1. CS 245Notes 091 CS 245: Database System Principles Notes 09: Concurrency Control Hector Garcia-Molina

2. CS 245Notes 092 Chapter 18 [18] Concurrency Control T1T2…Tn DB (consistency constraints)

3. CS 245Notes 093 Example: T1:Read(A)T2:Read(A) A  A+100A  A  2Write(A)Read(B) B  B+100B  B  2Write(B) Constraint: A=B

4. CS 245Notes 094 Schedule A T1T2 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);

5. CS 245Notes 095 Schedule A T1T2 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); AB25 125 250 250

6. CS 245Notes 096 Schedule B T1T2 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);

7. CS 245Notes 097 Schedule B T1T2 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); AB25 50 150 150

8. CS 245Notes 098 Schedule C T1T2 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);

9. CS 245Notes 099 Schedule C T1T2 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); AB25 125 250 125 250250

10. CS 245Notes 0910 Schedule D T1T2 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);

11. CS 245Notes 0911 Schedule D T1T2 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); AB25 125 250 50 150 250150

12. CS 245Notes 0912 Schedule E T1T2’ 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); Same as Schedule D but with new T2’

13. CS 245Notes 0913 Schedule E T1T2’ 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); AB25 125 25 125125 Same as Schedule D but with new T2’

14. CS 245Notes 0914 Want schedules that are “good”, regardless of –initial state and –transaction semantics Only look at order of read and writes Example: Sc=r 1 (A)w 1 (A)r 2 (A)w 2 (A)r 1 (B)w 1 (B)r 2 (B)w 2 (B)

15. CS 245Notes 0915 Sc’=r 1 (A)w 1 (A) r 1 (B)w 1 (B)r 2 (A)w 2 (A)r 2 (B)w 2 (B) T 1 T 2 Example: Sc=r 1 (A)w 1 (A)r 2 (A)w 2 (A)r 1 (B)w 1 (B)r 2 (B)w 2 (B)

16. CS 245Notes 0916 However, for Sd: 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) as a matter of fact, T 2 must precede T 1 in any equivalent schedule, i.e., T 2  T 1

17. CS 245Notes 0917 T 1 T 2 Sd cannot be rearranged into a serial schedule Sd is not “equivalent” to any serial schedule Sd is “bad” T 2  T 1 Also, T 1  T 2

18. CS 245Notes 0918 Returning to Sc Sc=r 1 (A)w 1 (A)r 2 (A)w 2 (A)r 1 (B)w 1 (B)r 2 (B)w 2 (B) T 1  T 2 T 1  T 2

19. CS 245Notes 0919 Returning to Sc Sc=r 1 (A)w 1 (A)r 2 (A)w 2 (A)r 1 (B)w 1 (B)r 2 (B)w 2 (B) T 1  T 2 T 1  T 2  no cycles  Sc is “equivalent” to a serial schedule (in this case T 1,T 2 )

20. CS 245Notes 0920 Concepts Transaction: sequence of r i (x), w i (x) actions Conflicting actions: r 1(A) w 2(A) w 1(A) w 2(A) r 1(A) w 2(A) Schedule: represents chronological order in which actions are executed Serial schedule: no interleaving of actions or transactions

21. CS 245Notes 0921 Is it OK to model reads & writes as occurring at a single point in time in a schedule? S=… r 1 (x) … w 2 (b) …

22. CS 245Notes 0922 What about conflicting, concurrent actions on same object? start r 1 (A)end r 1 (A) start w 2 (A) end w 2 (A) time

23. CS 245Notes 0923 Assume equivalent to either r 1 (A) w 2 (A) orw 2 (A) r 1 (A)  low level synchronization mechanism Assumption called “atomic actions” What about conflicting, concurrent actions on same object? start r 1 (A)end r 1 (A) start w 2 (A) end w 2 (A) time

24. CS 245Notes 0924 Definition S 1, S 2 are conflict equivalent schedules if S 1 can be transformed into S 2 by a series of swaps on non-conflicting actions.

25. CS 245Notes 0925 Definition A schedule is conflict serializable if it is conflict equivalent to some serial schedule.

26. CS 245Notes 0926 Nodes: transactions in S Arcs: Ti  Tj whenever - p i (A), q j (A) are actions in S - p i (A) < S q j (A) - at least one of p i, q j is a write Precedence graph P(S) (S is schedule )

27. CS 245Notes 0927 Exercise: What is P(S) for S = w 3 (A) w 2 (C) r 1 (A) w 1 (B) r 1 (C) w 2 (A) r 4 (A) w 4 (D) Is S serializable?

28. CS 245Notes 0928 Another Exercise: What is P(S) for S = w 1 (A) r 2 (A) r 3 (A) w 4 (A) ?

29. CS 245Notes 0929 Lemma S 1, S 2 conflict equivalent  P(S 1 )=P(S 2 )

30. CS 245Notes 0930 Lemma S 1, S 2 conflict equivalent  P(S 1 )=P(S 2 ) Proof: Assume P(S 1 )  P(S 2 )   T i : T i  T j in S 1 and not in S 2  S 1 = …p i (A)... q j (A)… p i, q j S 2 = …q j (A)…p i (A)... conflict  S 1, S 2 not conflict equivalent

31. CS 245Notes 0931 Note: P(S 1 )=P(S 2 )  S 1, S 2 conflict equivalent

32. CS 245Notes 0932 Note: P(S 1 )=P(S 2 )  S 1, S 2 conflict equivalent Counter example: S 1 =w 1 (A) r 2 (A) w 2 (B) r 1 (B) S 2 =r 2 (A) w 1 (A) r 1 (B) w 2 (B)

33. CS 245Notes 0933 Theorem P(S 1 ) acyclic  S 1 conflict serializable (  ) Assume S 1 is conflict serializable   S s : S s, S 1 conflict equivalent  P(S s ) = P(S 1 )  P(S 1 ) acyclic since P(S s ) is acyclic

34. CS 245Notes 0934 (  ) Assume P(S 1 ) is acyclic Transform S 1 as follows: (1) Take T 1 to be transaction with no incident arcs (2) Move all T 1 actions to the front S 1 = ……. q j (A)……. p 1 (A)….. (3) we now have S 1 = (4) repeat above steps to serialize rest! T 1 T 2 T 3 T 4 Theorem P(S 1 ) acyclic  S 1 conflict serializable

35. CS 245Notes 0935 How to enforce serializable schedules? Option 1: run system, recording P(S); at end of day, check for P(S) cycles and declare if execution was good

36. CS 245Notes 0936 Option 2: prevent P(S) cycles from occurring T 1 T 2 …..T n Scheduler DB How to enforce serializable schedules?

37. CS 245Notes 0937 A locking protocol Two new actions: lock (exclusive):l i (A) unlock:u i (A) scheduler T 1 T 2 lock table

38. CS 245Notes 0938 Rule #1: Well-formed transactions T i : … l i (A) … p i (A) … u i (A)...

39. CS 245Notes 0939 Rule #2 Legal scheduler S = …….. l i (A) ………... u i (A) ……... no l j (A)

40. CS 245Notes 0940 What schedules are legal? What transactions are well-formed? S1 = l 1 (A)l 1 (B)r 1 (A)w 1 (B)l 2 (B)u 1 (A)u 1 (B) r 2 (B)w 2 (B)u 2 (B)l 3 (B)r 3 (B)u 3 (B) S2 = l 1 (A)r 1 (A)w 1 (B)u 1 (A)u 1 (B) l 2 (B)r 2 (B)w 2 (B)l 3 (B)r 3 (B)u 3 (B) S3 = l 1 (A)r 1 (A)u 1 (A)l 1 (B)w 1 (B)u 1 (B) l 2 (B)r 2 (B)w 2 (B)u 2 (B)l 3 (B)r 3 (B)u 3 (B) Exercise:

41. CS 245Notes 0941 What schedules are legal? What transactions are well-formed? S1 = l 1 (A)l 1 (B)r 1 (A)w 1 (B)l 2 (B)u 1 (A)u 1 (B) r 2 (B)w 2 (B)u 2 (B)l 3 (B)r 3 (B)u 3 (B) S2 = l 1 (A)r 1 (A)w 1 (B)u 1 (A)u 1 (B) l 2 (B)r 2 (B)w 2 (B)l 3 (B)r 3 (B)u 3 (B) u 2 (B)? S3 = l 1 (A)r 1 (A)u 1 (A)l 1 (B)w 1 (B)u 1 (B) l 2 (B)r 2 (B)w 2 (B)u 2 (B)l 3 (B)r 3 (B)u 3 (B) Exercise:

42. CS 245Notes 0942 Schedule F T1 T2 l 1 (A);Read(A) A A+100;Write(A);u 1 (A) l 2 (A);Read(A) A Ax2;Write(A);u 2 (A) l 2 (B);Read(B) B Bx2;Write(B);u 2 (B) l 1 (B);Read(B) B B+100;Write(B);u 1 (B)

43. CS 245Notes 0943 Schedule F T1 T2 25 25 l 1 (A);Read(A) A A+100;Write(A);u 1 (A) 125 l 2 (A);Read(A) A Ax2;Write(A);u 2 (A) 250 l 2 (B);Read(B) B Bx2;Write(B);u 2 (B) 50 l 1 (B);Read(B) B B+100;Write(B);u 1 (B) 150 250 150 A B

44. CS 245Notes 0944 Rule #3 Two phase locking (2PL) for transactions T i = ……. l i (A) ………... u i (A) ……... no unlocks no locks

45. CS 245Notes 0945 # locks held by Ti Time Growing Shrinking Phase Phase

46. CS 245Notes 0946 Schedule G delayed

47. CS 245Notes 0947 Schedule G delayed

48. CS 245Notes 0948 Schedule G delayed

49. CS 245Notes 0949 Schedule H (T 2 reversed) delayed

50. CS 245Notes 0950 Assume deadlocked transactions are rolled back –They have no effect –They do not appear in schedule E.g., Schedule H = This space intentionally left blank!

51. CS 245Notes 0951 Next step: Show that rules #1,2,3  conflict- serializable schedules

52. CS 245Notes 0952 Conflict rules for l i (A), u i (A): l i (A), l j (A) conflict l i (A), u j (A) conflict Note: no conflict,,...

53. CS 245Notes 0953 Theorem Rules #1,2,3  conflict (2PL) serializable schedule

54. CS 245Notes 0954 Theorem Rules #1,2,3  conflict (2PL) serializable schedule To help in proof: Definition Shrink(Ti) = SH(Ti) = first unlock action of Ti

55. CS 245Notes 0955 Lemma Ti  Tj in S  SH(Ti) < S SH(Tj)

56. CS 245Notes 0956 Lemma Ti  Tj in S  SH(Ti) < S SH(Tj) Proof of lemma: Ti  Tj means that S = … p i (A) … q j (A) …; p,q conflict By rules 1,2: S = … p i (A) … u i (A) … l j (A)... q j (A) …

57. CS 245Notes 0957 Lemma Ti  Tj in S  SH(Ti) < S SH(Tj) Proof of lemma: Ti  Tj means that S = … p i (A) … q j (A) …; p,q conflict By rules 1,2: S = … p i (A) … u i (A) … l j (A)... q j (A) … By rule 3: SH(Ti) SH(Tj) So, SH(Ti) < S SH(Tj)

58. CS 245Notes 0958 Proof: (1) Assume P(S) has cycle T 1  T 2  …. T n  T 1 (2) By lemma: SH(T 1 ) < SH(T 2 ) <... < SH(T 1 ) (3) Impossible, so P(S) acyclic (4)  S is conflict serializable Theorem Rules #1,2,3  conflict (2PL) serializable schedule

59. CS 245Notes 0959 2PL subset of Serializable 2PL Serializable

60. CS 245Notes 0960 S1: w1(x) w3(x) w2(y) w1(y) S1 cannot be achieved via 2PL: The lock by T1 for y must occur after w2(y), so the unlock by T1 for x must occur after this point (and before w1(x)). Thus, w3(x) cannot occur under 2PL where shown in S1 because T1 holds the x lock at that point. However, S1 is serializable (equivalent to T2, T1, T3).

