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{\Large \bf DIMACS EDUCATIONAL MODULE SERIES}

\vspace{0.7in}

{\Large \bf MODULE 08-4}\\

{\Large \bf Modeling Biological Populations}\\

{\large \bf Date prepared: December 21, 2007}



\vspace{.3in}


{\bf Teresa Contenza}\\
{\bf Case Western Reserve University\\
Cleveland, OH 44106}\\
{\bf email: teresa.contenza@case.edu}\\

\vspace{.2in}

{\bf Maureen Cox}\\
{\bf St. Bonaventure University\\
St. Bonaventure, NY 14778}\\
{\bf email: mcox@sbu.edu}\\

\vspace{.2in}

{\bf Jodie Novak}\\
{\bf University of Northern Colorado\\
Greeley, CO 80639}\\
{\bf email: jodie.novak@unco.edu}\\

\vspace{.2in}

{\bf Miranda Teboh-Ewungkem}\\
{\bf Lafayette College\\
Easton, PA 18042}\\
{\bf email: tebohewm@lafayette.edu}\\

\vspace{.2in}

{\bf Mary Vanderschoot}\\
{\bf Wheaton College\\
Wheaton, IL 60187}\\
{\bf email: mary.vanderschoot@wheaton.edu}\\


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\begin{center}

{\large \bf Module Description Information}

\end{center}



\begin{itemize}

\item {\bf Title:}

Modeling Biological Populations

\item {\bf Authors:}

Teresa Contenza, Case Western Reserve University\\
Maureen Cox, St. Bonaventure University\\
Jodie Novak,  University of Northern Colorado\\
Miranda Teboh-Ewungkem, Lafayette College\\
Mary Vanderschoot, Wheaton College


\item {\bf Abstract:}

This module introduces three discrete-time population models: the
logistic model, the Ricker model, and the Beverton-Holt model. These
models are studied using real data sets from biological populations.
Students use techniques from dynamical systems (iterations, cobwebs,
fixed points, stability, and sensitivity analysis) to analyze
population models.


\item {\bf Informal Description:}

In this module we use mathematical models to study biological
populations such as the pine-looper moth, coho salmon, the common
wasp, and the paramecium aurelia. Exercises throughout the module
allow students to work with a model and a data set to determine how
well the model fits the data, and what it predicts about the
population's long-term survival.  The exercises range from simple
computations requiring a graphing calculator or Microsoft Excel
(assumes Office 2003), to more challenging problems requiring
calculus.

\item {\bf Target Audience:}

This module is aimed at high school students in their junior or
senior years, and at freshman college students.  It would be an
appropriate module for an applied mathematics course.

\item {\bf Prerequisites:}

This module assumes some familiarity with composing functions, and
with using exponential functions and logarithms. Some exercises
require first semester calculus.


\item {\bf Mathematical Field:}

Mathematical Biology, Dynamical Systems, Difference Equations

\item {\bf Applications Areas:}

This module uses difference equations and technology to model
population dynamics.

\item {\bf Mathematics Subject Classification:}

37N25, 39A11, 92D25


\item {\bf Contact Information:}

Mary Vanderschoot\\
Mathematics and Computer Science Department\\
Wheaton College \\
Wheaton, IL 60187 \\
email: mary.vanderschoot@wheaton.edu\\


\item {\bf Other DIMACS modules related to this module:}

Module 07--3: Using Population Models in the Teaching of Eigenvalues

\end{itemize}

\newpage

\renewcommand{\contentsname}{Table of Contents} \tableofcontents

\newpage

\section{Introduction and Definitions}

The population of a biological species depends on many factors, such
as birth and death rates, immigration and emigration rates,
competition, harvesting, and environmental influences.   The
simplest types of populations to study are those with
non-overlapping generations.  In a population with non-overlapping
generations, one generation dies out before the next generation is
born.  This is the case for some species of fish and for seasonally
breeding insects.

A \emph{basic population model} for a population with
non-overlapping generations is an equation that attempts to describe
how the population in one generation depends on the population in
the previous generation.  Say $x(n+1)$ is the population in
generation $n+1$, and $x(n)$ is the population in generation $n$. A
basic population model is an equation of the form $x(n+1)=f(x(n))$
for some function $f$.  We call $f$ a \emph{generating function}.
Different choices of generating functions produce different
population models. The choice of generating function depends on the
assumptions that are made about how a population grows. In this
module we will consider four different population models, i.e., four
different generating functions.


In order to simplify the notation, we will write $x_n$ for $x(n)$.
If we have a basic population model $x_{n+1}=f(x_n)$, with initial
population $x_0$, then $x_1=f(x_0)$ is the population in generation
1, $x_2=f(x_1)=f(f(x_0))=f^2(x_0)$ is the population in generation
2, and, in general, $x_n=f^n(x_0)$ is the population in generation
$n$. Here $f^n$ represents the composition or \emph{iteration} of
$f$ with itself $n$ times.   The sequence of iterations $\{x_0, x_1,
x_2, x_3, \ldots\}$ is called the \emph{orbit} of $x_0$ under $f$.

\begin{example} {\bf Example:} Suppose $f(x)=x^2$ and $x_0=1/2$. Then
\begin{eqnarray*}
x_1 = (x_0)^2 &=& (1/2)^2 = 1/4\\
x_2 = (x_1)^2 &=& (1/4)^2 = 1/16\\
x_3 = (x_2)^2 &=& (1/16)^2 = 1/256\\
& \vdots&\\
\end{eqnarray*}
and so forth, and the orbit of $x_0=1/2$ is $\{ 1/2, 1/4, 1/16,
1/256, \ldots\}$. We can see that as $n$ increases, the values of
$x_n$ are getting closer and closer to $0$. Similar limiting
behavior occurs for any initial value $x_0$ with $|x_0|<1$. The
values of $x_n$ would increase without bound (as $n$ increases) if
$x_0>1$.
\end{example}


\begin{exercise} {\bf Exercise 1.1:} What happens if $x_0<-1$?
\end{exercise}


Since $f(0)=0$, the orbit of $x_0=0$ is $\{0,0,0,\ldots\}$.  That
is, the orbit of $x_0=0$ remains fixed at $0$.  We say that $x_0$ is
a \emph{fixed point} of a generating function $f$ if $f(x_0)=x_0$.
In order to find the fixed points of $f(x)=x^2$ we solve $f(x)=x$.
Solving $x^2=x$,
 we find $f$ has two fixed points: $x=0$ and
$x=1$.

In some of the exercises in this module you will be asked to
determine if population values over time move toward a fixed point
or away from a fixed point. In our example, we've seen that for any
initial value $x_0$ near the fixed point 0, the values of $x_n$
approach the fixed point 0 as $n$ increases. In this case, we say
that 0 is an {\it attracting} fixed point of $f$. Similarly, for any
initial population value near (but not equal to) the fixed point 1,
the values of $x_n$ either approach 0 (for $x_0=0.9$, for instance)
or grow without bound (for $x_0=1.1$, for instance) as $n$
increases. In either case, the values of $x_n$ move away from 1.
Therefore we say that 1 is a {\it repelling} fixed point of $f$.
(It's worth noting that we cannot classify every fixed point of
every generating function as either attracting or repelling.)

One of the first population models was introduced by Thomas Malthus
in 1798 to describe the population of Great Britain.  The Malthus
model is based on the simplifying assumption that the population at
time $n+1$ is proportional to the population at time $n$.   For
example, suppose that a population at time $n$ is given by $x_n$ and
that in any given time period, the number of births is equal to 20\%
of the population, and the number of deaths is equal to 13\% of the
population. The net change in population between times $n$ and $n+1$
would be the difference between the number of births and the number
of deaths during that time period. Mathematically, we have
$x_{n+1}-x_n = 0.20 x_n - 0.13 x_n$, or simply $x_{n+1}=1.07 x_n$.
In general, we can write the Malthus model as
\begin{eqnarray} x_{n+1} &=& k x_n
\label{Malthus}\end{eqnarray} for some constant $k$.  We can also
write the Malthus model as $x_{n+1}=f(x_n)$, where $f(x)=kx$ is the
generating function.  The constant $k$ will always be \mbox{1 +
birth rate $-$ death rate.}

\begin{exercise} {\bf Exercise 1.2:} What can you say about the birth rate and
death rate for the population if $k>1$? What implication would that
have for the population?
\end{exercise}

\begin{exercise}
\noindent {\bf Exercise 1.3:} What can you say about the birth rate
and death rate for the population if  $k=1$? And if $0<k<1$? What
implications would that have for the population?
\end{exercise}

You can verify your answers above with the following observation. If
$x_0$ is the initial population, then \begin{eqnarray*}
x_1 & = & k x_0, \\
x_2 &=& kx_1 = k^2 x_0,\\
\mbox{and, in general, ~} x_n & = & k^n x_0. \end{eqnarray*}

We can see from this equation, for example, that if $k>1$, then
$k^n$ increases without bound as $n$ increases.  Over time the
population would increase without bound.


