1 Chapter 21 Query Processing Transparencies ownerNoclient © Pearson Education Limited 1995, 2005.

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Chapter 21

Query Processing

Transparencies

© Pearson Education Limited 1995, 2005

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Chapter 21 - Objectives

Objectives of query processing and optimization. Static versus dynamic query optimization. How a query is decomposed and semantically

analyzed. How to create a R.A.T. to represent a query. Rules of equivalence for RA operations. How to apply heuristic transformation rules to

improve efficiency of a query.


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Types of database statistics required to estimate cost of operations.

Different strategies for implementing selection. How to evaluate cost and size of selection. Different strategies for implementing join. How to evaluate cost and size of join. Different strategies for implementing projection. How to evaluate cost and size of projection.


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How to evaluate the cost and size of other RA operations.

How pipelining can be used to improve efficiency of queries.

Difference between materialization and pipelining.

Advantages of left-deep trees. Approaches to finding optimal execution

strategy. How Oracle handles QO.


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Introduction

In network and hierarchical DBMSs, low-level procedural query language is generally embedded in high-level programming language.

Programmer’s responsibility to select most appropriate execution strategy.

With declarative languages such as SQL, user specifies what data is required rather than how it is to be retrieved.

Relieves user of knowing what constitutes good execution strategy.


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Introduction

Also gives DBMS more control over system performance.

Two main techniques for query optimization:– heuristic rules that order operations in a query; – comparing different strategies based on relative

costs, and selecting one that minimizes resource usage.

Disk access tends to be dominant cost in query processing for centralized DBMS.


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Query Processing

Activities involved in retrieving data from the database.

Aims of QP:– transform query written in high-level language

(e.g. SQL), into correct and efficient execution strategy expressed in low-level language (implementing RA);

– execute strategy to retrieve required data.


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Query Optimization

Activity of choosing an efficient execution strategy for processing query.

As there are many equivalent transformations of same high-level query, aim of QO is to choose one that minimizes resource usage.

Generally, reduce total execution time of query. May also reduce response time of query. Problem computationally intractable with large

number of relations, so strategy adopted is reduced to finding near optimum solution.


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Example 21.1 - Different Strategies

Find all Managers who work at a London branch.

SELECT *

FROM Staff s, Branch b

WHERE s.branchNo = b.branchNo AND

(s.position = ‘Manager’ AND b.city = ‘London’);


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Three equivalent RA queries are:

(1) (position='Manager') (city='London')

(Staff.branchNo=Branch.branchNo) (Staff X Branch)

(2) (position='Manager') (city='London')(

Staff Staff.branchNo=Branch.branchNo Branch)

(3) (position='Manager'(Staff)) Staff.branchNo=Branch.branchNo

(city='London' (Branch))


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Assume:– 1000 tuples in Staff; 50 tuples in Branch;– 50 Managers; 5 London branches;– no indexes or sort keys;– results of any intermediate operations stored

on disk;– cost of the final write is ignored;– tuples are accessed one at a time.


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Example 21.1 - Cost Comparison

Cost (in disk accesses) are:

(1) (1000 + 50) + 2*(1000 * 50) = 101 050 (2) 2*1000 + (1000 + 50) = 3 050 (3) 1000 + 2*50 + 5 + (50 + 5) = 1 160

Cartesian product and join operations much more expensive than selection, and third option significantly reduces size of relations being joined together.


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Phases of Query Processing

QP has four main phases:

– decomposition (consisting of parsing and validation);

– optimization;– code generation;– execution.


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Phases of Query Processing


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Dynamic versus Static Optimization

Two times when first three phases of QP can be carried out:– dynamically every time query is run;– statically when query is first submitted.

Advantages of dynamic QO arise from fact that information is up to date.

Disadvantages are that performance of query is affected, time may limit finding optimum strategy.


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Dynamic versus Static Optimization

Advantages of static QO are removal of runtime overhead, and more time to find optimum strategy.

