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Given the principal and term payment and number of terms, how can I calculate the interest rate of this mortgage?

Been searching the internet for formulas, but to no avail.

What is being calculated on the page linked to below is what I would like to do:

http://www.calcamo.net/loancalculator/quickcalculations/loan-rate.php5

Thanks,

Thomas Christensen

totoro
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  • Hope that my answer to problem #710331 can help. – Mick Mar 24 '14 at 11:39
  • I got all the functions in place as long as know the interest rate. The problem is a loan with a principal of P and n payments of A, which is very common where I live. People don't know the APR, and I want to calculate it so they can consider if it's worth borrowing. I have a solution programatically to approximate it iteratively by knowning A and trying different r (somewhat cleverly) until the function returning A ~= A. It takes about 60 iterations. – totoro Mar 25 '14 at 05:11
  • Let the scheme be [Principal = P, Rate = R% p.a., Installment = S, No. of installments = N months]. Their relation is given by S = PX^N (X–1) / (X^N – 1) where X = 1+ R/1200. It is not an easier job & no suitable method to find X (hence R) even if P, S, N are known. No wonder you need more than 60 iterations. However, if we re-write the formula as S = P(X–1) / [1–(1/X^N)], then R is approximately equal to 1200S/P, provided N is very large (like 40 years of repayments). That approximation can, however, be used as a closer initial guess so that the no. of iterations can be greatly reduced. – Mick Mar 25 '14 at 18:24
  • Yes, I am down to less than 10 iterations now. – totoro Mar 26 '14 at 14:20
  • Glad to hear that. – Mick Mar 26 '14 at 15:15

1 Answers1

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Let

  1. $A$ be the periodic amortization payment
  2. $P$ be the principal amount borrowed
  3. $r$ be the percentage rate per period
  4. $n$ be the number of payments

Then, the formula is $$A = P \frac{r (r+1)^n}{(r+1)^n-1}$$ This gives an equation in which the only unknown is $r$ and you need to solve it.

For the example you give, we have $A=700$, $P=100,000$, $n=15 \times 12=180$ so the solution is $r=0.00266252$ per month, that is to say $r=0.0319503$ per year; in more practical terms, $r=3.19$%.

If you want me to elaborate the solution technique for finding $r$, just post.

Added later to this answer

In my opinion, it is not possible to extract the analytical value of $r$ from the equation and only numerical techniques could be used. The simplest Newton method, which just uses the values of the function and of its first derivative, would do a very reasonable job. So, let us consider $$f(r)= P \frac{r (r+1)^n}{(r+1)^n-1}-A$$ $$f'(r)=P\frac{(r+1)^{n-1} \left((r+1)^{n+1}-n r-r-1\right)}{\left((r+1)^n-1\right)^2}$$

Starting from a guess $r_{old}$, the iterate will be given by $$r_{new}=r_{old}-\frac{f(r_{old})} {f'(r_{old})}$$

The problem is to find a first guess for $r_{old}$. Since we know that the number we look for is small, let us start with $r_{old}=0$. For this particular value, the is an indetermination for $\frac{f(r_{old})} {f'(r_{old})}$ but going to limit we can find that the first iterate is given by $$r_{new}=\frac{2 (A n-P)}{(n+1) P}$$. Compute this value, update $r_{old}$ by $r_{new}$ and repeat until convergence.

For the numerical case you gave, starting from $r_{old}=0$, the successive iterates are $0.00287293$, $0.00266362$, $0.00266252$ which is the solution for six significant figures.

You should notice that the very first estimate is not too bad.

We could get a better approximation expanding the equation as a Taylor series built at $r=0$. Limited to the second order, this leads to an expression which can be simplified and which reduces to $$P(n^2-1) r^2+6P(n+1)r-12(An-P)=0$$ Solving the quadratic equation for $r$ and retaining the positive root, in your case we immediately get $r=0.00266159$ which is quite better.

Do not forget to multiply the value of $r$ by $12$ and then by $100$ to get the usual result in %.

Added later to this answer

I have been trying to get a better estimate of the solution applying an higher Newton iterative sheme. Using a quartical method, what I obtained as a first estimate is

$$r_{old}=\frac{2 (n-k) (2 k (n+2)+(n-1) n)}{k^2 (n+2) (n+3)+2 k n \left(n^2+n-2\right)+(1-n) n^2}$$ in which $k=\frac{P}{A}$.

Applied to the test case, this leads to a value equal to $0.00266300$ while the exact solution is $0.00266252$.

Starting with the new definition of $r_{old}$, you would probably converge in a single iteration using the iterative scheme $$r_{new}=r_{old}-\frac{f(r_{old})} {f'(r_{old})}$$

Claude Leibovici
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