The Malthus model assumes that a population is always growing at a
rate of $k$.  This is a reasonable assumption for short-term growth
of small populations with vast resources, such as the first bacteria
in a bacteria colony. However, the Malthus model is not reasonable
for describing long-term population growth. For instance, if
$k=1.07$ and $x_0=1$, then $x_{325}$ is over one billion!



No generating function can possibly capture all the  factors
affecting a population's growth.  A mathematical population model
attempts to capture some of the main factors that govern a
population.    In the following sections we will consider three
population models (logistic, Ricker, and Beverton-Holt) which,
unlike the Malthus model, assume that the rate at which a population
grows decreases as the population increases.

\newpage

\section{The Logistic Model}

\subsection{Introduction}
The maximum population that an environment can sustain is called the
\emph{carrying capacity} of the environment.  The simplest
population model which takes into account the carrying capacity is
the logistic model. We will denote the carrying capacity for a
population by $L$, and the population in generation $n$ by $y_n$.

The logistic model is given by the difference equation
\begin{eqnarray}y_{n+1}&=& ry_n \left( 1-\frac{y_n}{L}\right),
\label {rL}\end{eqnarray} where $L$ is the carrying capacity and $r$
is a constant combining the birth and death rates.  We will assume
$r>0$.

Note the following two features of the logistic model.  First, if
the current population $y_n$ is relatively small compared to the
carrying capacity $L$, so that $(1-\frac{y_n}{L})$ is close to 1,
the environment contains an abundance of resources for the current
population.  As a result, the population behaves in the short-term
as though its growth were unrestricted. That is, when $y_n$ is much
smaller than $L$, the logistic model is approximately the Malthus
model because  $y_{n+1}$ is approximately $r y_n$.  The second
feature to note is that  $(1-\frac{y_n}{L})$ approaches 0 as $y_n$
approaches $L$.  So the growth rate of the population slows down as
the population nears the carrying capacity $L$.



Though the logistic model makes unrealistic predictions (such as
negative populations if $y_n>L$), it is useful for describing how a
population would grow in the absence of outside intervention. The
model can be modified to incorporate intervening factors such as
harvesting or migration (see Exercise 2.4). From a mathematical
perspective, the logistic model is interesting to study. Though it
seems very simple, it displays rather complicated behaviors. (For a
classic paper with examples of simple models that have very complex
behaviors see [\ref{May}].)



If we let $x_n=\frac{y_n}{L}$, it can be shown that the equation
\begin{eqnarray} x_{n+1} &=& r x_n \left( 1-x_n\right)
\label{alpha}\end {eqnarray}

\noindent is equivalent to Equation (\ref{rL}). Note that in
Equation (\ref{alpha}), the population values are scaled as a
fraction of the carrying capacity, so that $x_n$ will take values
between zero and one. In making this adjustment, the ``1'' now
represents carrying capacity as 100\%, and the factor $(1-x)$ tells
us how close the population is to the carrying capacity as $x$ is a
fraction of the carrying capacity.  For the remainder of this
module, we will refer to Equation (\ref{alpha}) as the logistic
model.  We can also write the logistic model as $x_{n+1}=f(x_n)$,
where $f(x)=r x(1-x)$ is the generating function for the logistic
model.

\begin{exercise} {\bf Exercise 2.1:} Show that if $x_n=\frac{y_n}{L}$ then Equation (\ref{rL}) can
be rewritten as Equation (\ref{alpha}).
\end{exercise}

\begin{exercise} {\bf Exercise 2.2:} Find the fixed points of the logistic model by solving the equation
$rx(1-x)=x$ for $x$.
\end{exercise}

\begin{exercise} {\bf Exercise 2.3:} Consider $x_{n+1}=f(x_n)$ where
$f(x)=rx(1-x)$.
\begin{itemize}
\item[(a)] Suppose $x_n>0$.  Show that the logistic model predicts
$x_{n+1}<0$ if and only if $x_n>1$.
\item[(b)] What is the maximum
value of $f$, and where does it occur?
\item[(c)] Show that
$x_{n+1}>1$ is possible only if $r>4$.
\item[(d)]  What are
conditions on $r$ and $x_0$ that guarantee $x_n>0$ for
$n=1,2,3,\ldots$?
\end{itemize}
\end{exercise}

\subsection{Iterations Using Excel}
In this section we give steps to easily compute iterations of a
function using Excel. Consider the logistic model $x_{n+1}=r x_n
(1-x_n)$.

\begin{itemize}
\item We will use the A column to keep track of the time (or
generation).   Enter {\bf 0} for the initial time in cell A1.  In A2
enter {\bf =A1+1}.  Next, drag the fill handle in A2 (the small box
in the lower right corner of the cell) downward until, say, A20.
This will copy and update the formula in A2 into the cells below.
(Each time 1 will be added to the entry in the preceding cell.)
\item We will keep track of the parameter $r$ in C1.  For
now let's enter {\bf 2} in C1.
\item Enter the initial population $x_0$ in B1.  For now
let's enter {\bf 0.1} in B1.
\item In the B column we will calculate the iterations of
the generating function $f(x)=r x (1-x)$ by entering {\bf
=\$C\$1*B1*(1-B1)} in B2. (The \$ symbol in the reference to C1
signifies that the reference is absolute and will not be updated as
we iterate.  The references to B1 are relative and will change in
subsequent formulas.) Now drag the fill handle in B2 downward until
B20.
\item Every time you enter a new value for $x_0$,
the initial population, in B1 or change the parameter $r$ in C1,
your iterations will be updated. (See Figure \ref{iterate1}.)
\end{itemize}

\begin{figure}[htbp]
\centering\includegraphics[width=4.5cm]{iterate1.eps}
\caption{Excel
iterations of the logistic model with $r=2$ and $x_0 = 0.1$.}
\label{iterate1}
\end{figure}

\begin{figure}[htbp] \begin{center}
\includegraphics[width=4.5cm]{plot1.eps} \caption{Excel plot of
iterations with $r=2$ and $x_0 = 0.1$.} \label{plot1} \end{center}
\end{figure}



In order to graph the data points as in Figure \ref{plot1}, select
{\bf Chart} under the {\bf Insert} menu.  Then select {\bf XY
(Scatter)} and choose the sub-chart ``Scatter with data points
connected by lines''. After clicking {\bf Next}, a {\bf Data Range}
box will appear. Click on the small square to the right  of this
box, and then highlight the data points in columns A and B that are
to be graphed. Finally, follow the steps in the {\bf Chart Wizard}
to label your chart and the axes, and choose the scale of your
graph. When you select {\bf Finish}, the data will be displayed on a
chart as shown in Figure \ref{plot1}. Your graph will be
automatically updated every time you enter a new value of $r$ in C1
or a new value of $x_0$, the initial population, in B1.










\subsection{Iteration Exercises}
In the exercises in this section you'll work with logistic
population models for biological populations. Keep in mind that the
predictions the model makes about the size of the population are
what we might expect to observe if outside conditions remain
approximately the same.

\begin{exercise} {\bf Exercise 2.4:} The wheat bulb fly is a common pest to wheat crops. Wheat bulb
flies have one brood per year and have non-overlapping generations.
Each fall the flies lay their eggs. The eggs hatch in mid-winter,
and the adult flies are present from mid-June through September.  In
[\ref{bulbfly}] W. Morris fit logistic models to previously
published life-table data of eight generations of the wheat bulb fly
from a wheat growing area of England.  It was estimated that the
carrying capacity for wheat bulb flies is $0.72 \times 10^6$ flies
per ha (hectare) and that $r=1.5$.

