\problemdata      
        {annuities-2} % latex-friendly label for the prob.
        {Annuities} % The topic(s) of the problem content.
        {?} % Source (if known)
        {S02 PS6; F00 PS6} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).


\begin{problem} 
Shady finance people often assert that a loan can be paid off more quickly
by making half-size payments twice as often.  The theory is that frequent
payments prevent interest from building up.  Let's analyze this.

Suppose that I take out a loan for $L$ dollars, and at the end of every
month, the bank charges me $m$\% monthly interest on the loan.
\begin{problemparts}

\problempart 
First, let's consider a ``frequent payment'' repayment
strategy.  Suppose that I pay $P$ dollars to the bank every month,
right after the bank charges me interest.  So after 1 month I owe $Ls -
P$ dollars, after 2 months I owe $(L s - P)s - P$ dollars, \etc, where $s
\eqdef 1 + m/100$.  Give a closed-form expression for the number
of months until the loan is repaid.

\noindent
(Assume that interest charges are smaller than my monthly payment; that
is, $L(1-s) < P$.  Do not worry about rounding to an integral number of
months.)

\solution{
After $k$ months, I owe
\[
L s^k - P \cdot s^{k-1} - P \cdot s^{k-2} - \cdots - P =
L s^k - P \cdot \frac{s^k - 1}{s-1}.
\]
The loan is paid when the amount that I owe equals zero; that is, when
\begin{eqnarray*}
L s^k - P \cdot \frac{s^k - 1}{s-1} & = & 0.
\end{eqnarray*}
Solving for $k$, the number of months until the loan is repaid, gives
\begin{eqnarray*}
k & = &	\frac{\ln\left(\frac{P}{P-(s-1)L}\right)}{\ln s}.
\end{eqnarray*}
}

\problempart Now let's consider an ``infrequent payment, late start''
repayment strategy.  Suppose that I pay $2P$ dollars to the bank every two
months, making the first payment two months after I take out the loan.
Thus, after 2 months I owe $L s^2 - 2P$ dollars, and after 4 months I owe
$(L s^2 - 2P) s^2 - 2P$ dollars.  Give a closed-form expression for the
number of months until the loan in repaid.

\solution{
After $k$ months, where $k$ is even, I owe
\[
L s^k - 2P \cdot s^{k-2} - 2P \cdot s^{k-4} - \ldots - 2P = 
L s^k - 2P \cdot \frac{s^k - 1}{s^2 - 1}.
\]
As before, the loan is repaid when the amount I owe equals zero.  This
happens when
\begin{eqnarray*}
k & = &	\frac{\ln\left(\frac{2P}{2P-(s^2-1)L}\right)}{\ln s}.
\end{eqnarray*}
}

\problempart Finally, let's consider an ``infrequent payment, early
start'' strategy.  As before, I pay $2P$ dollars to the bank every two
months, but now I make the first payment one month after I take out the
loan.  Thus, after 1 month I owe $L s - 2P$ dollars, and after 3 months I
owe $(L s - 2P) s^2 - 2P$ dollars.  Give a closed-form expression for the
number of months until the loan is repaid.

\solution{
After $k$ months, where $k$ is odd, I owe
\[
L s^k - 2P \cdot s^{k-1} - 2P \cdot s^{k-3}- \ldots - 2P =
L s^k - 2P \cdot \frac{s^{k+1} - 1}{s^2 - 1}.
\]
The loan is repaid when
\begin{eqnarray*}
k & = &	\frac{\ln\left(\frac{2P}{2Ps-(s^2-1)L}\right)}{\ln s}.
\end{eqnarray*}
}

\problempart Suppose that I take a loan of $L = \$100,000$.  The monthly
interest rate is $m = (1/2)\%$.  I either pay the bank $P = \$600$ every
month or $2P = \$1,200$ every two months, depending on the repayment
strategy that I choose.  How long will repayment take for each of the
three strategies?

\solution{
Substitution gives the following results:
\begin{center}
\begin{tabular}{lcr}
frequent payment & \hspace{1in} & 360 months \\
infrequent payment, late start & & 362 months \\
infrequent payment, early start & & 356 months
\end{tabular}
\end{center}
}

\problempart 
In 50 words or fewer, explain the merits of frequent and
infrequent payment strategies.

\solution{Getting money to the bank early is more important than the
frequency of payments.}

\end{problemparts}

\end{problem}

