
\problemdata     
{S01-PS5-7}           
{recurrences, annuities}              
{}              
{S01-PS5} 		
{S01 Karger}


\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.  Suppose
that I take out a loan for $L$ dollars, and at the end of every month,
the bank charges me interest on the loan at rate $i$.

\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 $L
(1 + i) - P$ dollars, and after 2 months I owe $L (1 + i)^2 - P \cdot
(1 + i) - P$ dollars.)  Give a closed-form expression for the number
of months until the loan is repaid.

\bigskip

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

\solution{

After $k$ months, I owe

\begin{eqnarray*}
\lefteqn{L (1+i)^k - P \cdot (1+i)^{k-1} - P \cdot (1+i)^{k-2}
                - \ldots - P}
        \hspace{1in} \\
        & = &   L (1+i)^k - P \cdot \frac{(1+i)^k - 1}{i}.
\end{eqnarray*}

The loan is paid when the amount that I owe equals zero; that is, when

\begin{eqnarray*}
L (1+i)^k - P \cdot \frac{(1+i)^k - 1}{i} & = & 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-iL}\right)}{\ln(1 + i)}.
\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 $L (1 + i)^2 - 2P$ dollars, and
after 4 months I owe $L (1+i)^4 - 2P \cdot (1 + i)^2 - 2P$ 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

\begin{eqnarray*}
\lefteqn{L (1+i)^k - 2P \cdot (1+i)^{k-2} - 2P \cdot (1+i)^{k-4}
                - \ldots - 2P}
        \hspace{1in} \\
        & = &   L (1+i)^k - 2P \cdot \frac{(1+i)^k - 1}{(1+i)^2 - 1}.
\end{eqnarray*}

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-((1+i)^2-1)L}\right)}{\ln(1 + i)}.
\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 $L (1 + i) - 2P$ dollars, and
after 3 months I owe $L (1 + i)^3 - 2P \cdot (1+i)^2 - 2P$ 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

\begin{eqnarray*}
\lefteqn{L (1+i)^k - 2P \cdot (1+i)^{k-1} - 2P \cdot (1+i)^{k-3}
                - \ldots - 2P}
        \hspace{1in} \\
        & = &   L (1+i)^k - 2P \cdot \frac{(1+i)^{k+1} - 1}{(1+i)^2 - 1}.
\end{eqnarray*}

The loan is repaid when

\begin{eqnarray*}
k & = & \frac{\ln\left(\frac{2P}{2P(1+i)-((1+i)^2-1)L}\right)}{\ln(1 + i)}.
\end{eqnarray*}
}

\problempart Suppose that I take a loan of $L = \$100,000$.  The
monthly interest rate is $i = 0.005$.  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{tabular}{lcr}
frequent payment & \hspace{1in} & 360 months \\
infrequent payment, late start & & 362 months \\
infrequent payment, early start & & 357 months (k is odd)
\end{tabular}
}

\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}