Disadvantages arise from fact that chosen execution strategy may no longer be optimal when query is run.

Could use a hybrid approach to overcome this.


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Query Decomposition

Aims are to transform high-level query into RA query and check that query is syntactically and semantically correct.

Typical stages are:– analysis, – normalization, – semantic analysis, – simplification, – query restructuring.


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Analysis

Analyze query lexically and syntactically using compiler techniques.

Verify relations and attributes exist. Verify operations are appropriate for object type.


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Analysis - Example

SELECT staff_no

FROM Staff

WHERE position > 10;

This query would be rejected on two grounds:– staff_no is not defined for Staff relation

(should be staffNo).– Comparison ‘>10’ is incompatible with type

position, which is variable character string.


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Analysis

Finally, query transformed into some internal representation more suitable for processing.

Some kind of query tree is typically chosen, constructed as follows:– Leaf node created for each base relation.– Non-leaf node created for each intermediate

relation produced by RA operation.– Root of tree represents query result.– Sequence is directed from leaves to root.


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Example 21.1 - R.A.T.


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Normalization

Converts query into a normalized form for easier manipulation.

Predicate can be converted into one of two forms:

Conjunctive normal form:(position = 'Manager' salary > 20000) (branchNo = 'B003')

Disjunctive normal form:(position = 'Manager' branchNo = 'B003' )

(salary > 20000 branchNo = 'B003')


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Semantic Analysis

Rejects normalized queries that are incorrectly formulated or contradictory.

Query is incorrectly formulated if components do not contribute to generation of result.

Query is contradictory if its predicate cannot be satisfied by any tuple.

Algorithms to determine correctness exist only for queries that do not contain disjunction and negation.


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Semantic Analysis

For these queries, could construct:– A relation connection graph. – Normalized attribute connection graph.

Relation connection graph

Create node for each relation and node for result. Create edges between two nodes that represent a join, and edges between nodes that represent projection.

If not connected, query is incorrectly formulated. © Pearson Education Limited 1995, 2005

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Semantic Analysis - Normalized Attribute Connection Graph

Create node for each reference to an attribute, or constant 0.

Create directed edge between nodes that represent a join, and directed edge between attribute node and 0 node that represents selection.

Weight edges a b with value c, if it represents inequality condition (a b + c); weight edges 0 a with -c, if it represents inequality condition (a c).

If graph has cycle for which valuation sum is negative, query is contradictory.


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Example 21.2 - Checking Semantic Correctness

SELECT p.propertyNo, p.streetFROM Client c, Viewing v, PropertyForRent pWHERE c.clientNo = v.clientNo AND

c.maxRent >= 500 AND

c.prefType = ‘Flat’ AND p.ownerNo = ‘CO93’;

Relation connection graph not fully connected, so query is not correctly formulated.

Have omitted the join condition (v.propertyNo = p.propertyNo) .


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Relation Connection graph

Normalized attribute

connection graph


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SELECT p.propertyNo, p.streetFROM Client c, Viewing v, PropertyForRent pWHERE c.maxRent > 500 AND

c.clientNo = v.clientNo AND v.propertyNo = p.propertyNo AND c.prefType = ‘Flat’ AND c.maxRent <

200;

Normalized attribute connection graph has cycle between nodes c.maxRent and 0 with negative valuation sum, so query is contradictory.


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Simplification

– Detects redundant qualifications, – eliminates common sub-expressions, – transforms query to semantically equivalent

but more easily and efficiently computed form. Typically, access restrictions, view definitions,

and integrity constraints are considered. Assuming user has appropriate access privileges,

first apply well-known idempotency rules of boolean algebra.


