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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?

Christoffer
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    Not an answer, but perhaps relevant. Borrowing money by paying less than the total balance due on your credit card is extraordinarily costly even when optimized. Consider asking about better strategies on the personal finance stack exchange. – Ethan Bolker Apr 01 '20 at 21:24

1 Answers1

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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?

Mark
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