Transactions Practice Questions

A schedule is conflict-serializable if its precedence graph, which shows dependencies between conflicting operations of transactions, is acyclic. If the graph is acyclic, a topological sort of its vertices (the transactions) can be performed. This topological ordering provides a valid serial schedule that is conflict-equivalent to the original concurrent schedule. Therefore, a topological order is guaranteed to yield a correct serial schedule.

The transaction transfers a value of 50 from account x to account y. The total sum of x and y must remain unchanged for the database to be in a valid state. This is an example of an integrity constraint. The 'Consistency' property in ACID ensures that any transaction brings the database from one valid state to another, preserving all database rules and constraints. Atomicity ensures the transaction is all-or-nothing, Isolation prevents interference from concurrent transactions, and Durability ensures changes are permanent.

In the given schedule, transaction T2 reads data item 'A' which was written by transaction T1. Subsequently, T2 commits at time instance 8, before T1 has committed. If T1 fails at time instance 9, the value of 'A' that T2 read and used for its commit is now invalid (a dirty read). Since T2 has already committed, its effects cannot be rolled back, leading to an inconsistent database state. This type of schedule is known as non-recoverable and cannot guarantee transaction atomicity.

At the time of the crash, we categorize the transactions based on their status relative to the checkpoint and their commit operations:

1. Undo List: Includes transactions that were active at the time of the crash and did not commit. Both $T1$ (started before the checkpoint, never committed) and $T3$ (started after the checkpoint, never committed) fall into this category.
2. Redo List: Includes transactions that committed after the checkpoint. $T2$ is the only transaction that committed after the checkpoint.
3. Excluded: $T4$ committed before the checkpoint, so its changes are already reflected in the stable database and it is excluded from recovery operations.

Thus, the Undo list is $\{T1, T3\}$ and the Redo list is $\{T2\}$.

Transaction $T1$ completed successfully (log record 4 exists), requiring its operations (log records 2 and 3) to be durable. Transaction $T2$ was interrupted by the crash before its commit record (7) could be written, making it incomplete. During recovery, the database must ensure the atomicity of $T2$ by undoing its effects (log record 6) to restore $B$ to \10,000$, and the durability of$T1$ by redoing its operations (log records 2 and 3). This ensures the database returns to a consistent state where committed changes are reflected and uncommitted changes are rolled back.

To determine if a schedule is conflict-serializable, we construct a precedence graph. Edges are drawn from $T_i$ to $T_j$ if a conflict operation (Read-Write, Write-Read, or Write-Write on the same data item) exists where $T_i$ performs the operation before $T_j$.

1. Conflicts:

   • $T_2$ Reads[X] then $T_1$ Writes[X]: $T_2 \to T_1$
   • $T_1$ Writes[X] then $T_4$ Reads[X]: $T_1 \to T_4$
   • $T_1$ Writes[X] then $T_3$ Writes[X]: $T_1 \to T_3$
   • $T_3$ Writes[X] then $T_4$ Reads[X]: $T_3 \to T_4$
   • $T_3$ Writes[Y] then $T_4$ Reads[Y]: $T_3 \to T_4$

2. Precedence Graph: The edges are $T_2 \to T_1$, $T_1 \to T_4$, $T_1 \to T_3$, $T_3 \to T_4$. There are no cycles in this graph ($T_2 \to T_1 \to T_3 \to T_4$ and $T_2 \to T_1 \to T_4$). Therefore, the schedule S is conflict-serializable.

3. Recoverability: A schedule is recoverable if, for every pair of transactions $T_i$ and $T_j$ such that $T_j$ reads a data item previously written by $T_i$, the commit operation of $T_i$ appears before the commit operation of $T_j$.

   • $T_2$ Reads[X] and $T_1$ Writes[X]. $T_1$ commits after $T_2$. No, this condition checks if $T_j$ reads data written by $T_i$. $T_2$ reads $X$ before $T_1$ writes it, so this is fine.
   • $T_4$ Reads[X] which $T_1$ wrote, and $T_1$ commits before $T_4$.
   • $T_4$ Reads[X] which $T_3$ wrote, and $T_3$ commits before $T_4$.
   • $T_4$ Reads[Y] which $T_3$ wrote, and $T_3$ commits before $T_4$.
     All reading transactions commit after the writing transactions. Thus, the schedule is recoverable.

To determine if a schedule is conflict serializable, we need to construct its precedence graph. A schedule is conflict serializable if and only if its precedence graph is acyclic. A conflict occurs when two operations by different transactions access the same data item, and at least one of them is a write operation.

