《數(shù)據(jù)庫系統(tǒng)》英文教學(xué)課件

《數(shù)據(jù)庫系統(tǒng)》英文教學(xué)課件,數(shù)據(jù)庫系統(tǒng),數(shù)據(jù)庫,系統(tǒng),英文,教學(xué),課件 Relational Query OptimizationR&G Chapters 12/15ReviewImplementation of single Relational OperationsChoices depend on indexes,memory,stats,JoinsBlocked nested loops:simple,exploits extra memoryIndexed nested loops:best if 1 rel small and one indexedSort/Merge Joingood with small amount of memory,bad with duplicatesHash Joinfast(enough memory),bad with skewed dataQuery Optimization OverviewSELECT S.snameFROM Reserves R,Sailors SWHERE R.sid=S.sid AND R.bid=100 AND S.rating5ReservesSailorssid=sidbid=100 rating 5snameQuery can be converted to relational algebraRel.Algebra converted to tree,joins as branchesEach operator has implementation choicesOperators can also be applied in different order!(sname)(bid=100 rating 5)(Reserves Sailors)Query Optimization Overview(cont.)Plan:Tree of R.A.ops(and some others)with choice of algorithm for each op.Three main issues:For a given query,what plans are considered?How is the cost of a plan estimated?How do we“search”in the“plan space”?Ideally:Want to find best plan.Reality:Avoid worst plans!Cost-based Query Sub-System Query ParserQuery OptimizerPlan GeneratorPlan Cost EstimatorQuery ExecutorCatalog ManagerUsually there is aheuristics-basedrewriting step beforethe cost-based steps.SchemaStatisticsSelect*From Blah BWhere B.blah=blahQueriesSchema for ExamplesAs seen in previous lecturesReserves:Each tuple is 40 bytes long,100 tuples per page,1000 pages.Assume there are 100 boatsSailors:Each tuple is 50 bytes long,80 tuples per page,500 pages.Assume there are 10 different ratings Assume we have 5 pages in our buffer pool!Sailors(sid:integer,sname:string,rating:integer,age:real)Reserves(sid:integer,bid:integer,day:dates,rname:string)Motivating ExampleCost:500+500*1000 I/OsBy no means the worst plan!Misses several opportunities:selections could have been pushed earlier,no use is made of any available indexes,etc.Goal of optimization:To find more efficient plans that compute the same answer.SELECT S.snameFROM Reserves R,Sailors SWHERE R.sid=S.sid AND R.bid=100 AND S.rating5SailorsReservessid=sidbid=100 rating 5sname(Page-Oriented Nested loops)(On-the-fly)(On-the-fly)Plan:500,500 IOsAlternative Plans Push Selects(No Indexes)SailorsReservessid=sidbid=100 rating 5sname(Page-Oriented Nested loops)(On-the-fly)(On-the-fly)SailorsReservessid=sidrating 5sname(Page-Oriented Nested loops)(On-the-fly)(On-the-fly)bid=100(On-the-fly)250,500 IOsAlternative Plans Push Selects(No Indexes)SailorsReservessid=sidrating 5sname(Page-Oriented Nested loops)(On-the-fly)(On-the-fly)bid=100(On-the-fly)SailorsReservessid=sidbid=100sname(Page-Oriented Nested loops)(On-the-fly)rating 5(On-the-fly)(On-the-fly)250,500 IOs250,500 IOsSailorsReservessid=sidrating 5sname(Page-Oriented Nested loops)(On-the-fly)(On-the-fly)bid=100(On-the-fly)6000 IOsSailorsReservessid=sidrating 5sname(Page-Oriented Nested loops)(On-the-fly)(On-the-fly)bid=100(On-the-fly)250,500 IOsAlternative Plans Push Selects(No Indexes)SailorsReservessid=sidrating 5sname(Page-Oriented Nested loops)(On-the-fly)bid=100(Scan&Write totemp T2)(On-the-fly)6000 IOsSailorsReservessid=sidrating 5sname(Page-Oriented Nested loops)(On-the-fly)(On-the-fly)bid=100(On-the-fly)Alternative Plans Push Selects(No Indexes)4250 IOsReservesSailorssid=sidbid=100sname(Page-Oriented Nested loops)(On-the-fly)rating5(Scan&Write totemp T2)(On-the-fly)Alternative Plans Push Selects(No Indexes)4010 IOs=500+1000+10+(250*10)SailorsReservessid=sidrating 5sname(Page-Oriented Nested loops)(On-the-fly)bid=100(Scan&Write totemp T2)(On-the-fly)4250 IOs=1000+500+250+(10*250)More Alternative Plans(No Indexes)Main