61. CS 245Notes 0961 S C : w1(A) w2(A) w1(B) w2(B) If you need a bit more practice: Are our schedules S C and S D 2PL schedules? S D : w1(A) w2(A) w2(B) w1(B)

62. CS 245Notes 0962 Beyond this simple 2PL protocol, it is all a matter of improving performance and allowing more concurrency…. –Shared locks –Multiple granularity –Inserts, deletes and phantoms –Other types of C.C. mechanisms

63. CS 245Notes 0963 Shared locks So far: S =...l 1 (A) r 1 (A) u 1 (A) … l 2 (A) r 2 (A) u 2 (A) … Do not conflict

64. CS 245Notes 0964 Shared locks So far: S =...l 1 (A) r 1 (A) u 1 (A) … l 2 (A) r 2 (A) u 2 (A) … Do not conflict Instead: S=... ls 1 (A) r 1 (A) ls 2 (A) r 2 (A) …. us 1 (A) us 2 (A)

65. CS 245Notes 0965 Lock actions l-t i (A): lock A in t mode (t is S or X) u-t i (A): unlock t mode (t is S or X) Shorthand: u i (A): unlock whatever modes T i has locked A

66. CS 245Notes 0966 Rule #1 Well formed transactions T i =... l-S 1 (A) … r 1 (A) … u 1 (A) … T i =... l-X 1 (A) … w 1 (A) … u 1 (A) …

67. CS 245Notes 0967 What about transactions that read and write same object? Option 1: Request exclusive lock T i =...l-X 1 (A) … r 1 (A)... w 1 (A)... u(A) …

68. CS 245Notes 0968 Option 2: Upgrade (E.g., need to read, but don’t know if will write…) T i =... l-S 1 (A) … r 1 (A)... l-X 1 (A) …w 1 (A)...u(A)… Think of - Get 2nd lock on A, or - Drop S, get X lock What about transactions that read and write same object?

69. CS 245Notes 0969 Rule #2 Legal scheduler S =....l-S i (A) … … u i (A) … no l-X j (A) S =... l-X i (A) … … u i (A) … no l-X j (A) no l-S j (A)

70. CS 245Notes 0970 A way to summarize Rule #2 Compatibility matrix Comp S X S true false Xfalse false

71. CS 245Notes 0971 Rule # 3 2PL transactions No change except for upgrades: (I) If upgrade gets more locks (e.g., S  {S, X}) then no change! (II) If upgrade releases read (shared) lock (e.g., S  X) - can be allowed in growing phase

72. CS 245Notes 0972 Proof: similar to X locks case Detail: l-t i (A), l-r j (A) do not conflict if comp(t,r) l-t i (A), u-r j (A) do not conflict if comp(t,r) Theorem Rules 1,2,3  Conf.serializable for S/X locks schedules

73. CS 245Notes 0973 Lock types beyond S/X Examples: (1) increment lock (2) update lock

74. CS 245Notes 0974 Example (1): increment lock Atomic increment action: IN i (A) {Read(A); A  A+k; Write(A)} IN i (A), IN j (A) do not conflict! A=7 A=5A=17 A=15 IN i (A) +2 IN j (A) +10 IN j (A) +2 IN i (A)

75. CS 245Notes 0975 CompSXI S X I

76. CS 245Notes 0976 CompSXI STFF XFFF IFFT

77. CS 245Notes 0977 Update locks

78. CS 245Notes 0978 Solution If T i wants to read A and knows it may later want to write A, it requests update lock (not shared)

79. CS 245Notes 0979 CompSXU S X U New request Lock already held in

80. CS 245Notes 0980 CompSXU STFT XFFF U TorF FF -> symmetric table? New request Lock already held in

81. CS 245Notes 0981 Note: object A may be locked in different modes at the same time... S 1 =...l-S 1 (A)…l-S 2 (A)…l-U 3 (A)… l-S 4 (A)…? l-U 4 (A)…?

82. CS 245Notes 0982 Note: object A may be locked in different modes at the same time... S 1 =...l-S 1 (A)…l-S 2 (A)…l-U 3 (A)… l-S 4 (A)…? l-U 4 (A)…? To grant a lock in mode t, mode t must be compatible with all currently held locks on object

83. CS 245Notes 0983 How does locking work in practice? Every system is different (E.g., may not even provide CONFLICT-SERIALIZABLE schedules) But here is one (simplified) way...

