# Multivariate regression and process control

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?

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 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? Aug 12, 2022 at 7:17