I am trying gradient descent, I wrote following however not getting any answer,
n=0; %initialize iteration counter
eps=1; %initialize error
a=0.8; %set iteration parameter
x=[1;1]; %set starting value
f=6*x(1)^2+8*x(2)^2-3*x(1)*x(2);
%Computation loop
while eps>1e-12||n<100
gradf=[12*x(1)-3*x(2); 16*x(2)-3*x(1)]; %gradf(x)
eps=(norm(gradf)/(1+abs(f))); %error
y=x-a*gradf; %iterate
x=y; %update x
n=n+1; %counter+1
end
n;x;eps; %display end values
When I add this file to path and type x it shows NaN, NaN. what is wrong?
There are several errors in your code. Consider this (I put comments where corrections were needed)
n=0;
eps=1;
a=0.1; %You need a way smaller parameter to converge!
x=[1;1];
A = [6 -3/2 ; -3/2 8]; %You have a bilinear positive definite form,
%you may use matrix form for convenience
while eps>1e-12 && n<100 %You had wrong termination conditions!!
gradf=2*A*x; %(gradf in terms of matrix)
f=x'*A*x; %you need to update f every iteration!!
eps=(norm(gradf)/(1+abs(f)))
disp(eps > 1e-12)
x=x-a*gradf;
%Now you can see the orbit towards minimum
plot(x(1),x(2),'o'); hold on
n=n+1;
end
n
x
eps
for instance with the current value a=.1
I get
n = 100
eps = 1.2308e-011
x =
1.0e-012 *
-0.2509
0.4688
That is I had to perform 100 iteration because my epsilon is still above the threshold. If I allow 200 iterations I get
n = 110
eps =
7.9705e-013
x =
1.0e-013 *
-0.1625
0.3036
Ie 110 iterations are sufficient.
Case of a general f
(ie not a quadratic form).
You can use, for instance, function handles , ie you define (before the while
)
foo = @(x) 6*x(1)^2+8*x(2)^2-3*x(1)*x(2);
foo_x = @(x) 12*x(1)-3*x(2);
foo_y = @(x) 16*x(2)-3*x(1);
then, in the while
you substitute
gradf = [foo_x(x);foo_y(x)];
f = foo(x);
PS for what concerns the while
cycle, please keep in mind that you keep on iterating while you are not satisfied with your precision ( eps>1e-12
) AND your total number of iteration is below a given threshold ( n<100
).
Consider also that you are working in finite precision : a numerical algorithm can never reach the analytic solution (ie what you have with infinite precision and infinite iterations), therefore, you always have to set a threshold ( eps
, which should be above the machine precision \\approx 1e-16
) and that is your 0
.
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