Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Consider the following transaction with data ...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
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.
Free Test
FREE
| Start Free Test |
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.
Attention Computer Science Engineering (CSE) Students!
To make sure you are not studying endlessly, EduRev has designed Computer Science Engineering (CSE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Computer Science Engineering (CSE).
|
Explore Courses for Computer Science Engineering (CSE) exam
|
|
Similar Computer Science Engineering (CSE) Doubts
Top Courses for Computer Science Engineering (CSE)View all
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?
Question Description
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? 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 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? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
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? 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 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? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
Solutions for 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? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution for 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice Computer Science Engineering (CSE) tests.
|
|
Explore Courses for Computer Science Engineering (CSE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup