\magnification=\magstep1
\centerline{MONT 100N -- Modeling the Environment}
\centerline{Solutions/Comments on Chapter 7 Project}
\centerline{November 21, 2017}
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\noindent
There was a bit of confusion about how to find and/or classify the 
equilibrium solutions of the difference equations representing the 
constant harvesting and the proportional harvesting strategies from 
parts B,C of the project.  Here's the way you can determine them 
algebraically.  Recall that an equilibrium solution should be a
constant value $E$ so that if $P(0) = E$, then $P(n) = E$ for
all times.  This means that if you compute 
$P(n+1) - P(n)$ at an equilibrium, then you should get $0$.
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\noindent
B.  For the constant harvesting equation:
$$P(n+1) = 1.71 P(n) - .00875 (P(n))^2 - h,$$
if you subtract $P(n)$ from both sides you get
$$P(n+1) - P(n) = 0.71 P(n) - .00875 (P(n))^2 - h.$$
Hence a population level $P = P(n)$ is an equilibrium if 
$$-0.00875 P^2 + 0.71 P - h = 0.$$
This quadratic equation has at most two roots, given by 
the quadratic formula:
$$P = {-0.71 \pm \sqrt{(0.71)^2 - 4(-0.00875)(-h)}\over 2(-0.00875)},$$
which simplify to approximately 
$$P = 40.57 \pm 0.57 \sqrt{5041 - 350h}.$$

For $h = 5$, for instance, this gives two positive real values, 
$$P \doteq 7.79, 73.35.$$
Now,
\bigskip
\item{$\bullet$} The $73.35$ is the one you ``see'' by plotting solutions of the 
difference equation in Excel -- if $P(0)$ is large enough, then
you get solutions tending to that equilibrium.  {\it That one is a stable
equilibrium, meaning that solutions that start close to that level
tend toward it.}
\item{$\bullet$} You don't see the $7.79$ so directly.  However, that is 
actually the ``cutoff'' value between the initial values  $P(0)$ that
produce solutions tending up to $73.35$, and the other solutions that
go to 0 and then become negative (the ones where the population ``crashes'').  
{\it That one is an unstable equilibrium, meaning that unless $P(0)$ equals
the actual root of the quadratic equation above, then the solution
will tend away from that equilbrium.}  

You should have noticed that the harvesting level $h = 20$ yields {\it 
only solutions that ``crash''} no matter what $P(0)$ is.  The reason for
this is that when $h = 20$, the roots of the quadratic equation above
are actually non-real because $5041 - 350 \cdot 20 < 0$ under the 
square root.  In fact, the largest value of $h$ for which there are 
real roots is the solution of 
$$5041 - 350 h = 0 \Rightarrow h \doteq 14.4$$
This is (about) the largest $h$ for which a stable equilibrium exists.  
If you harvest at this constant level, and $P(0)$ is large enough (say $> 40.57$), then
you can keep doing this forever, so the strategy is {\it sustainable} in that sense.
The equilibrium population level in this case is about $40.57$.  
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\noindent
C.  The proportional harvesting equation is
$$P(n+1) - P(n) = 0.71 P(n) - 0.00875 (P(n))^2 - p P(n),$$
or 
$$P(n+1) - P(n) = (0.71 - p)\cdot P(n) - 0.00875 (P(n))^2.$$
The equilibrium values are determined by setting the right side equal
to zero, just as in question B.  This gives equilibrium levels at
$$P = 0, P = {0.71 - p\over 0.00875}.$$
Note that if $p > .71$, then the second one becomes negative and hence
unrealistic as a population value.  

If $p = 0.3$, then we get a positive equilibrium at 
$P = {.71 - 0.3\over 0.00875} \doteq 46.86$.  This can be seen to be
stable by looking at the solutions in Excel.  If we harvest over
a long period then we will eventually tend to $(.3) \cdot 46.86 \doteq 14.1$
million kg of fish harvested per year.  

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\noindent
D.  To compare the three harvesting methods, we might first want to take into 
account whether the method is realistic.  That rules out A pretty quickly.
After all, who would want to eat halibut that has been sitting around
in a freezer for up to 11 years??  Between B and C, the choice could be
made on which one would produce the largest yearly take.  We saw above that
we cannot do better than $14.4$ million kg per year on a sustainable
basis using constant harvesting.  The constant level $h = 14.4$
is also problematic because any $P(0)$ less than $40.57$ leads to a crash, and
that level is sustainable only if $P(0) > 40.57$.  With proportional harvesting, though
if $p = .355$ we can take the same amount of fish per year ($14.4 \times 10^6$ kg)
but the equilbrium level population level ${.71 - .355\over 0.00875} = 40.57$ 
is still a stable equilibrium.  In other words, ``proportional harvesting is the 
way to go.''
\bye  

