Minimization of two variables

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Joshua Woodard
Joshua Woodard 2019 年 6 月 27 日
コメント済み: infinity 2019 年 6 月 28 日
I cant seem to figure this out on my own. I need to minimize the following function where xi and yi are given data sets (points in 2D),
f(x,y) = minimize{maximize[sqrt((xi-x)^2+(yi-y)^2)]-minimize[sqrt((xi-x)^2+(yi-y)^2)]}
I will need to find x,y.

採用された回答

infinity
infinity 2019 年 6 月 27 日
編集済み: infinity 2019 年 6 月 27 日
Hello,
Here is an example for your problem, which you can refer
clear
xiyi = [0,1;
1,3;
-1 2];
x0 = zeros(size(xiyi));
fun = @(x) max(sqrt((x(:,1)-xiyi(:,1)).^2 + (x(:,2)-xiyi(:,2)).^2))...
- min(sqrt((x(:,1)-xiyi(:,1)).^2 + (x(:,2)-xiyi(:,2)).^2));
xsol = fminsearch(fun,x0)
It is assumed that xi and yi are the first and second column of vector xiyi (in the code).
The solution (x,y) will be stored in xsol.
  3 件のコメント
Joshua Woodard
Joshua Woodard 2019 年 6 月 28 日
I am looking for an x and y that is the center of two concentric circles where the zone of the two circles contain the data. One circle is bigger and contains all the data whereas the small circle does not. I am maximizing the small circle and minimizing the big circle.
infinity
infinity 2019 年 6 月 28 日
Hello,
In the case of searching only one index, for example,
max(sqrt((x(1,1)-xiyi(1,1)).^2 + (x(1,2)-xiyi(1,2)).^2))...
- min(sqrt((x(1,1)-xiyi(1,1)).^2 + (x(1,2)-xiyi(1,2)).^2))
the usage of max and min functions are not necessary since only one value in these functions. I am still unclear with your description of the problem. You may illustrate by picture, which will be more easy.

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