Q1: What are the potential problems when a DBMS executes multiple transactions concurrently?

Question:
What are the potential problems when a DBMS executes multiple transactions concurrently?

1. Phantom problem
2. The dirty read problem
3. Unrepeatable read problem
4. All of the above

Answer: Option (d) All of the above

Solution:

The above-mentioned problems and the lost update problem are effects of executing multiple transactions concurrently in a DBMS. Let’s break them down:

• Dirty Read: A transaction reads uncommitted changes from another transaction. If the other transaction rolls back, the read data becomes invalid.
• Unrepeatable Read: A transaction reads the same data twice but gets different results due to updates by another transaction.
• Phantom Problem: A transaction reads a set of rows that satisfy a condition, but another transaction inserts/deletes rows, causing inconsistent results.

Example:

• Transaction T1 updates a record but does not commit. Transaction T2 reads this uncommitted data. If T1 rolls back, T2’s data is inconsistent (Dirty Read).
• Transaction T1 reads all rows where salary > 50,000. If Transaction T2 inserts a new row with salary > 50,000, T1’s query returns inconsistent results (Phantom Problem).

Q2: Assume transaction A holds a shared lock on R. If transaction B also requests a shared lock on R, it will:

Question:
Assume transaction A holds a shared lock on R. If transaction B also requests for a shared lock on R, it will:

1. Result in a deadlock situation
2. Immediately be granted
3. Immediately be rejected
4. Be granted as soon as released by A

Answer: Option (b) Immediately be granted

Solution:

Shared locks allow multiple transactions to read the same data item simultaneously without causing inconsistencies. Since both transactions only request a shared lock, they can coexist without any conflict or deadlock.

Example:

• Transaction A locks record R1 for reading. Transaction B also requests a shared lock for R1. Since shared locks do not block read operations, the request is granted immediately.

Q3: Consider the following schedules involving 2 transactions. Which of the following statements are true?

Question:
Schedules:

S1: r1(X); r1(Y); r2(X); r2(Y); w2(Y); w1(X)
S2: r1(X); r2(X); r2(Y); w2(Y); r1(Y); w1(X)

1. Both S1 and S2 are conflict serializable
2. S1 is conflict serializable and S2 is conflict serializable
3. S1 is not conflict serializable and S2 is conflict serializable
4. Both S1 and S2 are not conflict serializable

Answer: Option (c) S1 is not conflict serializable, and S2 is conflict serializable

Solution:

To determine conflict serializability, we construct a precedence (or dependency) graph for both schedules:

Schedule S1:

• r1(X) → w2(Y): Transaction T1 reads X before T2 writes to Y. No conflict.
• r2(X) → w1(X): Transaction T2 reads X before T1 writes to X. Conflict.

The dependency graph for S1 has a cycle (T1 → T2 → T1). Thus, S1 is not conflict serializable.

Schedule S2:

• r1(X) → w2(Y): Transaction T1 reads X before T2 writes to Y. No conflict.
• r2(X) → w1(X): Transaction T2 reads X before T1 writes to X. No conflict.

The dependency graph for S2 has no cycles. Thus, S2 is conflict serializable.

Example:

• If T1 writes a value to X, and T2 reads X afterward, a dependency exists.
• To avoid cycles, ensure no overlapping read/write operations between transactions.

Q4: Conflict Serializable Schedules

Question:
Consider the 2 transactions T₁ and T₂, and four schedules S₁, S₂, S₃, and S₄ are given below:

• T₁: R₁[X], W₁[X], W₁[Y]
• T₂: R₂[X], R₂[Y], W₂[Y]

Schedules:

S1: R1[X] R2[X] R2[Y] W1[X] W1[Y] W2[Y]
S2: R1[X] R2[X] R2[Y] W1[X] W2[Y] W1[Y]
S3: R1[X] W1[X] R2[X] W1[Y] R2[Y] W2[Y]
S4: R2[X] R2[Y] R1[X] W1[X] W1[Y] W2[Y]

Which of the following schedules are conflict serializable?

