Logistic Growth with Harvesting

Explore how a constant harvest rate affects logistic population growth: below the critical harvest the population stabilizes, above it the population collapses to extinction.

Worked example — Logistic Growth with Harvesting

Final population	473.607
Max. sustainable yield	50
Stable equilibrium N₊	473.607
Collapse threshold N₋	26.393

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Logistic Growth with Harvesting

Logistic growth with constant-rate harvesting is the equation dN/dt = rN(1 − N/K) − H: a population that grows fastest when it is half full, minus a fixed number taken away every year regardless of how many are left. That last clause is the whole difficulty. The catch does not shrink when the stock does.

What each symbol means

r: the intrinsic growth rate: how fast the population grows when it is small and unconstrained.
K: the carrying capacity, where growth falls to zero.
H: the harvest — a number of individuals per unit time, not a percentage.
N₋, N₊: the two equilibria. N₊ is stable and is where a healthy stock settles; N₋ is unstable and is the line below which the stock cannot recover.

Where the formula comes from

An equilibrium is a population that does not change, so set dN/dt = 0: rN(1 − N/K) = H. Growth on the left, catch on the right — the stock holds steady exactly when what it grows equals what you take.
The left side is a downward parabola in N, zero at N = 0 and at N = K, with its peak at N = K/2 where growth is rK/4. So the equation is a horizontal line at height H cutting a hill: two crossings, one, or none.
Solving gives N± = (K/2)(1 ± √(1 − 4H/(rK))). The square root is real only while H ≤ rK/4, which is where the maximum sustainable yield comes from — not as a policy choice but as the height above which the line clears the hill entirely.
The two roots are not equivalent. Just above N₋ the hill is higher than the line, so growth beats the catch and the stock climbs; just below it, the catch wins and the stock falls further, which makes the next year worse. N₋ repels and N₊ attracts. As H rises the two move together and meet at K/2, and past that there is no crossing at all.

How to read what you see

The blue line is the population over time. The grey dashed line is the carrying capacity, the green dashed line the stable equilibrium the stock is heading for, and the red dotted line the collapse threshold. The red-shaded band beneath it is the region from which no recovery is possible at this catch — if the blue line starts in there, or is ever pushed into it, it reaches zero. When the catch passes the maximum sustainable yield both dashed lines disappear, because there is nothing left for the stock to settle at.

Assumes: One species with no age structure, no predators and no environment that varies from year to year. Growth is logistic, so the surplus depends on the current stock alone. And the catch is truly constant — it continues at the same absolute size no matter how few individuals remain, which is what makes extinction reachable in finite time rather than merely approached.
Breaks when: The unstable equilibrium is dangerous in a way this deterministic picture cannot show. A real stock fluctuates, and it only has to be pushed below N₋ once, by one bad recruitment year, for the model to send it to zero — even with the quota untouched. So the closer the catch sits to the maximum sustainable yield, the higher N₋ climbs and the smaller the shock needed to cross it. Managing to the theoretical maximum is therefore not a slightly risky policy; it is the policy that makes the fatal shock smallest.

Maximum sustainable yield is the most fragile place to fish 🖖

The logistic curve grows fastest when the population sits at exactly half the carrying capacity, so the largest catch you can take forever is rK/4, harvested at K/2. That makes MSY sound like the obvious target, and for decades it was the stated goal of fisheries management. The problem is what MSY costs you in safety: at that point the stock is already producing all it can, so there is no spare growth left to absorb a bad year, a bad estimate of r, or a warmer ocean. Push slightly past it and the population does not settle at a lower level — it falls all the way, because above MSY no equilibrium exists at all. The Grand Banks cod fishery collapsed in 1992 and has not recovered in thirty years.

A sustainable catch is not enough; you also need enough fish 🖖

Set r = 0.4, K = 500 and a catch of 10. The maximum sustainable yield is 50, so you are taking a fifth of what the stock could bear, and the readout duly reports a stable equilibrium at 473.607. Now start the population at 26 instead of 100. It goes to zero. The threshold row explains why: this model has two equilibria, and the lower one at 26.393 is unstable — a dividing line, not a resting place. Above it the stock climbs to 473.607; below it, a constant catch removes more each year than a small population can regrow, and the shortfall compounds. Starting at 26.4 recovers completely and starting at 26 ends in extinction, on the same catch. Sustainability is a property of the pair, never of the quota alone.

