Secondary Index Size

[Liam1]

Hello,

I would like to know a method to calculate the size of a secondary index/indices.

Is this possible without calculating the size of the data held within the table and then jiggering about with figures?

The reason being I'm trying to explain to some one that they might not need 13 SI's on their table....

Thanks,

Liam1

 

[tdatgod]

start up Ferret from the DBW window.

scope table "databasename.tablename" 0
showblocks /m

there are 2 columns over on the right hand side of the display ( probably off the screen ).

Average Datablock size
Number of datablocks

multiply the 2 of of these together and divide by 2 ( because the value is in 512 bytes units ) to get The size of the subtables in K.

1024 is the primrary data table
1028 - 2047 are your secondary indexes.
2048 and beyond is fall back

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[dnoeth]

The next time they should query dbc.tablesize before and after creating an index ;-)

If you don't have access to ferret, you'll have to calculate.
There are some formulas in the manuals:
Database Design -> Capacity Planning -> Database Level Considerations -> Sizing Base Tables and Index Subtables

Dieter