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Concurrency Control

Concurrency Control. Complete Schedule. T 1. R 1 (X). W 1 (X). C 1. T 2. R 2 (X). W 2 (X). C 2. Example 1. T 1 : {R 1 (x), x = x+1, W 1 (x), C 1 } T 2 : {R 2 (x), x = x+1, W 2 (x), C 2 } Conflicting operations between T 1 and T 2 :

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Concurrency Control

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  1. Concurrency Control

  2. Complete Schedule T1 R1(X) W1(X) C1 T2 R2(X) W2(X) C2 Example 1 T1: {R1(x), x = x+1, W1(x), C1} T2: {R2(x), x = x+1, W2(x), C2} Conflicting operations between T1 and T2: { (R1(x), W2(x)) , (R2(x), W1(x)) , (W2(x),W1(x)) } We need to order these conflicting operations, e.g., R1(x)  W2(x) R2(x)  W1(x) W2(x)  W1(x)

  3. Another Way of Representing Schedules T1 R1(X) W1(X) C1 T2 R2(X) W2(X) C2 Note the explicit ordering of R1(X) and R2(X)

  4. T2 R2(x)  W2(y)  W2(x)  C2 T3 R3(y)  W3(x)  W3(y)  W3(z)  C3 T1 R1(x)  W1(x)  C1 Example 2 T1 = R1(x)  W1(x)  C1 T2 = R2(x)  W2(y)  W2(x)  C2 T3 = R3(y)  W3(x)  W3(y)  W3(z)  C3

  5. T2 R2(x)  W2(y)  W2(x)  C2 T3 R3(y)  W3(x)  W3(y)  W3(z)  C3 T1 R1(x)  W1(x)  C1 Example 2 in Table Form

  6. Precedence Graph • A set of transactions may have many possible schedules, which ones are correct? • Answer: use precedence graph • Precedence graph is a directed graph: • nodes/vertices as transactions • edges denoting precedence relationship between conflicting operations of different transactions. • For a pair of conflicting operations Oij Tiand Okl  Tk, • if Oij Okl then have an edge from Ti to Tk, else • if Okl Oij then have an edge from Tk to Ti • If there are more than one edge between two nodes, retain only one of the edges

  7. R1(X) W1(X) C1 T1 T2 R2(X) W2(X) C2 Precedence Graph of Example 1 Since there are two transactions, we have two nodes T1 and T2 R1(X)  W2(X)  edge T1 T2; R2(X)  W1(X)  edge T2 T1, T1 T2 W2(X)  W1(X)  edge T2 T1 A cycle in Precedence Graph implies potential for database inconsistency

  8. R2(x)  W2(y)  W2(x)  C2 T2 T3 R3(y)  W3(x)  W3(y)  W3(z)  C3 T1 R1(x)  W1(x)  C1 Precedence Graph of Example 2 T1 = R1(x)  W1(x)  C1 T2 = R2(x)  W2(y)  W2(x)  C2 T3 = R3(y)  W3(x)  W3(y)  W3(z)  C3 T1 T3 T2 T1  T3 because W1(x)  W3(x); T3  T2 because W3(x)  R2(x); T1  T2 because W1(x)  W2(x) No cycle in precedence graph so no database inconsistency.

  9. R2(x)  W2(y)  W2(x)  C2 T1  T3  T2 R3(y)  W3(x)  W3(y)  W3(z)  C3 R1(x)  W1(x)  C1 Using Another Schedule for Example 2 T1 = R1(x)  W1(x)  C1 T2 = R2(x)  W2(y)  W2(x)  C2 T3 = R3(y)  W3(x)  W3(y)  W3(z)  C3 PG has a cycle; the database may not be consistent

  10. T1 T2 T3 Another Example on Serializability Schedule is NOT serializable

  11. R2(x)  W2(y) Schedule = R3(y)  W3(x)  W3(y) R1(x)  W1(x)  C1 Run-time Construction of Precedence Graph T1 = R1(x)  W1(x)  C1 T2 = R2(x)  W2(y)  W2(x)  C2 T3 = R3(y)  W3(x)  W3(y)  W3(z)  C3 • The scheduler keeps building the Precedence Graph as the operations of active transactions are scheduled so that the PG is acyclic. • The scheduler delays scheduling those operation that can cause a cycle in PG

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