Consider the following schedules involving two transactions. S 1: r 1(X) ; r 1(Y) ; r 2(X) ; r 2(Y) ; w 2(Y) ; w 1(X) S2 : r 1(X) ; r 2(X) ; r 2(Y) ; w 2(Y) ; r 1(Y) ; w 1(X)
S 1is not conflict serializable and S 2is conflict serializable
Understanding how concurrent transactions in a database are handled is crucial for maintaining data consistency. Serializability is a key concept here, ensuring that the result of running multiple transactions at the same time is the same as running them one after another in some order. Conflict serializability is one way to check for this equivalence.
A schedule is conflict serializable if it is conflict equivalent to a serial schedule. Two schedules are conflict equivalent if they involve the same transactions, and for any two conflicting operations (read-write, write-read, or write-write on the same data item by different transactions), their relative order is the same in both schedules.
To check for conflict serializability, we construct a precedence graph. Each transaction is a node in the graph. An edge is drawn from transaction $T_i$ to transaction $T_j$ ($T_i \rightarrow T_j$) if an operation in $T_i$ conflicts with an operation in $T_j$, and the operation in $T_i$ occurs before the operation in $T_j$ in the schedule. If the precedence graph contains no cycles, the schedule is conflict serializable.
Schedule S1 is given as: r1(X) ; r1(Y) ; r2(X) ; r2(Y) ; w2(Y) ; w1(X)
Here we have two transactions, T1 and T2, accessing data items X and Y. To check for conflict serializability, we identify pairs of conflicting operations involving different transactions and determine the dependencies. Conflict operations occur when two operations from different transactions access the same data item, and at least one is a write operation (RW, WR, WW).
r1(X) and r2(X): T1 reads X, T2 reads X. Same data item, different transactions, but both are reads (RR). Read-Read operations do not conflict.r1(Y) and w2(Y): T1 reads Y, T2 writes Y. Same data item, different transactions. r1(Y) appears before w2(Y) in S1. This is a Read-Write (RW) conflict. The dependency is T1 → T2.r2(X) and w1(X): T2 reads X, T1 writes X. Same data item, different transactions. r2(X) appears before w1(X) in S1. This is a Write-Read (WR) conflict. The dependency is T2 → T1.r1(Y)/r2(Y) (RR), r2(Y)/w2(Y) (same transaction), w2(Y)/w1(X) (different data item) do not create conflicts according to the definition.The conflicts involving operations on the same data item by different transactions are:
| Operation 1 | Operation 2 | Data Item | Type | Order in S1 | Dependency (Precedence Graph Edge) |
|---|---|---|---|---|---|
r1(Y) |
w2(Y) |
Y | RW | r1(Y) before w2(Y) |
T1 → T2 |
r2(X) |
w1(X) |
X | WR | r2(X) before w1(X) |
T2 → T1 |
Based on these dependencies, the precedence graph for S1 has the edges: T1 → T2 and T2 → T1. This creates a cycle: T1 → T2 → T1.
Since the precedence graph for S1 contains a cycle, S1 is not conflict serializable.
Schedule S2 is given as: r1(X) ; r2(X) ; r2(Y) ; w2(Y) ; r1(Y) ; w1(X)
Let's identify pairs of conflicting operations involving different transactions in S2:
r1(X) and r2(X): T1 reads X, T2 reads X. RR conflict. No conflict.r2(X) and w1(X): T2 reads X, T1 writes X. r2(X) appears before w1(X) in S2. This is a WR conflict. Dependency: T2 → T1.r2(Y) and r1(Y): T2 reads Y, T1 reads Y. RR conflict. No conflict.w2(Y) and r1(Y): T2 writes Y, T1 reads Y. w2(Y) appears before r1(Y) in S2. This is an RW conflict. Dependency: T2 → T1.w2(Y)/w1(X) (different data item) do not create conflicts.The conflicts involving operations on the same data item by different transactions are:
| Operation 1 | Operation 2 | Data Item | Type | Order in S2 | Dependency (Precedence Graph Edge) |
|---|---|---|---|---|---|
r2(X) |
w1(X) |
X | WR | r2(X) before w1(X) |
T2 → T1 |
w2(Y) |
r1(Y) |
Y | RW | w2(Y) before r1(Y) |
T2 → T1 |
Based on these dependencies, the precedence graph for S2 has only one edge: T2 → T1. There is no cycle in the graph.
Since the precedence graph for S2 contains no cycle, S2 is conflict serializable. It is conflict equivalent to the serial schedule where T2 runs completely before T1.
Based on our analysis:
Therefore, S1 is not conflict serializable and S2 is conflict serializable.
| Concept | Description | Relevance to Serializability |
|---|---|---|
| Concurrency | Running multiple database transactions concurrently. | Increases performance but introduces potential consistency issues. |
| Serial Schedule | Transactions executing sequentially, one after another. | Provides a correct, consistent baseline for results. |
| Serializability | A concurrent schedule's result is same as some serial schedule. | The goal of concurrency control mechanisms to maintain database consistency. |
| Conflict Operations | Operations from different transactions accessing same data, at least one being a write (RW, WR, WW). | Used to define conflict equivalence and build the precedence graph. |
| Precedence Graph | Directed graph showing dependencies based on conflict operations order. | A cycle in the graph means the schedule is NOT conflict serializable. |
While conflict serializability is a common way to ensure serializability, there are other aspects and types to consider in database concurrency control.
Understanding these concepts helps in designing and analyzing systems that handle concurrent transactions effectively while preserving data integrity.
Consider the following statements:
Statement I: Conservative 2 PL is a deadlock-free protocol.
Statement II: Thomas's write rule enforces conflict serializability.
Statement III: Timestamp ordering protocol ensures serializability based on the order of transaction timestamps.
Which of the following is correct?
Consider the following four schedules due to three transactions (indicated by the subscript) using read and write on a data item X, denoted by r(X) and w(X) respectively. Which one of them is conflict serializable ?
S 1 : r 1(X); r 2(X); w 1(X); r 3(X); w 2(X)
S2: r 2(X); r 1(X); w 2(X); r 3(X); w 1(X)
S 3 : r 3(X); r 2(X); r 1(X); w 2(X); w 1(X)
S4 : r 2(X); w 2(X); r 3(X); r 1(X); w 1(X)
Suppose a database schedule 5 involves transactions T 1, T 2, .............T nConsider the precedence graph of S with vertices representing the transactions and edges representing the conflicts. If S is serializable, which one of the following orderings of the vertices of the precedence graph is guaranteed to yield a serial schedule ?