## Osp

about. First, we can consider the intersection of the parabola and the curve. At this intersection point R(E) = 0 from which we conclude that the net revenue of the fishery is 0 (economists say that the rent is dissipated). H. Scott Gordon called this the ''bionomic equilibrium'' (Gordon 1954). It is a marine version of the famous tragedy of the commons, in which effort increases until there is no longer any money to be made.

Alternatively, we might imagine that somehow we can control effort, in which case we find the value of effort that maximizes the revenue. If we write the revenue as R(E) = pY(E)— cE then the value of effort that maximizes revenue is the one that satisfies p(A/dE)Y(E) = c, so that the level of effort that makes the line tangent to pY(E) have slope c is the one that we want (Figure 6.7).

Show that the bionomic level of effort (which makes total revenue equal to 0) is Eb = (r/q) [1 — (c/pqK)] and that the corresponding population size is Nb = N(E) = c/pq. What is frightening, from a biological perspective, about this deceptively beautiful equation? Does the former equation make you feel any more comfortable?

Next, we consider the dynamics of effort. Suppose that we assume that effort will increase as long as R(E) > 0, since people perceive that money can be made and that effort will decrease when people are losing money. Assuming that the rate of increase of effort and the rate of decrease of effort is the same, we might append an equation for the dynamics of effort to Eqs. (6.8) and write which can be analyzed by phase plane methods (and which will be déjà vu all over again if you did Exercise 2.12). One steady state of Eqs. (6.13) is N = 0, E = 0; otherwise the first equation gives the steady state condition E = (r/q)[i — (N/K)] and the second equation gives the condition N = c/pq. These are shown separately in Figure 6.8a and then combined. We conclude that if K > c/pq (the condition for bionomic equilibrium and the economic persistence of the fishery), then the system will show oscillations of effort and stock abundance.

Now, you might expect that there are differences in the rate at which effort is added and at which effort is reduced. I agree with you and the following exercise will help sort out this idea.

Figure 6.8. Phase plane analysis of the dynamics of stock and effort. (a, b) The isoclines for population size and effort are shown separately. (c) If K < c/pq, the isoclines do not intersect and the fishery will be driven to economic extinction (N — K, E — 0). (d) If K> c/pq, then the isoclines intersect (at the bionomic equilibrium) and a phase plane analysis shows that the system will oscillate.

In this exercise, you will explore the dynamics of the Schaefer model when the effort responds to profit. For simplicity, you will use parameter values chosen for ease of presentation rather than values for a real fishery. In particular, set r — 0.1 and K — 1000 (say tons, if you wish). Assume discrete logistic growth, written like this

N(t + 1)— N(t)+ rN(t)(^ 1 - N^) -(1 - e-qE(t))N(t)) (6.14)

where E(t) is effort in year t and q is catchability. Set q — 0.05 and E(0) — 0.2 and assume that this is a developing fishery so that N(0) — K. (a) Use a Taylor expansion of e-qE(t) to show that this formulation becomes the Schaefer model in Eq. (6.8) when qE(t) ^ 1. Use this to explain the form of Eq. (6.14), rather than simply qEN for the harvest. (b) Next assume that the dynamics of effort are determined by profit and set n(t) — p(1 - e-qE(t))N(t)- cE(t) (6.15)

where n(t) is profit in year t; for calculations, setp — 0.1 and c — 2. Assume that in years when profit is positive effort increases by an amount AE+ and that in years when profit is negative it decreases by an amount AE- For computations, set AE+ — 0.2 and AE- — 0.1, to capture the idea that fishing capacity is often irreversible (boats are more rather than less specialized). The effort dynamics are thus

Figure 6.8. Phase plane analysis of the dynamics of stock and effort. (a, b) The isoclines for population size and effort are shown separately. (c) If K < c/pq, the isoclines do not intersect and the fishery will be driven to economic extinction (N — K, E — 0). (d) If K> c/pq, then the isoclines intersect (at the bionomic equilibrium) and a phase plane analysis shows that the system will oscillate.

E(t + 1)=E(t) + AE+ if n(t) > 0 E(t + 1)=E(t) if n(t)=0

Include the rule that if E(t + 1) is predicted by Eqs. (6.16) to be less than 0 then E(t + 1) = 0 and that if E(t) = 0, then E(t + 1) = AE+. Iterate Eqs. (6.15) and (6.16) for 100 years and interpret your results; using at least the following three plots: effort versus population size, catch versus time, and profit versus time. Interpret these plots. A more elaborate version of these kinds of ideas, using differential equations, is found in Mchich et al. (2002).

There is one final complication that we must discuss, whether we like its implications or not. This is the notion of discounting, which is the preference for an immediate reward over one of the same value but in the future (Souza, 1998). The basic concept is easy enough to understand: would you rather receive 100 dollars today or one year from today, given that you can do anything you want with that money between now and one year from today except spend it? It does not take much thinking to figure out that you'd take it today and put it in a bank account (if you are risk averse), a mutual fund (if you are less risk averse), or your favorite stock (if you really like to gamble). We can formalize this idea by introducing a rate S at which future returns are devalued relative to the present in the sense that one dollar t years in the future is worth e—St dollars today. That is, all else being equal, when the discount rate is greater than 0 you would always prefer rewards now rather than in the future. Thus, discounting compounds the effects of the tragedy of the commons.

Let us now think about the problem of harvesting a renewable resource when the returns are discounted. We will conduct a fairly general analysis, following the example of Colin Clark (Clark 1985, 1990). Instead of logistic dynamics, we assume a general biological growth function g(N), and instead of C(t) = qEN(t) we assume a general harvest function h(t), so that the dynamics for the stock are dN/dt = g(N)— h(t). A harvest h(t) obtained in the time interval t to t + dt years in the future has a present-day value h(t)e—Stdt, so that the present value, PV, of all future harvest is and our goal is to find the pattern of harvest that maximizes the present value, given the stock dynamics. In light of those dynamics, we write h(t) = g(N) — (dN/dt) so that the present value becomes oo

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