Concurrency Control

Example Schedules Constraint: The sum of A+B must be the same Before: After: T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B) Transactions: T1: transfers $50 from A to B T2: transfers 10% of A to B =150, consistent Example 1: a “serial” schedule

Example Schedule Another “serial” schedule: T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B) Before: After: Consistent but not the same as previous schedule.. Either is OK! =150, consistent

Example Schedule (Cont.) Another “good” schedule: T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B) Effect: Before After A B Same as one of the serial schedules Serializable

Example Schedules (Cont.) A “bad” schedule Before: = 150 After: = 110 !! Not consistent T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B) Non Serializable

Equivalence by Swapping T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B) T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B)

Equivalence by Swapping T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B) T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B)

Equivalence by Swapping T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B) T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B)

Example Schedules (Cont.) A “bad” schedule T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B) X Y Can’t move Y below X read(B) and write(B) conflict

Example Schedules (Cont.) A “bad” schedule T1 read(A) A = A -50 write(A) read(B) B=B+50 write(B) T2 read(A) tmp = A*0.1 A = A – tmp write(A) read(B) B = B+ tmp write(B) X Y Can’t move Y below X read(B) and write(B) conflict Other options don’t work either So: Not Conflict Serializable

Lock instructions New instructions - lock-S: shared lock request - lock-X: exclusive lock request - unlock: release previously held lock Example: lock-X(B) read(B) B  B-50 write(B) unlock(B) lock-X(A) read(A) A  A + 50 write(A) unlock(A) lock-S(A) read(A) unlock(A) lock-S(B) read(B) unlock(B) display(A+B) T1 T2

Locking Issues No xction proceeds: Deadlock - T1 waits for T2 to unlock A - T2 waits for T1 to unlock B T1T2 lock-X(B) read(B) B  B-50 write(B) lock-X(A) lock-S(A) read(A) lock-S(B) Rollback transactions Can be costly...

Locking Issues Does not ensure serializability by itself: lock-X(B) read(B) B  B-50 write(B) unlock(B) lock-X(A) read(A) A  A + 50 write(A) unlock(A) lock-S(A) read(A) unlock(A) lock-S(B) read(B) unlock(B) display(A+B) T1 T2 T2 displays 50 less!!

2PhaseLocking Example: T1 in 2PL T1 lock-X(B) read(B) B  B - 50 write(B) lock-X(A) read(A) A  A - 50 write(A) unlock(B) unlock(A) Growing phase Shrinking phase

2PL Issues 2PL does not prevent deadlock > 2 xctions involved? - Rollbacks expensive T1T2 lock-X(B) read(B) B  B-50 write(B) lock-X(A) lock-S(A) read(A) lock-S(B)

Strict 2PL T1T2T3 lock-X(A) read(A) lock-S(B) read(B) write(A) unlock(A) lock-X(A) read(A) write(A) unlock(A) lock-S(A) read(A) Strict 2PL will not allow that

Dealing with Deadlocks How do you detect a deadlock?  Wait-for graph  Directed edge from Ti to Tj  Ti waiting for Tj T1T2T3T4 S(V) X(V) S(W) X(Z) S(V) X(W) T1 T2 T4 T3 Suppose T4 requests lock-S(Z)....

Example of Granularity Hierarchy The highest level in the example hierarchy is the entire database. The levels below are of type area, file or relation and record in that order.

Compatibility Matrix with Intention Lock Modes The compatibility matrix for all lock modes is: IS IX S S IX X IS IX S S IX X          holder requestor

ParentChild can belocked in IS IX S SIX X P C IS, S IS, S, IX, X, SIX [S, IS] not necessary X, IX, [SIX] none

Example R1 t1t1 t2t2 t3t3 t4t4 T 1 (IS) T 1 (S), T 2 (IX) T 2 (X)

Examples R t1 t3 t4 t2 f2.1 f2.2 f4.2 T1(IX) T1(X) R t1 t3 t4 t2 f2.1 f2.2 f4.2 T1(IS) T1(S) R t1 t3 t4 t2 f2.1 f2.2 f4.2 T1(SIX) T1(IX) T1(X) Can T2 access object f2.2 in X mode? What locks will T2 get?

Examples T1 scans R, and updates a few tuples:  T1 gets an SIX lock on R, then repeatedly gets an S lock on tuples of R, and occasionally upgrades to X on the tuples. T2 uses an index to read only part of R:  T2 gets an IS lock on R, and repeatedly gets an S lock on tuples of R. T3 reads all of R:  T3 gets an S lock on R.  OR, T3 could behave like T2; can use lock escalation to decide which. -- ISIX -- IS IX      SX   S X      