# Recent questions tagged transactions

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Consider the following schedule: The number of serial schedules which are view equal to schedule (S) ___________. Answer is 10 as per the given solution. Please explain how to solve these type of questions
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As commit operations are ordered in each transcations, this schedule is recoverable right? Also please explain how to check for view serializabilty?
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No. Of serial schedules view equal to S S: R1(A), R3(D), W1(B), R2(B), W3(B), R4(B), W2(C), R5(C), W4(E), R5(E), W5(B) . If possible please provide detailed solutions. Thank you in advance
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We have 2 Schedule S1 and S2 ... There are no Conflict Operation in S1 and S2. So can we say they are conflict equivalent ?
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Hi Guys, Please tell following statement is T/F --> If there is a schedule which is conflict serializable. But there may not exist any 2 PL with lock upgrade scheme to execute it. PS: ping @Krish__, @rahul sharma 5, @Red_devil, @Shivam Chauhan, @Tuhin Dutta, @Anu007, @Ashwin Kulkarni @reena_kandari and @srestha ji.
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Assume basic timestamp ordering protocol and that time starts from $1$, each operation takes unit amount of time and start of transaction $T_i$ is denoted as $S_i$. The table of timestamp is given below: Find $rts(a), wts(a), rts(b)$ and $wts(b)$ at the end 1, 5, 2, 5 1, 7, 3, 3 3, 7, 3, 7 1, 7, 3, 7
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S:- r1(a) , w2(a) c2 w1(a) c1 w3(a) c3 Is this strict recoverable and serializable ? c1 means commit of transaction T1
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A Phantom Read is when two SELECT queries within the same transaction return a different number of rows. But if the number of rows returned is the same and the data is different, is that also considered as Phantom?
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does graph based and time stamp ordering protocol ensure freedom from cascadeless rollback?
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what is the difference btw ABORT AND FAIL IN TRANSACTION? PLEASE EXPLAIN WITH A SMALL EXAMPLE!
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Please explain the concept of DIRTY READ In a simplified manner w/ a small example!
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Please explain the concept of Blind Write in a simplified manner w/ a small example!
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Consider the following schedule: S:R2(A) W1(B) W1(C) R3(B) R2(B) R1(A) C1 R2(C) C2 W3(A)C3 Please explain how it is allowed in 2 PL
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Consider the 2 transactions T1: R(A) W(A) W(B) T2: R(A) W(A) R(B) W(B) How many view serializable schedules are possible which are not conflict serializable? (A) 0 (B) 1 (C) 2 (D) 3
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Consider the following schedule: S:r1(a) r3(d) w1(B) r2(B) r4(B) w2(C) r5(C) w4(E) r5(E) w5(B) How many serial schedules conflict equal to schedules (S)?
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How many concurrent schedules are conflict serializable of given transactions T1 and T2: T1 = r1(A) W1(A)R1(B)W1(B) T2 = R2(B)W2(B)R2(A)W2(A)
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Deadlock prevention(wait-kill && wound-wait) is free from both starvation as well as deadlock? True or false?
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In below schedule (S) it has cascading abort or not?
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In 2PL protocol if all exclusive locks are acquired by transaction in increasing order of their accesses then is the 2PL starvation free ?
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This is strict schedule or not?
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