Multiple linear regression to fit data to a third degree polynomial equation with interaction terms

Hello,

I have data for two independent variables and one dependent variable (obtained from experiment). I need to fit this data using linear regression to a 10 coefficient third degree polynomial equation - for the engineers among you, this is the standard equation for specifying refrigeration compressor performance. This can be done quite easily in EES by entering data into a table and selecting the option "Linear Regression" from the Tables menu. How do I achieve the same in Matlab? My equation is of the form X = C1 + C2.(S) + C3.(D) + C4·(S2) + C5 · (S·D) + C6 · (D2 ) + C7 · (S3 ) + C8 · (D·S2 ) +C9 · (S·D2 ) + C10 · (D3 ) Here, X is the dependent variable and S and D are the independent variables. The numbers next to S and D indicate the power to which they are raised. C1 to C10 are the coefficients that need to be calculated.

Using the 'regress' function gives me a constant of 0, and warns me that my design matrix is rank deficient. Using the curve fitting toolbox (cftool - polynomial option) gives me ridiculous values for the coefficients (p00 = -6.436e15). When I try to input a custom equation in the cftool, it is switching to non-linear regression and asks me to input guess values for the coefficients, which I don't want to do. What other functions are available that I might use to perform this regression and how do I implement them. Any suggestions/ help/ recommendations would be greatly appreciated.

Thanks very much.

Gautam.

[Star Strider]

Please post your regression equation and 15-10 rows of your X, S and D data.

It’s difficult to determine what the problem may be without that information.

[Gautam]

>> X = [-10 101; -10 114; -10 129; -10 144; -5 96; -5 106; -5 119; -5 133; 5 92; 5 97; 5 108; 5 122];

>> Y = [72.03; 67.90; 62.77; 57.55; 88.53; 84.70; 79.91; 73.73; 123.43; 120.91; 115.44; 107.83];

>> x1 = X(:,1);

>> x2 = X(:,2);

>> cftool

I used a custom equation, Y = f(x1, x2) and the custom equation is:

a + b*x1 + c*x2 + d*(x1^2) + e*(x1*x2) + f*(x2^2) + g*(x1^3) + h*(x1^2*x2) + i*(x1*x2^2) + j*(x2^3)

What turned up in the Results box was this:

Complex value computed by model function, fitting cannot continue. Try using or tightening upper and lower bounds on coefficients.

Any help?

Answer 1

[Tom Lane]

You have only three distinct values in the first column of your X matrix, so you won't be able to estimate constant, linear, quadratic, and cubic effects for that column. Here's how you can use all terms in the cubic model except x1^3:

fitlm(X,Y,[0 0;1 0;0 1;1 1;2 0;0 2;0 3;1 2;2 1])

Alternatively, if you know that a third order term is appropriate, you will need to collect some data at another value of x1.

[Gautam]

Thank you very much Mr. Lane, for the comment. I certainly need to have the x1^3 term - I am not allowed to modify the equation - it is the AHRI 540-2004 Standard. And, the data is indistinct because it is a measurement of evaporator temperature (at the suction of a compressor) and that is how compressor data is taken and performance evaluated. Either the evaporator or the condenser temperature will have to be held constant, while the other is measured. Is there any way I can get around that?

[Roger Stafford]

To Gautam: Tom has told you the essential difficulty with your problem as you have presented it. The difficulty is not with your ten-coefficient third degree polynomial. It is with the shortage of data to be fitted to it. The quantity you called 'x1' occurs in only three distinct values: -10, -5, and +5. That is just not enough, given the complexity of your polynomial.

Look at it this way. Suppose your problem involves only a single variable and is approximated with a third degree polynomial with four adjustable coefficients. It is obvious that three data values of that single variable would not be enough to uniquely determine the four coefficients. You would have only three equations and four unknowns, and there are infinitely many coefficient combinations that would give you an exact fit to that limited data.

