Finding x and y values of minimum z in 2-variable function.

I'm using for loops to find a minimum of z(x,y) as well as the x and y of that value.

With a step size of 1 it finds the right values, but with a step size of 0.1 the y value is 10 when it should be around 2.

Any help would be appreciated.

[x,y] = meshgrid(0:0.1:10, 0:0.1:10);
z = (x.^2).*(y.^3)-1.4*(x.^2).*(y.^2)-3.91*(x.^2).*y+5.78*(x.^2)-...
    11*x.*(y.^3)+15.4*x.*(y.^2)+43.01*x.*y-63.58*x+30.25*(y.^3)-...
    42.35*(y.^2)-118.278*y+174.845;
min = 100;

for i=1:101
    for j=1:101
        if z(i,j)<min
            min = z(i,j);
            xmin = x(i,j);
            ymin = y(i,j);
        end
    end
end

[John D'Errico]

Note that it is a really bad idea to use a variable neamed min, as then later on when you need to use the function min, MATLAB will not know how to distinguish between the function and variable.

Answer 1

[Matt J]

The continuous minimum can't be at (5,2) as you claim. Direct evaluation shows that the value of z at (5.5,10) is lower.

loopingMin(1)

xmin = 5

ymin = 2

minval = 0.0890

loopingMin(0.1)

xmin = 5.5000

ymin = 10

minval = -0.0050

function loopingMin(stepsize)

    [x,y] = meshgrid(0:stepsize:10, 0:stepsize:10);
    z = (x.^2).*(y.^3)-1.4*(x.^2).*(y.^2)-3.91*(x.^2).*y+5.78*(x.^2)-...
        11*x.*(y.^3)+15.4*x.*(y.^2)+43.01*x.*y-63.58*x+30.25*(y.^3)-...
        42.35*(y.^2)-118.278*y+174.845;
    minval = inf;
    
    for i=1:height(x)
        for j=1:width(y)
            if z(i,j)<minval
                minval = z(i,j);
                xmin = x(i,j);
                ymin = y(i,j);
            end
        end
    end
    xmin,ymin,minval
end

Answer 2

[Star Strider]

Is there a particular reason to use that sort of iteration? MATLAB has a number of optimization funcitons you can use, with fminsearch being a part of core MATLAB (no Toolboxes required) or one of the Optimization Toolbox functions, such as fsolve

That aside, I keep getting different results betweeen runs and between functions —

z = @(x,y) (x.^2).*(y.^3)-1.4*(x.^2).*(y.^2)-3.91*(x.^2).*y+5.78*(x.^2)-...
    11*x.*(y.^3)+15.4*x.*(y.^2)+43.01*x.*y-63.58*x+30.25*(y.^3)-...
    42.35*(y.^2)-118.278*y+174.845;

B0 = randn(2,1)

B0 = 2×1

-0.6770 0.9243

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[B,fv] = fminsearch(@(b) norm(z(b(1),b(2))), B0)

B = 2×1

0.0204 1.6972

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fv = 3.3765e-11

zval = z(B(1),B(2))

zval = -3.3765e-11

[B,fv] = fsolve(@(b) norm(z(b(1),b(2))), B0)

Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle non-square systems; using Levenberg-Marquardt algorithm instead.

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

B = 2×1

-0.6004 1.6975

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fv = 8.6173e-07

zval = z(B(1),B(2))

zval = 8.6173e-07

[B,fv] = fminunc(@(b) norm(z(b(1),b(2))), B0)

Local minimum possible.

fminunc stopped because it cannot decrease the objective function
along the current search direction.

B = 2×1

-0.5319 1.7025

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fv = 4.0937e-09

[X,Y] = ndgrid(linspace(min(B)-0.1*abs(min(B)), max(B)+0.1*abs(max(B)), 250));

figure
surfc(X, Y, z(X,Y), EdgeColor='interp', FaceAlpha=0.1)
hold on
stem3(B(1), B(2), z(B(1),B(2))+5, 'vr', MarkerFaceColor='r')
hold off
grid
zlim([min([zval zlim]) max(zlim)])
xlabel('X')
ylabel('Y')
zlabel('Z')
colormap(turbo)
view(60,30)
grid on

.

Answer 3

[Torsten]

z = @(x,y)(x.^2).*(y.^3)-1.4*(x.^2).*(y.^2)-3.91*(x.^2).*y+5.78*(x.^2)-...
          11*x.*(y.^3)+15.4*x.*(y.^2)+43.01*x.*y-63.58*x+30.25*(y.^3)-...
          42.35*(y.^2)-118.278*y+174.845;
sol = fmincon(@(u)z(u(1),u(2)),[1 1],[],[],[],[],[0 0],[10 10])

Local minimum possible. Constraints satisfied.

fmincon stopped because the size of the current step is less than
the value of the step size tolerance and constraints are 
satisfied to within the value of the constraint tolerance.

sol = 1×2

5.5000 9.9840

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z(sol(1),sol(2))

ans = -0.0050