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Consider the transactions T1, T2, and T3 and the schedules S1 and S2 given below.
T1: r1(X); r1(Z); w1(X); w1(Z)
T2: r2(Y); r2(Z); w2(Z)
T3: r3(Y); r3(X); w3(Y)
S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z);
w3(Y); w2(Z); r1(Z); w1(X); w1(Z)
S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z);
r2(Z); w3(Y); w1(X); w2(Z); w1(Z)
T1: r1(X); r1(Z); w1(X); w1(Z)
T2: r2(Y); r2(Z); w2(Z)
T3: r3(Y); r3(X); w3(Y)
S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z);
w3(Y); w2(Z); r1(Z); w1(X); w1(Z)
S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z);
r2(Z); w3(Y); w1(X); w2(Z); w1(Z)
- a)Only S1 is conflict-serializable.
- b)Only S2 is conflict-serializable.
- c)Both S1 and S2 are conflict-serializable.
- d)Neither S1 nor S2 is conflict-serializable.
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Consider the transactions T1, T2, and T3 and the schedules S1 and S2 g...
For conflict serializability of a schedule( which gives same effect as a serial schedule ) we should check for conflict operations, which are Read-Write, Write-Read and Write-Write between each pair of transactions, and based on those conflicts we make a precedence graph, if the graph contains a cycle, it\'s not a conflict serializable schedule. To make a precedence graph: if Read(X) in Ti followed by Write(X) in Tj ( hence a conflict ), then we draw an edge from Ti to Tj ( Ti -> Tj) If we make a precedence graph for S1 and S2 , we would get directed edges for S1 as T2->T1, T2->T3, T3->T1, and for S2 as T2->T1, T2->T3, T3->T1, T1->T2. In S1 there is no cycle, but S2 has a cycle. Hence only S1 is conflict serializable. Note : The serial order for S1 is T2 -> T3 -> T1.
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Consider the transactions T1, T2, and T3 and the schedules S1 and S2 g...
Explanation:
Conflict Serializable Schedules:
- Conflict serializability is a property of a schedule that ensures that the final result of the schedule is the same as the result of some serial execution of the transactions in the schedule.
- A schedule is conflict-serializable if it is equivalent to a serial schedule of the same transactions, where the order of non-conflicting operations is preserved.
Checking for Conflict Serializability:
To determine if a schedule is conflict-serializable, we can use the precedence graph method. We draw a directed edge from Ti to Tj if there is a conflicting operation between Ti and Tj in the schedule.
Analysis:
- For schedule S1:
- The precedence graph for S1 is T1 -> T3, T3 -> T2, T2 -> T1.
- As there is a cycle in the precedence graph, the schedule S1 is not conflict-serializable.
- For schedule S2:
- The precedence graph for S2 is T1 -> T3, T3 -> T2, T2 -> T1.
- As there is a cycle in the precedence graph, the schedule S2 is not conflict-serializable.
Therefore, only schedule S1 is conflict-serializable, according to the precedence graph method.
Conflict Serializable Schedules:
- Conflict serializability is a property of a schedule that ensures that the final result of the schedule is the same as the result of some serial execution of the transactions in the schedule.
- A schedule is conflict-serializable if it is equivalent to a serial schedule of the same transactions, where the order of non-conflicting operations is preserved.
Checking for Conflict Serializability:
To determine if a schedule is conflict-serializable, we can use the precedence graph method. We draw a directed edge from Ti to Tj if there is a conflicting operation between Ti and Tj in the schedule.
Analysis:
- For schedule S1:
- The precedence graph for S1 is T1 -> T3, T3 -> T2, T2 -> T1.
- As there is a cycle in the precedence graph, the schedule S1 is not conflict-serializable.
- For schedule S2:
- The precedence graph for S2 is T1 -> T3, T3 -> T2, T2 -> T1.
- As there is a cycle in the precedence graph, the schedule S2 is not conflict-serializable.
Therefore, only schedule S1 is conflict-serializable, according to the precedence graph method.
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Consider the transactions T1, T2, and T3 and the schedules S1 and S2 given below.T1: r1(X); r1(Z); w1(X); w1(Z)T2: r2(Y); r2(Z); w2(Z)T3: r3(Y); r3(X); w3(Y)S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z); w3(Y); w2(Z); r1(Z); w1(X); w1(Z)S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z)a)Only S1 is conflict-serializable.b)Only S2 is conflict-serializable.c)Both S1 and S2 are conflict-serializable.d)Neither S1 nor S2 is conflict-serializable.Correct answer is option 'A'. Can you explain this answer?
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Consider the transactions T1, T2, and T3 and the schedules S1 and S2 given below.T1: r1(X); r1(Z); w1(X); w1(Z)T2: r2(Y); r2(Z); w2(Z)T3: r3(Y); r3(X); w3(Y)S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z); w3(Y); w2(Z); r1(Z); w1(X); w1(Z)S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z)a)Only S1 is conflict-serializable.b)Only S2 is conflict-serializable.c)Both S1 and S2 are conflict-serializable.d)Neither S1 nor S2 is conflict-serializable.Correct answer is option 'A'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider the transactions T1, T2, and T3 and the schedules S1 and S2 given below.T1: r1(X); r1(Z); w1(X); w1(Z)T2: r2(Y); r2(Z); w2(Z)T3: r3(Y); r3(X); w3(Y)S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z); w3(Y); w2(Z); r1(Z); w1(X); w1(Z)S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z)a)Only S1 is conflict-serializable.b)Only S2 is conflict-serializable.c)Both S1 and S2 are conflict-serializable.d)Neither S1 nor S2 is conflict-serializable.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the transactions T1, T2, and T3 and the schedules S1 and S2 given below.T1: r1(X); r1(Z); w1(X); w1(Z)T2: r2(Y); r2(Z); w2(Z)T3: r3(Y); r3(X); w3(Y)S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z); w3(Y); w2(Z); r1(Z); w1(X); w1(Z)S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z)a)Only S1 is conflict-serializable.b)Only S2 is conflict-serializable.c)Both S1 and S2 are conflict-serializable.d)Neither S1 nor S2 is conflict-serializable.Correct answer is option 'A'. Can you explain this answer?.
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