# Math Insight

### Harvest of natural populations

#### Video introduction

Harvest of natural populations, part 1.

Harvest of natural populations, part 2.

#### Overview of harvesting

If a population is undergoing exponential growth, the population will grow without bound. In the problem of controlling a rabbit population, we explored how different harvesting strategies could be used to keep the population in check.

Given that an environment cannot sustain an infinite population of a species, population growth will slow down as the population size increases. We modeled the effect of these environmental factors with a carrying capacity that limited growth, leading to a logistic equation for population growth of the form \begin{align*} P_{t+1}-P_{t} = r P_t\left(1-\frac{P_t}{M}\right), \end{align*} where $r$ was the low density growth rate and $M$ was the carrying capacity. If the evolution of a population is following such logistic growth and $r$ isn't too large, then the population size will evolve to a steady state of $M$ even without harvesting. As is true in general for natural populations, a population following logistic growth doesn't require harvesting to maintain a steady population.

In most cases, the goal of harvesting a natural population is not population control, but simply to yield of a substantial harvest from the population. In such cases, one may ask what harvest strategy will provide the maximum long term yield, what is the maximum allowable harvest that will still retain the population, and what are the stable equilibrium sizes of the population under harvesting? In this page, we investigate these questions using the logistic model where the harvest is proportional to the population size, i.e., with the model \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} where $h_t$ is the fraction of the population harvested at time step $t$.

#### Example

We'll illustrate the effects of harvesting with an example. We begin with the equation \begin{align*} P_0 & = 1000 \\ P_{t+1} - P_t & = 0.2 P_t \left(1 - \frac{P_t}{1000} \right) - h P_t \end{align*} of a population with an initial population of 1000 individuals, low density growth rate of 0.2 per time interval, carrying capacity 1000 individuals. We let the harvesting rate $h$ be constant for all time. Our goal is to determine what fraction $h$ of individuals present may be harvested while still retaining the population.

To begin, we'll determine the equilibrium population. We'll denote the equilibrium population size as $P_e$. Setting both $P_{t+1}=P_e$ and $P_t=P_e$, we solve for $P_e$: \begin{align*} P_{e} - P_e &= 0.2 P_e \left(1 - \frac{P_e}{1000} \right) - h P_e\\ 0 & = 0.2 P_e \left(1 - \frac{P_e}{1000} \right) - h P_e \\ 0 & = P_e \left( 0.2 \left(1 - \frac{P_e}{1000} \right) - h \right) \end{align*} The product is zero if either \begin{gather*} P_e = 0 \qquad\text{or}\qquad 0.2 \left(1 - \frac{P_e}{1000} \right) - h = 0. \end{gather*} We can solve the second option for $P_e$: \begin{gather*} 1 - \frac{P_e}{1000} = h/0.2 = 5h\\ P_e = 1000 (1 - 5h). \end{gather*} Therefore, the two possible equilibrium values are \begin{gather} P_e = 0 \qquad\text{and}\qquad P_e = 1000 (1 - 5h). \label{equilibria} \end{gather}

Since $P_e$ is a population size, the equilibria make physical sense only if they are positive. If $1- 5h$ is negative, then the second equilibrium is negative and doesn't make sense; in those cases, the only realistic equilibrium is $P_e=0$. If we harvest more than 20% of the population present at each time, then $h>0.2$ and $1-5h$ is negative. Such a large harvest rate will cause the population to die out. This conclusion makes sense. The low density growth rate (births minus natural deaths) is 20%, and the low density growth rate is the maximum growth rate (achieved only when the population is small and is far from the carrying capacity). If the harvest rate exceeds the low density growth rate, the population will dwindle away and disappear. In fact, even if the harvest rate is exactly 20%, then the second equilibrium is $P_e = 1000 (1 - 5(0.2)) = 0$, and the population will still die out.

To simplify the further analysis, let's normalize the equation by dividing equation \eqref{harvest_fraction} by the carrying capacity 1000.

We next normalize the initial equation by dividing by the carrying capacity: $\frac{P_{t}}{1000} - \frac{P_t}{1000} = 0.2 \frac{P_t}{1000} \left(1 - \frac{P_t}{1000} \right) - h \frac{P_t}{1000}.$ Then, letting $p_t=P_t/1000$, we can write the dynamical system for $p_t$ as \begin{align} p_0 & = 1 \notag\\ p_{t+1} - p_t & = 0.2 p_t \left(1 - p_t \right) - h p_t \label{harvest_fraction_normalized} \end{align}

This normalization just replaces the carrying capacity by the number 1. We perform this normalization to show that the value of the carrying capacity doesn't play an important role in determining the dynamics. If the carrying capacity had been a number other than 1000, we could have divided by that number, and the end result would be the same equation.

