1 votes 1 votes T1T2R(A) R(A) R(B) R(B) W(A) COMMIT R(A) W(A) COMMIT in the above problem identify the concurrency in the following Schedule involving 2 Transaction T1 AND T2 (r: read and w : write )a) Dirty Read problem b) Unrepeatable Read problem c) Lost update Problem d) Both a and b Databases concurrency + – spriti1991 asked May 6, 2015 spriti1991 1.0k views answer comment Share Follow See 1 comment See all 1 1 comment reply spriti1991 commented May 6, 2015 reply Follow Share According to me my answer is Lost update Problem , because whatever T2 update its final value of A it is replaced By T1 . Hence Lost update problem was one of my option . And Second would be Unrepeatable Problem because when T1 read value of A first time it would differ with second Read of same data item (A) because in between T2 has update it . So this doesnt preserve Isolation Property . Hence i opted for this answer . Am i right ? 0 votes 0 votes Please log in or register to add a comment.
3 votes 3 votes Dirty read ( uncommitted W-R ) : does not exist here.Unrepeatable read ( R-W ) : exists. As R(A) is read in T1 and W(A) is done in T2.Lost update ( W-W ) : does not exist here' So , Unrepeatable read exists here. worst_engineer answered May 15, 2015 worst_engineer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes every write commit before next operation.. no Write Write conflict so no Lost update problem. no Dirty read also.. Digvijay Pandey answered May 6, 2015 Digvijay Pandey comment Share Follow See 1 comment See all 1 1 comment reply spriti1991 commented May 6, 2015 i edited by spriti1991 May 6, 2015 reply Follow Share Okay should i say "WRITE-WRITE CONFLICT " happen between any 2 consecutive 2 write operation if there is no operation in between them . Surely the above question is NOT DIRTY READ BUT IT is "UNREPEATABLE READ PROBLEM " ! 1 votes 1 votes Please log in or register to add a comment.