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Concurrency control In production environments, it is unlikely that we can limit our system to just one user at a time. – Consequently, it is possible.

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Presentation on theme: "Concurrency control In production environments, it is unlikely that we can limit our system to just one user at a time. – Consequently, it is possible."— Presentation transcript:

1 Concurrency control In production environments, it is unlikely that we can limit our system to just one user at a time. – Consequently, it is possible for multiple queries or transactions to be submitted at approximately the same time. If all of the queries were very small (i.e., in terms of time), we could probably just execute them on a first- come-first-served basis. However, many queries are both complex and time consuming. – Executing these queries would make other queries wait a long time for a chance to execute. So, in practice, the DBMS may be running many different transactions at about the same time.

2 The problem Transactions consist of multiple actions (reads, writes, updates, etc.) Since transactions may access or modify the same elements as other transactions (e.g., fields), conflicts may result. – Specifically, arbitrarily interleaving the individual actions of various transactions may produce unexpected results. A primary goal of the DBSM is to maintain the database in a consistent and predictable state. – Therefore we must take steps to ensure or guarantee this consistency. We assume that each transaction in isolation takes the database from a consistent state to a consistent state

3 It is the job of the Transaction Manager to convert queries into actions To maintain database consistency, restrictions must be placed on the order of the transactions – The Transaction Manager does not do this. Instead it passes requests to the Scheduler sub-system. Its job is to create an order that preserves the consistency of the DB. Scheduler may: – delay a request from a transaction – abort a transaction The Scheduler

4 Serial and serializable schedules Example with two transactions Assume A = B is required for consistency. We deal only with Reads and Writes in the main memory buffers. T 1 and T 2 individually preserve DB consistency.

5 An Acceptable (serial) schedule S 1 Assume initially A = B = 25. Here is one way to execute S 1 = (T 1 ; T 2 ) so they do not interfere.

6 Another acceptable (serial) schedule S 2 Here, transactions are executed as S 2 = (T 2 ; T 1 ). The result is different, but consistency is maintained.

7 Interleaving Doesn't Necessarily Hurt (S 3 )

8 But Then Again, It Might!

9 The Semantics of transactions is also important. Here T2 adds 200 to A and to B

10 We Need a Simpler Model Assume that whenever a transaction T writes X, it changes X in some unique way. – Arithmetic coincidence never happens Thus, we focus on the reads and writes only, assuming that whenever transaction T reads X and then writes it, X has changed in some unique way. – Notation: r T (X) denotes T reads the DB element X w T (X) denotes T writes X If transactions are T 1,…,T k, then we will simply use r i and w i, instead of r Ti and w Ti

11 Transactions and Schedules A transaction (model) is a sequence of r and w actions on database elements. A schedule is a sequence of reads/writes actions performed by a collection of transactions. Serial Schedule = All actions for each transaction are consecutive. r1(A); w1(A); r1(B); w1(B); r2(A); w2(A); r2(B); w2(B); … Serializable Schedule: A schedule whose “effect” is equivalent to that of some serial schedule. We will introduce a sufficient condition for serializability.

12 Conflicts Suppose for fixed DB elements X & Y, r i (X); r j (Y) is part of a schedule, and we flip the order of these operations. – r i (X); r j (Y) ≡ r j (Y); r i (X) … In what sense? – This holds always (even when X=Y) We can flip r i (X); w j (Y), as long as X≠Y That is, r i (X); w j (X)  w j (X); r i (X) – In the RHS, T i reads the value of X written by T j, whereas it is not so in the LHS.

13 Conflicts (Cont’d) We can flip w i (X); w j (Y); provided X≠Y However, w i (X); w j (X) ≢ w j (X); w i (X); – The final value of X may be different depending on which write occurs last. There is a conflict if 2 conditions hold. A read and a write of the same X, or Two writes of X conflict in general and may not be swapped in order. All other events (reads/writes) may be swapped without changing the effect of the schedule (on the DB).

14 Example Two scheduless are conflict-equivalent if they can be converted into the other by a series of non-conflicting swaps of adjacent elements A schedule is conflict-serializable if it can be converted into a serializable schedule in the same way r 1 (A); w 1 (A); r 2 (A); w 2 (A); r 1 (B); w 1 (B); r 2 (B); w 2 (B) r 1 (A); w 1 (A); r 2 (A); r 1 (B); w 2 (A); w 1 (B); r 2 (B); w 2 (B) r 1 (A); w 1 (A); r 1 (B); r 2 (A); w 2 (A); w 1 (B); r 2 (B); w 2 (B) r 1 (A); w 1 (A); r 1 (B); r 2 (A); w 1 (B); w 2 (A); r 2 (B); w 2 (B) r 1 (A); w 1 (A); r 1 (B); w 1 (B); r 2 (A)w 2 (A); r 2 (B); w 2 (B) This final result is a serial schedule.

15 Precedence Graphs Non-swappable pairs of actions represent potential conflicts between transactions. The existence of non-swappable actions enforces an ordering on the transactions that perform these actions. We say that T 1 takes precedence over T 2 in the schedule S, written T 1 < S T 2, if there are actions A 1 of T 1 and A 2 of T 2 such that 1.A 1 is ahead of A 2 in S AND 2.Both A 1 and A 2 involve the same element AND 3.At least one of A 1 and A 2 is a write.

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17 If there is a cycle in the graph – Then, there is no serial schedule which is conflict­ equivalent to S. Each arc represents a requirement on the order of transactions in a conflict­ equivalent serial schedule. A cycle puts too many requirements on any linear order of transactions. If there is no cycle in the graph – Then any topological order of the graph suggests a conflict­equivalent schedule.