Suppose that there are $0.06 \times 10^6$ flies per ha. Because we
have scaled the population values to be a fraction of the carrying
capacity, to find $x_0$ for our model we must divide the initial
population $0.06 \times 10^6$ by $0.72 \times 10^6$ to get the
fraction $x_0 = 0.06/0.72 = 0.0833$.

\begin{itemize}
\item[(a)] Find the first 20 iterations of this model and show that the
model predicts that the population stabilizes at 0.333 of the
carrying capacity. What is the actual number of wheat bulb flies
that this fraction represents?
\item[(b)] Suppose $0.5 \times 10^5$ flies
per ha migrated to this region each year.  How would you modify the
model to take into account this migration?
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 2.5:} Understanding how tumors grow is an important part of
understanding how to treat cancer.  In [\ref{tumor}], Cross and
Cotton use a logistic model $x_{n+1}=r x_n (1-x_n)$ to predict tumor
growth. Assume that the tumor cells are in a container in a
laboratory and that the container can support a maximum number of
cells, which is the carrying capacity of the container.  Instead of
specifying a container carrying capacity of, say, 2 million cells,
and also the actual number of cells, we instead represent
populations as fractions of the carrying capacity.  Therefore the
population $x_n$ is always between 0 and 1. For each of the
following values of $r$ find the first 30 iterations of three
different initial populations between 0 and 1, and plot your
results. Based on your results, what (if anything) does the model
predict will happen to the tumor cell population over time?
\begin{itemize}
\item[(a)] $r = 2.5$
\item[(b)] $r = 3$
\item[(c)] $r = 3.5$
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 2.6:} Consider a population modeled by the logistic equation
$x_{n+1}=2x_n(1- x_n)$.
\begin{itemize}
\item[(a)] Find the fixed points.
\item[(b)] For each value of $x_0$ given below, determine the corresponding
values of $x_n$ for times $n=1, 2, 3, 4, 5$. Describe the patterns
you observe and interpret them in terms of the population.
\begin{itemize}
\item[i.] $x_0=0.1$
\item[ii.] $x_0=1.1$
\item[iii.] $x_0=0.6$
\item[iv.]  $x_0=0.5$
\end{itemize}
\item[(c)] Based on your calculations in part (b), determine whether each
fixed point is attracting or repelling.
\end{itemize}
\end{exercise}


\subsection{Cobweb Diagrams}

A cobweb diagram is a graphical representation of the orbit of a
point $x_0$ under a generating function $f(x)$ that does not require
any computations. To start, we graph both $y=f(x)$ and the line
$y=x$. Note that the $x-$coordinates of the intersection points are
precisely the fixed points of $f$ since they are the solutions to
the equation $f(x)=x$.

The coordinates of any point on the graph of the generating function
$f(x)$ represent successive population values. That is, a point on
the graph of $f(x)$ has coordinates $(x_n, x_{n+1})$. Also note that
a point on the graph of $y=x$ has coordinates $(x_n, x_n)$. The
cobweb diagram is constructed by alternately drawing a sequence of
vertical and horizontal line segments.

Let's consider the logistic model $x_{n+1}=2x_n(1-x_n)$ with initial
population $x_0=0.2$.   We first graph the generating function
$y=2x(1-x)$ and the line $y=x$. (The intersection points occur at
$x=0$ and $x=0.5$, the fixed points of $f$.)  Locate the initial
population $x_0=0.2$ on the horizontal axis.  Sketch a vertical line
segment up to the graph of $y=2x(1-x)$. The $y$-coordinate of this
point on the parabola is the value of the population at time 1, or
$x_1$. (See Figure \ref{cobwb1}.) \begin{figure}[htbp]
\begin{center} \includegraphics[width=4.5cm]{cobwb1.eps}
\caption{Starting a cobweb diagram.} \label{cobwb1} \end{center}
\end{figure}


From the point $(x_0, x_1)$ on the parabola, sketch a horizontal
line segment over to the line $y=x$. The horizontal line segment
intersects the line $y=x$ at the point ($x_1, x_1$). We repeat this
process of drawing a vertical line segment from $(x_1,x_1)$ up to
the parabola and a horizontal line segment over to the line $y=x$.
We should now be at the point $(x_2, x_2)$. (See Figure
\ref{cobwb2}.) In repeating this process we are graphically
iterating $x_0$ under the function $f$. The result is a diagram that
looks like a cobweb or staircase.  We can visualize the orbit
$\{x_0, x_1, x_2, \ldots\}$ and its behavior, in this case
increasing toward the fixed point at 0.5, without doing any
calculations! (See Figure \ref{cobwb3}.)

A cobweb diagram provides a way to visualize the long-term behavior
of orbits.  From the cobweb diagram of $x_{n+1}=2 x_n(1-x_n)$ we see
that the orbit of any point $x_0$ near 0.5 always approaches the
fixed point at 0.5. This verifies your observation in Exercise 2.6
that 0.5 is an attracting fixed point. However, the orbit of any
point $x_0$ near 0 (but not equal to 0) always goes away from 0,
which verifies that 0 is a repelling fixed point.

\begin{figure}[htbp]
\begin{center}
\includegraphics[width=4.5cm]{cobweb2.eps} \caption{Two iterations} \label{cobwb2}
\end{center}
\end{figure}


\begin{figure}[htbp]
\begin{center}
\includegraphics[width=4.5cm]{cobweb3.eps} \caption{The orbit of $x_0$ approaches the fixed point $0.5$.}
\label{cobwb3}
\end{center}
\end{figure}


\begin{exercise} {\bf Exercise 2.7:} Draw cobweb diagrams for the Malthus model $x_{n+1}=k x_n$, for
$0<k<1$, $k>1$ and for $k=1$.  (Here the generating function is the
line with slope $k$ that passes through the origin.) Based on your
cobweb diagrams, is the fixed point $x=0$ attracting, repelling or
neither attracting nor repelling?
\end{exercise}


The following theorem is useful for classifying fixed points. It
says that the slope of the generating function at a fixed point
determines whether a fixed point is attracting or repelling,
confirming what you observed in the previous exercise with the
Malthus model. (The theorem is proved using the Mean Value Theorem
and can be found in [\ref{devaney}])

\begin{them} \label{fixedpoint} Let $f$ be a continuous function
whose derivative is also continuous, and let $p$ be a fixed point of
$f$.  If $|f'(p)|<1$, then $p$ is an attracting fixed point. If
$|f'(p)|>1$, then $p$ is a repelling fixed point.
\end{them}

\begin{exercise} {\bf Exercise 2.8:} Consider the logistic model $x_{n+1}=r x_n(1-x_n)$.  Use Theorem
\ref{fixedpoint} to determine the values of $r$ for which each fixed
point is attracting.
\end{exercise}



\subsection{Cobweb Diagrams using Excel}


We next describe how to use Excel to draw a cobweb diagram (as in
Figure \ref{3logisticweb}) for the logistic function $x_{n+1} = r
x_n (1-x_n)$.

\begin{figure}[htbp]
\begin{center}
\includegraphics{3logisticweb.eps} \caption{Cobweb of the logistic model using Excel with $r=3$}
\label{3logisticweb}
\end{center}
\end{figure}


\begin{itemize}
\item Enter the parameter(s) of the model and the initial population in column A.  For
the logistic model, the only parameter is $r$, so enter a value of
$r$ in A1, and enter a value of $x_0$ in A2.
\item The first point on the cobweb diagram lies on the $x-$axis at
$(x_0,0)$.  Enter {\bf = \$A\$2} in B1, and enter {\bf 0} in C1.
\item The second point lies on the graph of the generating function
at $(x_0, x_1)$.  Enter {\bf =B1} in B2 and the formula for the
generating function in C2.  For the logistic example we would enter
{\bf =\$A\$1*B2*(1-B2)} in C2.
\item The third point lies on the line $y=x$ at $(x_1,x_1)$.  Enter
{\bf =C2} in both B3 and C3.
\item Highlight the four cells B2, C2, B3 and C3.  Drag the fill
handle down until you've filled up to B50, C50.  The formula pattern
from the four cells will be repeated.  (Note that the
$x-$coordinates for the cobweb diagram are in column B, and the $y-$
coordinates are in column C.)
\item Next we generate data points that will be used to sketch the
generating function $f(x)=r x (1-x)$.  The $x-$coordinates will be
in column D. Enter {\bf 0} in D1 and {\bf =D1+0.1} in D2. Drag D2
down to D11.  To generate the $y-$coordinates, we need to enter the
generating function in E1.  For the logistic example, we enter {\bf
= \$A\$1*D1*(1-D1)} in E1.  Then drag this down to E11.
\item Select {\bf Chart} under the {\bf Insert} menu.  Next, select
{\bf XY (Scatter)} and choose the sub-chart ``Scatter with data
points connected by lines''. After clicking {\bf Next}, a {\bf Data
Range} box will appear. Click on the small square to the right  of
this box, and then highlight the data points in columns B and C that
 are to be graphed. Finally, follow the steps in the Chart Wizard
to label your chart and the axes, and choose the scale of your
graph. You now should have a chart with the cobweb, but it still
needs the graph of $y=x$ and the generating function.  Right click
on the chart and select {\bf Source data} and then select {\bf Add}
under {\bf Series}.  Click on the small square to the right  of the
x-values box and then highlight the data in column D.  Next, click
on the small square to the right of the y-values box and highlight
the data in column E.  This produces a graph of the generating
function. (If you right click on its graph you can change the
sub-type to ``scatter with data points connected by smooth lines
without markers''.) Finally, to sketch the line $y=x$, repeat these
instructions but use the data in column D for both the x and y
values on your graph.

\item Now you can change both $r$ and $x_0$ to construct new
cobweb graphs.

\end{itemize}




\begin{exercise} {\bf Exercise 2.9:} Find the fixed points of the logistic model $x_{n+1} = 2.5x_n (1-x_n)$ and use a cobweb
diagram to classify each fixed point as attracting or repelling.
\end{exercise}

\begin{exercise} {\bf Exercise 2.10:} Let $r>0$ and let $x_0$ be in $(0,1)$.  Use cobweb diagrams on
$x_{n+1}=r x_n (1-x_n)$ to find
\begin{itemize}
\item[(a)]  an interval of values of $r$ for which the fixed point $x=0$ is
attracting.
\item[(b)] an interval of values of $r$ for which the second
(nonzero)fixed point is attracting.
\item[(c)] a value of $r$ for
which the long-term behaviors of orbits cannot be determined.
\end{itemize}
\end{exercise}


A \emph{period k} point $p$ is a point such that $f^k(p)=p$ but
$f^j(p)\neq p$ for $0<j<k$.  The orbit of a period k point is called
a $k$\emph{-cycle}.  In Exercise 2.5 you saw that for some choice of
$r$ there was an attracting 2-cycle or 4-cycle.  In fact, it can be
shown that there are values of $r$ that give cycles of every period.

\begin{exercise} {\bf Exercise 2.11:} Find the values of $r$ for which there will there be a 2-cycle.
(Hint:  To find the period 2 points, you need to solve the equation
$f^2(x)=x$. Note that $f^2(x)-x$ is degree 4 polynomial, and two of
its roots will be the fixed points of $f$. Why? Factoring out the
fixed points of $f$ will simplify your calculations.)
\end{exercise}

\begin{exercise} {\bf Exercise 2.12:} For what values of $r$ will the 2-cycle found in the previous
exercise be attracting? Use Theorem \ref{fixedpoint} and the chain
rule.
\end{exercise}

\newpage

\section{The Ricker Model}


\subsection{Introduction}
When determining the optimal stocking rates for fisheries,
mathematical biologists often use a mathematical model developed in
1954 by W.E. Ricker to study the salmon population in the Pacific
northwest.  In this section, we'll examine the Ricker model and some
of its applications to mathematical biology.

The Ricker model is given by the difference equation
\begin{eqnarray} \label{ricker} x_{n+1}= \alpha e^{-\beta x_n} x_n.
\end{eqnarray}

The constants $\alpha$ and $\beta$ are positive and are estimated
from population data.  We can regard $\alpha$ as the maximum value
of the growth rate, and $\beta$ as a constant that scales the
carrying capacity.  We will see how to estimate these constants in
Section 3.2.

Recall that the logistic model can give unrealistic predictions,
such as negative populations. This situation does not occur in the
Ricker model because its generating function $f(x)=\alpha e^{-\beta
x}$ is positive for $x>0$. In the exercises you'll show that, like
the logistic function, the the graph of the Ricker generating
function has one-hump. That is, $f$ is increasing for $0<x<L$ for
some $L$, and $f$ is decreasing for $x>L$.

A population model of the form $x_{n+1}=g(x_n) x_n$ is called
\emph{density dependent} because $g(x)$, the per-capita growth rate,
depends on the size of the population.  Generally, the greater the
population, the more competition there is for resources, so $g$ will
be a decreasing function.  Both the logistic and Ricker models are
density dependent.  In the logistic model $g(x)=r(1-x)$. In the
Ricker model $g(x)=\alpha e^{-\beta x}$.  The growth rate in the
Malthus model does not depend on the size of the population; it is
always $k$.

Note, in the logistic model the population values were scaled to
take values between 0 and 1.  The population values are not scaled
in the Ricker model.

\begin{exercise} {\bf Exercise 3.1:}
The Ricker model has been used to study the population of the
Mexican bean beetle in the soybean fields of North Carolina.  (See
[\ref{beetle}].)  These beetles can have three generations in one
year. Occasionally the beetle causes serious damage to a soybean
crop. By fitting the beetle population data with a Ricker model it
was found that, for one variety of soybean, $\alpha = 8.86$ and
$\beta=0.148$.
\begin{itemize}
\item[(a)] Suppose that initially there are 20 beetles per square
meter. Determine the first 10 iterations of the Ricker model. What
patterns for the beetle population does this model predict? If you
are using the Excel directions in Section 2.2, you'll need to keep
track of two parameters, say, $\alpha$ in C1 and $\beta$ in D1. The
A column would be the same as described in Section 2.2, and B1 would
contain the initial population. The generating function in B2 would
be written as ${\bf =\$C\$1*Exp(-\$D\$1*B1)*B1}$.
\item[(b)] Suppose that
other varieties of soybeans yield different values of $\alpha$. For
each of the following values of $\alpha$ find the first 30
iterations and plot your results.  Based on your results, what (if
anything) does the model predict will happen to the beetle
population over time?
\begin{itemize}
\item[i.]  $\alpha = 0.5$
\item[ii.] $\alpha = 1.5$
\item[iii.] $\alpha = 10$
\end{itemize}
\item[(c)] For one of the values of $\alpha$ in part (b) you should have
observed the population bouncing back and forth between two values,
i.e., a two-cycle. Experiment with different values of $\alpha$
until you find a value that gives a 4-cycle.
\end{itemize}
\end{exercise}

\subsection{Finding a Ricker Model}

If we have a data set for a population with non-overlapping
generations that we want to fit it with a Ricker model, we need a
method of finding $\alpha$ and $\beta$.  In this section we give
directions for using Excel to find these constants.  We'll
illustrate the method with population data of the pine looper moth
found in [\ref{PineLooper}] that is taken from annual pupal surveys
conducted in Great Britain by the British Forestry service since the
1950s.

The pine looper moth is a threat to many species of conifers.  It is
found in the western United States, in western Canada, and in
northern Europe.  Adult moths emerge in early summer and live
without feeding for 10--14 days.  The females lay their eggs on the
pine needles. After three weeks the larvae hatch and feed on new and
old pine needles for four to five months.  The needles are often
eaten down to the sheath, which can lead to needle loss and the
death of trees. Sometime between mid October and December, the
larvae drop to the ground and burrow under the leaves, needles, and
twigs on the forest floor.  It overwinters in the pupa stage. In the
chart below are the mean totals of pupal densities in the Cannock
Chase forest from 1960 to 1969.  The densities are given per square
meter.
\medskip

\hspace*{-.25in}\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|} \hline
\label{Pine Looper Data}

Year & 1960 & 1961 & 1962 & 1963 & 1964 & 1965 & 1966 & 1967 & 1968 & 1969 \\

\hline

pupae per sq. m. & 15.09 & 9.15 & 9.32 & 1.02 & 0.35 & 0.26 & 1.75 &
9.09 & 5.93 & 1.53 \\

\hline
\end{tabular}

\medskip


\begin{exercise} {\bf Exercise 3.2:} Show that if we divide both sides of the Ricker equation
$x_{n+1}=\alpha e^{-\beta x_n} x_n$ by $x_n$ and take the natural
log of both sides,
 we
 obtain \begin{eqnarray} \label{regression}
 \ln \left[ \frac{x_{n+1}}{x_n} \right] = \ln(\alpha) - \beta x_n
 \end{eqnarray}
\end{exercise}

We can treat Equation (\ref{regression}) as a linear regression  $y
= b + mx$ where $y = \ln \left[ \frac{x_{n+1}}{x_n} \right], b=
\ln(\alpha),$ and $m = -\beta$.  Here $y$ is the dependent variable
and $x_n$ the independent variable.  Say $n=0$ corresponds to 1960.
We next compute the $y$ values, which are displayed in the following
table.
\medskip

\begin{tabular}{|l|l|l|l|}
\hline $n$ & {$x_n$} & $x_{n+1}$ & $y=\ln
\left[\frac{x_{n+1}} {x_n}\right]$ \\ \hline 0& 15.09 & 9.15 & -0.500 \\
\hline 1& 9.15 & 9.32 & 0.018\\ \hline 2& 9.32 & 1.02 & $-$2.212\\
\hline 3& 1.02 & 0.35 & $-$1.070 \\ \hline 4& 0.35 & 0.26 & $-$0.297 \\
\hline 5& 0.26 & 1.75 & 1.907 \\ \hline 6& 1.75 & 9.09 & 1.648 \\
\hline 7& 9.09 & 5.93 & $-$0.427 \\ \hline 8& 5.93 & 1.53 & $-$1.355 \\
 \hline
\end{tabular}

