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\begin{centering}
{\Large Mortgage Interest \& Principal Payments}
\bigskip
{\Large Seth Teller}
\end{centering}
\vspace*{0.25in}
A few years ago, I bought a house, and got frustrated when
the lending representatives were unable to explain
how they had arrived at particular mortgage payments
(all they could do was table lookups). I was curious about
how the amount of the fixed monthly payment is determined,
and how it gets partitioned into principal
and interest -- so I derived it directly, as follows.
Suppose we wish to take a loan of $A$ dollars at yearly
interest rate $R$ (equivalently, monthly interest rate
$M = R / 12$) to be paid monthly over $Y$ years (in
$N$ monthly payments, where $N = Y \cdot 12$).
The object of this problem is to determine the (constant)
total monthly payment $T$, and its (varying) apportionment
between principal $P_k$ and interest $I_k$ (for the $k$-th
monthly payment).
\bigskip
{\bf \noindent Problem:}
Express $T$, $P_k$, and $I_k$ as functions of
$M$, $N$, and the loan amount $A$.
\bigskip
{\bf \noindent Solution:}
Define the balance after the $k^{th}$ payment as $B_k$.
Then
$$
B_0 = A \mbox{ \phantom{XXXX} and \phantom{XXXX} } B_N = 0
$$
Moreover, the first interest payment
will be on the entire loan amount:
$$
I_0 = M A
$$
and the sum of all the principal payments must
equal the loan amount:
$$
\sum_{k=1}^{N}{P_k} = A
$$
The monthly payment $T$ is constant;
thus $T_{k-1} = T_k$ for any $k > 1$, and
$$
M(A - \sum_{j = 1}^{k-2}{P_j}) + P_{k-1} =
M(A - \sum_{j = 1}^{k-1}{P_j}) + P_{k}
$$
Cancelling common terms yields
$$
P_k = (1+M) P_{k-1}
$$
so that
$$
P_N = (1+M)^{N-1} P_1
$$
\noindent
Now, consider the initial balance $A$ and sequence of $N$ payments
\bigskip \noindent
\begin{tabular}{|r|c|c|c|c|}
\hline
k & $I_k$ & $P_k$ & $T$ & $B_k$ \\ \hline
0 & & & & A \\ \hline
1 & $M A$ & $P_1$ & $M A + P_1$ & $A - P_1$ \\ \hline
2 & $M (A - P_1)$ & $P_2$ & $M A - M P_1 + P_2$ & $A - P_1 - P_2$ \\ \hline
$\cdots$ & & & & \\ \hline
$N$ & $M ( A - \sum_{k = 1}^{N-1}{P_k})$
& $P_N$ & $M ( A - \sum_{k = 1}^{N-1}{P_k}) + P_N$
& $A - \sum_{k = 1}^{N}{P_k}$ \\ \hline
\end{tabular}
\bigskip \noindent Since $\sum_{k = 1}^{N}{P_k} = A$, we can rewrite the last row
(i.e., the $N^{th}$ payment) as
\bigskip % \noindent
\begin{center}
\begin{tabular}{|r|c|c|c|c|}
\hline
k & $I_N$ & $P_N$ & $T$ & $B_N$ \\ \hline
% $N$ & $M ( A - \sum_{k = 1}^{N-1}{P_k})$
% & $P_N$ & $M ( A - \sum_{k = 1}^{N-1}{P_k}) + P_N$
% & $A - \sum_{k = 1}^{N}{P_k}$ \\ \hline
$N$ & $M P_N$ & $P_N$ & $(1 + M) P_N$ & 0 \\ \hline
\end{tabular}
\end{center}
\bigskip \noindent
Equating $T = T_1 = T_N$ and substituting for $P_N$ yields
$$
P_1 = \frac{M A}{(1 + M)^N - 1}
$$
\bigskip \noindent
We can now -- as specified by the problem statement -- express $T$, $P_k$, and $I_k$ solely as
functions of $M$, $N$, and $A$:
$$
T = (1+M)^N P_1 = \frac{M A (1+M)^N}{(1 + M)^N - 1}
$$
\bigskip
$$
P_k = (1+M)^{k-1} P_1 = (1+M)^{k-1} \frac{M A}{(1 + M)^N - 1}
$$
\bigskip
$$
I_k = T - P_k
$$
\bigskip \noindent
Lenders, to avoid evaluating so many exponentiations, probably compute mortgage tables using the above
closed-form expression for $T$, the recurrence
$$
B_0 = A
\mbox{;}\quad I_k = M B_{k-1}
\mbox{;}\quad P_k = T - I_k
\mbox{;}\quad B_k = B_{k-1} - P_k
$$
and some rule for rounding fractional values to whole pennies.
\bigskip
\noindent{(Note: see \verb+C+ code \verb+mortgage.c+ for an implementation.)}
\end{document}