We should actually like to solve this equation for the reserve fraction, hence obtaining a( f), which is the fraction of habitat needed to be reserve to maintain the population steady state at fK, once f is specified. This can be done (Mangel 1998); you might set it as an optional exercise. One interesting question arises if we set f= 0; why we will do this becomes clear momentarily.

Exercise 6.15 (E)

Set f= 0 in the previous equation and show that a(0) = UU_L1__r (6.33)

We conclude from Eq. (6.33) that if the reserve fraction is greater than a(0) then the steady state stock size will be greater than 0. Interpret Eq. (6.33) for the case of very large r. (Of course, to assert that we sustain a fish stock as long as one individual remains is kind of a silly idea; we should like many more individuals than 1.) Interpret the case of modest or small r.

There are a number of ways that one can present the information contained in Eq. (6.32) (Figure 6.13).

There are also many ways in which stochastic effects could enter into what we have done. One possibility is that the catch fraction in the harvest region is not fixed but is a random variable U(t). An example of the distribution of this random variable is shown in Figure 6.14a; the mean and mode of the catch fraction are about 0.25, but the actual fraction varies from about 0.1 to 0.45. This should remind us that in operational situations such as fisheries, fishing mortality can be targeted but it cannot be controlled (Mangel 2000b).

When there are stochastic effects, the whole notion of sustainability must change and we have to think in terms of probabilities (Mangel 2000a). We understand now that the population size after fishing but before

Figure 6.13. The reserve fraction needed to achieve steady state population sizes that are 20%, 35%, or 60% of carrying capacity as a function of the harvest fraction outside of the maximum per capita growth rate r = 0.5 (panel a) or r = 1 (panel b). (c) An alternative way to view the information is to fix reserve size (say at 20%) and see how steady state population size varies with maximum per capita growth rate and harvest fraction.

Figure 6.13. The reserve fraction needed to achieve steady state population sizes that are 20%, 35%, or 60% of carrying capacity as a function of the harvest fraction outside of the maximum per capita growth rate r = 0.5 (panel a) or r = 1 (panel b). (c) An alternative way to view the information is to fix reserve size (say at 20%) and see how steady state population size varies with maximum per capita growth rate and harvest fraction.

Figure 6.14. (a) The distribution of fishing mortality on herring Clupea harengus (from Patterson (1999)). (b) The beta density with mean 0.25 and three different values of the coefficient of variation.

reproduction, aN(t) + (1 - a)(1 - U(t))N(t) = [1 - U(t)(1 - a)] N(t), is a random variable because harvest fraction U(t) is a random variable. Let us fix the reserve fraction at a, a time horizon T, and a critical population size Nc and define p(n, t|a, Nc) = PrfN(s) exceeds Ncfor all s, t < s < T|

reserve fraction a and that N(t)=ng (6.34)

For example, in the case of a developing fishery, we might assume that the population starts at carrying capacity and as a target that we do not want it to fall below 60% of carrying capacity, so that Nc = 0.6 K. For the case of an exploited fishery, we might change the goal so that the population does not fall below 35% of carrying capacity, so that Nc = 0.35 K.

We evaluate the probability defined in Eq. (6.33) by methods similar to, and actually easier than, stochastic dynamic programming. We have the end condition

At any earlier time t, when N(t) = n, assume that the random variable U(t) takes the value u (which will occur with probability determined by the density function that describes U(t)). Then the population size at the start of time period t + 1 will be g[(1 - u(1 - a))n] where g( •) is given by the right hand side of Eq. (6.31) and we are interested in the probability of staying above the critical level, starting at this new time and new population size. (In the next chapter, we will call this reasoning ''thinking along sample paths.'') Since the value of U(t) = u is determined by a probability density we have p(n,t|a,Nc)=Eufp(g[(1 - u(1 - a))n],t + 1|a,Nc)g (6.36)

In order to make Eq. (6.36) operational, we need to pick a distribution for U(t). Since U(t) is a catch fraction, a natural choice for the probability distribution is the beta density, some examples of which are shown in Figure 6.14b.

In year t, when N(t) = n and UU(t) = n, the harvest is (1 - a)un, so that if we define C(n, t|a, Nc) to be the accumulated catch between t and T, given that N(t) = n and the reserve fraction a, reasoning similar to that leading to Eq. (6.36) gives us

C(n, t|a, Nc)=Euf(1 - a)un + C(g[(1 - u(1 - a))n], t + 1|a, Nc)g (6.37)

The first term on the right hand side is a linear function of u; we take its expectation and write (using u to denote the mean U(t))

C(n, t|a,Nc) = (1 - a)un + EufC(g[(1 - u(1 - a))n], t + 1|a,Nc)g (6.38)

1000 r

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