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\begin{document}

\title{Math 242, Lecture 11}
\maketitle

\section{Values of revenue streams}

Often in business and economics, money is paid (back)
in installments,   as with student
loans, mortgages and retirement accounts.    
Standard application of the Fundamental Theorem allows
us to extract a total amount paid given the rate.  

\begin{theorem}
If money is paid at a rate of $R(t)$ dollars per unit of time, the
total amount paid between time $a$ and $b$ is 
$\int_a^b R(t) dt$.  
\end{theorem}

\begin{example}[Stupid example]
What does this theorem say when $R(t)$ is constant (say \$100 per month between
the second and sixth months of some year)? 
\end{example}

But as any pensioner or lottery winner will tell you, getting a constant amount
of money means that your purchasing power will go down over time.  
How can we determine the current or future value
of installment payments?  

\begin{example}
\$100 deposited per month in an account earning 5\% interest
has a present value of 
$100 + 100 e^{-0.05 \frac{1}{12}} + \cdots + 100 e^{-0.05 \frac{11}{12}}$
dollars.  
\end{example}


What if, instead of deposited as \$100 per month,   \$1200 was 
deposited as a continuous income stream, evenly over the year?  
Then, over a short  fraction of the year  
$\Delta t = \frac{1}{N}$th of a year,   we would deposit
$1200 \Delta t$ dollars.    Its present value would be 
$(1200 \Delta t) e^{-0.5 t}$, where $t$ is a time at which this
money is deposited.    The sum of all of these pieces
would be 
$$ 1200 \Delta t  e^{-0.5 t_0} + 1200 \Delta t  e^{-0.5 t_1} + \cdots.$$   

 We have gone through the steps of setting up a Riemann
 sum, leading to an integral;   by passing to a limit and using
the definite integral, we see that the present value 
will be $\int_0^1 1200 e^{0.5 t}$.    (Note how
much this income stream is worth   - making payments more
frequently can make a big difference!)  

\begin{theorem}
The present value of a revenue stream $R(t)$ between times $a$ and
$b$ is $\int_a^b R(t) e^{r(a-t)} dt$.   

The future value of a revenue stream $R(t)$ between times $a$ and $b$
is $\int_a^b R(t) e^{r(b-t)} dt$.  
\end{theorem}

Let's justify this again by analyzing the present and future value of the
money deposited over a short period of time.  

Then we're ready for
an example.    Note that the term $e^{ra}$ or $e^{rb}$ which occurs
can be ``pulled out'' of the integral.   

\begin{example}
What is the present value of a million dollars paid constantly over twenty years, 
assuming a prevailing inflation rate of four percent per year?  
\end{example}

But what happens if the revenue function isn't constant, but is even a linear
function?    We run into integrals like that of $t e^{rt}$.   
These are done in section 7.1, but we will just use a general formula.  

\begin{theorem}
\begin{multline*}
\int t^n e^{rt} dt = \\ \frac{1}{r} t^n e^{rt} - \frac{n}{r^2} t^{n-1} e^{rt} + \frac{n(n-1)}{r^3} t^{n-2} e^{rt} - \\ 
\cdots \pm \frac{n(n-1)(n-2)\cdots 1}{r^{n+1}} e^{rt} + C.
\end{multline*}
\end{theorem}

We are mainly concerned in the cases of $n = 0$ or $1$, in which case we can verify this
by taking the derivative.  

We can use this for more complicated revenue stream problems.

\begin{example}
If you invest  $R(t) = 2000 + 400t$ dollars per year in a retirement account 
earning nine percent per year, how much will you have after forty years?  
How much of that is principal and how much is interest?  
\end{example}

The other class of revenue streams for which exact answers are within
are reach are those described by exponential functions.  

\begin{example}
The revenues of Startup.com is modeled by $10 e^{1.2 t} - 20$ millions of dollars.
Adjusting for inflation of 4\%, estimate the present value of its revenues over its
first five years.
\end{example}

\end{document}
 