"Transactions & Concurrency Control" — Practice

Question 1

Under what condition do two operations conflict?

Accepted Answer

They belong to different transactions, access the same data item, and at least one of them is a write.

Question 2

State what two-phase locking guarantees and two things it does not.

Accepted Answer

It guarantees conflict serializability. It does not guarantee freedom from deadlock, and it does not guarantee recoverability.

Question 3

Which SQL isolation level permits non-repeatable reads but not dirty reads?

Accepted Answer

Read committed.

Question 4

For the schedule r1(A), w2(A), r3(B), w1(B), r2(C), w3(C), build the precedence graph and decide conflict serializability.

Accepted Answer

Item A gives an edge from T1 to T2, item B gives T3 to T1, and item C gives T2 to T3. Those three edges form a cycle, so the schedule is not conflict serializable.

Question 5

Three transactions have 2, 3 and 3 operations. How many schedules exist in total, and how many are serial?

Accepted Answer

Total operations are 8, so the count is 8 factorial divided by 2 factorial times 3 factorial times 3 factorial, which is 40320 over 72, or 560. Serial schedules number 3 factorial, which is 6.

Question 6

Classify the schedule w1(A), r2(A), w2(B), c1, c2 as recoverable, cascadeless or strict.

Accepted Answer

It is recoverable, because T2 commits after T1, whom it read from. It is not cascadeless, because T2 read A while T1 was still uncommitted, and it therefore cannot be strict either.

Question 7

Transaction T with timestamp 20 writes item X whose read timestamp is 25 and write timestamp is 10. What happens under basic timestamp ordering?

Accepted Answer

The read timestamp 25 exceeds 20, meaning a later transaction has already read the older value and would have needed to see T's write. The write is rejected and T is rolled back with a fresh timestamp.

Question 8

T1 reads A then writes B; T2 reads B then writes A. Show that strict 2PL can deadlock, then resolve it under wound-wait with T1 older.

Accepted Answer

T1 takes a shared lock on A and T2 on B, then each requests an exclusive lock on the other's item and must wait, and neither can commit to release. Under wound-wait, the older T1 requesting from the younger T2 wounds it, so T2 aborts, T1 completes, and T2 restarts keeping its original timestamp.

Question 9

Show that r1(A), w2(A), w1(A), w3(A) is view serializable but not conflict serializable.

Accepted Answer

The graph has T1 to T2 from the read preceding w2(A) and T2 to T1 from w2(A) preceding w1(A), a cycle, so it is not conflict serializable. Compared with the serial order T1, T2, T3, both have T1 reading the initial value of A and T3 performing the final write, and no other read of A exists, so the two are view equivalent.

Question 10

Explain Thomas's write rule and why it produces schedules outside the conflict serializable class.

Accepted Answer

When only the item's write timestamp exceeds the writer's, the write is discarded rather than causing a rollback, because a later transaction has already superseded it and the value would be immediately overwritten in any consistent serial order. Discarding a write breaks the write-write conflict ordering, so the schedule is not conflict serializable, yet the reads-from and final-write conditions still match a serial order, making it view serializable.

Practice — "Transactions & Concurrency Control"

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