Forum Discussion
ipmt and ppmt calculation has wrong data type or the result is too large/small
Was following a youtube video on PPMT and IPMT functions but encountered this error and don't know how to fix it.
Hi Gutts_Zero ,
According to your description, here are my steps you can follow as a solution.
(1) We can create measures.
Measure = var _a=CALCULATE(MAX('Fact'[Period]),FILTER(ALL('Fact'),'Fact'[Period]=MAX('Fact'[Period]))) return IF([Is within Period]=0,1,_a)ipmt = ipmt(0.06/12,[Measure],'Loan Period'[Loan Period Value]*12,200000,0,0)(2) Then the result is as follows.
Best Regards,
Neeko Tang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
3 Replies
Hi Gutts_Zero ,
I'm sorry I don't have access to your pbix file.
Double-check the data types and values of the arguments to be passed to the function to ensure that they are correct and within acceptable limits.
According to the documentation, the per argument must be between 1 and nper (inclusive). However, in your measure, you have max(fact[period]) as the per argument, which may be greater than loan period [loan period value]*12, which is the nper argument. This could cause the error that you see.
If the above one can't help you get the desired result, please provide some sample data in your tables (exclude sensitive data) with Text format and your expected result with backend logic and special examples.
Best Regards,
Neeko Tang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Hi Gutts_Zero ,
According to your description, here are my steps you can follow as a solution.
(1) We can create measures.
Measure = var _a=CALCULATE(MAX('Fact'[Period]),FILTER(ALL('Fact'),'Fact'[Period]=MAX('Fact'[Period]))) return IF([Is within Period]=0,1,_a)ipmt = ipmt(0.06/12,[Measure],'Loan Period'[Loan Period Value]*12,200000,0,0)(2) Then the result is as follows.
Best Regards,
Neeko Tang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
