Biodiversity and Optimal Multi-species Ecosystem Management

Abstract

We analyze optimal multi-species management in a dynamic bio-economic model taking into account both harvesting profit and biodiversity value. Within an analytical model, we show that extinction is never optimal when a global biodiversity value is taken into account. Moreover, a stronger preference for species diversity leads to a more even distribution of stock sizes in the optimal steady state, and a higher value of biodiversity increases steady state stock sizes for all species when species are ecologically independent or symbiotic. For a predator–prey ecosystem, the effects may be positive or negative depending on relative prices and the strength of species interaction. The analytical results are illustrated and extended using an age-structured three-species predator–prey model for the Baltic cod, sprat, and herring fisheries. In this quantitative application, we find that using stock biomass or stock numbers as abundance indicators in the biodiversity index may lead to opposite results.

Appendices

Appendix 1: Sufficiency Conditions

It is straightforward to verify that the sufficiency conditions are always fulfilled in the absence of species interactions, i.e., if Assumption A.1 holds, and if the biomass growth functions are concave.

For problems of optimal harvesting in multi-species systems however, non-concavities can easily arise, thus potentially violating the Arrow/Kurz (1970) sufficiency condition. Since the biodiversity index we are considering here is a concave function of stock sizes, the only way how including a biodiversity index may affect the sufficiency conditions is that the maximized Hamiltonian tends to become “more” concave. We formally illustrate this in the following by considering optimal harvesting of an ecosystem with competing species, based on Flaaten (1991). Consider the optimal harvesting of two competing species, where biomass dynamics are described by

$$\begin{aligned} {\dot{x}}_{1t}&=r_1\,x_{1t}\,\left( 1-x_{t1}-\gamma \,x_{2t}\right) -h_{1t}\end{aligned}$$

(30)

$$\begin{aligned} {\dot{x}}_{2t}&=r_2\,x_{2t}\,\left( 1-x_{t2}-\gamma \,x_{1t}\right) -h_{2t} \end{aligned}$$

(31)

with no harvesting costs and prices $p_1$ and $p_2$ for the harvest of the two species. To keep the analysis simple, we assume symmetric competition coefficients, $\gamma$. From the first-order conditions (9) for this case, we obtain that the maximized Hamiltonian is given by

$$\begin{aligned} H^{\max }=p_1\,r_1\,x_{1t}\,\left( 1-x_{t1}-\gamma \,x_{2t}\right) +r_2\,x_{2t}\,\left( 1-x_{t2}-\gamma \,x_{1t}\right) -h_{2t}+v\,B(x_{1t},x_{2t}). \end{aligned}$$

(32)

For $v=0$, the leading principal minors of the Hessian matrix for $H^{\max }$ are

$$\begin{aligned} M_1&=-2\,p_1\,r_1\end{aligned}$$

(33)

$$\begin{aligned} M_2&=-\gamma ^2\,(p_1\,r_1+p_2\,r_2)^2+4\,p_1\,r_1\,p_2\,r_2. \end{aligned}$$

(34)

The Hessian of the maximized Hamiltonian is negative definite if the leading principal minors have alternating signs. $M_1$ is always negative. $M_2$ is positive if

$$\begin{aligned} \gamma&<\frac{\sqrt{p_1\,r_1\,p_2\,r_2}}{\frac{1}{2}\,\left( p_1\,r_1\,\gamma +p_2\,r_2\,\gamma \right) }, \end{aligned}$$

(35)

i.e., if the competition coefficient $\gamma$ is smaller than the ratio of the geometric to the arithmetic mean of $p_1\,r_1$ and $p_2\,r_2$.

For $v>0$, the leading principal minors of the Hessian matrix for $H^{\max }$ are

$$\begin{aligned} M_1&=-2\,p_1\,r_1-B_{x_1x_1}(x_1,x_2)\end{aligned}$$

(36)

$$\begin{aligned} M_2&=-\gamma ^2\,(p_1\,r_1+p_2\,r_2)^2+4\,p_1\,r_1\,p_2\,r_2\nonumber \\&\;+ v\,\left( \underbrace{2\,\gamma \,\left( p_1\,r_1+p_2\,r_2\right) \,B_{x_1x_2}(x_1,x_2)}_{>0}\underbrace{-2\,p_1\,r_1\,B_{x_2x_2} (x_1,x_2)}_{>0}\underbrace{-2\,p_2\,r_2\,B_{x_1x_1}(x_1,x_2)}_{>0}\right) \nonumber \\&\;+ v^2\,\left( \underbrace{B_{x_1x_1}(x_1,x_2)\,B_{x_1x_1}(x_1,x_2) -\left( B_{x_1x_2}(x_1,x_2)\right) ^2}_{>0}\right) . \end{aligned}$$

(37)

Consequently, a value $v>0$ increases the range of values for $\gamma$ for which the second principal minor of the Hessian is positive.

Appendix 2: Comparative Statics w.r.t. v

Under Assumption A.2, but with ecological interactions, condition (10b) simplifies to

$$\begin{aligned} \left( \rho -\sum \limits _{j=1}^n\frac{p_j}{p_i}\,G_{jx_i}\right) \,p_i&=v \,B_{x_i}. \end{aligned}$$

(38)

Differentiating (38) with respect to v, we obtain

$$\begin{aligned}&-\left( p_1\,G_{1\bar{x}_1\bar{x}_1}+p_2\,G_{2\bar{x}_1\bar{x}_1} +v\,B_{\bar{x}_1\bar{x}_1}\right) \,\frac{d\bar{x}_1}{dv}\nonumber \\&\quad -\left( p_1\,G_{1\bar{x}_1\bar{x}_2}+p_2\,G_{2\bar{x}_1\bar{x}_2} +v\,B_{\bar{x}_1\bar{x}_2}\right) \,\frac{d\bar{x}_2}{dv}=B_{\bar{x}_1} \end{aligned}$$

(39)

$$\begin{aligned}&-\left( p_1\,G_{1\bar{x}_2\bar{x}_1}+p_2\,G_{2\bar{x}_2\bar{x}_1} +v\,B_{\bar{x}_2\bar{x}_1}\right) \,\frac{d\bar{x}_1}{dv}\nonumber \\&\quad -\left( p_1\,G_{1\bar{x}_2\bar{x}_2}+p_2\,G_{2\bar{x}_2\bar{x}_2} +v\,B_{\bar{x}_2\bar{x}_2}\right) \,\frac{d\bar{x}_2}{dv}=B_{\bar{x}_2}. \end{aligned}$$

(40)

Solving yields

$$\begin{aligned} \frac{d\bar{x}_1}{dv}&=\frac{1}{\Delta }\,\left( B_{\bar{x}_2}\,\left( p_1\,G_{1\bar{x}_1 \bar{x}_2}+p_2\,G_{2\bar{x}_1\bar{x}_2} +v\,B_{\bar{x}_1\bar{x}_2}\right) \right. \nonumber \\&\left. \quad -B_{\bar{x}_1}\,\left( p_1\,G_{1\bar{x}_2\bar{x}_2}+p_2\,G_{2\bar{x}_2\bar{x}_2} +v\,B_{\bar{x}_2\bar{x}_2}\right) \right) \end{aligned}$$

(41)

$$\begin{aligned} \frac{d\bar{x}_2}{dv}&=\frac{1}{\Delta }\,\left( B_{\bar{x}_1}\,\left( p_1\,G_{1\bar{x}_2 \bar{x}_1}+p_2\,G_{2\bar{x}_2\bar{x}_1}+v\,B_{\bar{x}_2\bar{x}_1}\right) \right. \nonumber \\&\left. \quad -B_{\bar{x}_2}\,\left( p_1\,G_{1\bar{x}_1\bar{x}_1}+p_2\,G_{2\bar{x}_1\bar{x}_1} +v\,B_{\bar{x}_1\bar{x}_1}\right) \right) \end{aligned}$$

(42)

with

$$\begin{aligned} \Delta= & {} p_1^2\,\left( G_{1\bar{x}_1\bar{x}_1}\,G_{1\bar{x}_2\bar{x}_2} -G_{1\bar{x}_1\bar{x}_2}^2\right) +p_2^2\,\left( G_{2\bar{x}_1\bar{x}_1} \,G_{2\bar{x}_2\bar{x}_2}-G_{2\bar{x}_1\bar{x}_2}^2\right) \nonumber \\&+\,p_1\,p_2\,\left( G_{1\bar{x}_1\bar{x}_1}\,G_{2\bar{x}_2\bar{x}_2} +G_{2\bar{x}_1\bar{x}_1}\,G_{1\bar{x}_2\bar{x}_2}+2\,G_{1\bar{x}_1\bar{x}_2} \,G_{2\bar{x}_1\bar{x}_2}\right) \nonumber \\&+\,v\,\bigg (B_{\bar{x}_1\bar{x}_1}\,\left( p_1\,G_{1\bar{x}_2\bar{x}_2} +p_2\,G_{2\bar{x}_2\bar{x}_2}\right) +B_{\bar{x}_2\bar{x}_2} \,\left( p_1\,G_{1\bar{x}_1\bar{x}_1}+p_2\,G_{2\bar{x}_1\bar{x}_1}\right) \nonumber \\&-2\,B_{\bar{x}_1\bar{x}_2}\,\left( p_1\,G_{1\bar{x}_1\bar{x}_2} +p_2\,G_{2\bar{x}_1\bar{x}_2}\right) \bigg )\nonumber \\&+v^2\,\left( B_{\bar{x}_1\bar{x}_1}\,B_{\bar{x}_2\bar{x}_2} -B_{\bar{x}_1\bar{x}_2}^2\right)>0\nonumber \\= & {} p_1^2\,\left( G_{1\bar{x}_1\bar{x}_1}\,G_{1\bar{x}_2\bar{x}_2} -G_{1\bar{x}_1\bar{x}_2}^2\right) +p_2^2\,\left( G_{2\bar{x}_1\bar{x}_1} \,G_{2\bar{x}_2\bar{x}_2}-G_{2\bar{x}_1\bar{x}_2}^2\right) \nonumber \\&+\,p_1\,p_2\,\left( G_{1\bar{x}_1\bar{x}_1}\,G_{2\bar{x}_2\bar{x}_2} +G_{2\bar{x}_1\bar{x}_1}\,G_{1\bar{x}_2\bar{x}_2}+2\,G_{1\bar{x}_1\bar{x}_2} \,G_{2\bar{x}_1\bar{x}_2}\right) \nonumber \\&+\,v\,\bigg (B_{\bar{x}_1\bar{x}_1}\,\left( p_1\,G_{1\bar{x}_2\bar{x}_2} +p_2\,G_{2\bar{x}_2\bar{x}_2}\right) +B_{\bar{x}_2\bar{x}_2} \,\left( p_1\,G_{1\bar{x}_1\bar{x}_1}+p_2\,G_{2\bar{x}_1\bar{x}_1}\right) \nonumber \\&-2\,B_{\bar{x}_1\bar{x}_2}\,\left( p_1\,G_{1\bar{x}_1\bar{x}_2} +p_2\,G_{2\bar{x}_1\bar{x}_2}\right) \bigg )>0. \end{aligned}$$

(43)

Appendix 3. Proof of Proposition 2

For $G_{i\bar{x}_i\bar{x}_j}=0$, the expressions (13) and (14) simplify to

$$\begin{aligned} \frac{d\bar{x}_1}{dv}&=\frac{1}{\Delta }\,\left( B_{\bar{x}_2}\,v\,B_{\bar{x}_1\bar{x}_2} -B_{\bar{x}_1}\,\left( p_2\,G_{2\bar{x}_2\bar{x}_2}+v\,B_{\bar{x}_2\bar{x}_2}\right) \right) >0 \end{aligned}$$

(44)

$$\begin{aligned} \frac{d\bar{x}_2}{dv}&=\frac{1}{\Delta }\,\left( B_{\bar{x}_1}\,v\,B_{\bar{x}_2\bar{x}_1} -B_{\bar{x}_2}\,\left( p_1\,G_{1\bar{x}_1\bar{x}_1}+v\,B_{\bar{x}_1\bar{x}_1} \right) \right) >0 \end{aligned}$$

(45)

with

$$\begin{aligned} \Delta =p_1\,p_2\,G_{1\bar{x}_1\bar{x}_1}\,G_{2\bar{x}_2\bar{x}_2} +v\,B_{\bar{x}_1\bar{x}_1}\,p_2\,G_{2\bar{x}_2\bar{x}_2} +v\,B_{\bar{x}_2\bar{x}_2}\,p_1\,G_{1\bar{x}_1\bar{x}_1}>0. \end{aligned}$$

(46)

This concludes the proof of part (i) of the proposition.

For $G_{i\bar{x}_i\bar{x}_j}>0$, $p_1\,G_{1\bar{x}_1\bar{x}_2}+p_2\,G_{2\bar{x}_1\bar{x}_2}>0$ such that the RHS of  (13) and  (14) are positive. This concludes the proof of part (ii) of the proposition.

Appendix 4: Comparative Statics w.r.t. p

Differentiating (38) with respect to $p_1$ for $i=1,2$ we obtain

$$\begin{aligned}&\left( \rho - G_{1\bar{x}_1} \right) -\left( p_1\,G_{1\bar{x}_1\bar{x}_1}+p_2\,G_{2\bar{x}_1\bar{x}_1} +v\,B_{\bar{x}_1\bar{x}_1}\right) \,\frac{d\bar{x}_1}{dp_1}\nonumber \\&\quad -\left( p_1\,G_{1\bar{x}_1\bar{x}_2}+p_2\,G_{2\bar{x}_1\bar{x}_2} +v\,B_{\bar{x}_1\bar{x}_2}\right) \,\frac{d\bar{x}_2}{dp_1}=0 \end{aligned}$$

(47)

$$\begin{aligned}&- G_{1\bar{x}_2} -\left( p_1\,G_{1\bar{x}_2\bar{x}_1}+p_2\,G_{2\bar{x}_2\bar{x}_1} +v\,B_{\bar{x}_2\bar{x}_1}\right) \,\frac{d\bar{x}_1}{dp_1}\nonumber \\&\quad -\left( p_1\,G_{1\bar{x}_2\bar{x}_2}+p_2\,G_{2\bar{x}_2\bar{x}_2} +v\,B_{\bar{x}_2\bar{x}_2}\right) \,\frac{d\bar{x}_2}{dp_1}=0. \end{aligned}$$

(48)

Solving yields

$$\begin{aligned} \frac{d\bar{x}_1}{dp_1}= & {} \frac{1}{\Delta }\,\left( (-G_{1 \bar{x}_2}\,\left( p_1\,G_{1\bar{x}_1\bar{x}_2}+p_2\,G_{2\bar{x}_1\bar{x}_2} +v\,B_{\bar{x}_1\bar{x}_2}\right) \right. \nonumber \\&\left. +(\rho - G_{1\bar{x}_1})\,\left( p_1\,G_{1\bar{x}_2\bar{x}_2} +p_2\,G_{2\bar{x}_2\bar{x}_2}+v\,B_{\bar{x}_2\bar{x}_2}\right) \right) \end{aligned}$$

(49)

$$\begin{aligned} \frac{d\bar{x}_2}{dp_1}= & {} \frac{1}{\Delta }\,\left( -(\rho - G_{1\bar{x}_1})\,\left( p_1\,G_{1\bar{x}_2\bar{x}_1}+p_2\,G_{2\bar{x}_2\bar{x}_1} +v\,B_{\bar{x}_2\bar{x}_1}\right) \right. \nonumber \\&\left. +\,G_{1\bar{x}_2}\,\left( p_1\,G_{1\bar{x}_1\bar{x}_1}+p_2\,G_{2\bar{x}_1\bar{x}_1} +v\,B_{\bar{x}_1\bar{x}_1}\right) \right) \end{aligned}$$

(50)

with $\Delta$ as above.

