数据库版本Oracle11gR2
SQL> select * from v$version where rownum=1;
BANNER
——————————————————————————–
Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 – Production
创建手动管理的表空间，blockssize 8k
SQL> create tablespace test datafile
‘/u01/app/oracle/oradata/ROBINSON/datafile/test.dbf’ size 50m autoextend on maxsize 200m
uniform size 1m segment space management manual blocksize 8k; 2 3
Tablespace created.
创建测试用户test,默认表空间 test
SQL> create user test identified by oracle default tablespace test;
User created.
为了简便，授权DBA给test
SQL> grant dba to test;
Grant succeeded.
创建测试表test
SQL> create table test as select * from dba_objects where 1=0 ;
Table created.
设置pctfree 99
SQL> alter table test pctfree 99 pctused 1;
Table altered.
SQL> insert into test select * from dba_objects where rownum<2;
1 row created.
确保一行一个block
SQL> alter table test minimize records_per_block;
Table altered.
SQL> insert into test select * from dba_objects where rownum<1000;
999 rows created.
SQL> commit;
Commit complete.
收集表统计信息
SQL> BEGIN
DBMS_STATS.GATHER_TABLE_STATS(ownname => ‘TEST’,
tabname => ‘TEST’,
estimate_percent => 100,
method_opt => ‘for all columns size 1’,
degree => DBMS_STATS.AUTO_DEGREE,
cascade=>TRUE
);
END;
/ 2 3 4 5 6 7 8 9 10
PL/SQL procedure successfully completed.
SQL> select owner,blocks from dba_tables where owner=’TEST’ and table_name=’TEST’;
OWNER BLOCKS
—————————— ———-
TEST 1000
SQL> show parameter db_file_multiblock_read_count
NAME TYPE VALUE
———————————— ———– ——————————
db_file_multiblock_read_count integer 16
全表扫描的成本等于220
SQL> select count(*) from test;
Execution Plan
———————————————————-
Plan hash value: 1950795681
——————————————————————-
| Id | Operation | Name | Rows | Cost (%CPU)| Time |
——————————————————————-
| 0 | SELECT STATEMENT | | 1 | 220 (0)| 00:00:03 |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | TABLE ACCESS FULL| TEST | 1000 | 220 (0)| 00:00:03 |
——————————————————————-
成本的计算方式如下：
Cost = (
#SRds * sreadtim +
#MRds * mreadtim +
CPUCycles / cpuspeed
) / sreadtime
#SRds – number of single block reads
#MRds – number of multi block reads
#CPUCyles – number of CPU cycles
sreadtim – single block read time
mreadtim – multi block read time
cpuspeed – CPU cycles per second
注意：如果没有收集过系统统计信息，那么Oracle采用非工作量统计， 如果收集了，，Oracle采用工作量统计的计算方法
SQL> select pname, pval1 from sys.aux_stats$ where sname=’SYSSTATS_MAIN’;
PNAME PVAL1
—————————— ———-
CPUSPEED
CPUSPEEDNW 2696.05568
IOSEEKTIM 10
IOTFRSPEED 4096
MAXTHR
MBRC
MREADTIM
SLAVETHR
SREADTIM
9 rows selected.
我这里因为MBRC 为0，所以CBO采用了非工作量(noworkload)来计算成本
#SRds=0,因为是全表扫描，单块读为0
#MRds=表的块数/多块读参数=1000/16
mreadtim=ioseektim+db_file_multiblock_count*db_block_size/iotftspeed
SQL> select (select pval1 from sys.aux_stats$ where pname = ‘IOSEEKTIM’) +
2 (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”
3 4 5 6 7 from dual;
mreadtim
———-
42
sreadtim=ioseektim+db_block_size/iotfrspeed
SQL> 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; 2 3 4
sreadtim
———-
12
CPUCycles 等于 PLAN_TABLE里面的CPU_COST
SQL> explain plan for select count(*) from test;
Explained.
SQL> select cpu_cost from plan_table;
CPU_COST
———-
7271440
cpuspeed 等于 CPUSPEEDNW= 2696.05568
那么COST=1000/16*42/12+7271440/2696.05568/12/1000
SQL> select ceil(1000/16*42/12+7271440/2696.05568/12/1000) from dual;
CEIL(1000/16*42/12+7271440/2696.05568/12/1000)
———————————————-
219
手工计算出来的COST用四舍五入等于219，和我们看到的220有差别， 这是由于隐含参数_tablescan_cost_plus_one参数造成的
SQL> 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%’
/ 2 3 4 5 6 7
NAME VALUE DESCRIB
—————————— ———- ——————————
_table_scan_cost_plus_one TRUE bump estimated full table scan
and index ffs cost by one
根据该参数的描述，在table full scan和index fast full scan的时候会将cost+1
那么我把改参数禁止了试一试
SQL> alter session set “_table_scan_cost_plus_one”=false;
Session altered.
SQL> set autot trace
SQL> select count(*) from test;
Execution Plan
———————————————————-
Plan hash value: 1950795681
——————————————————————-
| Id | Operation | Name | Rows | Cost (%CPU)| Time |
——————————————————————-
| 0 | SELECT STATEMENT | | 1 | 219 (0)| 00:00:03 |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | TABLE ACCESS FULL| TEST | 1000 | 219 (0)| 00:00:03 |
——————————————————————-
这次得到的Cost等于219，与计算值正好匹配，现在更改db_file_multiblock_read_count参数
SQL> alter session set db_file_multiblock_read_count=32;
Session altered.
这个时候 sreadtim=12
SQL> 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; 2 3 4
sreadtim
———-
12
mreadtim=74
SQL> select (select pval1 from sys.aux_stats$ where pname = ‘IOSEEKTIM’) +
(select value
2 3 from v$parameter
4 where name = ‘db_file_multiblock_read_count’) *
5 (select value from v$parameter where name = ‘db_block_size’) /
6 (select pval1 from sys.aux_stats$ where pname = ‘IOTFRSPEED’) “mreadtim”
7 from dual;
mreadtim
———-
74
那么cost等于
SQL> select ceil(1000/32*74/12+7271440/2696.05568/12/1000) from dual;
CEIL(1000/32*74/12+7271440/2696.05568/12/1000)
———————————————-
193
SQL> set autot trace
SQL> select count(*) from test;
Execution Plan
———————————————————-
Plan hash value: 1950795681
——————————————————————-
| Id | Operation | Name | Rows | Cost (%CPU)| Time |
——————————————————————-
| 0 | SELECT STATEMENT | | 1 | 193 (0)| 00:00:03 |
| 1 | SORT AGGREGATE | | 1 | | |
| 2 | TABLE ACCESS FULL| TEST | 1000 | 193 (0)| 00:00:03 |
——————————————————————-
与计算的Cost相匹配，从实验种可以得出，在11gR2中，全表扫描计算Cost的方式依然和9i/10g一样，没有变化。
相关链接