Find independent variables that minimize function

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I have a mathematical function with around 5 independent variables. I want to find the values of the independent variables for which the function is minimal. All the independent variables have certain boundaries that need to be satisfied.
Up until now I use the Excel plugin "Solver" for this. But my task becomes a bit too complex for Excel, thus I want to switch to Matlab.
This is my function:
st1_l = ((a-real(0.5*(a-(sqrt(a*(a-4*b))))))*(b^2))/((b^2)-(1.22*(0.0004)*c*(1-((-d/(a - b1)) * e)/d))*c*((a - real(0.5*(a-(sqrt(a*(a-4*b))))))+b))
Boundaries are something like this:
The parameters should now be changed within these boundaries, so that st1_l becomes minimal. I already tried fmincon. However, I can't find a way to pass all the parameters and boundaries to fmincon. Just in case it makes clearer what I want to do, here an screenshot from the Excel solver I am using at the moment:
Thank you for any help.
dev3011 on 3 Mar 2020
Edited: dev3011 on 3 Mar 2020
Thank you for your help. I tried to follow your suggenstion and the documentation and produced this code (the formula is shortened to only 2 IVs):
fun = @(x) ((x(1)-real(0.5*(x(1)-(sqrt(x(1)*(x(1)-4*x(2)))))))*(x(2)^2))/((x(2)^2)-(1.22*(0.0004)*(1-(10)/(x(1)))));
A = [1,0;-1,0;0,1;0,-1];
b = [1000;-318;100;-10];
I assume that with this I included the boundaries
However, this way seems to be overly complicated and also, when I try to introduce more than 2 independent variables by writing
fun = @(x) ((x(1)-real(0.5*(x(1)-(sqrt(x(1)*(x(1)-4*x(2)))))))*(x(2)^2))/((x(2)^2)-(1.22*(wellenl)*(1-(x(3))/(x(1)))));
A = [1,0,0;-1,0,0;0,1,0;0,-1,0];
b = [1000;-318;100;-10];
I get the error "A must have 2 column(s)". Any suggestions on how to use more than 2 IVs? Also, do I really have to use x(1), x(2), etc. or is there a way to use my own variable names in the formula?

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Accepted Answer

Fabio Freschi
Fabio Freschi on 3 Mar 2020
If you have only lower and upper bounds, use them directly in fmincon
% your function
fun = @(x) ((x(1)-real(0.5*(x(1)-(sqrt(x(1)*(x(1)-4*x(2)))))))*(x(2)^2))/((x(2)^2)-(1.22*(0.0004)*(1-(x(3))/(x(1)))));
% inequality constraints (none);
A = [];
b = [];
% equality constraints (none)
Aeq = [];
beq = [];
% lower and upper bounds (I assume x(3) unbounded
lb = [318; 10; -Inf];
ub = [1000; 100; Inf];
% inital guess (here: mid point)
x0 = (ub-lb)/2;
% remove Inf
x0(x0 == Inf) = 0;
% run minimization
[x,fval] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub)
Regarding the name of the variables, if you use an anonymous function you should stick with x(1), x(2), etc. If you write your function in a file, you can rename teh variables inside the function
function f = fun(x)
a = x(1);
b = x(2);
c = x(3);
f = ((a-real(0.5*(a-(sqrt(a*(a-4*b))))))*(b^2))/((b^2)-(1.22*(0.0004)*(1-(c)/(a))));
  1 Comment
dev3011 on 4 Mar 2020
Awesome, thanks a lot for this extensive answer. It works very well.
While implementing it, however, I found that I oversimplified my problem. I will accept your answer as it solves the question I asked perfectly and extend my question below.

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More Answers (1)

dev3011 on 4 Mar 2020
Edited: dev3011 on 4 Mar 2020
Fabios answer solves the problem I posted very well. However, I realized that I oversimplified my problem. Actually, some of my boundary conditions are functions rather than independent variables.
Let's assume I have the two following functions:
f = ((a-real(0.5*(a-(sqrt(a*(a-4*b))))))*(b^2))/((b^2)-(1.22*(0.0004)*(1-(c)/(a))));
d = ((a * b)^2)*g % In reality this function is more complicated
and I now want the boundary
to be satisfied. Thus, my boundary must be satisfied for a function that is not directly included in my initial function to be minimized but depends on independent variables of this function.
I guess I could solve every one of these boundary functions for the independent variable and then pass the result to fmincon() as a boundary condition. However, as the boundary functions are quite complex this would not be feasible. Is there any way I can pass a linear boundary function like above to fmincon()?
  1 Comment
Fabio Freschi
Fabio Freschi on 4 Mar 2020
fmincon allows the use of nonlinear constraints
look at the documentation
x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon)

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