buyu2.m

捕鱼问题源代码

close all;
clc;clear;
%{
该程序用于求出五年后鱼群恢复情况
%}
len=10;
time=1:1:len+1;
k=0;f=0;
r=0.8;
n=zeros(1,len+1);
N1=[1.1941e11,zeros(1,len)];
N2=[5.3640e10,zeros(1,len)];
N3=[2.4127e10,zeros(1,len)];
N4=[0.0777e9,zeros(1,len)];

for i=1:len
     n(i+1)=0.5*1.109e5*N3(i)*exp(-(0.42*k+r)*2/3)+1.109e5*N4(i)*exp(-(k+r)*2/3);
     N1(i+1)=(1.22e11*n(i+1))/(1.22e11+n(i+1));
     N2(i+1)=exp(-r)*N1(i);
     N3(i+1)=exp(-r)*N2(i);
     N4(i+1)=N3(i)*exp(-(0.42*k*2/3+r))+N4(i)*exp(-(k*2/3+r));
 %    f=f+17.86*N3(i)*0.42*k(j)*(1-exp(-(r+0.42*k(j))*2/3))/(r+0.42*k(j))+22.99*N4(i)*k(j)*(1-exp(-(r+k(j))*2/3))/(r+k(j));
end
disp('各年龄鱼群的数量变化：');
subplot(2,2,1);  plot(time,N1);title('N1');
N1(end)
subplot(2,2,2);  plot(time,N2);title('N2');
N2(end)
subplot(2,2,3);  plot(time,N3);title('N3');
N3(end)
subplot(2,2,4);  plot(time,N4);title('N4');
N4(end)