Chapter 13: Query Optimization
■ Introduction
■ Transformation of Relational Expressions
■ Catalog Information for Cost Estimation
■ Statistical Information for Cost Estimation
■ Cost-based optimization
■ Dynamic Programming for Choosing Evaluation Plans
■ Materialized views
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Query Optimization in a Nutshell
■ Cost-based versus rule-based query optimization
■ Cost-based: look at all legal/correct execution plans, pick cheapest one
■ Problem: how to correctly estimate cost of a plan
■ Cost depends in input size, data properties, available space, CPU, disk
■ Cost of higher-level operators depends on output of lower levels
■ Rule-based: transform into a good plan using rules
■ Challenge: there are exceptions to most rules
■ Example rules: do selection before join when possible, use an index for a selection if possible, use an index for a join if possible
■ State of the art: combine rules and cost estimation – use rules to restrict to “reasonable” execution plans, and choose the cheapest of those
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Query Optimization in a Nutshell
■ Need to be able to enumerate all legal (or reasonable) plans
■ This is done via equivalence rules that tell us when operators can be
switched and thus, which other plans can be generated from initial one
■ For example, we can push selection from after to before the join, assuming attributes are from only one table
■ Pseudocode for combined cost/rule-based optimizer
● For all legal plans p for the query Q
! If plan p satisfies a restricted set of rules
– Estimate cost(p), and our confidence interval for the estimate ● Choose a plan with minimum or close to minimum cost
■ Restricted set of rules to prune search space
■ Ideally, want to directly enumerate all legal plans satisfying set of rules
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Introduction
■ Alternativewaysofevaluatingagivenquery ● Equivalent expressions
● Different algorithms for each operation
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Introduction (Cont.)
■ Anevaluationplandefinesexactlywhatalgorithmisusedforeach
operation, and how the execution of the operations is coordinated.
■ Find out how to view query execution plans on your favorite database
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Introduction (Cont.)
■ Cost difference between evaluation plans for a query can be enormous
● E.g. seconds vs. days in some cases
■ Steps in cost-based query optimization
1. Generate logically equivalent expressions using equivalence rules
2. Annotate resultant expressions to get alternative query plans
3. Choose the cheapest plan based on estimated cost
■ Estimation of plan cost based on:
● Statistical information about relations. Examples:
! numberoftuples,numberofdistinctvaluesforanattribute
● Statistics estimation for intermediate results ! tocomputecostofcomplexexpressions
● Cost formulae for algorithms, computed using statistics
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Transformation of Relational Expressions
■ Two relational algebra expressions are said to be equivalent if the two expressions generate the same set of tuples on every legal database instance
● Note: order of tuples is irrelevant
● we don’t care if they generate different results on databases
that violate integrity constraints
■ In SQL, inputs and outputs are multisets of tuples
● Two expressions in the multiset version of the relational algebra are said to be equivalent if the two expressions generate the same multiset of tuples on every legal database instance.
■ Anequivalencerulesaysthatexpressionsoftwoformsare equivalent
● Can replace expression of first form by second, or vice versa
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Equivalence Rules
1. Conjunctiveselectionoperationscanbedeconstructedintoa sequence of individual selections.
σθ1∧θ2 (E)=σθ1 (σθ2 (E))
2. Selectionoperationsarecommutative.
σθ1 (σθ2 (E))=σθ2 (σθ1 (E))
3. Onlythelastinasequenceofprojectionoperationsis needed, the others can be omitted.
ΠL (ΠL (…(ΠLn (E))…)) = ΠL (E) 121
4. SelectionscanbecombinedwithCartesianproductsand theta joins.
a. σθ(E1XE2)= E1 θE2 b. σθ1(E1 θ2 E2) = E1
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θ1∧ θ2 E2
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Equivalence Rules (Cont.)
5. Theta-join operations (and natural joins) are commutative.
E1 θ E2 = E2 θ E1
6. (a) Natural join operations are associative: (E1 E2) E3 = E1 (E2 E3)
(b) Theta joins are associative in the following manner:
(E1 θ1 E2) θ2∧ θ3 E3 = E1 θ1∧ θ3 (E2 θ2 E3)
where θ2 involves attributes from only E2 and E3.
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Pictorial Depiction of Equivalence Rules
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Equivalence Rules (Cont.)
7. Theselectionoperationdistributesoverthethetajoinoperation under the following two conditions:
(a) When all the attributes in θ0 involve only the attributes of one
of the expressions (E1) being joined.