Appendix 5: Proof of Proposition 3

For $G_{i \bar{x}_j}=0$ and $G_{i\bar{x}_i\bar{x}_j}=0$, the expressions (16) and  (17) simplify to

$$\begin{aligned} \frac{d\bar{x}_1}{dp_1}&=\frac{1}{\Delta }\, (\rho - G_{1\bar{x}_1}) (p_2\, G_{2 \bar{x}_2 \bar{x}_2 } +v\,B_{\bar{x}_2 \bar{x}_2} ) <0 \end{aligned}$$

(51)

$$\begin{aligned} \frac{d\bar{x}_2}{dp_1}&=\frac{-1}{\Delta }\, (\rho - G_{1\bar{x}_1}) v\,B_{\bar{x}_2 \bar{x}_1} <0 \end{aligned}$$

(52)

with

$$\begin{aligned} \Delta =p_1\,p_2\,G_{1\bar{x}_1\bar{x}_1}\,G_{2\bar{x}_2\bar{x}_2} +v\,B_{\bar{x}_1\bar{x}_1}\,p_2\,G_{2\bar{x}_2\bar{x}_2}+v\,B_{\bar{x}_2\bar{x}_2} \,p_1\,G_{1\bar{x}_1\bar{x}_1}>0. \end{aligned}$$

(53)

This concludes the proof of part (i) of the proposition.

For $G_{i \bar{x}_j}>0$ and $G_{i\bar{x}_j\bar{x}_i}>0$, $p_1\,G_{1\bar{x}_1\bar{x}_2}+p_2\,G_{2\bar{x}_1\bar{x}_2}>0$ and $(\rho -G_{1 \bar{x}_1})>0$ such that the RHS of  (16) and  (17) are negative. This concludes the proof of part (ii) of the proposition.

Appendix 6: Proof of Proposition 4

We first determine the sign of the following term in (19):

$$\begin{aligned}&\frac{\partial B_{\bar{x}_2}}{\partial \omega }\,B_{\bar{x}_1\bar{x}_2}-\frac{\partial B_{\bar{x}_1}}{\partial \omega }\,B_{\bar{x}_2\bar{x}_2}\nonumber \\&\quad =-2^{-2\,\frac{\omega }{\omega -1}}\,\frac{{\hat{x}}^{\frac{1}{\omega }} \,\left( 1+{\hat{x}}^{\frac{1-\omega }{\omega }}\right) ^{\frac{2}{\omega -1}}}{x_2\,\omega ^2}\, \underbrace{\frac{1}{1-\omega }\,\left( \frac{{\hat{x}}^{\frac{1-\omega }{\omega }}}{1 +{\hat{x}}^{\frac{1-\omega }{\omega }}}\,\ln \left( {\hat{x}}\right) +\ln \left( \frac{1 +{\hat{x}}^{\frac{1-\omega }{\omega }}}{2}\right) ^{\frac{\omega }{\omega -1}} \right) }_{\equiv \Omega } \le 0.\nonumber \\ \end{aligned}$$

(54)

Lemma 1

$\Omega \ge 0$ with $\Omega =0$ only for ${\hat{x}}=1$.

Proof

$\Omega$ has a global minimum $\Omega =0$ at ${\hat{x}}=1$, as

$$\begin{aligned} \frac{d\Omega }{d{\hat{x}}}&=\frac{{\hat{x}}^{\frac{1-\omega }{\omega }}\,\ln ({\hat{x}})}{\omega \,{\hat{x}}\,\left( 1+{\hat{x}}^{\frac{1-\omega }{\omega }}\right) ^2} \end{aligned}$$

(55)

is zero if and only if ${\hat{x}}=1$, negative for all ${\hat{x}}<1$, and positive for all ${\hat{x}}>1$. $\square$

