78 lines
2.0 KiB
Markdown
78 lines
2.0 KiB
Markdown
---
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id: 2026-01-29T17:57:00-0500
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title: 2026-01-29 17:57:??
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tags: []
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daily: "[[2026-01-29]]"
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---
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# 2026-01-29 17:57:??
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### Calculating Utility of Above-Minimum Mortgage Payment
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Follow-up to [[2026-01-25#Calculating Monthly Principal & Interest Payment]]:
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Homeowners are often advised to make elective mortgage payments
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to reduce the total interest paid on the loan,
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but an unrelated investment with a sufficient return
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could outweigh the reduced loss.
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Suppose you have a budget surplus of $E$ dollars
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and are deciding whether to make an elective payment on your mortgage
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or to invest in a promising opportunity.
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The return on electing to pay $E$ to the mortgage
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is the ...
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(i.e. interest that will no longer accrue)
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at the end of the loan is:
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$$
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R_{\text{mortgage}} = E(1+i)^{n}
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$$
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> [!info]- Explanation
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> This formula may seem suspiciously straightforward,
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> but suppose you did _not_ contribute $E$.
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> That portion of the principle would accrue interest
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> every month at rate $i$.
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> After $n$ months, the interest accrued by that portion is given by:
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>
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> $$
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> E(1+i)^{n}
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> $$
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If the same $E$ is invested elsewhere at monthly return $j$,
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its future value after $n$ months takes the same form:
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$$
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\text{FV}_{\text{investment}} = E(1+j)^{n}
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$$
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Therefore, $j$ must exceed $i$
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for the alternative investment to be preferable to elective payment.
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Note that $i$ and $j$ are adjusted rates,
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including respect for taxes and utility.
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On second thought, in a utility context,
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time preference could make $j$ preferable
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even when slightly lower.
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Short-term investments may be favored
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when liquidity is needed during the term,
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and tax deferred investments (IRA)
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are strongly favored over elective payment
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since interest is deductible
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(effective interest < nominal).
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### Calculating Effect of Elective Payment on Term Length
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The monthly payment and interest rate are fixed,
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so the term length must decrease
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$$
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\begin{align}
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A &= P \cdot \frac{i(1+i)^n}{(1+i)^n-1} \\
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P &= A \cdot \frac{(1+i)^n-1}{i(1+i)^n} \\
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n &= \frac{\ln\left(\frac{A}{A-Pi}\right)}{\ln(1+i)} \\
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\end{align}
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$$
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