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Transformation Rules for RA Operations

Conjunctive Selection operations can cascade into individual Selection operations (and vice versa).

pqr(R) = p(q(r(R)))

Sometimes referred to as cascade of Selection.

branchNo='B003' salary>15000(Staff) = branchNo='B003'(salary>15000(Staff))


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Commutativity of Selection.

p(q(R)) = q(p(R))

For example:

branchNo='B003'(salary>15000(Staff)) = salary>15000(branchNo='B003'(Staff))


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In a sequence of Projection operations, only the last in the sequence is required.

LM … N(R) = L (R)

For example:

lNamebranchNo, lName(Staff) = lName (Staff)


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Commutativity of Selection and Projection.

If predicate p involves only attributes in projection list, Selection and Projection operations commute:

Ai, …, Am(p(R)) = p(Ai, …, Am(R)) where p {A1, A2, …, Am}

For example:

fName, lName(lName='Beech'(Staff)) = lName='Beech'(fName,lName(Staff))


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Commutativity of Theta join (and Cartesian product).

R p S = S p R

R X S = S X R

Rule also applies to Equijoin and Natural join. For example:

Staff staff.branchNo=branch.branchNo Branch =

Branch staff.branchNo=branch.branchNo Staff


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Commutativity of Selection and Theta join (or Cartesian product).

If selection predicate involves only attributes of one of join relations, Selection and Join (or Cartesian product) operations commute:

p(R r S) = (p(R)) r S

p(R X S) = (p(R)) X S

where p {A1, A2, …, An}


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If selection predicate is conjunctive predicate having form (p q), where p only involves attributes of R, and q only attributes of S, Selection and Theta join operations commute as:

p q(R r S) = (p(R)) r (q(S))

p q(R X S) = (p(R)) X (q(S))


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For example:

position='Manager' city='London'(Staff

Staff.branchNo=Branch.branchNo Branch) =

(position='Manager'(Staff)) Staff.branchNo=Branch.branchNo (city='London' (Branch))


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Commutativity of Projection and Theta join (or Cartesian product).

If projection list is of form L = L1 L2, where L1 only has attributes of R, and L2 only has attributes of S, provided join condition only contains attributes of L, Projection and Theta join commute:

L1L2(R r S) = (L1(R)) r (L2(S))


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If join condition contains additional attributes not in L (M = M1 M2 where M1 only has attributes of R, and M2 only has attributes of S), a final projection operation is required:

L1L2(R r S) = L1L2( (L1M1(R)) r (L2M2(S)))


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For example:

position,city,branchNo(Staff Staff.branchNo=Branch.branchNo Branch) =

(position, branchNo(Staff)) Staff.branchNo=Branch.branchNo (

city, branchNo (Branch))

and using the latter rule:

position, city(Staff Staff.branchNo=Branch.branchNo Branch) =

position, city ((position, branchNo(Staff)) Staff.branchNo=Branch.branchNo ( city, branchNo (Branch)))


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Commutativity of Union and Intersection (but not set difference).

R S = S R

R S = S R


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Commutativity of Selection and set operations (Union, Intersection, and Set difference).

p(R S) = p(S) p(R)

p(R S) = p(S) p(R)

p(R - S) = p(S) - p(R)


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Commutativity of Projection and Union.

L(R S) = L(S) L(R)

Associativity of Union and Intersection (but not Set difference).

(R S) T = S (R T)

(R S) T = S (R T)


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Associativity of Theta join (and Cartesian product). Cartesian product and Natural join are always

associative:

(R S) T = R (S T)

(R X S) X T = R X (S X T)

If join condition q involves attributes only from S and T, then Theta join is associative:

(R p S) q r T = R p r (S q T)


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For example:

(Staff Staff.staffNo=PropertyForRent.staffNo PropertyForRent)

ownerNo=Owner.ownerNo staff.lName=Owner.lName Owner =

Staff staff.staffNo=PropertyForRent.staffNo staff.lName=lName

(PropertyForRent ownerNo Owner)


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Example 21.3 Use of Transformation Rules

For prospective renters of flats, find properties that match requirements and owned by CO93.

