Question

4. In class you discussed a model for fishery management based on the logistic equation with a parameterization of harvesting, N =EN (1-)-mN, N(0) = No where m is the fishing rate ("m" for mortality). With m = 0, there are two fixed points: Ni = 0 (unstable) and N = K (stable). With m > 0, the second fixed point becomes N = K(1 - m/r) <K (a) At what critical fishing rate, me, will the population die out? You can just set N = 0, but you might also notice that the critical value occurs when d/dNmN] =d/dNrN(1 - N/K) at N = 0). Now notice that, with m<me, the stable population size can be anything between 0 and K, including arbitrarily small populations. This is unrealistic. As an example, sardines tend to cluster into large schools, making it hard for predators to get to many of them. But at small populations, this strategy is not possible and the mortality rate is higher, leading effectively to a lower growth rate. To account for this dependence of the population growth rate on N, we modify the logistic equation to a "depensation model" N=rn (. - 1) (--) - mN, N(0) = No (b) Graph the right-hand side with m = 0 as a function of N. Notice first that the curve should be negative for N <Ne, positive for Ne<N<K, and negative for N > K. Thus there are now three fixed points. What are they and which are stable? (c) Now consider m >0 and add the line mN to your graph. Different values of m will lead to different numbers and types of fixed points. Classify the regimes, i.e., make a plot of N* vs. m for each N*, with dotted lines for unstable points, solid lines for stable points (this is called a 'bifurcation diagram'). What is the critcial value of m above which the population dies? What is happening geometrically? Give as complete an analysis as possible.

Answer 1

Page-1 Answers . As per the given question a model for fishery management based on logistic equation with a parameterization of harvesting. Ň = N(1-N-mN, N(O)= No. here q = fraction . E= effort/ time The equation is, dw of C) =rn (1-N) hno .given equation derivative to zero, then - rn (1-N) - hNzo NN (1-N2 - 2) =0 50, Sn=O of = K.(-5)

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