8.6 Logistic Growth

In our basic exponential growth scenario, we had a recursive equation of the form

$P_{n} = P_{n - 1} + r P_{n - 1}$

In a confined environment, however, the growth rate may not remain constant. In a lake, for example, there is some maximum sustainable population of fish, also called a carrying capacity.

For our fish, the carrying capacity is the largest population that the resources in the lake can sustain. If the population in the lake is far below the carrying capacity, then we would expect the population to grow essentially exponentially. However, as the population approaches the carrying capacity, there will be a scarcity of food and space available, and the growth rate will decrease. If the population exceeds the carrying capacity, there won’t be enough resources to sustain all the fish and there will be a negative growth rate, causing the population to decrease back to the carrying capacity.

Line graph with the horizontal axis Population marked at 0, 5000 and 10000 and the vertical axis Growth Rate marked at minus 0.1, 0 and 0.1. One straight line falls steadily from a growth rate of 0.1 when the population is 0, crosses zero exactly at the carrying capacity of 5000, and reaches minus 0.1 when the population is 10000. If the carrying capacity was 5000, the growth rate might vary something like that in the graph shown. Note that this is a linear equation with intercept at $0.1$ and slope $- \frac{0.1}{5000}$ , so we could write an equation for this adjusted growth rate as:

$r_{adjusted} = 0.1 - \frac{0.1}{5000} P = 0.1 (1 - \frac{P}{5000})$

Substituting this in to our original exponential growth model for $r$ gives

$P_{n} = P_{n - 1} + 0.1 (1 - \frac{P_{n - 1}}{5000}) P_{n - 1}$

Note

Logistic Growth

If a population is growing in a constrained environment with carrying capacity $K$ , and absent constraint would grow exponentially with growth rate $r$ , then the population behavior can be described by the logistic growth model:

$P_{n} = P_{n - 1} + r (1 - \frac{P_{n - 1}}{K}) P_{n - 1}$

Unlike linear and exponential growth, logistic growth behaves differently if the populations grow steadily throughout the year or if they have one breeding time per year. The recursive formula provided above models generational growth, where there is one breeding time per year (or, at least a finite number); there is no explicit formula for this type of logistic growth.

Example 1

A forest is currently home to a population of 200 rabbits. The forest is estimated to be able to sustain a population of 2000 rabbits. Absent any restrictions, the rabbits would grow by 50% per year. Predict the future population using the logistic growth model.

# Ch 8.6 - logistic growth. One extra factor (1 - P/K) turns exponential growth
# into something that levels off, overshoots, oscillates, or goes chaotic.
import matplotlib.pyplot as plt

def logistic(P0, r, K, steps):
    out = [float(P0)]
    for _ in range(steps):
        p = out[-1]
        nxt = p + r * (1 - p / K) * p
        if nxt <= 0:
            out.append(0.0)      # the model has crashed the population to zero
            break
        out.append(nxt)
    return out

P0, R, K, STEPS = 200, 0.50, 2000, 10     # the book's rabbits
# TRY IT: R = 1.9, then 2.05, then 2.5, then 2.9. Watch the last panel.
rabbits = logistic(P0, R, K, STEPS)
print(f"{P0} rabbits, forest carrying capacity {K}, unconstrained growth {100 * R:g}%")
print(" n" + "".join(f"{n:>7}" for n in range(STEPS + 1)))
print("Pn" + "".join(f"{p:>7.0f}" for p in rabbits))
print("\n n   Pn      growth this year   the brake (1 - Pn/K)")
for n in range(6):
    p = rabbits[n]
    print(f"{n:>2}   {p:>6.0f}   {rabbits[n + 1] - p:>15.1f}   {1 - p / K:>18.3f}")
print("Early on the brake is near 1 and growth looks exponential. As Pn nears the")
print("carrying capacity the brake goes to 0 and growth stops.\n")

cases = [(0.50, "r = 0.5: settles smoothly onto K"),
         (1.50, "r = 1.5: overshoots, then comes back"),
         (2.05, "r = 2.05: never settles, a 2-cycle"),
         (2.70, "r = 2.7: chaos")]
fig, axes = plt.subplots(2, 2, figsize=(10, 6), sharex=True)
for ax, (r, title) in zip(axes.ravel(), cases):
    series = logistic(600, r, 1000, 30)
    ax.plot(range(len(series)), series, "o-", ms=3, color="tab:blue")
    ax.axhline(1000, ls="--", color="tab:red", lw=1)
    ax.set_title(title, fontsize=10)
    ax.set_ylim(0, 1500)
for ax in axes[1]:
    ax.set_xlabel("generation n")
for ax in axes[:, 0]:
    ax.set_ylabel("population")
plt.suptitle("All four start at 600 with carrying capacity 1000 (dashed). Only r differs.")
plt.tight_layout()
plt.show()
final = {r: logistic(600, r, 1000, 60)[-4:] for r, _ in cases}
for r, tail in final.items():
    print(f"r = {r}: last four generations {[round(v) for v in tail]}")

