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1 Query Processing + Optimization: Outline Operator Evaluation Strategies Query processing in general Selection Join Query Optimization Heuristic query optimization Cost-based query optimization Query Tuning

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  • 1

    Query Processing + Optimization: Outline Operator Evaluation Strategies

    Query processing in generalSelectionJoin

    Query OptimizationHeuristic query optimizationCost-based query optimization

    Query Tuning

  • 2

    Query Processing + Optimization Operator Evaluation Strategies


    Query Optimization Query Tuning

  • 3

    Architectural Context


    DDL Statements

    Privileged Commands




    Application Programmers

    Host LanguageCompiler




    Query ExecutionPlan


    Transaction andData Manager

    Data Dictionary Data Files


    DBA Staff Casual Users ParametricUsers

  • 4

    Evaluation of SQL Statement The query is evaluated in a different order.

    The tables in the from clause are combined using Cartesian products.The where predicate is then applied.The resulting tuples are grouped according to the group by clause.The having predicate is applied to each group, possibly eliminating some groups.The aggregates are applied to each remaining group.The select clause is performed last.

  • 5

    Overview of Query Processing

    Scanning, Parsing, andSemantic Analysis

    Query Optimization

    Query Code Generator

    Runtime DatabaseProcessor

    Intermediate form of query

    Execution Plan

    Code to execute the query

    Result of query

    Query in high-level language1. Parsing and translation2. Optimization3. Evaluation4. Execution

  • 6

    Selection Queries Primary key, point

    FilmID = 2 (Film) Point

    Title = Terminator (Film) Range

    1 < RentalPrice < 4 (Film) Conjunction

    Type = M Distributor = MGM (Film) Disjunction

    PubDate < 2004 Distributor = MGM (Film)

  • 7

    Selection Strategies Linear search

    Expensive, but always applicable. Binary search

    Applicable only when the file is appropriately ordered. Hash index search

    Single record retrieval; does not work for range queries.Retrieval of multiple records.

    Clustering index searchMultiple records for each index item.Implemented with single pointer to block with first associated record.

    Secondary index searchImplemented with dense pointers, each to a single record.

  • 8

    Selection Strategies for Conjunctive Queries Use any available indices for attributes involved in

    simple conditions.If several are available, use the most selective index. Then check each record with respect to the remaining conditions.

    Attempt to use composite indices. This can be very efficient.

    Do intersection of record pointers.If several indices with record pointers are applicable to the selection predicate, retrieve and intersect the pointers. Then retrieve (and check) the qualifying records.

    Disjunctive queries provide little opportunity for smart processing.

  • 9

    Joins Join Strategies

    Nested loop joinIndex-based joinSort-merge joinHash join

    Strategies work on a per block (not per record) basis.Need to estimate #I/Os (block retrievals)

    Relation sizes and join selectivities impact join cost.Query selectivity = #tuples in result / #candidates

    More selective means smaller selectivity valueFor join, #candidates is the size of Cartesian product

  • 10

    Nested Loop Join and Index-Based Join Nested loop join

    Exhaustive comparison (i.e., brute force approach)The ordering (outer/inner) of files and allocation of buffer space is important.

    Index-based joinRequires (at least) one index on a join attribute.At times, a temporary index is created for the purpose of a join.The ordering (outer/inner) of files is important.

  • 11


    Nested Loop Basically, for each block of the outer table (r), scan the

    entire inner table (s). Requires quadratic time, O(n2)Improved when buffer is used.


    Main MemoryBuffer



    spill when full




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    Example of Nested-Loop JoinCustomer nC.CustomerID = CO.EmpId CheckedOut

    ParametersrCheckedOut = 40.000 rCustomer = 200bCheckedOut = 2.000 bCustomer = 10nB = 6 (size of main memory buffer)

    Algorithm:repeat: read (nB - 2) blocks from outer relation

    repeat: read 1 block from inner relationcompare tuples

    Cost: bouter + ( bouter/ (nB -2) ) binner CheckedOut as outer: 2.000 + 2.000/4 10 = 7.000 Customer as outer: 10 + 10/4 2.000 = 6.010

  • 13


    Index-based Join Requires (at least) one index on a join attribute

    A temporary index can be created




    spill when full


    index on r

    if joins

    Main Memory

    for each record ofs, query in indexOuter

  • 14

    Example of Index-Based JoinCustomer nC.CustomerID = CO.EmpId CheckedOut

    Cost: bouter + router cost use of index Assume that the video store has 10 employees.