Let's analyze each option by drawing its precedence graph:

Option A: r1(x); r2(x); w1(x); r3(x); w2(x)
Conflicts:

• $r2(x)$ and $w1(x)$: $T_2 \to T_1$
• $w1(x)$ and $w2(x)$: $T_1 \to T_2$
  This creates a cycle $T_1 \leftrightarrow T_2$. Hence, not conflict serializable.

Option B: r2(x); r1(x); w2(x); r3(x); w1(x)
Conflicts:

• $r1(x)$ and $w2(x)$: $T_1 \to T_2$
• $w2(x)$ and $w1(x)$: $T_2 \to T_1$
  This creates a cycle $T_1 \leftrightarrow T_2$. Hence, not conflict serializable.

Option C: r3(x); r2(x); r1(x); w2(x); w1(x)
Conflicts:

• $r3(x)$ and $w2(x)$: $T_3 \to T_2$
• $r2(x)$ and $w1(x)$: $T_2 \to T_1$
• $r1(x)$ and $w2(x)$: $T_1 \to T_2$ (due to $r1(x)$ then $w2(x)$)
• $r1(x)$ and $w1(x)$: No conflict (same transaction)
  This creates a cycle $T_1 \leftrightarrow T_2$ (from $T_1 \to T_2$ and $T_2 \to T_1$). Hence, not conflict serializable.

Option D: r2(x); w2(x); r3(x); r1(x); w1(x)
Conflicts:

• $r2(x)$ and $w1(x)$: $T_2 \to T_1$
• $w2(x)$ and $r3(x)$: $T_2 \to T_3$
• $w2(x)$ and $r1(x)$: $T_2 \to T_1$ (This is redundant as $T_2 \to T_1$ is already established)
• $w2(x)$ and $w1(x)$: $T_2 \to T_1$ (This is redundant as $T_2 \to T_1$ is already established)
  The precedence graph has edges $T_2 \to T_1$ and $T_2 \to T_3$. There are no cycles. Hence, this schedule is conflict serializable.

The final answer is $\boxed{D}$

To determine if a schedule is conflict-serializable, we construct its precedence graph. An edge $Ti \to Tj$ exists if $Ti$ accesses a data item before $Tj$ and at least one of these accesses is a write operation, and they are conflicting operations (read-write or write-write). If the precedence graph is acyclic, the schedule is conflict-serializable.

For Schedule S1:

1. $r1(X)$ from T1 before $w1(X)$ from T1.
2. $r3(Y)$ from T3 before $w3(Y)$ from T3.
3. $r2(Z)$ from T2 before $w2(Z)$ from T2.
4. $r1(Z)$ from T1 before $w1(Z)$ from T1.
   No cycles are formed, thus S1 is conflict-serializable.

For Schedule S2:

1. $r1(X)$ from T1 before $w1(X)$ from T1.
2. $r3(Y)$ from T3 before $r2(Y)$ from T2, then $r2(Y)$ from T2 before $w3(Y)$ from T3. This indicates a conflict over $Y$ where T3 reads then T2 reads then T3 writes.
3. $r1(Z)$ from T1 before $w1(Z)$ from T1.
4. $r2(Z)$ from T2 before $w2(Z)$ from T2.
   There is a read-write conflict on $Y$ between T2 and T3 ($r2(Y)$ before $w3(Y)$ implies $T2 \to T3$), and then another conflict on $Y$ between T3 and T2 ($w3(Y)$ before $r2(Y)$ implies $T3 \to T2$). This forms a cycle $T2 \leftrightarrow T3$. Thus S2 is not conflict-serializable.

Therefore, only S1 is conflict-serializable.

Transaction isolation levels define the degree to which transaction operations are shielded from concurrent operations. They are ranked based on the anomalies they prevent:

1. $Read\ Uncommitted$ is the lowest level, allowing dirty reads.
2. $Read\ Committed$ prevents dirty reads.
3. $Repeatable\ Read$ prevents both dirty reads and non-repeatable reads.
4. $Serializable$ is the highest isolation level, which guarantees that the outcome of concurrent transactions is equivalent to executing them sequentially. It prevents all concurrency anomalies, including dirty reads, non-repeatable reads, and phantom reads, ensuring maximum data integrity.

Let's analyze the transactions $T_1$ and $T_2$ and their potential for conflict serializability.
$T_1$: $R(P)$, $R(Q)$, $W(Q)$ (if $P=0$, then $Q := Q+1$)
$T_2$: $R(Q)$, $R(P)$, $W(P)$ (if $Q=0$, then $P := P+1$)

Consider a non-serial schedule where $T_1$ starts before $T_2$, but $T_2$ interleave.

