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
Let
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 $$r_{new}=r_{old}-\frac{f(r_{old})} {f'(r_{old})}$$