<<<<<<< Annuities.tex
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  \paragraph{Problem}}
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  \paragraph{Solution}}

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\begin{problems}

%BEGIN Kavita's problems


% Fall 1998

%**************
%   PSET 4    *
%**************


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\problem {\bf (10 points)} A banker loans \$100,000 to the Addams
family so that they can purchase a new dilapidated mansion.  At the
end of each year for the next 20 years, Morticia Addams will drop by
the bank to make a payment on the loan.  The banker will immediately
invest this returned money at a 7\% annual rate of return.

What annual payment should the banker require of the Addams family so
that his return after 20 years is the same as if he had invested the
whole amount at 7\% today instead of making the loan?

\solution{

Suppose that Morticia pays $p$ dollars at the end of each year for 20
years.  The banker can invest the first payment for 19 years, the
second payment for 18 years, etc.  Therefore, after 20 years his total
cash is:

\begin{eqnarray*}
\sum_{i=0}^{19} p \cdot 1.07^i
	& = &	p \cdot \sum_{i=0}^{19} 1.07^i \\
	& = &	p \cdot \frac{1.07^{20} - 1}{1.07 - 1}
\end{eqnarray*}

The second equation uses the formula for the sum of a finite
geometric series.

On the other hand, if the banker invests the money at 7\% today, then
after 20 years his total cash is $100,000 \cdot 1.07^{20}$.

Equating these two expressions and solving for $p$, Morticia's annual
payment, we find:

\begin{eqnarray*}
p \cdot \frac{1.07^{20} - 1}{1.07 - 1}
	& = & 100,000 \cdot 1.07^{20} \\
p	& = &	100,000 \cdot 1.07^{20} \cdot
		\frac{1.07 - 1}{1.07^{20} - 1} \\
	& \approx & \$9439.29
\end{eqnarray*}
}

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%END   Kavita's problems

%BEGIN Joe's problems
%END   Joe's problems

%BEGIN Ray's problems

%%%%%%%%%%%%%%%  S99 ps5


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\problem {\bf (10 points)}

Recently, David Bowie, a British music star, sold a portion of the
future profits from his music for \$55 million dollars to Prudential
Insurance.  This deal spawned others.

\begin{problemparts}

\problempart Billy Joel wants to sell his future profits for \$200
million today.  Suppose that he can generate a fixed profit at the end
of every year for the next 20 years.  If the inflation rate is $p =
0.03$, how large must Billy Joel's annual profit be for investors to
make a $10\%$ return?

\solution{Suppose that Billy Joel makes $M$ dollars at the end of
every year for the next 20 years.  Then his contract is worth

\begin{eqnarray*}
\sum_{i=1}^{20} \frac{M}{(1+p)^i}
	& = &	\frac{M}{1+p} \cdot \sum_{i=0}^{19} \frac{1}{(1+p)^i} \\
	& = &	\frac{M}{1+p} \cdot
		\frac{\left(\frac{1}{1+p}\right)^{20} - 1}
			{\frac{1}{1+p} - 1} \\
	& = &	M \cdot 14.877\ldots.
\end{eqnarray*}

To make his contract worth \$200 million + 10\%, Billy Joel must make
$M \approx \$14.7$ million in profit every year.}

\problempart The Eagles figure that they can generate at least \$25
million from their music at the end of each year for eternity.  What
should an investor who wants a 15\% return pay today for the Eagles
contract?

\solution{Let $M = \$25,000,000$.  The Eagles are worth

\begin{eqnarray*}
\sum_{i=1}^{\infty} \frac{M}{(1+p)^i}
	& = &	\frac{M}{1+p} \cdot \frac{1}{1 - \frac{1}{1+p}} \\
	& = &	\frac{M}{p}
\end{eqnarray*}

Since the investor wants a 15\% return, he should pay

\begin{eqnarray*}
\frac{M}{1.15p}	& \approx &	\$724 \text{ million}
\end{eqnarray*}
}

\end{problemparts}

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\problem {\bf (24 points)} 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 & & 356 months
\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   Ray's problems

%BEGIN Ivan's problems (spring '98)
%END   Ivan's problems

%BEGIN George's problems
%END   George's problems

%BEGIN Josh's problems
%END   Josh's problems

%BEGIN Tina's problems
%END   Tina's problems


\end{problems}

\end{document}