1. $T_1$: $R(P)$
2. $T_2$: $R(Q)$
3. $T_1$: $R(Q)$
4. $T_2$: $R(P)$
5. $T_1$: $W(Q)$ (assuming $P=0$ initially, $Q$ becomes 1)
6. $T_2$: $W(P)$ (assuming $Q=0$ initially, $P$ becomes 1)

Now, let's identify conflicts:

• $R_1(P)$ and $R_2(P)$ are not conflicting.
• $R_1(Q)$ and $R_2(Q)$ are not conflicting.
• $W_1(Q)$ and $R_2(Q)$ is a R-W conflict. Since $R_2(Q)$ executes before $W_1(Q)$, there is an edge $T_1 \to T_2$ in the precedence graph.
• $W_2(P)$ and $R_1(P)$ is a R-W conflict. Since $R_1(P)$ executes before $W_2(P)$, there is an edge $T_2 \to T_1$ in the precedence graph.

With edges $T_1 \to T_2$ and $T_2 \to T_1$, a cycle exists in the precedence graph ($T_1 \leftrightarrow T_2$). Therefore, this non-serial schedule is not conflict serializable. Any non-serial interleaving is likely to create such a cycle due to the read/write patterns on shared data items P and Q.

To find the equivalent serial schedule, we construct a precedence graph based on conflicting operations (Write-Read, Read-Write, Write-Write) on the same data items:

1. Variable $Y$: $T_3$ performs $W(Y)$ before $T_1$ performs $R(Y)$ and $W(Y)$, implying $T_3 \rightarrow T_1$. Subsequently, $T_1$ performs $W(Y)$ before $T_2$ performs $R(Y)$ and $W(Y)$, implying $T_1 \rightarrow T_2$.
2. Variable $Z$: $T_3$ performs $W(Z)$ before $T_2$ performs $W(Z)$, confirming $T_3 \rightarrow T_2$.
3. Variable $X$: $T_1$ performs $W(X)$ before $T_2$ performs $R(X)$ and $W(X)$, confirming $T_1 \rightarrow T_2$.

Combining these dependencies gives the topological order $T_3 \rightarrow T_1 \rightarrow T_2$.

To determine the correct serialization order of transactions, we analyze the schedule for conflicting operations and construct a precedence graph. An edge is drawn from transaction Ti to Tj (Ti → Tj) if Ti performs an operation that conflicts with a subsequent operation in Tj on the same data item. A conflict occurs between two transactions if they access the same data item and at least one of the operations is a write.

The resulting graph for the given schedule must be acyclic for it to be conflict-serializable. The topological sort of this graph gives the equivalent serial schedule. The correct answer, T1 → T3 → T2, implies that the precedence graph contains a path from T1 to T3 and from T3 to T2. This means:

1. There is a conflict where T1 acts on a data item before T3 does (e.g., T1 reads/writes X, and later T3 writes X), creating the edge T1 → T3.
2. There is a conflict where T3 acts on a data item before T2 does (e.g., T3 reads/writes Y, and later T2 writes Y), creating the edge T3 → T2.

For the schedule to be valid and not have a deadlock, there must be no cycles in the graph (e.g., no edge from T2 back to T1).

Let's analyze the properties of each concurrency control protocol:

I. 2-Phase Locking (2PL):

• Conflict Serializability: 2PL guarantees conflict serializability. By enforcing that a transaction cannot acquire new locks after it has released any lock, it ensures that the resulting schedule is equivalent to a serial schedule.
• Freedom from Deadlock: 2PL does not guarantee freedom from deadlock. Transactions can acquire locks and then block while waiting for other locks held by other transactions. This can lead to a circular wait condition, resulting in a deadlock. For example, Transaction T1 locks item A and waits for B, while T2 locks B and waits for A. Deadlock prevention or detection mechanisms must be used alongside 2PL.

II. Time-stamp Ordering (TO):

• Conflict Serializability: The TO protocol ensures conflict serializability by ordering transactions based on their timestamps. Conflicting operations are processed in timestamp order, ensuring the schedule is equivalent to a serial schedule ordered by timestamps.
• Freedom from Deadlock: The TO protocol is free from deadlock. In this protocol, if a transaction requests an operation that violates the timestamp order, it is aborted and restarted, rather than being forced to wait. Since transactions do not wait for each other to release resources, the circular wait condition required for deadlock cannot occur.

Therefore, only Time-stamp ordering ensures both conflict serializability and freedom from deadlock.

A schedule is conflict-serializable if its conflict precedence graph is acyclic. A conflict exists between two operations from different transactions if they access the same data item and at least one is a write operation. We assume $T_2$ is $R_2[x]R_2[y]W_2[y]$ to be consistent with the given schedules, as $T_2:R_2[x]R_2[x]W_1[y]$ contains a typo ($W_1[y]$ is an operation of $T_1$, and all schedules use $W_2[y]$).

