Different results in SQL vs. VBA using POWER function and VBA equivalent - sql

Question

I'm hoping someone who is actually good at math can help me out with this. I get a different result in SQL than I do in VBA. In SQL, I have this function that calculates a payment. Variables for both VBA and SQL are:
Principal = 239762.05
Rate = 0.03 (aka 3%) ETA: in both cases, 0.03 is divided by 12, so really this is .0025
Period = 268
#principal / (power(1+#rate,#period)-1) * (#rate*power(1+#rate,#period))
In SQL this gives a value of 1228.76 (rounded)
In VBA, I do not have the POWER function. So I copied this public function from the internet:
Public Function Power(ByVal number As Double, ByVal exponent As Double) As Double
Power = number ^ exponent
End Function
And I am calling it in a sub here like so:
NewPI = PrinBal / (Power(1 + IntCalc, Term) - 1) * (IntCalc * Power(1 + IntCalc, Term))
but here, the answer I get is 1228.63 (rounded). Only 13 cents off!
I have tried lots of adjustments and either ended up with this same figure, or a much worse result. I am thinking it's some sort of Order of Operations mistake, but I'm not sure.
EDIT
I am adding this to possibly get to the bottom of the problem, which might be the data types in the SQL version. This is the full function
create function [dbo].[PMT] (#rate numeric(15,9), #periods smallint, #principal numeric(20,2) )
returns numeric(16,2)
as
begin
declare #pmt numeric (38,9)
select #pmt = #principal / (power(1+#rate,#periods)-1) * (#rate*power(1+#rate,#periods))
return #pmt
end

Answer 1

Short answer: Data types in SQL matter.
Worse, you can experience some implicit data conversion. Check this out...
DECLARE #RealPrincipal real
SET #RealPrincipal = 239762.05
DECLARE #RealRate real
SET #RealRate = 0.0025 --0.03 --(aka 3%) ETA: in both cases, 0.03 is divided by 12, so really this is .0025
DECLARE #Period int
SET #Period = 268
SELECT #RealPrincipal / (power(1.0+#RealRate,#Period)-1.0) * (#RealRate*power(1.0+#RealRate,#Period))
Result = 1228.61333410069
Compare that to your formula from your OP comment with all literals and no variables...
SELECT 239762.05 / (power(1+0.0025,268)-1) * (0.0025*power(1+0.0025,268)) as 'Value!'
Result = 1228.761629
Now, use the same exact structure as the first code block, but replace the real type variables with money type...
DECLARE #moneyPrincipal money
SET #moneyPrincipal = 239762.05
DECLARE #moneyRate money
SET #moneyRate = 0.0025 --(aka 3%) ETA: in both cases, 0.03 is divided by 12, so moneyly this is .0025
DECLARE #Period int
SET #Period = 268
SELECT #moneyPrincipal / (power(1+#moneyRate,#Period)-1) * (#moneyRate*power(1+#moneyRate,#Period))
Result = 1233.2921
Now, using the money data types, watch what happens when you replace the literal 1 values in the formula with 1.0 ...
SELECT #moneyPrincipal / (power(1.0+#moneyRate,#Period)-1.0) * (#moneyRate*power(1.0+#moneyRate,#Period))
Result = 1228.761629

Answer 2

This is a simply explanation. VBA is simply using a higher precision then SQL. Not saying this is best practice and i am sure someone else it going to SLAP / downvote me but if you change it too something like the below you will see it forces SQL to use more decimal places and you get the same answer.
select
#principal/(POWER((#rate+1)*1.000000000,#period*1.00000000000000000)-1)(#ratePOWER((#rate+1)*1.0000000000000000000,#period*1.00000000000000000))
Again there are better ways to do this, this is just the quick and dirty.

Answer 3

I think you should simplify your calculation:
In VBA:
function payment(principal as double, rate as double, term as integer) as double
' Inputs:
' principal: Ammount of the loan
' rate: Effective interest rate per period
' term: Number of periods / payments
dim v as double, a as double
v = 1 / (1 + rate)
a = (1 - (v ^ term) / rate)
payment = principal / a
end function
or, if you want it squeezed into a single line:
function payment(principal as double, rate as double, term as integer) as double
' Inputs:
' principal: Ammount of the loan
' rate: Effective interest rate per period
' term: Number of periods / payments
payment = principal * rate / (1 - ((1 + rate) ^ (-term)))
end function
and, in SQL:
#principal * #rate / (1 - power(1 + #rate, -#term))
-- #rate is the effective rate per period (in your example: 0.0025)
Minimize the number of arithmetic operations.
If you're having problems with data type (as LDMJoe mentions in his answer), you should try casting each variable to an appropriate data type. In Access you can do something like this (assuming you're writing the expression in a query):
CDbl(principal) * CDbl(rate) / (1 - power(1 + CDbl(rate), -CInt(term))
I'm also assuming that principal, rate and term are columns in a table.