From b396980d4699eb12b161fa09de712d775f22b053 Mon Sep 17 00:00:00 2001
From: Zane Meyers <zanemeyers@gmail.com>
Date: Thu, 29 Jan 2026 22:10:26 -0500
Subject: [PATCH] vault backup: 2026-01-29 22:10:26

---
 2026-01-29.md | 56 +++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 56 insertions(+)

diff --git a/2026-01-29.md b/2026-01-29.md
index 2be57fa..fdc96a2 100644
--- a/2026-01-29.md
+++ b/2026-01-29.md
@@ -29,3 +29,59 @@ would be unlikely to be able to provide.
 Their dismissal of legitimate _instrumental_ complaints
 (misplaced effort, and deviation from more widely accepted standards)
 speaks to a fundamental misunderstanding of the purpose of estimating.
+
+## 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 had $E$ dollars to
+
+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.
+
+### Calculating Effect of Elective Payment on Term Length
+
+The monthly payment and interest rate are fixed,
+so the term length must decrease
+
+$$
+\begin{align}
+A &= P \cdot \frac{i(1+i)^n}{(1+i)^n-1} \\
+P &= A \cdot \frac{(1+i)^n-1}{i(1+i)^n} \\
+n &= \frac{\ln\left(\frac{A}{A-Pi}\right)}{\ln(1+i)} \\
+\end{align}
+$$