Query Processing

Query Processing Overview

(Translation, Optimisation, Execution)

Query Evaluation 2/154

... Query Evaluation 3/154

The most common form of interaction with a DBMS involves:

supplying some input in the form of an SQL query
getting back the results as a set of tuples

Overall approach clearly applies to select ... but also to delete and update.

... Query Evaluation 4/154

A query in SQL:

states what answers are required (declaratively)
does not say how they should be computed

A query evaluator/processor :

takes declarative description of query (in SQL)
determines plan for answering query (expressed as DBMS ops)
executes plan via DBMS engine (to produce result tuples)

Typically: translate → plan → execute ... in a pipeline.

Some DBMSs can save query plans for later re-use
(because the planning step is potentially quite expensive).

... Query Evaluation 5/154

Internals of the query evaluation "black-box":

... Query Evaluation 6/154

Three phases of query evaluation:

parsing/compilation
- input: SQL query, catalog
- using: parsing techniques, mapping rules
- output: relational algebra (RA) expression
query optimisation
- input: RA expression, DB statistics
- using: cost models, search strategy
- output: query execution plan (DB engine ops)
query execution
- input: query execution plan
- using: database engine
- output: tuples that satisfy query

We ignore the "display" step; simply maps result tuples into appropriate format

Intermediate Representations 7/154

SQL query text is not easy to manipulate/transform.

Need a query representation formalism that ...

is powerful enough to express all queries
has a well-defined, formal basis
simplifies the query transformation process

Relational algebra (RA) expressions:

are easy to transform and have a procedural interpretation

Thus, RA typically used as "target language" for SQL compilation.

... Intermediate Representations 8/154

In principle, could base RA engine on { σ, π, ∪, -, ×} (completeness).

In practice, having only these operators makes execution inefficient.

The standard set of RA operations in a DBMS includes:

filtering and combining (select, project, join (inner,outer))
set operations (union, intersection, difference)

Other operations typically provided in extended RA engines:

grouping (group by) and group-based selection (having)
aggregates (count, sum, avg, max, min)
sorting (order by), uniq (distinct, sets)

RA rename operator is "hidden" in table/tuple representation.

... Intermediate Representations 9/154

In practice, DBMSs provide several versions of each RA operation.

For example:

several "versions" of selection (σ) are available

each version is effective for a particular kind of selection, e.g

select * from R where id = 100  -- hashing
select * from S                 -- Btree index
where age > 18 and age < 35
select * from T                 -- grid file
where a = 1 and b = 'a' and c = 1.4

Similarly, π and ⋈ have versions to match specific query types.

... Intermediate Representations 10/154

We call these specialised version of RA operations RelOps.

One major task of the query processor:

given a set of RA operations to be executed
find a combination of RelOps to do this efficiently

Requires the query translator/optimiser to consider

information about relations (e.g. sizes, primary keys, ...)
information about operations (e.g. select reduces size)

RelOps are realised at execution time

as a collection of inter-communicating nodes
communicating either via pipelines or temporary relations

... Intermediate Representations 11/154

Terminology variations ...

Relational algebra expression of SQL query

intermediate query representation
logical query plan

Execution plan as collection of RelOps

query evaluation plan
query execution plan
physical query plan

Query Translation

... Intermediate Representations 13/154

Query translation: SQL statement text → RA expression

Query Translation 14/154

Translation step is a mapping

from the text of an SQL statement
to a useful internal representation for optimisation

Standard internal representation is relational algebra (RA).

Mapping from SQL to RA may including some optimisations.

Mapping processes: lexer/parser, mapping rules, rewriting rules.

Example:

SQL: select name from Students where id=7654321;
-- is translated to
RA:  Proj[name](Sel[id=7654321]Students)

Parsing SQL 15/154

Parsing task is similar to that for programming languages.

Language elements:

keywords: create, select, from, where, ...
identifiers: Students, name, id, CoursCode, ...
operators: +, -, =, <, >, AND, OR, NOT, IN, ...
constants: 'a string', 123, 19.99, '01-jan-1970'

One difference to parsing PLs:

PLs define objects as part of program definition (e.g. in *.h)
SQL references tables/types/functions stored in the DBMS

Therefore, SQL parser needs access to catalog for definitions.

... Parsing SQL 16/154

PostgreSQL dialect of SQL ...

large set of keywords (> 700 of them) (e.g. select)
parser is implemented via lex/yacc (src/backend/parser)
handles multiple languages (i.e. Unicode strings)
maps all identifiers to lower-case (A-Z → a-z)
needs to handle user-extendable operator set
makes extensive use of catalog (src/backend/catalog)
- pg_class hold pre- and user-defined tables
- pg_type hold pre- and user-defined types
- pg_proc hold pre- and user-defined functions

Catalog and Parsing 17/154

Consider the following simple University schema:

Staff(id, name, position, office, phone, ...)
Students(id, name, birthday, degree, avg, ...)
Subjects(id, title, prereqs, syllabus, ...)
Enrolments(sid, subj, session, mark, ...)

DBMS catalog stores following kinds of information:

Staff, Students, ... are relations owned by some user
id, name ... are fields of the Staff relation
id has type INTEGER and contains a unique value
there are 1000 tuples in Staff, 20000 tuples in Students, ...
Enrolments.sid is a foreign key referencing Students.id
a Student is associated to <40 Subjects via Enrolments(?)

... Catalog and Parsing 18/154

The schema/type information is used for

checking that the named tables and attributes exist
resolving attribute references (e.g. is id from Students or Subjects?)
checking that attrs are used appropriately (e.g. not id='John')

The statistical information is used for

choosing appropriate operators for query execution plans

Examples:

a query with exactly one solution:
select * from Students where id=12345
a query with thousands of solutions:
select * from Students where degree='BSc'

Query Blocks 19/154

A query block is an SQL query with

no nesting
exactly one SELECT, FROM clause
at most one WHERE, GROUP-BY, HAVING clause

Query optimisers typically deal with one query block at a time ...

⇒ SQL compilers need to decompose queries into blocks

Interesting kinds of blocks:

non-correlated query returning a set of tuples/values
non-correlated query returning a single result (treat as constant)
correlated sub-query (query changes for each outer tuple)

... Query Blocks 20/154

Consider the following example query:

SELECT s.name FROM Students s
WHERE  s.avg = (SELECT MAX(avg) FROM Students)

which consists of two blocks:

Block1: SELECT MAX(avg) FROM Students
Block2: SELECT s.name FROM Students s
        WHERE  s.avg = <<Block1>>>

Query processor arranges execution of each block separately and transfers result of Block1 to Block2.

Mapping SQL to Relational Algebra 21/154

A naive query compiler might use the following translation scheme:

SELECT clause → projection
FROM clause → product
WHERE clause → selection

Example:

SELECT s.name, e.subj
FROM   Students s, Enrolments e
WHERE  s.id = e.sid AND e.mark > 50;

is translated to

π_{s.name,e.subj}( σ_{s.id=e.sid ∧ e.mark>50} ( Students × Enrolments ) )

... Mapping SQL to Relational Algebra 22/154

A better translation scheme would be something like:

SELECT clause → projection
WHERE clause on single relation → selection
WHERE clause on two relations → join

Example:

SELECT s.name, e.subj
FROM   Students s, Enrolments e
WHERE  s.id = e.sid AND e.mark > 50;

is translated to

π_{s.name,e.subj}( σ_e.mark>50 ( Students ⋈_s.id=e.sid Enrolments ) ) )

... Mapping SQL to Relational Algebra 23/154

In fact, many SQL compilers ...

produce the cross-product version as an intermediate result
map it into the join version by applying rewriting rules

This is one instance of a general query translation process

expression rewriting via algebraic laws on RA expressions

Aim of rewriting: convert a given RA expression

into an equivalent RA expression
that is guaranteed/likely to be more efficient

(Rewriting is discussed later)

Mapping Rules 24/154

Mapping from an SQL query to an RA expression requires:

a collection of templates for particular kinds of queries
a matching process to ...
- determe what kind of query we have (i.e. choose a template)
- bind components of actual query to slots in the template
mapping rules to ...
- convert the matched query into relational algebra
- filling slots in RA expression from matched components

May need to use multiple templates to map whole SQL statement.

