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I’m solving some finance problems and faced the equation below. $$PV = C(\frac{1-\frac{1}{(1+r)^n}}{r}) + \frac{FV}{(1+r)^n}$$ Since I don’t have the finance calculator, I want to solve that for r in hand. Assume all PV, FV, n, C are constant.

Is there anyway to solve it, not using a trial and error approach?

I’ve tweaked that equation a couple of times. And this is what I’ve reached $$(1+r)^n\{rPV - C\} = rFV - C$$ But I don’t know how to go further. Is there anyway to solve it?

Joseph Hwang
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2 Answers2

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No, you can't solve it algebraically as far as I know, but your last equation is close to a good fixed point expression. Write it as $$r=\frac {(1+r)^n\{rPV - C\} +C}{FV}$$ now pick some starting $r$, plug into the right and calculate the left. Plug that into the right and keep going to convergence. If your starting guess is close enough and the derivative of the right side is less than $1$ in absolute value it will converge. If $nr \ll 1$ you can replace $(1+r)^n$ with $1+nr$ and have a linear equation to solve. That can give you a start if you don't have a better (experience based) guess for $r$.

Ross Millikan
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  • Thanks for your kind answer! $(1+r)^n$ to $1+nr$ approximation idea is very helpful! – Joseph Hwang May 29 '22 at 13:35
  • @JosephHwang it comes from the binomial expansion. $(1+r)^n=1+{n \choose 1}n+{n\choose 2}r^2+\ldots$. If $nr \ll 1$ the later terms are smaller than $nr$ and it may be accurate enough to ignore them. It is very useful in estimation. In this case it says the interest is small compared to the principal. – Ross Millikan May 29 '22 at 13:38
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You can obtain approximate solutions since, usually $r \ll 1$. For brevity, I shall use $P=PV$ and $F=FV$ $$P=C \,\frac{1-(1+r)^{-n}}{r}+F\,(1+r)^{-n}$$ Expand the rhs as an infinite sum $$P=\sum_{k=0}^\infty \Bigg[F \binom{-n}{k}-C \binom{-n}{k+1}\Bigg]\,r^n$$ Truncate to some order and use series reversion to obtain $$r \sim t + a_1\, t^2+a_2\, t^3+O(t^4)\qquad \text{with} \qquad t=\frac{2 (C n+F-P)}{n (C (n+1)+2 F)}$$ $$3 (C (n+1)+2 F)\,a_1=C n^2+3 (C+F)n+(2 C+3 F) $$ $$36 (C (n+1)+2 F)^2\,a_2=5 C^2 n^4+3 C (9 C+10 F)n^3+ \left(53 C^2+108 C F+48 F^2\right)n^2+3 (3 C+4 F) (5 C+6 F)n+\left(14 C^2+36 C F+24 F^2\right) $$

Try it with problems of your choice and, please, let me know how it works for you.

Edit

To make it more legible, consider the function $$f(r)=C \,\frac{1-(1+r)^{-n}}{r}+F\,(1+r)^{-n}-P$$ We have $$f(0)=C n+F-P $$ $$ f'(0)=-\frac{1}{2} n (C (n+1)+2 F)$$ $$f''(0)=\frac{1}{3} n (n+1) (C (n+2)+3 F)$$ $$f^{(3)}=-\frac{1}{4} n (n+1) (n+2) (C (n+3)+4 F)$$ $$f^{(4)}(0)=\frac{1}{5} n (n+1) (n+2) (n+3) (C (n+4)+5 F)$$ which show a very simple pattern.

Now, use the first iteration of the method which is one level above Householder's one (that is to say of order $5$) to get the following estimate of $r$ $$\frac{4 f(0) \left(f(0)^2 f^{(3)}(0)+6 f'(0)^3-6 f(0) f'(0) f''(0)\right)}{-24 f'(0)^4+f(0)^2 \left(f(0) f^{(4)}(0)-6 f''(0)^2\right)-8 f(0)^2 f^{(3)}(0) f'(0)+36 f(0) f'(0)^2 f''(0)}$$

Claude Leibovici
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  • For the approximations you need $nr \ll 1$, not just $r \ll 1$. They are pretty poor with a $5%$ mortgage that lasts $30$ years. It still works for many cases. – Ross Millikan May 29 '22 at 03:48
  • @RossMillikan. For sure but this is the start of the inverse series. On the site, I made much better; here the problem is that the formulae of the $a_i$ are really nasty. I shall try to improve. Cheers :-) – Claude Leibovici May 29 '22 at 03:54
  • That seems very interesting. I'll try your suggestion and see how it goes for me! – Joseph Hwang May 29 '22 at 13:36