SELECT p.propertyNo, p.streetFROM Client c, Viewing v, PropertyForRent pWHERE c.prefType = ‘Flat’ AND

c.clientNo = v.clientNo AND v.propertyNo = p.propertyNo ANDc.maxRent >= p.rent AND c.prefType = p.type AND p.ownerNo = ‘CO93’;


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Heuristical Processing Strategies

Perform Selection operations as early as possible.– Keep predicates on same relation together.

Combine Cartesian product with subsequent Selection whose predicate represents join condition into a Join operation.

Use associativity of binary operations to rearrange leaf nodes so leaf nodes with most restrictive Selection operations executed first.


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Heuristical Processing Strategies

Perform Projection as early as possible.

– Keep projection attributes on same relation together.

Compute common expressions once.

– If common expression appears more than once, and result not too large, store result and reuse it when required.

– Useful when querying views, as same expression is used to construct view each time.


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Cost Estimation for RA Operations

Many different ways of implementing RA operations.

Aim of QO is to choose most efficient one. Use formulae that estimate costs for a number of

options, and select one with lowest cost. Consider only cost of disk access, which is usually

dominant cost in QP. Many estimates are based on cardinality of the

relation, so need to be able to estimate this.


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Database Statistics

Success of estimation depends on amount and currency of statistical information DBMS holds.

Keeping statistics current can be problematic. If statistics updated every time tuple is changed,

this would impact performance. DBMS could update statistics on a periodic basis,

for example nightly, or whenever the system is idle.


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Typical Statistics for Relation R

nTuples(R) - number of tuples in R.

bFactor(R) - blocking factor of R.

nBlocks(R) - number of blocks required to store R:

nBlocks(R) = [nTuples(R)/bFactor(R)]


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Typical Statistics for Attribute A of Relation R

nDistinctA(R) - number of distinct values that

appear for attribute A in R.

minA(R),maxA(R)

– minimum and maximum possible values for attribute A in R.

SCA(R) - selection cardinality of attribute A in R.

Average number of tuples that satisfy an equality condition on attribute A.


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Statistics for Multilevel Index I on Attribute A

nLevelsA(I) - number of levels in I.

nLfBlocksA(I) - number of leaf blocks in I.


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Selection Operation

Predicate may be simple or composite. Number of different implementations, depending

on file structure, and whether attribute(s) involved are indexed/hashed.

Main strategies are:– Linear Search (Unordered file, no index).– Binary Search (Ordered file, no index). – Equality on hash key.– Equality condition on primary key.


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Selection Operation

– Inequality condition on primary key.– Equality condition on clustering (secondary)

index. – Equality condition on a non-clustering

(secondary) index. – Inequality condition on a secondary B+-tree

index.


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Estimating Cardinality of Selection

Assume attribute values are uniformly distributed within their domain and attributes are independent.

nTuples(S) = SCA(R)

For any attribute B A of S, nDistinctB(S) =

nTuples(S) if nTuples(S) < nDistinctB(R)/2

nDistinctB(R) if nTuples(S) > 2*nDistinctB(R)

[(nTuples(S) + nDistinctB(R))/3] otherwise


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Linear Search (Ordered File, No Index)

May need to scan each tuple in each block to check whether it satisfies predicate.

For equality condition on key attribute, cost estimate is:

[nBlocks(R)/2] For any other condition, entire file may need to be

searched, so more general cost estimate is:

nBlocks(R)


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Binary Search (Ordered File, No Index)

If predicate is of form A = x, and file is ordered on key attribute A, cost estimate:

[log2(nBlocks(R))] Generally, cost estimate is:

[log2(nBlocks(R))] + [SCA(R)/bFactor(R)] - 1 First term represents cost of finding first tuple

using binary search. Expect there to be SCA(R) tuples satisfying

predicate.


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Equality of Hash Key

If attribute A is hash key, apply hashing algorithm to calculate target address for tuple.

