Oracle预估的基数算法

SQL> create table t as select * from dba_objects;

Table created.

SQL> create index idx_t on t(object_id);

Index created.

SQL> BEGIN

  2    DBMS_STATS.GATHER_TABLE_STATS(ownname          => 'TEST',

  3                                  tabname          => 'T',

  4                                  estimate_percent => 100,

  5                                  method_opt       => 'for all columns size auto',

  6                                  degree           => DBMS_STATS.AUTO_DEGREE,

  7                                  cascade          => TRUE);

  8  END;

  9  /

SQL> alter session set optimizer_features_enable='9.2.0';

Session altered.

SQL> explain plan for  select owner from t where object_id<1000;

Explained.

SQL> select * from table(dbms_xplan.display());

PLAN_TABLE_OUTPUT

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

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

Plan hash value: 1594971208

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

| Id  | Operation		    | Name  | Rows  | Bytes | Cost  |

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

|   0 | SELECT STATEMENT	    |	    |	958 | 10538 |	 26 |

|   1 |  TABLE ACCESS BY INDEX ROWID| T     |	958 | 10538 |	 26 |

|*  2 |   INDEX RANGE SCAN	    | IDX_T |	958 |	    |	  4 |

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

Predicate Information (identified by operation id):

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

   2 - access("OBJECT_ID"<1000)

Note

-----

   - cpu costing is off (consider enabling it)

18 rows selected.

那么这个958 Oracle是怎么估算的呢？

Oracle预估的基数等于有效选择性*(num_rows-num_nulls)

其中 有效选择性 ，< 的有效选择性算法为:

(limit-low_value)/(high_value-low_value)

set linesize 200

SQL> select b.num_rows,

       a.num_distinct,

       a.num_nulls,

       utl_raw.cast_to_number(high_value) high_value,

       utl_raw.cast_to_number(low_value) low_value,

       (b.num_rows - a.num_nulls) "NUM_ROWS-NUM_NULLS",

       utl_raw.cast_to_number(high_value) -

       utl_raw.cast_to_number(low_value) "HIGH_VALUE-LOW_VALUE"

  from dba_tab_col_statistics a, dba_tables b

 where a.owner = b.owner

   and a.table_name = b.table_name

   and a.owner = 'TEST'

   and a.table_name = upper('T')

   and a.column_name = 'OBJECT_ID';  2    3    4    5    6    7    8    9   10   11   12   13   14

  NUM_ROWS NUM_DISTINCT  NUM_NULLS HIGH_VALUE  LOW_VALUE NUM_ROWS-NUM_NULLS HIGH_VALUE-LOW_VALUE

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

     73964	  73964 	 0	77085	       2	      73964		   77083

那么估算为:

SQL>

select ceil((1000-2)/77083*73964) from dual;SQL>

CEIL((1000-2)/77083*73964)

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

		       958

Oracle 就是根据这个算法的