1. S₁ and S₂
2. S₂ and S₃
3. S₃ only
4. S₄ only

Answer: Option (b) S₂ and S₃

Solution:

Step 1: Understanding Conflict Serializable Schedules
A schedule is conflict serializable if its precedence graph (dependency graph) has no cycles. Let’s analyze each schedule:

• S₁: Creates a precedence graph with a cycle. Not conflict serializable.
• S₂: Precedence graph has no cycles. Conflict serializable.
• S₃: Precedence graph has no cycles. Conflict serializable.
• S₄: Creates a precedence graph with a cycle. Not conflict serializable.

Thus, S₂ and S₃ are conflict serializable.

Example:
If T₁ writes to X, and T₂ reads X afterward, a dependency exists (T₁ → T₂). Ensure no overlapping operations create cycles.

Q5: Scenarios Leading to Irrecoverable Errors

Question:
Which of the following scenarios may lead to an irrecoverable error in a database system?

1. A transaction writes a data item after it is read by an uncommitted transaction
2. A transaction reads a data item after it is read by an uncommitted transaction
3. A transaction reads a data item after it is written by a committed transaction
4. A transaction reads a data item after it is written by an uncommitted transaction

Answer: Option (d)

Solution:

Explanation:

• (a): Reading and writing after another transaction is allowed.
• (b): Reading after reading is allowed.
• (c): Reading after a committed write is allowed.
• (d): Reading after an uncommitted write can lead to an irrecoverable error if the writing transaction rolls back. This is because the reading transaction has already used invalid data.

Example:

• Transaction T₁ writes a value to X but has not committed. Transaction T₂ reads X. If T₁ rolls back, T₂ uses invalid data.

This leads to inconsistency, making the database state irrecoverable.

Q6: Concurrency Control Protocols

Question:
Which of the following concurrency control protocols ensures both conflict serializability and freedom from deadlock?

1. 2-phase locking
2. Timestamp ordering

Options:

1. (i) only
2. (i) and (ii)
3. (ii) only
4. None of these

Answer: Option (c) (ii) only

Solution:

Step 1: Understanding 2-phase locking (2PL)
2-phase locking ensures conflict serializability but does not guarantee freedom from deadlock, as locks acquired during transactions can lead to circular wait conditions.

Step 2: Understanding Timestamp Ordering
Timestamp ordering protocol ensures:

• Conflict serializability by maintaining a logical order of transactions based on timestamps.
• Freedom from deadlock as it does not require locking mechanisms.

Hence, timestamp ordering satisfies both requirements.

Q8: R-Timestamp in Timestamp Ordering

Question:
In the case of timestamp ordering, R-timestamp(Q) denotes:

1. The largest timestamp of any transaction that executes read(Q) successfully
2. The average timestamp of any transaction that executes read(Q) successfully
3. The average timestamp of any transaction that executes read(Q) unsuccessfully
4. The smallest timestamp of any transaction that executes read(Q) successfully

Answer: Option (a)

Solution:

Explanation: R-timestamp(Q) stores the largest timestamp of the transaction that has successfully read data item Q. It ensures consistency by maintaining the order of operations on Q.

Example:

• Transaction T1 reads Q at timestamp 5.
• Transaction T2 reads Q at timestamp 7.

The R-timestamp(Q) is updated to 7 after T2.

Q9: Purpose of Locking in Databases

Question:
Locking was introduced into databases so that:

1. Keys can be provided to maintain security
2. All simultaneous transactions are prevented
3. Passwords can be provided
4. Consistency can be enforced

Answer: Option (d)

Solution:

Explanation: Locking mechanisms are used in databases to:

• Prevent data corruption by controlling access to data.
• Ensure consistency during concurrent transaction execution.
• Prevent dirty reads, lost updates, and other anomalies.

Example:

• Transaction T1 locks a data item X for writing. Any other transaction attempting to read or write X will wait until T1 releases the lock.

