Logistic growth calculator
Enter the starting population, the intrinsic growth rate and the carrying capacity of the environment. Get the population after a given time, or the time it takes to reach a target size, together with the current growth rate, when growth is fastest and how far plain exponential growth would have got.
Runs in your browser: nothing you enter is sent anywhere.
How to use it
- Choose to find the population at a given time or the time to reach a given size.
- Enter the initial population N₀, the carrying capacity K and the growth rate r.
- Enter the time t, or the target population.
- Read N(t) or the time, the share of K, the growth rate and the inflection point.
- N(t) = K ÷ (1 + ((K − N₀) ÷ N₀) × e^(−r·t))
- Time to reach N: t = ln((K − N₀) × N ÷ (N₀ × (K − N))) ÷ r
- Growth rate: dN/dt = r × N × (1 − N ÷ K), at most r × K ÷ 4, at N = K ÷ 2
Worked example
N₀ = 10, K = 1,000, r = 0.5 per day, t = 10 days.
(K − N₀) ÷ N₀ = 99; e^(−5) = 0.0067379; 1 + 99 × 0.0067379 = 1.66706.
N = 1,000 ÷ 1.66706 = 599.86 — 59.99 % of K. The population is then growing by 0.5 × 599.86 × (1 − 0.59986) = 120.014 per day. Half the capacity was passed at t = ln 99 ÷ 0.5 = 9.1902 days, when growth peaked at 125 per day.
To reach 900: t = ln(990 × 900 ÷ (10 × 100)) ÷ 0.5 = ln 891 ÷ 0.5 = 13.5847 days.
Tips
- Above K the population falls. Starting with N₀ > K, the curve decreases towards K; the calculator handles that too.
- Estimating r. Early on, while N is far below K, r is close to the exponential rate: r ≈ ln(N₂ ÷ N₁) ÷ (t₂ − t₁).
- A model, not a forecast. Real populations fluctuate, overshoot or have changing capacities; use the result as a first approximation.
Frequently asked questions
What is the logistic growth equation?
dN/dt = r × N × (1 − N ÷ K). Its solution is N(t) = K ÷ (1 + ((K − N₀) ÷ N₀) × e^(−r × t)). Growth starts almost exponentially, slows as N approaches the carrying capacity K, and levels off at K.
What is the carrying capacity?
The largest population the environment can sustain with its food, space and other resources. In the model it is the level where births and deaths balance and growth stops.
When is the population growing fastest?
At half the carrying capacity, the inflection point of the S-shaped curve. There dN/dt reaches its maximum, r × K ÷ 4. It is reached at t = ln((K − N₀) ÷ N₀) ÷ r.
How is this different from exponential growth?
Exponential growth, N₀ × e^(r × t), never slows. In the default example the logistic population after 10 time units is 599.86, while unlimited exponential growth would already be at 1,484.13 — more than the carrying capacity.
Which time unit should I use?
Any, as long as r and t use the same one: r per day with t in days, r per year with t in years.