Help with fmincon Optimization for a Function Involving Matrices in MATLAB

Hello everyone,

I'm currently working on an optimization problem in MATLAB and could use some help with the fmincon function.

I'm trying to minimize an objective function defined as myfunction = x/(y*z). In my case, y and z are constants and do not change during the optimization process. The variable x is defined by the relationship x = 0.5*a*b + (0.34*c), and I have an additional constraint m = a*b*c.

Here are my constraints:

• If 1 < y < 5, then:
• 0.5 < a/b < 2
• 0.1 < b/c < 1
• 0.2 < y/m < 1
• Else if 5 < y < 10, then:
• 0.2 < a/b < 1.3
• 0.11 < b/c < 1.31
• 0.3 < y/m < 1.4

The optimization needs to be run on 50 matrices, each of size 6, I have a excel file, which contain the values for a, b, c, x, y, and z. I want to find out a,b,c to minimize myfunction, actually x value.

My question is two-fold:

1. How can I set up my fmincon optimization function to exclude y and z from the optimization process since they are fixed values in my large dataset?
2. How should I approach applying the optimization to the 50 matrices? Should I loop through each one, or is there a more efficient method?

Any help or pointers would be greatly appreciated!

Thank you!

[Dyuman Joshi]

What have you tried yet?

[Shariful Islam]

@Dyuman Joshi, thanks for your response. Yes, I've made some attempts. Please review the updated post for the full context of my question.

reagrds

[John D'Errico]

Sorry, but this makes little sense. You say that y and z are constants. But then have a branch on the value of y.

Next, you say you have a "constraint" that m=a*b*c. But a,b,c are all fixed constants from what you say. But next you say you want to optimize the values of a,b,c. So are a,b,c constant, or are they fixed? What is the point of having values for a,b,c if they are not fixed?

You ask how you can set up the optimization to exclude y and z. That is trivial. Don't vary them! You tell the optimizer what to vary. It makes no decisions for you.

I'm sorry, but your question is totally confusing, which probably means you are confused. But we cannot easily help you if you do not explain the problem clearly enough.

[Shariful Islam]

Dear @John D'Errico,

Apologies for the confusion earlier. Let me clarify: y and z were indeed mentioned as constants, but that was in reference to their role in a particular part of the problem, not that their values would change during the optimization process.

Regarding a, b, and c, they are not fixed constants. I provided initial values for these variables as a starting point for the optimization process. These values are meant to be adjusted to find the optimal solution. The aim is to vary a, b, and c to satisfy this constraint while optimizing the outcome.

To exclude y and z from the optimization, I meant that they should remain same after the optimization, while the optimizer adjusts the values of a, b, and c. I hope this clarifies my question and I'm open to any suggestions on how to properly set up this optimization.

If you have any queries or need further clarification, please feel free to ask me.

Rageds

Answer 1

[Torsten]

y = 3;
if y > 1 & y < 5
    obj = @(p)0.5*p(1)*p(2)+0.34*p(3);
    nonlcon = @(p) deal([],p(4)-p(1)*p(2)*p(3));
    A = [0 0 0 0.2;0 0 0 -1;-1 0.5 0 0;1 -2 0 0;0 -1 0.1 0;0 1 -1 0];
    b = [y;-y;0;0;0;0];
    Aeq = [];
    beq = [];
    lb = [0;0;0;0];
    ub = [Inf;Inf;Inf;Inf];
    p0 = [2 4 6 2*4*6];
    p = fmincon(obj,p0,A,b,Aeq,beq,lb,ub,nonlcon,optimoptions(@fmincon,'ConstraintTolerance',1e-12))
    0.5 <= p(1)/p(2) & p(1)/p(2)<=2
    0.1 <= p(2)/p(3) & p(2)/p(3)<=1
    0.2 <= y/p(4) & y/p(4)<=1
    abs(p(4)-p(1)*p(2)*p(3))<=1e-11
    obj(p)
elseif y > 5 & y < 10
% fill in according to case 1
end

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.

p = 1×4

1.2173 1.1733 2.1004 3.0000

ans = logical
   1

ans = logical
   1

ans = logical
   1

ans = logical
   1

ans = 1.4283

[Shariful Islam]

@Torsten, Thanks for your awesome response. Kindly explain how you exclude y and z from the optimization process?

and how should I approach applying the optimization to the 50 matrices? Should I loop through each one, or is there a more efficient method?

Regards

[Torsten]

You wrote that y and z are given positive constants during optimization. So it doesn't matter whether you minimize x or x/(y*z). But when you study the code, you will see that y appears in the constraints - so you cannot exclude it completely for the optimization.

And if you have 50 different y values, you will have to loop over them.

[Shariful Islam]

But, y are in constraints. How can I exclude variables frrom the optimization process, if they also are in constraints?

[Torsten]

They are constants. As such their numerical values can appear in constraints although they are not variables to be optimized. Look at the code and find the answer.