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I have a question regarding process control with the use of multivariate regression.

The setup is as follows: say we have some data, representing the results of a plant process. Specifically, several continuous variables $\{x_1,\ldots,x_n\}$ represent inputs of the process (and we can control directly only $x_1, x_2, x_3$) and there is a continuous dependent variable $y$ being the result of this process. Our goal is to model the dependence and to control $y$ with the help of $x_1, x_2, x_3$. So, given real data, to model this dependence we can build the multiple regression with $y=\sum \alpha_i x_i$.

However, to control $y$ with $x_1, x_2, x_3$, it seems better to "reverse" the problem, i.e. to build multivariate regression: the independent variables (inputs) are now $\{x_4,\ldots,x_n,y\}$ and we predict the values of $x_1, x_2, x_3$ to be applied in order to obtain $y$.

And now for the main part: suppose there are some constraints on the inputs $x_1, x_2, x_3$, e.g. we may be forced to put $x_3$ or $\frac{x_1}{x_2}$ equal to some values. In other words, sometimes we might not be able to freely adjust the inputs in order to obtain given value of $y$, but still we want to be as close to $y$ as possible, so it seems like an optimization problem. The question: how such constraints could be introduced in the model?

EDIT:

Let me rephrase the problem by means of a particular constraint $x_1/x_2=u$. Given the model $(x_1, x_2, x_3) = f(x_4,\ldots,x_n, y)$, it has been obviously trained on samples with various values of $x_1/x_2$; for simplicity, suppose that a half of samples has $x_1/x_2\approx 1$ and the other half $x_1/x_2\approx 2$, and these ratios were set during production due to some process-specific constraints, such as availability of quantities $x_1, x_2$ through time.

Now, let's assume the tuple $(x_4,\ldots,x_n, y)$ is such that the model gives $f(x_4,\ldots,x_n, y)=(x_1, x_2, x_3)$ with $x_1/x_2\approx 1$, but due to practical reasons we insist on input $(x'_1, x'_2, x'_3)$ with $x'_1/x'_2=2$.

In that case, can we find $(x'_1, x'_2, x'_3)$ that is optimal, i.e. such that $(x'_1, x'_2, x'_3, x_4, \ldots, x_n)$ gives $y'$ that minimizes $(y-y')^2$? A wild guess would be to use a multiple regression $y=\sum \alpha_i x_i$ and then minimize an objective function $(y-y')^2$ w.r.t. $(x_1,x_2,x_3)$ with $x_1/x_2=2$, or perhaps there is another useful strategy?

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1 Answer 1

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Sounds like you might already have a model $(x_1,x_2,x_3)=f(x_4,\ldots,x_n,y;\hat\theta)$, so I'll just talk about the constraints.

If you have a constraint like $x_1/x_2 = u = \text{const}$, then you actually only need to model $(x_2,x_3) = f(\ldots)$, since you can recover $x_1$ with $x_1=ux_2$. Similarly if $x_3$ is fixed then this is even easier.

If you're constraints are more complicated than this (i.e. nasty and non-linear), you may be able to parameterize your problem, but failing that you could try nonlinear modelling with something like JuMP https://jump.dev/

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  • $\begingroup$ If I got it right, you propose to train a new model each time we face a constraint. My idea is a little bit different: suppose the multivariate regression for $(x_1, x_2, x_3)$ has been trained on data with various ratios $x_1/x_2$. Then, based on that single model, I would like to find an optimal $(x_1, x_2, x_3)$ given some particular $x_1/x_2=u=\mathrm{const}$. Does this make sense? $\endgroup$
    – krzysiekb
    Aug 12, 2022 at 7:17
  • $\begingroup$ I was thinking you would bake all of your constraints into a single model, not train a new one with each constraint. If you are having to add new constraints frequently or constraints are changing frequently then I suppose that may not be practical. $\endgroup$
    – BenA
    Aug 12, 2022 at 7:42
  • $\begingroup$ I edited the question, hope it narrows down the problem. $\endgroup$
    – krzysiekb
    Aug 18, 2022 at 12:07

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