\medskip

To find $b$ and $m$, enter the data from the $x_n$ column in cells
A1-A9, and enter the data from the $y$ column in cells B1-B9.  To
find $m$ enter ${\bf =Slope(B1:B9, A1:A9)}$ in an empty cell. To
find $b$ enter ${\bf =Intercept(B1:B9, A1:A9)}$ in another empty
cell. Then set $\beta=-m$ and $\alpha=e^{b}$.  For this data set, we
find that $\beta=0.111$ and $\alpha=e^{0.388}=1.474$.  So the Ricker
model would be
\begin{eqnarray} \label{moth}
x_{n+1}=1.474 e^{0.111 x_n}x_n
\end{eqnarray}

\begin{exercise} {\bf Exercise 3.3:} Find the first 15 iterations of Equation (\ref{moth}). What does
this model predict for the pine looper pupae population per square
meter in the Cannock Chase forest for 1975?
\end{exercise}

\begin{exercise} {\bf Exercise 3.4:} Below are data of the pine looper moth from the the Moray forest
district. Find a Ricker model for this data. Then find the first 8
iterations. What does the model predict for the population of pine
looper pupae in 2000?
\end{exercise}

\medskip


\begin{tabular}{|l|l|l|l|l|l|l|}
\hline

Year & 1992 & 1993 & 1994 & 1995 & 1996 & 1997 \\

\hline

pupae per sq. m. & 6.0& 3.2 & 4.4 & 10.0 & 24.0 & 30.0 \\

\hline
\end{tabular}

\medskip

\begin{exercise} {\bf Exercise 3.5:} The Ricker model was recently used to study the density of wasp
nests in New Zealand beech forests. (See [\ref{Wasps}].) Each year
wasps build nests in the trees of New Zealand's South Island. The
wasps compete with rare birds for honeydew sap, and an overabundance
of wasps threatens both the birds and the trees. In the model, $x_n$
is the number of wasp nests per hectare at each site.

\medskip

\hspace*{-.75in} {\small{
\begin{tabular} {|l|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline Site & 1988 &  1989 & 1990 & 1991 & 1992 & 1993 & 1994 & 1995  & 1996  & 1997 &
1998 &1999 & 2000\\
\hline Mt Misery & 16.0 &  11.8 & 10.0 & 15.7 & 25.3 & 28.3 & 30.5 &
4.4  & 16.5  &
11.4 & 17.1 &11.4 & 11.9\\
\hline Watson & 20.5 &  19.8 & 12.9 & 15.7 & 10.6 & 18.3 & 12.9 &
6.8  & 1.1  & 12.2 & 9.0 &6.9 & 8.0\\ \hline \end{tabular}}}

\medskip
\begin{itemize}
\item[(a)] Use the Watson data to find a Ricker model for the number of
wasp nests, and sketch the graph of the generating function.  What
does your model predict about the number of wasp nests in 2002?
\item[(b)] If we use a model $x_{n+1}=f(x_n)$ to make predictions about a
population, then ordered pairs of actual population values $(p_n,
p_{n+1})$ should lie approximately on the graph of the generating
function.  Recall that if $(x,y)$ is a point on the graph of the
generating function $f(x)$, the $x-$coordinate represents the
current population value $x_n$, and the $y-$coordinate represents
$x_{n+1}$ the predicted population value at the next time period.
 Plot the ordered pairs $(p_n, p_{n+1})$ of actual Watson population
values given in the chart below together with the generating
function you found in part (a).


\medskip

\begin{tabular} {|l|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
$p_n$ & 20.5 &19.8 & 12.9&15.7 &10.6 &18.3 &12.9 &6.8 &1.1 &12.2
&9.0&6.9 \\
\hline

\medskip


$p_{n+1}$ & 19.8 &12.9 &15.7 &10.6 &18.3 &12.9 &6.8 &1.1 &12.2 &9.0
&6.9& 8.0 \\
\hline
\end{tabular}

\medskip

\item[(c)] Based on part (b), does it seem reasonable to fit this data with a Ricker model?
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 3.6:} Coho salmon are one of the main fish varieties caught off the coast
of southeastern Alaska, and they have a life span of about three
years. When the adult salmon are ready to spawn, they swim from the
ocean to the river bed where they were born.  After this arduous
journey, the females lay their eggs, which the males then fertilize.
Several months later, the eggs hatch. In about a year, the young
salmon start their migration to the ocean, where they live between
one and two years before they begin their own journey to the river
bed spawning grounds.

The following data set can be found in [\ref{coho}]. The fish
counted here are called \emph{recruits}.  Recruits are the salmon
that are ready to begin migration from the river to the ocean, and
which will be available for fishing the following year.

\medskip

\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline Year & 1980 & 1983 & 1986 & 1989 & 1992 &1995 & 1998 & 2001
\\ \hline Coho Population & 10714 & 6901 & 5617 & 7011 & 6596 & 4884 &
7420 & 5980 \\
in Auke Creek fishery & & & & & & & & \\
\hline \end{tabular}

\medskip
\begin{itemize}
\item[(a)] Fit this data with a Ricker model, and find the first 8
iterations. What does this model predict the population of Coho
salmon will be in the Auke Creek fishery in 2004 (when $n=8$)?
\item[(b)] As in Exercise 3.5, plot the generating function and the
actual population points $(p_n, p_{n+1})$.  Compare your plot with
the Watson-wasp plot in Exercise 3.5.  In which case does the Ricker
model appear to better fit the data?
\item[(c)] If $f(x_n)>x_n$ (that is, the graph of $f$ lies above the diagonal line $y=x$) then the
fishery could catch the surplus $f(x)-x$ and be left with a fish
population of the same size as the last generation. The largest
harvest of this type is called the \emph{maximum sustainable yield}.
Use technology to find the maximum sustainable yield if $\alpha=10$
and $\beta=0.05$, and  determine the population size that gives this
harvest size.
\end{itemize}
\end{exercise}

\subsection{Additional Exercises}




\begin{exercise} {\bf Exercise 3.7:} Find the fixed points of the Ricker model $x_{n+1}= \alpha
e^{-\beta x_n} x_n$.
\end{exercise}

\begin{exercise} {\bf Exercise 3.8:} In this exercise you will show that the graph of the Ricker
model's generating function has a ``one hump'' shape.
\begin{itemize}
\item[(a)] Find $f'(x)$ and determine where $f$ is increasing or
decreasing.
\item[(b)] Show that $\displaystyle \lim_{x \to \infty} f(x)=0$.
\item[(c)]  At what value of $x$ is $f(x)$ the greatest?
\item[(d)] What is the maximum value of $f(x)$?
\item[(e)] Based on the information in parts (a) through (c), sketch
the graph of $f$.
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 3.9:} In the beetle exercise (Exercise 3.1) we saw that when $\alpha =
0.5$, the model predicted that the population would become extinct.
When $\alpha = 1.5$, the model showed that the population would
approach 2.73, which was (approximately) the fixed point $\frac{\ln
\alpha}{\beta}$.  So for some values of $\alpha$ the fixed points
are attracting, and for other values they are repelling. In this
exercise you will find the values of $\alpha$ for which the fixed
points are attracting or repelling.  To simplify the calculations,
let $\beta = 1$ so that $f(x)=\alpha x e^{-x}$.
\begin{itemize}
\item[(a)] Show that the fixed points of $f$ are $x=0$ and $x=\ln \alpha$.
\item[(b)] Use cobweb diagrams to find an interval of values of $\alpha$
for which $x=0$ is attracting.
\item[(c)] Use cobweb diagrams to find an interval of values of $\alpha$
for which $x=\ln \alpha$ is attracting.
\item[(d)] Use Theorem \ref{fixedpoint} to confirm your answers to parts (b) and (c).
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 3.10:} As with the logistic model, the Ricker model can exhibit very
complicated behaviors.  Let $\beta=1$.  Use cobwebbing to find a
value of $\alpha$ for which an attracting 2-cycle appears.  Also,
find a value of $\alpha$ for which an attracting 4-cycle appears.
\end{exercise}

\newpage

\section{The Beverton-Holt Model}

\subsection{Introduction}

The Beverton-Holt model, introduced by Beverton and Holt in 1957
(see [\ref{BVHolt}]) is one of the most commonly used stock and
recruitment models in fishery management.  It is used to describe
the population of insect or fish species, such as the pacific
salmon, which reproduce just once during their lifetime after which
death is inevitable. The Beverton-Holt model depicts density
dependent recruitment of a population in which resources are limited
and unequally shared.