84. CS 245Notes 0984 (1) Don’t trust transactions to request/release locks (2) Hold all locks until transaction commits # locks time Sample Locking System:

85. CS 245Notes 0985 Ti Read(A),Write(B) l(A),Read(A),l(B),Write(B)… Read(A),Write(B) Scheduler, part I Scheduler, part II DB lock table

86. CS 245Notes 0986 Lock table Conceptually A  B C ... Lock info for B Lock info for C If null, object is unlocked Every possible object

87. CS 245Notes 0987 But use hash table: A If object not found in hash table, it is unlocked Lock info for A A... H

88. CS 245Notes 0988 Lock info for A - example tran mode wait? Nxt T_link Object:A Group mode:U Waiting:yes List: T1 S no T2 U no T3X yes  To other T3 records

89. CS 245Notes 0989 What are the objects we lock? ? Relation A Relation B... Tuple A Tuple B Tuple C... Disk block A Disk block B... DB

90. CS 245Notes 0990 Locking works in any case, but should we choose small or large objects?

91. CS 245Notes 0991 Locking works in any case, but should we choose small or large objects? If we lock large objects (e.g., Relations) –Need few locks –Low concurrency If we lock small objects (e.g., tuples,fields) –Need more locks –More concurrency

92. CS 245Notes 0992 We can have it both ways!! Ask any janitor to give you the solution... hall Stall 1Stall 2Stall 3Stall 4 restroom

93. CS 245Notes 0993 Example R1 t1t1 t2t2 t3t3 t4t4

94. CS 245Notes 0994 Example R1 t1t1 t2t2 t3t3 t4t4 T 1 (IS) T 1 (S)

95. CS 245Notes 0995 Example R1 t1t1 t2t2 t3t3 t4t4 T 1 (IS) T 1 (S), T 2 (S)

96. CS 245Notes 0996 Example (b) R1 t1t1 t2t2 t3t3 t4t4 T 1 (IS) T 1 (S)

97. CS 245Notes 0997 Example R1 t1t1 t2t2 t3t3 t4t4 T 1 (IS) T 1 (S), T 2 (IX) T 2 (IX)

98. CS 245Notes 0998 Multiple granularity CompRequestor IS IX S SIX X IS Holder IX S SIX X

99. CS 245Notes 0999 Multiple granularity CompRequestor IS IX S SIX X IS Holder IX S SIX X TTTTF F F F FFFFF FFFT FTFT FFTT

100. CS 245Notes 09100 ParentChild can belocked in IS IX S SIX X P C

101. CS 245Notes 09101 ParentChild can be locked locked inby same transaction in IS IX S SIX X P C IS, S IS, S, IX, X, SIX none X, IX, [SIX] none not necessary

102. CS 245Notes 09102 Rules (1) Follow multiple granularity comp function (2) Lock root of tree first, any mode (3) Node Q can be locked by Ti in S or IS only if parent(Q) locked by Ti in IX or IS (4) Node Q can be locked by Ti in X,SIX,IX only if parent(Q) locked by Ti in IX,SIX (5) Ti is two-phase (6) Ti can unlock node Q only if none of Q’s children are locked by Ti

103. CS 245Notes 09103 Exercise: Can T 2 access object f 2.2 in X mode? What locks will T 2 get? R1 t1t1 t2t2 t3t3 t4t4 T 1 (IX) f 2.1 f 2.2 f 3.1 f 3.2 T 1 (IX) T 1 (X)

104. CS 245Notes 09104 Exercise: Can T 2 access object f 2.2 in X mode? What locks will T 2 get? R1 t1t1 t2t2 t3t3 t4t4 T 1 (X) f 2.1 f 2.2 f 3.1 f 3.2 T 1 (IX)

105. CS 245Notes 09105 Exercise: Can T 2 access object f 3.1 in X mode? What locks will T 2 get? R1 t1t1 t2t2 t3t3 t4t4 T 1 (S) f 2.1 f 2.2 f 3.1 f 3.2 T 1 (IS)

106. CS 245Notes 09106 Exercise: Can T 2 access object f 2.2 in S mode? What locks will T 2 get? R1 t1t1 t2t2 t3t3 t4t4 T 1 (IX) f 2.1 f 2.2 f 3.1 f 3.2 T 1 (SIX) T 1 (X)

107. CS 245Notes 09107 Exercise: Can T 2 access object f 2.2 in X mode? What locks will T 2 get? R1 t1t1 t2t2 t3t3 t4t4 T 1 (IX) f 2.1 f 2.2 f 3.1 f 3.2 T 1 (SIX) T 1 (X)

108. CS 245Notes 09108 Insert + delete operations Insert A Z ...

109. CS 245Notes 09109 Modifications to locking rules: (1) Get exclusive lock on A before deleting A (2) At insert A operation by Ti, Ti is given exclusive lock on A

110. CS 245Notes 09110 Phantoms Still have a problem: Phantoms Example: relation R (E#,name,…) constraint: E# is key use tuple locking RE#Name…. o155Smith o275Jones

111. CS 245Notes 09111 T 1 : Insert into R T 2 : Insert into R T 1 T 2 S 1 (o 1 ) S 2 (o 1 ) S 1 (o 2 ) S 2 (o 2 ) Check Constraint Insert o 3 [08,Obama,..] Insert o 4 [08,McCain,..]...

