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Consider the following transactions with data items P and Q initialized to zero:
Any non-serial interleaving of T1 and T2 for concurrent execution leads to
- a)A serializable schedule
- b)A schedule that is not conflict serializable
- c)A conflict serializable schedule
- d)A schedule for which a precedence graph cannot be drawn
Correct answer is option 'B'. Can you explain this answer?
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Consider the following transactions with data items P and Q initialize...
Answer (B)
Two or more actions are said to be in conflict if:
1) The actions belong to different transactions.
2) At least one of the actions is a write operation.
3) The actions access the same object (read or write).
The schedules S1 and S2 are said to be conflict-equivalent if the following conditions are satisfied:
1) Both schedules S1 and S2 involve the same set of transactions (including ordering of actions within each transaction).
2) The order of each pair of conflicting actions in S1 and S2 are the same.
A schedule is said to be conflict-serializable when the schedule is conflict-equivalent to one or more serial schedules.
In the given scenario, there are two possible serial schedules:
1) T1 followed by T2
2) T2 followed by T1.
In both of the serial schedules, one of the transactions reads the value written by other transaction as a first step. Therefore, any non-serial interleaving of T1 and T2 will not be conflict serializable.
Two or more actions are said to be in conflict if:
1) The actions belong to different transactions.
2) At least one of the actions is a write operation.
3) The actions access the same object (read or write).
The schedules S1 and S2 are said to be conflict-equivalent if the following conditions are satisfied:
1) Both schedules S1 and S2 involve the same set of transactions (including ordering of actions within each transaction).
2) The order of each pair of conflicting actions in S1 and S2 are the same.
A schedule is said to be conflict-serializable when the schedule is conflict-equivalent to one or more serial schedules.
In the given scenario, there are two possible serial schedules:
1) T1 followed by T2
2) T2 followed by T1.
In both of the serial schedules, one of the transactions reads the value written by other transaction as a first step. Therefore, any non-serial interleaving of T1 and T2 will not be conflict serializable.
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Consider the following transactions with data items P and Q initialized to zero:Any non-serial interleaving of T1 and T2 for concurrent execution leads toa)A serializable scheduleb)A schedule that is not conflict serializablec)A conflict serializable scheduled)A schedule for which a precedence graph cannot be drawnCorrect answer is option 'B'. Can you explain this answer?
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