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Consider the following transaction with data items P and Q initialized to zero:
T1 : read (P) ;
read (Q) ;
if P = 0 then Q : = Q + 1;
write (Q) ;
T2 : read (Q) ;
read (P) ;
if Q = 0 then P : = P + 1 ;
write (P) ;
read (Q) ;
if P = 0 then Q : = Q + 1;
write (Q) ;
T2 : read (Q) ;
read (P) ;
if Q = 0 then P : = P + 1 ;
write (P) ;
Q. Any non-serial interleaving of T1 and T2 for concurrent execution leads to
- a)A serializable schedule
- b)A conflict serializable schedule
- c)A schedule for which a precedence graph cannot be drawn
- d)A schedule that is not conflict serializable
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Consider the following transaction with data items P and Q initialized...
Concept:
Two or more actions are said to be in conflict if:
- The actions are associated with various transactions.
- A write operation is involved in at least one of the operations.
- The actions all refer to the same thing (read or write).
If the following conditions are met, the schedules S1 and S2 are said to be conflict-equivalent:
- The transactions in schedules S1 and S2 are the same (including ordering of actions within each transaction).
- In both S1 and S2, the order of each pair of conflicting actions is the same.
Conflict-serializable:
A schedule is said to be conflict-serializable when the schedule is conflict-equivalent to one or more serial schedules. Conflict-serializability is that a schedule is conflict-serializable if and only if its precedence graph/serializability graph, when only committed transactions are considered, is acyclic.
There are two possible serial schedules:
- T1 followed by T2.
- T2 followed by T1.
As an initial step in both serial schedules, one of the transactions reads the value written by the other transaction. As a result, any non-serial T1 and T2 interleaving will not be conflict serializable.
Hence the correct answer is a schedule that is not conflict serializable.
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Community Answer
Consider the following transaction with data items P and Q initialized...
Explanation:
Non-serial interleaving:
- In the given transaction T1 and T2, the operations are interleaved in a non-serial manner, meaning they are executed concurrently or in an order different from the original sequence.
Conflict Serializable Schedule:
- A schedule is said to be conflict serializable if it is equivalent to a serial schedule, meaning the final result is the same as if the transactions were executed in serial order.
- In this case, the given non-serial interleaving of T1 and T2 does not result in a conflict serializable schedule.
Precedence Graph:
- In conflict serializability, a precedence graph can be drawn to determine if the schedule is conflict serializable.
- However, in this case, a precedence graph cannot be drawn for the given schedule due to conflicting operations.
Conclusion:
- The non-serial interleaving of T1 and T2 leads to a schedule that is not conflict serializable. The conflicting operations and lack of a precedence graph indicate that the schedule does not maintain the same final result as a serial execution.
Non-serial interleaving:
- In the given transaction T1 and T2, the operations are interleaved in a non-serial manner, meaning they are executed concurrently or in an order different from the original sequence.
Conflict Serializable Schedule:
- A schedule is said to be conflict serializable if it is equivalent to a serial schedule, meaning the final result is the same as if the transactions were executed in serial order.
- In this case, the given non-serial interleaving of T1 and T2 does not result in a conflict serializable schedule.
Precedence Graph:
- In conflict serializability, a precedence graph can be drawn to determine if the schedule is conflict serializable.
- However, in this case, a precedence graph cannot be drawn for the given schedule due to conflicting operations.
Conclusion:
- The non-serial interleaving of T1 and T2 leads to a schedule that is not conflict serializable. The conflicting operations and lack of a precedence graph indicate that the schedule does not maintain the same final result as a serial execution.
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Consider the following transaction with data items P and Q initialized to zero:T1 : read (P) ;read (Q) ;if P = 0 then Q : = Q + 1;write (Q) ;T2 : read (Q) ;read (P) ;if Q = 0 then P : = P + 1 ;write (P) ;Q. Any non-serial interleaving of T1 and T2 for concurrent execution leads toa)A serializable scheduleb)A conflict serializable schedulec)A schedule for which a precedence graph cannot be drawnd)A schedule that is not conflict serializableCorrect answer is option 'D'. Can you explain this answer?
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