Solving for $n$ in Future Value Growing Annuity formula

Question

Iam trying to solve for the $n$ parameter in the Future Value Growing Annuity formula:

$$FV = \frac{C}{r-g}\big[(1+r)^n - (1+g)^n\big],$$ where

• $C$ is the periodic payment.
• $r$ is the interest rate.
• $g$ is the growth rate.
• $FV$ is the future value of payments $C$ at interest rate $r$ and growth rate $g$ over $n$ periods.
• $n$ is the number of periods.

I am using the following parameter values.

• $C = 2,800$.
• $r=.04$.
• $g=.03$.
• $FV=100,000.$

The answer should equate to $19.123$ periods.

However I am not able to compute this mathematically, since I am not too savvy with logarithms.

I do know that for example $(1+r)^n$ is equal to $n\log(1+r)$, but I am having difficulty with the FV Growing Annuity formula since there is two $n$ parameters.

Please help me out with the full mathematical solution for parameter $n$ using the above noted values.

Thank you.

Answer 1

Someone can correct me if I'm wrong, but I don't believe this can be solved analytically. This is an instance where you will need to use approximation techniques such as linear interpolation, Newton's method, or the bisection method. If you have a financial calculator such as the BAII Plus it might be able to solve this for you: http://www.tvmcalcs.com/index.php/calculators/apps/ti-baii-plus-graduated-annuities.

Answer 2

It is not possible to solve it algebraically. It seems that you know this already since you wanted to calculate the derivative. The derivative is necessary to apply the Newton´s method. The function is

$$f(x)=\frac{1.04^x-1.03^x}{0.01}-\frac{100000}{2800}$$

And its derivative is

$$f^{'}(x)=\frac{\log(1.04)\cdot 1.04^x-\log(1.03)\cdot 1.03^x}{0.01}$$

According to the newton method we have $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$. You have to choose a n initial point $x_0$. I´ve chosen $x_0=10$. It is some work to do the calculation for all the iterations. To avoid that I used a spreadsheet program. See the picture below

Since $n$ is an integer you probably need three or four steps only. If something is unclear, feel free to ask.

Answer 3

Starting from @callculus'answer, you want to find the zero of function

$$f(x)=\frac{1.04^x-1.03^x}{0.01}-\frac{100000}{2800}$$ which, using whole numbers, write $$f(x)=\left(\frac{26}{25}\right)^x-\left(\frac{103}{100}\right)^x-\frac{5}{14} $$What you can do is to consider instead function $$g(x)=x \log \left(\frac{26}{25}\right)-\log \left(\frac{5}{14}+\left(\frac{103}{100}\right)^x\right)$$ which is almost a straight line.

Just make a series expansion around $x=0$ to get $$f(x)=-\log \left(\frac{19}{14}\right)+x \left(\log \left(\frac{26}{25}\right)-\frac{14}{19} \log \left(\frac{103}{100}\right)\right)+O\left(x^2\right)$$Ignoring the higher order terms, this gives, as an estimate, $$x=\frac{19 \log \left(\frac{19}{14}\right)}{19 \log \left(\frac{26}{25}\right)-14 \log \left(\frac{103}{100}\right)}\approx 17.51$$ This is in fact the first iterate of Newton method applied to $g(x)$ with $x_0=0$.

Let us look at all iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.000000 \\ 1 & 17.50987 \\ 2 & 19.11198 \\ 3 & 19.12295 \end{array} \right)$$

If we do the same with $f(x)$, they would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.000000 \\ 1 & 36.96400 \\ 2 & 25.28435 \\ 3 & 20.04064 \\ 4 & 19.14579 \\ 5 & 19.12296 \\ 6 & 19.12295 \end{array} \right)$$

Notice that in a second case, we have a serious overshoot of the solution (by Darboux theorem).