Let $x_n$ represent the total population at time $n$ of a species
with initial population $x_0$.  (Time can be given in minutes, days,
years, etc.)  Then $x_{n+1}$ is the population at time $n+1$. The
Beverton-Holt model is given by

\begin{eqnarray} \label{BH}
x_{n+1}=\frac{\alpha x_n}{1+\beta x_n  }. \end{eqnarray}

The constants $\alpha$ and $\beta$ are estimated from population
data, and $\alpha$ is positive.  (We regard $\alpha$ as the maximum
growth rate and $\beta$ as a measure of growth inhibition.) Equation
(\ref{BH}) can be written as $x_{n+1}=f(x_n)$ where
$f(x)=\frac{\alpha x}{1+\beta x}$ is the generating function for the
Beverton-Holt model.

\begin{exercise} {\bf Exercise 4.1:} Equation (\ref{BH}) can also be written as $x_{n+1}=g(x_n)x_n$
where $\displaystyle g(x)=\frac{\alpha}{1+\beta x}$ is the
per-capita growth rate.  What are some reasons why we need $g$ to be
positive and decreasing?
\end{exercise}

\begin{exercise} {\bf Exercise 4.2:} Suppose that for a species of Atlantic salmon in the Western Arm
Brook River it was estimated that $\alpha=84.2$ and $\beta =
0.0036$. Graph the generating function.  If $x_0=500,000$, find the
first 4 iterations of this model.
\end{exercise}

\begin{exercise} {\bf Exercise 4.3:} Find the fixed points of the Beverton-Holt model.
\end{exercise}

\begin{exercise} {\bf Exercise 4.4:} In this exercise you will examine the graph of the generating
function $f(x) = \frac{\alpha x}{1+\beta x}$.  Assume $x \geq 0$.
\begin{itemize}
\item[(a)] Find $f'(x)$.
\item[(b)] Determine the intervals of increase and decrease of $f$.
\item[(c)] Find $\displaystyle \lim_{x \to \infty} f(x)$.
\item[(d)] Find $f''(x)$.
\item[(e)] Determine the values of $\beta$ and intervals of $x$ where $f$
is concave up or concave down.
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 4.5:} Consider a Beverton-Holt model with $\alpha = 2.5$ and
$\beta=1.5$.
\begin{itemize}
\item[(a)] Show that the fixed points are $x=0$ and $x=1$.
\item[(b)] Suppose $x_0=0.55$. Find the first 20 iterations of $f$ and
describe the limiting behavior of the iterations.
\item[(c)] Repeat part (b) for $x_0 = 1.4$.
\item[(d)] Draw cobweb diagrams for three different values of $x_0$. Based
on your diagrams and calculations, is $x=0$ attracting or repelling?
Is $x=1$ attracting or repelling?
\end{itemize}
\end{exercise}


If $K$ is the nonzero fixed point, it can be shown that
$\displaystyle \beta = \frac{\alpha -1}{K}$ and that the generating
function for the Beverton-Holt model has the form
\begin{eqnarray} \label{closedBH2} f(x) =\frac{\alpha x}{1+\left(\frac{\alpha - 1}{K}\right) x}
  \end{eqnarray}

\begin{exercise} {\bf Exercise 4.6:} Verify that if $K$ is the nonzero fixed point, then the
generating
 function has the form in Equation (\ref{closedBH2}).
\end{exercise}

\begin{exercise} {\bf Exercise 4.7:} Consider the Beverton-Holt model given by Equation
(\ref{closedBH2}), and suppose $K=1$.
\begin{itemize}
\item[(a)] Construct a cobweb diagram for the following $\alpha$
values and initial conditions.
\begin{itemize}
\item[i.] $\alpha=2,$ $x_{0}=0.4$
\item[ii.] $\alpha=2$, $x_{0}=1.4$
\item[iii.] $\alpha=0.7$, $x_{0}=0.4$
\item[iv.] $\alpha=0.7$, $x_{0}=1.1$
\end{itemize}
\item[(b)] From your cobweb diagrams, make a conjecture about when the
fixed point $x=0$ will be attracting or repelling.  Make a
conjecture about when the fixed point $x=1$ will be attracting or
repelling.
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 4.8:} Consider the generating function $f$,  given by Equation
(\ref{closedBH2}).   Use Theorem \ref{fixedpoint} to find values of
$\alpha$ for which $x=0$ is attracting.  Repeat for the nonzero
fixed point $K$.
\end{exercise}




\subsection{Finding a Beverton-Holt Model}




In this section we describe how to find a model of the form
$x_{n+1}=f(x_n)$ where $f$ is given by Equation (\ref{closedBH2}).
That is, given a set of data, how does one find the parameters
$\alpha$ and $K$ that best fit it? The data set in Table
\ref{parameciumaureliadata} is from experiments conducted by G. F.
Gauss (1934a) on a single cell organism, \emph{paramecium aurelia},
a protozoa found in a number of fresh water ponds. In the data set
in Table \ref{parameciumaureliadata}, time $n$ is measured in days
and $x_n$ is the mean population density of paramecium, measured in
numbers of individuals per 0.5 cm$^{3}$.

\begin{table}[t]
\caption{growth of paramecium aurelia in isolation} \vspace*{.2in}
\centering \footnotesize{\begin{tabular}  {|c|c|} \hline






Day (n) & Mean Density of P. Aurelia $x_n$ \\
\hline
0 & 2  \\
\hline
1 & --  \\
\hline
2 & 14  \\
\hline
3 & 34 \\
\hline
4 & 56  \\
\hline
5 & 94  \\
\hline
6 & 189  \\
\hline
7 & 266 \\
\hline
8 & 330  \\
\hline
9 & 416  \\
\hline
10 & 507 \\
\hline
11 & 580  \\
\hline
12 & 610  \\
\hline
13 & 513  \\
\hline
14 & 593  \\
\hline
15 & 557  \\
\hline
16 & 560  \\
\hline
17 & 522  \\
\hline
18 & 565  \\
\hline
19 & 517  \\
\hline
20 & 500  \\
\hline
21 & 585  \\
\hline
22 & 500  \\
\hline
23 & 495  \\
\hline
24 & 525  \\
\hline
25 & 510 \\
\hline

\end{tabular}}
\label{parameciumaureliadata}
\end{table}


\begin{exercise} {\bf Exercise 4.9:} Construct a scatter plot of the data in Table
\ref{parameciumaureliadata}. Does it look like the graph of a
possible generating function $f(x)$ for the Beverton-Holt model?
\end{exercise}

We can use Excel as we did in Section 3.2 to estimate the parameters
$\alpha$ and $K$.

\begin{exercise} {\bf Exercise 4.10:} Show that if we invert both sides of the Beverton-Holt equation
$x_{n+1}=\frac{\alpha x_n}{1+\left(\frac{\alpha-1}{K}\right)x_n}$
and then multiply by $x_n$ and simplify, we obtain $\displaystyle
\frac{x_n}{x_{n+1}}=\frac{1}{\alpha}+\left( \frac{\alpha-1}{\alpha
K}\right)  x_n.
$
\end{exercise}

We can treat the equation in Exercise 4.10 as a linear regression $y
= b + mx$ where $\displaystyle y = \frac{x_{n}}{x_{n+1}}, b=
\frac{1}{\alpha}$ and $\displaystyle m = \frac{\alpha - 1}{\alpha
K}$. Here $y$ is the dependent variable and $x_n$ is the independent
variable.
\newpage

\begin{exercise} {\bf Exercise 4.11:} Use linear regression to find a Beverton-Holt model for the
paramecium in Table \ref{parameciumaureliadata}.  What does the
Beverton-Holt model predict about the number of paramecium on day
50?
\end{exercise}





\begin{exercise} {\bf Exercise 4.12:} Recall that in Exercise 3.6 we found a Ricker
model for the salmon population in the Auke Creek fishery from the
following data set found in [\ref{coho}]:

\medskip

\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline Year & 1980 & 1983 & 1986 & 1989 & 1992 &1995 & 1998 & 2001
\\ \hline Population & 10714 & 6901 & 5617 & 7011 & 6596 & 4884 &
7420 & 5980 \\ \hline \end{tabular} \medskip

Fit this data with a Beverton-Holt model and find the first 8
iterations of this model. What does this model predict the
population of Coho salmon will be in the Auke Creek fishery in 2004?
Sketch the graph of the generating function and the ordered pairs of
actual population values $(p_n, p_{n+1})$. Is it realistic to make
predictions about the salmon population with this model? Why or why
not? How do your results compare with those you found using the
Ricker model?
\end{exercise}

\subsection{Finding a Closed Form}

Unlike the Ricker model, the Beverton-Holt model may be rewritten in
closed form.  That is, we can find an expression for $x_n$ in terms
of $x_0$,  namely,

\begin{eqnarray} \label{closedBH1} x_n=\frac{\alpha^{n}\left(
\alpha-1\right)  x_{0}}{\left( \alpha-1\right)  +\beta\left(
\alpha^{n}-1\right)  x_{0}}.
\end{eqnarray}

\begin{exercise} {\bf Exercise 4.13:} This exercise will guide you through the derivation of Equation
(\ref{closedBH1}).
\begin{itemize}
\item[(a)]  Use mathematical induction to show that if Equation (\ref{BH})
is iterated starting at $x_0$, we obtain the closed form solution
\begin{eqnarray} \label{closedBH}
x_n=\frac{\alpha^{n}x_{0}}{1+\beta\left(  \sum_{i=0}^{n-1}
\alpha^{i}\right)  x_{0}}
\end{eqnarray}

\item[(b)] Show that $\displaystyle \sum_{i=0}^{n-1} \alpha^{i} =
\frac{\alpha^n - 1}{\alpha - 1}$.  (Hint: Let
$s_n=1+\alpha+\cdots+\alpha^{n-1}$.  Then consider $\alpha s_n -
s_n.)$

\item[(c)] Use part (b) to show that Equation (\ref{closedBH}) can be
rewritten as Equation (\ref{closedBH1}).
\end{itemize}
\end{exercise}


\begin{exercise} {\bf Exercise 4.14:} The closed-form solution allows us to readily determine the
long-term behavior of populations predicted by the model.  Suppose
$x_0 > 0$.
\begin{itemize}
\item[(a)] Let $\alpha = 0.5$. Use Equation (\ref{closedBH1}) to determine
the behavior of the population over time predicted by the model.
\item[(b)] Repeat if $\alpha = 2$.
\item[(c)] In general, what can you say if $\alpha
>1$ or if $\alpha < 1$ about the population over long periods of
time?
\item[(d)] Will the Beverton-Holt model ever have unpredictable
results like the Ricker model?  Why or why not?
\end{itemize}
\end{exercise}

\newpage

\section{Selected Solutions}

\begin{exercise} {\bf Exercise 1.1:} If $x_0<-1$, then $|x_n|$ grows without bound with $x_n$
alternating between positive and negative values.
\end{exercise}



\begin{exercise} {\bf Exercise 1.2:} If $k>1$, then the birth rate is greater than the death rate. The
population would grow without bound.
\end{exercise}

\begin{exercise} {\bf Exercise 1.3:} If $k=1$ the birth rate equals the death rate.  The population
would remain at $x_0$ forever.  If $0<k<1$, then the population
would die out over time.
\end{exercise}

\begin{exercise} {\bf Exercise 2.1:} Substitute $y_n = L x_n$ into Equation (\ref{rL}) and
simplify.
\end{exercise}

\begin{exercise} {\bf Exercise 2.2:} The fixed points are $x=0$ and $x=\frac{r-1}{r}$.
\end{exercise}

\begin{exercise} {\bf Exercise 2.3:} Consider $x_{n+1}=f(x_n)$ where $f(x) = r x (1-x)$ and
$r>0$.
\begin{itemize}
\item[(a)] Note that $x_n>1$
if and only if $(1-x_n)<0$. Moreover, $(1-x_n)<0$ if and only if
$rx_n(1-x_n)<0$.
\item[(b)] The graph of $f$ is a parabola that opens
downward with vertex at $(\frac{1}{2}, \frac{r}{4})$.  Thus the
maximum value of $f$ is $\frac{r}{4}$ and it occurs at
$x=\frac{1}{2}$, half the carrying capacity.
\item[(c)] If $r>4$ then
we can show that there is some $x_0 \in (0,1)$ such that the
logistic model predicts a negative population.  If
$x_0=\frac{1}{2}$, then by part (b) $x_1=
f(\frac{1}{2})=\frac{r}{4}>1$. By part (a), $x_2 < 0$.
\item[(d)]  If $r<4$ then $0<f(x)<1$ for all $x \in (0,1)$.  (Note: if $r=4$ and
$x_0=\frac{1}{2}$, then $x_2=0$.)  The two conditions $0<x_0<L$ and
$r<4$ guarantee that $x_n>0$ for $n=1,2,3, \ldots$.
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 2.4:}
\begin{itemize}
\item[(a)] About 240,000 flies.
\item[(b)] $x_{n+1}=1.5 x_n(1-x_n) + \frac{0.5 \times 10^6}{0.72 \times
10^6}$
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 2.5:}
\begin{itemize}
\item[(a)] It approaches 60\% of the carrying capacity.
\item[(b)] In this case the population alternates between  62\% and 70\%
of the carrying capacity.
\item[(c)] Over time, the population
alternates between  38\%, 83\%, 50\%, and 87\% of the carrying
capacity.
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 2.6:} Logistic model
$x_{n+1}=2x_n(1-x_n)$.
\begin{itemize}
\item[(a)] The fixed points are $x=0$ and $x=0.5$.
\item[(b)] Iterations:
\begin{itemize}
\item[i.] For the initial value $x_0 = 0.1$, we have the following
population sequence: $x_1=0.18, x_2=0.2952, x_3 = 0.4161,
x_4=0.4859, x_5=0.4996$.  The population appears to approach 50\% of
the carrying capacity.
\item[ii.]  For the initial value $x_0 = 1.1$,
we have the following population sequence: $x_1=-0.22, x_2=-0.5368,
x_3 = -1.6499, x_4=-8.7442,x_5=-170.411$. The model is giving
negative values.
\item[iii.] For the initial value $x_0 = 0.6$, we
have the following population sequence: $x_1=0.48, x_2=0.4992, x_3 =
0.4999, x_4=0.5,x_5=0.5$.  As in part (i), the population approaches
50\% of the carrying capacity.
\item[iv.] For the initial value $x_0
= 0.5$, we have the following population sequence: $x_1=0.5,
x_2=0.5, x_3 = 0.5, x_4=0.5,x_5=0.5$. Again, the population is
approaching 50\% of the carrying capacity.
\end{itemize}
\item[(c)] It appears that $x=0$ is a repelling fixed point and $x=0.5$ is
an attracting fixed point.
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 2.7:} The fixed point $x=0$ is attracting when $0<k<1$ and repelling
when $k>1$.  If $k=1$, then every point is fixed, and $x=0$ is
neither attracting nor repelling.
\end{exercise}

\begin{exercise} {\bf Exercise 2.8:} Note that $f'(x)=r-2rx$.  So $|f'(0)|=|r|$.  Thus $x=0$ is
attracting when $0<r<1$.  For the nonzero fixed point we have
$|f'(\frac{r-1}{r})|=|2-r|$.  We set $|2-r|<1$ to find that
$x=\frac{r-1}{r}$ is attracting when $1<r<3$.
\end{exercise}

\begin{exercise} {\bf Exercise 2.9:} The fixed points are $x=0$ and $x=0.6$.  From cobweb diagrams,
$x=0$ is repelling and $x=0.6$ is attracting.
\end{exercise}

\begin{exercise} {\bf Exercise 2.10:} From cobweb diagrams we find
\begin{itemize}
\item[(a)] if $0 < r < 1$, the fixed point $x=0$ is attracting,

\item[(b)] if $1 < r < 3$, the second fixed point $x=\frac{r-1}{r}$ is
attracting,

\item[(c)] When $r=3.7$ we cannot determine the long-term behavior using
cobwebbing.
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 2.11:} We find that $f^2(x)=r^2 x (1-x)(1-rx(1-x))$.  The next step is to
factor $f^2(x)-x$.  Using the hint, we find that $f^2(x)-x=
-x(rx+1-r)(r^2x^2-r^2x-rx+r+1)$.  The period 2 points are found by
solving $r^2 x^2 -r^2 x -rx +r+1=0$.  Using the quadratic formula
and simplifying, we find the roots are  $\displaystyle x = \frac{ r
+ 1 \pm \sqrt{r^2-2r-3}}{2r}$.  Therefore, there are period two
points when $\sqrt{r^2-2r-3} > 0$, which is the case when $r>3$.
\end{exercise}

\begin{exercise} {\bf Exercise 2.12:} Let $p= \frac{ r + 1 + \sqrt{r^2-2r-3}}{2r}$  and $q = \frac{ r +
1 - \sqrt{r^2-2r-3}}{2r}$ be the period-2 points. By the Chain Rule,
$(f^2)'(p)=f'(f(p))f'(p) = f'(q)f'(p)$.  After some algebra we find
$f'(p)f'(q)= -[(r-1)^2 -5]$.  We then find $|f'(p)f'(q)|<1$ if $3 <
r < \sqrt{6}+1$.
\end{exercise}


\begin{exercise} {\bf Exercise 3.1:}
\begin{itemize}
\item[(a)] From ten iterations of the Ricker model with $\alpha=8.86$,
$\beta=0.148$ and an initial population of 20 beetles, the model
predicts that the beetle population will oscillate between
approximately 21.8 and 7.6 beetles.
\item[(b)]
\begin{itemize}
\item[i.] For $\alpha=0.5$, the model predicts the beetle would become
extinct.
\item[ii.] For $\alpha=1.5$ the model predicts the beetle
population would reach 2.7 at generation 9 and will remain there.
\item[iii.] For $\alpha = 10$, the model predicts the population will
oscillate between 24.8 and 6.3 beetles.
\end{itemize}
\item[(c)] One answer is $\alpha = 14$.
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 3.2:} After dividing both sides by $x_n$ and taking the natural log of
both sides we have $\displaystyle \ln
\left[\frac{x_{n+1}}{x_n}\right]=\ln(\alpha e^{-\beta x_n} x_n)$.
Simplifying the right hand side using properties of the natural log
gives the desired result.
\end{exercise}

\begin{exercise} {\bf Exercise 3.3:} It predicts the population will approach 3.5 per square
meter. \end{exercise}

\begin{exercise} {\bf Exercise 3.4:} Using the linear regression process, we find that $\alpha = 1.396$
and $\beta = 0.001$.  The first 8 iterations  using this model are

