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Relational Query Optimization

Chapter 15

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Query Blocks: Units of Optimization

An SQL query is parsed into a collection of query blocks :

An SQL query with no nesting and exactly one SELECT, FROM, WHERE, GROUP BY, and HAVING clause.

WHERE is in conjunctive normal form.

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Query Blocks: Units of Optimization Nested blocks are usually treated as calls

to a subroutine, made once per outer tuple.

Optimization is one block at a time.

SELECT S.snameFROM Sailors SWHERE S.age IN (SELECT MAX (S2.age) FROM Sailors S2 GROUP BY S2.rating)

Nested blockOuter block

SELECT S.snameFROM Sailors SWHERE S.age IN Reference to nested block

SELECT MAX (S2.age) FROM Sailors S2 GROUP BY S2.rating

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Query Block

For each block, the plans considered are:– All available access methods

– for each relation in FROM clause.

– All left-deep join trees – all ways to join the relations one-at-a-time, with inner relation in FROM clause, considering all relation permutations and join methods.

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Cost Estimation

For each plan considered, must estimate cost: Must estimate cost of each operation in plan tree.

• Depends on input cardinalities.• Depends on algorithm (sequential scan, index scan).

Must also estimate size of result for each operation.

• Use information about the input relations.• Must make assumptions about effect of predicates.

Cost of plan = sum of cost of each operator in tree.

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Size Estimation and Reduction Factors

Goal : Estimate result size !

SELECT attribute listFROM relation listWHERE term1 AND ... AND termk

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Size Estimation and Reduction Factors

Consider a query block:

Given maximum # tuples in result : product of cardinalities of relations in FROM clause.

Reduction factor (RF) associated with each term : reflects impact of term in reducing result size.

Result size = product of cardinalities of involved relations (FROM) *

product of reduction factors (WHERE).

SELECT attribute listFROM relation listWHERE term1 AND ... AND termk

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Assumptions

Uniform distribution of values in domain Independent distribution of values in different columns. For selections and joins, assume independence of predicates.

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Reduction Factors Column = value.

- Given the index I on column, assume uniform distribution. • 1/Nkeys(I).• Otherwise, fixed reduction factor 1/10

Column1 = column2- Given indexes I1 and I2 on column1 and column2,

assuming each key value in I1 (smaller one) has a matching value in I2.

• 1/MAX(Nkeys(I1), Nkeys(I2)).• One index I, 1/Nkeys(I)• otherwise, 1/10

Column > values- Given an index I on column, arithmetic type.

• High(I) – value / High(I) – Low(I). • Not arithmetic type, or no index. • A fraction Less than half is arbitrarily chosen.

Column IN (list of values)- reduction factor for (column = value) * number of items in

list.

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More on Estimation

Uniform distribution is not accurate since real data is not uniformly distributed.

Histogram: a data structure maintained by a DBMS to approximate a data distribution.

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Estimation Equi-width: divide range of column values into subranges

(buckets). Assuming the distribution within the histogram bucket is uniform.

Equi-depth: number of tuples within each bucket is equal (almost).

Compressed equi-depth: maintain separate counts for a small number of very frequent values, and maintain equi-depth histogram to cover the remaining values.

2 3 3 1 2 1 3 4 8 2 0 1 2 4 9

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

2.67

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Equiwidth

1.33

5.0

1.0

5.0

Bucket 1 2 3 4 5Count 8 4 15 3 15

2.25

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Equidepth

2.5

5.0

1.75

9.0

Bucket 1 2 3 4 5Count 9 10 10 7 9

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Relational Algebra Equivalences

Allow us to choose different join orders

Allow us to `push’ selections and projections ahead of joins.

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Relational Algebra Equivalences

Selections: (Cascade)

c cn c cnR R1 1 ... . . .

c c c cR R1 2 2 1 (Commute)

Combine several selections into one selection. Evaluate conjunctive condition to each of the tuple.Separate conjunctions into smaller selection operations, Able to combine as Join.

