Oracle 11g R2 全表扫描成本计算(非工作量模式-noworkload)-mysql教程-PHP中文网

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Oracle 11g R2 全表扫描成本计算(非工作量模式-noworkload)

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Jun 07, 2016 pm 05:02 PM

数据库版本Oracle11gR2SQLgt; select * from v$version where rownum=1;BANNEROracle Database 11g Enterprise Edition Release

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

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

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。