Can Using a Christmas Bonus Annually Shorten Your Mortgage Repayment Time?

[issacnewton]

Homework Statement

Your friend tells you he has a very simple trick for taking one-third off the time it
takes to repay your mortgage: Use your Christmas bonus to make an extra payment
on January 1 of each year (that is, pay your monthly payment due on that day twice).
If you take out your mortgage on July 1, so your first monthly payment is due August
1, and you make an extra payment every January 1, how long will it take to pay off
the mortgage? Assume that the mortgage has an original term of 30 years and an
APR of 12%.

Homework Equations

Annuity formula

The Attempt at a Solution

I am bit confused by the statement of the problem. Its saying that we need to reduce the time period by one third. And the original term is for 30 years. So new term would be 20 years I guess. But I don't think the problem is as simple as that. Maybe the statement is a bit misleading. Any guidance ?

 

[andrewkirk]

I expect they mean that the strategy will reduce the period by 'approximately one third' or maybe 'at least one third'.

What they want the student to do is to use discounting formulas to calculate exactly what the reduced period will be, in months. Most likely it will not be exactly 240, but it is expected to be near there (if the problem has been set up correctly).

 

[Ray Vickson]

Homework Statement

Your friend tells you he has a very simple trick for taking one-third off the time it
takes to repay your mortgage: Use your Christmas bonus to make an extra payment
on January 1 of each year (that is, pay your monthly payment due on that day twice).
If you take out your mortgage on July 1, so your first monthly payment is due August
1, and you make an extra payment every January 1, how long will it take to pay off
the mortgage? Assume that the mortgage has an original term of 30 years and an
APR of 12%.

Homework Equations

Annuity formula

The Attempt at a Solution

I am bit confused by the statement of the problem. Its saying that we need to reduce the time period by one third. And the original term is for 30 years. So new term would be 20 years I guess. But I don't think the problem is as simple as that. Maybe the statement is a bit misleading. Any guidance ?

Since all we are interested in are relative amounts, let's assume you currently pay $1 per month for 30*12 = 360 months, at a monthly interest rate of r = 0.01. Just to keep things clear for a while, let's proceed symbolically: your mortgage has a present value of
$$\text{PV} = \sum_{n=1}^N 1/g^n, $$
where $g = 1.01$ and $N = 360$.

In your new scheme you start paying an extra $1, starting in 6 months and repeating every 12 months. If you make $K$ such extra payments, the NPV (now) is
$$\text{extra PV} = \sum_{n=0}^{K-1} 1/g^{6+12n}$$
If you pay off your mortgage in $M$ months your new present value is
$$ \text{new PV} = \sum_{n=1}^M 1/g^n + \sum_{n=0}^{K-1} 1/g^{6+12 n}.$$
You can use standard annuity-type formulas to evaluate both sums, so arrive at an expression $P(M,K)$. Now you want to equate the new PV to the old PV, so you must solve an equation
$$ P(M,K) = PV $$
to find $M$ and $K$.

We can get a good approximate solution by taking $K = M/12$, so we have an equation in $M$ alone, which is solvable numerically.

This solution will not be exact, because it is possible that we finish paying off the mortgage part-way through a year, so the number of annual payments might not be exactly one twelfth of the number of monthly payments. However, the error will not be great, and we can always fiddle with the final payment to make things come out right.

 

[issacnewton]

I think your second equation should be $$\text{extra PV} = \sum_{n=1}^K \Bigg[ \frac{1}{g^{6+12(n-1)}} \Bigg]$$ Also you are not taking into account the extra information given to us. That the new total period will be one third of the old period. I find the question phrased poorly.

 

[Ray Vickson]

I think your second equation should be $$\text{extra PV} = \sum_{n=1}^K \Bigg[ \frac{1}{g^{6+12(n-1)}} \Bigg]$$ Also you are not taking into account the extra information given to us. That the new total period will be one third of the old period. I find the question phrased poorly.

I made a correction, to write $\sum_{n=0}^{K-1} 1/g^{6+12n}$ in place of what I had before.

