Re: storage parameters

  • From: "Jay Hostetter" <jhostetter@xxxxxxxxxxxxxxxxxxxx>
  • To: <oracle-l@xxxxxxxxxxxxx>
  • Date: Tue, 16 Mar 2004 08:17:04 -0500

This may or may not be flawed logic, but what I would do is this:

150 (sum of columns) * 300 (number of records) * 1.10 (allow 10% or so for 
overhead) = 49500 or 49.5k.

I always use multiples of the database block size.  I usually don't go below 
128k.  So I would use initial and next of 128k.  Then load up the table in a 
test database to see if the table grew.  You don't have to size the table 
perfectly.  I just use this formula to get in the ballpark.  I used to spend 
the time with those formulas they had in the Oracle 7 manuals, but with uniform 
tablespaces, lots of disk space, and the debunking of the "extent" myth, I 
think it is just a waste of time.

Jay

>>> oracledbam@xxxxxxxxxxx 03/16/04 02:16AM >>>
what would be value of storage parameters of following tables?
create  table search_types (
        modifier_type_key   varchar2(50)  not null    PK ,
        modifier_pretty_type varchar2(100)  not null
);

assumption: total#of rows=300 , pctfree=5 ,inittrans=1 
,pctincrease=0,pctused=40
what would be good value of INITIAL and next ?
thx-seema

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