Harvest of natural populations exercises

In the page on harvesting natural populations, we introduced the logstic equation with a harvesting rate of the form

$\begin{align} P_{t+1}-P_{t} = r P_t\left(1-\frac{P_t}{M}\right) -h_t P_t, \label{harvest_fraction} \end{align}$ and analyzed a specific example with

$r=0.2$ . In these exercises, you will analyze the general from with constant

$h_t=h$ and

$0 < r < 2$ .

Divide each term of equation $\eqref{harvest_fraction}$ by $M$ to obtain

$\begin{align} p_{t+1} - p_t = r p_t \left(1 - p_t \right) - h p_t \label{harvest_fraction_normalized} \end{align}$ where

$p_t = P_t/M$ .

Show that the equilibria of equation $\eqref{harvest_fraction_normalized}$ are $p_{e} = 0$ and $p_{e} = 1 - h/r$ .
Show that in order for there to be a positive equilibrium, the fractional harvest rate, $h$ , must be less than the low density growth rate, $r$ .

Convert equation $\eqref{harvest_fraction_normalized}$ (which is in difference form) to function iteration form $p_{t+1}=F(P_t)$ . Show that $\begin{align} p_{t+1} = F(p_t) = p_t + r p_t (1 - p_t) - h p_t. \label{harvest_fraction_iteration} \end{align}$
Show that the equilibria of iteration $\eqref{harvest_fraction_iteration}$ are $p_{e} = 0$ and $p_{e} = 1 - h/r$ .
Assume that $h < r < 2$ . Compute $F'(p)$ and evaluate $F'(0)$ and $F'(1-h/r)$ . Conclude that $0$ is an unstable equilibrium and $1-h/r$ is a stable equilibrium.

Assume that $h < r < 2$ so that $1-h/r$ is a positive, stable equilibrium of the iteration of equation $\eqref{harvest_fraction_iteration}$ .

The harvest at that equilibrium will be

$h (1 - h/r)$ .

Find the value of $h$ for which $h (1-h/r)$ is the largest. Conclude that in order to maximize harvest, the harvest fraction $h$ should be set at one-half the low density growth rate, $r$ .
Find the positive equilibrium $p_e$ for this value of $h$ .
The variable $p_t$ was the actual population size $P_t$ normalized by the carrying capacity $p_t=P_t/M$ . What is the equilibrium value of the population size $P_t$ when the harvest rate is set at the value that maximizes harvest? Your answer will be in terms of $M$ .

Once you've worked out some of these exercises, you can check your work with the answers to selected problems.

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