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key figure calculation past-current-future logic and from-to period logic

Daniele_Cuozzo
Discoverer
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138

we need to create a calculated key figure (KF) that incorporates the following elements:

  1. Past values of one key figure (KF1).
  2. Present and future values of another key figure (KF2).

This calculated key figure should take in account for these components in its calculation logic. For instance:

At time t, the value of the calculated KF may depend on:

 PASTCURRENTFUTURE
monthmar-24apr-24mag-24giu-24lug-24ago-24set-24ott-24nov-24dic-24gen-25feb-25mar-25apr-25mag-25giu-25lug-25
kf1 1010101010101010101010101010101010
kf2 1010101010101010202020202020202020
k3kf1kf1kf1kf1kf1kf1kf1kf1kf2kf2kf2kf2kf2kf2kf2kf2kf2

for example:

  • KF1(t−1): Past value of KF1 (one period before).
  • KF2(t): Present value of KF2.
  • KF2(t+1): Future value of KF2 (one period ahead)
  • kf3= IF(MONTH>= TODAY; KF2; KF1)

how is possible to use this logic in ibp calculation?

in addiction, is it possible to create ibp calculation with "FROM\TO PERIOD" logic?

 20232024
month123456789101112123456789101112
kf1 101010101010101010101010101010101010101010101010
kf2 202020202020202020202020202020202020202020202020
kf3kf1kf1kf1kf1kf1kf1kf2kf2kf2kf2kf2kf2kf1kf1kf1kf1kf1kf1kf2kf2kf2kf2kf2kf2

the kf3 logic is:

every years from month 1 to month 6 take in account kf1; from moth 7 to moth 12 take in account kf2

How can we create  an IBP calculation like the kf3, that take in account the value of KF1 from month 1 to  month 6 ;from moth 7 to moth 12 take in account kf2 , independent of the year?

thanks in advance

 

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peter_casper
Contributor
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Hi @Daniele_Cuozzo,

this can be achieved by adding an IF statement to your calculated KF3 in the following way:

IF(("PERIODID3" => "$$PERIODID3CU$$"), "KF2", "KF1")

The number within the PERIODID represents the time level of your time profile. In this case 3 stands for months. However you need to reveal the correct number based on your indiviually used time profile.

Your other questions are related to the same logic and can be solved with small additions.

BR, Peter