how can I set the options for Genetic Algorithm and SQP optimization? objective function and constraints are defined but I can not get the result.

    function f=objfun(x)
Rd = x(1); Rb = x(2); bd = x(3);
dR=(Rd-Rb)
Td=4/3*pi*40000*dR^3+pi*3.8*45/0.001*dR^4+2*pi*Rd^2*bd*[15+0.1*(45*Rd/0.001)]
f=-Td

function [c ceq]=constraint(x)
Rd = x(1); Rb = x(2); bd = x(3);
dR=(Rd-Rb)
Td=4/3*pi*40000*dR^3+pi*3.8*45/0.001*dR^4+2*pi*Rd^2*bd*[15+0.1*(45*Rd/0.001)]
Td0=4/3*pi*15*dR^3+pi*0.1*45/0.001*dR^4+2*pi*Rd^2*bd*[15+0.1*(45*Rd/0.001)]
muJ=Rd^4*bd/[(0.069)^4*0.022-(0.065)^4*(bd+0.002)]
f=-Td

g1=1-Td/(50*Td0)
g2=1-muJ/0.25
g3=muJ/0.5-1
g4=Rb/Rd-1


c(1)=g1
c(2)=g2
c(3)=g3
c(4)=g4

ceq=[]

clc
clear all

%GA optimization
% seting lower bounds and upper bouds
LB = [0.030;0.010;0.003];
UB = [0.068;0.020;0.012];
% starting point
% x0 = [0.050;0.0010;0.010];
options = gaoptimset('MutationFcn',@mutationadaptfeasible);
[x,fval,exitflag,output] = ga(@objfun,3,[],[],[],[],LB,UB,@constraint,options);

%SQP optimization
clc
clear all
close all
LB = [0.030;0.010;0.003];
UB = [0.068;0.020;0.012];
x0 = [0.060;0.012;0.010];
options = optimoptions('fmincon','display','iter','Algorithm','sqp','PlotFcn','optimplotfval');
%options = struct('MaxFunctionEvaluations',100000);
[x,fval,exitflag,output,lambda,grad,hessian] = fmincon(@objfun,x0,[],[],[],[],LB,UB,@constraint,options)

Answer 1

[Walter Roberson]

It works for me.

Do not expect these kinds of optimizers to output a message that says something like "Optimization terminated, global minima found!". Some of the linear and quadratic problem solvers can prove that they have found a global minima, but none of the optimizers that accept a function handle can prove that they have found a global minima.

When you have an arbitrary function that is not known to be linear or polynomial , the only way to be sure that you have found a global minima is if you do a mathematical analysis of the function. Otherwise how could you be sure that there is no hidden statement along the lines of if x == sqrt(352395.231135); y = -324932523532; end ?

Therefore the best that the optimizers can do is say that they have found as good of a local minima as your configuration asks them to look at.

%GA optimization
% seting lower bounds and upper bouds
LB = [0.030;0.010;0.003];
UB = [0.068;0.020;0.012];
% starting point
% x0 = [0.050;0.0010;0.010];
options = gaoptimset('MutationFcn',@mutationadaptfeasible);
[x,fval,exitflag,output] = ga(@objfun,3,[],[],[],[],LB,UB,@constraint,options)

Optimization terminated: average change in the fitness value less than options.FunctionTolerance
 and constraint violation is less than options.ConstraintTolerance.

x = 1×3

0.0680 0.0100 0.0076

fval = -38.8418

exitflag = 1

output = struct with fields:
      problemtype: 'nonlinearconstr'
         rngstate: [1×1 struct]
      generations: 3
        funccount: 7400
          message: 'Optimization terminated: average change in the fitness value less than options.FunctionTolerance↵ and constraint violation is less than options.ConstraintTolerance.'
    maxconstraint: 0
       hybridflag: []

%SQP optimization
LB = [0.030;0.010;0.003];
UB = [0.068;0.020;0.012];
x0 = [0.060;0.012;0.010];
options = optimoptions('fmincon','display','iter','Algorithm','sqp','PlotFcn','optimplotfval');
%options = struct('MaxFunctionEvaluations',100000);
[x,fval,exitflag,output,lambda,grad,hessian] = fmincon(@objfun,x0,[],[],[],[],LB,UB,@constraint,options)

 Iter  Func-count            Fval   Feasibility   Step Length       Norm of   First-order  
                                                                       step    optimality
    0           4   -2.144608e+01     0.000e+00     1.000e+00     0.000e+00     1.399e+03  
    1           8   -3.881789e+01     0.000e+00     1.000e+00     8.660e-03     7.137e+02  
    2          12   -3.884198e+01     1.023e-03     1.000e+00     3.178e-04     8.001e-01  
    3          16   -3.884193e+01     3.089e-07     1.000e+00     5.510e-06     1.495e-02  
    4          20   -3.884193e+01     2.784e-13     1.000e+00     1.665e-09     2.800e-07  

Feasible point with lower objective function value found.

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.

x = 3×1

0.0680 0.0100 0.0076

fval = -38.8419

exitflag = 1

output = struct with fields:
         iterations: 4
          funcCount: 20
          algorithm: 'sqp'
            message: '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.↵↵<stopping criteria details>↵↵Optimization completed: The relative first-order optimality measure, 1.324947e-10,↵is less than options.OptimalityTolerance = 1.000000e-06, and the relative maximum constraint↵violation, 2.784439e-13, is less than options.ConstraintTolerance = 1.000000e-06.'
    constrviolation: 2.7844e-13
           stepsize: 1.6647e-09
       lssteplength: 1
      firstorderopt: 2.8000e-07
       bestfeasible: [1×1 struct]

lambda = struct with fields:
         eqlin: [0×1 double]
      eqnonlin: [0×1 double]
       ineqlin: [0×1 double]
    ineqnonlin: [4×1 double]
         lower: [3×1 double]
         upper: [3×1 double]

grad = 3×1

1.0e+03 * -2.1133 2.1102 -0.0093

hessian = 3×3

1.0e+03 * 0.0206 0.0007 0.1437 0.0007 0.0017 -0.0000 0.1437 -0.0000 1.0199

function f=objfun(x)
    Rd = x(1); Rb = x(2); bd = x(3);
    dR=(Rd-Rb);
    Td=4/3*pi*40000*dR^3+pi*3.8*45/0.001*dR^4+2*pi*Rd^2*bd*[15+0.1*(45*Rd/0.001)];
    f=-Td;
end

function [c ceq]=constraint(x)
    Rd = x(1); Rb = x(2); bd = x(3);
    dR=(Rd-Rb);
    Td=4/3*pi*40000*dR^3+pi*3.8*45/0.001*dR^4+2*pi*Rd^2*bd*[15+0.1*(45*Rd/0.001)];
    Td0=4/3*pi*15*dR^3+pi*0.1*45/0.001*dR^4+2*pi*Rd^2*bd*[15+0.1*(45*Rd/0.001)];
    muJ=Rd^4*bd/[(0.069)^4*0.022-(0.065)^4*(bd+0.002)];
    f=-Td;
    
    g1=1-Td/(50*Td0);
    g2=1-muJ/0.25;
    g3=muJ/0.5-1;
    g4=Rb/Rd-1;
    
    
    c(1)=g1;
    c(2)=g2;
    c(3)=g3;
    c(4)=g4;
    
    ceq=[];
end

[Misi]

Thanks for your answer.