Vtgyjfy

$$p=frac{c}r*(1-frac1{(r+1)^t} )$$

I'm stuck. I'm building an excel model where I will be able to put in inputs for p, c, and t... but it will need to solve for "r."

For some reason, I can't figure out the algebra.

Can you help me solve for "r"?

As Hello_World answered, the simplest way would be to use Newton method for finding the zero of function
$$f(r) = frac{c}r left(1-frac1{(r+1)^t}right)-p$$

If you start using $r_0=0$, the first iterate would be $r_1=frac{2 (c t-p)}{ct left(t+1right)}$ and you just need to continue until convergence to the desired precision.

The is another thing you can do since $r ll 1$. Build the Taylor series at $r=0$ to get
$$f(r)+p=c t-frac{1}{2} r (c t (t+1))+frac{1}{6} c r^2 t (t+1) (t+2)-frac{1}{24} r^3 (c t
(t+1) (t+2) (t+3))+frac{1}{120} c r^4 t (t+1) (t+2) (t+3) (t+4)-frac{1}{720}
r^5 (c t (t+1) (t+2) (t+3) (t+4) (t+5))+Oleft(r^6right)$$ and now use series reversion to get
$$color{blue}{r=x+frac{t+2}{3} x^2+frac{5 t^2+17 t+14}{36} x^3+frac{17 t^3+78 t^2+117 t+58}{270}
x^4+frac{193 t^4+1094 t^3+2301
t^2+2144 t+748 }{6480}x^5+Oleft(x^{6}right)}$$ where $color{blue}{x=frac{2 (c t-p)}{ct left(t+1right)}}$ (this is $r_1$)

Let us try using $p=100000$, $c=1000$ and $t=120$. This gives $x=frac 1 {363}$. Using the expansion above, this leads to $r=frac{31795409217001}{10210526568429660}approx 0.00311398$ while the "exact" solution obtained using Newton method would be obtained after the following iterates
$$left(
begin{array}{cc}
n & r_n \
0 & 0.000000000000 \
1 & 0.002754820937 \
2 & 0.003161445197 \
3 & 0.003114792911 \
4 & 0.003114182051 \
5 & 0.003114181946
end{array}
right)$$

As you can see, without any iterative procedure we can get a very close solution of the problem (for the working case, the relative error is about $0.0065$% !).

Adding more terms will give a better result (if you want them, just ask).

Edit

Working a bit more, the blue formula may be written in a more compact form (for an even better accuracy) using a Padé approximant
$$r=x ,frac{1+a_1x+a_2x^2}{1+b_1x+b_2x^2}$$ where
$$a_1=-frac{2 left(2 t^2+3 t+1right)}{11 t+13}qquad a_2=frac{4 t^3-3 t-1}{45 (11 t+13)}$$
$$b_1=-frac{23 t^2+53 t+32}{3 (11 t+13)}qquad b_2=frac{67 t^3+240 t^2+291 t+122}{60 (11 t+13)}$$ For the work example, this would give $r=frac{13916770468}{4468838772519}approx 0.0031141805$

Update

Asking a banker friend of mine, he mentioned an approximation he saw somewhere in the past (he does not remember when and/or where. I just found where). I is
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ Applied to the worked example, this gives $0.00310743$.

Using this estimate as $r_0$, Newton iterates would be
$$left(
begin{array}{cc}
n & r_n \
0 & 0.003107429977 \
1 & 0.003114180156 \
2 & 0.003114181946
end{array}
right)$$ which is real fast. I suppose that one single iteration of Newton method will be more than sufficient.

You could use as a very safe solution
$$r=r_0+frac{2,f(r_0) ,f'(r_0)}{f(r_0), f''(r_0)-2, f'(r_0)^2}$$ where $r_0$ is the result of the green formula.

Applied to the worked example, this would lead to
$r=0.00311418194589$ while th exact solution would be
$r=0.00311418194600$

You could try Newton's method to solve for $r.$ Take,
$$F(r) = p-frac{c}rcdot left(1-frac1{(r+1)^t}right)$$
then your goal is to find $r$ such that $F(r)=0.$ For this you start with some choice $r_0$ and then use the following recursive definition:
$$r_{n+1}= r_n - frac{F(r_n)}{F'(r_n)}.$$
This will converge to the root of $F.$

I prefer to add a second answer.

Being just fascinated by David W. Cantrell's approximation
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ totally inspired by it, I tried something in the same spirit
$$frac{c}r left(1-frac1{(r+1)^t}right)=pimplies 1+frac{c}{p}=1+frac{r}{1-(r+1)^{-t}}$$ Taking logarithms of both sides
$$log left(1+frac{c}{p}right)=log left(1+frac{r}{1-(r+1)^{-t}}right)$$ Now, expanding the rhs as a Taylor series at $r=0$
$$log left(1+frac{c}{p}right)=log left(1+frac{1}{t}right)+frac{r}{2}+frac{2t-5}{24} r^2 +Oleft(r^3right)$$
Neglecting the second order term, we the obtain the first approximation
$$color{blue}{r_1 =2 log left(frac{t ,(c+p)}{p, (t+1)}right)}$$
Using the complete expansion to $Oleft(r^3right)$, we then have the second approximation
$$color{blue}{r_2=frac{sqrt{1+4, alpha, r_1}-1}{2 alpha }}qquad text{where}qquad color{blue}{alpha=frac {2t-5}{12}}$$

We could even avoid quadratic equations building the simplest Padé approximant instead of the Taylor series. This gives
$$log left(1+frac{c}{p}right)=frac{log left(1+frac{1}{t}right)+frac{1}{12} left((5-2 t) log
left(1+frac{1}{t}right)+6right) r} {1+ frac{5-2t}{12} r }implies
color{blue}{r_3=frac{12,r_1}{12+(2t-5),r_1}}$$

Applied to the worked example, this would give
$$r_1=2 log left(frac{606}{605}right)approx 0.00330306$$
$$r_2=frac{2}{235} left(sqrt{9+1410 log left(frac{606}{605}right)}-3right)approx 0.00311325$$
$$r_3=frac{12 log left(frac{606}{605}right)}{6+235 log
left(frac{606}{605}right)}approx 0.00310238$$
while the exact solution is $$r=0.00311418$$