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Consider the three transactions T1, T2, and T3, and the schedules S1 and S2given below. State whether each schedule is serializable or not. If a schedule is serializable, write down the equivalent serial schedule(s). T1: r1 (X); r1 (Z); w1 (X); T2: r2 (Z); r2 (Y); w2 (Z); w2 (Y); T3: r3 (X); r3 (Y); w3 (Y); S1: r1 (X); r2 (Z); r1 (Z); r3 (X); r3 (Y); w1 (X); w3 (Y); r2 (Y); w2 (Z); w2 (Y); S2: r1 (X); r2 (Z); r3 (X); r1 (Z); r2 (Y); r3 (Y); w1 (X); w2 (Z); w3 (Y); w2 (Y);

Question

Consider the three transactions T1, T2, and T3, and the schedules S1 and S2given below. State whether each schedule is serializable or not. If a schedule is serializable, write down the equivalent serial schedule(s). T1: r1 (X); r1 (Z); w1 (X); T2: r2 (Z); r2 (Y); w2 (Z); w2 (Y); T3: r3 (X); r3 (Y); w3 (Y); S1: r1 (X); r2 (Z); r1 (Z); r3 (X); r3 (Y); w1 (X); w3 (Y); r2 (Y); w2 (Z); w2 (Y); S2: r1 (X); r2 (Z); r3 (X); r1 (Z); r2 (Y); r3 (Y); w1 (X); w2 (Z); w3 (Y); w2 (Y);

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Solution

To determine whether a schedule is serializable, we need to check if it is conflict serializable. Conflict serializability is determined by constructing a precedence graph and checking for cycles. If there are no cycles, the schedule is conflict serializable.

Let's start with Schedule S1:

  1. Construct the precedence graph for S1. The nodes of the graph are the transactions T1, T2, and T3. There is a directed edge from Ti to Tj if a data item is first written by Ti and then read by Tj, or if it is first written by Ti and then written by Tj.

  2. In S1, T1 writes X before T3 reads X, so there is an edge from T1 to T3. T1 also reads Z before T2 writes Z, so there is an edge from T1 to T2. T2 reads Y before T3 writes Y, so there is an edge from T2 to T3.

  3. The precedence graph for S1 is: T1 -> T3, T1 -> T2, T2 -> T3. There are no cycles in this graph, so S1 is conflict serializable.

  4. The equivalent serial schedules for S1 are T1 -> T2 -> T3 or T1 -> T3 -> T2.

Now let's check Schedule S2:

  1. Construct the precedence graph for S2. The nodes of the graph are the transactions T1, T2, and T3.

  2. In S2, T1 writes X before T3 reads X, so there is an edge from T1 to T3. T1 also reads Z before T2 writes Z, so there is an edge from T1 to T2. T2 reads Y before T3 writes Y, so there is an edge from T2 to T3.

  3. The precedence graph for S2 is: T1 -> T3, T1 -> T2, T2 -> T3. There are no cycles in this graph, so S2 is conflict serializable.

  4. The equivalent serial schedules for S2 are T1 -> T2 -> T3 or T1 -> T3 -> T2.

So, both S1 and S2 are serializable and their equivalent serial schedules are T1 -> T2 -> T3 or T1 -> T3 -> T2.

This problem has been solved

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