Ricker Stock-Recruitment Curve

The Ricker curve gives the number of young fish that join a population, the recruits $R$, from the number of parents that spawned, $S$. Two parameters shape it. The first, $a$, is the productivity at low stock: when spawners are few, each produces about $a$ recruits, so $R$ is close to $aS$ and recruitment grows in proportion. The second, $b$, is the crowding: the factor $e^{-bS}$ shrinks as spawners increase, standing in for eggs and young that compete and bury each other's nests. $S$ and $R$ must be in the same units, whether fish, eggs or spawning biomass, and for salmon an $a$ of 2 to 8 is typical.

A rising line multiplied by a falling exponential makes a dome. Recruitment peaks at $S = 1/b$, where it reaches $a/(b\,e)$, and it falls on the far side, so too many spawners can give fewer recruits. That falling side is what marks the Ricker curve, and it is the usual choice for salmon, shad and other fish whose young suffer from crowding. W. E. Ricker worked it out for Pacific salmon in 1954 (Journal of the Fisheries Research Board of Canada).

Take $a = 3$ and $b = 0.001$. Then 500 spawners give $3\times500\times e^{-0.5} = 909.796$ recruits, the page's example. At the peak, $S = 1/b = 1000$, the curve reaches $a/(b\,e) = 3000/e = 1103.638$. Raise the stock to 1500 and recruitment drops to 1004.086, fewer than at 1000. Because of the dome, a recruit count below the peak can be reached from either side, and solving for $S$ uses the principal branch $W_0$ of the Lambert W function, which returns the rising-side stock: 909.796 recruits from this $a$ and $b$ needs 500 spawners.

The same curve is the Ricker population model written the fisheries way. With growth rate $r$ and carrying capacity $k$, $a = e^{r}$ and $b = r/k$. With $r = 1$, $k = 1000$ and $N = 500$, the population form $N\,e^{r(1 - N/k)}$ gives $500\,e^{0.5} = 824.361$, and entering $a = 2.718281828$ and $b = 0.001$ here returns the same 824.36.