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<chapter id="planner-stats">

<title>How the Planner Uses Statistics</title>

<para>

This chapter builds on the material covered in <xref linkend="using-explain">

and <xref linkend="planner-stats">, and shows how the planner uses the

system statistics to estimate the number of rows each stage in a query might

return. This is a significant part of the planning / optimizing process,

providing much of the raw material for cost calculation.

</para>

<para>

The intent of this chapter is not to document the code &mdash;

better done in the code itself, but to present an overview of how it works.

This will perhaps ease the learning curve for someone who subsequently

wishes to read the code. As a consequence, the approach chosen is to analyze

a series of incrementally more complex examples.

</para>

<para>

The outputs and algorithms shown below are taken from version 8.0.

The behaviour of earlier (or later) versions may vary.

</para>

<sect1 id="row-estimation-examples">

<title>Row Estimation Examples</title>

<indexterm zone="row-estimation-examples">

<primary>row estimation</primary>

<secondary>planner</secondary>

</indexterm>

<para>

Using examples drawn from the regression test database, let's start with a

very simple query:

<programlisting>

EXPLAIN SELECT * FROM tenk1;

QUERY PLAN

-------------------------------------------------------------

Seq Scan on tenk1 (cost=0.00..445.00 rows=10000 width=244)

</programlisting>

How the planner determines the cardinality of <classname>tenk1</classname>

is covered in <xref linkend="using-explain">, but is repeated here for

completeness. The number of rows is looked up from

<classname>pg_class</classname>:

<programlisting>

SELECT reltuples, relpages FROM pg_class WHERE relname = 'tenk1';

relpages | reltuples

----------+-----------

345 | 10000

</programlisting>

The planner will check the <structfield>relpages<structfield> estimate

(this is a cheap operation) and if incorrect may scale

<structfield>reltuples<structfield> to obtain a row estimate. In this case it

does not, thus:

<programlisting>

rows = 10000

</programlisting>

</para>

<para>

let's move on to an example with a range condition in its

<literal>WHERE</literal> clause:

<programlisting>

EXPLAIN SELECT * FROM tenk1 WHERE unique1 &lt; 1000;

QUERY PLAN

------------------------------------------------------------

Seq Scan on tenk1 (cost=0.00..470.00 rows=1031 width=244)

Filter: (unique1 &lt; 1000)

</programlisting>

The planner examines the <literal>WHERE</literal> clause condition:

<programlisting>

unique1 &lt; 1000

</programlisting>

and looks up the restriction function for the operator

<literal>&lt;</literal> in <classname>pg_operator</classname>.

This is held in the column <structfield>oprrest</structfield>,

and the result in this case is <function>scalarltsel</function>.

The <function>scalarltsel</function> function retrieves the histogram for

<structfield>unique1</structfield> from <classname>pg_statistics</classname>

- we can follow this by using the simpler <classname>pg_stats</classname>

view:

<programlisting>

SELECT histogram_bounds FROM pg_stats

WHERE tablename='tenk1' AND attname='unique1';

histogram_bounds

------------------------------------------------------

{1,970,1943,2958,3971,5069,6028,7007,7919,8982,9995}

</programlisting>

Next the fraction of the histogram occupied by <quote>&lt; 1000</quote>

is worked out. This is the selectivity. The histogram divides the range

into equal frequency buckets, so all we have to do is locate the bucket

that our value is in and count <emphasis>part</emphasis> of it and

<emphasis>all</emphasis> of the ones before. The value 1000 is clearly in

the second (970 - 1943) bucket, so by assuming a linear distribution of

values inside each bucket we can calculate the selectivity as:

<programlisting>

selectivity = (1 + (1000 - bckt[2].min)/(bckt[2].max - bckt[2].min))/num_bckts

= (1 + (1000 - 970)/(1943 - 970))/10

= 0.1031

</programlisting>

that is, one whole bucket plus a linear fraction of the second, divided by

the number of buckets. The estimated number of rows can now be calculated as

the product of the selectivity and the cardinality of

<classname>tenk1</classname>:

<programlisting>

rows = rel_cardinality * selectivity

= 10000 * 0.1031

= 1031

</programlisting>

</para>

<para>

Next let's consider an example with equality condition in its

<literal>WHERE</literal> clause:

<programlisting>

EXPLAIN SELECT * FROM tenk1 WHERE stringu1 = 'ATAAAA';

QUERY PLAN

----------------------------------------------------------

Seq Scan on tenk1 (cost=0.00..470.00 rows=31 width=244)

Filter: (stringu1 = 'ATAAAA'::name)

</programlisting>

Again the planner examines the <literal>WHERE</literal> clause condition:

<programlisting>

stringu1 = 'ATAAAA'

</programlisting>

and looks up the restriction function for <literal>=</literal>, which is

<function>eqsel</function>. This case is a bit different, as the most

common values &mdash; <acronym>MCV</acronym>s, are used to determine the

selectivity. Let's have a look at these, with some extra columns that will

be useful later:

<programlisting>

SELECT null_frac, n_distinct, most_common_vals, most_common_freqs FROM pg_stats

WHERE tablename='tenk1' AND attname='stringu1';

null_frac | 0

n_distinct | 672

most_common_vals | {FDAAAA,NHAAAA,ATAAAA,BGAAAA,EBAAAA,MOAAAA,NDAAAA,OWAAAA,BHAAAA,BJAAAA}

most_common_freqs | {0.00333333,0.00333333,0.003,0.003,0.003,0.003,0.003,0.003,0.00266667,0.00266667}

</programlisting>

The selectivity is merely the most common frequency (<acronym>MCF</acronym>)

corresponding to the third <acronym>MCV</acronym> &mdash; 'ATAAAA':

<programlisting>

selectivity = mcf[3]

= 0.003

</programlisting>

The estimated number of rows is just the product of this with the

cardinality of <classname>tenk1</classname> as before:

<programlisting>

rows = 10000 * 0.003

= 30

</programlisting>

The number displayed by <command>EXPLAIN</command> is one more than this,

due to some post estimation checks.

</para>

<para>

Now consider the same query, but with a constant that is not in the

<acronym>MCV</acronym> list:

<programlisting>

EXPLAIN SELECT * FROM tenk1 WHERE stringu1 = 'xxx';

QUERY PLAN

----------------------------------------------------------

Seq Scan on tenk1 (cost=0.00..470.00 rows=15 width=244)

Filter: (stringu1 = 'xxx'::name)

</programlisting>

This is quite a different problem, how to estimate the selectivity when the

value is <emphasis>not</emphasis> in the <acronym>MCV</acronym> list.

The approach is to use the fact that the value is not in the list,

combined with the knowledge of the frequencies for all of the

<acronym>MCV</acronym>s:

<programlisting>

selectivity = (1 - sum(mvf))/(num_distinct - num_mcv)

= (1 - (0.00333333 + 0.00333333 + 0.003 + 0.003 + 0.003

+ 0.003 + 0.003 + 0.003 + 0.00266667 + 0.00266667))/(672 - 10)

= 0.001465

</programlisting>

That is, add up all the frequencies for the <acronym>MCV</acronym>s and

subtract them from one &mdash; because it is <emphasis>not</emphasis> one

of these, and divide by the <emphasis>remaining</emphasis> distinct values.

Notice that there are no null values so we don't have to worry about those.

The estimated number of rows is calculated as usual:

<programlisting>

rows = 10000 * 0.001465

= 15

</programlisting>

</para>

<para>

Let's increase the complexity to consider a case with more than one

condition in the <literal>WHERE</literal> clause:

<programlisting>

EXPLAIN SELECT * FROM tenk1 WHERE unique1 &lt; 1000 AND stringu1 = 'xxx';

QUERY PLAN

-----------------------------------------------------------

Seq Scan on tenk1 (cost=0.00..495.00 rows=2 width=244)

Filter: ((unique1 &lt; 1000) AND (stringu1 = 'xxx'::name))

