用MATLAB實(shí)現(xiàn)最速下降法-牛頓法和共軛梯度法求解實(shí)例

1、精品好資料學(xué)習(xí)推薦 實(shí)驗(yàn)的題目和要求1、 所屬課程名稱：最優(yōu)化方法2、 實(shí)驗(yàn)日期： 3、 實(shí)驗(yàn)?zāi)康恼莆兆钏傧陆捣ǎｎD法和共軛梯度法的算法思想，并能上機(jī)編程實(shí)現(xiàn)相應(yīng)的算法。二、實(shí)驗(yàn)要求用MATLAB實(shí)現(xiàn)最速下降法，牛頓法和共軛梯度法求解實(shí)例。四、實(shí)驗(yàn)原理最速下降法是以負(fù)梯度方向最為下降方向的極小化算法，相鄰兩次的搜索方向是互相直交的。牛頓法是利用目標(biāo)函數(shù)在迭代點(diǎn)處的Taylor展開式作為模型函數(shù)，并利用這個二次模型函數(shù)的極小點(diǎn)序列去逼近目標(biāo)函數(shù)的極小點(diǎn)。共軛梯度法它的每一個搜索方向是互相共軛的，而這些搜索方向僅僅是負(fù)梯度方向與上一次接待的搜索方向的組合。五運(yùn)行及結(jié)果如下: 最速下降法：題目：f

2、=(x-2)2+(y-4)2M文件：function R,n=steel(x0,y0,eps)syms x;syms y;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;syms kk;while (tempeps) d=-T; f1=x1+kk*d(1);f2=y1+kk*d(2); fT=subs(j(1),x,f1),subs(j(2),y,f2); fun=sqrt(fT(1)2+(fT(2)2); Mini=G

3、old(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調(diào)用黃金分割法：M文件：function Mini=Gold(f,a0,b0,eps)syms x;format long;syms kk;u=a0+0.382*(b0-a0);v=a0+0.618*(b0-a0);k=0;a=a0;b=b0;array(k+1,1)=a;array(k+1,2)=b;whil

4、e(b-a)/(b0-a0)=eps) Fu=subs(f,kk,u); Fv=subs(f,kk,v);if(FuFv) a=u; u=v; v=a+0.618*(b-a); k=k+1;end array(k+1,1)=a;array(k+1,2)=b;endMini=(a+b)/2;輸入：R,n=steel(0,1,0.0001)R = 1.99999413667642 3.99999120501463R = 1.99999413667642 3.99999120501463n = 1牛頓法：題目：f=(x-2)2+(y-4)2M文件：syms x1x2; f=(x1-2)2+(x2-4

5、)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_count+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

6、,x2,x0(1,1),x0(2,1); k=k+1; mul_count=mul_count+16;sum_count=sum_count+11; end; k x0 mul_count sum_count結(jié)果：k = 1x0 =2 4mul_count = 28sum_count = 17共軛梯度法：題目：f=(x-2)2+(y-4)2M文件：function f=conjugate_grad_2d(x0,t)x=x0;syms xiyiaf=(xi-2)2+(yi-4)2;fx=diff(f,xi);fy=diff(f,yi);fx=subs(fx,xi,yi,x0);fy=subs(f

7、y,xi,yi,x0);fi=fx,fy;count=0;while double(sqrt(fx2+fy2)t s=-fi;if count=0 s=-fi;else s=s1;end x=x+a*s; f=subs(f,xi,yi,x); f1=diff(f); f1=solve(f1);if f1=0 ai=double(f1);elsebreak x,f=subs(f,xi,yi,x),countend x=subs(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輸入：conjugate_grad_2d(0,0,0.0001)結(jié)果：x = 0.24998825499785 -0.24999998741273f = 0.12499999986176count = 10ans