18 Why the Precedence-Graph Test Works Idea: if the precedence graph is acyclic, then we can swap actions to form a serial schedule consistent with some topological order; Proof: By induction on n, number of transactions. Basis: n = 1. That is, S={T1}; then S is already serial. Induction: S={T1,T2,…,Tn}. Given that SG(S) is acyclic, then pick Ti in S such that Ti has no incoming arcs – We swap all actions of Ti to the front (of S). – (Actions of Ti)(Actions of the other n-1 transactions) – The tail is a precedence graph that is the same as the original without Ti, i.e. it has n-1 nodes and it is acyclic.  By the induction hyposthesis, we can reorder the actions of the other transactions to turn it into a serial schedule

19 Schedulers A scheduler takes requests from transactions for reads and writes, and decides if it is “OK” to allow them to operate on DB or defer them until it is safe to do so. Ideal: a scheduler forwards a request iff it cannot lead to inconsistency of DB – Too hard to decide this in real time. Real: it forwards a request if it cannot result in a violation of conflict­serializability. We thus need to develop schedulers which ensure conflict- serializablility.

20 To prevent conflicts, a Scheduler employs Lock Tables. – The idea is to record information about the current or requested locks. Consequently, by blocking access to certain elements, the Scheduler can enforce an access order – Basically, we can prevent actions on the same element from being improperly swapped – This essentially enforces the order of the precedence graph. Lock tables

21 Lock Actions Before reading or writing an element X, a transaction Ti requests a lock on X from the scheduler. The scheduler can either grant the lock to Ti or make Ti wait for the lock. If granted, Ti should eventually unlock (release) the lock on X. Shorthands: – l i (X) = “transaction Ti requests a lock on X” – u i (X) = “Ti unlocks/releases the lock on X”

22 The use of lock must be proper in 2 senses: – Consistency of Transactions: Read or write X only when hold a lock on X. –r i (X) or w i (X) must be preceded by some l i (X) with no intervening u i (X). If Ti locks X, Ti must eventually unlock X. –Every l i (X) must be followed by u i (X). – Legality of Schedules: Two transactions may not have locked the same element X without one having first released the lock. –A schedule with l i (X) cannot have another l j (X) until u i (X) appears in between

23 Legal Schedule Doesn’t Mean Serializable

24 Two Phase Locking There is a simple condition, which guarantees confict-serializability: In every transaction, all lock requests (phase 1) precede all unlock requests (phase 2).

25 Why 2PL Works Precisely: a legal schedule S of 2PL transactions is conflict­ serializable. Proof is an induction on n, the number of transactions. Remember, conflicts involve only read/write actions, not locks, but the legality of the transaction requires that the r/w's be consistent with the l/u's.

26 Why 2PL Works (Cont’d) Basis: if n=1, then S={T 1 }, and hence S is conflict-serializable. Induction: S={T 1,…,T n }. Find the first transaction, say T i, to perform an unlock action, say u i (X). Can we show that the r/w actions of T i can be moved to the front of the other transactions without conflict? Consider some action such as w i (Y). Can it be preceded by some conflicting action w j (Y) or r j (Y)? In such a case we cannot swap them. – If so, then u j (Y) and l i (Y) must intervene, as w j (Y)...u j (Y)...l i (Y)...w i (Y). – Since T i is the first to unlock, u i (X) appears before u j (Y). – But then l i (Y) appears after u i (X), contradicting 2PL. Conclusion: w i (Y) can slide forward in the schedule without conflict; similar argument for a r i (Y) action.

27 Risk of deadlocks The scheduler must perform deadlock resolution (usually by aborting one of the locked transactions)

28 Shared/Exclusive Locks Problem: while simple locks + 2PL guarantee conflict­serializability, they do not allow two readers of DB element X at the same time. But having multiple readers is not a problem for conflict­ serializability (since read actions commute).

29 Shared/Exclusive Locks (Cont’d) Solution: Two kinds of locks: 1. Shared lock sl i (X) allows T i to read, but not write X. – It prevents other transactions from writing X but not from reading X. 2. Exclusive lock xl i (X) allows T i to read and/or write X; no other transaction may read or write X.

30 Consistency of transaction conditions: – A read r i (X) must be preceded by sl i (X) or xl i (X), with no intervening u i (X). – A write w i (X) must be preceded by xl i (X), with no intervening u i (X). Legal schedules: – No two exclusive locks. If xl i (X) appears in a schedule, then there cannot be a xl j (X) until after a u i (X) appears. – No exclusive and shared locks. If xl i (X) appears, there can be no sl j (X) until after u i (X). If sl i (X) appears, there can be no xl j (X) until after u i (X). 2PL condition: – No transaction may have a sl(X) or xl(X) after a u(Y).

31 Scheduler Rules When there is more than one kind of lock, the scheduler needs a rule that says “if there is already a lock of type A on DB element X, can I grant a lock of type B on X?” The compatibility matrix answers the question. Compatibility Matrix for Shared/Exclusive Locks is:

32 Upgrading Locks Instead of taking an exclusive lock immediately, a transaction can take a shared lock on X, read X, and then upgrade the lock to exclusive so that it can write X.

33 Upgrading Locks (Cont.) Had T1 asked for an exclusive lock on B before reading B, the request would have been denied, because T2 already has a shared lock on B.

34 Deadlocks Problem: when we allow upgrades, it is easy to get into a deadlock situation. Example:T1 and T2 each reads X and later writes X.

35 Solution: Update Locks Update lock ul i (X) with asymetric compatibility matrix. – Only an update (not read) can be upgraded to write (If there are no shared locks anymore). – Legal schedules: read action permitted when there is either a shared or update lock. – An update can be granted while there is a shared­ lock, but the scheduler will not grant a shared­lock when there is an update.

36 Compatibility Matrix for Shared, Exclusive, and Update Locks

37 Example: T 1 and T 2 each read X and later write X.


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