How to define constraint in Optimization such that difference in value between two consecutive unknowns is not greater than 50%

Hello everyone,

I have prepared code in matlab for genetic algorithm from toolbox

The function code is a bit long. In summary, There are 30 unknowns with upper and lower bounds as 1.

How to apply a constraint while running such that difference in value between two consecutive unknowns obtained is not greater than 50% ?.

[Matt J]

50% of what? Do you mean |x(i+1)-x(i)|<=0.5*x(i) ?

[Ankur Shah]

Yes, 50% greater than the previous.

Answer 1

[Matt J]

Here is a way to set up the constraint matrices using prob2matrices from,

https://www.mathworks.com/matlabcentral/fileexchange/74481-prob2matrices-a-selective-version-of-prob2struct

x=optimvar('x',30,'Lower',0,'Upper',1);
Constraints.diffUB=diff(x)<=+0.5*x(1:end-1);
Constraints.diffLB=diff(x)>=-0.5*x(1:end-1);

p=prob2matrices({x},'Constraints',Constraints)

p = struct with fields:
    intcon: []
        lb: [30×1 double]
        ub: [30×1 double]
     Aineq: [58×30 double]
     bineq: [58×1 double]
       Aeq: []
       beq: []

[Ankur Shah]

How can it be done using toolbox ? Is it inequality constraint ?

[Matt J]

Yes, it is a linear inequality constraint of the form Aineq*x<=bineq. The matrices Aineq, bineq are given in my solution above.

[Ankur Shah]

Is it something like this ?, Apologies I am a bit new to defining inequalities.

function [c,ceq] = confuneq(x)
c = +0.5*x(1:end-1);
ceq = -0.5*x(1:end-1);
end

[Matt J]

You don't want to implement the constraint using the nonlinear constraint function inputs. That will make it unecessarily harder for the solver. The constraint as you stated it is,

|x(i+1)-x(i)|<=0.5*x(i) 

which is equivalent to linear inequalities,

-0.5*x(i) <= x(i+1)-x(i) <= 0.5*x(i) 

or,

 0.5*x(i)-x(i+1)<=0
-1.5*x(i)+x(i+1)<=0 

[Ankur Shah]

I get your point, but i am not sure how to apply in code

for example if linear equalities are as below

x(1) -x(2) <= -1

-x(1) + x(2) <= 5

We define

A = [-1,-1;

-1,1];

b = [-1;5];

because above is 2 variables, it is easy to apply as follows

fun = @ps_example;

x = ga(fun,2,A,b)

How to apply the solution which you gave in coding ?

[Matt J]

The code in my original answer generates the matrices for you. You can use them directly in the call to ga:

x=optimvar('x',30,'Lower',0,'Upper',1);
Constraints.diffUB=diff(x)<=+0.5*x(1:end-1);
Constraints.diffLB=diff(x)>=-0.5*x(1:end-1);

p=prob2matrices({x},'Constraints',Constraints);

x=ga(fun,30,p.Aineq,p.bineq,p.Aeq,p.beq,p.lb,p.ub);

Answer 2

[Bjorn Gustavsson]

Perhaps a constraint-function like this would get the job done (might be used with fmincon for example):

function [c,ceq] = your_con(x,a,b)

if nargin < 3
  b = 0.1;
end
if nargin < 2
  a = 1/2;
end
dx = diff(x);
c = 2*abs(dx)./(max(abs(x(1:end-1))+abs(x((2:end))),...
                    max(b,...
                        max(abs(x(1:end-1)),abs(x(2:end)))...
                        )...
                   ))-a;
ceq = [];

To read up on the details of using this read the help and documentation of fmincon.

HTH