    There are 10 distinct EmpIDs in CheckedOut. Assume 1-level index on CustomerID of Customer. Iterate through all 40.000 tuples in CheckedOut (outer

    rel.)2.000 disk reads (bCheckedOut) to scan CheckedOutFor each CheckedOut tuple, search for matching Customertuples using index.

    0 disk reads for index (in main memory) + 1 disk read for actual data block

    Cost: 2.000 + 40.000 (0 + 1) = 42.000

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    Sort-Merge Join







    Sort each relation using multiway merge-sort Perform merge-join


    Main Memory





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    External or Disk-based Sorting Relation on disk often too large to fit into memory Sort in pieces, called runs


    Main MemoryBuffer (N blocks)

    Scan Output when sorted


    Unsorted relation(M blocks)



    rM div N


    Runs(M div N runs each of

    size N blocks, and maybe one last run of < N leftover blocks)

  • 17

    Output when merged

    External or Disk-based Sorting, Cont. Runs are now repeatedly merged One memory buffer used to collect output

    Main MemoryBuffer (N blocks)


    sorted runs(N blocks each)



    mlast - 1


    Merged runs((M div N) div N-1) runs each of

    size N*N-1 blocks, and maybe one last run of < N*N-1 leftover blocks)



    rM div N


    rN - 1

    Main MemoryBuffer (N blocks)


  • 18

    External Sorting (Multiway Merge Sort)

    32 5 1712 1 14 8 3 3130232629 2 2511 6 7 151628 4 9 221821102013272419

    5 121732 1 3 8 14 23263031 2 112529 6 7 1516 4 9 2228 10182021 13192427

    1 3 5 8 1214172326303132 2 4 6 7 9 11151622252829 1013181920202427

    1 2 3 4 5 6 7 8 9 1011121314151617181920212223242526272829303132

    Orginal rel.

    Initial runs

    First merge

    Second merge

    Buffer size is nB = 4 (N)

    Cost: 2 brelation + 2 brelation lognB - 1 (brelation/nB) 2 32 + 2 32 log3(32/4) = 192

  • 19

    Example of Sort-Merge Join Cost to sort CheckedOut (bCheckedOut= 10)

    CostSort ChecedOut = 2 2.000 + 2 2.000 log5(2.000/6) = 20.000 Cost to sort Customer relation (bCustomer= 10)

    CostSort Customer= 2 10 + 2 10 log5(10/6) = 40 Cost for merge join

    Cost to scan sorted Customer + cost to scan sorted CheckedOutCostmerge join= 10 + 2.000 = 2.010

    Costsort-merge join = CostSort Customer + CostSort ChecedOut + Costmerge join Costsort-merge join= 20.000 + 40 + 2.010 = 22.050

  • 20

    Hash Join Hash each relation on the join attributes Join corresponding buckets from each relation














    r2 n s2

    r1 n s1

    rn n sn

    Buckets from r Buckets from s

    Hash r (same for s) Join corresponding r and s buckets

  • 21

    Partitioning Phase Partitioning phase: r divided into nh partitions. The

    number of buffer blocks is nB. One block used for reading r. (nh = nB -1)

    Similar with relation sI/O cost: 2 (br + bs)

    nB main memory buffers DiskDisk

    Relationr or s





    hnB -1




    nB -1

    . . .

  • 22

    Joining Phase Joining (or probing) phase: nh iterations where rin si.

    Load ri into memory and build an in-memory hash index on it using the join attribute. (h2 needed, ri called build input)Load si, for each tuple in it, join it with ri using h2. (si called probe input)I/O cost: br + bs + 4 nh (each partition may have a partially filled block)

    One write and one read for each partially filled block

    Partitionsof r & s

    Input bufferfor si

    In-memory hash table for partition ri (< nB -1 blocks)

    nB main memory buffersDisk

    Output buffer


    Join ResultHashfunc.h2


  • 23

    Hash Join Cost CostTotal = Cost Partitioning + CostJoining

    = 3 (br + bs) + 2 nh Cost = 3 (2000 + 10) + 2 5 = 6040 Any problem not considered?

