Let us assume that transaction T1 has arrived before transaction T2. Consider the schedule S=r1(A); r2(B) ; w2(A); w1(B) Which of the following is true?
Let us assume that transaction T1 has arrived before transaction T2. Consider the schedule S=r1(A); r2(B) ; w2(A); w1(B) Which of the following is true? Correct Answer Not allowed under basic timestamp protocols because T1 is rolled
Transaction Table:
|
T1 |
T2 |
|
r(A) |
|
|
|
r(B) |
|
|
w(A) |
|
w(B) |
|
Explanation:
In basic timestamp ordering protocol, the conflicting operations should be executed in the timestamp order.
T2 performing read on B, then its timestamp will be newer and when older transaction performs Write on B, that is, T1 then it violated our condition. Therefore option 2 is the correct answer
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Feb 20, 2025