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CS4432: Database Systems IILecture #20

Concurrency Control – Chapter 18

Professor Elke A. Rundensteiner

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Chapter 9 Concurrency Control

T1 T2 … Tn

DB(consistencyconstraints)

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Example:

T1: Read(A) T2: Read(A)A A+100 A A2Write(A) Write(A)Read(B) Read(B)B B+100 B B2Write(B) Write(B)

Constraint: A=B

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A schedule

An ordering of operations inside one or more transactions over

time

What is correct outcome ?

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Schedule A

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

Read(A);A A2;

Write(A);

Read(B);B B2;

Write(B);

A B25 25

125

125

250

250250 250

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Schedule B

T1 T2

Read(A);A A2;

Write(A);

Read(B);B B2;

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

A B25 25

50

50

150

150150 150

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Serial Schedules !

• Any serial schedule is “good”.

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Yet Interleave Transactions

ina Schedule?

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Schedule C

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 D

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

Read(A);A A2;

Write(A);

Read(B);B B2;

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

A B25 25

125

250

50

150250 150

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Schedule E

T1 T2’Read(A); A A+100Write(A);

Read(A);A A1;

Write(A);

Read(B);B B1;

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

A B25 25

125

125

25

125125 125

Same as Schedule Dbut with new T2’

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What is a ‘good’ schedule?

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• We want schedules that are “good” regardless of :

• initial state and• transaction semantics

• Hence we consider only :– order of read/writes

Example: Sc=r1(A)w1(A)r2(A)w2(A)r1(B)w1(B)r2(B)w2(B)

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Remember Schedule C

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);

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Sc’=r1(A)w1(A) r1(B)w1(B)r2(A)w2(A)r2(B)w2(B)

T1 T2

Example: Sc=r1(A)w1(A)r2(A)w2(A)r1(B)w1(B)r2(B)w2(B)

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Now Let’s Try that for Schedule D !!!

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

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

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

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

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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)

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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)

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In short, Schedule D cannot be “fixed” :Sd=r1(A)w1(A)r2(A)w2(A)

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

Observation : there seems to be no save way to

transform this schedule Sd into an equivalent serial schedule !?!

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Sd cannot be rearrangedinto a serial

scheduleSd is not “equivalent” to

any serial scheduleSd is “bad”

We note about schedule D:• T2 T1

• T1 T2 T1 T2

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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 aserial schedule,I.e., in this case (T1,T2).

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

schedule.