1、 用外点法求下列问题的最优解
方法一：外点牛顿法：
clc
m=zeros(1,50);a=zeros(1,50);b=zeros(1,50);f0=zeros(1,50);%a b为最优点坐标，f0为最优点函数值，f1 f2最优点梯度。
syms x1 x2 e; %e为罚因子。
m(1)=1;c=10;a(1)=0;b(1)=0; %c为递增系数。赋初值。
f=x1^2+x2^2+e*(1-x1)^2;f0(1)=1;
fx1=diff(f,'x1');fx2=diff(f,'x2');fx1x1=diff(fx1,'x1');fx1x2=diff(fx1,'x2');fx2x1=diff(fx2,'x1');fx2x2=diff(fx2,'x2');%求偏导、海森元素。
for k=1:100 %外点法e迭代循环.
x1=a(k);x2=b(k);e=m(k);
for n=1:100 %梯度法求最优值。
f1=subs(fx1); %求解梯度值和海森矩阵
f2=subs(fx2);
f11=subs(fx1x1);
f12=subs(fx1x2);
f21=subs(fx2x1);
f22=subs(fx2x2);
if(double(sqrt(f1^2+f2^2))<=0.001) %最优值收敛条件
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) %罚因子迭代收敛条件
a(k+1) %输出最优点坐标，罚因子迭代次数，最优值
b(k+1)
k
f0(k+1)
break;
else
m(k+1)=c*m(k);
end
end
方法二：外点梯度法：
clc
m=zeros(1,50);a=zeros(1,50);b=zeros(1,50);f0=zeros(1,50);
syms d x1 x2 e;
m(1)=1;c=10;a(1)=0;b(1)=0;
f=x1^2+x2^2+e*(1-x1)^2; f0(1)=1;
fx1=diff(f,'x1');
fx2=diff(f,'x2');
for k=1:100
x1=a(k);x2=b(k);e=m(k);
for n=1:100
f1=subs(fx1);
f2=subs(fx2);
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
D=(x1-d*f1)^2+(x2-d*f2)^2+e*(1-(x1-d*f1))^2;
Dd=diff(D,'d'); dd=solve(Dd); x1=x1-dd*f1; x2=x2-dd*f2;
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
2、用内点法求下列问题的最优解
内点牛顿法
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
转：https://blog.csdn.net/soaringlee_fighting/article/details/72885138