... Mapping Rules 25/154

Projection:

SELECT f₁, f₂, ... f_n FROM ...

⇒ Project_{[f₁, f₂, ... f_n]}(...)

SQL projection extends RA projection with renaming and assignment:

SELECT a+b AS x, c AS y FROM R ...

⇒ Project_{[x ← a+b, y ← c]}(R)

... Mapping Rules 26/154

Join: (e.g. on R(a,b,c,d) and S(c,d,e))

SELECT ... FROM ... R, S ... WHERE ... R.a op S.e ... , or

SELECT ... FROM ... R JOIN S ON (R.a op S.e) ... WHERE ...

⇒ Join_{[R.a op S.e]}(R,S)

SELECT ... FROM ... R NATURAL JOIN S, or

SELECT ... FROM ... R JOIN S USING (c,d) ... WHERE ...

⇒ Proj_[a,b,e](Join_{[R.c=S.c ∧ R.d=S.d]}(R,S))

... Mapping Rules 27/154

Selection:

SELECT ... FROM ... R ... WHERE ... R.f op val ...

⇒ Select_{[R.f op val]}(R)

SELECT ... FROM ... R ... WHERE ... Cond_1,R AND Cond_2,R ...

⇒    Select_{[Cond_1,R ∧ Cond_2,R]}(R)
or
⇒    Select_{[Cond_1,R]}(Select_{[Cond_2,R]}(R))
or
⇒    Select_{[Cond_2,R]}(Select_{[Cond_1,R]}(R))

... Mapping Rules 28/154

Aggregation operators (e.g. MAX, SUM, ...):

add as new operators in extended RA
e.g. SELECT MAX(age) FROM ... ⇒ max(Project_[age](...))
incorporate into projection operator
e.g. SELECT MAX(age) FROM ... ⇒ Project_[age](max,...)
add new projection operators
e.g. SELECT MAX(age) FROM ... ⇒ ProjectMax_[age](...)

... Mapping Rules 29/154

Sorting (ORDER BY):

add Sort operator into extended RA

Duplicate elimination (DISTINCT):

add Uniq operator into extended RA (e.g. Uniq(Project(...)))
or, extend RA ops with uniq parameter (e.g. Project(uniq,...)))

Grouping (GROUP BY, HAVING):

add operators into extended RA (e.g. GroupBy, GroupSelect )

... Mapping Rules 30/154

View definitions produce:

mapping of view statement to RA expression
association between view name and RA expression

Example: assuming Employee(id,name,birthdate,salary)

create view OldEmps as
select * from Employees
where birthdate < '01-01-1960';

yields

OldEmps = Select_{[birthdate<'01-01-1960']}(Employees)

... Mapping Rules 31/154

General case:

CREATE VIEW V AS SQLstatement

⇒ V = mappingOf(SQLstatement)

Special case: views with attribute renaming:

CREATE VIEW V(a,b,c) AS SELECT h,j,k FROM R WHERE C ...

⇒ V = Proj_{[a ← h, b ← j, c ← k]}(Select_[C](R))

... Mapping Rules 32/154

Views used in queries:

references to views are replaced by the view definition
introduces a new subexpression into the RA expression
may require renaming of some attributes

Example:

select name from OldEmps; -- using OldEmps as defined above

⇒ Proj_name(mappingOf(OldEmps))

⇒ Proj_name(Select_{[birthdate<'01-01-1960']}(Employees))

Mapping Example 33/154

The query

List the names of all subjects with more than 100 students in them

can be expressed in SQL as

select   distinct s.code
from     Course c, Subject s, Enrolment e
where    c.id = e.course and c.subject = s.id
group by s.id
having   count(*) > 100;

... Mapping Example 34/154

In the SQL compiler, the query

select   distinct s.code
from     Course c, Subject s, Enrolment e
where    c.id = e.course and c.subject = s.id
group by s.id
having   count(*) > 100;

might be translated to the relational algebra expression

Uniq(Project_[code](
    GroupSelect_{[groupSize>100]}(
        GroupBy_[id] (
            Enrolment ⋈ Course ⋈ Subjects
))))

... Mapping Example 35/154

The join operations could be done in two different ways:

[Diagram:Pics/qproc/join-trees-small.png]

The query optimiser determines which has lower cost.

Note: for a join involving n tables, there are O(n!) possible trees.

Expression Rewriting Rules 36/154

Since RA is a well-defined formal system

there exist many algebraic laws on RA expressions
which can be used as a basis for expression rewriting
in order to produce equivalent (more-efficient) expressions

Expression transformation based on such rules can be used

to simplify/improve SQL→RA mapping results
to generate new plan variations during query optimisation

Relational Algebra Laws 37/154

Commutative and Associative Laws:

R × S ↔ S × R, (R × S) × T ↔ R × (S × T)
R ⋈ S ↔ S ⋈ R, (R ⋈ S) ⋈ T ↔ R ⋈ (S ⋈ T) (natural join)
R ∪ S ↔ S ∪ R, (R ∪ S) ∪ T ↔ R ∪ (S ∪ T)
R ∩ S ↔ S ∩ R, (R ∩ S) ∩ T ↔ R ∩ (S ∩ T)
R ⋈_Cond S ↔ S ⋈_Cond R (theta join)

But it is not true in general that

(R ⋈_Cond1 S) ⋈_Cond2 T ↔ R ⋈_Cond1 (S ⋈_Cond2 T)

Example: R(a,b), S(b,c), T(c,d), (R Join_[R.b>S.b] S) Join_[a<d] T

Cannot rewrite as R Join_[R.b>S.b] (S Join_[a<d] T) because neither S.a nor T.a exists.

... Relational Algebra Laws 38/154

Selection commutativity (where c is a condition):

σ_c ( σ_d (R)) ↔ σ_d ( σ_c (R))

Selection splitting (where c and d are conditions):

σ_{c ∧ d}(R) ↔ σ_c ( σ_d (R))
σ_{c ∨ d}(R) ↔ σ_c(R) ∪ σ_d(R) (but only if R is a set)

Selection pushing ( σ_c(R op S) ):

σ_c(R ∪ S) ↔ σ_cR ∪ σ_cS
(must be pushed into both arguments of union)
σ_c(R - S) ↔ σ_cR - S, σ_c(R - S) ↔ σ_cR - σ_cS
(must be pushed into left branch of difference, may be pushed to right branch)

... Relational Algebra Laws 39/154

Selection pushing with join, cross-product and intersection ...

If c refers only to attributes from R:

σ_c (R ⋈ S) ↔ σ_c(R) ⋈ S (similarly for × and ∩)

If c refers only to attributes from S:

σ_c (R ⋈ S) ↔ R ⋈ σ_c(S) (similarly for × and ∩)

If c refers to attributes from both R and S:

σ_c (R ⋈ S) ↔ σ_c(R) ⋈ σ_c(S)
above is always true for ∩ (union-compatible)

... Relational Algebra Laws 40/154

Rewrite rules for projection ...

All but last projection can be ignored:

π_L1 ( π_L2 ( ... π_Ln (R))) ↔ π_L1 (R)

Projections can be pushed into joins:

π_L (R ⋈_c S) ↔ π_L ( π_M(R) ⋈_c π_N(S) )

where

M and N must contain all attributes needed for c
M and N must contain all attributes used in L (L ⊂ M∪N)

Mapping Subqueries 41/154

Two varieties of sub-query: (sample schema R(a,b), S(c,d))

independent: sub-query makes no reference to outer query

select * from R
where a in (select c from S where d>5)

select * from R
where a = (select max(c) from S)

correlated: sub-query depends on data from outer query

select * from R
where a in (select c from S where d=R.b)

Standard strategy: convert to a join.

... Mapping Subqueries 42/154

Example of mapping independent subquery ...

select * from R
where a in (select c from S where d>5)

is mapped as

⇒ Sel_{[a in Proj_[c](Sel_[d>5]S)]}(R)

⇒ Sel_[a=c](R × Proj_[c](Sel_[d>5](S)))

⇒ R Join_[a=c] Proj_[c](Sel_[d>5](S))

Query Optimisation

Query Optimisation 44/154

Query optimiser: RA expression → efficient evaluation plan

... Query Optimisation 45/154

Query optimisation is a critical step in query evaluation.

The query optimiser

takes a relational algebra expression from SQL compiler
produces a sequence of RelOps to evaluate the expression
the query execution plan yields an efficient evaluation

"Optimisation" is necessary because there can be enormous differences in costs between different plans for evaluating a given RA expression.

E.g. tmp ← A × B; res ← σ_x(tmp) vs res ← A ⋈_x B

Query Evaluation Example 46/154

Example database schema (multi-campus university):

Employee(eid, ename, status, city)
Department(dname, city, address)
Subject(sid, sname, syllabus)
Lecture(subj, dept, empl, time)

Example database instance statistics:

Relation	r	R	C	b
`Employee`	1000	100	10	100
`Department`	100	200	5	20
`Subject`	500	95	10	100
`Lecture`	2000	100	10	200

... Query Evaluation Example 47/154

Query: Which depts in Sydney offer database subjects?


select dname
from   Department D, Subject S, Lecture L
where  D.city = 'Sydney' and S.sname like '%Database%'
       and D.dname = L.dept and S.sid = L.subj

Additional information (needed to determine query costs):

5 departments are located in Sydney
80 subjects are about databases
300 lectures are on databases
100 lectures are in Sydney
3 of these are on databases

... Query Evaluation Example 48/154

Consider total I/O costs for five evaluation strategies.

Assumptions in computing costs:

database tables are unsorted and have no indexes
intermediate tables are written to disk and then re-read

These are worst-case scenarios and could be improved by

having indexes on database tables
writing intermediate results as e.g. sorted/hashed

... Query Evaluation Example 49/154

Strategy #1 for answering query:

TMP1 ← Subject × Lecture
TMP2 ← TMP1 × Department
TMP3 ← Select[check](TMP2)
check = (city='Sydney' & sname='Databases' & dname=dept & sid=subj)
RESULT ← Project[dname](TMP3)

... Query Evaluation Example 50/154

Costs involved in using strategy #1:

Reln	r	R	C	b
`TMP1`	1000000	195	5	20000
`TMP2`	100000000	395	2	50000000
`TMP3`	3	395	2	2
`RESULT`	3	100	10	1

Total I/Os
= Cost_Step1 + Cost_Step2 + Cost_Step3 + Cost_Step4
= (100+100*200+20000) + (20+20*20000+5*10⁶) + (5*10⁶+2) + 2
= 100,440,124

... Query Evaluation Example 51/154

Strategy #2 for answering query:

TMP1 ← Join[sid=subj](Subject,Lecture)
TMP2 ← Join[dept=dname](TMP1,Department)
TMP3 ← Select[city='Sydney' & sname='Databases'](TMP2)
RESULT ← Project[dname](TMP3)

... Query Evaluation Example 52/154

Costs involved in using strategy #2:

Reln	r	R	C	b
`TMP1`	2000	195	5	400
`TMP2`	2000	395	2	1000
`TMP3`	3	395	2	2
`RESULT`	3	100	10	1

Total I/Os
= Cost_Step1 + Cost_Step2 + Cost_Step3 + Cost_Step4
= (100+100*200+400) + (20+20*400+1000) + (1000+2) + 2
= 30,524

... Query Evaluation Example 53/154

Strategy #3 for answering query:

TMP1 ← Join[sid=subj](Subject,Lecture)
TMP2 ← Select[sname='Databases'](TMP1)
TMP3 ← Join[dept=dname](TMP2,Department)
TMP4 ← Select[city='Sydney'](TMP3)
RESULT ← Project[dname](TMP4)

... Query Evaluation Example 54/154

Costs involved in using strategy #3:

Reln	r	R	C	b
`TMP1`	2000	195	5	400
`TMP2`	20	195	5	4
`TMP3`	20	395	2	10
`TMP4`	3	395	2	2
`RESULT`	3	100	10	1

Total I/Os
= Cost_Step1 + Cost_Step2 + Cost_Step3 + Cost_Step4 + Cost_Step5
= (100+100*200+400) + (400+4) + (4+4*20+10) + (10+2) + 1
= 21,011

... Query Evaluation Example 55/154

Strategy #4 for answering query:

TMP1 ← Select[sname='Databases'](Subject)
TMP2 ← Select[city='Sydney'](Department)
TMP3 ← Join[sid=subj](TMP1,Lecture)
TMP4 ← Join[dept=dname](TMP3,TMP2)
RESULT ← project[dname](TMP4)

... Query Evaluation Example 56/154

Costs involved in using strategy #4:

Reln	r	R	C	b
`TMP1`	80	95	5	16
`TMP2`	5	200	5	1
`TMP3`	300	195	5	60
`TMP4`	3	395	2	2
`RESULT`	3	100	10	1

Total I/Os
= Cost_Step1 + Cost_Step2 + Cost_Step3 + Cost_Step4 + Cost_Step5
= (100+16) + (20+1) + (16+16*200+60) + (1+1*60+2) + 2
= 3478

... Query Evaluation Example 57/154

Strategy #5 for answering query:

TMP1 ← Select[sname='Databases'](Subject)
TMP2 ← Select[city='Sydney'](Department)
TMP3 ← Join[dept=dname](TMP2,Lecture)
TMP4 ← Join[sid=subj](TMP3,TMP1)
RESULT ← project[dname](TMP4)

... Query Evaluation Example 58/154

Costs involved in using strategy #5:

Reln	r	R	C	b
`TMP1`	80	95	5	16
`TMP2`	5	200	5	1
`TMP3`	100	295	3	34
`TMP4`	3	395	2	2
`RESULT`	3	100	10	1

Total I/Os
= Cost_Step1 + Cost_Step2 + Cost_Step3 + Cost_Step4 + Cost_Step5
= (100+16) + (20+1) + (1+1*200+34) + (16+16*34+2) + 2
= 936

Query Optimisation Problem 59/154

Given:

a query Q, a database D, a database "engine" E

Determine a sequence of relational algebra operations that:

produces the answer to Q in D
executes Q efficiently on E (minimal I/O)

The term "query optimisation" is a misnomer:

not just for queries (e.g. also updates)
not necessarily optimal ("reasonably efficient")

... Query Optimisation Problem 60/154

Why do we not generate optimal query execution plans?

Finding an optimal query plan ...

requires an exhaustive search of a space of possible plans
for each possible plan, need to estimate cost (not cheap)

Even for a small query (e.g. 4-5 joins, 4-5 selects)

the space of possible query plans is very large
(choices: order of operations, access methods, intermediate results, etc.)
cost of searching plan space ≅ cost of executing query

Compromise:

do limited searching of query plan space (guided by heuristics)
quickly choose a reasonably efficient execution plan

Cost Models and Analysis 61/154

The cost of evaluating a query is determined by:

size of relations (database relations and temporary relations)
access mechanisms (indexing, hashing, sorting, join algorithms)
size/number of main memory buffers (and replacement strategy)

Analysis of costs involves estimating:

size of intermediate results
number of secondary storage accesses

Approaches to Optimisation 62/154

Three main classes of techniques developed:

algebraic (equivalences, rewriting, heuristics)
physical (execution costs, search-based)
semantic (application properties, heuristics)

All driven by aim of minimising (or at least reducing) "cost".

