# Issues in the constraint functions of the Optimization Problem solving

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I want to add more than two constraints which are linear, nonlinear and with both equality and inequality signs. As per the MATLAB documentation the examples shown are mostly of same variables which are written as array Examples.

I am using the following shown constraint for the optimisation. The constraint function is named as "area", this same function is called for the 'nonlcon' (default function for constraints).

#1 The equations commented are also to be considered as Constraints for the Optimisation. Do MATLAB optimisation considers this kind of constraint equations?

#2 How to check whether all constraints are satisfied.

function [c,ceq]= area(x0)

M=5;

x0=[5,8];

M2=m(M,x0(1));

M3=m(M2,x0(2));

M4=m(M3,(x0(1)+x0(2)));

beta1=b(M,(x0(1)));

beta2=b(M2,(x0(2)));

beta3=b(M3,((x0(1)+x0(2))));

% M*sin(beta1)==M2*sin(beta2) & M2*sin(beta2)==M3*sin(beta3)

% (M4/M) <= 0.38

% beta1,beta2,beta3 < 62 degrees (This can be written in terms of Thetas)

% Some more constraints will be added

c=[];

ceq = log(2.4/((2.8*M*(sin(beta1)^2))-0.4)) + log(2.4/((2.8*M2*(sin(beta2)^2))-0.4)) + log(P(x0))

end

Below code shows the use of constraint function 'area' in the optimisation algorithm as "nonlcon".

%Initial Condition

% x0=[5,5] ;

% Lower bounds

lb=[1,1];

% Upper Bounds

ub=[30,40];

%constraint

nonlcon=@area

opts = optimoptions('fmincon','PlotFcn',["optimplotx","optimplotfunccount","optimplotfvalconstr","optimplotfval"],'Display','iter')

opts1 = optimoptions(opts,'MaxIterations',50,'StepTolerance',1e-9,'ConstraintTolerance',1e-9)

[x,fval,exitflag,output,lambda,grad,hessian]= fmincon(@f,x0,[],[],[],[],lb,ub,nonlcon,opts1)

The optimisation is stopped between saying the below shown statement and then after enabling Feasibility Mode also the error is shown which is also mentioned in the end.

Converged to an infeasible point.

fmincon stopped because the size of the current step is less than

the value of the step size tolerance but constraints are not

satisfied to within the value of the constraint tolerance.

<stopping criteria details>

Consider enabling the interior point method feasibility mode.

%Initial Condition

x0=[5,5] ;

% Lower bounds

lb=[1,1];

% Upper Bounds

ub=[30,40];

%constraint

nonlcon=@area

opts = optimoptions('fmincon','PlotFcn',["optimplotx","optimplotfunccount","optimplotfvalconstr","optimplotfval"],'Display','iter', ...

'Algorithm','interior-point','EnableFeasibilityMode',true)

opts1 = optimoptions(opts,'MaxIterations',50,'StepTolerance',1e-9,'ConstraintTolerance',1e-9); % Recommended

[x,fval,exitflag,output,lambda,grad,hessian]= fmincon(@f,x0,[],[],[],[],lb,ub,nonlcon,opts1);

Converged to an infeasible point.

fmincon stopped because it is unable to find a point locally that satisfies

the constraints within the value of the constraint tolerance.

<stopping criteria details>

Even after changing the Constraint Tolerance it is not helping. Tried in the range 1e-9 to 1e-3

Thank you!

##### 4 Comments

### Answers (2)

Sulaymon Eshkabilov
on 31 Jan 2023

Edited: Sulaymon Eshkabilov
on 31 Jan 2023

There should be elementwise operations in ceq formulation if M2, beta1, beta2 and P are vectors.

##### 0 Comments

Matt J
on 31 Jan 2023

Edited: Matt J
on 31 Jan 2023

function [c,ceq]= area(x0)

M=5;

% x0=[5,8];

M2=m(M,x0(1));

M3=m(M2,x0(2));

M4=m(M3,(x0(1)+x0(2)));

beta1=b(M,(x0(1)))*180/pi;

beta2=b(M2,(x0(2)))*180/pi;

beta3=b(M3,((x0(1)+x0(2))))*180/pi;

c(1)=M4-0.38*M;

c(2:4)=[beta1,beta2,beta3]-62;

ceq(1) = log(2.4/((2.8*M*(sin(beta1)^2))-0.4)) + log(2.4/((2.8*M2*(sin(beta2)^2))-0.4)) + log(P(x0));

ceq(2) = M*sin(beta1)-M2*sin(beta2) ;

ceq(3) = M2*sin(beta2)-M3*sin(beta3) ;

end

##### 10 Comments

Matt J
on 2 Feb 2023

Torsten
on 2 Feb 2023

Edited: Torsten
on 2 Feb 2023

Using exponential function will change the range of all numbers and it cant be used in further calculations.

Taking the exponential of a constraint does not change anything in the range of the parameters.

It doesn't matter whether you set

log(x)-2 = 0

or

exp(log(x)-2) = exp(0)

as a constraint.

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