Find independent variables that minimize function

Hey,

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:

318<a<1000

10<b<100

...

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.

[Mario Malic]

fmincon(fun,x0,A,b)

Refer to documentation, with ~~variables A and b, you supply inequalities.~~

Edit: I am sorry, I made a silly mistake suggesting you inequalities. Better option is to use bounds as Fabio suggested.

[dev3011]

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

318<a<1000

10<b<100.

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?

Answer 1

[Fabio Freschi]

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)))); 
end

[dev3011]

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.

Answer 2

[dev3011]

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

10<d<100

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()?

[Fabio Freschi]

fmincon allows the use of nonlinear constraints

look at the documentation

x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon)