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\begin{document}

\newpage
\title{\ifnum\TeacherEdition=1 TEACHER VERSION \else STUDENT VERSION \fi\\Fish Mixing}


\markboth{Fish Mixing}{Fish Mixing}






\author{
    Eric Sullivan \& Elizabeth Carlson\\
Carroll College\\
Helena  MT 59625 USA\\
\teacher{esullivan@carroll.edu \, \& \, ecarlson10@uvic.ca}
}

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\begin{abstract}
    This activity gives students a chance to build the underlying differential equation and/or difference equation
    for a mixing problem using tangible objects (fish) and a student-designed restocking and
    fishing plan in a lake.  The mixture is of two species of fish, one being the current
    sole species in the lake and the other being introduced via restocking. The students
    generate data for the mixing problem via a brief game and then conjecture forms of the
    differential or difference equation model.  
\end{abstract}



\section*{SCENARIO DESCRIPTION}
A lake in northern Montana is dominated by Arctic Grayling (henceforth called ``species
A'') but the Department of Fish, Wildlife, and Parks is planning to slowly introduce Bull
Trout (``species B'').  The lake is popular with sport fishermen who remove both species of
fish from the lake regularly.

The Department of Fish, Wildlife, and Parks has carefully estimated the number of fish
taken by sport fishing each week, and they have decided to keep the fish population as
constant as possible, by replacing the fish lost by an equal number of Arctic Grayling and
Bull Trout.  For example, if there are $N=50$ fish in the lake at the beginning of the
week and fishermen remove $M=10$ fish during that week, then the fish and wildlife people
will restock the lake with $5$ Arctic Grayling and $5$ Bull Trout. Hence the population of
the lake will remain $N=50$ fish at the end of each week, assuming no new fish are born.  Both fish species swim freely
throughout the lake and both are targeted by similar bait used by sport fisherman.

\vspace*{.4 true cm}
In your lake you will use $N = \underline{\hspace{1in}}$ and $M =
\underline{\hspace{1in}}$. 

\vspace{0.1in}

\noindent In summary:\vspace{-0.1in}
\begin{itemize}
    \item The week starts with $N=\underline{\hspace{1in}}$ fish.
    \item The fish swim freely around the lake.
    \item $M=\underline{\hspace{1in}}$ fish are removed from the lake at random during the week.
    \item $M=\underline{\hspace{1in}}$ fish are restocked at the end of the week.  $M/2=\underline{\hspace{1in}}$ of those fish are Arctic
        Grayling and $M/2=\underline{\hspace{1in}}$ of those fish are Bull Trout.
\end{itemize}


\begin{enumerate}
    \item {\bf Conjecture:}
        \begin{enumerate}
            \item What do you think will happen to the populations of species A and B over
                a long period of time?
            \item Is it possible that species A will be eliminated from the lake with the
                restocking plan? Explain.
        \end{enumerate}

    \item {\bf Simulate:}
        \begin{enumerate}
            \item Use pennies to represent your $N$ fish and decide with your partner(s) which
                coin face represents which species (e.g., heads up is species A). Start your lake with 100\% species A.
            \item Decide with your partner(s) how to simulate the swimming of fish, the fishermen,
                and the Department of Fish, Wildlife, and Parks' restocking plan.  Simulate
                roughly 15 weeks of the fish population representing species A and B with coins.
                Be sure to let the fish swim thoroughly around the lake and keep track of the
                proportions of species A and B.
\vspace*{.2 true cm}
                \begin{center}
                    \begin{tabular}{|c||c|c||c|c|}
                        \hline
                        Week \# & \multicolumn{2}{|c||}{{\bf Number in population}} & \multicolumn{2}{|c|}{{\bf Proportion of population}}\\
                        & species A & species B & species A & species B \\
                        \hline
                        0 &  &  & 1 & 0 \\ \hline
                        1 & & & & \\ \hline
                        2 & & & & \\ \hline
                        3 & & & & \\ \hline
                        4 & & & & \\ \hline
                        \vdots & \vdots & \vdots & \vdots & \vdots \\ \hline
                    \end{tabular}
                \end{center}
        \end{enumerate}

    \item {\bf Model:} 
        \begin{enumerate}
            \item Propose a verbal model for the rate of change of species B in the lake.
                \[ \text{rate at which species B changes} = \underline{\hspace{1.5in}} +
                \underline{\hspace{1.5in}} \]
        % B' = -\alpha B + beta M/N

            \item Explicitly state any assumptions that you are using in your verbal model.

            \item Introduce mathematical notation for your proposed model and write your verbal model mathematically.  Be sure to include any necessary condition(s).  
                \begin{flalign*}
                    &\text{model: } \underline{\hspace{3in}} \\
                    &\text{condition(s): } \underline{\hspace{3in}} 
                \end{flalign*}
        \end{enumerate}
    \item {\bf Analyze:} 
        \begin{enumerate}
            \item According to your model, what is the long term effect on the fish population in the lake?  Use your model to justify your answer algebraically and graphically.
            \item Solve your mathematical model (either numerically or analytically) and compare with your data.
            \item (Extension) Suppose now that the Department of Fish, Wildlife, and Parks does not attempt to keep the population in the lake constant.  That is, suppose that fishing reduces the population by $M_1$ fish each week and the Department of Fish, Wildlife, and Parks restocks $M_2$ fish each week.  Fully explore this scenario.
        \end{enumerate}
\end{enumerate}


\teacher{
\section*{NOTES FOR TEACHERS}
This scenario is meant to be a lead in to classic mixing problems via a biological
situation, but the teacher should keep in mind that this is not the only model that could
arise from this scenario.  The teacher comments for this modeling scenario are organized
into the following categories:
\begin{itemize}
    \item an analytic solution to the typical first-order linear differential equation that
        arises from this scenario;
    \item an analytic solution to the typical first-order linear difference equation that
        arises from this scenario;
    \item practical considerations for running the simulation in the classroom;
    \item suggestions for guiding students through the modeling process; and
    \item a presentation of three sample student data sets.
\end{itemize}


\subsection*{An Analytic Solution: First-Order Linear Differential Equation Model}
If $b(t)$ is the proportion of species B in the lake's fish population at time $t$  then the intended
(and likely the simplest) mixing model could be
\begin{flalign}
    b'(t) = -\alpha b(t) + \frac{1}{2} \cdot \frac{M}{N} \quad \text{with} \quad b(0) = 0,
    \label{eqn:ode1}
\end{flalign}
where $\alpha$ is a rate parameter ($[\alpha]=$1/time), $[M]=$fish/week, $[N]=$fish, and
the factor of $1/2$ represents the fraction of the restock that is species B.  Equation
\eqref{eqn:ode1} is a linear first-order differential equation that lends itself to
separation of variables, integrating factors, or the method of undetermined coefficients.
Using any of these methods the solution to 
\eqref{eqn:ode1} is
\begin{flalign}
    b(t) &= \frac{M}{2 \alpha N} \left( 1- e^{-\alpha t} \right).
    \label{eqn:ode1_soln}
\end{flalign}
Note that $a(t)$ (the proportion of species A) can be modeled as $a(t) = 1-b(t)$ since
these are the only two species under consideration and the population remains constant.
The value of $\alpha$ is the percent of the population replaced at any given time.  


To provide a concrete example consider the case where $N=50$ and $M=10$.  In this case we
have $\alpha = 10/50 = 0.20$ and the proportion of species B populating the lake is given
(after some simplification) by 
\begin{flalign}
    b(t) &= \frac{1}{2} \left( 1- e^{-0.2 t} \right).
    \label{eqn:ode_sample_soln}
\end{flalign}

The equilibrium solution for the differential equation \eqref{eqn:ode1} occurs when
$b'(t)=0$.  Hence, the equilibrium is $b(t) = 1/2$.  This can also be seen in equation
\eqref{eqn:ode_sample_soln} by considering that $\lim_{t\to\infty} b(t) = 1/2$. 
Furthermore, we can see the equilibrium of $b=0.5$ graphically by using a slope
field (see Figure \ref{fig:1-34-SlopeFieldImage}).


\begin{figure}[ht!]
    \begin{center}
        \includegraphics[width=0.5\columnwidth]{1-34-SlopeFieldImage.eps}
    \end{center}
\vspace*{-.2 true cm}
    \caption{Slope field for $b' = -\alpha b + \frac{M}{2N}$ with several solutions shown.}
    \label{fig:1-34-SlopeFieldImage}
\vspace*{-.2 true cm}
\end{figure}


\subsection*{An Analytic Solution: First-Order Linear Difference Equation Model}
If students choose to model this scenario with a difference equation, the analogous
equation to \eqref{eqn:ode1} is
\begin{flalign}
    b_{n+1} = b_n - \alpha b_n + \frac{1}{2} \cdot \frac{M}{N} \quad \text{with} \quad b_0
    = 0 \label{eqn:dds1}
\end{flalign}
where $b_n$ is the proportion of the lake populated by species B at the end of week $n$.
Using the method of undetermined coefficients and conjecturing that the analytic solution takes the
form $b_n = C_1 (1-\alpha)^n + C_2$, we arrive at the solution 
\begin{flalign}
    b_n = \frac{M}{2\alpha N} \left( 1- (1-\alpha)^n \right).
    \label{eqn:dds1_soln}
\end{flalign}
If $N=50$ and $M=10$ then $\alpha = 10/50=0.20$ and the solution to \eqref{eqn:dds1} is
\begin{flalign}
    b_n = \frac{1}{2} \left( 1 - 0.8^n \right).
    \label{eqn:dds_sample_soln}
\end{flalign}
Similar to the differential equation model, the difference equation model predicts an
equilibrium of $b_{eq} = 1/2$.

Solutions \eqref{eqn:ode_sample_soln} and \eqref{eqn:dds_sample_soln} are given in Figure~\ref{fig:1-34-SolnPlotN50M10} to show that the continuous differential equation and the discrete difference equation give qualitatively similar predictions. Notice, though, that the difference equation predicts a larger proportion of population for species B when the curvature of the solution is greatest.
\begin{figure}[ht!]
    \begin{center}
        \includegraphics[width=0.5\columnwidth]{1-34-SolnPlotN50M10.eps}
   
    \parbox{9 true cm}{\caption{\label{fig:1-34-SolnPlotN50M10} Solutions to equations \eqref{eqn:ode_sample_soln} and
    \eqref{eqn:dds_sample_soln} with $N=50$ and $M=10$ ($\alpha = 0.20$). Note the slight differences in curvature between the discrete and continuous models.}}
 \end{center}
 
\end{figure}

\subsection*{Practical Considerations for the Simulation}
We use coins to model the fish, but this could be done equally well with M\&M's, two-sided
poker chips, or any two easily identifiable objects of the same size. When mixing the {\it
fish} some students find that it is possible to inadvertently flip one of the coins.
Calling the students' attention to this fact at least encourages them to beware of this
possible issue. 

Students will come up with many creative ways to remove fish from the lake. The methods of
fishing, restocking, and swimming are purposefully left vague in the statement of the
problem to give the students some room for creativity.   A nice discussion comes from
asking students why it is important to let the fish {\it swim} and how this affects the
choice of their fishing technique.  Some groups will inevitably realize that if the mixing is
thorough enough then the fishing can take place from anywhere in the lake without
additional bias.  Allowing the students to discuss this fact leans on their potential
knowledge of randomization and random sampling from statistics.  Some examples of student
techniques are:
\begin{itemize}
    \item Place several different coins in the lake that act as boats. Assume that the
        boats all draw only exactly their alloted number of fish.
    \item Thoroughly mix the fish and then have the fishermen draw the $M$ closest fish from one {\it shore}.
        Given thorough mixing this method does not introduce any bias.
    \item Drop an object (such as a pen or a needle) into the lake and draw the fish that
        are touching the point of the object.
    \item Use coins or chips with different colors to represent the different species,
        place them in a jar, mix them thoroughly, and {\it fish} blindly from the top of
        the mixture.
\end{itemize}

Finally, the choices of $M$ and $N$ determine the length of time that the simulation will
take during class.  If $N=50$ then it may be best to use $M \approx 10$.  This choice
should show approximate convergence to an equilibrium within 15 iterations (see Figure
\eqref{fig:1-34-SolnPlotN50M10}). We have run the experiment with $M =4$ and with $M=6$
but the convergence is not obvious in a timely manner and leads to some confusion with the
students.  


\subsection*{Suggestions for Guidance with the Modeling Process}
Since this modeling scenario can lend itself to a first-order linear differential equation we expect that it will likely be done with students early in a semester.  In our experience, students are more likely to begin modeling the scenario as a difference equation earlier in the course since modeling discrete change is often more comfortable.

If students are having trouble with the verbal and mathematical model then we find that directing them to a partial model such as
\[ B_{new} - B_{old} = \underline{\hspace{1in}} + \underline{\hspace{1in}} \]
gives the students a more firm starting point, where $B$ is the total number of species B fish in the lake.  We intend for students to give a model such as
\[ B_{new} - B_{old} = \left( \text{B removed via fishing} \right) + \left( \text{B added via restocking} \right). \]
Given that this is truly a discrete scenario (where time is measured weekly) it may be
sufficient to allow the students to stick with the discrete difference equation, but this
can also be used to build the differential equation by taking $dB/dt \approx
B_{new}-B_{old}$.  The extent of the emphasis on the discrete model versus the continuous
model is at the discretion of the instructor.

When the students begin building their verbal and mathematical models they often don't
have a firm grasp that the growth rate might be proportional to the amount in the current
population. After giving the students several minutes to struggle with the model, we allow
them to write their ideas and partial models on the board for everyone to see and
critique.  Three sample student models are as follows:
\begin{subequations}
    \begin{flalign}
        B_{new} - B_{old} &= \left( \text{some function of B} \right) + M/2,
        \label{eqn:student1} \\
        \Delta B &= \text{some linear function of B}\label{eqn:student2} \\
        \frac{dB}{dt} &= \text{inversely proportional to time} \label{eqn:student4}.
    \end{flalign}
\end{subequations}
After further exploration, the students that generated equation \eqref{eqn:student1}
explained that they expected the data to be approximated by the model $B_{new} - B_{old} =
-\alpha(B) \cdot B + M/2$; explicitly stating that the constant of proportionality depends
on the concentration of species B. This lends itself to several families of possible
models. In particular, one group decided to try and fit the data to the differential
equation model
\[ \frac{dB}{dt} = -\alpha B^r + C \]
where $\alpha$, $r$, and $C$ were unknown parameters to be fit with software (e.g., Excel's Solver). Unfortunately, this type of model can potentially lead to an equilibrium value that contradicts the students' conjectures about the equilibrium. 

The students who generated equation \eqref{eqn:student2} observed that when plotting $\Delta B$ vs. $B$ the result was approximately linear for their data. This suggested to them that the scenario could be modeled by the difference equation  
\[ \Delta B = -\alpha B + C. \]  
Upon further investigation, they realized that a linear regression on $\Delta B$
vs $B$ could be used to estimate $\alpha$ and $C$ without regard to the context of the
problem.  The danger of this approach is that the inherently random nature of the
simulation will not yield data that allows an easy tie from the estimated parameters to
the context of the problem.  In other words, if $\alpha$ is estimated as $\alpha \approx
0.17$ instead of $\alpha = 0.20$ in Figure \ref{fig:1-34-SolnPlotN50M10} then students may
fail to make the connection between the value of $\alpha$ and the ratio $M/N$. A second
danger to this approach is that the small data sets may not lend themselves to observing a
linear relationship in the $\Delta B$ vs $B$ data.

The students who generated equation \eqref{eqn:student4} indicated that their data and
intuition suggested the change was inversely proportional to time.  Unfortunately this
leads to a logarithmic model for $B(t)$ with no steady-state solution.  

Students initially seem to prefer working with the total population instead of the
proportion of the population.  As the problem evolves it may be beneficial to discuss the
advantages to using total population vs. population proportions.  When working with proportions, the value of the initial population isn’t relevant since it cancels out of the analytic solution to both the difference and differential equation.  Furthermore, the
parameter $\alpha$ should approximate the growth rate of the species B population, but if
we are dealing with total populations the value of $\alpha$ will be divided by the
size of the population; making the physical meaning of the constant more difficult to see
and understand.  Even with these advantages, the tangible nature of working with
total populations is potentially far more comfortable for students and should be
encouraged until the model is nearly complete.


\subsection*{Sample Student Data}
Three sample student data sets are given in Table \ref{tab:sample_data_sets} with
corresponding plots in Figures~\ref{fig:1-34-DataPlot1}, \ref{fig:1-34-DataPlot2}, and \ref{fig:1-34-DataPlot3}. These
data sets were built using $N=50$ total fish and $M=6$ fish replaced at each iteration.
Each figure shows solutions to both the differential equation \eqref{eqn:ode1} and
difference equation \eqref{eqn:dds1} with $\alpha = 0.12$.  The value of $\alpha$ is given
by the physical problem, but this is not the value of the parameter that fits in the least
squares sense.  A value for $\alpha$ that leads to a least squares fit between the data
and the model can be found using technology (e.g., Excel's Solver), but the reader should be
cautioned that if the students don't use a physically meaningful parameter then the
equilibrium point can be affected and the ultimate meaning of the model may lose its
impact.  Lastly, we suggest using $M>6$ in future implementations since Figures
\ref{fig:1-34-DataPlot1} and \ref{fig:1-34-DataPlot2} do  not show a clear convergence to an equilibrium point
in a short amount of time.

\begin{table}[th]
    \centering
    \begin{tabular}{|c||c|c||c|c||c|c|}
        \hline
        &\multicolumn{2}{|c||}{Set \#1} & \multicolumn{2}{|c||}{Set \#2} &
        \multicolumn{2}{|c|}{Set
        \#3} \\ 
        Week ($n$) & $a_n$ & $b_n$ & $a_n$ & $b_n$ & $a_n$ & $b_n$ \\ \hline
        \hline
        %
0 & 1     & 0    &  1    &  0    &  1    &  0 \\
1 & 0.94  & 0.06 &  0.96 &  0.04 &  0.94 &  0.06 \\
2 & 0.88  & 0.12 &  0.92 &  0.08 &  0.88 &  0.12 \\
3 & 0.82  & 0.18 &  0.88 &  0.12 &  0.84 &  0.16 \\
4 & 0.8   & 0.2  &  0.84 &  0.16 &  0.82 &  0.18 \\
5 & 0.78  & 0.22 &  0.8  &  0.2  &  0.78 &  0.22 \\
6 & 0.74  & 0.26 &  0.76 &  0.24 &  0.74 &  0.26 \\
7 & 0.72  & 0.28 &  0.74 &  0.26 &  0.68 &  0.32 \\
8 & 0.74  & 0.26 &  0.72 &  0.28 &  0.68 &  0.32 \\
9 & 0.72  & 0.28 &  0.7  &  0.3  &  0.64 &  0.36 \\
10& 0.7   & 0.3  &  0.68 &  0.32 &  0.66 &  0.34 \\
11&      &      & 0.68 &  0.32 &  0.64 &  0.36\\
12&      &      & 0.68 &  0.32 &  0.64 &  0.36\\
13&      &      & 0.64 &  0.36 &  0.64 &  0.36\\
14&      &      & 0.62 &  0.38 &  0.6  &  0.4\\
15&      &      & 0.6  &  0.4  &  0.6  &  0.4\\
16&      &      & 0.62 &  0.38 & &  \\
17&      &      & 0.62 &  0.38 & &  \\      
18&      &      & 0.6  &  0.4  &  &  \\ 
19&      &      & 0.58 &  0.42 &       & \\ 
20&      &      & 0.56 &  0.44 &       & \\ 
21&      &      & 0.58 &  0.42 &       & \\\hline 
    \end{tabular}
    \caption{Three sample student data sets with $N=50$ total fish and $M=6$ fish replaced
    at each iteration.  Each data set shows the proportion of the population inhabited by
the given species.}
    \label{tab:sample_data_sets}
\end{table}


\begin{figure}[ht!]
    \begin{center}
        \includegraphics[width=0.73\columnwidth]{1-34-DataPlot1.eps}
   
       
    \parbox{11.5 true cm}{\caption{\label{fig:1-34-DataPlot1}Data Set \#1 with $N=50$ and $M=6$ along with the analytic
        solutions to the differential and difference equations (\eqref{eqn:ode1} and
        \eqref{eqn:dds1}) with $\alpha = 0.12$.}}
 \end{center}
    
\vspace*{-.6 true cm}
\end{figure}


\begin{figure}[ht!]
    \begin{center}
        \includegraphics[width=0.73\columnwidth]{1-34-DataPlot2.eps}
    
    \parbox{11.5 true cm}{\caption{ \label{fig:1-34-DataPlot2} Data Set \#2 with $N=50$ and $M=6$ along with the analytic
        solutions to the differential and difference equations (\eqref{eqn:ode1} and
        \eqref{eqn:dds1}) with $\alpha = 0.12$.}}
\end{center}
   

\vspace*{-.6 true cm}
\end{figure}

\begin{figure}[ht!]
    \begin{center}
        \includegraphics[width=0.73\columnwidth]{1-34-DataPlot3.eps}
   \parbox{11.5 true cm}{\caption{ \label{fig:1-34-DataPlot3}Data Set \#3 with $N=50$ and $M=6$ along with the analytic
        solutions to the differential and difference equations (\eqref{eqn:ode1} and
        \eqref{eqn:dds1}) with $\alpha = 0.12$.}}
 \end{center}
     
   
\vspace*{-.6 true cm}
\end{figure}

\newpage
\subsection*{Final Note}
This modeling scenario was created during the July 2015 SIMIODE Developer's Conference at
Carroll College in Helena, MT. Eric Sullivan is an Assistant Professor of mathematics at
Carroll and Elizabeth Carlson was a senior mathematics major during the 2015-2016 school year.


}



\end{document}
