优化函数补充说明

优化函数补充说明求解函数f(x1,x2)=100(x2-x12)2+(1-x1)2的无约束最优化问题。>>formatlong>>x=fminunc('cfun',[0;0],ff)Warning:Gradientmustbeprovidedfortrust-regionmethod;usingline-searchmethodinstead.>Infminuncat241Optimizationterminated:relativeinfinity-normofgradientlessthanoptions.TolFun.x=0.999995588493430.99999116722602fval=1.947095564687865e-011求解函数f(x1,x2)=100(x2-x12)2+(1-x1)2的无约束最优化问题。求目标函数的梯度>>symsx1x2>>f=100*(x2-x1^2)^2+(1-x1)^2;>>J=jacobian(f,[x1,x2])J=[-400*(x2-x1^2)*x1-2+2*x1,200*x2-200*x1^2]function[y,Gy]=cfun(x)y=100*(x(2)-x(1)^2)^2+(1-x(1))^2;Gy=[-400*(x(2)-x(1)^2)*x(1)-2+2*x(1);200*x(2)-200*x(1)^2];%梯度x=fminunc('cfun',[0;0],ff)Optimizationterminated:RelativefunctionvaluechangingbylessthanOPTIONS.TolFun.x=0.999999885712490.99999976187545fval=2.218103884282714e-014参数传递方法全局变量法

functionexampleglobalbcb=2;c=3.5;x0=0;options=optimset(‘display’,’off’)y=fsolve(@poly,x0,options)functiony=poly(x)globalbcy=x^3+b*x+c;直接传递参数法

functionexampleb=2;c=3.5;x0=0options=optimset(‘display’,’off’)y=fsolve(@poly,x0,options,b,c)functiony=poly(x)y=x^3+b*x+c;采用evalin和assignin函数functionexampleb=2;c=3.5;x0=0;assigin(‘base’,’b’,b)assigin(‘base’,’c’,c)y=fsolve(@poly,x0)functiony=poly(x)b=evalin(‘base’,’b’,b)c=evalin(‘base’,’c’,c)y=x^3+b*x+c;匿名函数法functionexampleb=2;c=3.5;x0=0;y=fsolve(@(x)poly(x,b,c),x0)functiony=poly(x,b,c)y=x^3+b*x+c;嵌套函数法functionexampleb=2;c=3.5;x0=0;y=fsolve(@poly,x0)functiony=poly(x)y=x^3+b*x+c;endend文件传递法functionexampleb=2;c=3.5;x0=0;sava

parasbcy=fsolve(@poly,x0)functiony=poly(x)loadparas.maty=x^3+b*x+c;%STEP0初始化

>0,x0∈Rn,k=0;%step1计算f(xk),g(xk)%step2检验||g(xk)||≤eps,若成立则输出xk

,STOP.%step3:计算dk=-g(xk)%step4求步长

k%step4.1令

k0=1，j=0;%step4.2若f(xk+

kj

dk

)≤f(xk)+

kj

gkTdk

(0,0.5),则

k=

kj

，否则

kj+1=

kj

/2;j=j+1;%step5xk+1=xk+

k

dk

;k=k+1;gotostep2;最速下降算法

f代表目标函数

g代表梯度函数%STEP0初始化

eps1=10^(-8);

rho=0.3;initmax=40;x0=(1;2;3);n=length(x0);k=0;x=x0;%mainloop(step1---step5)done1=true;whiledone%step1计算f(xk),g(xk)f1=fun1(x);g1=grad1(x);%step2检验||g(xk)||≤eps,若成立则输出xk

,STOP.ifnorm(g1)<=eps

disp(‘finish,ok’);xbreakend%step3:计算dk=-g(xk)dir=-g1;%step4求步长

k%step4.1令

k0=1，j=0;alpha=1;j=0;done2=true;rho=0.3;%step4.2若f(xk+

kj

dk

)≤f(xk)+

kj

gkTdk

(0,0.5),则

k=

kj

,否则

kj+1=

kj

/2,j=j+1;

xn=x+alpha*dir;f2=fun(xn);iff2<=f1+rho*alpha*g1’*dir;breakendalpha=0.5*alpha;j=j+1;ifj>=initmaxdone=false;endend%whiledone2%step5xk+1=xk+

k

dk

;k=k+1;goto