difference:Sort Merge JoinWith 5 buffers,cost of plan:Scan Reserves(1000)+write temp T1(10 pages,if we have 100 boats,uniform distribution)=1010.Scan Sailors(500)+write temp T2(250 pages,if have 10 ratings)=750.Sort T1(2*2*10)+sort T2(2*4*250)+merge(10+250)=2300Total:4060 page I/Os.If use BNL join,join=10+4*250,total cost=2770.Can also push projections,but must be careful!T1 has only sid,T2 only sid,sname:T1 fits in 3 pgs,cost of BNL under 250 pgs,total 5(Scan;write to temp T1)(Scan;write totemp T2)(Sort-Merge Join)More Alt Plans:IndexesWith clustered index on bid of Reserves,we get 100,000/100=1000 tuples on 1000/100=10 pages.INL with outer not materialized.v Decision not to push rating5 before the join is based on availability of sid index on Sailors.v Cost:Selection of Reserves tuples(10 I/Os);then,for each,must get matching Sailors tuple(1000*1.2);total 1210 I/Os.v Join column sid is a key for Sailors.At most one matching tuple,unclustered index on sid OK.Projecting out unnecessary fields from outer doesnt help.(On-the-fly)(Use hashIndex,donot writeto temp)ReservesSailorssid=sidbid=100 snamerating 5(Index Nested Loops,with pipelining)(On-the-fly)What is needed for optimization?A closed set of operators Relational ops(table in,table out)Encapsulation(e.g.based on iterators)Plan spaceBased on relational equivalences,different implementationsCost Estimation,based onCost formulasSize estimation,based on Catalog information on base tablesSelectivity(Reduction Factor)estimationA search algorithmTo sift through the plan space based on cost!SummaryQuery optimization is an important task in a relational DBMS.Must understand optimization in order to understand the performance impact of a given database design(relations,indexes)on a workload(set of queries).Two parts to optimizing a query:Consider a set of alternative plans.Must prune search space;typically,left-deep plans only.Must estimate cost of each plan that is considered.Must estimate size of result and cost for each plan node.Key issues:Statistics,indexes,operator implementations.Query OptimizationQuery can be dramatically improved by changing access methods,order of operators.Iterator interfaceCost estimationSize estimation and reduction factorsStatistics and CatalogsRelational Algebra EquivalencesChoosing alternate plansMultiple relation queriesWill focus on“System R”-style optimizersHighlights of System R OptimizerImpact:Most widely used currently;works well for=2 For each sailor with at least two reservations for red boats,find the sailor id and the earliest date on which the sailor has a reservation for a red boat.pS.sid,MIN(R.day)(HAVING COUNT(*)2(GROUP BY S.Sid(sB.color=“red”(Sailors Reserves Boats)Translating SQL to Relational AlgebraSELECT S.sid,MIN(R.day)FROM Sailors S,Reserves R,Boats BWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=“red”GROUP BY S.sidHAVING COUNT(*)=2 Allow us to choose different join orders and to push selections and projections ahead of joins.Selections:c1cn(R)c1(cn(R)(cascade)c1(c2(R)c1(c1(R)(commute)Projections:a1(R)a1(a1,an(R)(cascade)Cartesian ProductR (S T)(R S)T (associative)R S S R (commutative)This means we can do joins in any order.Butbeware of cartesian product!Relational Algebra EquivalencesMore EquivalencesEager projectionCan cascade and“push”some projections thru selectionCan cascade and“push”some projections below one side of a joinRule of thumb:can project anything not needed“downstream”Selection between attributes of the two arguments of a cross-product converts cross-product to a join.A selection on just attributes of R commutes with R S.