1 / 25

Concurrency Control 18.1 Serial and Serializable Schedules ~ 18.2 Conflict-Serializability

Concurrency Control 18.1 Serial and Serializable Schedules ~ 18.2 Conflict-Serializability. 2008. 5. 21 황원석 데이터 마이닝 및 지식공학 연구실 한양대학교. Concurrency Control. 여러 Transactions 의 상호작용에 의해 DB 가 inconsistent 해 질 수 있음 Transaction User 나 application 에 의해 실행되는 action 의 시리즈

lleach
Download Presentation

Concurrency Control 18.1 Serial and Serializable Schedules ~ 18.2 Conflict-Serializability

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Concurrency Control18.1 Serial and Serializable Schedules~18.2 Conflict-Serializability 2008. 5. 21 황원석 데이터 마이닝 및 지식공학 연구실 한양대학교

  2. Concurrency Control • 여러 Transactions의 상호작용에 의해 DB가 inconsistent해 질 수 있음 • Transaction • User나 application에 의해 실행되는 action의 시리즈 • DB의 컨텐츠를 접근(access)하거나, 변경(change)함 • Concurrency Control • DB를 consistent하게 유지할 수 있도록 제어하는 방법 • 다른 transactions에 속하는 여러 actions의 실행 순서를 제약하는 방법 • Scheduler • DBMS의 컴포넌트 • Concurrency Control을 수행 • Transactions의액션을 실행하거나, 보류시킴

  3. Concurrency Control • DB의 consistency가 유지되기 위하여 필요한 요소 • 추상적으로 필요한 조건(abstract requirement) • Serializability • 중요한 구체적인 제약 조건(important, stronger condition) • Conflict-serializability • Scheduler에 구현되어야 할 techniques • Locking • Timestamping • Validation

  4. Concurrency Control • Correctness Principle • Transactions는 각각 독립적으로 실행됨 • Consistent state에서 consistent state로 변형됨 • 실제로는 여러 Transactions이 동시에 함께 실행되므로 지켜지기 어려움 • “schedules”를 고려해야 함 • Schedules • Transaction이 하나씩 실행된 결과와 같은 결과를 같은 결과를 가지도록 보장 T1 T2 … Tn DB (consistency constraints)

  5. Example T1: Read(A) T2: Read(A) A  A+100 A  A2 Write(A) Write(A) Read(B) Read(B) B  B+100 B  B2 Write(B) Write(B) Constraint: A=B (A와 B가 최초에는 같은 값을 가지고 있다고 가정)

  6. A B 25 25 125 125 250 250 250 250 Schedule A T1 T2 Read(A); A  A+100; Write(A); Read(B); B  B+100; Write(B); Read(A);A  A2; Write(A); Read(B);B  B2; Write(B); Schedule A is serial (각 transaction들의 action이 섞이지 않은 상태로 실행됨)

  7. A B 25 25 50 50 150 150 150 150 Schedule B T1 T2 Read(A);A  A2; Write(A); Read(B);B  B2; Write(B); Read(A); A  A+100 Write(A); Read(B); B  B+100; Write(B); Schedule B is serial

  8. A B 25 25 125 250 125 250 250 250 Schedule C T1 T2 Read(A); A  A+100 Write(A); Read(A);A  A2; Write(A); Read(B); B  B+100; Write(B); Read(B);B  B2; Write(B); Schedule C is serializable but not serial (어떤 schedule이 실행된 결과가 어떤 serial schedule이 수행된 결과와 동일한 경우)

  9. A B 25 25 125 250 50 150 250 150 Schedule D T1 T2 Read(A); A  A+100 Write(A); Read(A);A  A2; Write(A); Read(B);B  B2; Write(B); Read(B); B  B+100; Write(B); Schedule D is not serializable but not serial

  10. A B 25 25 125 125 25 125 125 125 Same as Schedule D but with new T2’ Schedule E T1 T2’ Read(A); A  A+100 Write(A); Read(A);A  A1; Write(A); Read(B);B  B1; Write(B); Read(B); B  B+100; Write(B); Schedule E is serializable (현실적으로 이러한 연산이 Serializable인지 파악하기 어려움)

  11. Want schedules that are “good”, regardless of • initial state • transaction semantics • Only look at order of read and writes • Example: • Sc = r1(A)w1(A)r2(A)w2(A)r1(B)w1(B)r2(B)w2(B)

