虚拟机测试全表扫描成本

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;

SQL> select owner,blocks from dba_tables where owner='TEST' and table_name='TEST';

OWNER                              BLOCKS
------------------------------ ----------
TEST                                32910

SQL>  select count(distinct dbms_rowid.rowid_block_number(rowid)) from test;

COUNT(DISTINCTDBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID))
---------------------------------------------------
                                              23563



SQL> show parameter multi

NAME                                 TYPE        VALUE
------------------------------------ ----------- ------------------------------
db_file_multiblock_read_count        integer     16
parallel_adaptive_multi_user         boolean     TRUE


SQL> explain plan for select * from test where owner='SYS';

已解释。

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

PLAN_TABLE_OUTPUT
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

-----------------------
Plan hash value: 1357081020

--------------------------------------------------------------------------
| Id  | Operation         | Name | Rows  | Bytes | Cost (%CPU)| Time     |
--------------------------------------------------------------------------
|   0 | SELECT STATEMENT  |      | 77444 |  7336K|  7223   (1)| 00:01:27 |
|*  1 |  TABLE ACCESS FULL| TEST | 77444 |  7336K|  7223   (1)| 00:01:27 |
--------------------------------------------------------------------------

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

   1 - filter("OWNER"='SYS')

已选择13行


COST=7223


 select count(*) from test where owner='SYS';
 --986880



全表扫描成本计算公式:

成本的计算方式如下：
Cost = (#SRds * sreadtim + #MRds * mreadtim + CPUCycles / cpuspeed) / sreadtime

#SRds - number of single block reads 单块读次数

#MRds - number of multi block reads  多块读次数

#CPUCyles - number of CPU cycles     CPU时钟周期数

sreadtim - single block read time    一次单块读耗时(单位milliseconds 毫秒，1000毫秒等于1秒)

mreadtim - multi block read time     一次多块读耗时(单位milliseconds 毫秒，1000毫秒等于1秒)

cpuspeed - CPU cycles per second     CPU频率(单位MHZ)--每秒钟CPU做多少个轮训


全表扫描多块读 那么#SRds=0

#MRds=16,每次I/O读16个块

mreadtim =32910/16

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.

这里因为MBRC 为0，所以CBO采用了非工作量(noworkload)来计算成本，所有的系统全是 用的 非工作量。

计算mreadtim - multi block read time     一次多块读耗时(单位milliseconds 毫秒，1000毫秒等于1秒)

mreadtim=ioseektim+db_file_multiblock_count*db_block_size/iotftspeed

ioseektim:寻道寻址的时间

db_file_multiblock_count*db_block_size=16*8K=128K ---一次I/O的数据量


db_file_multiblock_count*db_block_size/iotftspeed=多块读耗时时间


iotftspeed:I/O传输速度

mreadtim(多块读耗时)=寻道寻址的时间+多块读耗时
SQL> 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;  2    3    4    5    6    7  

  mreadtim
----------
	42

多块读的耗时有42毫秒

sreadtim(单块读耗时)=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

单块读耗时12毫秒


操作系统CPU 和磁盘信息Oracle都可以获取到

CPUCycles 等于 PLAN_TABLE里面的CPU_COST---这个ORACLE未解密，无法知道怎么计算的



SQL> explain plan for select  * from test where owner='SYS';

已解释。

SQL> select cpu_cost from plan_table;          

  CPU_COST
----------
 720716510
 720716510

cpuspeed 等于 CPUSPEEDNW= 2696.05568
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.



成本的计算方式如下：
Cost = (
       #SRds * sreadtim +
       #MRds * mreadtim +
       CPUCycles / cpuspeed
       ) / sreadtime

#SRds * sreadtim =0 单块读次数为0

#SRds * sreadtim=(number of single block reads 单块读次数) * 12


#MRds * mreadtim =(number of multi block reads) * 42=32910/16 * 42=86388.75


sreadtime=12


SQL> select ceil(32910/16*42/12 +  720716510/2696.05568/1000/12) from dual;

CEIL(32910/16*42/12+720716510/2696.05568/1000/12)
-------------------------------------------------
                                             7222


Cost = (
             #MRds * mreadtim +
       CPUCycles / cpuspeed
       ) / sreadtime


Cost = 
             #MRds  +（
       CPUCycles / cpuspeed
       ) / sreadtime


最终的成本计算公式=Cost = #MRds (忽略CPU的情况下,就是多块读的次数=1000/16)

说明减少 物理 io扫描次数,是SQL优化的核心思想,

Cost =        #MRds * mreadtim/ sreadtime


#MRds - number of multi block reads  多块读I/O次数