    What if nh > nB -1? I.e., more partitions than available buffer blocks!How to solve it?

    nB main memory buffersDisk

    Relationr or s





    hnB -1

    . . . nB

  • 24

    Recursive Partitioning Required if number of partitions nh is greater than

    number of available buffer blocks nB -1.instead of partitioning nh ways, use nB 1 partitions for sFurther partition the nB 1 partitions using a different hash functionUse same partitioning method on rRarely required: e.g., recursive partitioning not needed for relations of 1GB or less with memory size of 2MB, with block size of 4KB.

    Costhjrp = 2 (br + bs) lognB - 1 (bs) -1 + br + bs Costhjrp = 2 (2000 + 10) log5 (10) -1 + 2000 + 10

    = 6030

  • 25

    Cost and Applicability of Join Strategies Nested-loop join

    Brute-forceCan handle all types of joins (=, )

    Index-based joinRequires minimum one index on join attributes

    Sort-merge joinRequires that the files are sorted on the join attributes.Sorting can be done for the purpose of the join.A variation is also applicable when secondary indices are available instead.

    Hash joinRequires good hashing functions to be available.Performance best if smallest relation fits in memory. ?

  • 26

    Query Processing + Optimization Operator Evaluation Strategies Query Optimization

    Heuristic Query OptimizationCost-based Query Optimization

    Query Tuning

  • 27

    Query Optimization Aim: Transform query into faster, equivalent query


    Heuristic (logical) optimizationQuery tree (relational algebra) optimizationQuery graph optimization

    Cost-based (physical) optimization

    equivalent query 1

    equivalent query 2

    equivalent query n


  • 28

    Query Tree Optimization Example What are the names of customers living on Elm Street

    who have checked out Terminator? SQL query:

    SELECT NameFROM Customer CU, CheckedOut CH, Film FWHERE Title = Terminator AND F.FilmId = CH.FilmIDAND CU.CustomerID = CH.CustomerID AND CU.Street = Elm

    Canonical query tree

    CU CH


    NameTitle = Terminator F.FilmId = CH.FilmID CU.CustomerID = CH.CustomerID CU.Street = Elm

    Note the use of Cartesian product!

  • 29

    Apply Selections Early





    Street = Elm

    CU.CustomerID = CH.CustomerID Title = Terminator

    F.FilmId = CH.FilmID

  • 30

    Apply More Restrictive Selections Early





    Title = Terminator

    F.FilmId = CH.FilmID Street = Elm

    CU.CustomerID = CH.CustomerID

  • 31

    Form Joins


    CH CU

    n F.FilmId = CH.FilmID

    n CU.CustomerID = CH.CustomerID

    Title = Terminator

    Street = Elm


  • 32

    Apply Projections Early




    Title = Terminator

    n F.FilmId = CH.FilmID

    Street = Elm

    n CU.CustomerID = CH.CustomerID

    FilmID FilmID, CustomerID

    Name, CustomerID

  • 33

    Some Transformation Rules Cascade of : c1 c2 ... cn(R) = c1(c2(...(cn(R))...)) Commutativity of : c1(c2(R)) = c2(c1(R)) Commuting with : L(c (R)) = c(L(R))

    Only if c involves solely attributes in L.

    Commuting with n : c (R n S) = c (R) n SOnly if c involves solely attributes in R.

    Commuting with set operations: c(R S) = c(R) c(S)Where is one of , , or -.

    Commutativity of , , and n: R S = S RWhere is one of , and n.

    Associativity of n, , : (R S) T = R (S T)

  • 34

    Transformation Algorithm Outline Transform a query represented in relational algebra to

    an equivalent one (generates the same result.)

    Step 1: Decompose operations. Step 2: Move as far down the query tree as possible. Step 3: Rearrange leaf nodes to apply the most

    restrictive operations first. Step 4: Form joins from and subsequent operations. Step 5: Decompose and move down the query tree as

    far as possible. Step 6: Identify candidates for combined operations.

  • 35

    Heuristic Query Optimization Summary Heuristic optimization transforms the query-tree by

    using a set of rules (Heuristics) that typically (but not in all cases) improve execution performance.

