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CS542

CS542. Concurrency Control: Theory and Protocol. Professor Elke A. Rundensteiner. For Schedule D: Sd=r 1 (A)w 1 (A)r 2 (A)w 2 (A) r 2 (B)w 2 (B)r 1 (B)w 1 (B). T 1  T 2 T 2  T 1. T 1 T 2 . D cannot be rearranged into serial schedule.

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CS542

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  1. CS542 Concurrency Control: Theory and Protocol Professor Elke A. Rundensteiner

  2. 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 D cannot be rearranged into serial schedule

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

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

  5. Definition A schedule is conflict serializable if it is conflict equivalent to some (any) serial schedule.

  6. How determine this ? One Answer: A Precedence Graph !

  7. 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

  8. 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?

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

  10. 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)

  11. Note: 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) P(S1)=P(S2)  S1, S2 conflict equivalent ??

  12. 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

  13. 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!

  14. How to enforce serializable schedules?

  15. How to enforce serializable schedules? Option 1 : try all possible swaps of non-conflicting operation pairs to determine if the schedule can be turned into a serial one, i.e., if the schedule is ‘good’.

  16. How to enforce serializable schedules? Option 2: a) run system, recording P(S); b) at end of day, check for P(S) cycles c) and declare if execution was good

  17. How to enforce serializable schedules? Option 3: prevent P(S) cycles from occurring T1 T2 ….. Tn Scheduler DB

  18. A locking protocol Two new actions: lock (exclusive): li (A) unlock: ui (A) T1 T2 lock table scheduler

  19. Rule #1: Well-formed transactions Ti: … li(A) … pi(A) … ui(A) ...

  20. Rule #2 Legal scheduler S = …….. li(A) ………... ui(A) ……... no lj(A)

  21. Exercise: • What schedules are legal?What transactions are well-formed? S1 = l1(A)l1(B)r1(A)w1(B)l2(B)u1(A)u1(B) r2(B)w2(B)u2(B)l3(B)r3(B)u3(B) S2 = l1(A)r1(A)w1(B)u1(A)u1(B) l2(B)r2(B)w2(B)l3(B)r3(B)u3(B) S3 = l1(A)r1(A)u1(A)l1(B)w1(B)u1(B) l2(B)r2(B)w2(B)u2(B)l3(B)r3(B)u3(B)

  22. Exercise: • What schedules are legal?What transactions are well-formed? S1 = l1(A)l1(B)r1(A)w1(B)l2(B)u1(A)u1(B) r2(B)w2(B)u2(B)l3(B)r3(B)u3(B) S2 = l1(A)r1(A)w1(B)u1(A)u1(B) l2(B)r2(B)w2(B)l3(B)r3(B)u3(B) S3 = l1(A)r1(A)u1(A)l1(B)w1(B)u1(B) l2(B)r2(B)w2(B)u2(B)l3(B)r3(B)u3(B)

  23. Schedule F : Adding locking ! How? T1 T2 l1(A);Read(A) A A+100;Write(A);u1(A) l2(A);Read(A) A Ax2;Write(A);u2(A) l2(B);Read(B) B Bx2;Write(B);u2(B) l1(B);Read(B) B B+100;Write(B);u1(B) Legal ? Well-formed ?

  24. A B 25 25 125 250 50 150 250 150 Schedule F : Adding locking ! How? T1 T2 l1(A);Read(A) A A+100;Write(A);u1(A) l2(A);Read(A) A Ax2;Write(A);u2(A) l2(B);Read(B) B Bx2;Write(B);u2(B) l1(B);Read(B) B B+100;Write(B);u1(B)

  25. Rule #3 Two phase locking (2PL)for transactions Ti = ……. li(A) ………... ui(A) ……... no unlocks no locks

  26. 2 Phase Locking Protocol # locks held by Ti Time Growing Shrinking Phase Phase

  27. Schedule F : Can 2PL fix it ? T1 T2 l1(A);Read(A) A A+100;Write(A);u1(A) l2(A);Read(A) A Ax2;Write(A);u2(A) l2(B);Read(B) B Bx2;Write(B);u2(B) l1(B);Read(B) B B+100;Write(B);u1(B)

  28. Schedule G T1 T2 l1(A);Read(A) A A+100;Write(A) l1(B); u1(A) l2(A);Read(A) A Ax2;Write(A);

  29. Schedule G T1 T2 l1(A);Read(A) A A+100;Write(A) l1(B); u1(A) l2(A);Read(A) A Ax2;Write(A);l2(B) delayed

  30. Schedule G T1 T2 l1(A);Read(A) A A+100;Write(A) l1(B); u1(A) l2(A);Read(A) A Ax2;Write(A);l2(B) Read(B);B B+100 Write(B); u1(B) delayed

  31. Schedule G T1 T2 l1(A);Read(A) A A+100;Write(A) l1(B); u1(A) l2(A);Read(A) A Ax2;Write(A);l2(B) Read(B);B B+100 Write(B); u1(B) l2(B); u2(A);Read(B) B Bx2;Write(B);u2(B); delayed

  32. A B 25 25 125 250 125 250 250 250 We Got the GOOD Schedule !!!! 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);

  33. Schedule H (T2 reversed) T1 T2 l1(A); Read(A) l2(B);Read(B) A A+100;Write(A) B Bx2;Write(B) . . . . . . What’s happening ?

  34. Schedule H (T2 reversed) T1 T2 l1(A); Read(A) l2(B);Read(B) A A+100;Write(A) B Bx2;Write(B) l1(B) l2(A) delayed delayed

  35. Deadlocked transactions can arise in ‘good’ schedules (generated due to 2PL locking). • We assume that when they are rolled back: • They have no effect • They do not appear in schedule E.g., Schedule H = This space intentionally left blank!

  36. Correctness of 2PC Protocol : Show that rules #1,2,3 • Conflict-serializable schedules

  37. Conflict rules for li(A), ui(A): • li(A), lj(A) conflict • li(A), uj(A) conflict

  38. Theorem Rules #1,2,3  conflict (2PL) serializable schedule To help in proof: Definition Shrink(Ti) = SH(Ti) = first unlock action of Ti

  39. Proof of lemma: Ti  Tj means that S = … pi(A) … qj(A) …; p,q conflict By rules 1,2: S = … pi(A) … ui(A) … lj(A) ... qj(A) … Lemma Ti  Tj in S  SH(Ti) <S SH(Tj) SH(Ti)SH(Tj) By rule 3: So, SH(Ti) <S SH(Tj)

  40. Theorem Rules #1,2,3  conflict (2PL) serializable schedule Proof: (1) Assume P(S) has cycle T1 T2 …. Tn T1 (2) By lemma: SH(T1) <SH(T2) < ... <SH(T1) (3) Impossible, so P(S) acyclic (4)  S is conflict serializable

  41. Beyond this simple 2PL protocol, it is all a matter of improving performance and allowing yet more concurrency…. • Shared locks • Multiple granularity • Inserts, deletes and phantoms • Other types of C.C. mechanisms

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