Replacement from the Value of an Attribute*

Question

Replacement from the Value of an Attribute
With formula variables you can set the processing type Replacement from the Value of an Attribute and create a reference to the reference characteristic for the variable. The attribute Reference to Characteristic (Constant 1) is a dummy attribute that is available with each characteristic. It serves to create a reference to the characteristic, by which it does not need to be aggregated. By choosing this attribute, you can influence the aggregation behavior of calculated key figures in a targeted way and can improve performance during calculation.
1)what does this do in the query?
2)what is the value of the variable created ?(like 1,2 etc {i.e.} it changes or is  a constant value like 1)?
E.g. i have created a formula variable zvar , (replacement path  type) ref. char is wbselement, then what will be the value of the zvar if its used in a formula
eg.  Formula 1 = kef1 kef2zvar.
what will be the value(s)  of the variables in the these cases
1. project = 1000,1000.30.
2.project = 2000,2000.30.01
3.proect = 3000,3000.30.01,3000.30.02.
3.) as per the def. its a  number which refer to a wbs, then in the FORMULA , will it consider the zvar as a VALUE or what would happen?
Edited by: jumboash on Oct 23, 2009 7:25 PM

Answer 1

Hi,
ZVAR is not considered as value in the calculation. It is used only to calculate the formula correctly without aggregation of the operands.
For example there are two materials m1 and m2 . total amount has to be calculated for these materials whose quantitys are 10 and 20 and whose prices are 200 and 300 respectively.
Case1:- if ZVAR is not used then the calculation of the fromula will be as follows
qty   Price    amount
10 * 200   = 2000
20 * 300   = 6000
30 * 500 =  15000     ->   The total amount shows is 15000 which is not correct
Case 2: -  if ZVAR is used in the calculation of the formula
qty   Price    amount
10 * 200   = 2000
20 * 300   = 6000
                   8000   -> The total amount shows is 8000 which is correct
so using the ZVAR in the formula shows the correct result without considering the aggregation . This is useful in the case if the material is not part of rows.