---
id: 2026-01-29T17:57:00-05:00
aliases: []
title: 2026-01-29 17:57:??
tags:
  - topic/finance
  - type/timestamped
dg-publish: true
---
# 2026-01-29 17:57:??

### Calculating Utility of Above-Minimum Mortgage Payment

See [[2026-01-25#Calculating Monthly Principal & Interest Payment]].

Homeowners are often advised to make elective mortgage payments
to reduce the total interest paid on the loan,
but an unrelated investment with a sufficient return
could outweigh the reduced loss.

Suppose you have a budget surplus of $E$ dollars
and are deciding whether to make an elective payment on your mortgage
or to invest in a promising opportunity.

The return on electing to pay $E$ to the mortgage
is the
(i.e., interest that will no longer accrue)
at the end of the loan is:

$$
R_{\text{mortgage}} = E(1+i)^{n}
$$

> [!info]- Explanation
> This formula may seem suspiciously straightforward,
> but suppose you did _not_ contribute $E$.
> That portion of the principle would accrue interest
> every month at rate $i$.
> After $n$ months, the interest accrued by that portion is given by:
>
> $$
> E(1+i)^{n}
> $$

If the same $E$ is invested elsewhere at monthly return $j$,
its future value after $n$ months takes the same form:

$$
\text{FV}_{\text{investment}} = E(1+j)^{n}
$$

Therefore, $j$ must exceed $i$
for the alternative investment to be preferable to elective payment.

Note that $i$ and $j$ are adjusted rates,
including respect for taxes and utility.

On second thought, in a utility context,
time preference could make $j$ preferable
even when slightly lower.

Short-term investments may be favored
when liquidity is needed during the term,
and tax deferred investments (IRA)
are strongly favored over elective payment
since interest is deductible
(effective interest < nominal).

### Calculating Effect of Elective Payment on Term Length

The monthly payment and interest rate are fixed,
so the term length must decrease

$$
\begin{align}
A &= P \cdot \frac{i(1+i)^n}{(1+i)^n-1} \\
P &= A \cdot \frac{(1+i)^n-1}{i(1+i)^n} \\
n &= \frac{\ln\left(\frac{A}{A-Pi}\right)}{\ln(1+i)} \\
\end{align}
$$
