Warning: Rank deficient, rank

Discussion in 'MATLAB' started by Marlene, Jan 3, 2011.

  1. Marlene

    Marlene Guest

    Dear Matlab users,

    I am running into a recurrent problem that I would like to understand and, if possible, solve.

    I am using the '\' function to estimate 21 model parameters( corresponding to the combination of 2 predictors variables up to the order 5).
    I have about 130 experiments (hence 130 {y,x1,x2}) to do this regression.

    If I run it up to the 4th order, I model runs smoothly and the obtained model is fairly satisfying.

    However, I need a 5th order model (to be coherent with previous work from coworkers).
    The model is then quite far off. The first 3 model parameters + another one are given equal to zero and I get the message:Warning: Rank deficient, rank 17 (I guess the 17 corresponds to the 21 parameters I asked for minus the 4 equal to zero)
    I don't and won t have additionnal experiments and I really would like to get a 5th order model.
    Is there a way to go around?

    Thanks a lot
     
    Marlene, Jan 3, 2011
    #1
    1. Advertisements

  2. Can you explain what you are doing in more detail? I assume that you have a some sort of a least-squares Vandermonde system you are trying to solve?

    Is your model a polynomial? If so, you can use preexisting data fitting tools such as Matlab's polyfit() function.

    If you have your heart set on using a matrix representation, take a look at the singular value decomposition of the matrix with svd() to see what its rank is. It would be unusual to have a 130x5 matrix that is rank deficient. I guess more information is needed.
     
    Greg von Winckel, Jan 3, 2011
    #2
    1. Advertisements

  3. Marlene

    proecsm Guest

    I agree, more information is needed. Post some sample code
     
    proecsm, Jan 3, 2011
    #3
  4. Marlene

    Marlene Guest

    Hi again,

    My model is indeed polynomial and my code looks like:

    %%%%%%%%%%

    DesignMat=[ones(Nbpres*Nbtemp,1),PhP,TC,PhP.^2,PhP.*TC,TC.^2,PhP.^3,(PhP.^2).*TC,PhP.*(TC.^2),TC.^3,PhP.^4,(PhP.^3).*TC,(PhP.^2).*(TC.^2),PhP.*(TC.^3),TC.^4,PhP.^5,PhP.^4.*TC,(PhP.^3).*(TC.^2),(PhP.^2).*(TC.^3),PhP.*(TC.^4),TC.^5]

    Fit_Params=DesignMat\Exp_values;
    %%%%%%%%%%
    where:
    Dim (Exp_values)=132*1
    Dim(DesignMat)=132*21

    If I run this code, I get:

    Warning: Rank deficient, rank = 17, tol = 6.9333e+002

    and
    Fit_Params =

    0
    0
    0
    0.005289404149152
    -0.001747748358411
    0
    -0.000009531368644
    0.000001359824964.........up to 21 parameters

    I am not expecting the first 3 parameters to be equal to zero.
    Also, althougth the model obtained is not completely off, it is not very satisfying.
    When I compare it to the experimental values of pressure it is derived from, I get en error of 60 Hpa.
    As a comparison, when I run this same code with a 4th order polynom, I do not get the warning message and the precision of the model is of about 0.02 Hpa.

    So it really looks to me like matlab does not handle the regression with that many paramaters.
    But again, thoses very same values have been modeled by a 5th order polynom in the past and it would be much easier for me to do just the same.

    I hope it makes more sens now..

    thanks
     
    Marlene, Jan 4, 2011
    #4
  5. Marlene

    Greg Heath Guest

    The second dimension is much larger than 5.
    Just the second order model terms take up
    6 columns: 1,x1,x2,x1^2,x2^2,x1*x2.

    If the matrix is rank deficient the solution
    is not unique. Backslash automatically
    zeros out some of the terms according to
    pivot ratio info. Instead, you can choose
    to keep all of the terms by using pinv.

    Alternatively, if backslash excluded 4
    terms, you can probably choose the alternative
    of excluding the 4 fifth order interaction
    terms x1^m*x2^n, 1<= m,n <=4, m+n = 5.

    Hope this helps.

    Greg
    ..ther terms that exclude the same number of terms
    arbitrarily. I would begin by eliminating
    the higher order interaction terms

    Hope this helps

    Greg.
     
    Greg Heath, Jan 4, 2011
    #5
  6. Marlene

    Marlene Guest

    Thanks Greg,

    I understand better what is going on and, also it would not have been my first choice,
    I will stick to a 4th order model.

    Your comment about the solution not being unique is very valuable and will probably save me a lot of time when interpreting some of the results.

    Thanks a lot
     
    Marlene, Jan 5, 2011
    #6
    1. Advertisements

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.