I'm trying to find the minimum of a function with thirteen variables. Am I doing everything right? Tell me, how do I add restrictions?

I'm trying to find the minimum of a function with thirteen variables. Am I doing everything right? Tell me, how do I add restrictions?

in short, I solve the problem of optimizing the electrical network by installing reactive power compensating devices.

function F = multiObjFcn(optimInput)

x(1) = optimInput(1);
x(2) = optimInput(2);
x(3) = optimInput(3);
x(4) = optimInput(4);
x(5) = optimInput(5);
x(6) = optimInput(6);
x(7) = optimInput(7);
x(8) = optimInput(8);
x(9) = optimInput(9);
x(10) = optimInput(10);
x(11) = optimInput(11);
x(12) = optimInput(12);
x(13) = optimInput(13);

F(1) = 23.37*0.05*((sqrt(100^2+(90-x(1))^2))/(sqrt(3)*10^2)...
    +(sqrt(150^2+(130-x(2))^2))/(sqrt(3)*10^2)...
    +(sqrt(210^2+(180-x(3))^2))/(sqrt(3)*10^2)...
    +(sqrt(600^2+(450-x(6))^2))/(sqrt(3)*10^2)...
    +(sqrt(400^2+(390-x(7))^2))/(sqrt(3)*10^2)...
    +(sqrt(100^2+(90-x(8))^2))/(sqrt(3)*10^2)...
    +(sqrt(150^2+(130-x(9))^2))/(sqrt(3)*10^2)...
    +(sqrt(180^2+(162-x(10))^2))/(sqrt(3)*10^2)...
    +(sqrt(100^2+(90-x(13))^2))/(sqrt(3)*10^2))+266.214+0.256*(x(1)+x(2)+x(3)+x(4)+x(5)+x(6)+x(7)+x(8)+x(9)+x(10)+x(11)+x(12)+x(13))...
    +23.37*0.05*((sqrt(180^2+(162-x(10))^2))/(sqrt(3)*10^2)...
+(sqrt(100^2+(90-x(13))^2))/(sqrt(3)*10^2))...
    +23.37*0.05*((sqrt(400^2+(390-x(7))^2))/(sqrt(3)*10^2)...
    +(sqrt(100^2+(90-x(8))^2))/(sqrt(3)*10^2)...
    +(sqrt(150^2+(130-x(9))^2))/(sqrt(3)*10^2)...
    +(sqrt(180^2+(162-x(10))^2))/(sqrt(3)*10^2)...
    +(sqrt(100^2+(90-x(13))^2))/(sqrt(3)*10^2))...
    +23.37*0.05*((sqrt(210^2+(180-x(3))^2))/(sqrt(3)*10^2)...
    +(sqrt(600^2+(450-x(6))^2))/(sqrt(3)*10^2)...
    +(sqrt(400^2+(390-x(7))^2))/(sqrt(3)*10^2)...
    +(sqrt(100^2+(90-x(8))^2))/(sqrt(3)*10^2)...
    +(sqrt(150^2+(130-x(9))^2))/(sqrt(3)*10^2)...
    +(sqrt(180^2+(162-x(10))^2))/(sqrt(3)*10^2)...
    +(sqrt(100^2+(90-x(13))^2))/(sqrt(3)*10^2))...
    +23.37*0.05*((sqrt(100^2+(90-x(1))^2))/(sqrt(3)*10^2)...
    +(sqrt(150^2+(130-x(2))^2))/(sqrt(3)*10^2)...
    +(sqrt(210^2+(180-x(3))^2))/(sqrt(3)*10^2)...
    +(sqrt(600^2+(450-x(6))^2))/(sqrt(3)*10^2)...
    +(sqrt(400^2+(390-x(7))^2))/(sqrt(3)*10^2)...
    +(sqrt(100^2+(90-x(8))^2))/(sqrt(3)*10^2)...
    +(sqrt(150^2+(130-x(9))^2))/(sqrt(3)*10^2)...
    +(sqrt(180^2+(162-x(10))^2))/(sqrt(3)*10^2)...
    +(sqrt(100^2+(90-x(13))^2))/(sqrt(3)*10^2));

end

[Walter Roberson]

You might as well code

function F = multiObjFcn(x)

and skip all of the

x(1) = optimInput(1);
x(2) = optimInput(2);
x(3) = optimInput(3);
x(4) = optimInput(4);
x(5) = optimInput(5);
x(6) = optimInput(6);
x(7) = optimInput(7);
x(8) = optimInput(8);
x(9) = optimInput(9);
x(10) = optimInput(10);
x(11) = optimInput(11);
x(12) = optimInput(12);
x(13) = optimInput(13);

[Walter Roberson]

You might as well factor out the (sqrt(3)*10^2), and add together the terms that need that, and do a final / (sqrt(3)*10^2)

[Sam Chak]

Hey @Ivan, it says that the optimization problem appears to be unbounded.

[x, fval] = fmincon(@multiObjFcn, zeros(1, 13))

Problem appears unbounded.

fmincon stopped because the objective function value is less than
the value of the objective function limit and constraints 
are satisfied to within the value of the constraint tolerance.

x = 1×13

1.0e+19 * -3.2485 -3.2512 -3.1941 -3.3679 -3.3679 -3.2080 -3.1197 -3.1291 -3.1345 -3.0701 -3.3679 -3.3679 -3.0694

fval = -1.0048e+20

function F = multiObjFcn(x)
    ...
end

[Walter Roberson]

The term

0.256*(x(1)+x(2)+x(3)+x(4)+x(5)+x(6)+x(7)+x(8)+x(9)+x(10)+x(11)+x(12)+x(13))

can be arbitarily small if the x inputs are arbitarily small.

[Ivan]

Tell me how to add restrictions, please.

[Walter Roberson]

You currently have

Constraints Nonlinear function Local Function multiObjfcn

You would change that to

Constraints Nonlinear function Local Function multiConsfcn

and add a function named multiConsfcn that expressed the nonlinear constraints. The function must return two outputs, usually referred to as c and ceq . The output c expresses linear inequalities, c <= 0 and the output ceq expresses linear equalities, ceq == 0. You can define either c or ceq as [] (empty), but you need to define them as something

[Ivan]

Does anyone know any training videos and literature on how to solve multi-criteria problems with a bunch of variables in matlab?

[Torsten]

https://uk.mathworks.com/discovery/multiobjective-optimization.html

For the general approach, it does not matter how many variables your problem has.

[Sam Chak]

Hey @Ivan, can you list out the multiple objectives (or criteria) that you want to achieve?

[Ivan]

Well, first of all, my variables must have a limit - no lower than a certain value. Secondly, these variables must take discrete values from a certain list with a certain numerical step. Then I want to add a component that will take a random value from a certain range.

[Walter Roberson]

I suggest you use Problem Based Optimization

Otherwise, you need to use ga() to handle the discrete values.

If the discrete values happen to be integers, then use lb and ub to establish natural bounds, and mark those variables as being restricted to integers.

If the discrete values do not happen to be integers, then use substitute integer variables with lower bound 0 and upper bound (number of discrete states minus one), and then in your code do a transformation (VALUE * NUMERIC_STEP + MINIMUM)