Department of Mathematics

IDEA: Internet Differential Equations Activities

Fishing for Profit


Dead Fish

Sport fishing is big business in the Western United States. People travel from all over the country, and indeed the world, to experience the thrills seen in such movies as "A river runs through it". Even though many public lakes and streams are stocked by the U.S. Fish and Wildlife Service, those are frequently fished out quickly. For that reason, it is not uncommon to find companies that lease lakes and manage them for fishermen. This exercise concerns such a company.

The corporation must decide whether to lease a lake that generates revenue through the sale of fishing permits. The Federal government owns the lake and the land around it, and will allow the corporation to determine the duration of the lease, but stipulates that the fish populations at the beginning and the end of the leasing period must be equal. This could requre the corporation to close the lake to fishing before the end of the lease for the fish population to recover back to initial levels. The corporation wants to determine the optimal duration of the lease, as well as the ensuing strategy for the number of permits to be sold, so as to satisfy the government's stipulation.

The lake's fish population x grows at a daily rate f(x) that is proportional to the population level: f(x) = g(x)x The proportional growth rate g(x) declines with x (due to the lake's limited capacity to sustain increasingly large fish populations) according to the relation g(x)=r(1-x/k), where r and k are constant positive parameters. Note that g(x) is a linear function that increases toward its maximum value as x approaches zero, and decreases toward zero as x approaches k.

Problem 1 Substitute the expression for g(x) into that for f(x), and graph f(x) with x on the horizontal axis. Calculate the population levels that generate zero daily growth and explain why they are expected to do so. Calculate the population level that maximizes daily growth and the associated maximum level of growth (called the maximum sustained yield). Mark these levels on the graph of f(x) and label them in terms of r and k.

The corporation models the demand for fish using the relation p(h) = a-bh where h is the number of fish harvested daily, p(h) is the price per fish, and a and b are positive constants.

Problem 2 Graph the demand curve p(h) with h on the horizontal axis, and label the intercepts in terms of the parameters a and b. What are the economic interpretations of the intercepts?

The daily revenue associated with the total daily catch h is given by

R(h) = p(h)h = ah - bh2

Problem 3 Graph R(h) with h on the horizontal axis. Calculate the harvest levels that return zero revenue, and explain why they are expected to do so. Further calculate the harvest level that maximizes daily revenue. Label these harvest levels on the graph in terms of the parameters a and b.

Using these functions, the corporation is ready to assemble an economic model for the lake. The harvest level h and the population x are both considered to be functions of time, so the corporation wants to model the way in which they evolve. The rate equations governing the harvesting problem are

h' = -R'(h)[f'(x)-d] / R''(h)
x' = f(x) - h = rx(1 - x/k) - h
where R'(h) denotes the first derivative of the daily revenue function with respect to the harvest level h (i.e. marginal revenue), R''(h) denotes the second derivative, f'(x) is the first derivative of the growth function with respect to the population level, (i.e. the marginal productivity of the fish stock), and d denotes the discount rate.

Problem 4 (Those who are unfamiliar with optimal control theory should skip this exercise) Demonstrate that the rate equations above are the necessary conditions for the optimal control problem

maxh 0T e-dt R(h) dt
subject to: x' = f(x) - h, x(0)=x0

The corporation needs to analyze this problem in order to determine the optimal period for which to lease the lakes, to find the optimal management strategy for the fish in the lakes, thereby to maximize its revenues.

They begin the analysis by finding the nullclines for the equations. In other words, they solve for the curves obtained by setting the right hand sides of each differential equation equal to zero:

-R'(h) [f'(x)-x] / R''(h) = 0
f(x) - h = r x (1-x/k) - h = 0.

The first equation has two nullclines , one of which occurs at the harvest level hmx maximizing the daily marginal revenue, i.e. R'(hmx) = 0. The other occurs at the population level xgr where the marginal productivity of the population is equal to the financial discount rate, i.e. f'(xgr) = d. This second nullcline has a nice economic interpretation. The benefit to the corporation of declining to harvest an additional fish is that the fish remains in the lake to reproduce at rate f'(x), and this increases the level of future harvests. The cost of declining to catch the fish is that the additional fish left in the lake is not converted into money and invested at the going interest rate d. Equilibrium is established (in other words, the corporation is indifferent to whether the fish is caught or not) when the "biological" rate of return f'(x) on fish left in the lake is equal to the "financial" rate of return d on a harvested fish converted into a financial asset. This leaves the fish in the unpleasant position of needing to know the going interest rate, so that they can reproduce at a rate that is high enough to make it worth the corporation's while to leave them in the lake.

