Database System Concepts, 7th Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
Chapter 11: Query Optimization
©Silberschatz, Korth and Sudarshan11.2Database System Concepts - 7th Edition
Outline
▪ 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
©Silberschatz, Korth and Sudarshan11.3Database System Concepts - 7th Edition
Introduction
▪ Alternative ways of evaluating a given query
• Equivalent expressions
• Different algorithms for each operation
©Silberschatz, Korth and Sudarshan11.4Database System Concepts - 7th Edition
Introduction (Cont.)
▪ An evaluation plan defines exactly what algorithm is used for each
operation, and how the execution of the operations is coordinated.
▪ Find out how to view query execution plans on your favorite database
©Silberschatz, Korth and Sudarshan11.5Database System Concepts - 7th Edition
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 the estimated cost
▪ Estimation of plan cost based on:
• Statistical information about relations. Examples:
▪ number of tuples, number of distinct values for an attribute
• Statistics estimation for intermediate results
▪ to compute costs of complex expressions
• Cost formulae for algorithms, computed using statistics
©Silberschatz, Korth and Sudarshan11.6Database System Concepts - 7th Edition
Viewing Query Evaluation Plans
▪ Most database support explain <query>
• Displays plan chosen by the query optimizer, along with cost estimates
• Some syntax variations between databases
▪ Oracle: explain plan for <query> followed by select * from table
(dbms_xplan.display)
▪ SQL Server: set showplan_text on
▪ Some databases (e.g. PostgreSQL) support explain analyse <query>
• Shows actual runtime statistics found by running the query, in addition
to showing the plan
▪ Some databases (e.g. PostgreSQL) show cost as f..l
• f is the cost of delivering the first tuple and l is the cost of delivering all
results
©Silberschatz, Korth and Sudarshan11.7Database System Concepts - 7th Edition
Generating Equivalent Expressions
©Silberschatz, Korth and Sudarshan11.8Database System Concepts - 7th Edition
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.
▪ An equivalence rule says that expressions of two forms are equivalent
• Can replace expression of the first form with the second, or vice versa
©Silberschatz, Korth and Sudarshan11.9Database System Concepts - 7th Edition
Equivalence Rules
1. Conjunctive selection operations can be deconstructed into a sequence of
individual selections.
σ1  2
(E) ≡ σ1
(σ2
(E))
2. Selection operations are commutative.
σ1
(σ2
(E)) ≡ σ2
(σ1
(E))
3. Only the last in a sequence of projection operations is needed, the others
can be omitted.
 L1
( L2
(…( Ln
(E))…)) ≡  L1
(E)
where L1 ⊆ L2 … ⊆ Ln
4. Selections can be combined with Cartesian products and theta joins.
a. σ (E1 x E2) ≡ E1 ⨝  E2
b. σ 1
(E1 ⨝2
E2) ≡ E1 ⨝ 1∧2
E2
©Silberschatz, Korth and Sudarshan11.10Database System Concepts - 7th Edition
Equivalence Rules (Cont.)
5. Theta-join operations (and natural joins) are commutative.
E1 ⨝ E2 ≡ E2 ⨝ E1
6. (a) Natural join operations are associative:
(E1 ⨝ E2) ⨝ E3 ≡ E1 ⨝ (E2 ⨝ E3)
(b) Theta joins are associative in the following manner:
(E1 ⨝ 1
E2) ⨝ 2  3
E3 ≡ E1 ⨝1  3
(E2 ⨝ 2
E3)
where 2 involves attributes from only E2 and E3.
©Silberschatz, Korth and Sudarshan11.11Database System Concepts - 7th Edition
Pictorial Depiction of Equivalence Rules
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Equivalence Rules (Cont.)
7. The selection operation distributes over the theta join operation under the
following two conditions:
(a) When all the attributes in 0 involve only the attributes of one
of the expressions (E1) being joined.
0
(E1 ⨝ E2) ≡ (0
(E1)) ⨝ E2
(b) When 1 involves only the attributes of E1 and 2 involves
only the attributes of E2.
