myfunplot.m

捕鱼问题源代码

function [fval,x]=myfunplot(len)
k=[0:0.1:16.9,17:0.000001:18,18.1:0.1:35];

m=length(k);
r=0.8;
n=zeros(1,len+1);
N1=[1.22e11,zeros(1,len)];
N2=[2.97e10,zeros(1,len)];
N3=[1.01e10,zeros(1,len)];
N4=[3.29e9,zeros(1,len)];
f=zeros(1,m);
for j=1:m
for i=1:len
     n(i+1)=0.5*1.109e5*N3(i)*exp(-(0.42*k(j)+r)*2/3)+1.109e5*N4(i)*exp(-(k(j)+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(j)*2/3+r))+N4(i)*exp(-(k(j)*2/3+r));
     f(j)=f(j)+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
end
[fval,pos]=max(f);
x=k(pos);
figure(1);
plot(k,f);
xlabel('捕捞强度系数')
ylabel('连续五年捕捞总量');
title('连续五年捕捞总量与捕捞强度系数的关系曲线');
w=zeros(1,len);
for i=1:len
     n(i+1)=0.5*1.109e5*N3(i)*exp(-(0.42*x+r)*2/3)+1.109e5*N4(i)*exp(-(x+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*x*2/3+r))+N4(i)*exp(-(x*2/3+r));
     w(i)=w(i)+17.86*N3(i)*0.42*x*(1-exp(-(r+0.42*x)*2/3))/(r+0.42*x)+22.99*N4(i)*x*(1-exp(-(r+x)*2/3))/(r+x);
end
disp('每年的捕鱼量：')
w
time=1:1:len;
figure(2);
bar(time,w);
xlabel('年份')
ylabel('捕捞量');
title('连续捕捞五年每年捕捞量');

disp('最后各年龄鱼群的数量：');
N1
N2
N3
N4