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UW CSE 444 - Overview of Query Optimization

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Overview of Query OptimizationHighlights of System R OptimizerSchema for ExamplesMotivating ExampleAlternative Plans 1 (No Indexes)Alternative Plans 2 With IndexesQuery Blocks: Units of OptimizationCost EstimationStatistics and CatalogsSize Estimation and Reduction FactorsRelational Algebra EquivalencesMore EquivalencesEnumeration of Alternative PlansCost Estimates for Single-Relation PlansExampleQueries Over Multiple RelationsEnumeration of Left-Deep PlansEnumeration of Plans (Contd.)Slide 19Nested QueriesSummarySummary (Contd.)Overview of Query OptimizationPlan: Tree of R.A. ops, with choice of alg for each op.–Each operator typically implemented using a `pull’ interface: when an operator is `pulled’ for the next output tuples, it `pulls’ on its inputs and computes them.Two main issues:–For a given query, what plans are considered?Algorithm to search plan space for cheapest (estimated) plan.–How is the cost of a plan estimated?Ideally: Want to find best plan. Practically: Avoid worst plans!We will study the System R approach.Highlights of System R OptimizerImpact:–Most widely used currently; works well for < 10 joins.Cost estimation: Approximate art at best.–Statistics, maintained in system catalogs, used to estimate cost of operations and result sizes.–Considers combination of CPU and I/O costs.Plan Space: Too large, must be pruned.–Only the space of left-deep plans is considered.Left-deep plans allow output of each operator to be pipelined into the next operator without storing it in a temporary relation.–Cartesian products avoided.Schema for ExamplesSimilar to old schema; rname added for variations.Reserves:–Each tuple is 40 bytes long, 100 tuples per page, 1000 pages.Sailors:–Each tuple is 50 bytes long, 80 tuples per page, 500 pages. Sailors (sid: integer, sname: string, rating: integer, age: real)Reserves (sid: integer, bid: integer, day: dates, rname: string)Motivating ExampleCost: 500+500*1000 I/OsBy 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.rating>5ReservesSailorssid=sidbid=100 rating > 5snameReservesSailorssid=sidbid=100 rating > 5sname(Simple Nested Loops)(On-the-fly)(On-the-fly)RA Tree:Plan:Alternative Plans 1 (No Indexes)Main difference: push selects.With 5 buffers, cost of plan:–Scan Reserves (1000) + write temp T1 (10 pages, if we have 100 boats, uniform distribution).–Scan Sailors (500) + write temp T2 (250 pages, if we have 10 ratings).–Sort T1 (2*2*10), sort T2 (2*3*250), merge (10+250), total=1830 –Total: 3560 page I/Os.If we used BNL join, join cost = 10+4*250, total cost = 2770.If we `push’ projections, T1 has only sid, T2 only sid and sname:–T1 fits in 3 pages, cost of BNL drops to under 250 pages, total < 2000.ReservesSailorssid=sidbid=100 sname(On-the-fly)rating > 5(Scan;write to temp T1)(Scan;write totemp T2)(Sort-Merge Join)Alternative Plans 2With IndexesWith clustered index on bid of Reserves, we get 100,000/100 = 1000 tuples on 1000/100 = 10 pages.INL with pipelining (outer is not materialized). Decision not to push rating>5 before the join is based on availability of sid index on Sailors. Cost: Selection of Reserves tuples (10 I/Os); for each, must get matching Sailors tuple (1000*1.2); total 1210 I/Os. Join column sid is a key for Sailors.–At most one matching tuple, unclustered index on sid OK.–Projecting out unnecessary fields from outer doesn’t help.ReservesSailorssid=sidbid=100 sname(On-the-fly)rating > 5(Use hashindex; donot writeresult to temp)(Index Nested Loops,with pipelining )(On-the-fly)Query Blocks: Units of OptimizationAn SQL query is parsed into a collection of query blocks, and these are optimized one block at a time.Nested blocks are usually treated as calls to a subroutine, made once per outer tuple. (This is an over-simplification, but serves for now.)SELECT S.snameFROM Sailors SWHERE S.age IN (SELECT MAX (S2.age) FROM Sailors S2 GROUP BY S2.rating)Nested blockOuter block For each block, the plans considered are:– All available access methods, for each relation in FROM clause.– All left-deep join trees (i.e., all ways to join the relations one-at-a-time, with the inner relation in the FROM clause, considering all relation permutations and join methods.)Cost EstimationFor each plan considered, must estimate cost:–Must estimate cost of each operation in plan tree.Depends on input cardinalities.We’ve 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.We’ll discuss the System R cost estimation approach.–Very inexact, but works ok in practice.–More sophisticated techniques known now.Statistics and CatalogsNeed information about the relations and indexes involved. Catalogs typically contain at least:–# tuples (NTuples) and # pages (NPages) for each relation.–# distinct key values (NKeys) and NPages for each index.–Index height, low/high key values (Low/High) for each tree 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 FactorsConsider 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 RF’s.–Implicit assumption that terms are independent!–Term col=value has RF 1/NKeys(I), given index I on col–Term col1=col2 has RF 1/MAX(NKeys(I1), NKeys(I2))–Term col>value has RF (High(I)-value)/(High(I)-Low(I))SELECT attribute listFROM relation listWHERE term1 AND ... AND termkRelational Algebra EquivalencesAllow us to choose different join orders and to `push’ selections and projections ahead of joins.Selections:


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UW CSE 444 - Overview of Query Optimization

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