CMPSC 274: Transac0on Processing Lecture #6: Concurrency Control Protocols

Transcription

1 CMPSC 274: Transac0on Processing Lecure #6: Concurrency Conrol Proocols Divy Agrawal Deparmen of Compuer Science UC Sana Barbara Timesamp Ordering Serializa0on Graph Tes0ng Op0mis0c Proocols 4/20/11 Transac0onal Informa0on Sysems 4 2 1

2 (Basic) Timesamp Ordering Timesamp ordering rule (TO rule): Each ransac0on i is assigned a unique Dmesamp s( i ) (e.g., he 0me of i s beginning). If p i (x) and q j (x) are in conflic, hen he following mus hold: p i (x) < s q j (x) iff s( i ) < s( j ) for every schedule s. Theorem 4.15: Gen (TO) CSR. Basic Dmesamp ordering proocol (BTO): For each daa iem x mainain max r (x) = max{s( j ) r j (x) has been scheduled} and max w (x) = max{s( j ) w j (x) has been scheduled}. Opera0on p i (x) is compared o max q (x) for each conflic0ng q: if s( i ) < max q (x) for some q hen abor i else schedule p i (x) for execu0on and se max p (x) o s( i ) 4/20/11 Transac0onal Informa0on Sysems 4 3 BTO Example s = r 1 (x) w 2 (x) r 3 (y) w 2 (y) c 2 w 3 (z) c 3 r 1 (z) c 1 r 1 (x) r 1 (z) abor 1 2 w 2 (x) w 2 (y) abor r 3 (y) w 3 (z) c 3 3 r 1 (x) w 2 (x) r 3 (y) a 2 w 3 (z) c 3 a 1 4/20/11 Transac0onal Informa0on Sysems 4 4 2

3 4.4.1 Timesamp Ordering SerializaDon Graph TesDng Op0mis0c Proocols 4/20/11 Transac0onal Informa0on Sysems 4 5 SerializaDon Graph TesDng (SGT) SGT proocol: For p i (x) creae a new node in he graph if i is he firs opera0on of i Inser edges ( j, i ) for each q j (x) < s p i (x) ha is in conflic wih p i (x) (i j). If he graph has become cyclic hen abor i (and remove i from he graph) else schedule p i (x) for execu0on. Theorem 4.16: Gen (SGT) = CSR. Node deledon rule: A node i in he graph (and is inciden edges) can be removed when i is erminaed and is a source node (i.e., has no incoming edges). Example: r 1 (x) w 2 (x) w 2 (y) c 2 r 1 (y) c 1 removing node 2 a he 0me of c 2 would make i impossible o deec he cycle. 4/20/11 Transac0onal Informa0on Sysems 4 6 3

4 4.4.1 Timesamp Ordering Serializa0on Graph Tes0ng OpDmisDc Proocols 4/20/11 Transac0onal Informa0on Sysems 4 7 MoDvaDon: conflics are infrequen OpDmisDc Proocols Approach: divide each ransac0on ino hree phases: read phase: execue ransac0on wih wries ino privae workspace validadon phase (cerdfier): upon s commi reques es if schedule remains CSR if is commihed now based on s read se RS() and wrie se WS() wrie phase: upon successful valida0on ransfer he workspace conens ino he daabase (deferred wries) oherwise abor (i.e., discard workspace) 4/20/11 Transac0onal Informa0on Sysems 4 8 4

5 Backward oriened OpDmisDc CC (BOCC) Execue a ransac0on s valida0on and wrie phase ogeher as a cridcal secdon: while i being in he val wrie phase, no oher k can ener is val wrie phase BOCC validadon of j : compare j o all previously commihed i accep j if one of he following holds i has ended before j has sared, or RS( j ) WS( i ) = and i has validaed before j Theorem 4.46: Gen (BOCC) CSR. Proof: Assume ha G(s) is acyclic. Adding a newly validaed ransac0on can inser only edges ino he new node, bu no ougoing edges (i.e., he new node is las in he serializa0on order). 4/20/11 Transac0onal Informa0on Sysems 4 9 BOCC Example read phase wrie phase 1 r 1 (x) r 1 (y) 2 val. w 1 (x) r 2 (y) r 2 (z) val. w 2 (z) 3 r 3 (x) r 3 (y) val. abor 4 r 4 (x) val. w 4 (x) 4/20/11 Transac0onal Informa0on Sysems

6 Forward oriened OpDmisDc CC (FOCC) Execue a ransac0on s val wrie phase as a srong cridcal secdon: while i being in he val wrie phase, no oher k can perform any seps. FOCC validadon of j : compare j o all concurrenly ac0ve i (which mus be in heir read phase) accep j if WS( j ) RS*( i ) = where RS*( i ) is he curren read se of i Remarks: FOCC is much more flexible han BOCC: upon unsuccessful valida0on of j i has hree op0ons: abor j abor one of he ac0ve i for which RS*( i ) and WS( j ) inersec wai and rery he valida0on of j laer (aler he commi of he inersec0ng i ) Read only ransac0ons do no need o validae a all. 4/20/11 Transac0onal Informa0on Sysems 4 11 Correcness of FOCC Theorem 4.18: Gen (FOCC) CSR. Proof: Assume ha G(s) has been acyclic and ha valida0ng j would creae a cycle. So j would have o have an ougoing edge o an already commihed k. However, for all previously commihed k he following holds: If k was commihed before j sared, hen no edge ( j, k ) is possible. If j was in is read phase when k validaed, hen WS( k ) mus be disjoin wih RS*( j ) and all laer reads of j and all wries of j mus follow k (because of he srong cri0cal sec0on); so neiher a wr nor a ww/rw edge ( j, k ) is possible. 4/20/11 Transac0onal Informa0on Sysems

7 FOCC Example read phase wrie phase r 1 (x) r 1 (y) val. w 1 (x) 1 r 2 (y) r 2 (z) val. w 2 (z) 2 3 r 3 (z) abor r 4 (x) r 4 (y) val. w 4 (y) 4 5 r 5 (x) r 5 (y) 4/20/11 Transac0onal Informa0on Sysems /20/11 Transac0onal Informa0on Sysems

8 Hybrid Proocols Idea: Combine differen proocols, each handling differen ypes of conflics (rw/wr vs. ww) or daa par00ons Cavea: The combina0on mus guaranee ha he union of he underlying local conflic graphs is acyclic. Example 4.15: use SS2PL for rw/wr synchroniza0on and TO or TWR for ww wih TWR (Thomas wrie rule) as follows: for w j (x): if s( j ) > max w (x) hen execue w j (x) else do nohing s 1 = w 1 (x) r 2 (y) w 2 (x) w 2 (y) c 2 w 1 (y) c 1 s 2 = w 1 (x) r 2 (y) w 2 (x) w 2 (y) c 2 r 1 (y) w 1 (y) c 1 boh acceped by SS2PL/TWR wih s( 1 ) < s( 2 ), bu s 2 is no CSR Problem wih s 2 : needs synch among he wo local serializa0on orders SoluDon: assign 0mesamps such ha he serializa0on orders of SS2PL and TWR are in line s(i) < s(j) c i < c j 4/20/11 Transac0onal Informa0on Sysems /20/11 Transac0onal Informa0on Sysems

9 Lessons Learned S2PL is he mos versa0le and robus proocol and widely used in prac0ce Knowledge abou specifically resriced access paherns faciliaes non wo phase locking proocols (e.g., TL, AL) O2PL and SGT are more powerful bu have more overhead FOCC can be ahrac0ve for specific workloads Hybrid proocols are conceivable bu non rivial 4/20/11 Transac0onal Informa0on Sysems