(i.e.,(R S)(R)S)Cost EstimationFor each plan considered,must estimate total cost:Must estimate cost of each operation in plan tree.Depends on input cardinalities.Weve already discussed how to estimate the cost of operations(sequential scan,index scan,joins,etc.)Must estimate size of result for each operation in tree!Use information about the input relations.For selections and joins,assume independence of predicates.In System R,cost is boiled down to a single number consisting of#I/O+factor*#CPU instructionsQ:Is“cost”the same as estimated“run time”?Statistics and CatalogsNeed information about the relations and indexes involved.Catalogs typically contain at least:#tuples(NTuples)and#pages(NPages)per reln.#distinct key values(NKeys)for each index.low/high key values(Low/High)for each index.Index height(IHeight)for each tree index.#index pages(INPages)for each index.Catalogs updated periodically.Updating whenever data changes is too expensive;lots of approximation anyway,so slight inconsistency ok.More detailed information(e.g.,histograms of the values in some field)are sometimes stored.Size Estimation and Reduction FactorsConsider a query block:Maximum#tuples in result is the product of the cardinalities of relations in the FROM clause.Reduction factor(RF)associated with each term reflects the impact of the term in reducing result size.Result cardinality=Max#tuples *product of all RFs.(RF=|output|/|input|)RF usually called“selectivity”only R&G seem to call it Reduction Factorbeware of confusion between“high selectivity”as defined here and“highly selective”in common English!SELECT attribute listFROM relation listWHERE term1 AND.AND termkResult Size Estimation Result cardinality=Max#tuples *product of all RFs.Term col=value(given index I on col)RF=1/NKeys(I)Term col1=col2(This is handy for joins too)RF =1/MAX(NKeys(I1),NKeys(I2)Term colvalueRF=(High(I)-value)/(High(I)-Low(I)(Implicit assumptions:values are uniformly distributed and terms are independent!)Note,if missing indexes,assume 1/10!Think through estimation for joinsTerm col1=col2 RF =1/MAX(NKeys(I1),NKeys(I2)Q:Given a join of R and S,what is the range of possible result sizes(in#of tuples)?If join is on a key for R(and a Foreign Key in S)?A common case,can treat it specially General case:join on A(A is key for neither)estimate each tuple r of R generates NTuples(S)/NKeys(A,S)result tuples,soNTuples(R)*NTuples(S)/NKeys(A,S)but can also consider it starting with S,yielding:NTuples(S)*NTuples(R)/NKeys(A,R)If these two estimates differ,take the lower one!Q:Why?rS.AREnumeration of Alternative PlansThere are two main cases:Single-relation plansMultiple-relation plansFor queries over a single relation,queries consist of a combination of selects,projects,and aggregate ops:Each available access path(file scan/index)is considered,and the one with the least estimated cost is chosen.The different operations are essentially carried out together(e.g.,if an index is used for a selection,projection is done for each retrieved tuple,and the resulting tuples are pipelined into the aggregate computation).Cost Estimates for Single-Relation PlansIndex I on primary key matches selection:Cost is Height(I)+1 for a B+tree.Clustered index I matching one or more selects:(NPages(I)+NPages(R)*product of RFs of matching selects.Non-clustered index I matching one or more selects:(NPages(I)+NTuples(R)*product of RFs of matching selects.Sequential scan of file:NPages(R).+Recall:Must also charge for duplicate elimination if requiredExampleIf we have an index on rating:Cardinality=(1/NKeys(I)*NTuples(R)=(1/10)*40000 tuples Clustered index:(1/NKeys(I)*(NPages(I)+NPages(R)=(1/10)*(50+500)=55 pages are retrieved.