Detached index extentsize formula

Last year an Informix Corp. engineer posted this formula to calculate (detached) index extent size: (keysize + 9 / rowsize) * extent size which I take to mean: Add 9 bytes to the column size of the table in question's primary key; divide it by table rowsize; multiply this by the (first) extent size; round up; this is size in kilobytes of table's index extents. So, a table with pk size of 16 and default extent size would have index extent size of 3? Is this correct? Is there any way to modify this behavior? tia

In article <37EA49BF.DED57C06@yahoo.com>, Red Valsen <red_valsen@yahoo.com> writes >Last year an Informix Corp. engineer posted this formula to calculate >(detached) index extent size: > > (keysize + 9 / rowsize) * extent size > >which I take to mean: Add 9 bytes to the column size of the table in >question's primary key; divide it by table rowsize; multiply this by the >(first) extent size; round up; this is size in kilobytes of table's >index extents. So, a table with pk size of 16 and default extent size >would have index extent size of 3? > >Is this correct? Is there any way to modify this behavior? > Check the FAQ at www.smooth1.demon.co.uk. Detached indexes are different as well, but the FAQ now covers it all with examples in section 8.42! >tia > -- David Williams