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I'm working on a model of a behavior of some numerical methods. I need the model to take two input variables - number of processes started and size of the problem.

Number of processes I've modeled by inverse regression for every single problem size I've measured, so I have function described by formula $A + \frac{B}{nprocs}$ for all those sizes.

Size, degrees of freedom and $A$, $B$ coefficients: enter image description here


My question is

How can I describe the dependency of $A$ and $B$ coefficients on number of degrees of freedom? I'm not very experienced with regression analysis, so I don't know about any method how to do this.


My attempt

I suppose, that both $A$ and $B$ coefficients can be approximately described as a geometric sequence, e.g.:

$$ \begin{align} x*q^{3993}_1 &= 0.066181\\ y*q^{3993}_2 &= 0.548915\\ x*q^{7623}_1 &= 0.095864\\ y*q^{7623}_2 &= 1.076260\\ \vdots \end{align} $$

Is this a right idea? Or is it completely wrong (i.e. geometric sequence can't describe sequence of values like that?)?

And if it is a right idea, how is it possible to choose $x,y$ and $q$ coefficients, so that resulting formula will fit as much as possible for all the data (I mean something like linear/polynomial/inverse regression for curve fitting etc.)?


Raw data

Here is part of raw data from my measurements (from size $10\times 10\times 10$ to $10\times 10\times 50$):

enter image description here

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In my opinion, the simplest model could be of the form $$A=\alpha \,\text{DOF}^\beta \qquad,\qquad B=\gamma\,\text{DOF}^\delta $$ which means that, if you plot $\log(A)$ and $\log(B)$ as a function of $\log(\text{DOF})$, you should observe linear trends.

This appears to be very true for $B$ but much less for $A$.

Accepting such models and performing nonlinear regression, you should arrive to $$A=1.943\times 10^{-3} \,\times \text{DOF}^{\,0.431}\qquad (R^2=0.988423)$$ $$B=9.621\times 10^{-5} \,\times \text{DOF}^{\,1.045}\qquad (R^2=0.999958)$$ The fit is very good for $B$ but quite poor for $A$ (as almost expected just looking at the data) even if the global trend is respected.

Now, at this point, what I should is to reuse all original data points and fit the full model as $$Y=\alpha \,\text{DOF}^\beta +\gamma\frac{\,\text{DOF}^\delta}{\text{nprocs}} $$ since we have good parameters to start with. This could be much better.

Edit

After our last exchanges in comments, let me precise : you have $10$ values of $\text{DOF}$; for each of them, you have $8$ values of $\text{nprocs}$. This means that you have $10\times 8=80$ data points on the form $$(\text{nprocs}_i,\text{DOF}_i,Y_i)\qquad , \qquad (i=1,2,\cdots,80)$$

Based on that, you perform now a complete regression based on model $$Y=\alpha \,\text{DOF}^\beta +\gamma\frac{\,\text{DOF}^\delta}{\text{nprocs}} $$ using all data.

I did it using the partial table you gave in your edit and found $$Y=0.00238917 \, \text{DOF}^{0.413463}+0.0000727346\frac{\, \text{DOF}^{1.07399}}{\text{nprocs}}$$ to which corresponds $R^2=0.981991$.

This last fit is more than likely better since all information was treated globally.

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  • $\begingroup$ Thank you very much for your answer. Just one question - what do you mean by " reuse all original data points and fit the full model"? I understand, that you put the whole formula together from $A+\frac{B}{nprocs}$ and $\alpha DOF^\beta$, but what would you do after that? Simply substitute computed $A$ and $B$ into the formula? I'm sorry for my silly question, but I've never used this before. $\endgroup$
    – Eenoku
    Nov 15 '16 at 18:14
  • $\begingroup$ You made a fit to get $A$ and $B$ for each $DOF$; this was using data. Now, forget $A$ and $B$ for ever. Use all the original data points and adjust $\alpha,\beta,\gamma,\delta$ in a single run. Is this better explained ? If it is not, you tell me : your job is to learn and my duty is to find the proper way to make you understanding. By the way, there is no stupid questions but answers can be Cheers. $\endgroup$ Nov 15 '16 at 19:05
  • $\begingroup$ I wouldn't call it your duty in any way here on StackExchange, but thank you very much for your willingness to explain me the topic. I understand the principle now, but I somehow don't see the way I could "calibrate" the formula (i.e. adjust its coefficients) with the measured data. I've added now part of my measured data into the question as an example. $\endgroup$
    – Eenoku
    Nov 15 '16 at 21:58
  • $\begingroup$ By "complete regression" you mean least-squares method? So, if I understood you well, you'll assemble system of 4 variables and 80 equations (or 50 eq. if you have only partial data) and then you'll minimize it. Is it right? $\endgroup$
    – Eenoku
    Nov 16 '16 at 20:00
  • $\begingroup$ Tomorrow I'll add my next attempt :-) $\endgroup$
    – Eenoku
    Nov 16 '16 at 20:36

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