(This is the cost.)Unclustered index:(1/NKeys(I)*(NPages(I)+NTuples(R)=(1/10)*(50+40000)=401 pages are retrieved.If we have an index on sid:Would have to retrieve all tuples/pages.With a clustered index,the cost is 50+500,with unclustered index,50+40000.Doing a file scan:We retrieve all file pages(500).SELECT S.sidFROM Sailors SWHERE S.rating=8Queries Over Multiple RelationsA heuristic decision in System R:only left-deep join trees are considered.As the number of joins increases,the number of alternative plans grows rapidly;we need to restrict the search space.Left-deep trees allow us to generate all fully pipelined plans.Intermediate results not written to temporary files.Not all left-deep trees are fully pipelined(e.g.,SM join).BACDBACDCDBAEnumeration of Left-Deep PlansLeft-deep plans differ only in the order of relations,the access method for each relation,and the join method for each join.Enumerated using N passes(if N relations joined):Pass 1:Find best 1-relation plan for each relation.Pass 2:Find best way to join result of each 1-relation plan(as outer)to another relation.(All 2-relation plans.)Pass N:Find best way to join result of a(N-1)-relation plan(as outer)to the Nth relation.(All N-relation plans.)For each subset of relations,retain only:Cheapest plan overall,plusCheapest plan for each interesting order of the tuples.The Dynamic Programming TableSubset of tables in FROM clauseInteresting-order columnsBest planCostR,Shashjoin(R,S)1000R,Ssortmerge(R,S)1500A Note on“Interesting Orders”An intermediate result has an“interesting order”if it is sorted by any of:ORDER BY attributesGROUP BY attributesJoin attributes of yet-to-be-added(downstream)joinsEnumeration of Plans(Contd.)An N-1 way plan is not combined with an additional relation unless there is a join condition between them,unless all predicates in WHERE have been used up.i.e.,avoid Cartesian products if possible.ORDER BY,GROUP BY,aggregates etc.handled as a final step,using either an interestingly ordered plan or an additonal sort/hash operator.In spite of pruning plan space,this approach is still exponential in the#of tables.Recall that in practice,COST considered is#IOs+factor*CPU InstExampleSailors:Hash,B+on sidReserves:Clustered B+tree on bid B+on sidBoats B+,Hash on colorSelect S.sid,COUNT(*)AS numberFROM Sailors S,Reserves R,Boats BWHERE S.sid=R.sid AND R.bid=B.bid AND B.color=“red”GROUP BY S.sidReservesSailorssid=sidBoats Sid,COUNT(*)AS numbesGROUPBY sidbid=bidColor=redPass1:Best plan(s)for accessing each relationReserves,Sailors:File ScanQ:What about Clustered B+on Reserves.bid?Boats:B+tree&Hash on colorPass 1Best plan for accessing each relation regarded as the first relation in an execution planReserves,Sailors:File ScanBoats:B+tree&Hash on colorPass 2For each of the plans in pass 1,generate plans joining another relation as the inner,using all join methods(and matching inner access methods)File Scan Reserves(outer)with Boats(inner)File Scan Reserves(outer)with Sailors(inner)File Scan Sailors(outer)with Boats(inner)File Scan Sailors(outer)with Reserves(inner)Boats hash on color with Sailors(inner)Boats Btree on color with Sailors(inner)Boats hash on color with Reserves(inner)(sort-merge)Boats Btree on color with Reserves(inner)(BNL)Retain cheapest plan for each pair of relationsPass 3 and beyondFor each of the plans retained from Pass 2,taken as the outer,generate plans for the next joineg Boats hash on color with Reserves(bid)(inner)(sortmerge)inner Sailors(B-tree sid)sort-mergeThen,add the cost for doing the group by and aggregate:This is the cost to sort the result by sid,unless it has already been sorted by a previous operator.Then,choose the cheapest planPoints to RememberMust understand optimization in order to understand the performance impact of a given database design(relations,indexes)on a workload(set of queries).Two parts