Logistic growth and the carrying capacity

Why growth cannot go on

If one bacterium doubled every $20$ minutes for two days, its descendants would weigh more than the Earth. That never happens. Food runs short, waste builds up, and space fills. Each population has a carrying capacity $K$: the largest count its surroundings can support.

The logistic rule

Exponential growth says the population grows at a rate $rN$ cells per unit time. The logistic rule multiplies that rate by a brake:

• $\dfrac{dN}{dt} = rN\left(1 - \dfrac{N}{K}\right)$

Read the brake, $1 - \dfrac{N}{K}$, at three counts:

• when $N$ is tiny compared with $K$, the brake is nearly $1$, so the growth is nearly exponential;
• when $N = \dfrac{K}{2}$, the brake is $\dfrac{1}{2}$;
• when $N = K$, the brake is $0$, and the growth stops.

Stepping in small time steps

A computer cannot follow the change continuously. It moves forward in small steps of time $dt$. In each step the count grows by its rate times the step:

• $N_{\text{new}} = N + rN\left(1 - \dfrac{N}{K}\right) dt$

This is Euler's method. In Python it is one line inside a loop:

Here $r = 0.5$ per hour, so $dt$ is in hours, and $240$ steps of $0.1$ h make $24$ hours.

The S-curve

Plot the counts and you see an S-curve. It starts like the exponential, bends at half of $K$, and levels off at $K$.

The growth is fastest at $N = \dfrac{K}{2}$. There the rate is $r \times \dfrac{K}{2} \times \dfrac{1}{2} = \dfrac{rK}{4}$. For $r = 0.5$ per hour and $K = 650$ cells, that is about $81$ cells per hour.

The exact curve

For this rule, mathematicians have found the exact answer:

• $N(t) = \dfrac{K}{1 + \left(\dfrac{K}{N_0} - 1\right) e^{-rt}}$

Put $N = \dfrac{K}{2}$ into it, and the time to reach half of $K$ comes out as $t = \dfrac{\ln(K/N_0 - 1)}{r}$. The exact formula is a good test of the stepped program. With $dt = 0.1$ h the two differ by a few cells near the middle. A smaller step brings them closer.

How small is small?

The step must be small compared with $\dfrac{1}{r}$, the time over which the population changes a lot. With $r = 0.5$ per hour, $\dfrac{1}{r}$ is $2$ hours, so $0.1$ h is fine. A step close to $\dfrac{1}{r}$ jumps too far. The count can even overshoot $K$, which the real population never does.

Practise with new numbers

Try it in Python

The program steps the yeast population, prints it next to the exact formula every $4$ hours, and plots it with the exponential and the line $N = K$.

Common mistakes

• Forgetting the time step. N = N + r * N * (1 - N / K) takes a step of one whole unit of time. Multiply by dt.
• The brake the wrong way round. (1 - K / N) is negative for every count below $K$. The brake is (1 - N / K).
• A step that is too large. If the count overshoots $K$ or swings up and down, make dt smaller.
• Mixing units. With $r$ per hour, $dt$ and $t$ are in hours.
• Stopping the loop too early. The number of steps is the total time divided by dt.

Remember these

• Logistic growth: $\dfrac{dN}{dt} = rN\left(1 - \dfrac{N}{K}\right)$. It is nearly exponential while $N$ is small and stops at $N = K$.
• An Euler step: N = N + r * N * (1 - N / K) * dt, repeated round(total_time / dt) times.
• The S-curve grows fastest at $\dfrac{K}{2}$, at a rate of $\dfrac{rK}{4}$.
• Exact curve: $N = \dfrac{K}{1 + (K/N_0 - 1)e^{-rt}}$. Use it to test the stepped program, and keep $dt$ much smaller than $\dfrac{1}{r}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A population has $K = 1000$ cells. What is the brake $1 - \dfrac{N}{K}$ when $N = 900$ cells, and what does it mean?

2. One Euler step: $N = 100$ cells, $r = 0.2$ per hour, $K = 500$ cells and $dt = 1$ h. What is the new count?

3. At which count does a logistic population grow fastest?

4. Your stepped count rises above $K$ and then falls back. What should you change?

5. Why does the logistic curve look exponential at the start?