Hi, there. I was wondering if someone could help me with something.
I have 3 data matrices, of which I show you the contour plots (these I show would be just an example, these are from functions but the real ones are not from any kind of functions).
What I need is to be able to define a region of minimum optimums by weighting the matrices, that is, that not all of them have the same importance. The thing is that I only find things related to functions and they are not useful to me...
Thank you very much for everything.
Z1.jpg
Z2.jpg
Z3.jpg

4 件のコメント

Turlough Hughes
Turlough Hughes 2020 年 2 月 2 日
imregionalmin may help.
Alejandro Fernández
Alejandro Fernández 2020 年 2 月 2 日
My data is not an image nor does it allow me to weigh...
Mohammad Sami
Mohammad Sami 2020 年 2 月 2 日
What if you just add the 3 weighted matrix and then use the minimum function
Z1 = load('Z1.mat');
Z2 = load('Z2.mat');
Z3 = load('Z3.mat');
w1 = 1;
w2 = 1;
w3 = 1;
Z = w1 * Z1.Z + w2 * Z2.Z + w3 * Z3.Z;
[,idx] = min(Z);
Alejandro Fernández
Alejandro Fernández 2020 年 2 月 2 日
Ok, I'll try, thank you so much.

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Thiago Henrique Gomes Lobato
Thiago Henrique Gomes Lobato 2020 年 2 月 2 日

1 投票

Could you explain a little more the problem you're having? In the end it really doesn't matter if your matrix are functions or not since you can define the matrix weighting as a function, as an example:
f = @(Z1,Z2,Z3)Z1*2+Z2*1+Z3*3;
But actually what you seems to want is the minimum of the matrix weighted sum, and if you already have the values, it is really as simple as sum them and calculate the min:
w1 = 1;
w2 = 2;
w3 = 3;
Zweighted = Z1*w1+Z2*w2+Z3*w3;
[MinVal,MinPos] = min(Zweighted)

6 件のコメント

Alejandro Fernández
Alejandro Fernández 2020 年 2 月 2 日
編集済み: Alejandro Fernández 2020 年 2 月 2 日
Yeah, I'll try to explain it to you a little better.
Logically, the area of minimums in the first matrix coincides with the area of minimums in the second, so I have to optimize all these data to know the area of minimum regions that form these three matrices, taking into account that the third is much less important than the other two.
The idea was to implement something similar to the Pareto front optimization and the idea was to implement that, but with the matrices I don't know how to do it?
Thank you very much for your help.
Thiago Henrique Gomes Lobato
Thiago Henrique Gomes Lobato 2020 年 2 月 2 日
You don't need the Pareto front in your case. Pareto front is just when you have multiple different objectives and then want to check how to compromise one in respect to the other. In your case all 3 matrix are actually just one variable, more specifically they all represent together a distribution of your result. In this way you will get non-sensical results if you just do a weigted matrix sum. In your case there's at least two main approaches that are reasonable:
  1. The minimum is the minimum of the average matrix: This will produce good results if your goal is to minimize a specific process that will repeat itself for many many times.
  2. The minimum is the minimum of the mean-std*confidence : This will produce good results if your goal is to find the best possible scenario it could possible happens. The idea is to look at the tails of the distributions with some statistic confidence interval and check the minimum. In code it will look like something like this:
Zweighted = Zaverage-(Zstd*Confidence)./sqrt(ZnumberOfParticipants)
Confidence can be, for example, 1.96 for 95% confidence (read here for a more exact definition if you want, http://sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/BS704_Confidence_Intervals/BS704_Confidence_Intervals_print.html)
Thiago Henrique Gomes Lobato
Thiago Henrique Gomes Lobato 2020 年 2 月 2 日
編集済み: Thiago Henrique Gomes Lobato 2020 年 2 月 2 日
My comment was for you answer before your edited it. What you asked before editing and now are two completely different things and if your problem is really the before editing one (which had more details), it doesn't make sense asking for Pareto front. What you want to do?
Alejandro Fernández
Alejandro Fernández 2020 年 2 月 2 日
Ok, thank you so much for all!! :D
Alejandro Fernández
Alejandro Fernández 2020 年 2 月 2 日
It's okey like that, dont worry.
Thiago Henrique Gomes Lobato
Thiago Henrique Gomes Lobato 2020 年 2 月 2 日
No problem :)

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