    Perform selection early (reduces the number of tuples)Perform projection early (reduces the number of attributes)Perform most restrictive selection and join operations (i.e. with smallest result size) before other similar operations.

    Generate initial query tree from SQL statement. Transform query tree into more efficient query tree, via

    a series of tree modifications, each of which hopefully reduces the execution time.

    A single query tree is involved.

  • 36

    Cost-Based Optimization Use transformations to generate multiple candidate

    query trees from the canonical query tree. Statistics on the inputs to each operator are needed.

    Statistics on leaf relations are stored in the system catalog.Statistics on intermediate relations must be estimated; most important is the relations' cardinalities.

    Cost formulas estimate the cost of executing each operation in each candidate query tree.

    Parameterized by statistics of the input relations.Also dependent on the specific algorithm used by the operator.Cost can be CPU time, I/O time, communication time, main memory usage, or a combination.

    The candidate query tree with the least total cost is selected for execution.

  • 37

    Relevant Statistics Per relation

    Tuple sizeNumber of tuples (records): rLoad factor (fill factor), percentage of space used in each blockBlocking factor (number of records per block): bfrRelation size in blocks: bRelation organizationNumber of overflow blocks

  • 38

    Relevant Statistics, cont. Per attribute

    Attribute size and typeNumber of distinct values for attribute A: dAProbability distribution over the valuesRepresentation, e.g., compressedSelection cardinality specifies the average size of A = a(R) for an arbitrary value a. (sA)

    Could be maintained for the average attribute value, or on a per-value basis, as a histogram.

  • 39

    Relevant Statistics, cont. Per Index

    Base relationIndexed attribute(s)Organization, e.g., B+-Tree, Hash, ISAMClustering index?On key attribute(s)?Sparse or dense?Number of levels (if appropriate)Number of first-level index blocks: b1

    GeneralAvailable main memory blocks: N

  • 40

    Cost Estimation Example




    Title = Terminator

    F.FilmId = R.FilmID

    Street = Elm

    CU.CustomerID = R.CustomerID

    FilmID FilmID, CustomerID

    FilmID, CustomerID, Name


    2 3


  • 41

    Operation 1: followed by a Statistics

    Relation statistics: rFilm= 5,000 bFilm= 50Attribute statistics: sTitle= 1Index statistics: Secondary Hash Index on Title.

    Result relation size: 1 tuple.

    Operation: Use index with Terminator, then project on FilmID. Leave result in main memory (1 block).

    Cost (in disk accesses): C1 = 1 +1 = 2

  • 42

    Operation 2: n followed by a Statistics

    Relation statistics: rCheckedOut= 40,000 bCheckedOut= 2,000Attribute statistics: sFilmID= 8Index statistics: Secondary B+-Tree Index for CheckedOut on FilmID with 2 levels.

    Result relation size: 8 tuples.

    Operation: Index join using B+-Tree, then project on CustomerID. Leave result in main memory (one block).

    Cost: C2 = 1 +1 + 8 = 10

  • 43

    Operation 3: followed by a Statistics

    Relation statistics: rCustomer= 200 bCustomer= 10Attribute statistics:sStreet= 10

    Result relation size: 10 tuples.

    Operation: Linear search of Customer. Leave result in main memory (one block).

    Cost: C3 = 10

  • 44

    Operation 4: n followed by a Operation: Main memory join on relations in main


    Cost: C4 = 0

    Total cost: 220101024



  • 45

    Comparison Heuristic query optimization

    Sequence of single query plansEach plan is (presumably) more efficient than the previous.Search is linear.

    Cost-based query optimizationMany query plans generated.The cost of each is estimated, with the most efficient chosen.Search is multi-dimensional, usually using dynamic programming. Still can be very expensive.

    Hybrid waySystems may use heuristics to reduce the number of choices that must be made in a cost-based fashion.

  • 46

    Query Processing + Optimization Operator Evaluation Strategies Query Optimization Query Tuning

  • 47

    Query Tuning Query optimization is a very complex task.

    Combinatorial explosion.The task is to find one good query evaluation plan, not the best one.

    No optimizer optimizes all queries adequately. There is a need for query tuning.

    All optimizers differ in their ability to optimize queries, making it difficult to prescribe principles.

