T1: read (P) ; read (Q) ; if P = 0 then Q : = Q + 1 ; write (Q) ; T2: read (Q) ; read (P) ; if Q = 0 then P : = P + 1 ; write (P) ;Any non-serial interleaving of T1 and T2 for concurrent execution leads to
A serializable schedule
A schedule that is not conflict serializable
A conflict serializable schedule
A schedule for which a precedence graph cannot be drawn
Both I and II
Neither I nor II
2 Phase Locking (2PL) is a concurrency control method that guarantees serializability. The protocol utilizes locks, applied by a transaction to data, which may block (interpreted as signals to stop) other transactions from accessing the same data during the transaction’s life. 2PL may be lead to deadlocks that result from the mutual blocking of two or more transactions. See the following situation, neither T3 nor T4 can make progress.
Timestamp-based concurrency control algorithm is a non-lock concurrency control method. In Timestamp based method, deadlock cannot occur as no transaction ever waits.
T1 can complete before T2 and T3 as there is no conflict between Write(X) of T1 and the operations in T2 and T3 which occur before Write(X) of T1 in the above diagram.
T3 should can complete before T2 as the Read(Y) of T3 doesn’t conflict with Read(Y) of T2. Similarly, Write(X) of T3 doesn’t conflict with Read(Y) and Write(Y) operations of T2.
Another way to solve this question is to create a dependency graph and topologically sort the dependency graph. After topologically sorting, we can see the sequence T1, T3, T2.
S is conflict-serializable but not recoverable
S is not conflict-serializable but is recoverable
S is both conflict-serializable and recoverable
S is neither conflict-serializable nor is it recoverable
T1: r1(X); r1(Z); w1(X); w1(Z) T2: r2(Y); r2(Z); w2(Z) T3: r3(Y); r3(X); w3(Y) S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z); w3(Y); w2(Z); r1(Z); w1(X); w1(Z) S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z)Which one of the following statements about the schedules is TRUE?
Only S1 is conflict-serializable.
Only S2 is conflict-serializable.
Both S1 and S2 are conflict-serializable.
Neither S1 nor S2 is conflict-serializable.
1. T1 start 2. T1 B old=12000 new=10000 3. T1 M old=0 new=2000 4. T1 commit 5. T2 start 6. T2 B old=10000 new=10500 7. T2 commitSuppose the database system crashes just before log record 7 is written. When the system is restarted, which one statement is true of the recovery procedure?
We must redo log record 6 to set B to 10500
We must undo log record 6 to set B to 10000 and then redo log records 2 and 3.
We need not redo log records 2 and 3 because transaction T1 has committed.
We can apply redo and undo operations in arbitrary order because they are idempotent
A transaction writes a data item after it is read by an uncommitted transaction
A transaction reads a data item after it is read by an uncommitted transaction
A transaction reads a data item after it is written by a committed transaction
A transaction reads a data item after it is written by an uncommitted transaction
The schedule is serializable as T2; T3; T1
The schedule is serializable as T2; T1; T3
The schedule is serializable as T3; T2; T1
The schedule is not serializable
read(x); x := x – 50; write(x); read(y); y := y + 50; write(y)The constraint that the sum of the accounts x and y should remain constant is that of