The error 'Matrix dimensions must agree' during optimization.How can it be solved?

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Nitin Samuel
Nitin Samuel 2013-1-8
I have the following code where 'Ymeas' is an array of measured values and must be fitted with estimated values-'Yest' meanwhile optimizing K1 using an ode.The objective function is the sum of squares of difference b/w Ymeas and Yest. Ymeas and Yest are 11X2 arrays. It optimizes for a while and then the following error is shown:
'Matrix dimensions must agree.'
function optimize
[K1] = deal(3.5);
Ymeas = [
1 1
0.8696 0.8696
0.7692 0.7692
0.6897 0.6897
0.625 0.625
0.5714 0.5714
0.5263 0.5263
0.4878 0.4878
0.4545 0.4545
0.4255 0.4255
0.4 0.4 ];
T=[0:10:100];
A0=Ymeas(1,1);
B0=Ymeas(1,2);
figure;
plot(T,Ymeas);
hold on;
h = plot(T,Ymeas,'k','linewidth',2);
minERR = Inf;
opts = optimset('fminunc');
opts.LargeScale = 'off';
Xest = fminunc(@(X)objfun(X),[0],opts);
Xest = num2cell(Xest);
[K1] = deal(Xest{:}),
function ERR = objfun(X);
X = num2cell(X);
[K1] = deal(X{:});
[T,Yest] = ode45(@(t,y)[-K1*y(1)*y(2);
-K1*y(1)*y(2)],[T],[A0,B0]);
ERR = sum((Ymeas(:) - Yest(:)).^2);
if ERR < minERR
minERR = ERR;
for n = 1:2; set(h(n),'Ydata',Yest(:,n)); end
drawnow;
end;
end;
end
  2 个评论
Walter Roberson
Walter Roberson 2013-1-8
What are you expecting from your line
[K1] = deal(X{:});
The only point of that I can see at the moment is to issue an error message if X was not a scalar.
Nitin Samuel
Nitin Samuel 2013-1-8
oh..but even without that , how can there be a problem with the matrix dimensions? Dimensions of both Ymeas and Yest are constant and only the values of Yest change...So why exactly is there an issue with the dimensions? Thanks...

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Teja Muppirala
Teja Muppirala 2013-1-8
The problem occurs when K1 is negative. The ODE solver blows up before it can integrate to all your time values. You should specify a constraint that forces K1 to be >= 0. For example, you might use FMINCON instead of FMINUNC.
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Alan Weiss
Alan Weiss 2013-1-8
Take a look at the syntax for bounds. You should write
Xest = fmincon(@(X)objfun(X),[0,0],[],[],[],[],[0,0],[],[],opts);
Or pick a better point than [0,0] for x0.
Alan Weiss
MATLAB mathematical toolbox documentation

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