In a schedule S having 4 transactions executed concurrently. The order of their operation is given below. Find out whether the given schedule is serializable or not?R1(A), R2(B), W2(A), W3(C), R4(C), R3(A), W3(B), R4(A), W2(B), W4(B), W3(A)
Question
In a schedule S having 4 transactions executed concurrently. The order of their operation is given below. Find out whether the given schedule is serializable or not?R1(A), R2(B), W2(A), W3(C), R4(C), R3(A), W3(B), R4(A), W2(B), W4(B), W3(A)
Solution
To determine whether a schedule is serializable or not, we need to construct a precedence graph. A schedule is serializable if its precedence graph is acyclic.
Here are the steps to construct the precedence graph:
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Identify the transactions in the schedule. In this case, we have T1, T2, T3, and T4.
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Draw a node for each transaction.
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Draw a directed edge from Ti to Tj if an operation in Ti affects an operation in Tj. The edge direction is from the transaction that executes first to the one that executes later.
- From T1 to T2: There is no operation in T1 that affects T2.
- From T1 to T3: There is no operation in T1 that affects T3.
- From T1 to T4: There is no operation in T1 that affects T4.
- From T2 to T1: There is an operation in T2 (W2(A)) that affects T1 (R1(A)). So, draw an edge from T2 to T1.
- From T2 to T3: There is an operation in T2 (W2(B)) that affects T3 (W3(B)). So, draw an edge from T2 to T3.
- From T2 to T4: There is an operation in T2 (W2(B)) that affects T4 (W4(B)). So, draw an edge from T2 to T4.
- From T3 to T1: There is no operation in T3 that affects T1.
- From T3 to T2: There is no operation in T3 that affects T2.
- From T3 to T4: There is an operation in T3 (W3(A)) that affects T4 (R4(A)). So, draw an edge from T3 to T4.
- From T4 to T1: There is no operation in T4 that affects T1.
- From T4 to T2: There is no operation in T4 that affects T2.
- From T4 to T3: There is no operation in T4 that affects T3.
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Check if there are any cycles in the graph. If there are no cycles, the schedule is serializable. If there are cycles, the schedule is not serializable.
In this case, there are no cycles in the graph, so the schedule is serializable.
Similar Questions
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