# fminunc crashes matlab if large number of input variables used

3 views (last 30 days)
Satyajit Ghosh on 23 Sep 2021
Edited: Matt J on 8 Oct 2021
I am working on a problem where input varibales with respect to which objective function (nonlinear) needs to be optimized is elements of a n*3 matrix where n is from 2000-3000. Here under I am providing essentional portion of the script.
clear all
clc
%n=2000;
n=3;k=3;normalized=0;periodic=0;
m=n+k+1;
[t,Range]=UniformKnotVector(k,n,periodic,normalized);
c=load('Data.txt');
p1=c(:,18:20);
p2=c(:,21:23);
v=zeros(size(c,1),3);
f=@(A)0;
for i=1:1:size(c,1)
disp(i)
v(i,1)=(p2(i,1)-p1(i,1))/(sqrt(((p2(i,1)-p1(i,1)).^2)+((p2(i,2)-p1(i,2)).^2)+((p2(i,3)-p1(i,3)).^2)));
v(i,2)=(p2(i,2)-p1(i,2))/(sqrt(((p2(i,1)-p1(i,1)).^2)+((p2(i,2)-p1(i,2)).^2)+((p2(i,3)-p1(i,3)).^2)));
v(i,3)=(p2(i,3)-p1(i,3))/(sqrt(((p2(i,1)-p1(i,1)).^2)+((p2(i,2)-p1(i,2)).^2)+((p2(i,3)-p1(i,3)).^2)));
x=c(i,24)*((n-k+2)/127.59);
N11=BSpline1(k,t,x,n,periodic);
N22=BSpline2(k,t,x,N11,periodic);
N33=BSpline3(k,t,x,N22,periodic);
fx=@(A)0;fy=@(A)0;fz=@(A)0;
for j=1:1:(n+1)
fx=@(A)(fx(A)+A(j,1)*N33(j,1));
fy=@(A)(fy(A)+A(j,2)*N33(j,1));
fz=@(A)(fz(A)+A(j,3)*N33(j,1));
end
f=@(A)(f(A)+((p1(i,1)-fx(A))^2+(p1(i,2)-fy(A))^2+(p1(i,3)-fz(A))^2-(((p1(i,1)-fx(A))*v(i,1))+((p1(i,2)-fy(A))*v(i,2))+((p1(i,3)-fz(A))*v(i,3)))^2));
end
A0=ones(n+1,3);
[Aopt,fopt]=fminunc(f,A0);
Script is working fine for smaller values of n (e.g., 3,4,5,...). With larger values (e.g., n=15,16,...) it is gradually taking more time. Finally for the actual n value for which I am interested (n=2000) script leads crash of MATLAB without any error.
What do I do?
##### 0 CommentsShowHide -1 older comments

Sign in to comment.

### Accepted Answer

Matt J on 23 Sep 2021
Edited: Matt J on 8 Oct 2021
Well, the practice you have of cumulatively nesting anonymous functions inside one another is asking for trouble. You should at the very least convert to the whole thing to a non-anonymous function like below. Also, you are missing opportunities to do matrix/vector operations in place of for-loops.
for i=1:1:size(c,1)
v(i,1)=(p2(i,1)-p1(i,1))/(sqrt(((p2(i,1)-p1(i,1)).^2)+((p2(i,2)-p1(i,2)).^2)+((p2(i,3)-p1(i,3)).^2)));
v(i,2)=(p2(i,2)-p1(i,2))/(sqrt(((p2(i,1)-p1(i,1)).^2)+((p2(i,2)-p1(i,2)).^2)+((p2(i,3)-p1(i,3)).^2)));
v(i,3)=(p2(i,3)-p1(i,3))/(sqrt(((p2(i,1)-p1(i,1)).^2)+((p2(i,2)-p1(i,2)).^2)+((p2(i,3)-p1(i,3)).^2)));
x=c(i,24)*((n-k+2)/127.59);
N11(:,i)=BSpline1(k,t,x,n,periodic);
N22(:,i)=BSpline2(k,t,x,N11,periodic);
N33(:,i)=BSpline3(k,t,x,N22,periodic);
end
[Aopt,fopt]=fminunc(@(A)myObjective(A,N33,p,v),A0);
function f=myObjective(A,N33,p,v)
fxyz=N33.'*A;
D=p1-fxyz;
f=sum(D.^2-(D.*v).^2);
end
##### 3 CommentsShowHide 2 older comments
Satyajit Ghosh on 24 Sep 2021
I checked the function value in two cases for a specific intial estimate matrix A0. The values are same in both cases.

Sign in to comment.

R2018a

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!