How to solve for effective interest rate including fees?

Question

I have gotten very stuck on a math problem involving interest rates when combined with flat fees and cashback incentives. I am looking to determine an "effective interest rate" so that two loans with different fees, rates, and incentives can be compared.

Here's what I've got so far...

$$ COST = \frac{r}{12} \cdot P \cdot t \cdot \frac{(1+\frac{r}{12})^t}{(1+\frac{r}{12})^t-1}+f-c $$

where r is the annual interest rate, P is the principal, t is the term, f is the fee, and c is the cashback incentive.

Of these variables, I know everything needed to calculate COST.

My question is... how can I "unwind" COST to get to an "effective" interest rate?

$$ COST = \frac{R}{12}\cdot P\cdot t\cdot \frac{(1+\frac{R}{12})^t}{(1+\frac{R}{12})^t-1}$$

Where P and t are known. I've gotten as far as this...

$$ \frac{12\cdot COST}{P\cdot t}=\frac{R\cdot(1+\frac{R}{12})^t}{(1+\frac{R}{12})^t-1}$$

But now I'm stuck. How can I solve for R from here?

Answer 1

Here's how I have done it when I was looking at loans:

First, convert the annual rate to a monthly rate $r$ (so we don't clutter the formulas with all those "divide by $12$"s). Set $t$ to the number of monthly payments.

Nominally, you are borrowing principal $P.$ That is how the lender computes the monthly payments. If the annual rate is $r$ and you make $t$ payments, starting one month after you receive the loan amount, the lender should ask for a monthly payment equal to $$ A = \frac{r}{1 - (1 + r)^{-t}} P. $$

But in actual reality, after paying $f$ fees and getting $c$ cash back, you end up receiving a net amount of only $P - f + c$ from the lender at the start of the loan. That is the amount the lender really lent to you; the nominal amount $P,$ the fees, and the cash back are all just window dressing around this amount. But you still have to pay the amount $A$ each month that the lender asked for. So the effective monthly interest rate is the rate $R$ that satisfies this equation: $$ A = \frac{R}{1 - (1 + R)^{-t}} (P - f + c). $$

I think this will usually need to be solved numerically, that is, there is not a neat "closed-form" formula to solve it.

Answer 2

Starting from @David K's answer, let $k=\frac A{P-f+c}$ to make the equation $$\frac{R}{1 - (1 + R)^{-t}}=k$$ which does not show explicit solutions and which will then require some numerical method (Newton being probably the simplest).

However, we can make quite good approximations since we know that $R \ll 1$.

Then, using Taylor series, this would give $$\frac{R}{1 - (1 + R)^{-t}}=\frac{1}{t}+\frac{ (t+1)}{2 t}R+\frac{ t^2-1}{12 t}R^2- \frac{t^2-1}{24 t}R^3-\frac{ t^4-20 t^2+19}{720 t}R^4+\frac{ t^4-10 t^2+9}{480 t}R^5+\frac{ 2 t^6-147 t^4+1008 t^2-863}{60480 t}R^6+O\left(R^7\right)$$ and using series reversion $$\color{blue}{R=x-\frac{t-1}{6} x^2+\frac{2 t^2-t-1}{36} x^3-\frac{22 t^3+3 t^2-18 t-7}{1080}x^4+\frac{52 t^4+44 t^3-39 t^2-46 t-11}{6480}x^5-\frac{300 t^5+484 t^4-43 t^3-447 t^2-253 t-41}{90720}x^6+O\left(x^7\right)}$$ where $\color{blue}{x=\frac{2 (k t-1)}{t+1}}$.

To make an example, let us use $k=\frac 1 {100}$ and $t=120$ and let $R_{(n)}$ be the computed interest rate obtained using the expansion to $O\left(x^n\right)$. The results would be $$\left( \begin{array}{cc} n & R_{(n)} \\ 1 & 0.0033057851 \\ 2 & 0.0030890422 \\ 3 & 0.0031178218 \\ 4 & 0.0031136135 \\ 5 & 0.0031142750 \\ 6 & 0.0031141662 \end{array} \right)$$ while the exact solution would be $R=0.0031141827$.

For more simplicity, we can transform (for the same accuracy) the long formula as a Padé approximant $$\color{blue}{R=x \frac{1+\frac{t+1}{2} x+\frac{2 t^2+5 t+2}{45} x^2 } {1+\frac{2 t+1}{3} x+\frac{6 t^2+5 t+1}{60} x^2 }}$$