用MATLAB实现最速下降法-牛顿法和共轭梯度法求解实例

1、WORD格式f ( x)实验的题目和要求一、所属课程名称：最优化方法二、实验日期：三、实验目的掌握最速下降法，牛顿法和共轭梯度法的算法思想，并能上机编程实现相应的算法。二、实验要求用 MATLAB 实现最速下降法，牛顿法和共轭梯度法求解实例。四、实验原理最速下降法是以负梯度方向最为下降方向的极小化算法，相邻两次的搜索方向是互相直交的。牛顿法是利用目标函数在迭代点xk处的Taylor展开式作为模型函数，并利用这个二次模型函数的极小点序列去逼近目标函数的极小点。共轭梯度法它的每一个搜索方向是互相共轭的，而这些搜索方向 d k仅仅是负梯度方向 gk与上一次接待的搜索方向 dk 1的组合。五运行及结果

2、如下 :最速下降法：题目： f=(x-2)2+(y-4)2M文件：functionR,n=steel(x0,y0,eps)symsx ;symsy ;f=(x-2)2+(y-4)2;v=x,y;j=jacobian(f,v);T=subs(j(1),x,x0),subs(j(2),y,y0);temp=sqrt(T(1)2+(T(2)2);x1=x0;y1=y0;n=0;symskk ;while(temp>eps)d=-T;专业资料整理WORD格式f1=x1+kk*d(1);f2=y1+kk*d(2);fT=subs(j(1),x,f1),subs(j(2),y,f2);fun=sqr

3、t(fT(1)2+(fT(2)2);Mini=Gold(fun,0,1,0.00001);x0=x1+Mini*d(1);y0=y1+Mini*d(2);T=subs(j(1),x,x0),subs(j(2),y,y0);temp=sqrt(T(1)2+(T(2)2);x1=x0;y1=y0;n=n+1;endR=x0,y0调用黄金分割法：M 文件：functionMini=Gold(f,a0,b0,eps)symsx ;formatlong ;symskk ;u=a0+0.382*(b0-a0);v=a0+0.618*(b0-a0);k=0;a=a0;b=b0;array(k+1,1)=a;

4、array(k+1,2)=b;while(b-a)/(b0-a0)>=eps)Fu=subs(f,kk,u);Fv=subs(f,kk,v);if(Fu<=Fv)b=v;v=u;u=a+0.382*(b-a);k=k+1;elseif(Fu>Fv)a=u;u=v;v=a+0.618*(b-a);k=k+1;endarray(k+1,1)=a;array(k+1,2)=b;endMini=(a+b)/2;输入：R,n=steel(0,1,0.0001)R =1.999994136676423.99999120501463R =1.999994136676423.99999120

5、501463n =1专业资料整理WORD格式牛顿法：专业资料整理WORD格式题目： f=(x-2)2+(y-4)2M 文件：symsx1x2 ;f=(x1-2)2+(x2-4)2；v=x1,x2;df=jacobian(f,v);df=df.'G=jacobian(df,v);epson=1e-12;x0=0,0'g1=subs(df,x1,x2,x0(1,1),x0(2,1);G1=subs(G,x1,x2,x0(1,1),x0(2,1);k=0;mul_count=0;sum_count=0;mul_count=mul_count+12;sum_count=sum_coun

6、t+6;while(norm(g1)>epson)p=-G1g1;x0=x0+p;g1=subs(df,x1,x2,x0(1,1),x0(2,1);G1=subs(G,x1,x2,x0(1,1),x0(2,1);k=k+1;mul_count=mul_count+16;sum_count=sum_count+11;end ;kx0mul_countsum_count结果：k =1x0 =24mul_count =28sum_count =17共轭梯度法：题目： f=(x-2)2+(y-4)2M 文件：functionf=conjugate_grad_2d(x0,t)x=x0;symsxi

7、yiaf=(xi-2)2+(yi-4)2;fx=diff(f,xi);专业资料整理WORD格式fy=diff(f,yi);fx=subs(fx,xi,yi,x0);fy=subs(fy,xi,yi,x0);fi=fx,fy;count=0;whiledouble(sqrt(fx2+fy2)>ts=-fi;ifcount<=0s=-fi;elses=s1;endx=x+a*s;f=subs(f,xi,yi,x);f1=diff(f);f1=solve(f1);iff1=0ai=double(f1);elsebreakx,f=subs(f,xi,yi,x),countendx=subs

8、(x,a,ai);f=xi-xi2+2*xi*yi+yi2;fxi=diff(f,xi);fyi=diff(f,yi);fxi=subs(fxi,xi,yi,x);fyi=subs(fyi,xi,yi,x);fii=fxi,fyi;d=(fxi2+fyi2)/(fx2+fy2);s1=-fii+d*s;count=count+1;fx=fxi;fy=fyi;endx,f=subs(f,xi,yi,x),count输入： conjugate_grad_2d(0,0,0.0001)结果：x =0.24998825499785 -0.24999998741273专业资料整理WORD格式f =0.12499999986176专业资料整理WORD格式count =1