CMSC424: Database
Design
Instructor: Amol Deshpande
amol@cs.umd.edu
Query Optimization
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Introduction
Example of a Simple Type of Query
Transformation of Relational Expressions
Statistics Estimation
Optimization Algorithms
Query Optimization
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Why ?
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Many different ways of executing a given query
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Huge differences in cost
Example:
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select * from person where ssn = “123”
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Size of person = 1GB
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Sequential Scan:
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Takes 1GB / (20MB/s) = 50s
Use an index on SSN (assuming one exists):
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Approx 4 Random I/Os = 40ms
Query Optimization
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Equivalent relational expressions
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Drawn as a tree
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List the operations and the order
Query Optimization
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Execution plans
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Evaluation expressions annotated with the methods used
Query Optimization
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Steps:
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Generate all possible execution plans for the query
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Figure out the cost for each of them
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Choose the best
Not done exactly as listed above
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Too many different execution plans for that
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Typically interleave all of these into a single efficient search
algorithm
Query Optimization
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Steps:
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Generate all possible execution plans for the query
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First generate all equivalent expressions
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Then consider all annotations for the operations
Figure out the cost for each of them
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Compute cost for each operation

Using the formulas discussed before

One problem: How do we know the number of result tuples for,
say,  balance2500 (account )
Add them !
Choose the best
Query Optimization
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Introduction
Example of a Simple Type of Query
Transformation of Relational Expressions
Statistics Estimation
Optimization Algorithms
A Simple Case
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Queries with only selections on a single relation with no indexes
select *
from person
where substr(name, 1, 1) in [‘A’, ‘B’, ‘C’] and
zipcode = 94720 and
date-of-birth > to_date('1978/05/31', 'yyyy/mm/dd')
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Relation contains:
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1,000,000 tuples
CPU Costs
100ns
1ns
1000ns
A Simple Case
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Possible execution plan:
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For each tuple
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Evaluate substr predicate
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If true,

Evaluate zipcode predicate
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If true, evaluate date-of-birth predicate
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If true, output the tuple
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6 different possibilities
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How to choose one ?
A Simple Case
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Compute cost of each possibility
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Say, substr()  zipcode  date-of-birth
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Need some more information
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selectivity: fraction of tuples expected to pass the predicates
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Let selectivity(substr predicate) = 3/26
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Let selectivity(zipcode predicate) = 1/100
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And, selectivity(date-of-birth predicate) = 1/3
How are selectivities computed ?
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Must keep track of some additional information about the
relations
A Simple Case
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Compute cost of each possibility
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Say, substr()  zipcode  date-of-birth
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Given that:
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Cost of the above plan =
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1,000,000 * 100ns
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+ 1,000,000 * 3/26 * 1ns
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+ 1,000,000 * 1/26 * 1/100 * 1000ns
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= approx 100.5 ms
Cost of the plan: zipcode  substr()  date-of-birth:
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Approx 12.92 ms
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About a factor of 10 better.
A Simple Case
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Compute cost of each possibility
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Say, substr()  zipcode  date-of-birth
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Cost of the plan: zipcode  substr()  date-of-birth:
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Approx 12.92 ms
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About a factor of 10 better.
General algorithm:
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Don’t need to check all n! Possibilities
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Sort the predicates in the increasing order by:
1 – selectivity(predicate)
cost of the predicate
Query Optimization
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General case:
Need:
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A way to enumerate all plans
A way to find the cost of each plan
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Sub problem: Estimating the selectivities of various
operations
A way to search through the plans efficiently
Query Optimization
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Introduction
Example of a Simple Type of Query
Transformation of Relational Expressions
Statistics Estimation
Optimization Algorithms
Equivalence of Expressions
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Two relational expressions equivalent iff:
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Equivalence rules:
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Their result is identical on all legal databases
Allow replacing one expression with another
Examples:
1.    ( E )    (  ( E ))
1
2
1
2
2. Selections are commutative
  (  ( E ))    (  ( E ))
1
2
2
1
Equivalence Rules
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Examples:
3.  L ( L ( ( Ln ( E )) ))   L ( E )
1
5.
E1
2

1
E2 = E2

E1
7(a). If 0 only involves attributes from E1
0E1
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
E2) = (0(E1))
And so on…
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Many rules of this type

E2
Pictorial Depiction
Example
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Find the names of all customers with an account at a Brooklyn branch
whose account balance is over $1000.
customer_name(branch_city = “Brooklyn” 
(branch