1  2
(E1 ⨝ E2) ≡ (1
(E1)) ⨝ (2
(E2))
©Silberschatz, Korth and Sudarshan11.13Database System Concepts - 7th Edition
8. The projection operation distributes over the theta join operation as follows:
(a) if  involves only attributes from L1  L2:
 L1  L2
(E1 ⨝ E2) ≡  L1
(E1) ⨝  L2
(E2)
(b) In general, consider a join E1 ⨝ E2.
• Let L1 and L2 be sets of attributes from E1 and E2, respectively.
• Let L3 be attributes of E1 that are involved in the join condition , but
are not in L1  L2, and
• let L4 be attributes of E2 that are involved in the join condition , but
are not in L1  L2.
 L1  L2
(E1 ⨝ E2) ≡  L1  L2
( L1  L3
(E1) ⨝  L2  L4
(E2))
Similar equivalences hold for outer-join operations: ⟕, ⟖, and ⟗
Equivalence Rules (Cont.)
©Silberschatz, Korth and Sudarshan11.14Database System Concepts - 7th Edition
Equivalence Rules (Cont.)
9. The set operations union and intersection are commutative
E1  E2 ≡ E2  E1
E1  E2 ≡ E2  E1
(the 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 –.
a.  (E1  E2) ≡  (E1)  (E2)
b.  (E1  E2) ≡  (E1)  (E2)
c.  (E1 – E2) ≡  (E1) – (E2)
d.  (E1  E2) ≡ (E1)  E2
e.  (E1 – E2) ≡ (E1) – E2
preceding equivalence does not hold for 
12.The projection operation distributes over the union
L(E1  E2) ≡ (L(E1))  (L(E2))
©Silberschatz, Korth and Sudarshan11.15Database System Concepts - 7th Edition
Equivalence Rules (Cont.)
13. Selection distributes over aggregation as below
(G 𝛾A(E)) ≡ G 𝛾A((E))
provided  only involves attributes in G
14. a. Full outer-join is commutative:
E1 ⟗ E2 ≡ E2 ⟗ E1
b. Left and right outer-join are not commutative, but:
E1 ⟕ E2 ≡ E2 ⟖ E1
15. Selection distributes over left and right outer-joins as below, provided 1
only involves attributes of E1
a. 1
(E1 ⟕ E2) ≡ (1
(E1)) ⟕ E2
b. 1
(E1 ⟖ E2) ≡ E2 ⟕ (1
(E1))
©Silberschatz, Korth and Sudarshan11.16Database System Concepts - 7th Edition
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
• name, title(dept_name= ‘Music’
(instructor ⨝ (teaches ⨝ course_id, title (course))))
▪ Transformation using rule 7a.
• name, title((dept_name= ‘Music’(instructor)) ⨝
(teaches ⨝ course_id, title (course)))
▪ Performing the selection as early as possible reduces the size of the
relation to be joined.
©Silberschatz, Korth and Sudarshan11.18Database System Concepts - 7th Edition
Multiple Transformations (Cont.)
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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
(r1 ⨝ r2) ⨝ r3
so that we compute and store a smaller temporary relation.
©Silberschatz, Korth and Sudarshan11.21Database System Concepts - 7th Edition
Join Ordering Example (Cont.)
▪ Consider the expression
name, title(dept_name= “Music” (instructor) ⨝ teaches)
⨝ course_id, title (course))))
▪ Could compute teaches ⨝ course_id, title (course) first, and join the result
with
dept_name= “Music” (instructor)
But the first join’s result 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
dept_name= “Music” (instructor) ⨝ teaches
first.