112. CS 245Notes 09112 Solution Use multiple granularity tree Before insert of node Q, lock parent(Q) in X mode R1 t1t1 t2t2 t3t3

113. CS 245Notes 09113 Back to example T 1 : Insert T 2 : Insert T 1 T 2 X 1 (R) Check constraint Insert U(R) X 2 (R) Check constraint Oops! e# = 08 already in R! delayed

114. CS 245Notes 09114 Instead of using R, can use index on R: Example: R Index 0<E#<100 Index 100<E#<200 E#=2E#=5 E#=107 E#=109...

115. CS 245Notes 09115 This approach can be generalized to multiple indexes...

116. CS 245Notes 09116 Next: Tree-based concurrency control Validation concurrency control

117. CS 245Notes 09117 Example A B C D EF all objects accessed through root, following pointers

118. CS 245Notes 09118 Example A B C D EF all objects accessed through root, following pointers T 1 lock

119. CS 245Notes 09119 Example A B C D EF all objects accessed through root, following pointers T 1 lock  can we release A lock if we no longer need A??

120. CS 245Notes 09120 Idea: traverse like “Monkey Bars” A B C D EF

121. CS 245Notes 09121 Idea: traverse like “Monkey Bars” A B C D EF T 1 lock

122. CS 245Notes 09122 Idea: traverse like “Monkey Bars” A B C D EF T 1 lock

123. CS 245Notes 09123 Why does this work? Assume all T i start at root; exclusive lock T i  T j  T i locks root before T j Actually works if we don’t always start at root Root Q T i  T j

124. CS 245Notes 09124 Rules: tree protocol (exclusive locks) (1) First lock by T i may be on any item (2) After that, item Q can be locked by T i only if parent(Q) locked by T i (3) Items may be unlocked at any time (4) After T i unlocks Q, it cannot relock Q

125. CS 245Notes 09125 Tree-like protocols are used typically for B-tree concurrency control E.g., during insert, do not release parent lock, until you are certain child does not have to split Root

126. CS 245Notes 09126 Tree Protocol with Shared Locks Rules for shared & exclusive locks? A B C D EF T 1 S lock(released) T 1 S lock (held) T 1 X lock (released) T 1 X lock (will get)

127. CS 245Notes 09127 Tree Protocol with Shared Locks Rules for shared & exclusive locks? A B C D EF T 1 S lock(released) T 1 S lock (held) T 1 X lock (released) T 1 X lock (will get) T 2 reads: B modified by T 1 F not yet modified by T 1

128. CS 245Notes 09128 Need more restrictive protocol Will this work?? –Once T 1 locks one object in X mode, all further locks down the tree must be in X mode Tree Protocol with Shared Locks

129. CS 245Notes 09129 Validation Transactions have 3 phases: (1) Read –all DB values read –writes to temporary storage –no locking (2) Validate –check if schedule so far is serializable (3) Write –if validate ok, write to DB

130. CS 245Notes 09130 Key idea Make validation atomic If T 1, T 2, T 3, … is validation order, then resulting schedule will be conflict equivalent to S s = T 1 T 2 T 3...

131. CS 245Notes 09131 To implement validation, system keeps two sets: FIN = transactions that have finished phase 3 (and are all done) VAL = transactions that have successfully finished phase 2 (validation)

132. CS 245Notes 09132 Example of what validation must prevent: RS(T 2 )={B} RS(T 3 )={A,B} WS(T 2 )={B,D} WS(T 3 )={C} time T 2 start T 2 validated T 3 validated T 3 start  = 

133. CS 245Notes 09133 T 2 finish phase 3 Example of what validation must prevent: RS(T 2 )={B} RS(T 3 )={A,B} WS(T 2 )={B,D} WS(T 3 )={C} time T 2 start T 2 validated T 3 validated T 3 start  =  allow T 3 start

