Why dont we use more than 10 InfoProviders in one InfoSet ?

Question

. It is better to create multiple InfoSets
depending on reporting needs. plz make me understand about this...
thanks

Answer 1

Hi,
Please check the below link:
http://help.sap.com/saphelp_nw04s/helpdata/en/05/7ce2416149c717e10000000a155106/frameset.htm
Constraints
●     For performance reasons, you cannot define an InfoCube as the right operand of a left outer join.
●     SAP does not generally support more than two InfoCubes in an InfoSet. If you include more than two InfoCubes in an InfoSet, the system produces a warning. There are several reasons for this limitation:
○     Generally, the application server cannot create SQL statements with more than 64kb (in Unicode systems 32k characters). The more InfoCubes you use in an InfoSet, the quicker this limit is reached.
○     In contrast to the star schema (for which the potentially useful database access plans are limited by the table structure), several InfoCubes exist for a join, and several fact tables or DataStore object tables exist if you join InfoCubes with DataStore objects. There is no longer one large table at the center of the schema, and choosing a good access plan is much more difficult. Therefore, the average response time increases exponentially with the number of InfoCubes included.
○     If all of the characteristics affected by the join condition are not in the drilldown of a query, the key figure values of InfoCubes and DataStore objects are duplicated when you join them to InfoProviders (see SAP Note 592785). Therefore, interpreting the results of joins with non-unique InfoProviders becomes more difficult the more InfoProviders you include.
Design Recommendations
●     To avoid problems caused by duplicated key figure values (see SAP Note 592785), we recommend that you only stage the key figures of one DataStore object or InfoCube of the InfoSet for the query (indicator in the first column in InfoSet maintenance).
●     We recommend that you only use one InfoSet object (DataStore object, InfoCube, or master data table) with ambiguous characteristic values. This means that when you join a DataStore object with an InfoCube, as long as the InfoCube contains the visible key figures, all the key characteristics of the DataStore object are used in the join condition for the InfoCube. Equally, when joining a master data table with compounding to an InfoCube, all of the key characteristics of the master data table are joined with the characteristics of the InfoCube.
Thanks,
Venkat