Real query optimisers use a combination of algrebraic+physical.

Semantic QO is good idea, but expensive/difficult to implement.

Optimisation Process 63/154

Start with RA tree representation of query, then ...

apply algebraic query transformations
- giving standardised, simpler, more efficient RA tree
generate possible access plans (physical)
- replace each RA operation by specific access method to implement it
- consider possible orders of join operations (left-deep trees)
analyse cost of generated plans and select cheapest

Result: tree of DBMS operations to answer query efficiently.

... Optimisation Process 64/154

Example of optimisation transformations:

[Diagram:Pics/qproc/qtransform-small.png]

For join, may also consider sort/merge join and hash join.

Algebraic Optimisation 65/154

Make use of algebraic equivalences:

examine query expression
search for applicable transformation rules (heuristics)
generate equivalent (and "better") algebraic expressions

Most commonly used heuristic:

Apply Select and Project before Join

Rationale: minimises size of intermediate relations.

Can potentially be done during the SQL→RA mapping phase.

Algebraic optimisation cannot assist with finding good join order.

Physical Optimisation 66/154

Makes use of execution cost analysis:

examine query evaluation plan
determine efficient join sequences
select access method for each operation (e.g. index for select)
for distributed DB, select best sites (closest, best bandwidth)
determine total cost for evaluation plan
repeat for all possible plans and choose best

Physical optimisation is also called query evaluation plan generation.

Semantic Optimisation 67/154

Make use of application-specific properties:

functional dependencies
attributes constraints
tuple constraints
database constraints

Can be applied in algebraic or physical optimisation phase.

Basis: exploit meta-data and other semantic info about relations.

(E.g. this field is a primary key, so we know there'll only be one matching tuple)

Stages in Algebraic Optimisation 68/154

Start with relational algebra (RA) expression.

Standardise
- construct normal form of expression (boolean algebra laws)
Simplify
- transform to eliminate redundancy (boolean algebra and RA laws)
Ameliorate
- transform to improve efficiency (RA laws and heuristics)

Result: an RA expression equivalent to the input, but more efficient.

Standardisation 69/154

Many query optimisers assume RA expression is in a "standard form".

E.g. select conditions may be stated in

conjunctive normal form
(A1 AND ... AND An) OR ... OR (Z1 AND ... AND Zm)
disjunctive normal form
(A1 OR ... OR An) AND ... AND (Z1 OR ... OR Zm)

RA expression is first transformed into one of these forms.

Simplification 70/154

Main aim of simplification is to reduce redundancy.

Makes use of tranformation rules based on laws of boolean algebra.

`A AND B`	↔	`B AND A`
`A OR (B AND C)`	↔	`(A OR B) AND (A OR C)`
`A AND A`	↔	`A`
`A OR NOT(A)`	↔	`True`
`A AND NOT(A)`	↔	`False`
`A OR (A AND B)`	↔	`A`
`NOT(A AND B)`	↔	`NOT(A) OR NOT(B)`
`NOT(NOT(A))`	↔	`A`

(Programmers don't produce redundant expressions; but much SQL is produced by programs)

Simplification Example 71/154

Consider the query:

select Title from Employee
where  (not (Title = 'Programmer')
        and (Title = 'Programmer'
             or Title = 'Analyst')
        and not (Title = 'Analyst'))
       or Name = 'Joe'

Denote the atomic conditions as follows:

P = (Title = 'Programmer')
A = (Title = 'Analyst')
J = (Name = 'Joe')

... Simplification Example 72/154

Selection condition can then be simplified via:

Cond	=	( P ∧ (P ∨ A) ∧ A) ∨ J
	=	(( P ∧ A) ∧ (P ∨ A)) ∨ J
	=	( (P ∨ A) ∧ (P ∨ A)) ∨ J
	=	False ∨ J
	=	J

Giving the equivalent simplified query:

select Title from Employee where Name = "Joe"

Amelioration 73/154

Aim of amelioration is to improve efficiency.

Transform query to equivalent form which is known to be more efficient.

Achieve this via:

relational algebra laws/transformations
heuristics to choose which to apply

Often describe RA expression as a tree and RA transformations as tree transformations.

Amelioration Process 74/154

break up σ_{a ∧ b ∧ c ...} into ``cascade'' of σ
(gives more flexibility for later transformations)
move σ as far down as possible
(reduces size of intermediate results)
move most restrictive σ down/left
(reduces size of intermediate results)
move π as far down as possible
(reduces size of intermediate results)
replace × then σ by equivalent ⋈
(reduces computation/size of intermediate results)
replace subexpressions by single operation
(e.g. opposite of 1.) (reduces computation overhead)

Amelioration Example 75/154

Consider school information database:

CSG(Course, StudentId, grade)
SNAP(StudentId, name, address, phone)
CDH(Course, Day, Hour)
CR(Course, Room)

And the query:

Where is Brown at 9am on Monday morning?

... Amelioration Example 76/154

An obvious translation of the query into SQL

select Room from CSG,SNAP,CDH,CR
where  name='Brown' and day='Mon' and hour='9am'

and an obvious mapping of the SQL into relational algebra gives

π_Room ( σ_N&D&H ( CSG ⋈ SNAP ⋈ CDH ⋈ CR))

N is name='Brown', D is Day='Mon', H is Hour='9am'

... Amelioration Example 77/154

π_Room ( σ_N&D&H ( CSG ⋈ SNAP ⋈ CDH ⋈ CR))

Push selection:

π_Room ( σ_N&D&H ( CSG ⋈ SNAP ⋈ CDH) ⋈ CR)

Split selection:

π_Room ( σ_N ( σ_D&H ( CSG ⋈ SNAP ⋈ CDH)) ⋈ CR)

Push two selections:

π_Room ( ( σ_N (CSG ⋈ SNAP) ⋈ σ_D&H (CDH)) ⋈ CR)

Push selection:

π_Room ( ((CSG ⋈ σ_N (SNAP)) ⋈ σ_D&H (CDH)) ⋈ CR)

Can we show that the final version is more efficient?

... Amelioration Example 78/154

Could use an argument based on size of intermediate results, as above.

But run into problems with expressions like:

CSG ⋈ SNAP ⋈ CDH ⋈ CR

Order of joins can make a significant difference?

How to decide "best" order?

Need understanding of physical characteristics of relations
(for example, selectivity and likelihood of join matches across tables)

Physical (Cost-Based) Optimisation 79/154

Need to determine:

order in which operations applied (execution plan)
precisely how each operation done (map to DBMS ops)
size of intermediate results (need to estimate these)

From these, estimate overall evaluation cost.

Do this for all possible execution plans.

Choose cheapest plan.

Execution Plans 80/154

Consider query execution plans for the RA expression:

σ_c (R ⋈_d S ⋈_e T)

Plan #1

tmp1   :=  HashJoin[d](R,S)
tmp2   :=  SortMergeJoin[e](tmp1,T)
result :=  BinarySearch[c](tmp2)

... Execution Plans 81/154

Plan #2

tmp1   := SortMergeJoin[e](S,T)
tmp2   := HashJoin[d](R,tmp1)
result := LinearSearch[c](tmp2)

Plan #3

tmp1   := BtreeSearch[c](R)
tmp2   := HashJoin[d](tmp1,S)
result := SortMergeJoin[e](tmp2)

All plans produce same result, but have quite different costs.

