Multiple linear regression to fit data to a third degree polynomial equation with interaction terms
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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.
2 Comments
Star Strider
on 1 Apr 2014
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
on 1 Apr 2014
Accepted Answer
Tom Lane
on 2 Apr 2014
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.
3 Comments
Gautam
on 2 Apr 2014
Roger Stafford
on 2 Apr 2014
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
on 2 Apr 2014
More Answers (1)
Shashank Prasanna
on 1 Apr 2014
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:
Using the Statistics Toolbox you can do that by specifying the polyijf 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
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