Consider the following schedule S of transactions T1, T2, T3, T4: Which one of…

Analysis summary: determine conflict-serializability via the precedence graph and check recoverability by verifying that no transaction commits after reading an uncommitted value.

Step 1 — Build the precedence (conflict) graph:

• T1 -> T2 because T1 performs a write(X) that is read later by T2.

• T2 -> T3 because T2 reads X before T3 performs a write(X) (read-write conflict).

• T3 -> T4 because T3 writes X before T4 reads X (write-read conflict).

• T2 -> T4 because T2 writes Y before T4 reads Y (write-read conflict).

Because the precedence graph has no cycles, the schedule is conflict-serializable. One equivalent serial order consistent with these edges is:

1. T1

2. T2

3. T3

4. T4

Step 2 — Check recoverability:

• T2 reads X that was written by T1, and T1 has already committed before T2 depends on that value.

• T4 reads X that was written by T3, and T3 commits before T4 uses that value.

• T4 reads Y that was written by T2, and T2 commits before T4 uses that value.

Since no transaction commits after reading an uncommitted value, the schedule is recoverable.

Conclusion: the schedule is both conflict-serializable and recoverable.