Mortgage calculator

I don’t like the mortgage calculators banks put on their website. You type in a bunch of numbers and then you get one number back.

Life moves much faster than this. Now you think it’s 30 years, then it’s 25, mmmh, and how about an interest rate of 5.5% p.a. instead of 4.8% p.a. You don’t want to type that in each time, do you?

These mortgage calculators are also not showing all the information. Rarely do you see straight away how much interest you have to pay. For a 30-year loan that can be as much as the principal loan you want to borrow.

Worst of all, you wouldn’t learn how amortization works. And that would be a shame.

Technical details

This is for those who want to see what’s going on under the hood.

Monthly repayments are conditional on the length of the loan term (horizontal axis) and annual interest rate (vertical axis). Click on any point in the graph and the sliders above for Loan term and Annual interest rate will be updated.

Have a go and see what happens when you move left and right or up and down. Pay attention to the Total interest. For example, if you try to follow one of the contour lines, say $4,000, from lower left to upper right, that means you’re still paying the same amount each month. But the the total paid depends on where you are on that $4,000 contour line.

Now, imagine you have a much higher loan and the loan term extends to infinity. This way you could repay a higher loan indefinitely with a reasonable monthly repayment. That’s what we call rent. 🤩

How it works

The contour lines show combinations of interest rate and loan term that produce the same monthly repayment for the selected loan amount. This makes the trade-off visible: for a fixed loan, a shorter term can have the same repayment effect as a higher interest rate, and vice versa.

The maths behind

The equation for a fixed scheduled repayment \(M\) of a principal loan \(P\) over a number of periods \(n\) with an interest rate \(r\) is

\[M = P \frac{r(1+r)^n}{(1+r)^n-1}.\]

In case of monthly repayments, we divide the annual interest rate (it’s usually given as % p.a., which is short for per annum, which is Latin for per year) by 12 and have to multiply \(n\) with 12 if the duration of the term is in years.

Let’s find out under what conditions the principal equals the interest paid.

The total amount repaid is \(n\,M\). The total interest is \(I = n\,M - P\). And we want:

\[I = P\]

A little algebra turns this into

\[n\,M - P = P\]

or:

\[n\,M = 2P\]

We can substitute the amortization formula back for \(M\):

\[n\,P \frac{r(1+r)^n}{(1+r)^n-1} = 2P\]

\(P\) cancels out. This means our answer doesn’t depend on the principal loan amount \(P\).

We are left with

\[n\frac{r(1+r)^n}{(1+r)^n-1} = 2\]

We can rewrite \(r\) and \(n\) in annual-rates-and-years form. With \(R\) as the annual interest rate and \(T\) as term in years we have \(r = \frac{R}{12}\) and \(n=12\,T\). So

\[n\,r = 12\,T \cdot \frac{R}{12} = R\,T\]

The exact condition becomes

\[\frac{R\,T(1+\frac{R}{12})^{12\,T}}{(1+\frac{R}{12})^{12\,T}-1} = 2\]

The quality of principal and interest depends mostly on the product \(R\,T\), or the annual rate \(\times\) years.

We can get a useful heuristic when we replace monthly compounding with annual compounding

\[\left(1+\frac{R}{12}\right)^{-12\,T} \approx e^{-R\,T}\]

Then the condition becomes

\[\frac{R\,T}{1-e^{-R\,T}} = 2\]

Let \(x = R\,T\), then

\[\frac{x}{1-e^{-x}} = 2\]

This equation has the solution \(x \approx 1.594\). So the mental rule is \(\boxed{\text{annual rate}\;\times\;\text{years} \approx 1.6}\).

Examples:

\[0.053\,\times\,30 \approx 1.59\]

So whenever you find a loan in the wild where the product of annual interest rate times its duration equals 1.6 you know that someone is paying the same amount on interest as their loan is worth.