Question

You are given the following general equation about the relationship between the population size and the growth rate of the fishery off the coast of Zantia:

                             g = rS(1 – S/k)

                             where

                             g = the growth rate of the fish population

                             r = the intrinsic growth rate of the fish species in the fishery

                             S = the size of the population of fish

                             k = the carry capacity of the fish habitat in the waters off the coast                                     of Zantia

     The amount of fish that can be caught in Zantia is given by the following equation:

                             H = q E S

                            

                             where

                             H = the amount of fish harvested

                             q = a “catchability” coefficient that tells how easy fish can be                                            caught (for Zantia, assume that q has been determined to equal                                            one (q = 1))

                        E = amount of fishing effort (each unit of fishing effort is equal                                     to one vessel and the crew to man the vessel)

    

Assume for simplicity that fish in Zantia can be sold for a price of $1each and that the total cost of fishing effort is equal to:

                        TC = aE

                        where

                        TC = the total cost of fishing effort

                        a   = the marginal cost of fishing effort

Derive the level of fishing effort that maximizes profits for Zantia’s fishery. How does the profit maximizing level of effort change with changes in r (e.g., the intrinsic growth rate of the fish species in the fishery), S (e.g., the size of the population of fish), and k (e.g., the carry capacity of the fish habitat)?

Answer 1

Given the growth rate of the population of fish and the harvest rate we can calculate the rate of change of fish population(dS/dt), where S(t) is population density at time t

(1) dS/dt = rS(1-S/k) - qES

By setting (1) equal to zero, we get two equilibriums. These occur when the growth rate of the population is equivalent to the harvest rate i.e. rS(1-S/k) = qES. First there is the equilibria

(2) S* = k(1-qE/r)

the other is extinction where the fish population has been depleted. The value of qE is called fishing mortality. When fishibg mortality is small (2) is stable; if the population increases past S*, harvest rate is greater than growth rate and the stock decreases back to equilibrium. If the population decreases past S*, harvest is less than growth, and the population increases to S*. In this case, since the equilibrium defined in (2) is stable it is sustainable, then in turn extinction is unstable. As fishing mortality increase the equilibrium (2) decreases. For large values of fishing mortality, S* has decreased so much that harvest exceeds growth for all positive population levels. Then S*=0 is the stable equilibrium.

Our goal is to maximise yield, for that equilibrium (2) is considered. Thus the equilibrium for sustainable yield is

Y = qES* = qEk(1-qE/r)

Suppose that catchability is fixed, since it is a constant, then increased fishing effort will increase sustainable yield up to a point. Then maximum sustainable yield(MSY) occurs when

dY/dE = qk(1-2qE/r) = 0

which by solving gives the corresponding optimum level of effort

E_MSY = r/2q

and maximum sustainable yield

MSY = rk/4

When r increases i.e. the intrinsic growth rate of the fish species in the fishery increases, then the optimum level of effort will increase and vice versa. When S i.e. the size of the population of fish increases, then the optimum level of effort will increase and vice versa. When k i.e. the carry capacity of the fish habitat increases, then the optimum level of effort increases and vice versa.