Question
Se: r1(x); r2(z); rl(2); 13(x); 13(y); w1(x); cl; w3(y); a3; r2(y); w2(2); w2(y);c2 is recoverable or not
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Answered by
GPT 3.5
The schedule is not recoverable since it exhibits a cascading rollback and write-after-write dependency.
Explanation:
- The schedule starts with r1(x), r2(z), rl(2), then proceeds with 13(x) and 13(y).
- Next, w1(x) writes to x, followed by cl which commits the transaction.
- After that, w3(y) writes to y and a3 aborts the transaction.
- Following that, r2(y) reads y, w2(2) writes to 2, and w2(y) writes to y.
- Finally, c2 commits the transaction.
The issue with this schedule is that w2(y) is dependent on r2(y) which reads y after it was written by w3(y) in the aborted transaction. This is a write-after-read dependency. Additionally, w2(2) writes to 2, which was previously read by rl(2) in the first transaction. Hence, the schedule is not recoverable.
Explanation:
- The schedule starts with r1(x), r2(z), rl(2), then proceeds with 13(x) and 13(y).
- Next, w1(x) writes to x, followed by cl which commits the transaction.
- After that, w3(y) writes to y and a3 aborts the transaction.
- Following that, r2(y) reads y, w2(2) writes to 2, and w2(y) writes to y.
- Finally, c2 commits the transaction.
The issue with this schedule is that w2(y) is dependent on r2(y) which reads y after it was written by w3(y) in the aborted transaction. This is a write-after-read dependency. Additionally, w2(2) writes to 2, which was previously read by rl(2) in the first transaction. Hence, the schedule is not recoverable.
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