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Recent questions tagged transactions

1 vote
1 answer
1
Which of the following protocols does not ensure conflict serializability and safety from deadlocks? Graph based protocols Two-phase locking protocol Time-stamp ordering protocol Both a & b
asked Nov 18, 2017 in Databases yogi_p 275 views
2 votes
2 answers
2
Two transactions T1 and T2 are given as follows: T1: R1(A) W1(A) R1(B) W1(B) T2: R2(B) W2(B) R2(C) W2(C) The total number of conflicts serializable schedules that can be formed by T1 and T2 are:
asked Nov 13, 2017 in Databases Parshu gate 490 views
1 vote
1 answer
3
The schedule is given below. Find out the category of the schedule S: R1(X),R2(Z)R1(Z),R3(X) W1(X) W3(Y) R2(Y) W2(Z) W2(Y) C1 C2 C3 Recoverable schedule Conflict Serializable Strict schedule None of these
asked Nov 10, 2017 in Databases techbrk3 645 views
0 votes
1 answer
4
The time stamp of two transactions T1 and T2 are 10 and 15 respectively. Consider the following schedule with T1 and T2 operations. S1:r1(x),w1(x),r2(x),w2(x),r1(y),w1(y),r2(y),w2(y) Find the schedules? S is serializable using basic timestamp protocol S is serializable using basic Thomas write rule S is serializable using basic multiversion timestamp All of these
asked Nov 10, 2017 in Databases techbrk3 581 views
1 vote
1 answer
5
(TS: time stamp) If TS (T) < Read-TS (Q), then write is rejected and T is rolled back If TS (T) < write-TS (Q), then write is rejected and T is rolled back Both (A) and (B) None of these
asked Nov 10, 2017 in Databases techbrk3 355 views
0 votes
1 answer
6
Why is this schedule not allowed under strict two phase locking?
asked Nov 7, 2017 in Databases Kishan Kumar 138 views
1 vote
0 answers
7
Wait die is non preemptive technique :So cannot suffer from starvation. Wound wait is preemptive technique : is it suffers from starvation? i think yes but not sure plz explain with some example:
asked Nov 3, 2017 in Databases Anu007 244 views
1 vote
2 answers
8
1. How many super key is possible if (a,bc,cd) is candidate key and R(ABCDEF) #DOUBT When we use veen digram why we are taking only intersections part only ??? Need help please explain ..
asked Oct 26, 2017 in Databases air1ankit 199 views
1 vote
1 answer
9
In basic time stamp ordering W-time stamp (Q) denotes the largest time stamp of any transaction that has executed write-(Q) successfully.Suppose that we define W-time stamp(Q) as the most recent transaction to execute Write(Q) successfully.Which is true? 1)both ... both with give same result in few cases 3)both with give different result in all cases 4)both with give different result insome cases
asked Oct 25, 2017 in Databases set2018 673 views
1 vote
1 answer
10
Some where I have read, that WR, RW, WW problem occurs only in un-serializable schedule, and Irrecoverable problem, cascade less rollback problem and lost update problems occur in serializable schedule. Is the above statement true or false?
asked Oct 22, 2017 in Databases Shubhanshu 149 views
2 votes
0 answers
11
1 vote
1 answer
13
Consider a Serial Schedule given- T1 T2 T3 w1(A) w1(B) r2(A) w2(B) r3(A) w3(B) How many schedules which are view equivalent to above schedule? How many schedules which are confilct equivalent to above schedule? Is this schedule conflict serializable/ view serializable? If yes How many serial schedules are possible which are conflict equivalent/ view equivalent to able schedule?
asked Sep 30, 2017 in Databases Durgesh Singh 423 views
1 vote
2 answers
14
4 votes
1 answer
15
Where is unrepeatable read and dirty read taking place here?
asked Aug 6, 2017 in Databases just_bhavana 416 views
1 vote
1 answer
16
T1: r1(x)w1(x)r1(y)w1(y) T2: r2(y)w2(y) L is xclusive lock and U is unlock L1(y)L1(X)R1(Y)W1(Y)R1(X)W1(X)U1(X)U1(Y)L2(Y)R2(Y)W2(Y)U2(Y)? IS THIS SCHEDULE VALID UNDER A 2 PHASE LOCKING SCHEDULER????
asked Aug 3, 2017 in Databases Shivam Gupta 3 245 views
1 vote
2 answers
17
I have been reading about the tree protocol as an approach in locking management in databases. I read that it is deadlock free but I am not sure how it works. Example: Assume that T1 has locked B, D, E in exclusive mode. According to the rules: T2 can lock H (allowed by ... lock G it has to lock D but D is locked by T1 so it has to wait. Hence deadlock. What am I misunderstanding in the algorithm?
asked Jul 25, 2017 in Databases shashank023 1.1k views
4 votes
1 answer
18
Does the conservative two phase locking protocol produce cascadeless schedules? Let us consider the following schedule : Since in the schedule the lock on a is released, it is possible for the transaction T2 to acquire the lock on a. Now the transaction T2 ... purposes. Does it mean that even though a conservative two phase protocol is used, the schedules may not necesssarily be cascadeless?
asked Jul 25, 2017 in Databases shashank023 2.5k views
2 votes
2 answers
19
How to check if a schedule is allowed in 2PL or not? S1:R2(A)W1(B)W1(C)R3(B)R2(B)R1(A)R2(C)W3(A) S2:W2(A)W1(A)W3(A)W2(B)W1(B)W3(B)
asked Jul 22, 2017 in Databases reena_kandari 2.1k views
1 vote
0 answers
20
Classify the best balanced tree DS from AVl and B+trees for below applications 1. Main memory application with 100 keys 2. Secondary memory application with 100 keys 3. Main memory application with 1 million keys 4. Secondary memory application with 1 million keys
asked Jul 16, 2017 in Databases rahul sharma 5 82 views
1 vote
1 answer
21
Consider thw following schedules :- S1 :- W1(A) W2(A) W1(B) W2(B) S2:- W1(A) R2(A) W2(A) W2(B) W1(B) Assume that schedule S1 is failed after performing W1(B). Which of the following options are true? a.) S1 and S2 has lost update problem and WW ... free from lost update problem and WW problem c) S1 has WW problem and S2 has lost update problem d) S1 has lost update problem and s2 has WW problem
asked Jul 16, 2017 in Databases rahul sharma 5 355 views
2 votes
0 answers
22
True / false? 1. Serial schedules don't have any conflicting pairs. 2. Serializable schedules don't have any conflicting pairs
asked Jul 16, 2017 in Databases rahul sharma 5 195 views
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