σθ0(E1 θ E2) = (σθ0(E1)) θ E2
(b) When θ 1 involves only the attributes of E1 and θ2 involves
only the attributes of E2.
σθ1∧θ2 (E1 θ E2) = (σθ1(E1)) θ (σθ2 (E2))
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Equivalence Rules (Cont.)
8. Theprojectionoperationdistributesoverthethetajoinoperation as follows:
(a) if θ involves only attributes from L1 ∪ L2:
(b) Consider a join E1 θ E2.
∏ (E E)=(∏(E)) (∏(E)) L∪L1θ2 L1θL2
1212
● Let L1 and L2 be sets of attributes from E1 and E2, respectively.
● Let L3 be attributes of E1 that are involved in join condition θ, but are not in L1 ∪ L2, and
● let L4 be attributes of E2 that are involved in join condition θ, but are not in L1 ∪ L2.
∏L∪L(E1 θE2)=∏L∪L((∏L∪L(E1)) θ(∏L∪L(E2))) 12121324
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Equivalence Rules (Cont.)
9. The set operations union and intersection are commutative
E1∪E2 =E2∪E1
E1∩E2 =E2∩E1
■ (set difference is not commutative).
10. Set union and intersection are associative. (E1 ∪E2)∪E3 =E1 ∪(E2 ∪E3)
(E1 ∩E2)∩E3 =E1 ∩(E2 ∩E3)
11. The selection operation distributes over ∪, ∩ and –.
σθ (E1 – E2) = σθ (E1) – σθ(E2)
and similarly for ∪ and ∩ in place of –
Also: σθ (E1 – E2) = σθ(E1) – E2
and similarly for ∩ in place of –, but not for ∪ 12. The projection operation distributes over union
ΠL(E1 ∪ E2) = (ΠL(E1)) ∪ (ΠL(E2))
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Transformation Example: Pushing Selections
■ Query: Find the names of all instructors in the Music department, along with the titles of the courses that they teach
● Πname, title(σdept_name= “Music”
(instructor (teaches Πcourse_id, title (course))))
■ Transformation using rule 7a.
● Πname, title((σdept_name= “Music”(instructor))
(teaches Πcourse_id, title (course)))
■ Performing the selection as early as possible reduces the size
of the relation to be joined.
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Example with Multiple Transformations
■ Query: Find the names of all instructors in the Music department who have taught a course in 2009, along with the titles of the courses that they taught
● Πname, title(σdept_name= “Music”∧year = 2009
(instructor (teaches Πcourse_id, title (course))))
■ Transformation using join associatively (Rule 6a): ● Πname, title(σdept_name= “Music”∧gear = 2009
((instructor teaches) Πcourse_id, title (course)))
■ Second form provides an opportunity to apply the “perform
selections early” rule, resulting in the subexpression σdept_name = “Music” (instructor) σ year = 2009 (teaches)
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Multiple Transformations (Cont.)
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Transformation Example: Pushing Projections
■ Consider: Πname, title(σdept_name= “Music” (instructor) teaches)
Πcourse_id, title (course))))
■ When we compute
(σdept_name = “Music” (instructor teaches)
we obtain a relation whose schema is:
(ID, name, dept_name, salary, course_id, sec_id, semester, year)
■ Push projections using equivalence rules 8a and 8b; eliminate unneeded attributes from intermediate results to get:
Πname, title(Πname, course_id (
σdept_name= “Music” (instructor) teaches))
Πcourse_id, title (course))))
■ Performing the projection as early as possible reduces the size
of the relation to be joined.
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Join Ordering Example
■ For all relations r1, r2, and r3,
(r1 r2) r3 =r1 (r2 r3)
(Join Associativity)
■ If r2 r3 is quite large and r1 r2 is small, we choose
so that we compute and store a smaller temporary relation.
(r1 r2) r3
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Join Ordering Example (Cont.)
■ Consider the expression
Πname, title(σdept_name= “Music” (instructor) teaches)
Πcourse_id, title (course))))
■ Could compute teaches Πcourse_id, title (course) first, and
join result with
σdept_name= “Music” (instructor)
but the result of the first join is likely to be a large relation.
■ Only a small fraction of the university’s instructors are likely to be from the Music department
● it is better to compute
σdept_name= “Music” (instructor)
teaches
first.
