇1 and 𝑇2 write 𝐴

Hint:

Two operations conflict only if at least one of them is a write on the same data item. A read-read pair never conflicts, while read-write, write-read, and write-write pairs all conflict.

Answer 1

The Correct Option is A

Conflict serializability theory classifies any pair of operations from different transactions on the same data item into one of four categories: read-read, read-write, write-read, and write-write. Only the read-read category is guaranteed to be conflict-free, since reading a value never modifies it and never depends on operation ordering.

Look at why the other three combinations do create conflicts. When one transaction reads \(A\) while the other writes \(A\), the outcome the reader sees changes depending on whether the write happened before or after the read, so swapping their order changes the result, which is the very definition of a conflict. This logic applies symmetrically whether \(T1\) is the reader and \(T2\) the writer, or vice versa. When both transactions write \(A\), the final stored value of \(A\) is whichever write executed last, so again the order matters and the operations conflict, a case usually called a lost-update risk if not properly serialized.

Only when neither transaction changes \(A\), that is, when both merely read it, does the order of execution become irrelevant to the outcome. Reading twice, in any order, always returns the same unchanged value to both transactions.

Therefore the only conflict-free scenario listed is both transactions reading \(A\).

Final answer: Option (A).