Oracle explain plan estimates incorrect cardinality for an index range scan_问答_开发者

I have an Oracle 10.2.0.3 database, and a query like this:

select count(a.id) 
from LARGE_PARTITIONED_TABLE a
join SMALL_NONPARTITIONED_TABLE b on a.key1 = b.key1 and a.key2 = b.key2
where b.id = 1000

Table LARGE_PARTITIONED_TABLE (a) has about 5 million rows, and is partitioned by a column not present in the query. Table SMALL_NONPARTITIONED_TABLE (b) is not partitioned, and holds about 10000 rows.

Statistics are up-to-date, and there are height balanced histograms in columns key1 and key2 of table a.

Table a has a primary key and a global, nonpartitioned unique index on columns key1, key2, key3, key4, and key5.

Explain plan for the query displays the following results:

---------------------------------------------------------------------------------------------------
| Id  | Operation          | Name                         | Rows  | Bytes | Cost (%CPU)| Time     |
---------------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT   |                              |     1 |    31 |     4   (0)| 00:00:01 |
|   1 |  SORT AGGREGATE    |                              |     1 |    31 |            |          |
|   2 |   NESTED LOOPS     |                              |   406 | 12586 |     4   (0)| 00:00:01 |
|*  3 |    INDEX RANGE SCAN| INDEX_ON_TABLE_B            |     1 |    19 |     2   (0)| 00:00:01 |
|*  4 |    INDEX RANGE SCAN| PRIMARY_KEY_INDEX_OF_TABLE_A |   406 |  4872 |     2   (0)| 00:00:01 |
---------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   3 - access("b"."id"=1000)
   4 - access("a"."key1"="b"."key1" and
              "a"."key2"="b"."key2")

Thus the rows (cardinality) estimated for step 4 is 406.

Now, a tkprof trace reveals the following:

Rows     Row Source Operation
-------  ---------------------------------------------------
      1  SORT AGGREGATE (cr=51 pr=9 pw=0 time=74674 us)
   7366   NESTED LOOPS  (cr=51 pr=9 pw=0 time=824941 us)
      1    INDEX RANGE SCAN INDEX_ON_TABLE_B (cr=2 pr=0 pw=0 time=36 us)(object id 111111)
   7366    INDEX RANGE SCAN PRIMARY_KEY_INDEX_OF_TABLE_A (cr=49 pr=9 pw=0 time=810173 us)(object id 222222)

So the cardinality in reality was 7366, not 406!

My question is this: From where does Oracle get the estimated cardinality of 406 in this c开发者_如何学编程ase, and how can I improve its accuracy, so that the estimate is more in line of what really happens during query execution?

Update: Here is a snippet of a 10053 trace I ran on the query.

NL Join
  Outer table: Card: 1.00  Cost: 2.00  Resp: 2.00  Degree: 1  Bytes: 19
  Inner table: LARGE_PARTITIONED_TABLE  Alias: a
  ...
  Access Path: index (IndexOnly)
    Index: PRIMARY_KEY_INDEX_OF_TABLE_A
    resc_io: 2.00  resc_cpu: 27093
    ix_sel: 1.3263e-005  ix_sel_with_filters: 1.3263e-005
    NL Join (ordered): Cost: 4.00  Resp: 4.00  Degree: 1
      Cost_io: 4.00  Cost_cpu: 41536
      Resp_io: 4.00  Resp_cpu: 41536
  ****** trying bitmap/domain indexes ******
  Best NL cost: 4.00
          resc: 4.00 resc_io: 4.00 resc_cpu: 41536
          resp: 4.00 resp_io: 4.00 resp_cpu: 41536
Using concatenated index cardinality for table SMALL_NONPARTITIONED_TABLE
Revised join sel: 8.2891-e005 = 8.4475e-005 * (1/12064.00) * (1/8.4475e-005)
Join Card:  405.95 = outer (1.00) * inner (4897354.00) * sel (8.2891-e005)
Join Card - Rounded: 406 Computed: 405.95

So that is where the value 406 is coming from. Like Adam answered, join cardinality is join selectivity * filter cardinality (a) * filter cardinality (b), as can be seen on the second to last line of above trace quote.

What I do not understand is the Revised join sel line. 1/12064 is the selectivity of the index used to find the row from table b (12064 rows on table, and select based on unique id). But so what?