If there is no overflow, expected cost is 1. If there is overflow, additional accesses may be

necessary.


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Equality Condition on Primary Key

Can use primary index to retrieve single record satisfying condition.

Need to read one more block than number of index accesses, equivalent to number of levels in index, so estimated cost is:

nLevelsA(I) + 1


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Inequality Condition on Primary Key

Can first use index to locate record satisfying predicate (A = x).

Provided index is sorted, records can be found by accessing all records before/after this one.

Assuming uniform distribution, would expect half the records to satisfy inequality, so estimated cost is:

nLevelsA(I) + [nBlocks(R)/2]


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Equality Condition on Clustering Index

Can use index to retrieve required records. Estimated cost is:

nLevelsA(I) + [SCA(R)/bFactor(R)]

Second term is estimate of number of blocks that will be required to store number of tuples that satisfy equality condition, represented as SCA(R).


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Equality Condition on Non-Clustering Index

Can use index to retrieve required records. Have to assume that tuples are on different

blocks (index is not clustered this time), so estimated cost becomes:

nLevelsA(I) + [SCA(R)]


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Inequality Condition on a Secondary B+-Tree Index

From leaf nodes of tree, can scan keys from smallest value up to x (< or <= ) or from x up to maximum value (> or >=).

Assuming uniform distribution, would expect half the leaf node blocks to be accessed and, via index, half the file records to be accessed.

Estimated cost is:

nLevelsA(I) + [nLfBlocksA(I)/2 + nTuples(R)/2]


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Composite Predicates - Conjunction without Disjunction

May consider following approaches:- If one attribute has index or is ordered, can use one of above selection strategies. Can then check each retrieved record.

- For equality on two or more attributes, with composite index (or hash key) on combined attributes, can search index directly.

- With secondary indexes on one or more attributes (involved only in equality conditions in predicate), could use record pointers if exist.


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Composite Predicates - Selections with Disjunction

If one term contains an (OR), and term requires linear search, entire selection requires linear search.

Only if index or sort order exists on every term can selection be optimized by retrieving records that satisfy each condition and applying union operator.

Again, record pointers can be used if they exist.


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Join Operation

Main strategies for implementing join:

– Block Nested Loop Join.– Indexed Nested Loop Join.– Sort-Merge Join.– Hash Join.


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Estimating Cardinality of Join

Cardinality of Cartesian product is:

nTuples(R) * nTuples(S)

More difficult to estimate cardinality of any join as depends on distribution of values.

Worst case, cannot be any greater than this value.


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Estimating Cardinality of Join

If assume uniform distribution, can estimate for Equijoins with a predicate (R.A = S.B) as follows:– If A is key of R: nTuples(T) nTuples(S)– If B is key of S: nTuples(T) nTuples(R)

Otherwise, could estimate cardinality of join as:

nTuples(T) = SCA(R)*nTuples(S) or

nTuples(T) = SCB(S)*nTuples(R)


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Block Nested Loop Join

Simplest join algorithm is nested loop that joins two relations together a tuple at a time.

Outer loop iterates over each tuple in R, and inner loop iterates over each tuple in S.

As basic unit of reading/writing is a disk block, better to have two extra loops that process blocks.

Estimated cost of this approach is:

nBlocks(R) + (nBlocks(R) * nBlocks(S))


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Block Nested Loop Join

Could read as many blocks as possible of smaller relation, R say, into database buffer, saving one block for inner relation and one for result.

New cost estimate becomes:

nBlocks(R) + [nBlocks(S)*(nBlocks(R)/(nBuffer-2))]

If can read all blocks of R into the buffer, this reduces to:

nBlocks(R) + nBlocks(S)


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Indexed Nested Loop Join

If have index (or hash function) on join attributes of inner relation, can use index lookup.

For each tuple in R, use index to retrieve matching tuples of S.

