1b_query_optimization_sil_7ed_ch16.ppt

Database System Concepts, 6th Ed.
©Silberschatz, Korth and Sudarshan
See www.db-book.com for conditions on re-use
These slides are a modified version of the slides provided with the book:
(however, chapter numeration refers to 7th Ed.)
The original version of the slides is available at: https://www.db-book.com/
Chapter 16: Query Optimization
Data Management for Big Data
2019-2020 (spring semester)
Dario Della Monica

16.2
Database System Concepts - 7th Edition
Chapter 16: Query Optimization
 Introduction
 Generating Equivalent Expressions
 Equivalence rules
 How to generate (all) equivalent expressions
 Estimating Statistics of Expression Results
 The Catalog
 Size estimation
 Selection
 Join
 Other operations (projection, aggregation, set operations, outer join)
 Estimation of number of distinct values
 Choice of Evaluation Plans
 Dynamic Programming for Choosing Evaluation Plans

16.3
Introduction
 Query optimization: finding the “best” query execution plan (QEP) among the
many possible ones
 User is not expected to write queries efficiently (DBMS optimizer takes care of that)
 Alternative ways to execute a given query – 2 levels
 Equivalent relational algebra expressions
 Different implementation choices for each relational algebra operation
 Algorithms, indices, coordination between successive operations, …
INSTR(i_id, name, dept_name, ...)
COURSE(c_id, title, ...)
TEACHES(i_id, c_id, ...)
The name of all instructors in the department of Music
together with the titles of all courses they teach
SELECT I.name, C.title
FROM INSTR I, COURSE C, TEACHES T
WHERE I.i_id = T.i_id
AND T.c_id = C.c_id
AND dept_name=“Music”
))
)
(
(
(
 COURSE
TEACHES
INSTR
 )
)
(
)
(
(
 COURSE
TEACHES
INSTR


16.4
Introduction (Cont.)
 A query evaluation plan (QEP) 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

16.5
Introduction (Cont.)
 Cost difference between query evaluation plans can be enormous
 E.g. seconds vs. days in some cases
 It is worth spending time in finding “best” QEP
 Steps in cost-based query optimization
1. Generate logically equivalent expressions using equivalence
rules
2. Annotate in all possible ways resulting expressions to get
alternative QEP
3. Evaluate/estimate the cost (execution time) of each QEP
4. Choose the cheapest QEP based on estimated cost
 Estimation of QEP cost based on:
 Statistical information about relations (stored in the Catalog)
 number of tuples, number of distinct values for an attribute
 Statistics estimation for intermediate results
 to compute cost of complex expressions
 Cost formulae for algorithms, computed using statistics

Generating Equivalent Expressions
 Equivalence rules
 How to generate (all) equivalent expressions

16.7
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 (and also order of attributes)
 We don’t care if they generate different results on databases that
violate integrity constraints (e.g., uniqueness of keys)
 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
 We focus on relational algebra and treat relations as sets
 An equivalence rule states that expressions of two forms are equivalent
 One can replace an expression of first form by one of the second form,
or vice versa

16.8
Equivalence Rules
1. Conjunctive selection operations can be deconstructed into a
sequence of individual selections.
2. Selection operations are commutative.
3. Only the last in a sequence of projection operations is
needed, the others can be omitted
where
4. Selections can be combined with Cartesian products and
theta joins.
a. (E1 x E2) = E1  E2
b. 1
(E1 2
E2) = E1 1
 2
E2
))
(
(
))
(
( 1
2
2
1
E
E q
q
q
q s
s
s
s =
))
(
(
)
( 2
1
2
1
E
E q
q
q
q s
s
s =
Ù
)
(
))
))
(
(
(
( 1
2
1
E
E L
L
L
L n




 

n
L
L
L 

 
2
1

16.9
Equivalence Rules (Cont.)
5. Theta-join (and thus natural joins) operations are commutative.
E1  E2 = E2  E1
(but the order is important for efficiency)
6. (a) Natural join operations are associative:
(b) Theta joins are associative in the following manner:
(E1 1
E2) 23
E3 = E1 1 3
(E2 2
E3)
where 1 involves attributes from only E1 and E2
and 2 involves attributes from only E2 and E3
(E1 E2) E3 = E1 (E2 E3)
(again, the order is important for efficiency)

