Hi,
your row size is 36, not 24.
add 2 bytes for VARCHAR data types.
size of INTEGER is 4 bytes.
size of DOUBLE is 8 bytes.
your index size is 14+6+4+4 = 28 bytes
index entries needs physical row address as 8 bytes + node address.
pages use free space.
Firebird uses MGA architecture.

Firebird isn't a file based record manager like Btrieve. its a RDBMS.

Try the oracle for 17M rows and see the result.

Ali

--- In firebird-support@yahoogroups.com, "drstanl" <drstanl@v...>
wrote:

> I have been evaluating Firebird for a client. They rejected it
> because the file size was too large.
> It *does* seem rather large. I'm wondering if anybody can tell

me

> why.
>
> Firebird 1.0.3 on Windows 98 se.
>
> File size Situation
> ------------- -------------------
> ~500,000 Database, empty.
> ~650,000 Database with empty table.
> 1,758,031,872 Table with 16,947,365 records.
>
> Table definition:
>
> CREATE TABLE T1 (
> ID1 VARCHAR(12) NOT NULL,
> ID2 VARCHAR(4) NOT NULL,
> D1 INTEGER NOT NULL,
> D2 INTEGER NOT NULL,
> V1 DOUBLE PRECISION,
> CONSTRAINT PK_T1
> PRIMARY KEY (ID1, ID2, D1, D2) );
>
> Thus, a record is 24 bytes, the fields for the PK comprise 20
> bytes, but the database averages some 103 bytes per record (after
> dropping the 650,000 bytes of the empty database and table).
>
> This per-record overhead seems significant, and affected my
> client's decision. Not that it would matter to them, but I'd like

to

> understand how Firebird produces this result.
>
> TIA,
> -- Stan