大连理工大学优化方法上机大作业

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上机大作业1：

1.最速下降法：

function f = fun(x)

f = (1-x(1))^2 + 100*(x(2)-x(1)^2)^2; end

function g = grad(x) g = zeros(2,1);

g(1)=2*(x(1)-1)+400*x(1)*(x(1)^2-x(2)); g(2) = 200*(x(2)-x(1)^2);

end

function x_star = steepest(x0,eps) gk = grad(x0); res = norm(gk); k = 0;

while res > eps && k<=1000 dk = -gk;

ak =1; f0 = fun(x0); f1 = fun(x0+ak*dk); slope = dot(gk,dk); while f1 > f0 + *ak*slope ak = ak/4; xk = x0 + ak*dk; f1 = fun(xk); end k = k+1; x0 = xk; gk = grad(xk); res = norm(gk);

fprintf('--The %d-th iter, the residual is %f\\n',k,res); end x_star = xk; end

>> clear >> x0=[0,0]'; >> eps=1e-4;

>> x=steepest(x0,eps)

2.牛顿法：

function f = fun(x)

f = (1-x(1))^2 + 100*(x(2)-x(1)^2)^2; end

function g = grad2(x) g = zeros(2,2);

g(1,1)=2+400*(3*x(1)^2-x(2)); g(1,2)=-400*x(1); g(2,1)=-400*x(1); g(2,2)=200;

end

function g = grad(x) g = zeros(2,1);

g(1)=2*(x(1)-1)+400*x(1)*(x(1)^2-x(2)); g(2) = 200*(x(2)-x(1)^2);

end

function x_star = newton(x0,eps) gk = grad(x0); bk = [grad2(x0)]^(-1); res = norm(gk); k = 0;

while res > eps && k<=1000 dk=-bk*gk; xk=x0+dk; k = k+1; x0 = xk; gk = grad(xk); bk = [grad2(xk)]^(-1);