Show solution

Modeling this with a logistic growth model, $r equals 0.50$ , $K equals 2000$ , and $P sub 0 equals 200$ . Calculating the next year:

$P_{1} = P_{0} + 0.50 (1 - \frac{P_{0}}{2000}) P_{0} = 200 + 0.50 (1 - \frac{200}{2000}) 200 = 290$

We can use this to calculate the following year:

$P_{2} = P_{1} + 0.50 (1 - \frac{P_{1}}{2000}) P_{1} = 290 + 0.50 (1 - \frac{290}{2000}) 290 \approx 414$

A calculator was used to compute several more values:

n	0	1	2	3	4	5	6	7	8	9	10
$P_{n}$	200	290	414	578	784	1022	1272	1503	1690	1821	1902

Scatter plot with the horizontal axis Years marked 0 to 10 and the vertical axis Population marked 0 to 2000. Eleven points joined by a line trace an S shape, taking the values 200, 290, 414, 578, 784, 1022, 1272, 1503, 1690, 1821 and 1902. The climb steepens through about year 5 and then flattens off as the population closes on the carrying capacity of 2000. Plotting these values, we can see that the population starts to increase faster and the graph curves upwards during the first few years, like exponential growth, but then the growth slows down as the population approaches the carrying capacity.

Example 2

On an island that can support a population of 1000 lizards, there is currently a population of 600. These lizards have a lot of offspring and not a lot of natural predators, so have very high growth rate, around 150%. Calculating out the next couple generations:

Show solution

$P_{1} = P_{0} + 1.50 (1 - \frac{P_{0}}{1000}) P_{0} = 600 + 1.50 (1 - \frac{600}{1000}) 600 = 960$

$P_{2} = P_{1} + 1.50 (1 - \frac{P_{1}}{1000}) P_{1} = 960 + 1.50 (1 - \frac{960}{1000}) 960 = 1018$

Scatter plot with the horizontal axis Years marked 0 to 10 and the vertical axis Population marked 0 to 1200. The population leaps from 600 at year 0 to about 960 at year 1, overshoots the carrying capacity to about 1018 at year 2, settles back just under 1000 at year 3, and from there sits essentially flat on 1000 for the remaining years. Interestingly, even though the factor that limits the growth rate slowed the growth a lot, the population still overshot the carrying capacity. We would expect the population to decline the next year.

$P_{3} = P_{2} + 1.50 (1 - \frac{P_{3}}{1000}) P_{3} = 1018 + 1.50 (1 - \frac{1018}{1000}) 101$

Calculating out a few more years and plotting the results, we see the population wavers above and below the carrying capacity, but eventually settles down, leaving a steady population near the carrying capacity.

Your Turn

Try it Now 5

A field currently contains 20 mint plants. Absent constraints, the number of plants would increase by 70% each year, but the field can only support a maximum population of 300 plants. Use the logistic model to predict the population in the next three years.

Answer

$P_{1} = P_{0} + 0.70 (1 - \frac{P_{0}}{300}) P_{0} = 20 + 0.70 (1 - \frac{20}{300}) 20 = 33$

$P sub 2 equals 54$

$P sub 3 equals 85$

Taking this even further, we get more and more extreme behaviors as the growth rate increases higher. It is possible to get stable 4-cycles, 8-cycles, and higher. Quickly, though, the behavior approaches chaos (remember the movie Jurassic Park?).

Line graph titled r = 2.46, a 4-cycle, with Years 0 to 10 along the horizontal axis and Population 0 to 1400 up the vertical axis. Starting near 690, the plotted points settle into a repeating four-value pattern of roughly 1205, 560, 1160 and 670, so years 1, 5 and 9 all sit near 1205 while years 2 and 6 sit near 560.

Line graph titled r = 2.9, Chaos!, with Years 0 to 30 along the horizontal axis and Population 0 to 1400 up the vertical axis. From a start near 790 the points jump erratically between about 110 and 1300 with no repeating cycle, sometimes swinging from one extreme to the other in a single year and sometimes drifting through several middling values.

[1] users.rcn.com/jkimball.ma.ult...ulations2.html

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.