I tried to show effect of uncertain parameters on bi-variate function with color map and aslo highlight changes to max and min,I am not sure the function I wrote is right!!!

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sogol bandekian 2022-5-21

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1724310-i-tried-to-show-effect-of-uncertain-parameters-on-bi-variate-function-with-color-map-and-aslo-highli

评论： sogol bandekian 2022-5-24

for i=1:length(x)   
    x=lhsdesign(10,1,"iterations",2);
    for j=1:length(y)
        y=lhsnorm(50,10,10,"on");
        Z=[x,y];
        
        Z(i,j)=2.*x(i).^3+y(j).^3-3.*x(i).^2-12.*x(i)-3.*y(j);    
    end
end
options = optimset('Display','iter','PlotFcns',@optimplotfval);
x0=[x(1),y(1)]
[xmin,fval]=fminsearch(@(xy)Z(xy(1),xy(2)),x0,options); %min 
[xmax ,fval]=fminsearch(@(xy) -1 *Z(xy(1),xy(2)),x0,options); %max
[X,Y]=meshgrid(x,y);
surf(X,Y,Z(i,j),'EdgeColor',"interp","FaceAlpha",0.5);
hold on
plot(xmin(1),xmin(2), f(xmin(1),xmin(2)),'MarkerSize',6);
hold on
plot(xmax(1),xmax(2), f(xmax(1),xmax(2)),'MarkerSize',6);
hold off
xlabel("optimum valeus for x")
ylabel("optimum valeus for y")
colormap(parula(6));%default colormap with 6 colors
colorbar

can anyone help me for correcting this function?

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sogol bandekian 2022-5-23

but they are not optimization solvers like fminsearch.are not they?

Torsten 2022-5-23

No. It's a search for a minimum or maximum of a functrion on a discrete set of (x/y) values. The function is evaluated on a (fine enough) grid and the point where this evaluation is minimal (or maximal) is taken as the minimum (or maximum) of the function. It's an alternative to fminsearch and other optimizers if the region where minimum (or maximum) is is approximately known and if the function is badly behaved (not differentiable etc).

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采纳的回答

Walter Roberson 2022-5-22

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1724310-i-tried-to-show-effect-of-uncertain-parameters-on-bi-variate-function-with-color-map-and-aslo-highli#answer_969180

在 MATLAB Online 中打开

y=lhsnorm(50,10,10,"on");
[xmin,fval]=fminsearch(@(x)Z(x,y),x0,options) %min 

The y referred to there is being "captured" from the workspace variable y which is the lhsnorm that you defined earlier.

With y being non-scalar, your multinomial Z function is going to return multiple values for each input x value, but for fminsearch you need to return a vector.

Perhaps what you need is

[xmin,fval]=fminsearch(@(xy)Z(xy(1),xy(2)),x0,options) %min

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Walter Roberson 2022-5-23

在 MATLAB Online 中打开

x = 0:0.1:1;
y = 0:0.1:1;
[X,Y] = meshgrid(x,y);
Z = 2.*X.^3+Y.^3-3.*X.^2-12.*X-3.*Y;
Z = Z + (-0.1+0.2*rand(size(Z)));
fun = @(x)interp2(X, Y, Z, x(1), x(2), 'spline');
options = optimset('Display','iter','PlotFcns',@optimplotfval);
x0=[x(1),y(1)]
lb = [min(x), min(y)];
ub = [max(x), max(y)];
[xmin,fval] = fmincon(fun, x0, [], [], [], [], lb, ub, [], options); %min 

You will get a location near (1,1) and a fval about -14.9 .

The actual minimum for the non-noisy function is -15 at (1,1) exactly, and with the noise might be slightly less than that. But it does not find the minima that closely, as the minima is exactly at the bounds.

Consider your interpolation for a moment. Is there ever a case where the interpolated value can be less than the minimum Z (noisy) Z value?

If you were using linear interpolation or bilinear interpolation, the answer would be NO -- but on the other hand linear and bilinear interpolation do not satisfy the continuity requirements.

Is it possible for cubic interpolation to interpolate a value that is less than the minimum Z value? The answer to that is Yes. If you consider an edge such as

     + +
         +
    +

then in order to fit a quadratic or cubic to that, the interpolated value needs to come up from the left, pass upwards though the second, pass downwards through the third, and then on through the fourth. That implies that between the second and third, the projection would be for a value above the maximum (second, third) point. Any time that the third point does not fall off as quickly as the rise between the first and second points, quadratic or cubic fit needs to project a maximum between the second and third points. This discussion is in terms of max, but flip it upside down and you can see the same issue applies to minima.

Thus, in order to have the possibility of a projected minima being less than the minima present in the array, you have to be counting on the details of how the interpolation algorithm overshoots beyond the actual points.

If you do not want to be exploring the details of interpolation overshoot between close points, then there would be no possibility of interpolation getting a value less than the minima in the array, and so there would be no point in doing a search -- the minima would have to be at one of the locations that is exactly equal to the minimum in the array.

Walter Roberson 2022-5-24

在 MATLAB Online 中打开

x = 0:0.2:1;
y = 0:0.2:1;
[X,Y] = meshgrid(x,y);
Z = 2.*X.^3+Y.^3-3.*X.^2-12.*X-3.*Y;
Z = Z + (-0.1+0.2*rand(size(Z)));
fun = @(x)interp2(X,Y,Z,x(1),x(2), 'spline');
fun2 = @(x)-fun(x);
options = optimset('Display','iter','PlotFcns',@optimplotfval);
x0=[x(1),y(1)]
lb = [min(x), min(y)];
ub = [max(x), max(y)];
[xmin,fvalmin] = fmincon(fun, x0, [], [], [], [], lb, ub, [], options); %min
[xmax,fvalmax]= fmincon(fun2, x0, [], [], [], [], lb, ub, [], options); %max
fvalmax = -fvalmax;
surf(X, Y, Z, 'edgecolor', 'none') 
hold on 
plot3(xmin(1), xmin(2), fvalmin, 'rv') 
plot3(xmax(1), xmax(2), fvalmax, 'r^') 
hold off