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Enumeration of Equivalent Expressions
■ Query optimizers use equivalence rules to systematically generate expressions equivalent to the given expression
■ Can generate all equivalent expressions as follows: ● Repeat
! apply all applicable equivalence rules on every subexpression of every equivalent expression found so far
! add newly generated expressions to the set of equivalent expressions
Until no new equivalent expressions are generated above
■ The above approach is very expensive in space and time
● Twoapproaches
! Optimized plan generation based on transformation rules
! Special case approach for queries with only selections, projections and joins
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Implementing Transformation Based Optimization
■ Space requirements reduced by sharing common sub-expressions:
● when E1 is generated from E2 by an equivalence rule, usually only the top level of the two are different, subtrees below are the same and can be shared using pointers
!
E.g. when applying join commutativity
E1 E2
● Same sub-expression may get generated multiple times ! Detect duplicate sub-expressions and share one copy
■ Time requirements are reduced by not generating all expressions ● Dynamic programming
! We will study only the special case of dynamic programming for join order optimization
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Cost Estimation
■ Cost of each operator computed as described in Chapter 12 ● Needstatisticsofinputrelations
! E.g. number of tuples, sizes of tuples
■ Inputs can be results of sub-expressions
● Needtoestimatestatisticsofexpressionresults ● Todoso,werequireadditionalstatistics
! E.g. number of distinct values for an attribute
■ More on cost estimation later
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Choice of Evaluation Plans
■ Must consider the interaction of evaluation techniques when choosing evaluation plans
● choosingthecheapestalgorithmforeachoperationindependently may not yield best overall algorithm. E.g.
! merge-join may be costlier than hash-join, but may provide a sorted output which reduces the cost for an outer level aggregation.
! nested-loop join may provide opportunity for pipelining
■ Practical query optimizers incorporate elements of the following two
broad approaches:
1. Search all the plans and choose the best plan in a
cost-based fashion.
2. Uses heuristics to choose a plan.
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Cost-Based Optimization
■ Consider finding the best join-order for r1 r2 . . . rn.
■ There are (2(n – 1))!/(n – 1)! different join orders for above expression. With n = 7, the number is 665280, with n = 10, the number is greater than 176 billion!
■ No need to generate all the join orders. Using dynamic programming, the least-cost join order for any subset of
{r1, r2, . . . rn} is computed only once and stored for future use.
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Dynamic Programming in Optimization
■ To find best join tree for a set of n relations:
● TofindbestplanforasetSofnrelations,considerallpossible plans of the form: S1 (S – S1) where S1 is any non-empty subset of S.
● RecursivelycomputecostsforjoiningsubsetsofStofindthecost of each plan. Choose the cheapest of the 2n – 2 alternatives.
● Base case for recursion: single relation access plan
! Apply all selections on Ri using best choice of indices on Ri
● Whenplanforanysubsetiscomputed,storeitandreuseitwhenit is required again, instead of recomputing it
! Dynamic programming
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Join Order Optimization Algorithm
procedure findbestplan(S)
if (bestplan[S].cost ≠ ∞)
return bestplan[S]
// else bestplan[S] has not been computed earlier, compute it now
if (S contains only 1 relation)
set bestplan[S].plan and bestplan[S].cost based on the best way
of accessing S /* Using selections on S and indices on S */
else for each non-empty subset S1 of S such that S1 ≠ S
P1= findbestplan(S1)
P2= findbestplan(S – S1)
A = best algorithm for joining results of P1 and P2
cost = P1.cost + P2.cost + cost of A
if cost < bestplan[S].cost
bestplan[S].cost = cost
bestplan[S].plan = “execute P1.plan; execute P2.plan;
join results of P1 and P2 using A”
* Some modifications to allow indexed nested loops joins on relations that have
return bestplan[S] selections (see book)
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Left Deep Join Trees
■ In left-deep join trees, the right-hand-side input for each join is a relation, not the result of an intermediate join.
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Cost of Optimization
■ With dynamic programming time complexity of optimization with bushy trees is O(3n).
● Withn=10,thisnumberis59000insteadof176billion!
■ Space complexity is O(2n)
■ To find best left-deep join tree for a set of n relations:
● Considernalternativeswithonerelationasright-handsideinput and the other relations as left-hand side input.
● Modifyoptimizationalgorithm:
! Replace “for each non-empty subset S1 of S such that S1 ≠ S”
! By: for each relation r in S
let S1 = S – r .
