In order to compare credit cards I would like to calculate the effective interest of each with the following variables:
- nominal interest = e.g. 10%
- startfee = e.g. 300
- monthly_fee = e.g. 25
- period = e.g. 36 (months, e.g. 3 years)
- credit_sum = 10000
The credit sum is paid monthly and Interest is compounded monthly
I have found thousands of pages showing the formula how to calculate the effective interest based on only nominal interest but nothing that shows the inclusion of fees.
From what I understand how this is calculated, I calculate the total fees for the loan (fees+interest) and then do a backwards calculation to calculate the interest rate?
For example, if the total cost of a loan for 3 years 100.000 is 115.000, what would then be the yearly effective rate?
The following equation calculates how much you'd have to pay if no monthly fees or payments were made:
$T(t) = (L+f)(1+\frac{r}{n})^{tn}$
where
T= total cost of loan after time t (years)
L = Inital Loan Amount
f = any starting fees
r = interest rate over 1 time unit (annual interest rate)
n = number of times compounded over 1 time unit (yearly) (so n=12 for monthly)
nt = what you call period, total number of compoundings
If you want to add a monthly fee $m$ and payments $p$. The formula needs to change a bit: Call $b=p-m$, the amount paid minus any fees, the balance difference after a month, not accounting for interest accrued.
$T(2/n) = [(L+f)(1+\frac{r}{n})-b](1+\frac{r}{n})-b$
So $T(3/n) = \{[(L+f)(1+\frac{r}{n})-b](1+\frac{r}{n})-b\}(1+\frac{r}{n})$
So $$T(t) = \{[(L+f)(1+\frac{r}{n})-b](1+\frac{r}{n})-b\}(1+\frac{r}{n})... = (L+f)(1+\frac{r}{n})^{tn}-b(1+\frac{r}{n})^{tn-1}-b(1+\frac{r}{n})^{tn-2}-...-b(1+\frac{r}{n})^{1}$$ $$=(L+f)(1+\frac{r}{n})^{tn}+b(\frac{1-(1+r)^{tn+1}}{r} +1) $$
So for your example: $115000 = (100,000+f)(1+r/12)^{36}+b(\frac{1-(1+r)^{37}}{r} +1)$
So in order to know $r$ the annual interest rate, you need to know any intial flat assessment fees $f$, and the difference between the amount paid each month and monthly fees. If you want to bundle in the fees into an effective interest rate the problem is much simpler.
$T = L(1+\frac{r}{n})^{tn} \implies n[(\frac{T}{L})^\frac{1}{nt}-1]=r$
This is a more reasonable question: what is the effective interest rate you are paying each year?