    Having to tune queries is a fact of life.Query tuning has a localized effect and is thus relatively attractive.It is a time-consuming and specialized task.It makes the queries harder to understand.However, it is often a necessity.This is not likely to change any time soon.

  • 48

    Query Tuning Issues Need too many disk accesses (eg. Scan for a point

    query)? Need unnecessary computation?

    Redundant DISTINTSELECT DISTINCT cpr#FROM EmployeeWHERE dept = computer

    Relevant indexes are not used? (Next slide) Unnecessary nested subqueries?

  • 49

    Join on Clustering Index, and IntegerSELECT Employee.cpr#FROM Employee, StudentWHERE Employee.name = Student.name-->SELECT Employee.cpr#FROM Employee, StudentWHERE Employee.cpr# = Student.cpr#

  • 50

    Nested Queries Nested block is optimized independently, with the outer tuple

    considered as providing a selection condition. Outer block is optimized with the cost of calling nested block

    computation taken into account. Implicit ordering of these blocks means that some good

    strategies are not considered. The non-nested version of the query is typically optimized better.


    (SELECT *FROM Reserves RWHERE R.bid=103 AND R.sid=S.sid)

    Nested block to optimize:SELECT *FROM Reserves RWHERE R.bid=103

    AND S.sid= outer valueEquivalent non-nested query:SELECT S.snameFROM Sailors S, Reserves RWHERE S.sid=R.sid

    AND R.bid=103

  • 51

    Unnesting Nested Queries Uncorrelated sub-queries with aggregates.

    Most systems would compute the average only once.SELECT ssnFROM empWHERE salary > (SELECT AVG(salary) FROM emp)

    Uncorrelated sub-queries without aggregates.SELECT ssnFROM empWHERE dept IN (SELECT dept FROM techdept)

    Some systems may not use emp's index on dept, so a transformation is desirable.SELECT ssnFROM emp, techdeptWHERE emp.dept = techdept.dept

    When is this acceptable?

  • 52

    Unnesting Nested Queries, cont. Watch out for duplicates! Consider a query and its

    rewritten counterpart.SELECT AVG(salary)FROM empWHERE manager IN (SELECT manager FROM techdept)

    Unnested version, with problems: (whats the problem?)SELECT AVG(salary)FROM emp, techdeptWHERE emp.manager = techdept.manager

    This query may yield wrong results! A solution:SELECT DISTINCT (manager) INTO tempFROM techdept

    SELECT AVG(salary)FROM emp, tempWHERE emp. manager = temp. manager

  • 53

    Summary Query processing & optimization is the heart of a

    relational DBMS. Heuristic optimization is more efficient to generate, but

    may not yield the optimal query evaluation plan. Cost-based optimization relies on statistics gathered on

    the relations (the default in most DBMSs). Until query optimization is perfected, query tuning will

    be a fact of life.

    Query Processing + Optimization: OutlineQuery Processing + OptimizationArchitectural ContextEvaluation of SQL StatementOverview of Query ProcessingSelection QueriesSelection StrategiesSelection Strategies for Conjunctive QueriesJoinsNested Loop Join and Index-Based JoinNested LoopExample of Nested-Loop JoinIndex-based JoinExample of Index-Based JoinSort-Merge JoinExternal or Disk-based SortingExternal or Disk-based Sorting, Cont.External Sorting (Multiway Merge Sort)Example of Sort-Merge JoinHash JoinPartitioning PhaseJoining PhaseHash Join CostRecursive PartitioningCost and Applicability of Join StrategiesQuery Processing + OptimizationQuery OptimizationQuery Tree Optimization ExampleApply Selections EarlyApply More Restrictive Selections EarlyForm JoinsApply Projections EarlySome Transformation RulesTransformation Algorithm OutlineHeuristic Query Optimization SummaryCost-Based OptimizationRelevant StatisticsRelevant Statistics, cont.Relevant Statistics, cont.Cost Estimation ExampleOperation 1: s followed by a pOperation 2: n followed by a pOperation 3: s followed by a p Operation 4: n followed by a pComparisonQuery Processing + OptimizationQuery TuningQuery Tuning IssuesJoin on Clustering Index, and IntegerNested QueriesUnnesting Nested QueriesUnnesting Nested Queries, cont.Summary