In other words, Gautam, your data is too easy to fit as far as the x1 variable is concerned. You need much more variation in x1 to produce a meaningful set of ten coefficients. Actually you are also not providing enough variation in the other variable, x2. You have only four values of x2 for each value of x1 and that is insufficient to give enough information to determine the coefficients with sufficient reliability. It is too easy to fit. You really need much greater variation in both variables than you have provided. That is what is behind the messages you have received about "rank deficiency". Ideally the messages ought to have complained with the declaration: "Data! Data! Give me more data! I am starving from too little data!"

[Gautam]

Thank you very much Mr. Stafford for clearing that up for me. I'm a newcomer to Matlab and could not grasp completely the meaning behind the warnings. I tried out fitlm with a much bigger data set and it did give me the exact same values EES did, which has around 2% variation with the published coefficients. I cannot thank you enough.

Answer 2

[Shashank Prasanna]

There are numerous ways to do this. The most easiest of them all is to use the Curve Fitting Tool with the correct options. Try this link for help:

http://www.mathworks.com/help/curvefit/polynomial.html

Using the Statistics Toolbox you can do that by specifying the polyijf modelspec:

http://www.mathworks.com/help/stats/fitlm.html#inputarg_modelspec

Another way is to solve a linear system in MATLAB. Assuming X, S and D are vectors in your MATLAB workspace. Create random data and compute the coefficients in C:

S = randn(100,1);
D = randn(100,1),
X = randn(100,1);

M = [ones(length(S),1), S, D, S.^2, S.*D, D.^2, S.^3, D.*S.^2, S.*D.^2, D.^3]
C = M\X

[Gautam]

Shashank, thanks for the prompt response. In addition to the cftool I posted above, I also tried the below:

Using the same data (X and Y), I used the fitlm function with 'poly33' as the argument:

mdl = fitlm(X,Y,'poly33')

It did work, but the coefficients it gave me are not the coefficients I should be getting. The actual values of the coefficients are:

beta = [119.61 4.37 -0.39 -0.04 0.09 -0.01 0.00 0.02 0.00 0.00];

However, when I used fitlm, the value of the constant (119.61 above) is obtained as 0, which should not be the case.

As for the last method you mentioned (using the backslash operator), it gave me a warning that the rank of my matrix was deficient and still gave me wrong values for the values of the coefficients (and different from those obtained using the other functions).

Some more suggestions please?

[Gautam]

I'm posting some of the attempts I made below:

X = [-10 101; -10 114; -10 129; -10 144; -5 96; -5 106; -5 119; -5 133; 5 92; 5 97; 5 108; 5 122];

>> Y = [72.03; 67.90; 62.77; 57.55; 88.53; 84.70; 79.91; 73.73; 123.43; 120.91; 115.44; 107.83];

>> x1 = X(:,1);

>> x2 = X(:,2);

>> mdl = fitlm(X,Y,'poly33')

Warning: Regression design matrix is rank deficient to within machine precision. > In TermsRegression>TermsRegression.checkDesignRank at 98 In LinearModel.LinearModel>LinearModel.fit at 868 In fitlm at 117

mdl =

Linear regression model: y ~ 1 + x1^2 + x1*x2 + x2^2 + x1^3 + (x1^2):x2 + x1:(x2^2) + x2^3

Estimated Coefficients: Estimate SE tStat pValue _________ ________ _______ ______

    (Intercept)              0             0          NaN         NaN
    x1                 -10.143        5.3752       -1.887     0.19979
    x2                 -0.3125        1.1432     -0.27335     0.81022
    x1^2                5.5413        1.6852       3.2882    0.081361
    x1:x2            0.0043522      0.026071      0.16694     0.88277
    x2^2            2.6105e-05      0.010348    0.0025228     0.99822
    x1^3               0.55158       0.16763       3.2905     0.08126
    x1^2:x2        -5.4944e-05    0.00021022     -0.26136     0.81827
    x1:x2^2        -8.3639e-05    0.00012017     -0.69599     0.55844
    x2^3             -4.12e-06     3.109e-05     -0.13252      0.9067

Number of observations: 12, Error degrees of freedom: 3 Root Mean Squared Error: 0.165 R-squared: 1, Adjusted R-Squared 1 F-statistic vs. constant model: 2.76e+04, p-value = 3.3e-07

Attempt 2:

 A = [1  -10  101  100  -1010  10201  -1000  10100  -102010  1030301
1  -5  96  25  -480  9216  -125  2400  -46080  884736
1  5  92  25  460  8464  125  2300  42320  778688
1  -10  114  100  -1140  12996  -1000  11400  -129960  1481544
1  -5  106  25  -530  11236  -125  2650  -56180  1191016
1  5  97  25  485  9409  125  2425  47045  912673
1  -10  129  100  -1290  16641  -1000  12900  -166410  2146689
1  -5  119  25  -595  14161  -125  2975  -70805  1685159
1  5  108  25  540  11664  125  2700  58320  1259712
1  -10  144  100  -1440  20736  -1000  14400  -207360  2985984
1  -5  133  25  -665  17689  -125  3325  -88445  2352637
1  5  122  25  610  14884  125  3050  74420  1815848
];

b = A\Y Warning: Rank deficient, rank = 9, tol = 4.463946e-09.

b =

         0
  -62.6154
   18.9421
  -35.6208
    2.9627
   -0.0958
   -3.8184
   -0.0236
   -0.0136
    0.0001

Attempt 3:

 sf = fit([x1 x2],Y, 'poly33', 'Robust', 'Bi')

Warning: Equation is badly conditioned. Remove repeated data points or try centering and scaling. > In Warning>Warning.throw at 30 In fit>iLinearFit at 690 In fit>iFit at 396 In fit at 108 Warning: Iteration limit reached for robust fitting. > In Warning>Warning.throw at 30 In fit>iRobustLinearFit at 796 In fit>iFit at 404 In fit at 108

 Linear model Poly33:
     sf(x,y) = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p30*x^3 + 
                    p21*x^2*y + p12*x*y^2 + p03*y^3
     Coefficients (with 95% confidence bounds):
       p00 =  -2.736e+14  (-6.205e+17, 6.199e+17)
       p10 =  -2.736e+13  (-6.205e+16, 6.199e+16)
       p01 =      0.3112  (-346.1, 346.7)
       p20 =   1.094e+13  (-2.48e+16, 2.482e+16)
       p11 =   -0.009459  (-5.569, 5.551)
       p02 =   -0.005727  (-3.308, 3.296)
       p30 =   1.094e+12  (-2.48e+15, 2.482e+15)
       p21 =  -0.0001193  (-0.4058, 0.4055)
       p12 =  -1.846e-05  (-0.02545, 0.02542)
       p03 =   1.358e-05  (-0.01007, 0.01009)

Attempt 4:

 sf = fit([x1 x2],Y, 'poly33', 'Robust', 'Bi','Normalize','on')
Warning: Equation is badly conditioned. Remove repeated data points or try centering and scaling. 
> In Warning>Warning.throw at 30
  In fit>iLinearFit at 690
  In fit>iFit at 396
  In fit at 108 
Warning: Iteration limit reached for robust fitting. 
> In Warning>Warning.throw at 30
  In fit>iRobustLinearFit at 796
  In fit>iFit at 404
  In fit at 108 

 Linear model Poly33:
     sf(x,y) = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p30*x^3 + 
                    p21*x^2*y + p12*x*y^2 + p03*y^3
       where x is normalized by mean -3.333 and std 6.513
       and where y is normalized by mean 113.4 and std 16.34
     Coefficients (with 95% confidence bounds):
       p00 =   2.802e+14  (-4.173e+15, 4.734e+15)
       p10 =    1.15e+15  (-1.712e+16, 1.942e+16)
       p01 =      -6.475  (-15.32, 2.371)
       p20 =       1.005  (-3.659, 5.67)
       p11 =      -1.507  (-6.289, 3.275)
       p02 =     -0.3495  (-5.084, 4.385)
       p30 =  -8.362e+14  (-1.413e+16, 1.245e+16)
       p21 =     -0.2157  (-8.185, 7.754)
       p12 =     -0.6978  (-12.13, 10.73)
       p03 =     -0.3148  (-7.039, 6.409)

Is there anything I haven't yet tried out or any mistakes I made? Anything at all would be helpful. Thank you.