Cost-based Query Optimiser 82/154

Overview of cost-based query optimisation process:

start with RA tree from compilation of SQL query
use RA laws as rewrite rules to generate new RA trees
for each node in RA tree, choose specific access method

For a given SQL query, there are

very many possible RA trees (large choice of re-write rules)
many possible execution plans (choice of access methods, params)

Too many combinations to enumerate all; prune via heuristics.

... Cost-based Query Optimiser 83/154

Approximate algorithm for cost-based optimisation:

translate SQL query to RAexp
for all transformations RA' of RAexp {
    for each node e of RA' (recursively) {
        select access method for e
        plan[i++] = access method for e
    }
    cost = 0
    for each op in plan[]
        { cost += Cost(op) }
    if (cost < MinCost) {
        MinCost = cost
        BestPlan = plan
}   }

As noted above, we do not consider *all* transformations.
Heuristics: push selections down, consider only left-deep join trees.

Choosing Access Methods 84/154

Inputs:

a single RA operation (σ, π, ⋈)
information about file organisation, data distribution, ...
list of operations available in the database engine

Output:

calls to database operations to implement this RA operation

... Choosing Access Methods 85/154

Example:

RA operation: σ_{name='John' ∧ age>21}(Student)
Student relation has B-tree index on name
database engine (obviously) has B-tree search method

giving

tmp[i]   := BtreeSearch[name='John'](Student)
tmp[i+1] := LinearSearch[age>21](tmp[i])

Where possible, use pipelining to avoid storing tmp[i] on disk.

... Choosing Access Methods 86/154

Rules for choosing σ access methods:

σ_A=c(R) and R has index on A ⇒ indexSearch[A=c](R)
σ_A=c(R) and R is hashed on A ⇒ hashSearch[A=c](R)
σ_A=c(R) and R is sorted on A ⇒ binarySearch[A=c](R)
σ_{A ≥ c}(R) and R has clustered index on A ⇒ indexSearch[A=c+1](R) then scan
σ_{A ≥ c}(R) and R is hashed on A ⇒ linearSearch[A>=c](R)

... Choosing Access Methods 87/154

Rules for choosing ⋈ access methods:

R ⋈ S and R fits in memory buffers ⇒ bnlJoin(R,S)
R ⋈ S and R,S sorted on join attr ⇒ smJoin(R,S) (merge only)
R ⋈ S and R has index on join attr ⇒ inlJoin(S,R)
R ⋈ S and no indexes, no sorting, |R| ≪ |S| ⇒ hashJoin(R,S)

(bnl = block nested loop; inl = index nested loop; sm = sort merge)

Example Plan Enumeration 88/154

Consider the query:

select name
from   Student s, Enrol e
where  s.sid = e.sid AND e.cid = 'COMP9315';

where

Relation	r	b	PrimKey	Clustered	Indexes
`Student`	10000	500	`sid`	No	B-tree on `sid`
`Enrol`	40000	400	`cid`	Yes	B-tree on `cid`
`S` ⋈ `E`	40000	800	`sid,cid`	No	-

... Example Plan Enumeration 89/154

First step is to map SQL to RA expression

σ₉₃₁₅ ( Student ⋈_sid Enrol )

then devise an execution plan based on this expression

index on Student.sid ⇒ index nested loop join
join result not sorted on Enrol.cid ⇒ linear scan

giving

tmp1   := inlJoin[s.sid=e.sid](Enrol,Student)
result := LinearSearch[e.id='COMP9315'](tmp1)

... Example Plan Enumeration 90/154

Estimated cost for this plan:

Cost = inl-join on (Enrol,Student) + linear scan of tmp1

= T_r(b_S + r_S.btree_E) + T_wb_tmp1 + T_rb_tmp1

= 500 + 10,000.4 + 800 + 800

= 42,100

Notes:

if we assume pipelining on tmp1, can remove (800+800) term
if we make Enrol the outer relation, join cost is 400 + 40,000.3

... Example Plan Enumeration 91/154

Apply rule for pushing select into join:

σ₉₃₁₅ ( Student ⋈_sid Enrol )

giving

Student ⋈_sid σ₉₃₁₅ ( Enrol )

and devise a plan based on

index on Enrol.cid ⇒ B-tree lookup
σ₉₃₁₅ output small ⇒ buffered nested loop join

giving

tmp1   := BtreeSearch[e.cid='COMP9315'](Enrol)
result := bnlJoin[s.sid=e.sid](tmp1,Student)

... Example Plan Enumeration 92/154

Estimated cost for this plan:

Cost = btree-search on Enrol + bnl-join of (tmp1,Student)

= search+scan on Enrol + T_r(b_tmp1 + b_S)

= 3T_r + 3T_r + T_r(1 + 500)

= 3 + 3 + 1 + 500 = 507

Notes:

assumes 200 students enrolled in COMP9315
with pipelining 1T_r would vanish
if Enrol no clustered, 3T_r scan would increase

Cost Estimation 93/154

Without actually executing it, cannot always know the cost of plan precisely.

(E.g. how big is the select result that feeds into the join?)

Thus, query optimisers need to estimate costs.

Two aspects to cost estimation:

cost of performing operation (dealt with extensively in earlier lectures)
size of result (which affects cost of performing next operation)

Cost Estimation Information 94/154

Cost estimation uses statistical information about relations:

r_S	cardinality of relation S
R_S	avg size of tuple in relation S
V(A,S)	# distinct values of attribute A
min(A,S)	min value of attribute A
max(A,S)	max value of attribute A

Estimating Projection Result Size 95/154

Straightforward, since we know:

all tuples from input table are included in result
all required attributes (and their sizes) ⇒ R_out

#bytes in result = r_S × R_out

If pipelining through buffers of size B, then

#buffers-full = r_S × ⌈B/R_out⌉

Estimating Selection Result Size 96/154

Selectivity factor = fraction of tuples expected to satisfy a condition.

Common assumption: attribute values uniformly distributed.

Example: Consider the query

select * from Parts where colour='Red'

and assume that there are only four possible colours.

If we have 1000 Parts, we would expect 250 of them to be Red.

In other words,

V(colour,R)=4, r_R=1000 ⇒ |σ_colour=K(R)|=250

In general, |σ_A=c(R)| ≅ r_R/V(A,R)

... Estimating Selection Result Size 97/154

Estimating size of result for e.g.

select * from Enrolment
where  year > 2003;

Could estimate by using:

uniform distribution assumption, r, min/max years

Assume: min(year)=1996, max(year)=2005, |Enrolment|=10⁵

10⁵ from 1996-2005 means approx 10000 enrolments/year
this suggests 20000 enrolments since 2003

Of course, we know that enrolments grow continually, so this underestimates.

Simpler heuristic used by some systems: selected tuples = r/3

... Estimating Selection Result Size 98/154

Estimating size of result for e.g.

select * from Enrolment
where  course <> 'COMP9315';

Could estimate by using:

uniform distribution assumption, r, #courses

Alternative, simpler way to estimate:

assume that most tuples are not equal to chosen value
# selected tuples = r

... Estimating Selection Result Size 99/154

Uniform distribution assumption is convenient, but often not realistic.

How to handle non-uniform attribute value distributions?

collect statistics about the values stored in the attribute/relation
store these as e.g. a histogram in the meta-data for the relation

So, for the part colour example, we might have a distribution like:

White: 35% Red: 30% Blue: 25% Silver: 10%

Use histogram as basis for determining # selected tuples.

Disadvantage: cost of storing/maintaining histograms.

Estimating Join Result Size 100/154

Analysis is not as simple as select.

Relies on semantic knowledge about data/relations.

