Logistic Growth Calculator
Project capped population growth with the logistic S curve.
What it does
Populations boom, then brake. With food and space to spare, each generation multiplies the last; as the ceiling nears, competition bites and growth bends flat. The logistic curve captures both phases in one S: exponential lift early, saturation late.
The tool above draws that S at your numbers. Enter starting population, carrying capacity, rate, and time to get the population, its share of K, and when half of K arrives. Defaults grow 100 cells toward 1000 at r = 0.2, so the page opens solved at 450.9, about 45.1% of the ceiling, with half K at 11 time units.
The example shows the machinery: A = 9 measures how far below the ceiling the start sits, e^(-2) discounts it over ten units, and the denominator lands at 2.218. At t = 5 the same culture reads 232, still climbing steeply; starting at 200 instead reaches 648.8 by t = 10 with half K at 6.9.
Early growth explains a common confusion. When P sits far below K, the logistic and exponential curves nearly touch, so short experiments cannot tell them apart. Only longer runs reveal the bend, which is why yeast tubes, sheep ranges, and seal counts all confirm the S over time.
Use the table for landmarks of the example culture, then run your population. The page holds r and K fixed with no delay, overshoot, or migration, so boom bust cycles need richer models. Within range it answers planning questions fast: how many, what share of the ceiling, and when half arrives.
How to use
- Enter the starting population P0.
- Enter carrying capacity K above P0.
- Enter the growth rate r.
- Enter the elapsed time t.
- Read population, share of K, and half K time.
- Compare early t against pure exponential.
Formula
- P0
- Starting population.
- K
- Ceiling the environment sustains.
- r
- Intrinsic growth rate per time unit.
- t
- Elapsed time in matching units.
K must exceed P0 and all inputs stay positive inside wide bounds. The closed form assumes constant r and K with no delay or overshoot, so oscillating populations need richer models. Time units follow r: per hour r gives hours for t.
Worked example
Example: culture toward 1000
A culture starts at 100 cells with K = 1000 and r = 0.2 per time unit. Find the population at t = 10.
- A = (1000 - 100)/100 = 9.
- Exponent = -0.2 x 10 = -2, so e^(-2) = 0.13534.
- Denominator = 1 + 9 x 0.13534 = 2.218.
- P = 1000 / 2.218 = 450.9.
- Share of K = 450.9/1000 = 45.1 %.
- Half of K arrives at ln(9)/0.2 = 11 time units.
Reference table
| t | Population | Share of K |
|---|---|---|
| 0 | 100 | 10.0 % |
| 5 | 232 | 23.2 % |
| 10 | 450.9 | 45.1 % |
| 15 | 690.6 | 69.1 % |
| 20 | 855.5 | 85.5 % |
Limits
- Constant r and K assumed.
- No delay, overshoot, or migration.
- K must exceed P0.
- Time units follow r.
Common mistakes
- Using exponential past the bend.
- Setting K below the start.
- Mixing rate units with time units.
- Reading share of K as share of start.
- Expecting cycles from one S.
Try it yourself
Quick checks. Answers sit tucked under each question.
Same culture at t = 5?
Population 232. Share 23.2 %. Half K still at 11 time units.
Start at 200 instead, t = 10?
Population 648.8. Share 64.9 %. Half K at 6.9 time units.
Check t = 0 returns P0 by hand.
A = (1000-100)/100 = 9. P(0) = 1000/(1+9) = 100, the start, as required.
Questions
What is logistic growth?
Capped growth: P(t) = K/(1 + A e^(-rt)) with A = (K - P0)/P0. Growth starts near exponential, then bends flat at carrying capacity K.
Logistic versus exponential?
Exponential growth assumes endless resources and climbs forever. Logistic growth adds the ceiling K, giving the S curve seen in real populations.
What is carrying capacity?
K is the maximum population the environment sustains. Start below it; the tool requires K above P0.
When does it reach half of K?
Solve K/2 for t: t = ln(A)/r. The example needs ln(9)/0.2 = 11 time units.
Is early growth exponential?
Early on, when P sits far below K, the curve hugs the exponential with the same r. The ceiling only bites later.
Does it fit real populations exactly?
No. It assumes one constant r and K, equal access to resources, and no delay, overshoot, or migration.