Lemma 1 implies that $d\bar{x}_1/d\omega <0$ if $\frac{\partial B_{x_1}}{\partial \omega }<0$. To determine the sign of this last expression, we define

$$\begin{aligned} \Gamma \equiv \frac{1}{1-\omega }\,\left( \frac{{\hat{x}}^{\frac{1-\omega }{\omega }}}{1+{\hat{x}}^{\frac{1-\omega }{\omega }}}\,\ln \left( {\hat{x}}\right) +\omega \,\ln \left( \frac{1+{\hat{x}}^{\frac{1-\omega }{\omega }}}{2}\right) ^{\frac{\omega }{\omega -1}}\right) . \end{aligned}$$

(56)

Note that $\frac{\partial B_{x_1}}{\partial \omega }\lesseqqgtr 0$ if and only if $\Gamma \lesseqqgtr 0$.

We have $\Gamma =0$ for ${\hat{x}}=1$, and furthermore

$$\begin{aligned} \frac{d\Gamma }{d{\hat{x}}}&=\frac{d\Omega }{d{\hat{x}}}+\frac{{\hat{x}}^{\frac{1-\omega }{\omega }}}{{\hat{x}}\,\left( 1+{\hat{x}}^{\frac{1-\omega }{\omega }}\right) }=\frac{{\hat{x}}^{\frac{1-\omega }{\omega }}}{\omega \,{\hat{x}}\,\left( 1+{\hat{x}}^{\frac{1-\omega }{\omega }}\right) ^2}\,\left( \ln ({\hat{x}})+\omega \,\left( 1+{\hat{x}}^{\frac{1-\omega }{\omega }}\right) \right) . \end{aligned}$$

(57)

Thus, for ${\hat{x}}\ge 1$, $d\Gamma /d{\hat{x}}>0$. Thus, $\Gamma >0$ for all ${\hat{x}}>1$. This shows that for the larger stock, the effect of $\omega$ on $\bar{x}$ is ambiguous.

If $x_1$ is the smaller stock, i.e., ${\hat{x}}<1$, the situation is more complicated. For the following lemma, we use $\hat{\omega }$ to denote the solution of $\hat{\omega }+\ln (\hat{\omega }-1)=0$, which is $\hat{\omega }\approx 1.28$.

Lemma 2

(a) If $\omega \ge \hat{\omega }$, $\Gamma <0$ for all $0<{\hat{x}}<1$.

(b) If $\omega <\hat{\omega }$, it exists an $0 \le \underline{\hat{x}} <1$ such that $\Gamma <0$ for all ${\hat{x}}\in (\underline{\hat{x}},1)$.

Proof

In the following we show that for $\omega \ge \hat{\omega }$, $d\Gamma /d{\hat{x}}>0$ for all ${\hat{x}}>0$, and that for $\omega <\hat{\omega }$, $\Gamma$ has a unique minimum where it assumes some negative value.

To this end, we consider the expression $\Sigma \equiv \ln ({\hat{x}})+\omega \,\left( 1+{\hat{x}}^{\frac{1-\omega }{\omega }}\right)$, which determines the sign of $d\Gamma /d{\hat{x}}$, i.e., $d\Gamma /d{\hat{x}}\gtreqqless 0$ if and only if $\Sigma \gtreqqless 0$. We show that the equation $\Sigma =0$ has no solution (i.e., $\Gamma$ is monotonic) if $\omega >{\hat{\omega }}$, and (at least) one solution if $\omega \le \hat{\omega }$.

Note that for ${\hat{x}}=1$, $\Sigma =2\,\omega >0$, and $\lim \limits _{{\hat{x}}\rightarrow 0}\Sigma =+\infty$ for $\omega >1$, as

$$\begin{aligned} \lim \limits _{{\hat{x}}\rightarrow 0}\frac{\ln ({\hat{x}})}{1+{\hat{x}}^{\frac{1-\omega }{\omega }}}=\lim \limits _{{\hat{x}}\rightarrow 0}\frac{\frac{1}{{\hat{x}}}}{\frac{1-\omega }{\omega }\,{\hat{x}}^{\frac{1-\omega }{\omega }-1}}=\lim \limits _{{\hat{x}}\rightarrow 0}\frac{\omega }{1-\omega }\,{\hat{x}}^{\frac{\omega -1}{\omega }}=0. \end{aligned}$$