Figure 1
The second equation has only one nullcline, which gives the sustained yield at each population level: hsy = rx(1-x/k). Figure 1 demonstrates how the nullclines intersect to create three steady states , which are indicated using triangles. The middle steady state is flanked by two outer steady states created by the intersection of the nullcline associated with the harvest level that maximizes the daily marginal revenue hmx, with the graph of the last equation above. The lower steady state is denoted (xL, hmx), and the upper is (xU, hmx).

The nullclines in Figure 1 divide phase space into eight sectors. The directions of motion of the harvest and population variables through time in each sector are indicated by the arrows placed around each number. For example, both variables decrease through time in sectors 1 and 8, and increase through time in sectors 4 and 6.

Problem 5 Do the necessary analytical work to verify the directions of motion in each sector.

Problem 6 Figure 1 assumes that hmx < hgr < hmsy, where hmxy is the maximum of the sustained yield nullcline. Redraw Figure 1 for the cases in which hgr < hmx < hmsy and hgr < hmsy < hmx.

Problem 7 Use the Java applet below (or the Ordinary Differential Equation solving program of your choice) to draw the phase diagram in Figure 1. In the applet, we have set r=0.1, k=1000, and d=0.05. For the moment, you should use a=1 and b=0.1. The solution forward in time is indicated in green, and the solution backward in time is in red. The nullclines are indicated in "rainbow trout pink". We will describe the blue curves later.

Problem 8 Make the following parameter changes in your phase diagram: a=4 and d=0.075. These changes generate the case in which hgr < hmx < hmsy. Regenerate the numerical phase diagram. Next reset a = 6 while holding all other parameters constant. This change generates the case in which hgr<hmsy<hmx. Regenerate the numerical phase diagram.

The corporation requires addition information to select the optimal duration of the lease and the optimal harvest strategy that results in the same fish population at the end of the lease as at the beginning. The optimal leasing period is determined via the following transversality condition:

H(T) = R(hT) + R'(hT)[f(xT) - hT] = 0,

where T denotes the terminal day of the lease initiated at t=0, and xT, and hT denote terminal population and harvest levels, respectively. The function H(t) measures the daily flow of benefits that the corporation captures by operating the lake fishery and is called the "Hamiltonian" function, in optimal control parlance. The transversality condition H(T)=0 requires that the corporation operate the lease until the day when the flow of benefits runs down to zero.

The requirement that the fish population return to levels existing at the beginning of the lease by the end of the lease is given by the boundary condition x0 = xT. The applet above superimposes the transversality and boundary conditions on the phase space generated by the parameters underlying Problem 7. Variables x and h satisfying the transversality condition generate a parabolic curve bounded below by hmx (i.e. the harvest level maximizing daily marginal revenue) with the maximum at the maximum sustained yield population level, xmsy. The boundary condition is given by the vertical line x0=xT.

The transversality and boundary conditions are both satisfied when the corporation, facing initial stock x0, selects the initial harvest rate h0 placing the system on a solution trajectory intersecting the transversality and boundary curves at (xT, hT) In other words, the corporation selects the harvest rate, i.e. the rate at which it sells fishing permits, so as to force the solution curve for its model to meet the point (xT, hT) at time T. This second portion of the solution trajectory depicted in Figure 2 does this.

Problem 9 Using the parameter values specified in exercise 7, numerically generate the solution trajectory that satisfies the transversality and endpoint conditions. [Hint: begin at the terminal values (xT,hT) and integrate backward through time. The optimal initial harvest level h0 is found where the backward orbit crosses the x0 line.

With the advent of HTML5, Javascript is now ready for prime time for mathematical applications. There are new Javascript demos illustrating how we might use interactive web objects to help students learn Calculus.

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