Oracle SQL Tuning_03 CBO算法
全表扫描成本计算法则实例
/img/2018/0725_sqltune_08.png
创建表空间、用户及授权
-- 查看数据库版本
select * from v$version where rownum<=1;
--=================================================================================
BANNER
--------------------------------------------------------------------------------
Oracle Database 11g Enterprise Edition Release 11.2.0.4.0 - 64bit Production
--=================================================================================
col FILE_NAME for a50
select file_name from dba_data_files;
--=================================================================================
FILE_NAME
--------------------------------------------------
/u01/app/oracle/oradata/orcl/system01.dbf
/u01/app/oracle/oradata/orcl/sysaux01.dbf
/u01/app/oracle/oradata/orcl/undotbs01.dbf
/u01/app/oracle/oradata/orcl/users01.dbf
/u01/app/oracle/oradata/orcl/disdb01.dbf
--=================================================================================
create tablespace cbo datafile '/u01/app/oracle/oradata/orcl/cbo.dbf'
size 100m autoextend on maxsize 300m uniform size 1m segment space management manual;
create user cbo identified by cbo default tablespace cbo;
grant dba to cbo;
|
设置多块读参数
col value for a50
select value from v$parameter where name='db_file_multiblock_read_count';
--=================================================================================
VALUE
--------------------------------------------------
128
--=================================================================================
alter system set db_file_multiblock_read_count=16;
show parameter db_file_multiblock_read_count
--=================================================================================
NAME TYPE VALUE
------------------------------------ ----------- ------------------------------
db_file_multiblock_read_count integer 16
--=================================================================================
|
建表并插入数据
-- 设置pctfree 99,一行一个block
conn cbo/cbo
create table fulltable as select * from dba_objects where 1=0 ;
alter table fulltable pctfree 99 pctused 1;
insert into fulltable select * from dba_objects where rownum<2;
alter table fulltable minimize records_per_block;
insert into fulltable select * from dba_objects where rownum<1000;
commit;
|
收集表统计信息
BEGIN
DBMS_STATS.GATHER_TABLE_STATS(ownname => 'CBO',
tabname => 'FULLTABLE',
estimate_percent => 100,
method_opt => 'for all columns size 1',
degree => DBMS_STATS.AUTO_DEGREE,
cascade=>TRUE);
END;
/
select owner,blocks from dba_tables where owner='CBO' and table_name='FULLTABLE';
--=================================================================================
OWNER BLOCKS
------------------------------ ----------
CBO 1000
--=================================================================================
|
查看SQL执行计划
explain plan for select count(*) from fulltable;
SELECT * FROM TABLE(DBMS_XPLAN.DISPLAY);
--=================================================================================
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------------------
Plan hash value: 2411201521
------------------------------------------------------------------------
| Id | Operation | Name | Rows | Cost (%CPU)| Time |
------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 221 (1)| 00:00:03 |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | TABLE ACCESS FULL| FULLTABLE | 1000 | 221 (1)| 00:00:03 |
------------------------------------------------------------------------
--=================================================================================
|
计算I/O COST
select pname, pval1 from sys.aux_stats$ where sname='SYSSTATS_MAIN';
--=================================================================================
PNAME PVAL1
------------------------------ ----------
CPUSPEEDNW 1200
IOSEEKTIM 10
IOTFRSPEED 4096
SREADTIM
MREADTIM
CPUSPEED
MBRC
MAXTHR
SLAVETHR
--=================================================================================
mreadtim = ioseektim + db_file_multiblock_read_count * db_block_size / iofrrspeed
= 10 + 16 * 8192 / 4096 = 42
sreadtim = ioseektim + db_block_size / iotfrspeed
= 10 + 8192 / 4096 = 12