Relevant conflicts are:

1. $W_1[x]$ and $R_2[x]$ (on $x$)
2. $W_1[y]$ and $R_2[y]$ (on $y$)
3. $W_1[y]$ and $W_2[y]$ (on $y$)

• S1: $R_1[x]R_2[x]R_2[y]W_1[x]W_1[y]W_2[y]$

  • $R_2[x]$ before $W_1[x]$ implies $T_2 \to T_1$.
  • $R_2[y]$ before $W_1[y]$ implies $T_2 \to T_1$.
  • $W_1[y]$ before $W_2[y]$ implies $T_1 \to T_2$.
  • The graph has a cycle $T_1 \leftrightarrow T_2$. Thus, S1 is not conflict-serializable.

• S2: $R_1[x]R_2[x]R_2[y]W_1[x]W_2[y]W_1[y]$

  • $R_2[x]$ before $W_1[x]$ implies $T_2 \to T_1$.
  • $R_2[y]$ before $W_1[y]$ implies $T_2 \to T_1$.
  • $W_2[y]$ before $W_1[y]$ implies $T_2 \to T_1$.
  • The graph is $T_2 \to T_1$, which is acyclic. Thus, S2 is conflict-serializable.

• S3: $R_1[x]W_1[x]R_2[x]W_1[y]R_2[y]W_2[y]$

  • $W_1[x]$ before $R_2[x]$ implies $T_1 \to T_2$.
  • $W_1[y]$ before $R_2[y]$ implies $T_1 \to T_2$.
  • $W_1[y]$ before $W_2[y]$ implies $T_1 \to T_2$.
  • The graph is $T_1 \to T_2$, which is acyclic. Thus, S3 is conflict-serializable.

• S4: $R_2[x]R_2[y]R_1[x]W_1[x]W_1[y]W_2[y]$

  • $R_2[x]$ before $W_1[x]$ implies $T_2 \to T_1$.
  • $R_2[y]$ before $W_1[y]$ implies $T_2 \to T_1$.
  • $W_1[y]$ before $W_2[y]$ implies $T_1 \to T_2$.
  • The graph has a cycle $T_1 \leftrightarrow T_2$. Thus, S4 is not conflict-serializable.

Therefore, S2 and S3 are conflict-serializable.

Here if transaction writing data commits , then transaction which read the data might get phantom tuple/ Unrepeatable error. Though there is no irrecoverable error possible even in this option. This is non issue. Both transaction reading data. This is non issue. This is dirty read. In case if transaction reading uncommitted data commits, irrecoverable error occurs of uncommitted transaction fails. So (D) is answer

A transaction log is a system file that maintains a sequential, chronological record of all modifications, updates, and transactions performed on a database. Its primary function is to ensure data integrity and support recovery processes, allowing the system to restore the database to a consistent state after a failure. In contrast, a query language (Option B) is used to retrieve data, a report writer (Option C) formats existing data for presentation, and data manipulation language (Option D) consists of the commands (e.g., INSERT, UPDATE) themselves rather than the persistent log of those actions. Therefore, only the transaction log provides the complete audit trail of all database activity over time.

Two schedules are conflict equivalent if they involve the same transactions, perform the same operations on the same data items, and maintain the same conflict dependencies. Analyzing the conflicts:

For $S1$ and $S2$, the dependencies are: on object $A$, $T2 \xrightarrow{W} T3$ (since $2WA$ precedes $3WA$), $T2 \xrightarrow{W} T1$ ($2WA$ before $1WA$), and $T3 \xrightarrow{W} T1$ ($3WA$ before $1WA$); on object $B$, $T2 \xrightarrow{W} T1$ ($2WB$ before $1WB$). Since $S1$ and $S2$ preserve these identical orderings, they are conflict equivalent.
In $S3$, the order for object $A$ is $3WA$ before $2WA$, resulting in the conflict $T3 \xrightarrow{W} T2$. This dependency contradicts the $T2 \xrightarrow{W} T3$ order found in $S1$ and $S2$.
Therefore, $S3$ is not conflict equivalent to $S1$ or $S2$, making $S1$ conflict equivalent only to $S2$.

The Two-Phase Commit (2PC) protocol is an atomic commitment protocol used in distributed systems to ensure that all participating nodes either commit or abort a transaction together. Options B, C, and D are incorrect because deadlock handling-prevention, detection, and recovery-is the responsibility of the underlying concurrency control manager, not the commit protocol itself. 2PC maintains the global atomicity and consistency of transaction states, which is a fundamental requirement for the transaction scheduler (often using techniques like $2PL$) to enforce serializability. Therefore, among the choices provided, 2PC is correctly associated with upholding the properties required for serializability in distributed transactions.