3 3 votes Consider the following database schedule with two transactions T1T1 and T2T2. S=r2(X);r1(X);r2(Y);w1(X);r1(Y);a1;a2 where a1 and a2 stands for abort. Is It a strict schedule? Please explain why or why not? Databases databases concurrency transaction-and-concurrency + – Tendua 1.0k views answer comment Share Follow Print See 1 comment 1 1 comment reply Prashant. commented Oct 26, 2016 reply Follow flag Since both transaction going to be abort at last then thier is no problem of lost update and irrecoverability . so yes it is strict recoverable. 1 1 replyShare Please log in or register to add a comment.
Best answer 10 10 votes A schedule is strict if : A value written by a transaction T is not read or overwritten by other transactions until T either commits or aborts.. so in given schedule W1 (X) is not read or written by other transactions... so it seems to be a strict schedule.. kirti singh answered Oct 26, 2016 • selected Oct 26, 2016 by Tendua kirti singh comment Share Follow 0 reply Please log in or register to add a comment.
–1 –1 vote It is not a strict schedule because a schedule is strict if a value written by a transaction cannot be read or over writtten by other transaction until the transaction either commits (or) aborts Anurag Lahon answered Oct 28, 2016 Anurag Lahon comment Share Follow 0 reply Please log in or register to add a comment.