The stock stays reassuring right up until it is gone 🖖

Raise the catch from 10 to 49 on the same fishery and the standing stock only falls from 473.607 to 285.355 — you are taking nearly five times as much for a loss of about 40% of the fish, which reads like a well-run fishery. Then at 50 there is no equilibrium at all. Watch the other row while you do it: the collapse threshold climbs from 26.393 to 214.645. The safe zone is being squeezed from both ends at once, and the number a manager is most likely to be watching — the stock itself — is the one that moves least. Nothing in the standing biomass announces that the margin has almost run out.

Problem solved in full

Take only 10 from a fishery with maximum sustainable yield 50 5 steps
A fishery grows logistically with r = 0.4 and a carrying capacity of 500, and its maximum sustainable yield is 50. Take only 10 — a fifth of what the stock could bear — and find how close it still sits to collapse.
1. The model is logistic growth minus a constant catch. Constant is the assumption doing the damage: the catch does not fall when the stock does.
  
  $\frac{d N}{d t} = r N (1 - \frac{N}{K}) - H$
2. Setting the rate of change to zero gives a quadratic in N, so there are two equilibria rather than one. That is already the story — an unharvested logistic population has a single stable point at K, and adding a fixed catch splits it in two.
  
  $r N - \frac{r}{K} N^{2} - H = 0 ⟹ N_{\pm} = \frac{K}{2} (1 \pm 1 - \frac{4 H}{r K})$
3. With H = 10 the discriminant is 0.8 and the roots are 473.6 and 26.4. The upper one is stable: nudge the stock either way and it comes back. The lower one is not — it is an edge.
  
  $1 - \frac{4 ( 10 )}{0.4 ( 500 )} = 0.8 = 0.89443 ⟹ N_{+} = 473.607, N_{-} = 26.393$
4. The square root vanishes when H reaches rK/4 = 50. That is the maximum sustainable yield, and it is precisely where the two equilibria merge and annihilate: above it there is no equilibrium at all and the stock runs to zero from any starting size.
  
  $root vanishes at H = \frac{r K}{4} = 50, where N_{\pm} merge at \frac{K}{2} = 250$
5. Which reframes the lower root entirely. It is not a population, it is a safety margin — the amount of bad luck the fishery can absorb before the catch outruns regrowth.
  
  $H = 10 = 0.2 H_{m a x} already puts N_{-} at 26.393 = 5.3% of K$
Answer

The tool prints a stable stock of 473.607, a collapse threshold of 26.393 and an MSY of 50. Taking a fifth of the sustainable maximum still leaves a hard floor at about 5% of carrying capacity: fall below it once, for any reason, and the same modest catch drives the population to zero. Fishing at the MSY is worse than it sounds — there the two roots meet at 250, the safety margin is exactly zero, and the smallest overestimate of r or K puts you on the far side of a boundary that no longer exists. The preset starting at N₀ = 26 shows the whole trap: a legal, modest, genuinely sustainable catch, and a stock that dies anyway.

$N_{\pm} = \frac{K}{2} (1 \pm 1 - \frac{4 H}{r K}) : 473.607 and 26.393; H_{m a x} = \frac{r K}{4} = 50$

Learning path

Growth with a limit

Leads to Predator and prey what a fixed annual catch does to that surplus.

References (1)

The paper that put maximum sustainable yield into fisheries management, and the model this tool integrates: M. B. Schaefer, "Some aspects of the dynamics of populations important to the management of the commercial marine fisheries", 1954 — reprinted in Bulletin of Mathematical Biology 53(1–2), 253–279, 1991.

By the 42Math team · Updated 2026-05-21

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Stable - Harvesting 10 against a maximum sustainable yield of 50 → the population settles at 473.607, below the capacity of 500
Collapse - Harvesting 50 against a maximum sustainable yield of 30 → the final population is 0
Near capacity - Starting at 450 of a 500 capacity ends at 456.155: near the top the surplus is small, so the equilibrium barely moves
Sustainable catch, doomed anyway - A catch of 10 against a maximum sustainable yield of 50 — comfortably sustainable — but starting at 26 against a collapse threshold of 26.393, so the final population is 0