Cost of scanning R is nBlocks(R), as before. Cost of retrieving matching tuples in S depends

on type of index and number of matching tuples. If join attribute A in S is PK, cost estimate is:

nBlocks(R) + nTuples(R)*(nlevelsA(I) + 1)


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Sort-Merge Join

For Equijoins, most efficient join is when both relations are sorted on join attributes.

Can look for qualifying tuples merging relations. May need to sort relations first. Now tuples with same join value are in order. If assume join is *:* and each set of tuples with

same join value can be held in database buffer at same time, then each block of each relation need only be read once.


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Sort-Merge Join

Cost estimate for the sort-merge join is:

nBlocks(R) + nBlocks(S)

If a relation has to be sorted, R say, add:

nBlocks(R)*[log2(nBlocks(R)]


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Hash Join

For Natural or Equijoin, hash join may be used. Idea is to partition relations according to some

hash function that provides uniformity and randomness.

Each equivalent partition should hold same value for join attributes, although it may hold more than one value.

Cost estimate of hash join as:

3(nBlocks(R) + nBlocks(S))


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Projection Operation

To implement projection need to:– remove attributes that are not required;– eliminate any duplicate tuples produced from

previous step. Only required if projection attributes do not include a key.

Two main approaches to eliminating duplicates:

– sorting;

– hashing.


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Estimating Cardinality of Projection

When projection contains key, cardinality is:

nTuples(S) = nTuples(R)

If projection consists of a single non-key attribute, estimate is:

nTuples(S) = SCA(R)

Otherwise, could estimate cardinality as:

nTuples(S) min(nTuples(R), im

=1(nDistinctai(R)))


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Duplicate Elimination using Sorting

Sort tuples of reduced relation using all remaining attributes as sort key.

Duplicates will now be adjacent and can be removed easily.

Estimated cost of sorting is:

nBlocks(R)*[log2(nBlocks(R))].

Combined cost is:

nBlocks(R) + nBlocks(R)*[log2(nBlocks(R))]


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Duplicate Elimination using Hashing

Two phases: partitioning and duplicate elimination.

In partitioning phase, for each tuple in R, remove unwanted attributes and apply hash function to combination of remaining attributes, and write reduced tuple to hashed value.

Two tuples that belong to different partitions are guaranteed not to be duplicates.

Estimated cost is: nBlocks(R) + nB


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Set Operations

Can be implemented by sorting both relations on same attributes, and scanning through each of sorted relations once to obtain desired result.

Could use sort-merge join as basis. Estimated cost in all cases is:

nBlocks(R) + nBlocks(S) + nBlocks(R)*[log2(nBlocks(R))] + nBlocks(S)*[log2(nBlocks(S))]

Could also use hashing algorithm.


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Estimating Cardinality of Set Operations

As duplicates are eliminated when performing Union, difficult to estimate cardinality, but can give an upper and lower bound as:

max(nTuples(R), nTuples(S)) nTuples(T) nTuples(R) + nTuples(S)

For Set Difference, can also give upper and lower bound:

0 nTuples(T) nTuples(R)


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Aggregate Operations

SELECT AVG(salary)

FROM Staff;

To implement query, could scan entire Staff relation and maintain running count of number of tuples read and sum of all salaries.

Easy to compute average from these two running counts.


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Aggregate Operations

SELECT AVG(salary)

FROM Staff

GROUP BY branchNo;

For grouping queries, can use sorting or hashing algorithms similar to duplicate elimination.

Can estimate cardinality of result using estimates derived earlier for selection.


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Enumeration of Alternative Strategies

Fundamental to efficiency of QO is the search space of possible execution strategies and the enumeration algorithm used to search this space.

Query with 2 joins gives 12 join orderings:R (S T) R (T S) (S T) R (T S) R

S (R T) S (T R) (R T) S (T R) S

T (R S) T (S R) (R S) T (S R) T

With n relations, (2(n – 1))!/(n – 1)! orderings. If n = 4 this is 120; if n = 10 this is > 176 billion. Compounded by different selection/join methods.