134. CS 245Notes 09134 Another thing validation must prevent: RS(T 2 )={A} RS(T 3 )={A,B} WS(T 2 )={D,E} WS(T 3 )={C,D} time T 2 validated T 3 validated finish T 2

135. CS 245Notes 09135 Another thing validation must prevent: RS(T 2 )={A} RS(T 3 )={A,B} WS(T 2 )={D,E} WS(T 3 )={C,D} time T 2 validated T 3 validated finish T 2 BAD: w 3 (D) w 2 (D)

136. CS 245Notes 09136 finish T 2 Another thing validation must prevent: RS(T 2 )={A} RS(T 3 )={A,B} WS(T 2 )={D,E} WS(T 3 )={C,D} time T 2 validated T 3 validated allow finish T 2

137. CS 245Notes 09137 Validation rules for T j : (1) When T j starts phase 1: ignore(T j )  FIN (2) at T j Validation: if check (T j ) then [ VAL  VAL U {T j }; do write phase; FIN  FIN U {T j } ]

138. CS 245Notes 09138 Check (T j ): For T i  VAL - IGNORE (T j ) DO IF [ WS(T i )  RS(T j )   OR T i  FIN ] THEN RETURN false; RETURN true;

139. CS 245Notes 09139 Check (T j ): For T i  VAL - IGNORE (T j ) DO IF [ WS(T i )  RS(T j )   OR T i  FIN ] THEN RETURN false; RETURN true; Is this check too restrictive ?

140. CS 245Notes 09140 Improving Check(T j ) For T i  VAL - IGNORE (T j ) DO IF [ WS(T i )  RS(T j )   OR ( T i  FIN AND WS(T i )  WS(T j )   )] THEN RETURN false; RETURN true;

141. CS 245Notes 09141 Exercise: T: RS(T)={A,B} WS(T)={A,C} V: RS(V)={B} WS(V)={D,E} U: RS(U)={B} WS(U)={D} W: RS(W)={A,D} WS(W)={A,C} start validate finish

142. CS 245Notes 09142 Is Validation = 2PL? 2PL Val 2PL Val 2PL Val 2PL

143. CS 245Notes 09143 S2: w2(y) w1(x) w2(x) Achievable with 2PL? Achievable with validation?

144. CS 245Notes 09144 S2: w2(y) w1(x) w2(x) S2 can be achieved with 2PL: l2(y) w2(y) l1(x) w1(x) u1(x) l2(x) w2(x) u2(y) u2(x) S2 cannot be achieved by validation: The validation point of T2, val2 must occur before w2(y) since transactions do not write to the database until after validation. Because of the conflict on x, val1 < val2, so we must have something like S2: val1 val2 w2(y) w1(x) w2(x) With the validation protocol, the writes of T2 should not start until T1 is all done with its writes, which is not the case.

145. CS 245Notes 09145 Validation subset of 2PL? Possible proof (Check!): –Let S be validation schedule –For each T in S insert lock/unlocks, get S’: At T start: request read locks for all of RS(T) At T validation: request write locks for WS(T); release read locks for read-only objects At T end: release all write locks –Clearly transactions well-formed and 2PL –Must show S’ is legal (next page)

146. CS 245Notes 09146 Say S’ not legal (due to w-r conflict): S’:... l1(x) w2(x) r1(x) val1 u1(x)... –At val1: T2 not in Ignore(T1); T2 in VAL –T1 does not validate: WS(T2)  RS(T1)   –contradiction! Say S’ not legal (due to w-w conflict): S’:... val1 l1(x) w2(x) w1(x) u1(x)... –Say T2 validates first (proof similar if T1 validates first) –At val1: T2 not in Ignore(T1); T2 in VAL –T1 does not validate: T2  FIN AND WS(T1)  WS(T2)   ) –contradiction!

147. CS 245Notes 09147 Conclusion: Validation subset 2PL 2PL Val 2PL Val 2PL Val 2PL

148. CS 245Notes 09148 Validation (also called optimistic concurrency control) is useful in some cases: - Conflicts rare - System resources plentiful - Have real time constraints

149. CS 245Notes 09149 Summary Have studied C.C. mechanisms used in practice - 2 PL - Multiple granularity - Tree (index) protocols - Validation