Fermer X

# Equilibria and Stability in the Management of a Renewable Resource

Michel De Lara and Luc Doyen
(last modiﬁcation date: March 7, 2018)
Version pdf de ce document
Version sans bandeaux

### Contents

Let time t be measured in discrete units (such as years). Let B(t) denote the biomass of a population at time t (beginning of time interval [t,t + 1[). We consider the so called Schaefer model

 (1)

where Biol is the population dynamics and h(t) is the harvesting. Notice that, in the time interval [t,t + 1[, growth of the stock occurs ﬁrst, followed by harvesting1 .

The sustainable yield he = Sust(Be) solves Be = Biol(Be) he, which gives:

 (2)

The carrying capacity of the habitat is the level K > 0 of positive biomass such that Biol(K) = K, that is Sust(K) = 0.

The maximum sustainable yield hmse and the corresponding maximum sustainable equilibrium Bmse are

 (3)

From [1, p. 258] and numerical simulations, we shall consider the Paciﬁc yellowﬁn tuna example as in Table 1.

 Paciﬁc yellowﬁn tuna yearly intrinsic growth R = 2.25 carrying capacity K = 250 000 metric tons catchability q = 0.0 000 385 per SFD price p = 600 \$ per metric ton cost c = 2 500 \$ per SFD

Table 1: Paciﬁc yellowﬁn tuna data for a discrete time logistic model (adapted from [1, p. 258]). SFD: standard ﬁshing day.

### 1 The Beverton-Holt model

The Beverton-Holt model is characterized by the discrete time dynamics mapping

 (4)

We have

 (5)

Question 1 Use the data in Table 1 to compute b in (4). Give the maximum sustainable biomass Bmse and the maximum sustainable yield hmse as in (5).

R_tuna = 2.25 ;
K_tuna = 250000 ; // metric tons

// BEVERTON-HOLT DYNAMICS
R_BH = R_tuna ;
b_BH = (R_BH-1) / K_tuna ;

function [y]=Beverton(B)
y=(R_BH*B)./(1 + b_BH*B) ;
y=maxi(0,y) ;
endfunction;

// SUSTAINABLE YIELD MACRO
function [SY]=sust_yield(dynamic,B), SY=dynamic(B)-B, endfunction;

// MAXIMUM SUSTAINABLE EQUILIBRIUM
B_MSE = (sqrt(R_BH) - 1)/b_BH ;
// maximum sustainable yield
h_MSE = sust_yield(Beverton,B_MSE) ;
// maximum sustainable yield

Question 2 Select one biomass level Be between the maximum sustainable biomass Bmse and the carrying capacity K. Compute the corresponding sustainable yield he.

Draw the corresponding steady trajectory of the Schaefer model (1) with the Beverton-Holt dynamics (4) and h(t) = he. Pick up two diﬀerent initial conditions in the neighborhood of the equilibrium biomass Be. Draw the corresponding trajectories. Does the ﬁgure conﬁrm or not the fact that the equilibrium biomass Be is attractive?

Recall that, for an equilibrium, being stable or attractive are unrelated notions.

Question 3 Does the ﬁgure conﬁrm or not the fact that the equilibrium biomass Be is stable? Be speciﬁc in your justiﬁcations. What can you say about asymptotic stability of the equilibrium biomass Be?

// SUSTAINABLE EQUILIBRIUM

alpha=rand();
Be= alpha*B_MSE + (1-alpha)*K_tuna ;
// selection of one of many possible equilibria
he=sust_yield(Beverton,Be) ;
// corresponding sustainable yield

function [y]=sequential(y0,time,f)
[one,two]=size(Binit) ;
y=zeros(one,prod(size(time))) ; // time is a vector t0, t0+1,...,T
// vector will contain the trajectories y(1),...,y(T-t0+1)
// for different initial conditions
for k=1:one
y(k,1)=y0(k);
// initialization
for s=time(1:(\$-1)) -time(1)+1
// runs from 1 to T-t0+1
y(k,s+1)=f(s,y(k,s));
end ;
end ;
endfunction

// STATE TRAJECTORY UNDER DYNAMICS

function [y]=Beverton_e(t,B)
y=Beverton(B) - he ;
y=maxi(0,y) ;
endfunction
// Beverton-Holt dynamics with harvesting at equilibrium (Be,he)

T=20;
years=1:(T+1);

xset("window",31); xbasc(31);
Binit=Be;
Bt=sequential(Binit,years,Beverton_e);
plot2d2(years',Bt',1);
//
Binit=0.9*Be ;
Bt=sequential(Binit,years,Beverton_e);
plot2d2(years',Bt',2);
//
Binit=1.1*Be;
// It seems there is a bug with the previous version of 'sequential'
Bt=sequential(Binit,years,Beverton_e);
plot2d2(years',Bt',3);
//
xtitle('Trajectories with Beverton-Holt dynamics (R=' +string(R_tuna)...
+' and K=' +string(K_tuna) +')', 'years (t)','B(t)')
legends(['equilibrium biomass'],[1],'ur')

Question 4 Find an equilibrium state Be which is not attractive. Illustrate that Be is not attractive with some trajectories. What can you say about asymptotic stability of the equilibrium biomass Be?

With price p, catchability coeﬃcient q and harvesting unitary costs c, the private property equilibrium (ppe) is the equilibrium solution (Bppe,hppe) = (Bppe,Sust(Bppe)) which maximizes the rent as follows:

 (6)

The common property equilibrium Bcpe makes the rent null and is given by

 (7)

Question 5 Study the stability around the two following equilibria:

• common property equilibrium Bcpe,
• private property equilibrium  (8)

Compare your observations with the theoretical results.

// Economic parameters
c_tuna=2500; // unit cost of effort
p_tuna=600; // market price
q_tuna=0.0000385; // catchability

c=c_tuna;
p=p_tuna;
q=q_tuna;

B_PPE= ( sqrt( R_BH * (1 + (b_BH*c/(p*q)) ) ) - 1 ) / b_BH;
// private property equilibrium

B_CPE=c/(p*q) ;
// common property equilibrium

### 2 The logistic model

The logistic model is characterized by the discrete time dynamics mapping

 (9)

where R 1 and r = R 1 0 is the per capita rate of growth (for small populations), and κ is related to the carrying capacity K (which solves Biol(K) = K) by:

 (10)

We have

 (11)

Question 6 Adapt the previous Scilab code to the logistic model, and compare the results.

### 3 The Ricker model

The Ricker model is characterized by the discrete time dynamics mapping

 (12)

Question 7 Adapt the previous Scilab code to the Ricker model, and compare the results. Try numerical procedures: type help fsolve to obtain information about Scilab solver.

### References

[1]   M. Kot. Elements of Mathematical Ecology. Cambridge University Press, Cambridge, 2001.

 L'École des Ponts ParisTech est membre fondateur de L'École des Ponts ParisTech est certifiée