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Pipelining

Materialization - output of one operation is stored in temporary relation for processing by next.

Could also pipeline results of one operation to another without creating temporary relation.

Known as pipelining or on-the-fly processing. Pipelining can save on cost of creating temporary

relations and reading results back in again. Generally, pipeline is implemented as separate

process or thread.


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Types of Trees


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Pipelining

With linear trees, relation on one side of each operator is always a base relation.

However, as need to examine entire inner relation for each tuple of outer relation, inner relations must always be materialized.

This makes left-deep trees appealing as inner relations are always base relations.

Reduces search space for optimum strategy, and allows QO to use dynamic processing.

Not all execution strategies are considered.


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Physical Operators & Strategies

Term physical operator refers to specific algorithm that implements a logical operation, such as selection or join.

For example, can use sort-merge join to implement the join operation.

Replacing logical operations in a R.A.T. with physical operators produces an execution strategy (or query evaluation plan or access plan).


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Physical Operators & Strategies


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Reducing the Search Space Restriction 1: Unary operations processed on-

the-fly: selections processed as relations are accessed for first time; projections processed as results of other operations are generated.

Restriction 2: Cartesian products are never formed unless query itself specifies one.

Restriction 3: Inner operand of each join is a base relation, never an intermediate result. This uses fact that with left-deep trees inner operand is a base relation and so already materialized.

Restriction 3 excludes many alternative strategies but significantly reduces number to be considered.


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Dynamic Programming

Enumeration of left-deep trees using dynamic programming first proposed for System R QO.

Algorithm based on assumption that the cost model satisfies principle of optimality.

Thus, to obtain optimal strategy for query with n joins, only need to consider optimal strategies for subexpressions with (n – 1) joins and extend those strategies with an additional join. Remaining suboptimal strategies can be discarded.


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Dynamic Programming

To ensure some potentially useful strategies are not discarded algorithm retains strategies with interesting orders: an intermediate result has an interesting order if it is sorted by a final ORDER BY attribute, GROUP BY attribute, or any attributes that participate in subsequent joins.


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Dynamic Programming

SELECT p.propertyNo, p.streetFROM Client c, Viewing v, PropertyForRent pWHERE c.maxRent < 500 AND

c.clientNo = v.clientNo AND v.propertyNo = p.propertyNo;

Attributes c.clientNo, v.clientNo, v.propertyNo, and p.propertyNo are interesting.

If any intermediate result is sorted on any of these attributes, then corresponding partial strategy must be included in search.


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Dynamic Programming

Algorithm proceeds from the bottom up and constructs all alternative join trees that satisfy the restrictions above, as follows:

Pass 1: Enumerate the strategies for each base relation using a linear search and all available indexes on the relation. These partial strategies are partitioned into equivalence classes based on any interesting orders. An additional equivalence class is created for the partial strategies with no interesting order.


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Dynamic Programming

For each equivalence class, strategy with lowest cost is retained for consideration in next pass.

Do not retain equivalence class with no interesting order if its lowest cost strategy is not lower than all other strategies.

For a given relation R, any selections involving only attributes of R are processed on-the-fly. Similarly, any attributes of R that are not part of the SELECT clause and do not contribute to any subsequent join can be projected out at this stage (restriction 1 above).


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Dynamic Programming

Pass 2: Generate all 2-relation strategies by considering each strategy retained after Pass 1 as outer relation, discarding any Cartesian products generated (restriction 2 above). Again, any on-the-fly processing is performed and lowest cost strategy in each equivalence class is retained.

Pass n: Generate all n-relation strategies by considering each strategy retained after Pass (n – 1) as outer relation, discarding any Cartesian products generated. After pruning, now have lowest overall strategy for processing the query.


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Dynamic Programming

Although algorithm is still exponential, there are query forms for which it only generates O(n3) strategies, so for n = 10 the number is 1,000, which is significantly better than the 176 billion different join orders noted earlier.