</programlisting>

An assumption of independence is made and the selectivities of the

individual restrictions are multiplied together:

<programlisting>

selectivity = selectivity(unique1 &lt; 1000) * selectivity(stringu1 = 'xxx')

= 0.1031 * 0.001465

= 0.00015104

</programlisting>

The row estimates are calculated as before:

<programlisting>

rows = 10000 * 0.00015104

= 2

</programlisting>

</para>

<para>

Finally we will examine a query that includes a <literal>JOIN</literal>

together with a <literal>WHERE</literal> clause:

<programlisting>

EXPLAIN SELECT * FROM tenk1 t1, tenk2 t2

WHERE t1.unique1 &lt; 50 AND t1.unique2 = t2.unique2;

QUERY PLAN

-----------------------------------------------------------------------------------------

Nested Loop (cost=0.00..346.90 rows=51 width=488)

-> Index Scan using tenk1_unique1 on tenk1 t1 (cost=0.00..192.57 rows=51 width=244)

Index Cond: (unique1 &lt; 50)

-> Index Scan using tenk2_unique2 on tenk2 t2 (cost=0.00..3.01 rows=1 width=244)

Index Cond: ("outer".unique2 = t2.unique2)

</programlisting>

The restriction on <classname>tenk1</classname>

<quote>unique1 &lt; 50</quote> is evaluated before the nested-loop join.

This is handled analogously to the previous range example. The restriction

operator for <literal>&lt;</literal> is <function>scalarlteqsel</function>

as before, but this time the value 50 is in the first bucket of the

<structfield>unique1</structfield> histogram:

<programlisting>

selectivity = (0 + (50 - bckt[1].min)/(bckt[1].max - bckt[1].min))/num_bckts

= (0 + (50 - 1)/(970 - 1))/10

= 0.005057

rows = 10000 * 0.005057

= 51

</programlisting>

The restriction for the join is:

<programlisting>

t2.unique2 = t1.unique2

</programlisting>

This is due to the join method being nested-loop, with

<classname>tenk1</classname> being in the outer loop. The operator is just

our familiar <literal>=<literal>, however the restriction function is

obtained from the <structfield>oprjoin</structfield> column of

<classname>pg_operator</classname> - and is <function>eqjoinsel</function>.

Additionally we use the statistical information for both

<classname>tenk2</classname> and <classname>tenk1</classname>:

<programlisting>

SELECT tablename, null_frac,n_distinct, most_common_vals FROM pg_stats

WHERE tablename IN ('tenk1', 'tenk2') AND attname='unique2';

tablename | null_frac | n_distinct | most_common_vals

-----------+-----------+------------+------------------

tenk1 | 0 | -1 |

tenk2 | 0 | -1 |

</programlisting>

In this case there is no <acronym>MCV</acronym> information for

<structfield>unique2</structfield> because all the values appear to be

unique, so we can use an algorithm that relies only on the number of

distinct values for both relations together with their null fractions:

<programlisting>

selectivity = (1 - null_frac1) * (1 - null_frac2) * min(1/num_distinct1, 1/num_distinct2)

= (1 - 0) * (1 - 0) * min(1/10000, 1/1000)

= 0.0001

</programlisting>

This is, subtract the null fraction from one for each of the relations,

and divide by the maximum of the two distinct values. The number of rows

that the join is likely to emit is calculated as the cardinality of

cartesian product of the two nodes in the nested-loop, multiplied by the

selectivity:

<programlisting>

rows = (outer_cardinality * inner_cardinality) * selectivity

= (51 * 10000) * 0.0001

= 51

</programlisting>

</para>

<para>

For those interested in further details, estimation of the number of rows in

a relation is covered in

<filename>src/backend/optimizer/util/plancat.c</filename>. The calculation

logic for clause selectivities is in

<filename>src/backend/optimizer/path/clausesel.c</filename>. The actual

implementations of the operator and join restriction functions can be found

in <filename>src/backend/utils/adt/selfuncs.c</filename>.

</para>

</sect1>

</chapter>

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