Consider equijoin on common attr: R ⋈_A S

Case 1: |R ⋈_A S| = 0

Useful if we know that dom(A,R) ∩ dom(A,S) = {}

Case 2: A is unique key in both R and S
⇒ can't be more tuples in join than in smaller of R and S.

max(|R ⋈_A S|) = min(|R|, |S|)

... Estimating Join Result Size 101/154

Case 3: A is primary key in R, foreign key in S
⇒ every S tuple has at most one matching R tuple.

max(|R ⋈_A S|) = |S|

Example:

select name from Students s, Enrol e
where  e.cid = 'COMP9315' AND e.sid = s.sid

|Students ⋈ σ₉₃₁₅(Enrol)| = |σ₉₃₁₅(Enrol)|

Cost Estimation: Postscript 102/154

Inaccurate cost estimation can lead to poor evaluation plans.

Above methods can (sometimes) give inaccurate estimates.

To get more accurate estimates costs:

more time ... complex computation of selectivity
more space ... storage for histograms of data values

Either way, optimisation process costs more (more than query?)

Trade-off between optimiser performance and query performance.

Semantic Query Optimisation (SQO) 103/154

Makes use of semantics of data to assist query optimisation process.

The discussion of join cost estimation above gives an example of this.

Kinds of information used:

knowledge about relations
nature of data
constraints within/between attributes/relations

SQO uses this to simplify queries and reduce search space.

Two examples of semantic transformation operations:

restriction elimination, index introduction

Semantic Equivalence 104/154

Semantic equivalence does not require syntactic equivalence.

Consider the two queries:

select * from Emp where sal > 80K

select * from Emp where sal > 80K and job = 'Prof'

They are equivalent under the semantic rule:

Emp.job = 'Prof' ↔ Emp.sal > 80K

(i.e. Profs are the only people earning more than $80,000)

Could use this to transform query to exploit index on salary.

Restriction Elimination 105/154

Query:

List all the departments that store benzene in quantities of more than 400

select dept from  Storage
where material = 'Benzene' and qty > 400

Use rule: material = 'Benzene' ⇒ qty > 500

select dept from Storage where material = 'Benzene'

Result: Unnecessary restriction on the attribute qty is eliminated.

Index Introduction 106/154

Query: Find all the employees who make more than 42K

select name from Employees where salary > 42K

Use rule: salary > 42K ⇒ job = 'manager'

select name from Employees
where salary > 42K and job = 'manager'

Result: A new constraint is obtained on the indexed attribute job.

Query Execution

Query Execution 108/154

Query execution: applies evaluation plan → set of result tuples

... Query Execution 109/154

Query execution

applies a query execution plan
and produces a set of result tuples

... Query Execution 110/154

Example of query translation:

select s.name, s.id, e.course, e.mark
from   Student s, Enrolment e
where  e.student = s.id and e.semester = '05s2';

maps to

π_{name,id,course,mark}(Stu ⋈_{e.student=s.id} (σ_{semester=05s2}Enr))

maps to

Temp1  = BtreeSelect[semester=05s2](Enr)
Temp2  = HashJoin[e.student=s.id](Stu,Temp1)
Result = Project[name,id,course,mark](Temp2)

... Query Execution 111/154

A query execution plan:

consists of a sequence of operations
each operation is a relational algebra operator
(a specific implementation with particular performance characteristics)

Results may be passed from one operator to the next:

materialization ... writing results to disk and reading them back
pipelining ... generating and passing results one-at-a-time

... Query Execution 112/154

Two ways of communicating results between query blocks ...

Materialization

first block writes all results to disk
next block reads tuples from disk to process
advantage: can exploit file structures (e.g. hashing)

Pipelining

blocks execute "concurrently" as producer/consumer pairs
structured as interacting iterators (open; while(next); close)
advantage: no requirement for disk access

Materialization 113/154

Steps in materialization between two operators

first operator reads input(s) and writes results to disk
next operator treats tuples results on disk as its input

Advantage:

intermediate results can be placed in a file structure
(which can be chosen to speed up execution of subsequent operators)

Disadvantage:

requires disk space/writes for intermediate results
requires disk access to read intermediate results

... Materialization 114/154

Example:

select s.name, s.id, e.course, e.mark
from   Student s, Enrolment e
where  e.student = s.id and
       e.semester = '05s2' and s.name = 'John';

might be executed as

Temp1  = BtreeSelect[semester=05s2](Enrolment)
Temp2  = BtreeSelect[name=John](Student)
         -- indexes on name and semester
	 -- produce sorted Temp1 and Temp2
Temp3  = SortMergeJoin[e.student=s.id](Temp1,Temp2)
         -- SMJoin especially effective, since
         -- Temp1 and Temp2 are already sorted
Result = Project[name,id,course,mark](Temp3)

Pipelining 115/154

How pipelining is organised between two operators:

blocks execute "concurrently" as producer/consumer pairs
first operator acts as producer; second as consumer
structured as interacting iterators (open; while(next); close)

Advantage:

no requirement for disk access (results passed via memory buffers)

Disadvantage:

each operator accesses inputs via linear scan

Iterators (reminder) 116/154

Iterators provide a "stream" of results:

iter = open(params)
- set up data structures for iterator (create state, open files, ...)
- params are specific to operator (e.g. reln, condition, #buffers, ...)
tuple = next(iter)
- get the next tuple in the iteration; return null if no more
close(iter)
- clean up data structures for iterator

Other possible operations: reset to specific point, restart, ...

... Iterators (reminder) 117/154

Implementation of single-relation selection iterator:

typedef struct {
    File   inf;  // input file
    Cond   cond; // selection condition
    Buffer buf;  // buffer holding current page
    int    curp; // current page during scan
    int    curr; // index of current record in page
} Iterator;

Iterator structure contains information:

related to operation being performed (e.g. cond)
information giving current execution state (e.g. curp, curr)

... Iterators (reminder) 118/154

Implementation of single-relation selection iterator (cont):

Iterator *open(char *relName, Condition cond) {
    Iterator *iter = malloc(sizeof(Iterator));
    iter->inf  = openFile(fileName(relName),READ);
    iter->cond = cond;
    iter->curp = 0;
    iter->curr = -1;
    readBlock(iter->inf, iter->curp, iter->buf);
    return iter;
}
void close(Iterator *iter) {
    closeFile(iter->inf);
    free(iter);
}

... Iterators (reminder) 119/154

Implementation of single-relation selection iterator (cont):

Tuple next(Iterator *iter) {
    Tuple rec;
    do {
        // check if reached end of current page
        if (iter->curr == nRecs(iter->buf)-1) {
            // check if reached end of data file
    	    if (iter->curp == nBlocks(iter->inf)-1)
    	        return NULL;
    	    iter->curp++;
            iter->buf = readBlock(iter->inf, iter->curp);
            iter->curr = -1;
        }
        iter->curr++;
        rec = getRec(iter->buf, iter->curr);
    } while (!matches(rec, iter->cond));
    return rec;
}
// curp and curr hold indexes of most recently read page/record

Pipelining Example 120/154

Consider the query:

select s.id, e.course, e.mark
from   Student s, Enrolment e
where  e.student = s.id and
       e.semester = '05s2' and s.name = 'John';

which maps to the RA expression

Proj[id,course,mark](Join[student=id](Sel[05s2](Enr),Sel[John](Stu)))

which could represented by the RA expression tree

... Pipelining Example 121/154

Modelled as communication between RA tree nodes:

... Pipelining Example 122/154

This query might be executed as

System:
    iter0 = open(Result)
    while (Tup = next(iter0)) { display Tup }
    close(iter0)
Result:
    iter1 = open(Join)
    while (T = next(iter1))
        { T' = project(T); return T' }
    close(iter1)
Sel1:
    iter4 = open(Btree(Enrolment,'semester=05s2'))
    while (A = next(iter4)) { return A }
    close(iter4)
...