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Semantic Query Optimization

Based on constraints specified on the database schema to reduce the search space.

For example, a constraint states that staff cannot supervise more than 100 properties, so any query searching for staff who supervise more than 100 properties will produce zero rows. Now consider:CREATE ASSERTION ManagerSalary

CHECK (salary > 20000 AND position = ‘Manager’)SELECT s.staffNo, fName, lName, propertyNoFROM Staff s, PropertyForRent pWHERE s.staffNo = p.staffNo AND

position = ‘Manager’;© Pearson Education Limited 1995, 2005

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Semantic Query Optimization

Can rewrite this query as:

SELECT s.staffNo, fName, lName, propertyNoFROM Staff s, PropertyForRent pWHERE s.staffNo = p.staffNo AND

salary > 20000 AND position = ‘Manager’;

Additional predicate may be very useful if only

index for Staff is a B+-tree on the salary attribute. However, additional predicate would complicate

query if no such index existed.


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Query Optimization in Oracle

Oracle supports two approaches to query optimization: rule-based and cost-based.

Rule-based 15 rules, ranked in order of efficiency. Particular

access path for a table only chosen if statement contains a predicate or other construct that makes that access path available.

Score assigned to each execution strategy using these rankings and strategy with best (lowest) score selected.


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QO in Oracle – Rule-Based When 2 strategies have same score, tie-break

resolved by making decision based on order in which tables occur in the SQL statement.


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QO in Oracle – Rule-based: Example

SELECT propertyNoFROM PropertyForRentWHERE rooms > 7 AND city = ‘London’

Single-column access path using index on city from WHERE condition (city = ‘London’). Rank 9.

Unbounded range scan using index on rooms from WHERE condition (rooms > 7). Rank 11.

Full table scan - rank 15. Although there is index on propertyNo, column does not appear

in WHERE clause and so is not considered by optimizer. Based on these paths, rule-based optimizer will choose to use

index based on city column.


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QO in Oracle – Cost-Based

To improve QO, Oracle introduced cost-based optimizer in Oracle 7, which selects strategy that requires minimal resource use necessary to process all rows accessed by query (avoiding above tie-break anomaly).

User can select whether minimal resource usage is based on throughput or based on response time, by setting the OPTIMIZER_MODE initialization parameter.

Cost-based optimizer also takes into consideration hints that the user may provide.


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QO in Oracle – Statistics

Cost-based optimizer depends on statistics for all tables, clusters, and indexes accessed by query.

Users’ responsibility to generate these statistics and keep them current.

Package DBMS_STATS can be used to generate and manage statistics.

Whenever possible, Oracle uses a parallel method to gather statistics, although index statistics are collected serially. EXECUTE

DBMS_STATS.GATHER_SCHEMA_STATS(‘Manager’);


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QO in Oracle – Histograms

Previously made assumption that data values within columns of a table are uniformly distributed.

Histogram of values and their relative frequencies gives optimizer improved selectivity estimates in presence of non-uniform distribution.


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(a) uniform distribution of rooms; (b) actual non-uniform distribution.

(a) can be stored compactly as low value (1) and high value (10), and as total count of all frequencies (in this case, 100).


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Histogram is data structure that can improve estimates of number of tuples in result.

Two types of histogram:– width-balanced histogram, which divides data into a

fixed number of equal-width ranges (called buckets) each containing count of number of values falling within that bucket;

– height-balanced histogram, which places approximately same number of values in each bucket so that end points of each bucket are determined by how many values are in that bucket.


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(a) width-balanced for rooms with 5 buckets. Each bucket of equal width with 2 values (1-2, 3-4, etc.)

(b) height-balanced – height of each column is 20 (100/5).


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QO in Oracle – Viewing Execution Plan


1 Chapter 21 Query Processing Transparencies ownerNoclient © Pearson Education Limited 1995, 2005.

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