> 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.

i forget that ( i think ) the windows and linux use a 4Kb low level
i/o write chunks. So if my row doesn't exceed the 4Kb limit i use the
4096 page size but not less ...

Anyway this is only my thoughts ...

Dimitris