Logistic Growth and Carrying Capacity

Exponential growth never stops

A population that doubles every year grows exponentially: $10\times 2^t$10 × 2ᵗ after t years. After 6 years there are 640. After 20 years there are $10\times 2^{20}$10 × 2²⁰, which is more than ten million, and after 30 years there are more than ten billion.

Nothing real can do that for long. A pond has only so much food and space, and a town only so many homes. Exponential growth assumes the room never runs out, and that is the part of the model that fails first.

$P=10\times 2^t$P = 10 × 2ᵗ over six years, from 10 to 640. Each year the rise is bigger than the year before, with nothing to stop it.

A ceiling: the carrying capacity

A real population in a fixed space, such as fish in a pond, behaves differently. At first, when there is plenty of room, it grows almost exponentially. As the room runs out it grows more and more slowly, and in the end it levels off.

The level it settles at is the largest population the space can support, called the carrying capacity. On a graph the population traces an S-shaped curve that flattens onto a horizontal line at the carrying capacity. The curve gets closer and closer to that line without crossing it, so the line is a horizontal asymptote. This kind of growth is called logistic growth.

$P=\frac{500}{\left(1+49e^{-t}\right)}$P = 500/(1 + 49e⁻ᵗ). The population starts at 10, climbs steeply in the middle, and flattens onto the carrying capacity, the line P = 500.

The logistic model

A logistic model is usually written $P=\frac{L}{\left(1+A{e}^{-kt}\right)}$P = L/(1 + Ae⁻ᵏᵗ), where L, A and k are positive numbers. The letter e stands for a fixed number, about 2.718. All that matters here is what happens to $e^{-kt}$e⁻ᵏᵗ as t grows. A negative index means one over the power, so $e^{-kt}=\frac{1}{e^{kt}}$e⁻ᵏᵗ = 1/eᵏᵗ. As t grows, $e^{kt}$eᵏᵗ grows without limit, and one over it shrinks toward 0. Any base bigger than 1 does the same: $2^{-t}=\frac{1}{2^t}$2⁻ᵗ = 1/2ᵗ is $\frac{1}{1024}$1/1024 by t = 10.

So as t grows, the bottom of the fraction, $1+A{e}^{-kt}$1 + Ae⁻ᵏᵗ, closes in on 1 + 0 = 1, and P closes in on $L/1=L$L / 1 = L. The bottom is always more than 1, so P is always less than L. L is the carrying capacity.

For $P=\frac{500}{\left(1+49e^{-t}\right)}$P = 500/(1 + 49e⁻ᵗ), the carrying capacity is 500. At t = 10, $49e^{-10}$49e⁻¹⁰ is about 0.002, so $P=500/1.002=499$P = 500 / 1.002 = 499, to the nearest whole number: almost at the ceiling, and still under it.

Where the model starts

At the start, t = 0. Any number to the power 0 is 1, so $e^0=1$e⁰ = 1, and the model gives $P=\frac{L}{\left(1+A\times 1\right)}=\frac{L}{\left(1+A\right)}$P = L/(1 + A × 1) = L/(1 + A). That is the starting population.

For $P=\frac{500}{\left(1+49e^{-t}\right)}$P = 500/(1 + 49e⁻ᵗ), the start is $\frac{500}{\left(1+49\right)}=\frac{500}{50}=10$500/(1 + 49) = 500/50 = 10. So A decides how far below the ceiling the population starts: here the ceiling is 50 times the start, and 1 + A = 50.

Fastest halfway up

Two things decide how fast the population grows. The more animals there are, the more offspring they produce, which speeds growth up. The less room is left, the harder it is to survive, which slows growth down.

Early on the population is small, so growth is slow even though there is plenty of room. Near the ceiling the population is large, but there is almost no room left, so growth is slow again. The fastest growth comes in between, when the population is half of the carrying capacity. Before that point the curve bends upward, like an exponential; after it, the curve bends over toward the ceiling.

Here half of 500 is 250. The model passes 250 between t = 3, when P = 145, and t = 4, when P = 264, to the nearest whole number. The rise in that year, 118, is the largest of any year.

The rise in each of the first eight years of $P=\frac{500}{\left(1+49e^{-t}\right)}$P = 500/(1 + 49e⁻ᵗ), to the nearest whole number: 16, 39, 80, 118, 112, 70, 33, 13. The rises grow until the fourth year, which carries the population past 250, and then shrink.

The marked point is where P = 250, half of the carrying capacity, at $t\approx 3.9$t ≈ 3.9. The curve is steepest there.

Reading a logistic model

Take $P=\frac{1200}{\left(1+5e^{-0.4t}\right)}$P = 1200/(1 + 5e^(−0.4t)). As t grows, $e^{-0.4t}$e^(−0.4t) fades to 0, so the carrying capacity is 1200. At t = 0 the population is $\frac{1200}{\left(1+5\right)}=\frac{1200}{6}=200$1200/(1 + 5) = 1200/6 = 200. It grows fastest when it reaches half of 1200, which is 600.

The usual mistakes

Reading L as the starting population. The number on top is where the curve ends. At t = 0 the bottom is 1 + A, not 1, so the start is $\frac{L}{\left(1+A\right)}$L/(1 + A).

Taking A as the carrying capacity. A sits with the exponential and fixes how far below the ceiling the population starts; it is not a population at all.

Expecting the fastest growth at the start. An exponential speeds up from the start, but a logistic curve is slowed by the shrinking room, and its steepest point is halfway up.

Expecting the population to pass the carrying capacity. The bottom of the fraction is always more than 1, so P always stays below L.