Question
Basic TO Algorithm question
Two transactions T1 and T2. Initially RTS = 0 and WTS = 0 for data items X, Y
Timestamps are as follows: TS (T1)=10 and TS (T2)=20
T1
1. A * 1 = Read(X)
2. A1=A1-k
3. Write(X, A1)
T2
1. A * 1 = Read(X)
2. A 1=A1^ * 1.01
3. Write(X, A1)
T1
4. A^ 2 =R aad (V)
5. A 2=A2+k
6. Write(Y, A2)
T2
4. A * 2 = Read(Y)
5. A 2=A2^ * 1.01
6. Write(Y, A2)
Discuss whether the following schedule is serializable or not? Justify your answer using Basic TO Algorithm.
Two transactions T1 and T2. Initially RTS = 0 and WTS = 0 for data items X, Y
Timestamps are as follows: TS (T1)=10 and TS (T2)=20
T1
1. A * 1 = Read(X)
2. A1=A1-k
3. Write(X, A1)
T2
1. A * 1 = Read(X)
2. A 1=A1^ * 1.01
3. Write(X, A1)
T1
4. A^ 2 =R aad (V)
5. A 2=A2+k
6. Write(Y, A2)
T2
4. A * 2 = Read(Y)
5. A 2=A2^ * 1.01
6. Write(Y, A2)
Discuss whether the following schedule is serializable or not? Justify your answer using Basic TO Algorithm.
Answers
GPT 3.5
The Basic Two-Phase Locking (TO) algorithm can be used to check if a schedule is serializable.
In the given schedule, the transactions T1 and T2 are executed in an interleaved manner. Let's analyze the schedule step by step using the Basic TO Algorithm:
1. T1: Read(X) - T1 acquires a shared (S) lock on X.
T2: Read(X) - T2 requests an S lock on X, but it cannot be granted as T1 already holds an S lock on X. T2 waits.
2. T1: A1 = A1 - k - This operation does not involve any data items.
T2: T2 is still waiting.
3. T1: Write(X, A1) - T1 upgrades its shared (S) lock on X to an exclusive (X) lock. The value of X is modified to A1.
T2: T2 is still waiting.
4. T1: A2 = Read(V) - This operation does not involve X or Y.
T2: T2 is still waiting.
5. T1: A2 = A2 + k - This operation does not involve X or Y.
T2: T2 is still waiting.
6. T1: Write(Y, A2) - This operation involves a new data item Y. T1 requests a shared (S) lock on Y, which is granted as there are no conflicting locks on Y.
T2: T2 is still waiting.
7. T2: Read(Y) - T2 requests a shared (S) lock on Y, but it cannot be granted as T1 already holds a shared (S) lock on Y. T2 waits.
8. T2: A2 = A2^ * 1.01 - This operation does not involve Y.
9. T2: Write(Y, A2) - This operation involves Y. T2 requests an exclusive (X) lock on Y, but it cannot be granted as T1 still holds a shared (S) lock on Y. T2 waits.
In the given schedule, T1 holds shared (S) locks on both X and Y, whereas T2 is waiting to acquire shared (S) and exclusive (X) locks on X and Y, respectively. This violates the Basic TO Algorithm, which states that no transaction can request a lock that is incompatible with a lock held by another transaction.
Therefore, the given schedule is not serializable according to the Basic TO Algorithm.
In the given schedule, the transactions T1 and T2 are executed in an interleaved manner. Let's analyze the schedule step by step using the Basic TO Algorithm:
1. T1: Read(X) - T1 acquires a shared (S) lock on X.
T2: Read(X) - T2 requests an S lock on X, but it cannot be granted as T1 already holds an S lock on X. T2 waits.
2. T1: A1 = A1 - k - This operation does not involve any data items.
T2: T2 is still waiting.
3. T1: Write(X, A1) - T1 upgrades its shared (S) lock on X to an exclusive (X) lock. The value of X is modified to A1.
T2: T2 is still waiting.
4. T1: A2 = Read(V) - This operation does not involve X or Y.
T2: T2 is still waiting.
5. T1: A2 = A2 + k - This operation does not involve X or Y.
T2: T2 is still waiting.
6. T1: Write(Y, A2) - This operation involves a new data item Y. T1 requests a shared (S) lock on Y, which is granted as there are no conflicting locks on Y.
T2: T2 is still waiting.
7. T2: Read(Y) - T2 requests a shared (S) lock on Y, but it cannot be granted as T1 already holds a shared (S) lock on Y. T2 waits.
8. T2: A2 = A2^ * 1.01 - This operation does not involve Y.
9. T2: Write(Y, A2) - This operation involves Y. T2 requests an exclusive (X) lock on Y, but it cannot be granted as T1 still holds a shared (S) lock on Y. T2 waits.
In the given schedule, T1 holds shared (S) locks on both X and Y, whereas T2 is waiting to acquire shared (S) and exclusive (X) locks on X and Y, respectively. This violates the Basic TO Algorithm, which states that no transaction can request a lock that is incompatible with a lock held by another transaction.
Therefore, the given schedule is not serializable according to the Basic TO Algorithm.