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3 votes
3 votes

Consider the following schedule:

$\begin{array}{|l|l|} \hline T1 & T2 \\ \hline {} & \text{Read (A)}  \\ \hline {} &  A = A + 10 \\ \hline {} &  \text{ Read (B)} \\ \hline {} &  A =B +10 \\ \hline {} & \text{Write (A)} \\ \hline \text{Read (B)} & {} \\ \hline B = B+ 10 & {} \\ \hline \text{Read (A)} & {} \\ \hline B = A+ 10 \\ \hline \text{write (B)} & {} \\ \hline \text{commit} & {} \\ \hline {} &  \text{commit} \\ \hline \end{array}$

Which of the following transaction problems is present in the given schedule ?

  1. Lost update
  2. Dirty Rea
  3. Unrepeatable read
  4. Both (A) & (B)
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3 Comments

what is unrepeatable read?
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when Ti reads data item A twice or more and in between any of the two reads of data item A of Ti , some other transaction Tj writes data item A then this situation is termed as unrepeatable read problem

for example: R1(A), W2(A), R1(A)

T1                     T2

R(A)

                        W(A)

R(A)
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when we read one value two time and both time we read different different values then that problem called unrepeatable problem.
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1 Answer

2 votes
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Best answer
A T1 transaction is reading uncommitted data that is written by transaction T2 on data item A, it is also known write read conflict or Dirty read problem...
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Sir ,

Values of B and A are updated, and again overwritten by transactions ..

Wouldnt this also be a case of LOST UPDATE?
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No, It's not a lost update because there are no W-W conflict.
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Answer:

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