(58)

Furthermore,

$$\begin{aligned} \frac{d\Sigma }{d{\hat{x}}}=\frac{1}{{\hat{x}}}\,\left( 1-(\omega -1)\,{\hat{x}}^{\frac{1-\omega }{\omega }}\right) . \end{aligned}$$

(59)

Case $\omega \ge {\hat{\omega }}$: $\Sigma$ has a minimum $\Sigma ^*=\frac{\omega }{\omega -1}\,\left( \omega +\ln \left( \omega -1\right) \right)$ at ${\hat{x}}^\star =(\omega -1)^{\frac{\omega }{\omega -1}}$. This minimum is non-negative if $\omega \ge {\hat{\omega }}$ and thus $\Sigma \ge 0$ for all ${\hat{x}}<1$ in this case. This implies that $\Gamma$ monotonically increases with ${\hat{x}}$ if $\omega \ge {\hat{\omega }}$. Since $\Gamma =0$ for ${\hat{x}} =1$, this implies that $\Gamma <0$ for all $0<{\hat{x}} < 1$. This concludes the proof of part (a) of the lemma.

Case $\omega <\hat{\omega }$: The minimum of $\Sigma$ is negative, $\Sigma ^\star <0$. As $\Sigma = 2\omega >0$ for ${\hat{x}}=1$, this implies that there exist a value $\underline{\underline{{\hat{x}}}}>\hat{x}^\star$ where $\Sigma =0$. At $\underline{\underline{{\hat{x}}}}$, $\Gamma$ assumes a minimum, i.e., for values of $\hat{x}<\underline{\underline{{\hat{x}}}}$, $\Gamma$ decreases with ${\hat{x}}$. Depending on the value of $\omega <\hat{\omega }$, $\Gamma$ may or may not intersect with zero for some value $\hat{x}<\underline{\underline{{\hat{x}}}}$. Let $\underline{x}$ be the maximum of that value of $\hat{x}$ where $\Gamma =0$ and zero. This concludes the proof of part (b) of the lemma. $\square$

Appendix 7: Parameter Values for the Age-Structured Bio-economic Model

Age	Maturity $\gamma _{is}$			Weight $w_{is}$ [g]			Catchability $q_{is}$			Mortality $M_2$			$\delta _{is}$ [$10^{-4}$]		Price $p_{Cs}$
	C	H	S	C	H	S	C	H	S	C	H	S	H	S	C
1	0.00	0.0	0.17	80	11	52	0.00	0.28	0.27	0.0	0.170	0.132	3.32	8.74	0.00
2	0.13	0.7	0.93	179	20	84	0.11	0.44	0.49	0.2	0.173	0.137	2.31	7.08	0.35
3	0.36	0.9	1.0	511	25	96	0.42	0.66	0.79	0.2	0.178	0.132	0.45	6.74	0.35
4	0.83	1.0	1.0	838	31	105	0.81	0.82	0.85	0.2	0.188	0.132	0.45	6.74	0.35
5	0.94	1.0	1.0	1204	37	111	1.00	0.97	1.00	0.2	0.188	0.132	0.45	6.74	0.48
6	0.96	1.0	1.0	1796	43	113	1.00	0.96	1.00	0.2	0.188	0.132	0.45	6.74	0.48
7	0.96	1.0	1.0	2596	48	111	1.00	1.00	1.00	0.2	0.188	0.132	0.45	6.74	0.64
8	0.98	1.0	1.0	4068	53	113	1.00	1.00	1.00	0.2	0.188	0.132	0.45	6.74	0.73

$c_C=55.2$ million euros, $p_S-c_S=0.039$ euros/kg, and $p_H-c_H=0.100$ euros/kg.

	Type of stock-recruitment function	$\phi _{i1}$	$\phi _{i2}$
Cod	Ricker	1.7	549.0
Herring	Ricker	30.33	2156.0
Sprat	Beverton-Holt	104.2	503.2