The equilibria for $p_t$ are the same as the equilibria of equation \eqref{equilibria} for $P_t$, with just the carry capacity 1000 changed to 1. If we denote the equilibria for $p_t$ as $p_e$, the equilibria are \begin{gather} p_e = 0 \qquad\text{and}\qquad p_e = 1 - 5h. \label{equilibria_normalized} \end{gather}

As long as the fractional harvest $h$ is set at a number less than 0.2, both equilibria $0$ and $1-5h$ are valid. You can show in the exercises that the equilibrium $p_e=0$ is unstable and the equilibrium $p_e=1-5h$ is stable. For now, just look at the following applet showing the graph of $p_{t+1}$ versus $p_t$ for $h=0.05$. From looking at the slopes of the graph at the equilibria, it is apparent that $p_e=0$ is an unstable equilibrium and $p_e=1-5h=0.75$ is stable. (You can click the equilibria to see the tangent lines.) You can can see that the stability doesn't change as the change $h$ as long as you keep it less than 0.2. Once you show in the exercises that this remains true for all $0 \le h < 0.2$, then you can be confident that as long as the harvest rate is less than the low density growth rate, the model predicts that the population won't die out.

Equilibria when harvesting natural populations. A plot showing the equilibria for logistic growth with harvesting. The updating function $f(x)=x+r(1-x)-hx$ is shown by the red curve along with the diagonal line $x=y$ in blue. The two equilibria, which are the intersections between the curves, are shown by the blue and green points. The green point corresponds to the equilibrium $E_1=0$. As you can observe by the slope of $f$ at the equilibria, as long as the harvesting rate $h$ is less than the low density growth rate $r$, the equilibrium $E_2$ corresponding to the blue point is positive and stable, and the equilibrium $E_1=0$ is unstable. If the harvest rate is larger than the low density growth rate, then the second equilibrium $E_2$ becomes negative an unphysical. The only physically relevant equilibrium is $E_1=0$, which is stable. You can change $r$ and $h$ by typing values in their boxes. To display the sustainable harvest, which is the stable equilibrium multiplied by the harvest rate $h$,check the “show harvest” box.

If our objective is to maximize the harvest, we don't want to harvest so much that the population is tiny. (Of course, this wouldn't be good from an ecological point of view, either.) The amount of the harvest will be $h$ times the population size $p_t$. To maximize the harvest we want both the population size $p_t$ to be large and the harvest rate $h$ to be large. Since the increasing the harvest rate will decrease the population size, it's not so simple to see what the optimal harvest rate should. Clearly $h=0.2$ is too large since the population will die out and the harvest will be zero. But, if $h=0$, the harvest will also be zero. Is there some intermediate value of the fractional harvest $h$ that will maximize the total harvest?

To optimize the harvest rate $h$, we will assume that we have harvested at the rate $h$ for a long time so that the population has reached equilibrium. It'd be too complicated to try to optimize over the full dynamical system of equation \eqref{harvest_fraction_normalized} when the population size $p_t$ is changing at every time step. As long as $h<0.2$, the population size will settle down to the stable equilibrium $p_e=1-5h$. With the population size is fixed at $p_e$, then the total harvest amount is $$h \times p_e = h(1-5h).$$ The maximum harvest at equilibrium will be obtained at a harvest level that maximizes $h\times(1-5h)$. Thus, we are left with the question: For what value of $h$ is $h(1-5h)$ the largest?

Let $G$ be the total harvest as a function of harveset rate $h$: $$G(h) = h(1-5h).$$ Our goal is to maximum $G$. We calculate the critical points of $G$. Since $G'(h) = 1-10h$, $G'(h)=0$ when $h=0.1$. Moreover, $G''(h)=-10<0$, so this critical point is a local maximum. In fact, $h=0.1$ yields the maximum value of $h (1 - 5h)$ for any value of the harvesting rate $h$.

In this example, we had set $r=0.2$. The optimal 10% harvest strategy is exactly one-half the low density growth rate. Futhermore, the equilibrium when $h=0.1$ is $p_e = 1- 5 h = 0.5$, which is exactly half the maximum supportable population (recall we normalized so that the carrying capacity was equal to 1).

We calculated this result for the special case when $r=0.2$, but it turns out that this is a general property of logistic models, as you can explore in the exercises.