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CS4432: Database Systems II

CS4432: Database Systems II. Lecture #21 Concurrency Control : Theory. Professor Elke A. Rundensteiner. Concepts. Transaction : sequence of r i (x), w i (x) actions Conflicting actions : read/write on same resource A: r 1(A) w 2(A) w 1(A)

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CS4432: Database Systems II

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  1. CS4432: Database Systems II Lecture #21 Concurrency Control : Theory Professor Elke A. Rundensteiner concurrency control

  2. Concepts Transaction: sequence of ri(x), wi(x) actions Conflicting actions: read/write on same resource A: r1(A) w2(A) w1(A) w2(A) r1(A) w2(A) Schedule: represents chronological order in which actions are executed Serial schedule: no interleaving of trans/actions concurrency control

  3. 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); concurrency control

  4. Example: Sc=r1(A)w1(A)r2(A)w2(A)r1(B)w1(B)r2(B)w2(B) Sc’=r1(A)w1(A) r1(B)w1(B)r2(A)w2(A)r2(B)w2(B) T1 T2 concurrency control

  5. Returning to Sc 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, I.e., in this case (T1,T2). concurrency control

  6. 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); concurrency control

  7. Now for Sd: Sd=r1(A)w1(A)r2(A)w2(A) r2(B)w2(B)r1(B)w1(B) Sd=r1(A)w1(A) r1(B)w1(B)r2(A)w2(A)r2(B)w2(B) T2 T1 concurrency control

  8. Or, let’s try for Sd: Sd=r1(A)w1(A)r2(A)w2(A) r2(B)w2(B)r1(B)w1(B) Sd=r2(A)w2(A)r2(B)w2(B) r1(A)w1(A)r1(B)w1(B) T2 T1 concurrency control

  9. In short, Schedule D cannot be “fixed” : Sd=r1(A)w1(A)r2(A)w2(A) r2(B)w2(B)r1(B)w1(B) • as a matter of fact, there seems to • be no save way to transform this Sd • into an equivalent serial schedule !? concurrency control

  10. For Schedule D: Sd=r1(A)w1(A)r2(A)w2(A) r2(B)w2(B)r1(B)w1(B) T1 T2 T2 T1 T1 T2 Sd cannot be rearranged into serial schedule concurrency control

  11. Idea: Swap non-conflicting operation pairs to see if you can go to a serial schedule. concurrency control

  12. Definition S1, S2 are conflict equivalentschedules if S1 can be transformed into S2 by a series of swaps of non-conflicting actions. concurrency control

  13. Definition A schedule is conflict serializable if it is conflict equivalent to some serial schedule. concurrency control

  14. How determine this ? Answer: A Precedence Graph ! concurrency control

  15. Precedence graph P(S) (S is schedule) Nodes: transactions in S Arcs: Ti  Tj whenever - pi(A), qj(A) are actions in S - pi(A) <S qj(A) - at least one of pi, qj is a write concurrency control

  16. Exercise: • What is P(S) forS = w3(A) w2(C) r1(A) w1(B) r1(C) w2(A) r4(A) w4(D) • Is S conflict-serializable? concurrency control

  17. Another Exercise: • What is P(S) forS = w1(A) r2(A) r3(A) w4(A) ? • Is S conflict-serializable? concurrency control

  18. Proof: Assume P(S1)  P(S2)  Ti: Ti  Tj in S1 and not in S2  S1 = …pi(A)... qj(A)… pi, qj S2 = …qj(A)…pi(A)... conflict  S1, S2 not conflict equivalent Lemma S1, S2 conflict equivalent  P(S1)=P(S2) concurrency control

  19. Note: P(S1)=P(S2)  S1, S2 conflict equivalent? concurrency control

  20. Counter example: S1=w1(A) r2(A) w2(B) r1(B) S2=r2(A) w1(A) r1(B) w2(B) Note: P(S1)=P(S2)  S1, S2 conflict equivalent concurrency control

  21. Theorem 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 concurrency control

  22. Theorem P(S1) acyclic  S1 conflict serializable T1 T2 T3 T4 () 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! concurrency control

  23. Theorem holds: P(S1) acyclic  S1 conflict serializable concurrency control

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