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A population of $N$ fish in a certain lake grow at rate $$N'(t) = aN(t) - bN^2(t)$$ if undisturbed by people. Fish can be withdrawn from the lake and consumed at rate $c(t)$, yielding utility $w(c(t))$ to the consuming community and reducing the fish growth rate accordingly: $$N'(t) = aN(t) - bN^2(t) - c(t)$$

Assume future utilities to the community are discounted at constant rate $r$. Characterize a fishing (consumption) plan to maximize the present value of the discounted stream of utilities. Assume that $N(0) = \frac{a}{b}$, and that $u' > 0$, $u'' < 0$.

Rodrigo de Azevedo
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Abbyss
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  • Nice problem. So what is the solution? Any hint? – Robert Z Oct 18 '19 at 11:58
  • are you working with finite or infinite horizon? – Chaos Oct 18 '19 at 12:13
  • @RScrlli, i suppose infinite horizon – Abbyss Oct 18 '19 at 12:25
  • @RobertZ Z,I found this problem in the textbook 'Dynamic Optimization' Morton I. Kamien, Nancy L.Schwartz (chapter 'Equilibria in Infinite Horizon Autonomous Problems'), maybe it will be useful. I tried to solve it in a similar way, but reasoning is at an impasse. – Abbyss Oct 18 '19 at 12:32
  • did you write the Hamiltonian ? – Chaos Oct 18 '19 at 12:42
  • @RScrlli, if I did everything right, then H = u(c) + lambda (a N - b N^2 - c). – Abbyss Oct 18 '19 at 12:44
  • You should add the discount factor $e^{-rt}$ – Chaos Oct 18 '19 at 13:37
  • @RScrlli, Well, in this case we get that the optimal c is obtained for $$e^{-rt} u' = \lambda $$ (because $$H_{cc}^{''} < 0$$ and $$ H_{c}^{'} = e^{-rt} u' - \lambda $$). My problem is that I do not understand what can be done next. – Abbyss Oct 18 '19 at 13:50
  • I think that the exercise is just asking you to characterise the solution, of course that without any parameters/functions you cannot solve the exercise – Chaos Oct 18 '19 at 14:12
  • Where does this come from? A book? – Rodrigo de Azevedo Oct 28 '25 at 18:40

1 Answers1

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Since the fish becomes cheaper at rate $r$, the Lagrangian: $$ \mathcal L=\int_0^\infty w(c(t))e^{-rt}dt $$

The Hamiltonian here is: $$ H(t, N, c,\lambda)=w(c)e^{-rt} +\lambda(a N-bN^2-c) %+\mu N \\ \frac{\partial H}{\partial c}=0=w'(c)e^{-rt}-\lambda\\ \lambda = w'(c) e^{-rt}\\ \dot \lambda =-\frac{\partial H}{\partial N}=-a\lambda+2b\lambda N\\ -rw'(c)e^{-rt}+w''(c)\dot c e^{-rt}=-e^{-rt}(a-2bN)\\[20pt] w''(c)\dot c=r w'(c)+2bN-a\\ \dot N=aN-bN^2-c $$

We have ODE with initial condition $N(0)=N_0$ and ending condition $\lambda(t_{end})=0$ ($t_{end}$ is the time when fish is depleted from the lake $N(t_{end})=0$).

There can be a case when the end condition cannot be satisfied and $\lambda(t_{end})>0$. That means the fish becomes cheaper faster than it can grow. So the optimal solution will be to fish all the fish as fast as you can $c=N_0\delta(t)$.

Vasily Mitch
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