Paying off a mortgage twice as fast? My brother has a 30 year fixed mortgage. He pays monthly. Every month my brother doubles his principal payment (so every month, he pays a little bit more, according to how much more principal he's paying).
He told me he'd pay his mortgage off in 15 years this way. I told him I though it'd take more than 15 years. Who's right? If I'm right (it'll take more than 15 years) how would I explain this to him?
CLARIFICATION: He doubles his principal by looking at his statement and doubling the "amount applied to principal this payment" field.
 A: I put together an Excel spreadsheet.  It depends upon what you mean by doubling the principal.  For a 5% loan the 30 year payment is 5.3682/1000.  If you look at the amortization schedule of the 30 year loan and increase the payment by the principal amount (with the additional applied to principal) it will take 20 years, not 15.  But if you look at the principal amount based on the current balance and double it, you do pay off in 15 years.  I would suggest you make your own spreadsheet and play with it.  Each month you charge interest (in my case .05/12 of the current balance) then subtract the payment made to find the new balance.  Excel has the PMT function to determine the payment.
A: The short answer to your question is if your brother pays double the monthly payment then he will pay off the mortgage in less than $15$ years.
The right way to analyze such problems from a mathematical frame-point is to look at the present value of the total amount he is paying.
Say your brother pays $\$x$ per month for $30$ years starting from the end of the first month. Let $r \%$ be the rate of interest per annum. We will compare the present value of the mortgage/loan.
(Note: that $\$1$ today is worth $\$\left(1+\frac{r}{12}\right)$ at the end of the first month).
The present value of the total amount he pays in $30$ years is $$\frac{x}{\left(1+\frac{r}{12}\right)} + \frac{x}{\left(1+\frac{r}{12}\right)^2} + \cdots + \frac{x}{\left(1+\frac{r}{12}\right)^{12 \times 30}} = \frac{12x}{r} \left (1- \left (\frac{12}{12+r}\right)^{360} \right )$$
Instead of $\$x$ if he were to pay $\$(2x)$ starting from the end of the first month, and let him pay for $n$ years, the present value of the total amount he pays in $n$ years is $$\frac{2x}{\left (1+\frac{r}{12} \right )} + \frac{2x}{\left (1+\frac{r}{12} \right)^2} + \cdots + \frac{2x}{\left (1+\frac{r}{12} \right)^{12 \times n}} = \frac{24x}{r} \left(1- \left (\frac{12}{12+r}\right)^{12n} \right)$$
For the mortgage to be fair, both the present values must be the same.
So find $n$ such that $$2 \left(1- \left (\frac{12}{12+r}\right)^{12n} \right) = \left(1- \left (\frac{12}{12+r}\right)^{360} \right)$$
Solving for $n$, we get $$n = \frac{\log \left (1+\left (\frac{12}{12+r} \right )^{360} \right)-\log \left(2 \right)}{12 \log \left(\frac{12}{12+r} \right)}$$.
So say plugging in $r = 6\%$ i.e. $r = 0.06$, we get $n = 9.01466$ years.
Whatever be the rate of interest, the total amount, the amount he is paying per month, if he decides to double the monthly payment he will pay it off in less than $15$ years.
In general, if a mortgage is for $N$ years and a person needs to pay $x$ per month, if the person decides to pay $k x$ per month instead of $x$ per month and $k>1$, then the person will pay the mortgage in less than $\frac{N}{k}$ years.
A: It seems your brother is essentially right.
In a standard amortization schedule, the amount applied to principal
each month increases geometrically, at the interest rate.
Doubling these amounts (or increasing them by any constant factor
over the amounts in the original amortization schedule) corresponds
to making payments at a higher constant level that amortizes
the loan over a shorter total time.
Here's the math: For month $j=0,1,2\ldots$ of the loan, let $P_j$ be
the principal remaining at the start of the month,
and $Y$ the payment, paid at the end of the month.
The amount paid toward interest is $I_j=rP_j$ with $r=0.05/12$,
and the amount paid toward principal is $A_j=Y-I_j=Y-rP_j$.
Then the new principal is
$$
P_{j+1}= P_j-A_j = P_j(1+r)-Y,
$$
so
$$
I_{j+1}= rP_{j+1}= (1+r)I_j-rY = (1+r)(I_j-Y)+Y.
$$
Hence $(1+r)A_j=A_{j+1}$ and therefore $A_j=(1+r)^jA_0$.
The standard payment $Y$ is rigged to make $P_N=I_N=0$ with
$N=360$ months. A higher (constant) payment $\hat Y$ corresponds to
principal payments $\hat A_j$ that are larger than $A_j$
by always the same proportion.
For a 30 year loan at 5 percent, the standard monthly payment
is \$5.3692 per \$1000.  Doubling the principal payment results
in a monthly \$6.5697 per \$1000, which amortizes the loan over
about 20 years.  Increasing the principal payments by 200 percent
(tripling them) amortizes the loan over a bit more than 15 years.
But from your description it seems your brother is doing something
different, something that increases his payments each month.
A spreadsheet calculation shows that he would indeed pay off
the loan in 15 years, if he adds, to the standard 30-year payment $Y$,
the amount of principal that the payment $Y$ would pay off this month.
(This amount may be listed on his statement.)
This means his payment at the end of month $j$
is $$
Y_j=Y+ (Y-rP_j).
$$
 As above, now his remaining principal satisfies
$$
P_{j+1}= P_j+I_j - Y_j = P_j(1+2r)-2Y.
$$
So effectively his principal is reduced as if he makes the constant
payment $2Y$ on a loan with interest rate $2r$.  As it happens,
with $Y$ being the original 30-year standard payment, $2Y$ is almost just the right value to amortize this loan over 15 years.
Any way you do it, paying principal off early is a great way to
save lots on interest later.
A: Let's look at two scenarios: two months of payment $P$ vs. one month of payment $2P$. Start with the second scenario. Assume the total amount to be payed is $X$ and the rate is $r > 1$, the total amount to be payed after one month would be $$r(X-2P).$$ Under the first scheme, the total amount to be payed after two months would be $$r(r(X-P)-P) = r(rX - (1+r)P).$$ Most of the time, $X$ is much larger than $P$, and so $X-2P$ is significantly smaller than $rX - (1+r)P$ (remember $r \approx 1$). So it should take your brother less than 15 years.
Note I first subtract the payment and then take interest, but it shouldn't really matter.
This all assumes the payments are fixed, but looking at the other answers this is not really the case... I guess my banking skills are lacking. Too young to take loans.