16.10
Equivalence Rules (Cont.)
More equivalences at Ch. 16.2 of the book ⋆
⋆ Silberschatz, Korth, and Sudarshan, Database System Concepts, 7° ed.
7. (a) Selection distribute over theta join in the following manner:
(b) Complex selection distribute over theta join in the following manner:
1
∧2
(E1 ⋈θ E2) = (1
(E1)) ⋈θ (2
(E2))
where 1 involves attributes from only E1
and 2 involves attributes from only E2
1
(E1 ⋈θ E2) = (1
(E1)) ⋈θ E2
where 1 involves attributes from only E1

16.11
Pictorial Depiction of Equivalence Rules

16.12
Exercise
 Disprove the equivalence
( R S ) T = R ( S T )
Definition (left outer join): the result of a left outer join T = R S is a super-set of the
result of the join T’ = R S in that all tuples in T’ appear in T. In addition, T preserve
those tuples that are lost in the join, by creating tuples in T that are filled with null
values
STUD TAKES
STUD stud_id name surname
1 gino bianchi
2 filippo neri
3 mario rossi
TAKES stud_id course grade
1 Math 30
2 DB 22
2 Logic 30
stud_id name surname course grade
1 gino bianchi Math 30
2 filippo neri DB 22
2 filippo neri Logic 30
3 mario rossi null null
TAKES STUD ???

16.13
Solution
 Disprove the equivalence
A AR
1 1
R
A AS
2 1
S
A AT
1 1
T
R S
A AR AS
1 1 null
S T
A AS AT
2 1 null
( R S ) T = R ( S T )
( R S ) T
A AR AS AT
1 1 null 1
R ( S T )
A AR AS AT
1 1 null null

16.14
Equivalence derivability and minimality
 Some equivalence can be derived from others
 example: 2 can be obtained from 1 (exploiting commutativity of conjunction)
7b can be obtained from 1 and 7a
 Optimizers use minimal sets of equivalence rules

16.15
Enumeration of Equivalent Expressions
 Query optimizers use equivalence rules to systematically generate
expressions equivalent to the given one
 Can generate all equivalent expressions as follows:
 Repeat (starting from the set containing only the given expression)
 apply all applicable equivalence rules on every sub-expression of
every equivalent expression found so far
 add newly generated expressions to the set of equivalent
expressions
Until no new equivalent expressions are generated
 The above approach is very expensive in space and time
 Space: efficient expression-representation techniques
 1 copy is stored for shared sub-expressions
 Time: partial generation
 Dynamic programming
 Greedy techniques (select best choices at each step)
 Heuristics, e.g., single-relation operations
(selections, projections) are pushed inside (performed earlier)
E1 E2

Estimating Statistics of Expression
Results
 The Catalog
 Size estimation
 Selection
 Join
 Other operations (projection, aggregation, set operations, outer join)
 Estimation of number of distinct values

16.17
Cost Estimation
 Cost of each operator computed as described in Chapter 15 ⋆
 Need statistics of input relations
 E.g. number of tuples, sizes of tuples
 Statistics are collected in the Catalog
 Inputs can be results of sub-expressions
 Need to estimate statistics of expression results
 Estimation of size of intermediate results
 # of tuple in input to successive operations
 Estimation of number of distinct values in intermediate results
 selectivity rate of successive selection operations
 Statistics are not totally accurate
 Information in the catalog might be not always up-to-date (delay)
 A precise estimate for intermediate results might be impossible to compute
⋆ Silberschatz, Korth, and Sudarshan, Database System Concepts, 7° ed.

16.18
Statistical Information for Cost Estimation
 Statistics information is maintained in the Catalog
 The catalog is itself stored in the database as relation(s)
 It contains:
 nr: number of tuples in a relation r
 br: number of blocks containing tuples of r
 lr: size of a tuple of r (in bytes)
 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 set of attributes A
 V(A, r) = A(r) – if A is a key, then V(A,r) = nr
 min(A,r): smallest value appearing in relation r for set of attribute A;
 max(A,r): largest value appearing in relation r for set of attribute A;
 statistics about indices (height of B+-trees, number of blocks for leaves, …)
 We assume tuples of r are stored together physically in a file; then: br = ⌈ nr / fr ⌉
 Information not always up-to-date
 Catalog is not updated to every DB change (done during periods of light system load)

16.19
Histograms
 Histogram on attribute age of relation person
 For each range
 Number of records (tuples) with value in the range
 Also, number of distinct values in the range
 Without histogram information, uniform distribution is assumed
 Little space occupation
 Histograms for many attributes on many relations can be stored
value
frequency
50
40
30
20
10
1–5 6–10 11–15 16–20 21–25