-- mreadtim
select (select pval1 from sys.aux_stats$ where pname = 'IOSEEKTIM') +
(select value from v$parameter where name = 'db_file_multiblock_read_count') *
(select value from v$parameter where name = 'db_block_size') /
(select pval1 from sys.aux_stats$ where pname = 'IOTFRSPEED') "mreadtim" from dual;
--=================================================================================
mreadtim
----------
42
--=================================================================================
-- sreadtim
select (select pval1 from sys.aux_stats$ where pname = 'IOSEEKTIM') +
(select value from v$parameter where name = 'db_block_size') /
(select pval1 from sys.aux_stats$ where pname = 'IOTFRSPEED') "sreadtim"
from dual;
--=================================================================================
sreadtim
----------
12
--=================================================================================
-- MRDs = +Blks / MBRC
-- 多块读次数 = 表的块数 / 多块读参数 = 1000 / 16
-- CEIL(n) 取大于等于数值n的最小整数
-- I/O Cost = 1 + CEIL(#MRDs * (mreadtim / sreadtim))
SELECT 1+CEIL((1000/16)*(42/12)) FROM DUAL;
--=================================================================================
1+CEIL((1000/16)*(42/12))
-------------------------
220
--=================================================================================
|
计算CPU COST
select pname, pval1 from sys.aux_stats$ where sname='SYSSTATS_MAIN';
--=================================================================================
PNAME PVAL1
------------------------------ ----------
CPUSPEEDNW 1200 # CPUSPEED
IOSEEKTIM 10
IOTFRSPEED 4096
SREADTIM
MREADTIM
CPUSPEED
MBRC
MAXTHR
SLAVETHR
--=================================================================================
-- CPUCycles(CPU周期)等于PLAN_TABLE里面的CPU_COST
explain plan for select count(*) from fulltable;
select cpu_cost from plan_table where object_name='FULLTABLE';
--=================================================================================
CPU_COST
----------
7271440
--=================================================================================
CPU Cost = ROUND(#CPUCycles / cpuspeed / 1000 / sreadtim)
select ROUND(7271440 / 1200 / 1000 / 12) CPU_COST from dual;
--=================================================================================
CPU_COST
----------
1
--=================================================================================
|
FTS COST
SELECT 1+CEIL((1000/16)*(42/12)) + ROUND(7271440/1200/1000/12) FTS_COST FROM DUAL;
--=================================================================================
FTS_COST
----------
221
--=================================================================================
SELECT * FROM TABLE(DBMS_XPLAN.DISPLAY);
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------------------
Plan hash value: 2411201521
------------------------------------------------------------------------
| Id | Operation | Name | Rows | Cost (%CPU)| Time |
------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 221 (1)| 00:00:03 |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | TABLE ACCESS FULL| FULLTABLE | 1000 | 221 (1)| 00:00:03 |
------------------------------------------------------------------------
|
隐含参数: _table_scan_cost_plus_one
-- 根据该参数的描述，在table full scan和index fast full scan的时侯会将cost + 1
col name for a30
col VALUE for a20
col DESCRIB for a60
SELECT x.ksppinm NAME, y.ksppstvl VALUE, x.ksppdesc describ
FROM x$ksppi x, x$ksppcv y
WHERE x.inst_id = USERENV ('Instance')
AND y.inst_id = USERENV ('Instance')
AND x.indx = y.indx
AND x.ksppinm LIKE '%_table_scan_cost_plus_one%';
--=================================================================================
NAME VALUE DESCRIB
------------------------------ -------------------- ------------------------------------------------------------
_table_scan_cost_plus_one TRUE bump estimated full table scan and index ffs cost by one
--=================================================================================
-- 棼用隐含参数
conn / as sysdba
alter system set "_table_scan_cost_plus_one" = false;
conn cbo/cbo
explain plan for select count(*) from fulltable;
