用MATLAB实现最速下降法-牛顿法和共轭梯度法求解实例

1、WOR格式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(x)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g k与上一次接待的搜索方向d k i的组合。五、 运行及结果如

2、下：最速下降法：题目：f=(x-2)A2+(y-4)八2M文件：function R,n=steel(xO,yO,eps)syms x ;syms y ;f=(x-2F2+(y-4)八2;v=x,y;j=jacobian(f,v);T=subs(j(1),x,x0),subs(j (2) ,y,y0);temp=sqrt(T(1)A2+(T(2)A2);x1=x0;y1=y0;n=0;syms kk ;while (temp>eps)专业资料整理WOR格式d=-T;专业资料整理WOR格式f1= x1+kk*d(1);f2=y1+kk*d(2); fT=subs(j(1),x,f1),su

3、bs(j (2),y,f2); fun=sqrt(fT(1)A2+(fT(2)A2);Mini=Gold(fun,0,1,0.00001); xO=x1+Mini*d(1);yO=y1+Mini*d(2); T=subs(j(1),x,x0),subs(j (2) ,y,y0); temp=sqrt(T(1)A2+(T(2)A2);x1= x0;y1=y0;n=n+1;endR=x0,y0调用黄金分割法：M文件：function Mini=Gold(f,aO,bO,eps)syms x format long ;syms kk ;u=a0+0.382*(b0-aO);v=a0+0.618*(b

4、0-aO);k=0;a=aO;b=bO;array(k+1,1)=a;array(k+1,2)=b;while (b-a)/(bO-aO)>=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;e ndarray(k+1,1)=a;array(k+1,2)=b; endMini=(a+b)/2;专业资料整理WOR格式专业资料整理输入：R,n=steel(O,1,O.OOO1)R = 1.9

5、99994136676423.99999120501463R = 1.999994136676423.99999120501463n =1牛顿法：WOR格式题目：f=(x-2)A2+(y-4)八2M文件：syms x1 x2 ;f=(x1 -2F2+(x2 -4)八2;v=x1,x2;df=jacobian(f,v);df=df.'G=jacobian(df,v);epson=1e-12;x0=0,0'g仁subs(df,x1,x2,xO(1,1),xO(2,1);G 1=subs(G,x1,x2,x0(1,1),x0(2,1);k=0;mul_count=0;sum_coun

6、t=0;mul_count=mul_count+12;sum_count=sum_count+6;while (norm(g1)>epson)p=-G1g1;xO=xO+p;g1= subs(df,x1,x2,x0(1,1),x0(2,1);G1= subs(G,x1,x2,xO(1,1),xO(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)A2+(y-4)A2M

7、文件：function f=conjugate_grad_2d(x0,t)x=x0;专业资料整理WOR格式syms xi yi a f=(xi-2F2+(yi -4)八2; fx=diff(f,xi);专业资料整理WOR格式fy=diff(f,yi);fx=subs(fx,xi,yi,xO);fy=subs(fy,xi,yi,xO);fi=fx,fy;count=0;while double(sqrt(fxA2+fyA2)>ts=-fi;if count<=0s=-fi;elses=s1;endx=x+a*s;f=subs(f,xi,yi,x);f1=diff(f);f仁 solv

8、e(f1);if f1=0ai=double(f1);elsebreakx,f=subs(f,xi,yi,x),countendx=subs(x,a,ai);f=xi-xiA2+2*xi*yi+yiA2;fxi=diff(f,xi);fyi=diff(f,yi);fxi=subs(fxi,xi,yi,x);fyi=subs(fyi,xi,yi,x);fii=fxi,fyi;d=(fxiA2+fyiA2)/(fxA2+fyA2);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)专业资料整理WOR格式结果:0.24998825499785 -0.24999998741273f =0.12499999986176专业资料整理WOR格式count