懲罰函數法（內點法、外點法）求解約束優化問題最優值 matlab

1、 用外點法求下列問題的最優解

方法一：外點牛頓法：
clc
m=zeros(1,50);a=zeros(1,50);b=zeros(1,50);f0=zeros(1,50);%a b爲最優點座標，f0爲最優點函數值，f1 f2最優點梯度。
syms x1 x2 e;                                            %e爲罰因子。
m(1)=1;c=10;a(1)=0;b(1)=0;                               %c爲遞增係數。賦初值。
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');%求偏導、海森元素。
for k=1:100                                                          %外點法e迭代循環.
    x1=a(k);x2=b(k);e=m(k);
    for n=1:100                                                      %梯度法求最優值。
        f1=subs(fx1); %求解梯度值和海森矩陣
        f2=subs(fx2);
        f11=subs(fx1x1);
        f12=subs(fx1x2);
        f21=subs(fx2x1);
        f22=subs(fx2x2);
        if(double(sqrt(f1^2+f2^2))<=0.001)                                  %最優值收斂條件
            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)   %罰因子迭代收斂條件
      a(k+1)   %輸出最優點座標，罰因子迭代次數，最優值
      b(k+1)
      k
      f0(k+1)
      break;
else
      m(k+1)=c*m(k);
end
end

方法二：外點梯度法：
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、用內點法求下列問題的最優解

內點牛頓法
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

轉：https://blog.csdn.net/soaringlee_fighting/article/details/72885138