16.20
Selection Size Estimation
 # of records that will satisfy the selection predicate (aka selection condition)
 A=v(r ) (we are assuming that v actually is present in A)
 nr / V(A,r) (no histogram, uniform distribution)
 1 if A is key
 A  v(r ) (case A  V(r) is symmetric)
 0 if v < min(A,r)
 nr if v >= max(A,r)
 otherwise (no histogram, uniform distribution)
 In absence of statistical information or when v is unknown at time of cost estimation
(e.g., v is computed at run-time by the application using the DB), the we assume
 nr / 2
 If histograms are available, we can do more precise estimates
 use values for restricted ranges instead of nr , V(A,r), min(A, r), max(A,r)
)
,
min(
)
,
max(
)
,
min(
*
r
A
r
A
r
A
v
nr



16.21
Complex Selection Size Estimation
 Conjunction E = θ1 ∧ θ2 ∧ … ∧ θn
(r )
 we compute si = size selection for θi (i = 1,…, n)
 selectivity rate (SR) of θi
(r): SR(θi
(r) ) = si / nr (i = 1,…, n)
 SR(E) = Πi (SR(θi
(r))) = s1 / nr * … * sn/ nr Πi is multiplication with i = 1,…,n
 # of record for E = nr * SR(E) = n
r
n
r
)
(n
*...*s
*s
s
*
n 2
1
 Disjunction E = θ1 ∨ θ2 ∨ … ∨ θn
(r ) = ¬(¬θ1 ∧ ¬θ2 ∧ … ∧ ¬θn) (r )
 SR(E) = 1 - SR(¬θ1 ∧ ¬θ2 ∧ … ∧ ¬θn
(r ))
 SR(¬θ1 ∧ ¬θ2 ∧ … ∧ ¬θn
(r )) = (1 - s1 / nr ) * … * (1 - sn/ nr )
 # of record for E = nr * SR(E) = 







 )
n
s
)*...*(
n
s
)*(
n
s
- (
*
n
r
n
r
r
r 1
1
1
1 2
1
 Negation E = ¬θ (r)
 # of record for E = nr - # of record for  θ (r)

16.22
Join Size Estimation
 # of records that will be included in the result
 (cartesian product) r x s: # of records = nr * ns
 (natural join on attribute A) r ⋈ s:
 for each tuple tr of r there are in average ns / V(A,s) many tuples of s selected
 thus, # of records = nr * ns / V(A,s)
 by switching the role of r and s we get # of records = nr * ns / V(A,r)
 lowest is more accurate estimation # of records = nr * ns / max{ V(A,r), V(A,s) }
 histograms can be used for more accurate estimations
 histograms must be on join attributes, for both relations, and with same ranges
 use values for restricted ranges instead of nr , ns , V(A,r), V(A,s) and then sum estimations for
each range
 if A is key for r, then # of records <= ns (and vice versa)
 in addition, if A is not null in s, then # of records = ns (and vice versa)
 (theta join) r ⋈θ s
 r ⋈θ s =  θ ( r x s) use formulas for cartesian product and selection

16.23
Size Estimation for Other Operations
 projection (no duplications): # of records = V(A,r)
 aggregation GγF (r) # of records = V(G,r)
 set operations
 between selections on same relation use formulas for selection
 es.: θ1
(r) ∪ θ2
(r) = θ1 ∨ θ2
(r)
 r ∪ s # of records = nr + ns
 r ∩ s # of records = min { nr , ns }
 r – s # of records = nr
 outer join
 left outer join # of records = # of records for inner join + nr
 right outer join # of records = # of records for inner join + ns
 full outer join # of records = # of records for inner join + nr + ns

16.24
Estimation for Number of Distinct Values
 # distinct values in the result for expression E and attribute (or set of attributes) A: V(A,E)
 Selection E = θ (r)
 V(A, E) is a specific value for some conditions – e.g., A=3 or 3 < A <= 6
 condition A < v (or A > v, A >= v, … ) V(A,E) = V(A,r) * selectivity rate of the selection
 otherwise V(A,E) = min { nE , V(A,r) }
 Join E = r ⋈ s
 A only contains attributes from r V(A,E) = min { nE , V(A,r) }
 A only contains attributes from s V(A,E) = min { nE , V(A,s) }
 A contains attributes A1 from r and attributes A2 from s
V(A,E) = min { nE , V(A1, r) * V(A2 – A1, s) , V(A2, s) * V(A1 – A2, r) }

Choice of Evaluation Plans
 Dynamic Programming for Choosing Evaluation Plans

16.26
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 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

16.27
Cost-Based Optimization
 A big part of a cost-based optimizer (based on equivalence rules) is
choosing the “best” order for join operations
 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 17.6 billion!
 No need to generate all the join orders. Exploiting some monotonicity
(optimal substructure property), the least-cost join order for any subset
of {r1, r2, . . ., rn} is computed only once.