I don't understand your remark about taking account of any extra information. The extra information (about the 1/3 reduction in time) is not a fact unless we have verified it; until then, it just remains somebody's opinion, which may or may not be true. You can test whether it is true by actually solving the equations and see what is the new total payment period. Is your friend telling you something that is true, or is he telling you nonsense? That is the question, as I interpret it.

BTW: the question did not claim that the new period is 1/3 of the old period; it claimed it is about 2/3 of the old period, since it claimed a 1/3 reduction.

 

[issacnewton]

So let the monthly payment be $x$. Then we have the following sums $$\sum_{n=1}^M x/g^n = \frac{x}{(g-1)}\left[ 1 - \frac{1}{g^M} \right]$$ $$\sum_{n=1}^K \Bigg[ \frac{x}{g^{6+12(n-1)}} \Bigg] = \frac{x g^6}{(g^{12} - 1)} \Bigg[ 1- \frac{1}{g^{12K}} \Bigg]$$ Now the original $\text{PV}$ would be $$\text{PV} = \frac{x}{(g-1)}\Bigg[ 1- \frac{1}{g^{360}} \Bigg] $$ And since the $\text{PV}$ should not change, we have (after canceling out $x$), $$\frac{1}{(g-1)}\Bigg[ 1- \frac{1}{g^{360}} \Bigg] = \frac{1}{(g-1)}\left[ 1 - \frac{1}{g^M} \right] + \frac{g^6}{(g^{12} - 1)} \Bigg[ 1- \frac{1}{g^{12K}} \Bigg] $$ Since $g = 1.01$, plugging this into the above equation and rearranging the terms, we have the following equation $$(0.9901)^M + (0.0836996)(0.9901)^{12K} = 0.1115163$$ Wolfram Alpha gets confused here

 

[Ray Vickson]

So let the monthly payment be $x$. Then we have the following sums $$\sum_{n=1}^M x/g^n = \frac{x}{(g-1)}\left[ 1 - \frac{1}{g^M} \right]$$ $$\sum_{n=1}^K \Bigg[ \frac{x}{g^{6+12(n-1)}} \Bigg] = \frac{x g^6}{(g^{12} - 1)} \Bigg[ 1- \frac{1}{g^{12K}} \Bigg]$$ Now the original $\text{PV}$ would be $$\text{PV} = \frac{x}{(g-1)}\Bigg[ 1- \frac{1}{g^{360}} \Bigg] $$ And since the $\text{PV}$ should not change, we have (after canceling out $x$), $$\frac{1}{(g-1)}\Bigg[ 1- \frac{1}{g^{360}} \Bigg] = \frac{1}{(g-1)}\left[ 1 - \frac{1}{g^M} \right] + \frac{g^6}{(g^{12} - 1)} \Bigg[ 1- \frac{1}{g^{12K}} \Bigg] $$ Since $g = 1.01$, plugging this into the above equation and rearranging the terms, we have the following equation $$(0.9901)^M + (0.0836996)(0.9901)^{12K} = 0.1115163$$ Wolfram Alpha gets confused here

Things are a lot easier if you look at the case where $12 K = M$, as I suggested.

 

[issacnewton]

With your suggestion, $M=228.555$, and so it will take about $M/12 = 19.046$ years, which is approximately $2/3$ of $30$ years. So the suggestion of the friend was to be tested. I was thinking that we need to take that information into account into solving the problem. So there is some ambiguity in the problem. In physics, we don't have such ambiguity.

 

[Ray Vickson]

With your suggestion, $M=228.555$, and so it will take about $M/12 = 19.046$ years, which is approximately $2/3$ of $30$ years. So the suggestion of the friend was to be tested. I was thinking that we need to take that information into account into solving the problem. So there is some ambiguity in the problem. In physics, we don't have such ambiguity.

There was no ambiguity: as you wrote it, the question asked "...how long will it take to pay off the mortgage? Assume that the mortgage has an original term of 30 years and an APR of 12%". That seems pretty clear to me.

Back in the Stone Age I did my undergraduate work and my PhD in physics, so I have been exposed to the world of physics teaching. All I can say is that if you have not faced ambiguity, you can count yourself lucky; you have had instructors who can communicate clearly.