  12. Conflict Serializability • Transaction • sequence of ri(x), wi(x) actions • Schedule • represents chronological order in which actions are executed • Serial schedule • no interleaving of actions or transactions • Conflict Serializability • A condition that is sufficient to assure that a schedule is serializable • Conflict • A pair of consecutive actions in a schedule • If their order is interchanged, then the behavior of at least one of the transactions involved can change • Conflicting actions • r1(A), w2(A) • w2(A), r1(A) • w1(A), w2(A)

  13. T2 issues write(B,S) input(B) completes System issues output(B) System issues input(B) output(B) completes B  S What about concurrent actions? Ti issues System Input(x) t  x read(x,t) issues completes input(x) time

  14. Notes 09 So net effect is either • S=…r1(x)…w2(B)… or • S=…w2(B)…r1(x)…

  15. What about conflicting, concurrent actions on same object? start r1(A) end r1(A) start w2(A) end w2(A) time • Assume equivalent to either r1(A) w2(A) or w2(A) r1(A) •  low level synchronization mechanism • Assumption called “atomic actions”

  16. Definition • S1, S2 are conflict equivalent schedules • if S1 can be transformed into S2 by a series of swaps on non-conflicting actions. • A schedule is conflict serializableif it is conflict equivalent to some serial schedule. • Example Serial schedule Conflict serializable Not conflict serializable

  17. Precedence graph, P(S) • Nodes • transactions in S • Arcs(Ti → Tj ) • pi(A), qj(A) are actions in S • pi(A) <S qj(A) • at least one of pi, qj is a write • P(S) is acyclic • P(S) is conflict-serializable • P(S) is cyclic • P(S) is not conflict-serializable

  18. Exercise • What is P(S) for S = w3(A)w2(C)r1(A)w1(B)r1(C)w2(A)r4(A)w4(D) • Is S serializable? • No • Precedence graph가 cyclic이면, conflict-serializable이 아님 T3 T2 T1 T4

  19. Exercise • What is P(S) for S = w1(A)r2(A)r3(A)w4(A) T1 T2 T4 T3

  20. Lemma • 두 schedule이 conflict equivalent하다는 것은 P(S1) = P(S2)하기 위한 충분조건임 • 증명 • S1, S2 conflict equivalent ⇒ P(S1) = P(S2) • Assume P(S1) ≠ P(S2) • ∃ Ti: Ti → Tj inS1 and not in S2 • S1 = …pi(A) … qj(A) … S2 = …qj(A) … pi(A) … (단, pi, qj는 conflicting actions) • S1, S2 not conflict equivalent • P(S1) = P(S2)  S1, S2 conflict equivalent • Counter example • S1 = w1(A)r2(A)w2(B)r1(B) • S2 = r2(A)w1(A)r1(B)w2(B) T2 T1

  21. Theorem • P(S1) acyclic ⇔ S1 conflict serializable • 증명 • P(S1) acyclic  S1 conflict serializable • Assume S1 is conflict serializable • ∃Ss: Ss, S1 conflict equivalent • P(Ss) = P(S1) • P(S1) acyclic since P(Ss) is acyclic • P(S1) acyclic ⇒ S1 conflict serializable • Assume P(S1) is acyclic • Transform S1 as follows: • (1) Take T1 to be transaction with no incident arcs • (2) Move all T1 actions to the front • S1 = ……. qj(A)…….p1(A)….. • (3) we now have S1 = < T1 actions ><... rest ...> • (4) repeat above steps to serialize rest! T1 T2 T3 T4

  22. Slide 10 Sc’=r1(A)w1(A) r1(B)w1(B)r2(A)w2(A)r2(B)w2(B) Sc = r1(A)w1(A)r2(A)w2(A)r1(B)w1(B)r2(B)w2(B) T1 T2

  23. Slide 11 However, for Sd: Sd = r1(A)w1(A)r2(A)w2(A) r2(B)w2(B)r1(B)w1(B) X X • as a matter of fact, • T2 must precede T1 • in any equivalent schedule, • i.e., T2 T1

  24. Slide 12 T1 T2 Sd cannot be rearranged into a serial schedule Sd is not “equivalent” to any serial schedule Sd is “bad” • T2 T1 • Also, T1 T2

  25. Returning to Sc Slide 13 Sc=r1(A)w1(A)r2(A)w2(A)r1(B)w1(B)r2(B)w2(B) T1 T2 T1 T2 no cycles  Sc is “equivalent” to a serial schedule (in this case (T1,T2) )

More Related