©Silberschatz, Korth and Sudarshan11.22Database System Concepts - 7th Edition
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
• Two approaches
▪ Optimized plan generation based on transformation rules
▪ Special case approach for queries with only selections, projections,
and joins
©Silberschatz, Korth and Sudarshan11.23Database System Concepts - 7th Edition
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
• 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
©Silberschatz, Korth and Sudarshan11.24Database System Concepts - 7th Edition
Cost Estimation
▪ Cost of each operator computed as described earlier
• Need statistics on input relations
▪ E.g., number of tuples, sizes of tuples
▪ Inputs can be results of sub-expressions
• Need to estimate statistics of expression results
• To do so, we require additional statistics
▪ E.g., the number of distinct values for an attribute
▪ More on cost estimation later
©Silberschatz, Korth and Sudarshan11.25Database System Concepts - 7th Edition
Choice of Evaluation Plans
▪ Must consider the interaction of evaluation techniques when choosing
evaluation plans
• choosing the cheapest algorithm for each operation independently may
not yield the best overall algorithm. E.g.
▪ merge-join may be costlier than hash-join, but may provide a sorted
output that reduces the cost for an outer level aggregation.
▪ Nested-loop join may provide an 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.
©Silberschatz, Korth and Sudarshan11.26Database System Concepts - 7th Edition
Cost-Based Optimization
▪ Consider finding the best join-order for r1 ⨝ r2 ⨝ . . . ⨝ rn.
▪ There are (2(n – 1))!/(n – 1)! different join orders for the 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.
©Silberschatz, Korth and Sudarshan11.27Database System Concepts - 7th Edition
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.
©Silberschatz, Korth and Sudarshan11.28Database System Concepts - 7th Edition
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:
• Perform selection early (reduces the number of tuples)
• Perform projection early (reduces the number of attributes)
• Perform the most restrictive selection and join operations (i.e., with the
smallest result size) before other similar operations.
• Some systems use only heuristics, while others combine heuristics with
partial cost-based optimization.
©Silberschatz, Korth and Sudarshan11.29Database System Concepts - 7th Edition
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
queries and do not depend on block structure
• Optimization cost budget to stop optimization early (if the cost of the
plan is less than the cost of optimization)
• Plan caching to reuse previously computed plan if the query is
resubmitted
▪ Even with different constants in the 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
©Silberschatz, Korth and Sudarshan11.30Database System Concepts - 7th Edition
Statistics for Cost Estimation
©Silberschatz, Korth and Sudarshan11.31Database System Concepts - 7th Edition
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:
ú
ú
ú
ú
ù
ê
ê
ê
ê
é
=
rf
rn
rb
©Silberschatz, Korth and Sudarshan11.32Database System Concepts - 7th Edition
Histograms
▪ Histogram on attribute age of relation person
▪ Equi-width histograms
▪ Equi-depth histograms break up ranges such that each range has
(approximately) the same number of tuples
• E.g. (4, 8, 14, 19)
▪ Many databases also store n most-frequent values and their counts
• Histogram is built on remaining values only
value
frequency
50
40
30
20
10
1–5 6–10 11–15 16–20 21–25
©Silberschatz, Korth and Sudarshan11.33Database System Concepts - 7th Edition
Histograms (cont.)
▪ Histograms and other statistics are usually computed based on a random
sample
▪ Statistics may be out of date
• Some databases require an analyze command to be executed to
update statistics
• Others automatically recompute statistics
▪ e.g., when the number of tuples in a relation changes by some
percentage
©Silberschatz, Korth and Sudarshan11.34Database System Concepts - 7th Edition
▪ A=v(r)
• nr / V(A,r) : number of records that will satisfy the selection
• Equality condition on a key attribute: size estimate = 1
▪ AV(r) (case of A  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 = 0 if v < min(A,r)
▪ C =
• If histograms are available, can refine the above estimate
• In the absence of statistical information c is assumed to be nr / 2.
Selection Size Estimation
),min(),max(
),min(
.
rArA
rAv
nr
-
-
©Silberschatz, Korth and Sudarshan11.35Database System Concepts - 7th Edition
Size Estimation of Complex Selections
▪ The selectivity of a condition i is the probability that a tuple in the relation
r satisfies i .
• If si is the number of satisfying tuples in r, the selectivity of i is given
by si /nr.
