有关内惩罚函数算法matlab程序

Question

我百度到障碍函数（内惩罚函数）的matlab算法程序 可是用matlab试很长时间跑不出来为什么啊 用改哪么 求求求 急急急！！

clc
m=zeros(1,50);
a=zeros(1,50);
b=zeros(1,50);
f0=zeros(1,50);
syms x1 x2 e;
m(1)=1;c=0.2;a(1)=2;b(1)=-3;
f=x1^2+x2^2-e*(1/(2*x1+x2-2)+1/(1-x1));
f0(1)=15; fx1=diff(f,'x1');fx2=diff(f,'x2');
fx1x1=diff(fx1,'x1');
fx1x2=diff(fx1,'x2');
fx2x1=diff(fx2,'x1');
fx2x2=diff(fx2,'x2');
for k=1:100
x1=a(k);x2=b(k);e=m(k);
for n=1:100
f1=subs(fx1);
f2=subs(fx2);
f11=subs(fx1x1);
f12=subs(fx1x2);
f21=subs(fx2x1);
f22=subs(fx2x2);
if(double(sqrt(f1^2+f2^2))<=0.002)
a(k+1)=double(x1);b(k+1)=double(x2);f0(k+1)=double(subs(f));
break;
else
X=[x1 x2]'-inv([f11 f12;f21 f22])*[f1 f2]';
x1=X(1,1);x2=X(2,1);
end
end
if(double(sqrt((a(k+1)-a(k))^2+(b(k+1)-b(k))^2))<=0.001)&&(double(abs((f0(k+1)-f0(k))/f0(k)))<=0.001)
a(k+1)
b(k+1)
k
f0(k+1)
break;
else
m(k+1)=c*m(k);
end
end

Answer 1

这个代码的问题在于没有将subs函数转换为数值，所以在下面求inv的时候会超级慢，修改过的代码如下：

clc;
clear;

m=zeros(1,50);
a=zeros(1,50);
b=zeros(1,50);
f0=zeros(1,50);

syms x1 x2 e;
m(1)=1;
c=0.2;
a(1)=2;
b(1)=-3;

f=x1^2+x2^2-e*(1/(2*x1+x2-2)+1/(1-x1));
f0(1)=15; 
fx1=diff(f,'x1');
fx2=diff(f,'x2');

fx1x1=diff(fx1,'x1');
fx1x2=diff(fx1,'x2');
fx2x1=diff(fx2,'x1');
fx2x2=diff(fx2,'x2');

%%
for k=1:length(a)-1
    k
    x1=a(k);
    x2=b(k);
    e=m(k);
    for n=1:length(a)-1
        n
        f1=double(subs(fx1));
        f2=double(subs(fx2));
        f11=double(subs(fx1x1));
        f12=double(subs(fx1x2));
        f21=double(subs(fx2x1));
        f22=double(subs(fx2x2));
        if double(sqrt(f1^2+f2^2)) <= 0.001
            a(k+1)=double(x1);
            b(k+1)=double(x2);
            f0(k+1)=double(subs(f));
            break
        else
            X=[x1 x2]'-inv([f11 f12;f21 f22])*[f1 f2]';
            x1=X(1,1);
            x2=X(2,1);
        end
    end
    if double(sqrt((a(k+1)-a(k))^2+(b(k+1)-b(k))^2))<=0.001 && double(abs((f0(k+1)-f0(k))/f0(k)))<=0.001
        a(k+1)
        b(k+1)
        k
        f0(k+1)
        break
    else
        m(k+1)=c*m(k);
    end
end追问

谢谢您帮我解决了算法问题 可是我又产生了新的问题

我需要用试验函数检验算法




问题


变量个数n=2

目标函数：fx=(x1-2)^2+(x2-1)^2

约束： -x1-x2+2>=0

-x1^2+x2>=0

初始点x0=（2,2），f（x0）=1

解：  x*=（1,1），f（x*）=1

我带到您改过的代码里    算的答案是（2,1） 可是它不符合约束条件呀     怎么办啊

麻烦您了！