... Pipelining Example 123/154

...
Join: -- nested-loop join
    iter2 = open(Sel1)
    while (R = next(iter2) {
        iter3 = open(Sel2)
        while (S = next(iter3))
            { if (matches(R,S) return (RS) }
        close(iter3) // better to reset(iter3)
    }
    close(iter2)
Sel2:
    iter5 = open(Btree(Student,'name=John'))
    while (B = next(iter5)) { return B }
    close(iter5)

Pipeline Execution 124/154

Piplines can be executed as ...

Demand-driven

producers wait until consumers request tuples

Producer-driven

producers generate tuples until output buffer full, then wait

In both cases, top-level driver is request for result tuples.

In parallel-processing systems, iterators could run concurrently.

Disk Accesses 125/154

Pipelining cannot avoid all disk accesses.

Some operations use multiple passes (e.g. merge-sort, hash-join).

data is written by one pass, read by subsequent passes

Thus ...

within an operation, disk reads/writes are possible
between operations, no disk reads/writes are needed

... Disk Accesses 126/154

Sophisticated query optimisers might realise e.g.

if operation X writes its results to a file with structure S,
the subsequent operation Y will proceed much faster
than if Y reads X's output tuple-at-a-time

In this case, it could materialize X's output in an S-file.

Produces a pipeline/materialization hybrid query execution.

Example:

selection writes output into an indexed file (Btree)
later join can then be implemented as efficient index-join

... Disk Accesses 127/154

Example: (pipeline/materialization hybrid)

select s.id, e.course, e.mark
from   Student s, Enrolment e
where  e.student = s.id and
       e.semester = '05s2' and s.name = 'John';

might be executed as

System:
    exec(Sel2)  -- creates Temp1
    iter0 = open(Result)
    while (Tup = next(iter0)) { display Tup }
    close(iter0)
Result:
    iter1 = open(Join)
    while (T = next(iter1))
        { T' = project(T); return T' }
    close(iter1)
...

... Disk Accesses 128/154

...
Join: -- index join
    iter2 = open(Sel1)
    while (R = next(iter2) {
        iter3 = open(Btree(Temp1,'id=R.student'))
        while (S = next(iter3)) { return (RS) }
        close(iter3)
    }
    close(iter2)
Sel1:
    iter4 = open(Btree(Enrolment,'semester=05s2'))
    while (A = next(iter4)) { return A }
    close(iter4)
Sel2:
    iter5 = open(Btree(Student,'name=John'))
    createBtree(Temp1,'id')
    while (B = next(iter5)) { insert(B,Temp1) }
    close(iter5)

PostgreSQL Execution 129/154

Defs: src/include/executor and src/include/nodes

Code: src/backend/executor

PostgreSQL uses pipelining ...

query plan is a tree of Plan nodes
each type of node implements one kind of RA operation
(node implements specific access methods and provides iterator interface)
node types e.g. Scan, Group, Indexscan, Sort, HashJoin
execution is managed via a tree of PlanState nodes
(mirrors the structure of the tree of Plan nodes; holds execution state)

... PostgreSQL Execution 130/154

Modules in src/backend/executor fall into two groups:

execXXX (e.g. execMain, execProcnode, execScan)

implement generic control of plan evaluation (execution)
provide overall plan execution and dispatch to node iterators

nodeXXX (e.g. nodeSeqscan, nodeNestloop, nodeGroup)

implement iterators for specific types of RA operators
typically contains ExecInitXXX, ExecXXX, ExecEndXXX

The "style" is OO (e.g. specialisations of Nodes), but implementation in C masks this

... PostgreSQL Execution 131/154

Top-level data/functions for executor ...

QueryDesc

contains plan and state information (e.g. pointer to root of plan tree)

ExecutorStart(QueryDesc *, ...)

initialises all dynamic state information (e.g. iterators)

ExecutorRun(QueryDesc *, ScanDirection, ...)

implements "get next result tuple" (via iterator tree)

ExecutorEnd(QueryDesc *)

cleans up all iterator and state information

... PostgreSQL Execution 132/154

Overview of query processing:

CreateQueryDesc

ExecutorStart
    CreateExecutorState -- creates per-query context
    switch to per-query context to run ExecInitNode
    ExecInitNode -- recursively scans plan tree
        CreateExprContext -- creates per-tuple context
        ExecInitExpr

ExecutorRun
    ExecutePlan -- invoke iterators from root
        ExecProcNode -- recursively called in per-query context
            ExecEvalExpr -- called in per-tuple context
            ResetExprContext -- to free memory

ExecutorEnd
    ExecEndNode -- recursively releases resources
    FreeExecutorState -- frees per-query and child contexts

FreeQueryDesc

... PostgreSQL Execution 133/154

More detailed view of plan execution (but still much simplified)

ExecutePlan(execState, planStateNode, ...) {
    process "before each statement" triggers
    for (;;) {
        tuple = ExecProcNode(planStateNode)
        check tuple validity // MVCC
        if (got a tuple) break
    }
    process "after each statement" triggers
    return tuple
}
ExecProcNode(node) {
    switch (nodeType(node)) {
    case SeqScan:
        result = ExecSeqScan(node); break;
    case NestLoop:
        result = ExecNestLoop(node); break;
    ...
    }
    return result;
}

... PostgreSQL Execution 134/154

Generic iterator interface is provided by ...

ExecInitNode

initialize a plan node and its subplans

ExecProcNode

get a tuple by executing the plan node

ExecEndNode

shut down a plan node and its subplans

Each calls corresponding function for specific node type
(e.g. for nested loop join ExecInitNestLoop(), ExecNestLoop(), ExecEndNestLoop())

Example PostgreSQL Execution 135/154

Consider the query:

-- get manager's age and # employees in Shoe department
select e.age, d.nemps
from   Departments d, Employees e
where  e.name = d.manager and d.name ='Shoe'

and its execution plan tree

... Example PostgreSQL Execution 136/154

This produces a tree with three nodes:

NestedLoop with join condition (Outer.manager = Inner.name)
IndexScan on Departments with selection (name = 'Shoe')
SeqScan on Employees

We ignore the top-level node here (it handles the projection via attrList)

... Example PostgreSQL Execution 137/154

Initially InitPlan() invokes ExecInitNode() on plan tree root.


ExecInitNode() sees a NestedLoop node ... 
    so dispatches to ExecInitNestLoop() to set up iterator
    and then invokes ExecInitNode() on left and right sub-plans
        in left subPlan, ExecInitNode() sees an IndexScan node
            so dispatches to ExecInitIndexScan() to set up iterator
        in right sub-plan, ExecInitNode() sees aSeqScan node
            so dispatches to ExecInitSeqScan() to set up iterator

Result: a plan state tree with same structure as plan tree.

... Example PostgreSQL Execution 138/154

Execution: ExecutePlan() repeatedly invokes ExecProcNode().


ExecProcNode() sees a NestedLoop node ...
    so dispatches to ExecNestedLoop() to get next tuple
    which invokes ExecProcNode() on its sub-plans
        in the left sub-plan, ExecProcNode() sees an IndexScan node
            so dispatches to ExecIndexScan() to get next tuple
            if no more tuples, return END
            for this tuple, invoke ExecProcNode() on right sub-plan
                ExecProcNode() sees a SeqScan node
                    so dispatches to ExecSeqScan() to get next tuple
                check for match and return joined tuples if found
            reset right sub-plan iterator

Result: stream of result tuples returned via ExecutePlan()

Performance Tuning

Performance Tuning 140/154

Schema design:

devise data structures to represent application information

Performance tuning:

devise data structures to achieve good performance

Good performance may involve any/all of:

making applications run faster
lowering response time of queries/transactions
improving overall transaction throughput

... Performance Tuning 141/154

Tuning requires us to consider the following:

which queries and transactions will be used?
(e.g. check balance for payment, display recent transaction history)
how frequently does each query/transaction occur?
(e.g. 90% withdrawals; 10% deposits; 50% balance check)
are there time constraints on queries/transactions?
(e.g. EFTPOS payments must be approved within 7 seconds)
are there uniqueness constraints on any attributes?
(define indexes on attributes to speed up insertion uniqueness check)
how frequently do updates occur?
(indexes slow down updates, because must update table and index)

