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Home -> Community -> Mailing Lists -> Oracle-L -> Re: storage parameters
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_at_hotmail.com 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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