(account
balance > 1000
depositor)))
Apply the rules one by one
customer_name((branch_city = “Brooklyn” 
(branch
account))
balance > 1000
depositor)
customer_name(((branch_city = “Brooklyn” (branch))
(account)))
depositor)
( balance > 1000
Example
Equivalence of Expressions
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The rules give us a way to enumerate all equivalent
expressions
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Note that the expressions don’t contain physical access methods,
join methods etc…
Simple Algorithm:
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Start with the original expression
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Apply all possible applicable rules to get a new set of
expressions
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Repeat with this new set of expressions
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Till no new expressions are generated
Equivalence of Expressions
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Works, but is not feasible
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Consider a simple case:
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R1
(R3
(…
Rn)))….)
Just join commutativity and associativity will give us:
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At least:
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n^2 * 2^n
At worst:
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(R2
n! * 2^n
Typically the process of enumeration is combined with the
search process
Evaluation Plans
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We still need to choose the join methods etc..
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Option 1: Choose for each operation separately
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Usually okay, but sometimes the operators interact
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Consider joining three relations on the same attribute:
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R1
a
(R2
a
R3)
Best option for R2 join R3 might be hash-join
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But if R1 is sorted on a, then sort-merge join is preferable
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Because it produces the result in sorted order by a
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Also, we need to decide whether to use pipelining or
materialization
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Such issues are typically taken into account when doing the
optimization
Query Optimization
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Introduction
Example of a Simple Type of Query
Transformation of Relational Expressions
Optimization Algorithms
Statistics Estimation
Optimization Algorithms
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Two types:
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Exhaustive: That attempt to find the best plan
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Heuristical: That are simpler, but are not guaranteed to find
the optimal plan
Consider a simple case
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Join of the relations R1, …, Rn
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No selections, no projections
Still very large plan space
Searching for the best plan
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Option 1:
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Enumerate all equivalent expressions for the original query
expression
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Using the rules outlined earlier
Estimate cost for each and choose the lowest
Too expensive !
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Consider finding the best join-order for r1
r2
. . . rn .
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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!
Searching for the best plan
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Option 2:
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Dynamic programming
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There is too much commonality between the plans
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Also, costs are additive
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Reduces the cost down to O(n3^n) or O(n2^n) in most
cases
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Interesting orders increase this a little bit
Considered acceptable
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Caveat: Sort orders (also called “interesting orders”)
Typically n < 10.
Switch to heuristic if not acceptable
Left Deep Join Trees
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In left-deep join trees, the right-hand-side input for each join
is a relation, not the result of an intermediate join
Early systems only considered these types of plans
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Easier to pipeline
Heuristic Optimization
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Dynamic programming is expensive
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Use heuristics to reduce the number of choices
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Typically rule-based:
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Perform selection early (reduces the number of tuples)
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Perform projection early (reduces the number of attributes)
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Perform most restrictive selection and join operations before other
similar operations.
Some systems use only heuristics, others combine heuristics
with partial cost-based optimization.
Steps in Typical Heuristic Optimization
1. Deconstruct conjunctive selections into a sequence of
single selection operations (Equiv. rule 1.).
2. Move selection operations down the query tree for the
earliest possible execution (Equiv. rules 2, 7a, 7b, 11).
3. Execute first those selection and join operations that will
produce the smallest relations (Equiv. rule 6).
4. Replace Cartesian product operations that are followed by a
selection condition by join operations (Equiv. rule 4a).
5. Deconstruct and move as far down the tree as possible lists
of projection attributes, creating new projections where
needed (Equiv. rules 3, 8a, 8b, 12).
6. Identify those subtrees whose operations can be pipelined,
and execute them using pipelining).
Query Optimization
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Introduction
Example of a Simple Type of Query
Transformation of Relational Expressions
Optimization Algorithms
Statistics Estimation
Cost Estimation
 Cost of each operator computer as described in Chapter 13
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Need statistics of input relations
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E.g. number of tuples, sizes of tuples
 Inputs can be results of sub-expressions
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Need to estimate statistics of expression results
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To do so, we require additional statistics
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E.g. number of distinct values for an attribute
 More on cost estimation later
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:




