惩罚函数法（内点法、外点法）求解约束优化问题最优值 matlab

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