select * from table(dbms_xplan.display);
--=================================================================================
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------------------
Plan hash value: 2411201521
------------------------------------------------------------------------
| Id | Operation | Name | Rows | Cost (%CPU)| Time |
------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 220 (1)| 00:00:01 |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | TABLE ACCESS FULL| FULLTABLE | 1000 | 220 (1)| 00:00:01 |
------------------------------------------------------------------------
--=================================================================================
|
索引范围扫描成本算法
创建表和索引
conn cbo/cbo
create table idx_range as select * from dba_objects;
create index idx_range_id on idx_range(object_id);
|
收集统计信息
BEGIN
DBMS_STATS.GATHER_TABLE_STATS(ownname=> 'CBO',
tabname => 'IDX_RANGE',
estimate_percent => 100,
method_opt => 'for all columns size auto',
degree => DBMS_STATS.AUTO_DEGREE,
cascade => TRUE);
END;
/
select leaf_blocks,blevel,clustering_factor from dba_indexes where index_name='IDX_RANGE_ID';
--=================================================================================
LEAF_BLOCKS BLEVEL CLUSTERING_FACTOR
----------- ---------- -----------------
31 1 244
--=================================================================================
|
计算有效选择率和计算基数
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 ='CBO'
and a.table_name = upper('IDX_RANGE')
and a.column_name='OBJECT_ID';
--=================================================================================
NUM_ROWS NUM_DISTINCT NUM_NULLS HIGH_VALUE LOW_VALUE NUM_ROWS-NUM_NULLS HIGH_VALUE-LOW_VALUE
---------- ------------ ---------- ---------- ---------- ------------------ --------------------
14329 14329 0 24246 2 14329 24244
--=================================================================================
有效选择率 = (limit-low_value) / (high_value-low_value)=(1000-2)/24244
计算基数 = 基本基数 * 有效选择率 = 14329 * (1000-2)/24244
|
查看执行计划
explain plan for select owner from idx_range where object_id<1000;
SELECT * FROM TABLE(DBMS_XPLAN.DISPLAY);
--=================================================================================
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------------------------------------
Plan hash value: 1377850575
--------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 590 | 5900 | 14 (0)| 00:00:01 |
| 1 | TABLE ACCESS BY INDEX ROWID| IDX_RANGE | 590 | 5900 | 14 (0)| 00:00:01 |
|* 2 | INDEX RANGE SCAN | IDX_RANGE_ID | 590 | | 3 (0)| 00:00:01 |
--------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
2 - access("OBJECT_ID"<1000)
--=================================================================================
|
I/O成本公式
-- Index Access I/O = LVLS + CEIL($LB * ix_sel)
SQL> select 1+ceil(31*(1000-2)/24244) from dual;
1+CEIL(31*(1000-2)/24244)
-------------------------
3
-- Table Access I/O Cost = CEIL(CLUF * ix_sel_with_filiters)
SQL> select ceil(244 * (1000-2)/24244) from dual;
CEIL(244*(1000-2)/24244)
------------------------
11
-- I/O Cost = Index Access I/O + Table Access I/O Cost
SQL> select 3 + 11 from dual;
3+11
----------
14
|
CPU成本公式
-- CPU Cost = ROUND(#CPUCycles / cpuspeed / 1000 / sreadtim)
explain plan for select owner from idx_range where object_id<1000;
select cpu_cost from plan_table where object_name = 'IDX_RANGE';
CPU_COST
----------
341600
select pname, pval1 from sys.aux_stats$ where sname='SYSSTATS_MAIN' and pname='CPUSPEEDNW';
PNAME PVAL1
------------------------------ ----------
CPUSPEEDNW 1194
select (select pval1 from sys.aux_stats$ where pname = 'IOSEEKTIM') +
(select value from v$parameter where name = 'db_block_size') /
(select pval1 from sys.aux_stats$ where pname = 'IOTFRSPEED') "sreadtim"
from dual;
sreadtim
----------
8.25704424
select ROUND(341600 / 1194 / 1000 / 8.25704424) CPU_COST from dual;
CPU_COST
----------
0
|
嵌套循环连接成本算法L
专业术语：
1、为什么执行计划中的COST很小，SQL却跑的很慢？（利用实验步骤证明）
统计信息不正确
2、根据以下的Hash Join的COST算法用实验过程证明？（写出整个计算过程）
HJ Cost = Outer Table Cost + Inner Table Cost + Join Cost
3、怎么判断NEST LOOP连接有问题，导致SQL执行缓慢？