# Exceptional control

Admittedly this isn’t the most interesting post. It is likely only interesting to Peter Berck. But, honestly, that’s the only person I care about for this post, since he has the power of the (red) pen. It may also be interesting to Tamma Carleton, but only because I worked with her to figure out how to model the difference between exceptional control and open access fisheries.

The objective of this post is to directly demonstrate how optimal management of a fishery where the owner controls total fishing effort differs from the extraction in open access. Formerly, we are looking for a solution to the following constrained optimization problem:

where $E$ is aggregate fishing effort normalized to number of trips, $x$ is the fish stock, $f(x)$ is the biological growth for a given fish stock, and $k$ is a “catchability” constant. Note that $pkx - c$ is a boat’s profits, catch less cost. For the open access problem, boats enter when profits are positive and exit when profits are negative, implying that profits must be zero in equilibrium. Our current problem is to maximize the present value of resource extraction by controlling the number of boat trips. To do this, note that the associated Hamiltonian is:

Let $x^{*}$ be the optimal stock for this problem and $x^{\text{open}}$ be the optimal stock for open access. Then, there are four equations to compare the dynamics for the two regimes.

The following is a recipe, of sorts, to find a closed-form solution for $x^{*}$ in terms of given parameters.

1. Solve for $\dot{\lambda}$ from Eqn. (4): $\dot{\lambda} = \frac{\dot{x}}{x}(p - \lambda)$
2. Solve for $(p - \lambda)$ from Eqn. (3): $(p - \lambda) = \frac{c}{kx}$
3. Combine the previous two steps: $\dot{\lambda} = \frac{\dot{x}}{x}\frac{c}{kx}$
4. Note that $r - f^{\prime}(x) = 0$ in open access. How does optimal control compare? The result requires some algebraic acrobatics.\

a. Isolate $r - f^{\prime}(x)$ in Eqn. (1): $\dot{\lambda} - (r - f^{\prime}(x))\lambda = - (p - \lambda)kE$
b. Solve for $kE$ from Eqn. (2): $kE = \frac{f(x) - \dot{x}}{x}$
c. Use Step 3 to find the control expression for $r - f^{\prime}(x)$

Note that there is just one unknown in the final expression, namely $x$ which will be our optimal stock. Actually finding the optimal stock is a heroic effort in algebra, since it can involve somewhat complicated expressions for $f(x)$, like the logistic growth function $f(x) = gx(1 - x/\kappa)$. However, there are some accessible insights from the existing expressions. Note that $r - f^{\prime}(x) > 0$ when $pkx - c > 0$, i.e., when $x > c/(pk)$. As such, when $x^{*} > c/(pk) = x^{\text{open}}$, $$r f^{\prime}(x)$$. When the optimal stock is greater than the open access stock, then the interest rate is greater than stock growth; and when the optimal stock is less than the open access stock, the interest rate is less than stock growth. With this, we have adequately constructed the exceptional control and sort of compared it to the open access solution.