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CS 440 Database Management Systems

Query Optimization

DBMS Architecture

Query Executor

Buffer Manager

Storage Manager

Storage

Transaction Manager

Logging & Recovery

Lock Manager

Buffers Lock Tables

Main Memory

User/Web Forms/Applications/DBAquery transaction

Query Optimizer

Query Rewriter

Query Parser

Files & Access Methods

Past lectures

Today’s lecture

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Many query plans to execute a SQL query

S UTRSR

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U

SR

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• Even more plans: multiple algorithms to execute each operation

SR

T

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Sort-merge

Sort-merge

hash join

Table-scan

index-scan

Table-scanindex-scan

• Compute the join of R(A,B) S(B,C) T(C,D) U(D,E)

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Query optimization: picking the fastest plan• Optimal approach plan– enumerate each possible plan– measure its performance by running it– pick the fastest one– What’s wrong?

• Rule-based optimization– Use a set of pre-defined rules to generate a fast plan • e.g. If there is an index over a table, use it for scan and join.

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Definitions• Statistics on table R:– T(R): Number of tuples in R– B(R): Number of blocks in R• B(R) = T(R ) / block size

– V(R,A): Number of distinct values of attribute A in R

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Review: Clustered index• The relation is stored on the disk according to the

order of index.

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DATAINDEX

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Plans to select tuples from R: sA=a(R)

• We have a clustered index on R• Plans: – (Clustered) indexed-based scan– Table-scan (sequential access)

• Statistics on R– B(R)=5000, T(R)=200,000– V(R,A) = 2, one value appears in 95% of tuples.

• Clustered indexed scan vs. table-scan ?

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Query optimization methods• Rule-based optimizer fails – It uses static rules– The rules do not consider the distribution of the data.

• Cost-based optimization– predict the cost of each plan– search the plan space to find the fastest one– do it efficiently• Optimization itself should be fast!

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Cost-based optimization• Plan space– which plans to consider?– it is time consuming to explore all alternatives.

• Cost estimator– how to estimate the cost of each plan without executing it?– we would like to have accurate estimation

• Search algorithm– how to search the plan space fast?– we would like to avoid checking inefficient plans

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Space of query plans• Selection– algorithms: sequential, index-based– ordering: why does it matter?

• Join– algorithms: nested loop, sort-merge, hash– ordering

• Ordering/ Grouping– can an “interesting order” be produced by join/

selection?– algorithms: sorting, hash-based

Reducing plan space• Multiple logical query plan for each SQL query Star(name, birthdate), StarsIn(movie, name, year) SELECT movie FROM Stars, StarsIn WHERE Star.name = StarsIn.name AND year = 1950

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Generally FasterStarsIn Star

StarsIn.name = Star.name

s year=1950

StarsIn

Star

StarsIn.name = Star.name

year=1950

moviemovie

Reducing plan space• Push selection down to reduce # of rows• Push projection down to reduce # of columns SELECT movie, name FROM Stars, StarsIn WHERE Star.name = StarsIn.name

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StarsIn Star

StarsIn.name = Star.name

movei, name

StarsIn Star

StarsIn.name = Star.name

movie, name

movie, name movie, name

Less effective than pushing down selection.

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• The algorithm requires exponential computation!• System-R style considers only left-deep joins

Reducing plan space

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• 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).

•

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• System R-style avoids the plans with Cartesian products– The size of a Cartesian product is generally larger

than (natural) joins.• Example: R(A,B), S(B,C), U(C,D)

(R U) S has a Cartesian product⋈ ⋈ pick (R S) U instead⋈ ⋈

• If cannot avoid Cartesian products, delay them.

Reducing plan space

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• Relative accuracy– Goal is to compare plans, not to predict exact cost– More of an art than an exact science

• Each operator: input size, cost, output size– estimate cost based on input size

• Example: sort-merge join of R ⋈ S is 3 B(R) + 3 B(S)

– estimate output size (for next operator) or selectivity• selectivity: ratio of output to input

Cost estimation

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Cost estimation: Selinger Style • Input: stats on each table– T(R): Number of tuples in R– B(R): Number of blocks in R• B(R) = T(R ) / block size

– V(R,A): Number of distinct values of attribute A in R

• Assumptions on attribute and predicate independence• When no estimate available, use magic numbers.• New alternative approach– Histogram of database

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Selectivity factors: selection• Point selection: S = sA=a(R)

– T(S) ranges from 0 to T(R) – V(R,A) + 1– consider its mean: F = 1 / V (R,A)

• Range selection: S = sA<a(R)– F = (max(A) – a) / (max(A) – min(A))– not-athematic inequality: use magic number

• F = 1 / 3

• Range selection: S = s b <A<a(R)– F = (a - b) / (max(A) – min(A))– If not athematic, use magic number

• F = 1 / 4

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Selectivity factors: selection

• Range selection: column in (set of values) – F: union of point selections

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Selectivity factors: selection

• S = sA=1 AND B<10(R)– multiply 1/V(R,A) for equality and 1/3 for inequality– T(R) = 10,000, V(R,A) = 50– T(S) = 10000 / (50 * 3) = 66

• S = sA=1 OR B<10(R)– sum of estimates of predicates minus their product– T(R) = 10,000, V(R,A) = 50– T(S) = 200 + 3333 – 66 = 3467

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• Containment of values assumption V(S,A) <= V (R,A): A values in S is a subset of A values in R

• Let’s assume V (S,A) <= V (R,A)– Each tuple t in S joins x tuple(s) in R– consider its mean: x = T(R) / V (R,A)– T(R ⋈A S) = T (S) * T(R) / V(R,A)

T(R ⋈A S) = T(R) * T(S) / max(V(R,A), V(S,A))

Selectivity factors: join predicates

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Search the plan space• Baseline: exhaustive search– enumerate all combinations and compare their costs – enormous space!

