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Edited

Given a computer model $F:\mathbb{R}^3 \to \mathbb{R}$, with inputs $x, w$ and $z$, and output $y=F(x,w,z)$, where for any input, we are able to evaluate the output, my goal is to tune the inputs to achieve a certain state change (change in output).

More precisely:

Goal: Given an input $x_0, w_0, z_0$ and an output $y_0$, I want to increase my output to a state $y_1$.

Question: How to change my inputs to achieve the new state $y_1$.

My approach: \begin{align} dy &= \frac{\partial F}{\partial x}(x_0, w_0, z_0)dx + \frac{\partial F}{\partial w}(x_0, w_0, z_0) dw + \frac{\partial F}{\partial z}(x_0, w_0, z_0)dz \\ y_1-y_0 &= \frac{\partial F}{\partial x}(x_0, w_0, z_0) (x_1 - x_0) + \frac{\partial F}{\partial w}(x_0, w_0, z_0) (w_1 - w_0)+ \frac{\partial F}{\partial z}(x_0, w_0, z_0) (z_1 - z_0) \end{align} Since $x_0, w_0, z_0, y_0, y_1$ are given, and assuming we are able to compute from the model $F$ the numerical approximation of the derivative of $F$ with respect to each input at the point $(x_0, w_0, z_0)$, the new values $x_1, w_1, z_1$ then should lie on the plane given by the second equation above.

I am also aware that I made a linear approximation that is only exact when $F$ is linear with respect to the inputs.

I am going in the right direction in formulation the problem ? If not, any suggestions?

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    $\begingroup$ This is a (possibly) nonlinear equation. There are many methods available to solve them: Newton iteration, Levenberg-Marquardt method, fixed-point iteration, pseudo-transient continuation, etc. If you have a good initial guess and you can evaluate the Jacobians, I suggest Newton's method. Otherwise (no good initial guess), Levenberg-Marquardt might work. $\endgroup$
    – cos_theta
    Oct 15 at 11:59
  • $\begingroup$ Is $y$ and input or an output? The first line is confusing in this regard, as is the statement "pair of inputs" when your function seems to be taking three arguments. $\endgroup$ Oct 16 at 1:14
  • $\begingroup$ @WolfgangBangerth Sorry for the confusion ... I edited the question $\endgroup$
    – outlaw
    Oct 16 at 8:19
  • $\begingroup$ @cos_theta The derivatives are evaluated numerically and hence the equation is always linear in $x_1, w_1, z_1$ .. $\endgroup$
    – outlaw
    Oct 16 at 14:48
  • $\begingroup$ the problem, as you have formulated it now, does not have a unique solution. What is this for exactly, that might help us suggest some extra conditions. $\endgroup$ Oct 16 at 16:44

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