How to solve for effective interest rate including fees? 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?
 A: 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.
