\problemdata      
        {?} % latex-friendly label for the prob.
        {Annuitites} % The topic(s) of the problem content.
        {?} % Source (if known)
        {Fall 2001, final} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).



\begin{problem}
 Theory Pig wants to borrow $d$ dollars so that he can purchase a
stylish new pigsty.  At the end of each year for the next $n$ years, Theory
Pig will drop by the bank to make a payment on the loan.  The banker will
immediately invest this returned money at an $r$\% annual rate of
return. The banker wants to wind up with at least as much money as a result
of Theory Pig's payments as he would if he had invested the whole $d$
dollars at $r$\% for $n$ years starting today.

Write a closed form formula in $d,n$ and $r$ for the least annual payment
the banker might ask from Theory Pig.

\solution{
Let $R \eqdef (1+r/100)$, so that after 1 year, a one dollar
investment is worth $R$ dollars.  The banker will require that
\begin{equation*}
p \geq \frac{dR^{n}(R-1)}{R^n - 1}.
\end{equation*}
We derive this bound as follows:

Suppose that Theory Pig pays $p$ dollars at the end of each year.  The
banker can invest the first payment for $n-1$ years, the second payment
for $n-2$ years, etc., so that after $n$ years his total cash is:
\begin{eqnarray*}
\sum_{i=1}^{n} p \cdot R^{i-1}
& = & \sum_{i=0}^{n-1} p \cdot R^i\\
& = & p \frac{R^n - 1}{R - 1}.
\end{eqnarray*}
On the other hand, if the banker could invested the $d$ dollars today, then
after $n$ years his total cash is $d R^{n}$.  The first amount must
be at least as large as this, so
\begin{eqnarray*}
p \frac{R^n - 1}{R - 1} & \geq & dR^{n}\\
p &\geq& \frac{dR^{n}(R-1)}{R^n - 1}.
\end{eqnarray*}
}

\end{problem}