This ensures consistency and prevents anomalies.

S₁: r₁(X); r₁(Y); r₂(X); r₂(Y); w₂(Y); w₁(X)
S₂: r₁(X); r₂(X); r₂(Y); w₂(Y); r₁(Y); w₁(X)

Conflict Serializability & Precedence Graph – Detailed DBMS Solutions

Question 1: Determine Conflict Serializability

Question: Consider the following schedules:

S₁: R₁(A), R₁(C), R₂(B), W₂(B), R₃(B), R₁(A), R₃(C), W₃(C), W₁(A)
S₂: R₂(A), R₁(C), R₂(B), R₃(B), W₂(B), R₁(A), R₃(C), W₃(C), W₁(A)

Which of the above schedules is conflict serializable?

• (a) S₁ only
• (b) S₂ only
• (c) Both S₁ and S₂
• (d) Neither S₁ nor S₂

Answer: Option (a)

Explanation:

Schedule S₁

For schedule S₁, the precedence graph is as follows:

• Transaction T₁ writes on A after reading it.
• Transaction T₂ writes on B after T₁ reads it.
• No cycle is present in the graph.

Conclusion: S₁ is conflict serializable.

Schedule S₂

For schedule S₂, the precedence graph contains a cycle:

• A dependency cycle exists between T₁, T₂, and T₃.

Conclusion: S₂ is not conflict serializable.

Real-World Example

Consider a banking database where:

• T₁: Reads and updates Account A.
• T₂: Reads and updates Account B.
• T₃: Consolidates balances of Accounts A and B.

If T₁ and T₃ overlap, and T₃ consolidates balances based on outdated values from T₁, inconsistencies may arise. The precedence graph helps identify such conflicts.

Key Takeaways

• Precedence graphs are crucial for detecting cycles and ensuring conflict serializability.
• Conflict serializability guarantees the correctness of concurrent transactions in a DBMS.
• Use precedence graphs to visually analyze dependencies between transactions.

Explore more DBMS questions and solutions on DigiiMento.

Conflict Serializability and Concurrency Control in DBMS

Question 1:

Consider the following schedule:

S₁: R₁(A); R₁(C); R₂(B); W₂(B); R₃(B); R₁(A); R₃(C); W₃(C); W₁(A)
S₂: R₂(A); R₁(C); R₂(B); R₃(B); W₂(B); R₁(A); R₃(C); W₃(C); W₁(A)

Which of the above schedules is conflict serializable?

1. a) S₁ only
2. b) S₂ only
3. c) Both S₁ and S₂
4. d) Neither S₁ nor S₂

Solution: Option (a)

Explanation:

S₁:

The precedence graph shows no cycle, which means S₁ is conflict serializable.

S₂:

The precedence graph has a cycle, making S₂ not conflict serializable.

Question 2:

Consider the following schedule:

S: r₂(A), r₁(B), w₂(A), r₂(B), r₃(A), w₁(B), w₃(A), w₂(B)

How many minimum moves are required to make this schedule conflict serializable?

1. a) 1
2. b) 2
3. c) 3
4. d) 4

Solution: Option (a)

Explanation:

The precedence graph for the schedule is as follows:

If w₁(B) is shifted after r₁(B), the precedence graph becomes:

Now the schedule is conflict serializable.

General Information:

In DBMS, Conflict Serializability ensures that a schedule produces the same results as some serial execution. It relies on analyzing the precedence graph of transactions to detect conflicts.

• A cycle in the graph indicates that the schedule is not serializable.
• Making the schedule serializable involves minimum moves to resolve conflicts.

For more details on transaction scheduling and serializability, check out our tutorials on DBMS topics.

Tags:

• Conflict Serializability
• DBMS Concurrency Control
• Transaction Scheduling
• Database Management Systems

Meta Description:

Learn about conflict serializability and concurrency control in DBMS with detailed examples, solutions, and explanation of precedence graphs. Ideal for GATE, UGC NET, and NIELIT preparation.