16.28
Cost-Based Optimization: An example
 Consider finding the best join-order for r1 r2 r3 r4 r5
 Number of possible different join orderings:
 The least-cost join order for any subset of { r1, r2, r3, r4, r5 } is computed only once
 Assume we want to compute N123/45 : number of possible different join orderings
where r1, r2, r3 sare grouped together, e.g.,
1680
!
4
!
8
)!
1
(
))!
1
(
2
(




n
n
(r1 r2 r3) r4 r5 (r2 r3 r1) (r5 r4 ) r4 (r5 (r1 (r2 r3)))
 The naïve approach
 N123/45 = N123 * N45
 N123 = (N123 : # ways of arranging r1, r2, and r3)
 N45 = N123 = 12 (N45 : # ways of arranging r4 and r5 wrt. block of r1, r2, and r3)
 N123/45 = 12 * 12 = 144
 Exploiting optimal substructure property:
 compute only once best ordering for r1 r2 r3 : 12 possibilities (N123)
 compute best ordering for R123 r4 r5 : 12 possibilities (N45)
 Therefore, N123/45 = 12 + 12 = 24
12
!
2
!
4

…

16.29
Dynamic Programming in Optimization
 To find best join tree (equivalently, best join order) for a set of n relations:
 Consider all possible plans of the form:
S’ ⋈ (S S’ )
for every non-empty subset S’ of S
 Recursively compute (and store) costs of best join orders for subsets
S’ and S S’. Choose the cheapest of the 2n – 2 alternatives
 Base case for recursion: find best algorithm for scanning relation
 When a plan for a subset is computed, store it and reuse it when it is
required again, instead of re-computing it
 Dynamic programming

16.30
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”
return bestplan[S]
* This is the algorithm shown in the 6th edition of the textbook.
It is slightly different from the algorithm we presented during our class, especially the way
the base case is handled.

16.31
Cost of Optimization
 With dynamic programming time complexity of optimization is O(3n).
 With n = 10, this number is 59000 instead of 17.6 billion!
 Space complexity is O(2n)
 Better time performance when considering only left-deep join tree O(n 2n)
Space complexity remains at O(2n) (heuristic approach)
 Cost-based optimization is expensive, but worthwhile for queries on
large datasets (typical queries have small n, generally < 10)

16.32
Cost Based Optimization with Equivalence
Rules
 Physical equivalence rules equates logical operations (e.g., join) to physical
ones (i.e., implementations – e.g., nested-loop join, merge join)
 Relational algebra expression are converted into QEP with implementation details
 Efficient optimizer based on equivalence rules depends on
 A space efficient representation of expressions which avoids making
multiple copies of sub-expressions
 Efficient techniques for detecting duplicate derivations of expressions
 Dynamic programming or memoization techniques, which store the “best”
plan for a sub-expression the first time it is computed, and reuses in on
repeated optimization calls on same sub-expression
 Cost-based pruning techniques that avoid generating all plans (greedy,
heuristics)

16.33
Heuristic Optimization
 Cost-based optimization is expensive, even with dynamic programming
 Systems may use heuristics to reduce the number of possibilities
choices that must be considered
 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 most restrictive selection and join operations (i.e. with
smallest result size) before other similar operations
 Only consider left-deep join orders (particularly suited for pipelining
as only one input has to be pipelined, the other is a relation)

16.34
Structure of Query Optimizers
 Some systems use only heuristics, others combine heuristics with partial
cost-based optimization.
 Many optimizers considers only left-deep join orders.
 Plus heuristics to push selections and projections down the query
tree
 Reduces optimization complexity and generates plans amenable to
pipelined evaluation.
 Heuristic optimization used in some versions of Oracle:
 Repeatedly pick “best” relation to join next
 it obtains and compares n plans (each starting with one relation.
In each plan, pick the best next relation for the join

End of Chapter

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