2016 年大连理工大学优化方法上机大作业 学院： 专业： 班级： 学号： 姓名： 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 + 0.1*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); 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; >> x1=newton(x0,eps) --The 1-th iter, the residual is 447.213595 --The 2-th iter, the residual is 0.000000 x1 = 1.0000 1.0000 3.BFGS 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 = bfgs(x0,eps) g0 = grad(x0); gk=g0; res = norm(gk); Hk=eye(2); k = 0; while...