▪ Conjunction: 1 2. . .  n (r). Assuming independence, the estimate
of the number of tuples in the result is:
▪ Disjunction:1 2 . . .  n (r). Estimated number of tuples:
▪ Negation: (r). Estimated number of tuples:
nr – size((r))
n
r
n
r
n
sss
n
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-**-*--* )1(...)1()1(1 21
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©Silberschatz, Korth and Sudarshan11.36Database System Concepts - 7th Edition
Join Operation: Running Example
Running example:
student ⨝ takes
Catalog information for join examples:
▪ nstudent = 5,000.
▪ fstudent = 50, which implies that
bstudent =5000/50 = 100.
▪ ntakes = 10000.
▪ ftakes = 25, which implies that
btakes = 10000/25 = 400.
▪ V(ID, takes) = 2500, which implies that on average, each student who has
taken a course has taken 4 courses.
• Attribute ID in takes is a foreign key referencing student.
• V(ID, student) = 5000 (primary key!)
©Silberschatz, Korth and Sudarshan11.37Database System Concepts - 7th Edition
Estimation of the Size of Joins
▪ The Cartesian product r x s contains nr .ns tuples; each tuple occupies sr +
ss bytes.
▪ If R  S = , then r ⋈ s is the same as r x s.
▪ If R  S is a key for R, then a tuple of s will join with at most one tuple from
r
• therefore, the number of tuples in r ⋈ s is no greater than the number
of tuples in s.
▪ If R  S in S is a foreign key in S referencing R, then the number of tuples
in r ⋈ s is exactly the same as the number of tuples in s.
▪ The case for R  S being a foreign key referencing S is symmetric.
▪ In the example query student ⋈ takes, ID in takes is a foreign key
referencing student
• hence, the result has exactly ntakes tuples, which is 10000
©Silberschatz, Korth and Sudarshan11.38Database System Concepts - 7th Edition
Estimation of the Size of Joins (Cont.)
▪ If R  S = {A} is not a key for R or S.
If we assume that every tuple t in R produces tuples in R S, the number
of tuples in R ⨝ S is estimated to be:
If the reverse is true, the estimate obtained will be:
The lower of these two estimates is probably the more accurate one.
▪ Can improve on above if histograms are available
• Use a formula similar to the above, for each cell of histograms on the
two relations
),( sAV
nn sr *
),( rAV
nn sr *
©Silberschatz, Korth and Sudarshan11.39Database System Concepts - 7th Edition
Size Estimation for Other Operations
▪ Projection: estimated size of A(r) = V(A,r)
▪ Aggregation : estimated size of G 𝛾A(r) = V(G,r)
▪ Set operations
• For unions/intersections of selections on the same relation: rewrite
and use size estimate for selections
▪ E.g., 1 (r)  2 (r) can be rewritten as 1 or 2 (r)
• For operations on different relations:
▪ estimated size of r  s = size of r + size of s.
▪ estimated size of r  s = minimum size of r and size of s.
▪ estimated size of r – s = r.
▪ All three estimates may be quite inaccurate, but provide upper
bounds on the sizes.
©Silberschatz, Korth and Sudarshan11.40Database System Concepts - 7th Edition
Size Estimation (Cont.)
▪ Outer join:
• Estimated size of r ⟕ s = size of r ⨝ s + size of r
▪ Case of the right outer join is symmetric
• Estimated size of r ⟗ s = size of r ⨝ s + size of r + size of s
©Silberschatz, Korth and Sudarshan11.41Database System Concepts - 7th Edition
Estimation of Number of Distinct Values
Selections:  (r)
▪ If  forces A to take a specified value: V(A, (r)) = 1.
▪ e.g., A = 3
▪ If  forces A to take on one of a specified set of values:
V(A, (r)) = number of specified values.
▪ (e.g., (A = 1 V A = 3 V A = 4 )),
▪ If the selection condition  is of the form A op r
estimated V(A, (r)) = V(A.r) * s
▪ where s is the selectivity of the selection.
▪ In all the other cases: use an approximate estimate of
min(V(A,r), n (r) )
• More accurate estimate can be got using probability theory, but this
one works fine generally
©Silberschatz, Korth and Sudarshan11.42Database System Concepts - 7th Edition
End of Chapter 11