Cardinality appears to be calculated by multiplying the filter cardinality of table b (4897354) with selectivity of table a (1/12064). Why? What does the selectivity on table a have to do with how much rows is expected to be found from table b, when a --> b join is not based on a.id?
Where does the number 8.4475e-005 come from (it does not appear anywhere else in the whole trace)? Not that it affects the output, but I'd still like to know.

I understand that the optimizer has likely chosen the correct path here. But still the cardinality is miscalculated - and that can have a major effect on the execution path that is chosen from that point onwards (as in the case I'm having IRL - this example is a simplification of that).

Generating a 10053 trace file will help show exactly what choices the optimizer's making regarding its estimation of cardinality and selectivity. Jonathan Lewis' excellect Cost Based Oracle Fundamentals is an excellent resource to understanding how the optimizer works, and the printing I have spans 8i to 10.1.

From that work:

Join Selectivity =   ((num_rows(t1) - num_nulls(t1.c1)) / num_rows(t1)) 
                   * ((num_rows(t2) - num_nulls(t2.c2)) / num_rows(t2))
                   / greater (num_distinct(t1.c1), num_distinct(t2.c2))

Join Cardinality =   Join Selectivity 
                   * filtered_cardinality (t1)
                   * filtered_cardinality (t2)

However, because we have a multi-column join, Join Selectivity isn't at the table level, it's the product (intersection) of the join selectivities on each column. Assuming there's no nulls in play:

Join Selectivity = Join Selectivity (key1) * Join Selectivity (key2)

Join Selectivity (key1) =   ((5,000,000 - 0) / 5,000,000)
                          * ((10,000 - 0)) / 10,000)
                          / max (116, ?)  -- distinct key1 values in B

                        = 1 / max(116, distinct_key1_values_in_B)

Join Selectivity (key2) =   ((5,000,000 - 0) / 5,000,000)
                          * ((10,000 - 0)) / 10,000)
                          / max (650, ?)  -- distinct key2 values in B

                        = 1 / max(650, distinct_key2_values in B)

Join Cardinality =  JS(key1) * JS(key2) 
                  * Filter_cardinality(a) * Filter_cardinality(b)

We know that there are no filters on A, so that's tables filter cardinality is the number of rows. We're selecting the key value from B, so that table's filter cardinality is 1.

So the best case for estimated estimated join cardinality is now

Join Cardinality  = 1/116 * 1/650 * 5,000,000 * 1

                  =~ 67

It might be easier to work backward. Your estimated cardinality of 406, given what we know, leads to a join selectivty of 406/5,000,000, or approximately 1/12315. That happens to be really, really close to 1 / (116^2), which is a sanity check within the optimizer to prevent it from finding too aggressive a cardinality on multi-column joins.

For the TL;DR crowd:

Get Jonathan Lewis' Cost Based Oracle Fundamentals.
Get a 10053 trace of the query whose behavior your can't understand.

The cardinality estimate would be based on the product of the selectivity of a.key1 and a.key2, which (at least in 10g) would each be based on the number of distinct values for those two columns as recorded in the column statistics.

For a table of 5M rows, a cardinality estimate of 406 is not significantly different to 7366. The question you have to ask yourself is, is the "inaccurate" estimate here causing a problem?

You can check what plan Oracle would choose if it were able to generate a perfectly accurate estimate by getting an explain plan for this:

select /*+CARDINALITY(a 7366)*/ count(a.id) 
from LARGE_PARTITIONED_TABLE a
join SMALL_NONPARTITIONED_TABLE b on a.key1 = b.key1 and a.key2 = b.key2
where b.id = 1000;

If this comes up with the same plan, then the estimate that Oracle is calculating is already adequate.

You might be interested to read this excellent paper by Wolfgang Breitling which has a lot of info on CBO calculations: http://www.centrexcc.com/A%20Look%20under%20the%20Hood%20of%20CBO%20-%20the%2010053%20Event.pdf.

As explained there, because you have histograms, the filter-factor calculation for these columns does not use number of distinct values (NDV) but density, which is derived from the histogram in some way.

What are the DENSITY values in USER_TAB_COLUMNS for a.key1 and a.key2?

Generally, the problem in cases like this is that Oracle does not gather statistics on pairs of columns, and assumes that their combined filter factor will be the product of their individual factors. This will produce low estimates if there is any correlation between values of the two columns.

If this is causing a serious performance issue, I suppose you could create a function-based index on a function of those columns, and use that to do the lookup. Then Oracle would gather statistics on that index and probably produce better estimates.