steepest descent using armijo rule

Question

Kendall Donahoe 2022-2-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1651370-steepest-descent-using-armijo-rule

回答： Brahmadev 2024-2-8

I need help using the armijo rule to find the steepest descent.

Here is the proplem I need to solve and the code I have so far. I know how to do it in 1-D but am having trouble translating that to 2-D.

Code a function to perform a generic steepest descent algorithm using the Armijo line-search rule. Your function should take as inputs, the number of iterations, the function to be minimized (fm), another function that returns the gradient of fm, some initial point x0, and the parameters needed for the line search. Turn in all of your code and the first five iterations when the algorithm is applied to minimizing the function

f(x,y)=(x-2y^2)(x-3y^2), starting at the initial point (-1,-1)

Use line-search parameters of s=1, beta=1/2, and sigma=0.1.

f=@(x) (x(1)-2*x(2).^2).*(x(1)-3*x(2).^2)

xk=[-1,-1]';

fg=@(x) [(x(1)-3*x(2).^2)+(x(1)-2*x(2).^2);4*x(2)*(x(1)-3*x(2).^2)+6*x(2)*(x(1)-2*x(2).^2)];

s=1;

beta=0.5;

sigma=0.1;

for j=1:6

xk(j+1)=armijostep(f,fg,-sign(fg(xk(j))),xk(j),s,beta,sigma); %inputs are f, fg, d, xk, s, beta, sigma in that order

end

hold on;

plot(xk,f(xk),'k*-')

xk

hold off;

2 个评论
显示无隐藏无

Torsten 2022-2-16

Use xk{j} instead of xk(j) if xk is n-dimensional (n>1) (already set the inital xk as xk{1} = [-1,-1].')

Kendall Donahoe 2022-2-16

I just tried that. I don't think it did anything. I'm not sure if I did it right. Before my main issue was getting an error on the number of array elements when trying to enter the gradient.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Brahmadev 2024-2-8

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1651370-steepest-descent-using-armijo-rule#answer_1405006

在 MATLAB Online 中打开

Hi @Kendall Donahoe,

Assuming that 'armijostep' function written by you takes the x and y co-ordinates (2x1 vector) from the current iteration and returns the x and y co-ordinates for the next iteration as a 2x1 vector, we can store pre-allocate 'xk' and modify the code as shown below:

% Defining the function f
f=@(x) (x(1)-2*x(2).^2).*(x(1)-3*x(2).^2);
no_of_iter = 6; % Defining no. of iterations
xk =  zeros([2, no_of_iter]); % Pre-allocating for each step
xk(:, 1)=[-1,-1]'; % Initial guess
% Defining the function fg
fg=@(x) [(x(1)-3*x(2).^2)+(x(1)-2*x(2).^2);4*x(2)*(x(1)-3*x(2).^2)+6*x(2)*(x(1)-2*x(2).^2)];
% Defining constants
s=1;
beta=0.5;
sigma=0.1;
% Calculating each new iteration and plotting it
hold on
for j=1:no_of_iter
    xk(:, j+1) = armijostep(f, fg, -sign(fg(xk(:, j))), xk(:, j), s, beta, sigma); %inputs are f, fg, d, xk, s, beta, sigma in that order
    plot(xk(:, j),f(xk(:, j)),'k*-'); % Plotting each iteration
end
hold off

Hope this helps in resolving your query!

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

steepest descent using armijo rule

2 个评论
显示无隐藏无

回答（1 个）

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

另请参阅

类别

标签

Community Treasure Hunt

steepest descent using armijo rule

2 个评论 显示 无隐藏 无

回答（1 个）

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

另请参阅

类别

标签

Community Treasure Hunt

2 个评论
显示无隐藏无

0 个评论
显示 -2更早的评论隐藏 -2更早的评论