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solving problem for gradient descent

Muhammad Kundi さんによって質問されました 2019 年 10 月 14 日
最新アクティビティ Prabhan Purwar さんによって 編集されました 2019 年 10 月 18 日
hi,
I am trying to solve the following question using gradient descent method.\
.Capture1.JPG
I wrote the following code but its giving error.
function [xopt,fopt,niter,gnorm,dx] = grad_descent(varargin)
if nargin==0
% define starting point
x0 = [3 3]';
elseif nargin==1
% if a single input argument is provided, it is a user-defined starting
% point.
x0 = varargin{1};
else
error('Incorrect number of input arguments.')
end
% termination tolerance
tol = 1e-6;
% maximum number of allowed iterations
maxiter = 10;
% minimum allowed perturbation
dxmin = 1e-6;
% step size ( 0.33 causes instability, 0.2 quite accurate)
alpha = 0.1;
% initialize gradient norm, optimization vector, iteration counter, perturbation
gnorm = inf; x = x0; niter = 0; dx = inf;
% define the objective function:
f = @(x1,x2,x3) 4*[x1.^2 + x2-x3].^2 +10;
% plot objective function contours for visualization:
figure(1); clf; ezcontour(f,[-5 5 -5 5]); axis equal; hold on
% redefine objective function syntax for use with optimization:
f2 = @(x) f(x(1),x(2),x(3));
% gradient descent algorithm:
while and(gnorm>=tol, and(niter <= maxiter, dx >= dxmin))
% calculate gradient:
g = grad(x);
gnorm = norm(g);
% take step:
xnew = x - alpha*g;
% check step
if ~isfinite(xnew)
display(['Number of iterations: ' num2str(niter)])
error('x is inf or NaN')
end
% plot current point
plot([x(1) xnew(1)],[x(2) xnew(2)],'ko-')
refresh
% update termination metrics
niter = niter + 1;
dx = norm(xnew-x);
x = xnew;
end
xopt = x;
fopt = f2(xopt);
niter = niter - 1;
%define the gradient of the objective
% function g = grad(x)
% g = [2*x(1) + x(2)
% x(1) + 6*x(2)];
function g = grad(x)
g = 4*(x(1).^2 + x(2)-x(3)).^2 +10;
.I saved this code in a file called steepest.m and then I try to run the following command
[xopt,fopt,niter,gnorm,dx]=steepest
.But I get error.
I have actually used the following code (which works) to solve my problem.
function [xopt,fopt,niter,gnorm,dx] = grad_descent(varargin)
if nargin==0
% define starting point
x0 = [3 3]';
elseif nargin==1
% if a single input argument is provided, it is a user-defined starting
% point.
x0 = varargin{1};
else
error('Incorrect number of input arguments.')
end
% termination tolerance
tol = 1e-6;
% maximum number of allowed iterations
maxiter = 10;
% minimum allowed perturbation
dxmin = 1e-6;
% step size ( 0.33 causes instability, 0.2 quite accurate)
alpha = 0.1;
% initialize gradient norm, optimization vector, iteration counter, perturbation
gnorm = inf; x = x0; niter = 0; dx = inf;
% define the objective function:
f = @(x1,x2) x1.^2 + x1.*x2 + 3*x2.^2;
% plot objective function contours for visualization:
figure(1); clf; ezcontour(f,[-5 5 -5 5]); axis equal; hold on
% redefine objective function syntax for use with optimization:
f2 = @(x) f(x(1),x(2));
% gradient descent algorithm:
while and(gnorm>=tol, and(niter <= maxiter, dx >= dxmin))
% calculate gradient:
g = grad(x);
gnorm = norm(g);
% take step:
xnew = x - alpha*g;
% check step
if ~isfinite(xnew)
display(['Number of iterations: ' num2str(niter)])
error('x is inf or NaN')
end
% plot current point
plot([x(1) xnew(1)],[x(2) xnew(2)],'ko-')
refresh
% update termination metrics
niter = niter + 1;
dx = norm(xnew-x);
x = xnew;
end
xopt = x;
fopt = f2(xopt);
niter = niter - 1;
% define the gradient of the objective
function g = grad(x)
g = [2*x(1) + x(2)
x(1) + 6*x(2)];
.
This code works perfectly but why my code is not working?
please help
thanks

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1 件の回答

回答者: Prabhan Purwar 2019 年 10 月 18 日
編集済み: Prabhan Purwar 2019 年 10 月 18 日
 採用された回答

Hi,
Following code Illustrates the working of Gradient Descent for 3 variables.
To eliminate error changes were made to:
  • Initial value
  • Maxiter value
  • Alpha value
function [xopt,fopt,niter,gnorm,dx] = grad_descent(varargin)
if nargin==0
% define starting point
x0 = [3 3 3]';
elseif nargin==1
% if a single input argument is provided, it is a user-defined starting
% point.
x0 = varargin{1};
else
error('Incorrect number of input arguments.')
end
% termination tolerance
tol = 1e-6;
% maximum number of allowed iterations
maxiter = 100000;
% minimum allowed perturbation
dxmin = 1e-6;
% step size ( 0.33 causes instability, 0.2 quite accurate)
alpha = 0.000001;
% initialize gradient norm, optimization vector, iteration counter, perturbation
gnorm = inf; x = x0; niter = 0; dx = inf;
% define the objective function:
f = @(x1,x2,x3) 4*(x1.^2 + x2-x3).^2 +10;
% plot objective function contours for visualization:
%figure(1); clf; contour3(f,[-5 5 -5 5 -5 5]); axis equal; hold on
% redefine objective function syntax for use with optimization:
f2 = @(x) f(x(1),x(2),x(3));
% gradient descent algorithm:
while and(gnorm>=tol, and(niter <= maxiter, dx >= dxmin))
% calculate gradient:
g = grad(x);
gnorm = norm(g);
% take step:
xnew = x - alpha*g;
% check step
if ~isfinite(xnew)
display(['Number of iterations: ' num2str(niter)])
error('x is inf or NaN')
end
% plot current point
%plot([x(1) xnew(1)],[x(2) xnew(2)],'ko-')
refresh
% update termination metrics
niter = niter + 1;
dx = norm(xnew-x);
x = xnew;
end
xopt = x;
fopt = f2(xopt);
niter = niter - 1;
%define the gradient of the objective
% function g = grad(x)
% g = [2*x(1) + x(2)
% x(1) + 6*x(2)];
function g = grad(x)
g = 4*(x(1).^2 + x(2)-x(3)).^2 +10;
ans =
0.3667
0.3667
0.3667
OR
Alternately make use of the following code for accurate result
fun = @(x) 4*(x(1).^2 + x(2)-x(3)).^2 +10;
x0 = [3,3,3];
x = fminsearch(fun,x0);

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