Hello Alexander,

i thought that it was

> Does it mean that if I have a frequently accessed table with:
>
> 5 NUMERIC(18,0)

INT64 -> 64bit
8 bytes per column 8*5 = 40 bytes

> 3 TIMESTAMP

Two 32 bit longwords -> 64bit
8 Bytes per column 8*3 = 24 bytes

> 2 VARCHAR(10)

plus 2 bytes for every varchar you use
( i read that in
http://www.ibphoenix.com/main.nfs?a=ibphoenix&page=ibp_data_types
not sure but i use that anyway ... )

10 + 2 Bytes per column 12*2 = 24 bytes

> My row structure will be: 5*64 + 3*64 + 20*1 = 212 bytes * 125 % = 265

88 bytes * 125 % = 198

> bytes meaning that a page size of 1024 bytes will be perfectly OK.

> Did I get you right?