Which of the following schedules is (conflict) serializable? For each serializable schedule, determine the equivalent serial schedules.

Question

(a) r1 (X); r3 (X); w1(X); r2(X); w3(X)

(b) r1 (X); r3 (X); w3(X); w1(X); r2(X)

(c) r3 (X); r2 (X); w3(X); r1(X); w1(X)

(d) r3 (X); r2 (X); r1(X); w3(X); w1(X)

Answer 1

Let there be three transactions T1, T2, and T3. They are executed concurrently and produce a schedule S. S is serializable if it can be reproduced as at least one serial schedule (T1 ??T2
??T3 or T1 ??T3 ??T2 or T2 ??T1 ??T3 or T2 ??T3 ??T1 or T3 ??T1 ??T2 or T3 ??T2
??T1).

(a) This schedule is not serializable because T1 reads X (r1(X)) before T3 but T3 reads X
(r3(X)) before T1 writes X (w1(X)), where X is a common data item. The operation
r2(X) of T2 does not affect the schedule at all so its position in the schedule is irrelevant. In a serial schedule T1, T2, and T3, the operation w1(X) comes after r3(X), which does not happen in the question.

(b) This schedule is not serializable because T1 reads X ( r1(X)) before T3 but T3 writes X
(w3(X)) before T1 writes X (w1(X)). The operation r2(X) of T2 does not affect the schedule at all so its position in the schedule is irrelevant. In a serial schedule T1, T3, and T2, r3(X) and w3(X) must come after w1(X), which does not happen in the question.

(c) This schedule is serializable because all conflicting operations of T3 happens before all conflicting operation of T1. T2 has only one operation, which is a read on X (r2(X)), which does not conflict with any other operation. Thus this serializable schedule is equivalent to r2(X); r3(X); w3(X); r1(X); w1(X) (e.g., T2 ??T3 ??T1) serial schedule.

(d) This is not a serializable schedule because T3 reads X (r3(X)) before T1 reads X (r1(X)) but r1(X) happens before T3 writes X (w3(X)). In a serial schedule T3, T2, and T1, r1(X) will happen after w3(X), which does not happen in the question.