■ If only left-deep trees are considered, time complexity of finding best join order is O(n 2n)
● SpacecomplexityremainsatO(2n)
■ Cost-based optimization is expensive, but worthwhile for queries on
large datasets (typical queries have small n, generally < 10)
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Interesting Sort Orders
■ Consider the expression (r1 r2) r3 (with A as common attribute)
■ An interesting sort order is a particular sort order of tuples that could
be useful for a later operation
● Using merge-join to compute r1 r2 may be costlier than hash join but generates result sorted on A
● Which in turn may make merge-join with r3 cheaper, which may reduce cost of join with r3 and minimizing overall cost
● Sortordermayalsobeusefulfororderbyandforgrouping
■ Not sufficient to find the best join order for each subset of the set of n
given relations
● mustfindthebestjoinorderforeachsubset,foreachinteresting
sort order
● Simpleextensionofearlierdynamicprogrammingalgorithms
● Usually,numberofinterestingordersisquitesmallanddoesn’t
affect time/space complexity significantly
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Cost Based Optimization with Equivalence Rules
■ Physical equivalence rules allow logical query plan to be converted to physical query plan specifying what algorithms are used for each operation.
■ Efficient optimizer based on equivalent rules depends on
● Aspaceefficientrepresentationofexpressionswhichavoids
making multiple copies of subexpressions
● Efficienttechniquesfordetectingduplicatederivationsof expressions
● Aformofdynamicprogrammingbasedonmemoization,which stores the best plan for a subexpression the first time it is optimized, and reuses in on repeated optimization calls on same subexpression
● Cost-basedpruningtechniquesthatavoidgeneratingallplans
■ Pioneered by the Volcano project and implemented in the SQL Server
optimizer
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Heuristic Optimization
■ Cost-based optimization is expensive, even with dynamic programming.
■ Systems may use heuristics to reduce the number of choices that must
be made in a cost-based fashion.
■ Heuristic optimization transforms the query-tree by using a set of rules that typically (but not in all cases) improve execution performance:
● Performselectionearly(reducesthenumberoftuples)
● Performprojectionearly(reducesthenumberofattributes)
● Performmostrestrictiveselectionandjoinoperations(i.e.with smallest result size) before other similar operations.
● Somesystemsuseonlyheuristics,otherscombineheuristicswith partial cost-based optimization.
■ So this is a hybrid of cost- and rule-based optimization
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Structure of Query Optimizers
■ Many optimizers considers only left-deep join orders.
● Plusheuristicstopushselectionsandprojectionsdownthequery
tree
● Reducesoptimizationcomplexityandgeneratesplansamenableto pipelined evaluation.
■ Heuristic optimization used in some versions of Oracle: ● Repeatedlypick“best”relationtojoinnext
! Starting from each of n starting points. Pick best among these
■ Intricacies of SQL complicate query optimization
● E.g.nestedsubqueries
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■
Structure of Query Optimizers (Cont.)
Some query optimizers integrate heuristic selection and the generation of alternative access plans.
● Frequently used approach
! heuristic rewriting of nested block structure and aggregation ! followed by cost-based join-order optimization for each block
● Some optimizers (e.g. SQL Server) apply transformations to entire query and do not depend on block structure
● Optimization cost budget to stop optimization early (if cost of plan is less than cost of optimization)
● Plan caching to reuse previously computed plan if query is resubmitted ! Even with different constants in query
Even with the use of heuristics, cost-based query optimization imposes a substantial overhead.
● But is worth it for expensive queries
● Optimizers often use simple heuristics for very cheap queries, and
perform exhaustive enumeration for more expensive queries
■
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Statistical Information for Cost Estimation
■ nr: number of tuples in a relation r.
■ br: number of blocks containing tuples of r.
■ lr: size of a tuple of r.
■ fr: blocking factor of r — i.e., the number of tuples of r that fit into one block.
■ V(A, r): number of distinct values that appear in r for attribute A; same as the size of ∏A(r).
■ If tuples of r are stored together physically in a file, then:
$n" br=# fr!
#! #r!
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Histograms
■ Histogram on attribute age of relation person
50
40
30
20 10
1–5 6–10 11–15 16–20 21–25
■ Equi-width histograms
■ Equi-depth histograms
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value
frequency
Selection Size Estimation
■ σA=v(r)
! nr / V(A,r) : number of records that will satisfy the selection ! Equality condition on a key attribute: size estimate = 1
■ σA≤V(r) (case of σA ≥ V(r) is symmetric)
● Let c denote the estimated number of tuples satisfying the condition.
●
If min(A,r) and max(A,r) are available in catalog ! c=0ifv