T USR SRT

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• Search method parameters– plan tree development

• construction: bottom-up, top-down• modification: improve a somehow-connected tree

– algorithms• heuristic selections: make choices based on heuristics• hill climbing: find “nearby” plans with lowest cost• Dynamic programming: construction by greedy selection

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Plan search: System-R style• A.K.A: Selinger style optimization• Bottom-up – start from the ground relation (in FROM) – work up the tree to form a plan– compute the cost of larger plans based on its sub-trees.

• Dynamic programming – greedily remove sub-trees that are costly (useless)

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• Step 1: For each {Ri}:– size({Ri}) = TCARD(Ri)– plan({Ri}) = Ri– cost({Ri}) = cost of access to Ri• e.g. TCARD(Ri) if no index on Ri

• Step 2: For each {Ri, Rj}:– size({Ri,Rj}) = estimate of the size of join– plan({Ri,Rj}) = join algorithm– cost = cost function of size of Ri and Rj

• #I/O access of the chosen join algorithm– plan({Ri,Rj}): the join algorithm with smallest cost

Dynamic programming

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• Step i: For each S ⊆ {R1, …, Rn} of cardinality i do:– Compute size(S) – for every S1 ,S2 s.t. S = S1 S2

c = cost(S1) + cost(S2) + cost(S1 S⋈ 2)– cost(S) = the smallest C– plan(S) = the plan for cost(S)

• Return Plan({R1, …, Rn})

Dynamic programming

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• Let’s assume that the cost of each join is the size of its intermediate results.– to simplify the example– other cost measures, #I/O access, are possible.

• cost(R) = 0 (no intermediate results)• cost(R ⋈ S) = 0 (no intermediate results)• cost( (R ⋈ S) ⋈ T)

= cost(R ⋈ S) + cost(T) + size( R ⋈ S ) = size(R ⋈ S)

Dynamic programming: example

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• Relations: R, S, T, U• Number of tuples: 2000, 5000, 3000, 1000• We use a toy size estimation method:– size (A B) = 0.01 * T(A) * T(B)⋈

Dynamic programming: example

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Query Size Cost Plan

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RST

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Query Size Cost Plan

RS 100k 0 RS

RT 60k 0 RT

RU 20k 0 UR

ST 150k 0 TS

SU 50k 0 US

TU 30k 0 UT

RST

RSU

RTU

STU

RSTU

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Query Size Cost Plan

RS 100k 0 RS

RT 60k 0 RT

RU 20k 0 UR

ST 150k 0 TS

SU 50k 0 US

TU 30k 0 UT

RST 3M 60k S(RT)

RSU 1M 20k S(UR)

RTU 0.6M 20k T(UR)

STU 1.5M 30k S(UT)

RSTU

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Query Size Cost Plan

RS 100k 0 RS

RT 60k 0 RT

RU 20k 0 UR

ST 150k 0 TS

SU 50k 0 US

TU 30k 0 UT

RST 3M 60k S(RT)

RSU 1M 20k S(UR)

RTU 0.6M 20k T(UR)

STU 1.5M 30k S(UT)

RSTU 30M 110k (US)(RT)

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Plan search: all operations• Base relations access– find all plans for accessing each base relations– push down selections and projections– choose good plans, discard bad ones• keep the cheapest plan for unordered and each interesting

order • Join ordering– use the bottom-up dynamic programming– consider only left-deep join trees: n! ordering for n tables– postpone Cartesian product

• Finally: grouping/ ordering– use interesting order– addition sorting

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Nested subqueries• Subqueries are optimized separately • Correlation: order of evaluation– uncorrelated queries

• nested subqueries do not reference outer subqueries• evaluate the most deeply nested subquery first

– correlated queries: nested subqueries reference the outer subqueries

Select name From employee XWhere salary > (Select salary

From employee Where employee_num

= X.manager)

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Nested subqueries – cont.• correlated queries: nested subqueries reference the outer

subqueriesSelect name From employee XWhere salary > (Select salary

From employee Where employee_num

= X.manager)• The nested subquery is evaluated once for each tuple in the

outer query.• If there are small number of distinct values in the outer

relation, it is worth sorting the tuples. – reduces the #evaluation of the nested query.

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Summary: optimization• Plan space– Huge number of alternatives, semantically equivalent

• Why important – Difference between good/bad plabs could be order of

magnitude • Idea goal – map a declarative query to the most efficient plan

• Conventional wisdom: at least avoid bad plans

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State of the art• Academic: always a core database research topic – Optimizing for interactive querying– Optimizing for novel parallel frameworks

• Industry: most optimizers use System-R style– They started with rule-based.• Oracle 7 and its prior versions used rule-based• Oracle 7 – 10: rule based and cost based• Oracle 10g (2003): cost-based

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• The importance of query optimization– difference between fast and slow plans

• Query optimization problem– find the fast plans efficiently.

• The components of a cost-based (system R style) query optimizer:– plan space definition– cost estimation– search algorithm

What you should know