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Relational Algebra Equivalences

Projections: a a anR R1 1 . . . (Cascade)Where ai ai+1 for i =1…n-1.

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Relational Algebra Equivalences

Joins: R (S T) (R S) T (Associative)

(R S) (S R) (Commute)

• When join several relations, we are free to join the relations in any order we choose.

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Selects, Projects and Joins

Case1: A projection commutes with a selection that only uses attributes retained by the projection.

a (c(R)) c (a(R))

Case2: Combine a selection with a cross-product to form a join. R c S c (RS)

Case3: A selection on just attributes of R commutes with Join. i.e., (R S) (R) S

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Selects, Projects and Joins Selection and Joins :

c1c2 c3 (R S) c1(c2 (R) c3

(S))

Project and Joins a (RS) a1 (R) a2 (S) a (R c S) a1 (R) c a2 (S) (Where c

appear in a)

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Enumeration of Alternative Plans

There are two main cases: Single-relation plans Multiple-relation plans

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Single Relation Plans Queries over a single relation: combination of

selects, projects, and aggregate operations:

Main decision : Which access path for retrieving tuples.

• Most selective access path (file scan / index) if only single operator considered.

The different operations 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.

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s.rating, COUNT(*)(HAVING COUNT DISTINCT(S.sname) > 2(GROUP Bys.rating( s.rating, S.sname (

s.rating>5 S.age=20 (Sailors)))))

Single Relation Plans without Index

SELECT S.rating, COUNT(*)FROM Sailors SWHERE S.rating > 5 AND S.age = 20GROUP BY S.ratingHAVING COUNT DISTINCT (S.sname) > 2

• File scan to retrieve tuples and apply the selections and projections.• Writing out tuples after the selections and projections.• Sorting these tuples to implement the GROUP BY clause. * Then GROUP BY and HAVING are done on-the-fly.

e.g., Cost = Cost1(scan) + cost2 (writing <S.rating, S.sname> pairs) + cost3 (sorting as per the GROUP BY clause).

cost1 = 500 pages cost2 = 500* ratio of tuple size * RFs = 20 pages

ratio of tuple size = pair size / tuple size = 0.8 RF(s.rating>5) = 0.5 RF(S.age=20) = 0.1

cost3 = 3*Npages = 60 (assuming two passes)

Sailors = 500 pages

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Single-Relation Plans with Index Single-index access path

• When several indexes match the selection conditions, choose the most selective access path.

Multiple-index access path• Several indexes using Alternatives (2) or (3) for data

entries match the selection condition.• Retrieve rids using them individually.• Intersect result set. Sort by page id.

Sorted-index access path• Grouping attributes is a prefix of a tree index.• Using index to retrieve tuples in order required by

GROUP BY Index-only access path

• All attributes mentioned in the query are included in the search key for some dense index on the relation in the FROM clause.

• Index-only scan to compute the answer.

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Single-Relation Plans with Index Index I on primary key matches selection:

Cost is Height(I)+1 for a B+ tree, about 1.2+1 for hash index.

Clustered index I matching one or more selects: (NPages(I)+NPages(R)) * product of RF’s of matching selects.

Non-clustered index I matching one or more selects: (NPages(I)+NTuples(R)) * product of RF’s of matching selects.

Sequential scan of file: NPages(R).

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Single-Relation Example

Sailors: 500 pages, 80 tuples/page Case2: index on rating.

Clustered index: (1/NKeys(I)) * (NPages(I)+NPages(S)) = (1/10) *

(50+500) pages Unclustered index: (1/NKeys(I)) * (NPages(I)+NTuples(S)) = (1/10) *

(50+40000) pages Case3: index on sid.

Would have to retrieve all tuples/pages. With a clustered index, the cost is 50+500, with unclustered index, 50+40000.

Case4: file scan: We retrieve all file pages (500).

