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Consider the following transactions 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 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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Most Upvoted Answer
Consider the following transactions with data items P and Q initialize...
P and Q are initialized to Zero, write operation on P and Q will be executed.
Consider any non-serial schedule like R1(P), R2(Q), R1(Q), R2(P), W1(Q), W2(P). R1(P) conflicts with W2(P) Hence T1 must be before T2. R2(Q) conflicts with W1(Q) Hence T2 must be before T1. There is no serial schedule that satisfies both of them hence it does not conflict with serializable.
Hence the correct answer is a schedule that is not conflict serializable.
Consider any non-serial schedule like R1(P), R2(Q), R1(Q), R2(P), W1(Q), W2(P). R1(P) conflicts with W2(P) Hence T1 must be before T2. R2(Q) conflicts with W1(Q) Hence T2 must be before T1. There is no serial schedule that satisfies both of them hence it does not conflict with serializable.
Hence the correct answer is a schedule that is not conflict serializable.
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Consider the following transactions 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 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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Consider the following transactions 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 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? 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 transactions 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 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? 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 transactions 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 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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