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Home -> Community -> Mailing Lists -> Oracle-L -> RE: ** SPAM scored: Low **Re: Clustering factor smaller than table blocks.
Thanks stephan, you had it right, I got the reason of the descrepancy:
it comes from the stats in the PK partitions following a first
dbms_stat with percent less than 100. subsequence dbms_stat with percent
= 100 where done only at partitions level.
I had :
1 select 'partitions' , sum(DISTINCT_KEYS)DISTINCT_KEYS,
2 sum(NUM_ROWS)NUM_ROWS 3 from dba_ind_partitions where index_name = 'CUSTOMERS_PK'and index_owner = 'SOE'
where index_name = 'CUSTOMERS_PK' and owner = 'SOE'
SQL> /
'PARTITION DISTINCT_KEYS NUM_ROWS
---------- ------------- ----------
global 36116 36112 partitions 36112 36112
So I made one at global level:
exec dbms_stats.gather_table_stats( ownname=>'SOE', tabname=>
'CUSTOMERS', Degree=> 4, estimate_percent=> 100, granularity=>'GLOBAL',
cascade=>TRUE) ;
And now:
'PARTITION DISTINCT_KEYS NUM_ROWS
---------- ------------- ----------
global 36112 36112 partitions 36112 36112
However the block count on the table is still 368 and CF of pk set to 345. I am investigating empty blocks not referenced in the free list.
B. Polarski
-----Original Message-----
From: Jonathan Lewis [mailto:jonathan_at_jlcomp.demon.co.uk]
Sent: Wednesday, 27 September, 2006 11:03 AM
To: oracle-l_at_freelists.org
Subject: ** SPAM scored: Low **Re: Clustering factor smaller than table
blocks.
One possibility relates to ASSM - you can have blocks below the highwater mark that are not yet formatted. I'd have to check, but I don't think they get recorded as EMPTY_BLOCKS.
The rather more surprising thing about your example is that you have done a compute, so the results should be accurate, but you have 36,364 rows in the table and only 36,112 entries in what we guess is the primary key index. Do you have a corrupt index perhaps ?
Regards
Jonathan Lewis
http://www.oracle.com/technology/community/oracle_ace/ace1.html#lewis
The Co-operative Oracle Users' FAQ
http://www.jlcomp.demon.co.uk/faq/ind_faq.html
Cost Based Oracle: Fundamentals
http://www.jlcomp.demon.co.uk/cbo_book/ind_book.html
I have a table with 368 blocks, no free blocks, and the clustering factor of the primary key is 345. how is it possible?
I thought that CF will give you the amount of table blocks that an index will need to access for a given rowset. Something like smallest( table blocks, ceil( (CF/tot table rows)*rowset/avg rows per block)) (real formula must be much complicated.
However if I have an avg ro
Here are the brute facts:
select EMPTY_BLOCKS,AVG_SPACE_FREELIST_BLOCKS ,NUM_FREELIST_BLOCKS
from dba_tables where table_name = 'CUSTOMERS' and owner =
'SOE' ;
EMPTY_BLOCKS AVG_SPACE_FREELIST_BLOCKS NUM_FREELIST_BLOCKS
------------ ------------------------- -------------------
0 0 0
The table was analysed with dbms_stat, percent = 100:
exec dbms_stats.gather_table_stats( ownname=>'SOE', tabname=>
'CUSTOMERS', Degree=> 4, estimate_percent=> 100, granularity=>'ALL',
cascade=>TRUE, stattab=>'sm_stattab', statid=>'CUSTOMERS09270906',
statown=>'SOE')
Table name Owner NUM_ROWS BLOCKSSize (m) LAST_ANALYSED
------------------------------ ---------------- ---------- ---------- -------- ------------------- CUSTOMERS SOE 36364 3683 09/27/2006 08:39:50
But still the clust factor reported by dba_indexes for CUSTOMERS_PK is inferiors to CUSTOMERS block count.
Table Index Column Clust Distinct Name Name U Name Factor Keys NUM_ROWS -------------------- ------------------------------ - -------------------- ---------- ---------- ---------- CUSTOMERS CUSTOMERS_PK U CUSTOMER_ID 345 36112 36112
This is oracle 10.1.0.4
B. Polarski
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-- http://www.freelists.org/webpage/oracle-l -- http://www.freelists.org/webpage/oracle-lReceived on Wed Sep 27 2006 - 06:37:31 CDT
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