Matrix is singular, RCOND=NAN

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dav 2013-3-17

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I am using the following code in a simulation and get a warning message. Is there any way to correct it?

       lb = 0.00001*ones(2,1);
       ub = 0.99998*ones(2,1);
       options =optimset('Display','off','LargeScale','off');
       coef = lsqlin(bC,by,[],[],[],[], lb ,ub ,[],options);

Warning: Matrix is singular, close to singular or badly scaled. Results may be inaccurate. RCOND = NaN. > In qpsub at 359 In lsqlin at 336 In seasonalarch1 at 330

Thanks

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Matt J 2013-3-17

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编辑：Matt J 2013-3-17

You are constructing C and y inadequately. You do not have enough linear equations, or they are not linearly independent enough, to uniquely identify the actual coef. Find additional and more informative equations and add them as new rows to C and y until cond(C) improves.

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Matt J 2013-3-22

编辑：Matt J 2013-3-22

在 MATLAB Online 中打开

As you can see ar coefficient is really small. this time it took the boundary value, but the actual value is 0.2 and sometimes it become equal to zero.

dav,

LSQLIN is giving you the correct values (for the problem data that you're providing it). Why do you think the correct value is 0.2? To see that LSQLIN is correct, you can generate contour plots of the function being minimized using the code below. You will see from the plots that as you descend through the isocontours, you converge precisely to the point on the boundary that LSQLIN is giving you.

    >> cc,cy
    cc =
      200.0000   27.5925
       27.5925   18.4254
    cy =
      107.8483
       11.7406
    [X,Y]=ndgrid(0.00001:.01:0.99998);
    xy=[X(:),Y(:)].';
    f=@(xy) sum((cc*xy).*xy)/2 -cy.'*xy; %objective function
    coef=[0.5392; 1e-5];  %alleged minimum
    fmin=f(coef);
    Z=reshape( f(xy)  , size(X));
    v=linspace(fmin+.05,fmin+5,10);
    contour(X,Y,Z,v); 
    title 'Contours of Objective Function'

could you also please let me know if I am using the fmincon code corresctly?

If you want to double check the results with FMINCON, you need to provide 2x1 initial guess. y will not be the correct size. A more sensible initial point would be C\y, so

coef = fmincon(fun,C\y,[],[],[],[], lb ,ub,[],options);

dav 2013-3-23

ok thanks..

one last question,

in an earlier comment you had asked why I said that the correct value is 0.2.

I said it because the data set we used for that lsqlin code was generated such that the ar coeff was 0.2.

If lsqlin works properly shouldn't it give values closer to 0.2 rather than very small values which course all these issues?

Thank you very much for you time, answers and comments!!!!

Matt J 2013-3-23

编辑：Matt J 2013-3-23

How close it will be to your simulated value depends on how much noise you have in your observation data.

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Youssef Khmou 2013-3-17

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hi,

The determinant of the matrix bC is close to zero, you have to alter bC .

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Walter Roberson 2013-3-17

How did you create bC ? There might not be a way to avoid the problem for that set of data.

dav 2013-3-17

编辑：dav 2013-3-17

在 MATLAB Online 中打开

I have a Time series data set (1000).

I took all the data corresponding to "day one" to one vector, day two to another vector and so on.. then I use lsqlin to estimate the parameters. However, the parameter estimates sometime are VERY CLOSE TO ZERO, but the actual parameters aren't that close to zero. this is why I am getting this warning.

Is there any way to correct it please?

CODE:
clc;
clear;
%warning off;
runs =100;
xxtt=[];
bxxtt=[];
p=1;
       yt1=[];
       yt2=[];
       yt3=[];
       yt4=[];
       yt5=[];
       xt1=[];
       xt2=[];
       xt3=[];
       xt4=[];
       xt5=[];
epsi=zeros(3000,1);
simsig=zeros(3000,1);
      a0(1)=0.5; a1(1)=0.2;
      for i=2:4
      a0(i)=0.1; a1(i)=0.3;
      end
      a0(5)=0.1;a1(5)=0.2;
for i = 1:3000
      m = mod(i,5);
      if  m==0
          m=5;
      end
                   if (i==1) 
                       simsig(i) = a0(m)/(1-a1(m));
                       s=(simsig(i))^0.5;
                       epsi(i) = normrnd(0,1) * s;
                   else                     
                       simsig(i) = a0(m)+ a1(m)*(epsi(i-1))^2;
                       s=(simsig(i))^0.5;
                       epsi(i) = normrnd(0,1) * s;
                   end
  end
epsi2=epsi.^2;
yt = epsi2(2001:3000);
ytinitial = epsi2(2001:3000);
sig = simsig(2001:3000);
ytlast=epsi(3000)^2;
ytlast2=epsi(2999)^2;
s=1;
for i=1:1000
       m = mod(i,5);
      if  m==0
          m=5;
      end
      if  m==1
          yt1(s) = yt(i);
          if i==1
          xt1(s)=0.00001;
          else
          xt1(s) = yt(i-1);
          end
      elseif m ==2
          yt2(s) = yt(i);
          xt2(s) = yt(i-1);
      elseif m==3
          yt3(s)= yt(i);
          xt3(s) = yt(i-1);
      elseif m==4
          yt4(s)= yt(i);
          xt4 (s)= yt(i-1);
      elseif m==5
          yt5(s)= yt(i);
          xt5(s)= yt(i-1);
      end
      s=s+1;
end
yt1(yt1==0) = [];
yt2(yt2==0) = [];
yt3(yt3==0) = [];
yt4(yt4==0) = [];
yt5(yt5==0) = [];
xt1(xt1==0) = [];
xt2(xt2==0) = [];
xt3(xt3==0) = [];
xt4(xt4==0) = [];
xt5(xt5==0) = [];
yt1=yt1';
yt2=yt2';
yt3=yt3';
yt4=yt4';
yt5=yt5';
xt1 = xt1';
xt2 = xt2';
xt3 = xt3';
xt4 = xt4';
xt5 = xt5';
ytt = [ yt1, yt2, yt3, yt4, yt5];
xtt = [ xt1, xt2, xt3, xt4, xt5];
%xxtt=[xxtt;xtt];
for m = 1:5
      y = ytt(:,m);
      x= xtt(:,m);
      len = length(y);
      C = zeros(len,p);
      C(:,1) = 1; % The first column is for a0
      %for i = 1:p %Then we create shifted columns ( p in number ) for a 
       %  C(1+i:len,1+i) = x(1:len-i,1);     
      %end
      C = [C x];
       lb = [0.00001 0.00001]';
       ub = [0.99998 0.99998]';
       options =optimset('Display','off','LargeScale','off');
       coef = lsqlin(C,y,[],[],[],[], lb ,ub,[],options);
      alpha0(m) = coef(1);
      ar1(m) = coef(2);
  end