nr 
br 
f r 
Histograms
 Histogram on attribute age of relation person
 Equi-width histograms
 Equi-depth histograms
Selection Size Estimation
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A=v(r)
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nr / V(A,r) : number of records that will satisfy the selection
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Equality condition on a key attribute: size estimate = 1
AV(r) (case of A  V(r) is symmetric)
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Let c denote the estimated number of tuples satisfying the condition.
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If min(A,r) and max(A,r) are available in catalog
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
c = 0 if v < min(A,r)

c=
nr .
v  min( A, r )
max( A, r )  min( A, r )
If histograms available, can refine above estimate
In absence of statistical information c is assumed to be nr / 2.
Size Estimation of Complex Selections
 The selectivity of a condition
relation r satisfies i .
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i is the probability that a tuple in the
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, estimate
of tuples in the result is:
s1  s2  . . .  sn
nr 
nrn
 Disjunction:1 2 . . .  n (r). Estimated number of tuples:

s 
s
s
nr  1  (1  1 )  (1  2 )  ...  (1  n ) 
nr
nr
nr 

 Negation: (r). Estimated number of tuples: nr – size((r))
Join Operation: Running Example
Running example:
depositor customer
Catalog information for join examples:
 ncustomer = 10,000.
 fcustomer = 25, which implies that
bcustomer =10000/25 = 400.
 ndepositor = 5000.
 fdepositor = 50, which implies that
bdepositor = 5000/50 = 100.
 V(customer_name, depositor) = 2500, which implies that , on
average, each customer has two accounts.
 Also assume that customer_name in depositor is a foreign key
on customer.

V(customer_name, customer) = 10000 (primary key!)
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
number of tuples in s.
s is no greater than the
 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 depositor
customer, customer_name in
depositor is a foreign key of customer

hence, the result has exactly ndepositor tuples, which is 5000
Estimation of the Size of Joins
 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
number of tuples in R S is estimated to be:
S, the
nr  ns
V ( A, s )
If the reverse is true, the estimate obtained will be:
nr  ns
V ( A, r )
The lower of these two estimates is probably the more accurate one.
 Can improve on above if histograms are available

Use formula similar to above, for each cell of histograms on the
two relations
Estimation of the Size of Joins (Cont.)
 Compute the size estimates for depositor
customer without using
information about foreign keys:

V(customer_name, depositor) = 2500, and
V(customer_name, customer) = 10000
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The two estimates are 5000 * 10000/2500 - 20,000 and 5000 *
10000/10000 = 5000
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We choose the lower estimate, which in this case, is the same as
our earlier computation using foreign keys.
Size Estimation for Other Operations
 Projection: estimated size of A(r) = V(A,r)
 Aggregation : estimated size of
A
gF(r) = V(A,r)
 Set operations
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For unions/intersections of selections on the same relation:
rewrite and use size estimate for selections

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E.g. 1 (r)  2 (r) can be rewritten as 1 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 the three estimates may be quite inaccurate, but provide
upper bounds on the sizes.
Size Estimation (Cont.)
 Outer join:

Estimated size of r


s = size of r
s + size of r
Case of right outer join is symmetric
Estimated size of r
s = size of r
s + size of r + size of s
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 approximate estimate of
min(V(A,r), n (r) )

More accurate estimate can be got using probability theory, but
this one works fine generally
Estimation of Distinct Values (Cont.)
Joins: r
s
 If all attributes in A are from r
estimated V(A, r
s) = min (V(A,r), n r
s)
 If A contains attributes A1 from r and A2 from s, then estimated
V(A,r
s) =
min(V(A1,r)*V(A2 – A1,s), V(A1 – A2,r)*V(A2,s), nr

s)
More accurate estimate can be got using probability theory, but
this one works fine generally
Estimation of Distinct Values (Cont.)
 Estimation of distinct values are straightforward for projections.

They are the same in A (r) as in r.
 The same holds for grouping attributes of aggregation.
 For aggregated values

For min(A) and max(A), the number of distinct values can be
estimated as min(V(A,r), V(G,r)) where G denotes grouping attributes

For other aggregates, assume all values are distinct, and use V(G,r)
Query Optimization






Introduction
Example of a Simple Type of Query
Transformation of Relational Expressions
Optimization Algorithms
Statistics Estimation
Summary
Query Optimization
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Integral component of query processing
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
Why ?
One of the most complex pieces of code in a
database system
Active area of research
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E.g. XML Query Optimization ?
What if you don’t know anything about the
statistics
Better statistics
Etc …