Question

Consider the following schedule S with two transactions T1 and T2:

S: R1(X); W1(X); R2(X); W2(X); R1(Y); W1(Y); Commit1; R2(Y); W2(Y); Commit2;

(i) Draw the precedence (serializability) graph for the schedule S.

(ii) Is the schedule S serializable? If yes, provide the equivalent serial schedule(s). If no, explain why.

iii) Identify any concurrency problems (e.g., Lost Update, Dirty Read) present in this schedule. Explain how the problem occurs.

12 Sep 2025
Answer :
Word Count : 585
The precedence (serializability) graph has two nodes: T1 and T2. We add a directed edge Ti → Tj whenever Ti performs an operation on some data item that conflicts with a later operation by Tj (read/write, write/read, or write/write on the same item), with the earlier action first in the schedule. Look at conflicts in S: * On X: W1(X) occurs before R2(X) and before W2(X). So there are conflicts W1(X) → R2(X) and W1(X) → W2(X), which produce an edge T1 → T2. * On Y: W1(Y) occurs before R2(Y) and before W2(Y). So there are conflicts W1(Y) → R2(Y) and W1(Y) → W2(Y), which also produce an edge T1 → T2. Thus the precedence graph contains a single directed edge from T1 to T2 (T1 → T2) and no edge from T2 back to T1. Because the graph is acyclic (just T1 → T2), _____ ____ ___ ___ ______ __________ _________ ________ __________ ____.
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