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CS4432: Database Systems IILecture #23

Concurrency Control

Professor Elke A. Rundensteiner

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Chapter 9 Concurrency Control

T1 T2 … Tn

DB(consistencyconstraints)

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ConceptsTransaction: sequence of ri(x), wi(x) actionsConflicting actions: read/write on same resource A.Schedule: represents chronological order in which

actions are executedSerial schedule: no interleaving of trans/actions S1, S2 are conflict equivalent schedules if S1

transform into S2 by swaps on non-conflicting actions.

A schedule is conflict serializable if it is conflict equivalent to some serial schedule.

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For a given schedule: Sd=r1(A)w1(A)r2(A)w2(A)

r2(B)w2(B)r1(B)w1(B)

T1 T2 T2 T1

T1 T2

If graph has cycles,

Then schedule is not conflict serializable.

Record dependencies in Precedence Graph:

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Theorem

P(S1) acyclic S1 conflict serializable

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How to enforce serializable schedules?

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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’.

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How to enforce serializable schedules?

Option 2: run system, recording P(S);

at end of day, check for P(S) cycles and declare if

execution was good

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Option 3: prevent P(S) cycles from occurring T1 T2 ….. Tn

Scheduler

DB

How to enforce serializable schedules?

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A locking protocol

Two new actions:lock (exclusive): li (A)

unlock: ui (A)

scheduler

T1 T2

locktable

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Rule #1: Well-formed transactions

Ti: … li(A) … pi(A) … ui(A) ...

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Rule #2 Legal scheduler

S = …….. li(A) ………... ui(A) ……...

no lj(A)

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• 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)

Exercise:

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• 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)

Exercise:

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Schedule F : Adding locking ???

T1 T2l1(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)

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Schedule F

T1 T2 25 25l1(A);Read(A)A A+100;Write(A);u1(A) 125

l2(A);Read(A)A Ax2;Write(A);u2(A) 250l2(B);Read(B)B Bx2;Write(B);u2(B) 50

l1(B);Read(B)B B+100;Write(B);u1(B) 150

250 150

A B

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Rule #3 Two phase locking (2PL)

for transactions

Ti = ……. li(A) ………... ui(A) ……...

no unlocks no locks

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# locksheld byTi

Time Growing Shrinking Phase Phase

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Schedule F : Does it follow 2PL ?

T1 T2l1(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)

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Schedule G

T1 T2l1(A);Read(A)A A+100;Write(A)l1(B); u1(A)

l2(A);Read(A) A Ax2;Write(A);

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Schedule G

T1 T2l1(A);Read(A)A A+100;Write(A)l1(B); u1(A)

l2(A);Read(A) A Ax2;Write(A);ll22(B)(B)

delayed

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Schedule G

T1 T2l1(A);Read(A)A A+100;Write(A)l1(B); u1(A)

l2(A);Read(A) A Ax2;Write(A);ll22(B)(B)

Read(B);B B+100Write(B); u1(B)

delayed

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Schedule G

T1 T2l1(A);Read(A)A A+100;Write(A)l1(B); u1(A)

l2(A);Read(A) A Ax2;Write(A);ll22(B)(B)

Read(B);B B+100Write(B); u1(B)

l2(B); u2(A);Read(B) B Bx2;Write(B);u2(B);

delayed

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We Got the GOOD Schedule !!!!

T1 T2Read(A); A A+100Write(A);

Read(A);A A2;Write(A);

Read(B); B B+100;Write(B);

Read(B);B B2;Write(B);

A B25 25

125

250

125

250250 250

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Schedule H (T2 reversed)

T1 T2l1(A); Read(A) l2(B);Read(B)A A+100;Write(A) B Bx2;Write(B)… …

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Schedule H (T2 reversed)

T1 T2l1(A); Read(A) l2(B);Read(B)A A+100;Write(A) B Bx2;Write(B)

ll11(B)(B) l l22(A)(A)delayeddelayed

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• 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!

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We need to show 2PL Protocol “works”:

Show that rules #1,2,3 conflict- serializable schedules

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Conflict rules for li(A), ui(A):

• li(A), lj(A) conflict • li(A), uj(A) conflict

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Theorem Rules #1,2,3 conflict (2PL) serializable

schedule

To help in proof:Definition Shrink(Ti) = SH(Ti) =

first unlock action of Ti

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LemmaTi Tj in S SH(Ti) <S SH(Tj)

Proof of lemma:Ti Tj means that

S = … pi(A) … qj(A) …; p,q conflictBy rules 1,2:

S = … pi(A) … ui(A) … lj(A) ... qj(A) …

By rule 3: SH(Ti) SH(Tj)

So, SH(Ti) <S SH(Tj)

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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

Theorem Rules #1,2,3 conflict (2PL) serializable

schedule

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• Beyond this simple 2PL protocol, it is all a matter of improving

performance and allowing more concurrency….

– Shared locks– Multiple granularity– Other types of C.C. mechanisms