1. cs4432concurrency control1 CS4432: Database Systems II Lecture #21 Concurrency Control Professor Elke A. Rundensteiner

2. cs4432concurrency control2 Chapter 9Concurrency Control T1T2…Tn DB (consistency constraints)

3. cs4432concurrency control3 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. cs4432concurrency control4 A schedule An ordering of operations inside one or more transactions over time Why interleave operations?

5. cs4432concurrency control5 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. cs4432concurrency control6 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

7. cs4432concurrency control7 Serial Schedules ! Any serial schedule is “good”.

8. cs4432concurrency control8 Interleave Transactions in a Schedule?

9. cs4432concurrency control9 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. cs4432concurrency control10 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

11. cs4432concurrency control11 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’

12. cs4432concurrency control12 What is a ‘good’ schedule?

13. cs4432concurrency control13 We want : –schedules that are “good” regardless of initial state and transaction semantics Hence we consider : –Only 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)

14. cs4432concurrency control14 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)

15. cs4432concurrency control15 Now 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) Sd=r 1 (A)w 1 (A) r 1 (B)w 1 (B)r 2 (A)w 2 (A)r 2 (B)w 2 (B)

16. cs4432concurrency control16 Or, let’s try 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) Sd=r 2 (A)w 2 (A)r 2 (B)w 2 (B) r 1 (A)w 1 (A)r 1 (B)w 1 (B)

17. cs4432concurrency control17 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, there seems to be no save way to transform this Sd into an equivalent serial schedule?

18. cs4432concurrency control18 T 1 T 2 Sd cannot be rearranged into a serial schedule Sd is not “equivalent” to any serial schedule Sd is “bad” We note, T 2  T 1 Also, T 1  T 2

19. cs4432concurrency control19 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, I.e., in this case (T 1,T 2 ).

20. cs4432concurrency control20 Idea: Swap non-conflicting operation pairs to see if you can go to a serial schedule.