... Performance Tuning 142/154

Performance can be considered at two times:

during schema design
- typically towards the end of schema design process
- requires schema transformations such as denormalisation
outside schema design
- typically after application has been deployed/used
- requires adding/modifying data structures such as indexes

Denormalisation 143/154

Normalisation minimises storage redundancy.

achieves this by "breaking up" data into logical chunks
requires minimal "maintenance" to ensure consistency

Problem: queries that need to put data back together.

need to use a (potentially expensive) join operation
if an expensive join is frequent, performance suffers

Solution: store some data redundantly

benefit: queries no longer need expensive joins
trade-off: extra maintenance effort to keep consistency
worthwhile if joins are frequent and updates are rare

... Denormalisation 144/154

Example ... consider the following normalised schema:

create table Subject (
   id        serial primary key,
   code      char(8), -- e.g. COMP9315
   title     varchar(60),
   syllabus  text, ... );
create table Term (
   id        serial primary key,
   name      char(4), -- e.g. 09s2
   starting  date, ... );
create table Course (
   subject   integer references Subject(id),
   term      integer references Term(id),
   lic       integer references Staff(id), ... );

... Denormalisation 145/154

Example: Courses = Course ⋈ Subject ⋈ Term

If we often need to refer to "standard" name (e.g. COMP9315 09s2)

add extra courseName column into Course table
cost: trigger before insert on Course to construct name
trade-off likely to be worthwhile: Course insertions infrequent


-- can now replace a query like:
select s.code||' '||t.name, avg(e.mark)
from   Course c, Subject s, Term t
where  c.subject = s.id and c.term = t.id
       and s.code='COMP9315' and t.name='09s2'
-- by a query like:
select c.courseName, e.grade, e.mark
from   Course c
where  c.courseName = 'COMP9315 09s2'

Indexes 146/154

Indexes provide efficient content-based access to tuples.

Can build indexes on any (combination of) attributes.

Definining indexes:

CREATE INDEX name ON table ( attr₁, attr₂, ... )

attr_i can be an arbitrary expression (e.g. upper(name)).

CREATE INDEX also allows us to specify

that the index is on UNIQUE values
an access method (USING btree, hash, gist, gin)

... Indexes 147/154

Indexes can significantly improve query costs.

Considerations in applying indexes:

is an attribute used in frequent/expensive queries?
(note that some kinds of queries can be answered from index alone)
should we create an index on a collection of attributes?
(yes, if the collection is used in a frequent/expensive query)
is the table containing attribute frequently updated?

should we use B-tree or Hash index?


-- use hashing for (unique) attributes in equality tests, e.g.
select * from Employee where id = 12345
-- use B-tree for attributes in range tests, e.g.
select * from Employee where age > 60

Query Tuning 148/154

Sometimes, a query can be re-phrased to affect performance:

by helping the optimiser to make use of indexes
by avoiding unnecessary/expensive operations

Examples which may prevent optimiser from using indexes:

select name from Employee where salary/365 > 100
       -- fix by re-phrasing condition to (salary > 36500)
select name from Employee where name like '%ith%'
select name from Employee where birthday is null
       -- above two are difficult to "fix"
select name from Employee
where  dept in (select id from Dept where ...)
       -- fix by using Employee join Dept on (e.dept=d.id)

... Query Tuning 149/154

Other factors to consider in query tuning:

select distinct requires a sort; is distinct necessary?

if multiple join conditions are available ...
choose join attributes that are indexed, avoid joins on strings


select ... Employee join Customer on (s.name = p.name)
vs
select ... Employee join Customer on (s.ssn = p.ssn)

sometimes or in condition prevents index from being used ...
replace the or condition by a union of non-or clauses


select name from Employee where dept=1 or dept=2
vs
(select name from Employee where dept=1)
union
(select name from Employee where dept=2)

PostgreSQL Query Tuning 150/154

PostgreSQL provides the explain statement to

give a representation of the query execution plan
with information that may help to tune query performance

Usage:

EXPLAIN [ANALYZE] Query

Without ANALYZE, EXPLAIN shows plan with estimated costs.

With ANALYZE, EXPLAIN executes query and prints real costs.

Note that runtimes may show considerable variation due to buffering.

EXPLAIN Examples 151/154

Example: Select on indexed attribute


ass2=# explain select * from Students where id=100250;
                            QUERY PLAN
-----------------------------------------------------------------
 Index Scan using student_pkey on student
                  (cost=0.00..5.94 rows=1 width=17)
   Index Cond: (id = 100250)

ass2=# explain analyze select * from Students where id=100250;
                            QUERY PLAN
-----------------------------------------------------------------
 Index Scan using student_pkey on student
                  (cost=0.00..5.94 rows=1 width=17)
                  (actual time=31.209..31.212 rows=1 loops=1)
   Index Cond: (id = 100250)
 Total runtime: 31.252 ms

... EXPLAIN Examples 152/154

Example: Select on non-indexed attribute


ass2=# explain select * from Students where stype='local';
                        QUERY PLAN
----------------------------------------------------------
 Seq Scan on student  (cost=0.00..70.33 rows=18 width=17)
   Filter: ((stype)::text = 'local'::text)

ass2=# explain analyze select * from Students
ass2-#                          where stype='local';
                         QUERY PLAN
---------------------------------------------------------------
 Seq Scan on student  (cost=0.00..70.33 rows=18 width=17)
             (actual time=0.061..4.784 rows=2512 loops=1)
   Filter: ((stype)::text = 'local'::text)
 Total runtime: 7.554 ms

... EXPLAIN Examples 153/154

Example: Join on a primary key (indexed) attribute


ass2=# explain
ass2-# select s.sid,p.name
ass2-# from Students s, People p where s.id=p.id;
                        QUERY PLAN
-----------------------------------------------------------
 Hash Join  (cost=70.33..305.86 rows=3626 width=52)
   Hash Cond: ("outer".id = "inner".id)
   -> Seq Scan on person p
               (cost=0.00..153.01 rows=3701 width=52)
   -> Hash  (cost=61.26..61.26 rows=3626 width=8)
       -> Seq Scan on student s
                   (cost=0.00..61.26 rows=3626 width=8)

... EXPLAIN Examples 154/154

Example: Join on a non-indexed attribute


ass3=> explain select s1.code, s2.code
ass2-# from Subjects s1, Subjects s2 where s1.offerer=s2.offerer;
                        QUERY PLAN
----------------------------------------------------------------
 Merge Join  (cost=2744.12..18397.14 rows=1100342 width=18)
   Merge Cond: (s1.offerer = s2.offerer)
   ->  Sort  (cost=1372.06..1398.33 rows=10509 width=13)
         Sort Key: s1.offerer
         ->  Seq Scan on subjects s1
                      (cost=0.00..670.09 rows=10509 width=13)
   ->  Sort  (cost=1372.06..1398.33 rows=10509 width=13)
         Sort Key: s2.offerer
         ->  Seq Scan on subjects s2
                      (cost=0.00..670.09 rows=10509 width=13)
(8 rows)

Produced: 28 Jul 2019

Cost	=	inl-join on (`Enrol,Student`) + linear scan of `tmp1`
	=	T_r(b_S + r_S.btree_E) + T_wb_tmp1 + T_rb_tmp1
	=	500 + 10,000.4 + 800 + 800
	=	42,100

Cost	=	btree-search on `Enrol` + bnl-join of (`tmp1,Student`)
	=	search+scan on `Enrol` + T_r(b_tmp1 + b_S)
	=	3T_r + 3T_r + T_r(1 + 500)
	=	3 + 3 + 1 + 500 = 507