Solving for i in mortgage payment formula in M = P [ i(1 + i)^n ] / [ (1 + i)^n – 1].

Question

This is my first post ont his website, so please excuse me if i missed some rules (please let me know and i'd be more than happy to edit my question accordingly).

I'm trying to figure out what level of annual interest rate would bring annual mortgage payments to a certain amount, call it $M.

My formula for annual mortgage payments is: $$M = \frac{P [ i(1 + i)^n ]}{[ (1 + i)^n – 1]}$$ where P is the price of the property, i is the annual interest rate and n is the term, call it 25 years in my case.

The idea is that i am making an assumption on M, n and P, and i'd like to solve for i. However, I have not done anything of this sort since my last year of university and am at a total loss while trying to isolate i. Essentially, I think I've forgotten the process and the tricks needed to be able to do this properly.

$$M = \frac {P*[ i(1 + i)^n ]}{[ (1 + i)^n – 1]}$$ $$\frac{M}{P} = \frac{[ i*(1 + i)^n ]}{[ (1 + i)^n – 1]}$$ $$\frac{M*[ (1 + i)^n – 1]}{P} = [ i*(1 + i)^n ]$$ $$\frac{[M+Mi]^n – M]}{P} = i+i^{n+1}$$ $$2Mi^n – M-i = P*[i^{n+1}]$$ $$\frac{2Mi^n-i}{i^{n+1}} = P+M$$ $$\frac{i^n-i}{i^{n+1}} = \frac{P+M}{2M}$$

Answer 1

Solving for $i$ amounts to computing the real root of a degree $n$ polynomial*, so unless $n \in \{1, 2, 3, 4\}$, there is in general no elementary closed form solution. For $n = 25$ as in your case, your best bet is to use a recursive numerical algorithm; e.g., Newton's method. It is convenient to let $v = 1/(1+i)$ or equivalently, $i = \frac{1}{v} - 1$, so that the equation of value is $$P\left(\frac{1}{v} - 1\right) = M(1-v^n). \tag{1}$$ Rewriting this as a polynomial in $v$, we obtain $$v^{n+1} - \left(1+\frac{P}{M}\right)v + \frac{P}{M} = 0. \tag{2}$$ Let $c = P/M > 1$ and we seek the unique real root $0 < v < 1$ of $(2)$. To this end, we set up the Newton's method recursion with initial guess $v_0$ by computing the derivative of $f(v) = v^{n+1} - (1+c)v + c$: $$f'(v) = (n+1)v^n - (1+c), \tag{3}$$ hence $$v_{k+1} = v_k - \frac{f(v_n)}{f'(v_n)} = \frac{n v_k^{n+1} - c}{(n+1)v_k^n - (c+1)}. \tag{4}$$

What is a suitable initial guess? Well, we know that if the payments were made in perpetuity, the equation of value would simply be $M = Pi$, hence $i = M/P$ is the periodic interest rate in the limiting case $n \to \infty$. So this furnishes a lower bound on $v$, namely $v > 1/(1+M/P)$. So we use this as our initial guess. Convergence is rapid: for instance, suppose the monthly mortgage is $M = 2500$ on a principal of $P = 375000$. The term of the loan is $n = 240$ months, or $20$ years. Payments are made at the end of each month. Then $c = 375000/2500 = 150$ and our initial estimate of $v$ is $$v_0 = \frac{1}{1 + M/P} = \frac{150}{151} \approx 0.993377.$$ We compute the iterates of $(4)$ until they stop changing, and summarize these in a table:

$$\begin{array}{c|c} k & v_k \\ \hline 0 & 0.993377 \\ 1 & 0.995353 \\ 2 & 0.995729 \\ 3 & 0.995750 \\ 4 & 0.995751 \\ 5 & 0.995751 \\ \end{array}$$

This gives us $i = 1/v - 1 \approx 0.00426763$, which is the effective periodic (monthly) rate of interest, and the effective annual rate is $(1+i)^{12} - 1 \approx 0.0524308$.

*Note. Although $(2)$ is degree $n+1$ in $v$, it should be observed that $v = 1$ is always a solution, thus the extra degree comes from an extraneous linear factor of the form $(v-1)$. However, factoring it out results in a cumbersome expression, which is why we employ $(2)$ instead.