\medskip

\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline Year & 1992 & 1993 & 1994 &1995 & 1996 & 1997 & 1998 & 1999 &
2000\\
\hline population & 6.0 & 8.3 & 11.5 & 15.9 & 21.9 & 29.9 & 40.4 &
54.2 & 71.7
\\
\hline
\end{tabular}
\end{exercise}

\medskip

\begin{exercise} {\bf Exercise 3.5:} (a) Using linear regression we find $\alpha = 2.112$ and $\beta =
0.067$.  The model predicts $11.156$ nests per hectare in 2002.
\end{exercise}



\begin{exercise} {\bf Exercise 3.6:} \\
(a) We find that $\alpha=2.367$ and $\beta=0.00013$.  The model
predicts the population will be 6627 salmon in 2004.\\
(c) Let $H(x)=f(x)-x$.  Then $H(x)$ is maximum when $H'(x)=0$ or
when $f'(x)=1$.  If solve this equation with the given $\alpha$ and
$\beta$ (using a numerical solver) we find $x$ is approximately
15.63.

\end{exercise}

\begin{exercise} {\bf Exercise 3.7:} We set $\alpha e^{-\beta x} x = x$ and solve for $x$ to find the
fixed points are $x=0$ and $x = \frac{\ln \alpha}{\beta}$ provided
$\alpha>1$.
\end{exercise}

\begin{exercise} {\bf Exercise 3.8:}
\begin{itemize}
\item[(a)] $f'(x)=\alpha e^{-\beta x}(1-\beta x)$.  $f$ is increasing on
$(0, \frac{1}{\beta})$ and decreasing for $x>\frac{1}{\beta}$.
\item[(b)] Use L'Hopital's rule.
\item[(c)] The maximum value occurs at $\frac{1}{\beta}$.
\item[(d)] The maximum value is $\frac{\alpha}{e \beta}$.
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 3.9:} (d) The fixed point $x=0$ is attracting for $\alpha < 1$ and
repelling for $\alpha>1$.  The nonzero fixed point is attracting for
$1<\alpha < e^2$. This is confirmed using Theorem \ref{fixedpoint},
since $|f'(0)|=\alpha$ and $|f'(\ln \alpha)|=|1-\ln \alpha|$.
\end{exercise}

\begin{exercise} {\bf Exercise 3.10:} From cobwebbing, it appears there's a 2-cycle when $\alpha=11$ and
a 4-cycle when $\alpha=14$.
\end{exercise}

\begin{exercise} {\bf Exercise 4.1:} If $g$ is positive, then the model will never predict
negative populations.  As the population increases, the availability
of resources decreases and so does the per capita growth rate.
\end{exercise}

\begin{exercise} {\bf Exercise 4.2:} $x_0=500,000$, $x_1=23,376$, $x_2=23,114$, $x_3=23,111$,
$x_4=23,111$.
\end{exercise}

\begin{exercise} {\bf Exercise 4.3:} There are two fixed points: $x=0$ and $x = \displaystyle
\frac{\alpha-1}{\beta}$.  In order for the non-zero fixed point to
be positive we need $\alpha>1$.
\end{exercise}

\begin{exercise} {\bf Exercise 4.4:}
\begin{itemize}
\item[(a)] $f'(x)=\frac{\alpha}{(1+\beta x)^2}$.
\item[(b)] $f$ is always increasing.
\item[(c)] $\lim_{x \to \infty}f(x)=\frac{\alpha}{\beta}$.
\item[(d)] $f''(x)=\frac{-2 \alpha \beta}{(1+\beta x)^3}$.
\item[(e)] If $\beta >0$, then $f$ is concave down.  If $\beta<0$ then $f$
is concave up for $x<\frac{-1}{\beta}$ and concave down for
$x>\frac{-1}{\beta}$.
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 4.5:}
\begin{itemize}
\item[(a)] To find the fixed points, solve $\frac{2.5x}{1+1.5x}=x$.
\item[(b)] The iterations increase and approach 1.
\item[(c)] The iterations decrease and approach 1.
\item[(d)] $x=0$ is repelling, and $x=1$ is attracting.
\end{itemize}
\end{exercise}

\begin{exercise} {\bf Exercise 4.6:} Set $K$ equal to the nonzero fixed point $\frac{\alpha-1}{\beta}$
and solve for $\beta$. Then substitution gives the desired result.
\end{exercise}

\begin{exercise} {\bf Exercise 4.8:} We find $|f'(0)|=\alpha$ and $|f'(K)|=\frac{1}{\alpha}$. Thus $x=0$
is attracting if $\alpha<1$ and $K$ is attracting if $\alpha>1$.
\end{exercise}

\begin{exercise} {\bf Exercise 4.11:} We find $x_{n+1}=\frac{2.074 x_n}{1+0.002 x_n}$.  The model
predicts there will be 537 paramecium on day 50.
\end{exercise}

\begin{exercise} {\bf Exercise 4.12:} We find $x_{n+1}=\frac{8.899 x_n}{1+0.0013 x_n}$.  The model
predicts there will be 6076 salmon in 2004.
\end{exercise}

\begin{exercise} {\bf Exercise 4.14:}
\begin{itemize}
\item[(a)] Substituting $\alpha=0.5$ and taking the limit as $n \to
\infty$ we have that $x_{n} \to 0$.
\item[(b)] Substituting $\alpha=2$
and taking the limit as $n \to \infty$ we have that $x_{n} \to
\frac{1}{\beta}$.
\item[(c)]  If $\alpha<1$, then $x_n \to 0$.  If
$\alpha>1$, then $x_n \to \frac{\alpha-1}{\beta}$.
\end{itemize}
\end{exercise}

\newpage

\section{References}

\begin{enumerate}
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\item \label{mmpbe} F. Brauer and C. Castillo-Chavez, \textit{Mathematical
Models in Population Biology and Epidemiology}. Springer, 2001.

\item \label{BVHolt}R.J.H. Beverton and S.J. Holt,
On the dynamics of exploited fish populations, \textit{Fisheries
Investment Series 2}, {\bf 19}, U.K. Ministry of Agriculture and
Fisheries, London (1957).

\item \label{PineLooper}N. Broekhuizen, M.P. Hassell and H.F. Evans,
Common mechanisms underlying contrasting dynamics in two populations
of the pine looper moth, \textit{The Journal of Animal Ecology},
{\bf 63} No.2 (1994), 245-255.

\item \label{tumor} S. Cross and D. Cotton, Chaos and
Antichaos in Pathology, \textit{Human Pathology}, {\bf 25} (1994),
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\item \label{devaney} R. Devaney, \textit{A First Course in Chaotic
Dynamical Systems, Theory and Experiment}. Perseus Books, 1992.

\item G.F. Gause, \textit{The Struggle for Existence}. Williams and Wilkins,
Baltimore, 1934.

\item \label{May} R.M. May, Simple mathematical models with very complicated dynamics,
\textit{Nature}, {\bf 261} (1977), 459-467.


\item \label{bulbfly} W. Morris, Problems in detecting chaotic behavior in natural
populations by fitting simple discrete population models,
\textit{Ecology}, {\bf 71} No.5 (1990), 1849-1862.



\item \label{coho}L. Shaul, S. McPherson, E. Jones, and K Crabtree.
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Alaska, Alaska Department of Fish and Game, Special Publication No.
03-02, Anchorage.



\item \label{beetle} N. Underwood and M. Rausher, Plant
genotype and herbivore dynamics, \textit{Ecology}, {\bf 81} No. 6
(2000), 1565-1576.

\end{enumerate}

\end{document}
