优化方法MATLAB编程-大连理工大学

优化方法上机大作业

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指导老师：肖现涛

第一题

源程序如下：

function zy_x = di1ti(x)

%di1ti是用来求解优化作业第一题的函数。x0=x; yimuxulong=0.000001;

g0=g(x0);s0=-g0;

A=2*ones(100,100);

k=0;

while k<100

lanmed=-(g0)'*s0/(s0'*A*s0);

x=x0+lanmed*s0;

g=g(x);

k=k+1;

if norm(g)

zy_x=x;

fprintf('After %d iterations,obtain the optimal solution.\n \n The optimal solution is \n %f.\n\nThe optimal "x" is "ans".',k,f(x) )

break;

end

miu=norm(g)^2/norm(g0)^2;

s=-g+miu*s0;

g0=g; s0=s;x0=x;

end

function f=f(x)

f=(x'*ones(100,1))^2-x'*ones(100,1);

function g=g(x)

g=(2*x'*ones(100,1))*ones(100,1)-ones(100,1);

代入x0,运行结果如下：

>> x=zeros(100,1);

>> di1ti(x)

After 1 iterations,obtain the optimal solution.

The optimal solution is

-0.250000.

The optimal "x" is "ans".

ans =0.005*ones(100,1).

第二题

1.最速下降法。

源程序如下：

function zy_x=di2titidu(x)

%该函数用来解大作业第二题。x0=x; yimuxulong=1e-5; k=0; g0=g(x0); s0=-g0;

while k>=0

if norm(g0)

break;

else

lanmed=10;c=0.1;i=0;

while i>=0&i<100

x=x0+lanmed*s0;

if f(x)>(f(x0)+c*lanmed*g0'*s0)

lanmed=lanmed/2;

i=i+1;

else

break;

end

end

x=x0+lanmed*s0;

x0=x;

g0=g(x);

s0=-g0;

k=k+1;

end

end

zy_x=x;

zyj=f(x);

fprintf('after %d iterations,obtain the optimal solution.\n\nThe optimal solution is %f.\n\n The optimal "x" is "ans".\n',k,zyj);

function f=f(x)

x1=[1 0 0 0]*x;

x2=[0 1 0 0]*x;

x3=[0 0 1 0]*x;

x4=[0 0 0 1]*x;

f=(x1-1)^2+(x3-1)^2+100*(x2-x1^2)^2+100*(x4-x3^2)^2;

function g=g(x)

x1=[1 0 0 0]*x;

x2=[0 1 0 0]*x;

x3=[0 0 1 0]*x;

x4=[0 0 0 1]*x;

g=[2*(x1-1)-400*x1*(x2-x1^2);200*(x2-x1^2);2*(x3-1)-400*x3*(x4-x3^2);200*( x4-x3^2)];

>> x=[-1.2 1 -1.2 1]';

>> di2titidu(x)

after 5945 iterations,obtain the optimal solution.

The optimal solution is 0.000000.

The optimal "x" is "ans".

ans =

1.0000

1.0000

1.0000

1.0000

2.牛顿法

源程序如下：

function zy_x=di2tinewton(x)

%该函数用来解大作业第二题。

x0=x; yimuxulong=1e-5; k=0;

g0=g(x0); h0=h(x0);s0=-inv(h0)*g0;

while k>=0&k<1000

if norm(g0)

break;

else

x=x0+s0;

x0=x;

g0=g(x);

h0=h(x);

s0=-inv(h0)*g0;

k=k+1;

end

end

zy_x=x;

zyj=f(x);

fprintf('after %d iterations,obtain the optimal solution.\n\nThe optimal solution is %f.\n\n The optimal "x" is "ans".\n',k,zyj);