SELECT S.sidFROM Sailors SWHERE S.rating=8

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Queries Over Multiple Relations

As the number of joins increases, the number of alternative plans grows rapidly

We need to restrict search space !

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Queries Over Multiple Relations

Fundamental decision in System R (IBM): Only left-deep join trees are considered. Left-deep trees can generate all fully pipelined

plans.• Intermediate results not written to temporary

files.• Not all left-deep trees are fully pipelined (e.g., SM

join).

BA

C

D

BA

C

D

C DBA

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Left-deep plans differ in : the order of relations, the access method for each relation, and the join method for each join.

Enumeration of Left-Deep Plans

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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 N’th relation. (All N-relation plans.)

For each subset of relations, retain : Cheapest plan overall, plus Cheapest plan for each interesting

order of the tuples.

Enumeration of Left-Deep Plans

BA C DPass 1

Pass 2

Pass 3

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Enumeration of Left-Deep Plans : Pass 1

Identify selection terms in WHERE clause that mention only attributes of A. (perform access of A, before Join)

Identify attributes of A not mentioned in SELECT or WHERE (project out when first access of A, before Join)

Keep cheapest overall plan for fetching all tuples : • a file scan

Keep cheapest plan with tuples in search key order :• B+ tree index

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Left-Deep Plans: All 2-relation Plans

Each of the single-relation plan from Pass 1 as the outer relation, and every other relation as the inner relation.

Examine WHERE clauses:• Selections involving only attributes of inner relation (apply

before Join).• Selections defining the Join.• Selections involving attributes of other relations (apply after

Join).

Only selections that are really applied before the join are those that match the chosen access paths for A and B.

Depending on the Join algorithm chosen, the cost may include materializing the outer relation.

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Enumeration of Plans (Contd.)

ORDER BY, GROUP BY, aggregates etc. handled as a final step, using either an `interestingly ordered’ plan or an additional sorting operator.

An N-1 way plan is not combined with another relation unless join condition between them : avoid Cartesian products if possible.

In spite of pruning plan space, this approach is still exponential in the # of tables.

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Example:Pass 1

Sailors: Choice1: B+ tree matches rating>5,

probably cheapest. If selection is expected to retrieve a lot of

tuples, and index is unclustered, file scan may be cheaper.

Decision: B+ tree plan kept (because tuples are in rating order).

Reserves: B+ tree on bid matches bid=100; cheapest.

Sailors: B+ tree on rating Hash on sidReserves: B+ tree on bid

Reserves Sailors

sid=sid

bid=100 rating > 5

sname

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ExamplePass 2

Pass1: Sailors:

• Choice1: B+ tree matches rating>5, Reserves:

• B+ tree on bid matches bid=100.

Sailors: B+ tree on rating Hash on sidReserves: B+ tree on bid

Pass 2:– each plan retained from Pass 1 as the outer.– how to join it with the (only) other relation.– e.g., Reserves as outer : Hash index can be used to get Sailors tuples

that satisfy sid = outer tuple’s sid value.

– e.g., Sailors as outer : Could possibly use sort-merge join, etc.

Reserves Sailors

sid=sid

bid=100 rating > 5

sname

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Highlights of System R Optimizer

Impact of R Optimizer: 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 next operator without storing it in temporary relation.

Cartesian products avoided.

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Other Approaches to Query Optimization Exhaustive search is not suitable for large number of

Joins.

Rule-based optimizers: a set of rules to guide the generation of candidate plans.

Randomized plan generation: probabilistic algorithms to explore a large space of plans.

Estimating the size of intermediate relations accurately.• Parametric query optimization: find good plans for a given query for

each of several different conditions that might be encountered at run-time.

• Multiple-query optimization: take concurrent execution of several queries into account.

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Summary

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.• Estimate size of result • Estimate cost for each plan node.• Key issues: Statistics, indexes, operator implementations.