to optimizing a query:Consider a set of alternative plans.Good to prune search space;e.g.,left-deep plans only,avoid Cartesian products.Must estimate cost of each plan that is considered.Output cardinality and cost for each plan node.Key issues:Statistics,indexes,operator implementations.Points to RememberSingle-relation queries:All access paths considered,cheapest is chosen.Issues:Selections that match index,whether index key has all needed fields and/or provides tuples in a desired order.More Points to RememberMultiple-relation queries:All single-relation plans are first enumerated.Selections/projections considered as early as possible.Next,for each 1-relation plan,all ways of joining another relation(as inner)are considered.Next,for each 2-relation plan that is retained,all ways of joining another relation(as inner)are considered,etc.At each level,for each subset of relations,only best plan for each interesting order of tuples is retained.Physical DB DesignQuery optimizer does what it can to use indices,clustering etc.DataBase Administrator(DBA)is expected to set up physical design well.Good DBAs understand query optimizers very well.One Key Decision:IndexesWhich tablesWhich field(s)should be the search key?Multiple indexes?Clustering?Index SelectionOne approach:Consider most important queries in turn.Consider best plan using the current indexesSee if better plan is possible with an additional index.If so,create it.But consider impact on updates!Indexes can make queries go faster,updates slower.Require disk space,too.Issues to Consider in Index SelectionAttributes mentioned in a WHERE clause are candidates for index search keys.Range conditions are sensitive to clustering Exact match conditions dont require clusteringOr do they?:-)Choose indexes that benefit many queries NOTE:only one index can be clustered per relation!So choose it wisely!Example 1B+tree index on D.dname supports Toy selection.Given this,index on D.dno is not needed.B+tree on E.dno allows us to get matching(inner)Emp tuples for each selected(outer)Dept tuple.What if WHERE included:.AND E.age=25?Could retrieve Emp tuples using index on E.age,then join with Dept tuples satisfying dname selection.Comparable to strategy that used E.dno index.So,if E.age index is already created,this query provides much less motivation for adding an E.dno index.SELECT E.ename,D.mgrFROM Emp E,Dept DWHERE E.dno=D.dno AND D.dname=ToyExample 2All selections are on Emp so it should be the outer relation in any Index NL join.Suggests that we build a B+tree index on D.dno.What index should we build on Emp?B+tree on E.sal could be used,OR an index on E.hobby could be used.Only one of these is needed,and which is better depends upon the selectivity of the conditions.As a rule of thumb,equality selections more selective than range selections.Have to understand optimizers to get this right!SELECT E.ename,D.mgrFROM Emp E,Dept DWHERE E.sal BETWEEN 10000 AND 20000 AND E.hobby=Stamps AND E.dno=D.dnoExamples of ClusteringB+tree index on E.age can be used to get qualifying tuples.How selective is the condition?Is the index clustered?Consider the GROUP BY query.If many tuples have E.age 10,using E.age index and sorting the retrieved tuples may be costly.Clustered E.dno index may be better!Equality queries and duplicates:Clustering on E.hobby helps!SELECT E.dnoFROM Emp EWHERE E.age40SELECT E.dno,COUNT(*)FROM Emp EWHERE E.age10GROUP BY E.dnoSELECT E.dnoFROM Emp EWHERE E.hobby=StampsSummaryOptimization is the reason for the lasting power of the relational systemBut it is primitive in some ways New areas:Smarter summary statistics(fancy histograms and“sketches”),auto-tuning statistics,adaptive runtime re-optimization(e.g.eddies)