%%%%Annuities.tex
\documentclass[12pt, twoside]{article}
\input{macros-course}


\begin{document}

\handout{annuities}{Problems: Annuities}{}

\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*}
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%END   Kavita's problems

%BEGIN Joe's problems
\problem %source: spring00 ps5
Suppose you were about to enter college today and a college loan
officer offered you the following deal: \$25,000 at the start of each
year for four years to pay for your college tuition and an option of
choosing one of the following repayment plans:

\begin{description}\itemsep 0\baselineskip
\item[Plan A:] Wait four years, then repay \$$20,000$ at the start of
  each year for the next ten years.
\item[Plan B:] Wait five years, then repay \$$30,000$ at the start of
  each year for the next five years.
\end{description}

Suppose the annual interest rate is currently 7\% and does not change
in the future.  

\begin{itemize}\itemsep 0\baselineskip
\item[a)] Which repayment plan would you choose?

\remove{

  
  \medskip
  \textbf{Solution:} 
  
  The use of the word `interest' above is slightly awkward. It refers to
the additional money you (or the loan officer) would make if your money
was just sitting in a bank. However, we can treat it just as we treat
inflation: $\$1$ today will be worth $\$1.07$ next year, and $\$1.07^2$ the
year after.

 Given this, set $r = \frac{1}{1.07}$. Then:
 
\begin{eqnarray*}
A & = & \sum_{y=0}^{9} 20000 \cdot r^{y+4}  \\
  & = & r^4 \cdot \sum_{y=0}^{9} 20000 \cdot r^y  \\
  & = & 20000r^4 \cdot \sum_{y=0}^{9} r^y \\
  & = & 20000r^4 \cdot \frac{1 - r^{10}}{1 - r} \\
  & = & \$114,66.69
\end{eqnarray*}

\begin{eqnarray*}
B & = & \sum_{y=0}^{4} 30000 \cdot r^{y+5}  \\
  & = & r^5 \cdot \sum_{y=0}^{4} 30000 \cdot r^y  \\
  & = & 30000r^5 \cdot \sum_{y=0}^{4} r^y \\
  & = & 30000r^5 \cdot \frac{1 - r^{5}}{1 - r} \\
  & = & \$93,840.63
\end{eqnarray*}

You should clearly take Plan B. You will be paying back much less in today's
dollars.

  \medskip


}

\item[b)] What is the loan officer's effective profit (in today's
  dollars) on the loan?

\remove{

  
  \medskip
  \textbf{Solution:} 

        The value of the money you are given is:
\begin{eqnarray*}
Loan & = & \sum_{y=0}^{3} 25000 \cdot r^{y}  \\
  & = & 25000 \cdot \sum_{y=0}^{3} r^y \\
  & = & 25000 \cdot \frac{1 - r^{4}}{1 - r} \\
  & = & \$90,607.90
\end{eqnarray*}

Therefore, the loan officer's profit is effectively $\$3,233$. (Or 
$\$24,059$ if we are not on the ball).


}

\end{itemize}
%END   Joe's problems

%BEGIN Ray's problems

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


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\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}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\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.)

\solnospace{
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)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%pset5 problem1 spring98
\problem 

A college loan officer offers prospective M.Eng students the following
deal.  He will give them \$23,100 at the start of each year for five years
starting today to pay for their college tuition.  He offers the following
plan for repayment:\\
Wait five years, then repay \$$x.00$ at the start
of each year for the next ten years.

Suppose the annual interest rate is 8\%.  What value of $x$ will make the 
present value of the repayment equal to the present value of the loan?

\solution{
The formula for finding the present value of an annuity (given in lecture)
is
\[V = m \sum_{j=0}^{n-1}{1 \over (1+p)^j}\]
where $V$ is the present value, $m$ is the dollar amount of each payment, $n$ is
the number of years of the annuity, and $p = 0.08$ is the current
interest rate.

So the present value of what the college loan officer offers is
\begin{eqnarray*}
V & = & 23100 \cdot {\displaystyle\sum_{j=0}^4(\frac{1}{1+p})^j} \\
 & = & 23100 \cdot {\displaystyle\frac {(1 - \frac{1}{(1 + p)^5})} {(1 -
 \frac{1}{1 + p})}} \\
 & = & 23100 \cdot 4.312\\
 & = & \$99,610.13.
\end{eqnarray*}

The value, $V'$, of the repayment is
\[V' = x \cdot \sum_{j=5}^{14} \frac{1}{(1+p)^j}.\]

Since we want the present values to be equal, let $V' = V$, and solve for
$x$:
\begin{eqnarray*}
x & = & {\$99,610.13 \over { (\frac{1}{1+p})^5} \cdot (\sum_{j=0}^9(\frac{1}{1+p})^j) }\\
 & = & \frac {\$99,610.13} { (\frac{1} {1+p})^5 \cdot { {(1 - \frac{1}{(1+p)^{10}})} \over {(1 - \frac{1}{1+p})}}}\\
 & = & \frac  {\$99,610.13}{0.6805 \cdot 7.246}\\
 & = & \$20,196.25.
\end{eqnarray*}

So, rounding off to the nearest dollar, $x=\$20, 196$.
}

%END   Ivan's problems


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

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

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%BEGIN  Jaime Problem 1
%SOURCE Fall 2001 Tutorial 7

\problem Suppose you deposit \$100 into your MIT Credit Union account
today, \$99 in one month from now, \$98 in two months from now, and so on.
Given that the interest rate is constantly 0.3\% per month, how long will
it take to save \$5,000?

\solnospace{
First note that you will certainly manage to have saved \$5,000
\emph{some} day, since, even without your earnings from the interest,
you will have $100+99+\cdots+1=100\times 101/2 = 5,050$ dollars after
99~months. 

But fewer months will be needed: After the first deposit you will have
\$100. After the second deposit, you will have \$$(100\times 1.003 +
99)$. After you third deposit, your saved money will be \$$\bigl(
(100\times 1.003 + 99)\times 1.003 + 98) =$ \$$\bigl( 100\times
(1.003)^2 + 99\times 1.003 + 98\bigr)$, and so on. So, after the $n$th 
deposit,  
$$
S_n = \sum_{i=0}^{n-1} (100 - i) (1.003)^{n-i-1}
$$
dollars will be in your account. Substituting $j=n-i-1$, we can
rewrite this as  
$$
\sum_{j=0}^{n-1} \bigl( 100 - (n-j-1)\bigr) (1.003)^{j};
$$
and then as 
$$
(101-n) 
\Bigl( \sum_{j=0}^{n-1} (1.003)^{j}   \Bigr)+
\Bigl( \sum_{j=0}^{n-1} j (1.003)^{j} \Bigr).
$$
We have closed forms for both of the sums, in Notes \#7. We can
substitute and then solve 
$$
S_n \geq 5,000
$$
for $n$.
}

%END    Jaime Problem 1

\end{problems}

\end{document}


