How to choose one between two constraint conditions

Question

ocean 2024-5-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2116776-how-to-choose-one-between-two-constraint-conditions

评论： ocean 2024-5-10

clear;clc;
x = optimvar('x',1,1,'LowerBound',0)
prob=optimproblem;
prob.Constraints.con1=x<=10 or prob.Constraints.con1=x^2<=10
[sol,faval,exit]=solve(prob,'Solver','ga')

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ocean 2024-5-8

Have you ever encountered the same problem? I have been troubled by this problem many times

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Answer 1

Matt J 2024-5-8

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2116776-how-to-choose-one-between-two-constraint-conditions#answer_1454706

编辑：Matt J 2024-5-9

在 MATLAB Online 中打开

Considering Walter's answer, your example may not have captured your real question. If you really do have a feasible set of the form region A or region B, where A and B are disjoint in the space of x, such as in this modified example,

clear;clc;
x = optimvar('x',1,'Lower',0);
prob=optimproblem;
prob.Constraints.con = (x<=1 | x>=5)
[sol,faval,exit]=solve(prob,'Solver',_____)

you normally have to deal with such situations by solving the optimization twice, once over A and once over B, and selecting the better of the two results. This is because local optimization solvers like fmincon can generally only search incrementally over a contiguous feasible set.

In the case of global, non-derivative-based solvers like 'ga', you might, however, be able to get away with the following,

prob.Constraints.con = fnc2optimexpr( @(z) min(z-1, 5-z) ,x)<=0;

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ocean 2024-5-9

在 MATLAB Online 中打开

for this problem i have tried a solution and it's successed!

clear;clc;
x = optimvar('x',2,2);
y=optimvar('y',2,2);
c=optimvar('c',2,2,'Type','integer','LowerBound',0,'UpperBound',1);
prob=optimproblem('Objective',sum(x.^2+y.^2,"all"));
%prob.Constraints.con1=x(1,1)>=10 or prob.Constraints.con1=y(1,1)>=15;
%prob.Constraints.con2=x(1,2)>=6 or prob.Constraints.con1=y(1,2)>=3;
%prob.Constraints.con3=x(2,1)>=10 or prob.Constraints.con1=y(2,1)>=15;
%prob.Constraints.con1=x(2,2)>=4 or prob.Constraints.con1=y(2,2)>=2;
%(i mean x>=some matirx or y>=some matrix)
%what if this problem what can i do?
xx=[10 6;10 4];yy=[15 3;15 2];
prob.Constraints.con=(c.*x+(1-c).*y)>=(c.*xx+(1-c).*yy); 
x0.x=xx;x0.y=yy;x0.c=ones(2,2);
[sol,faval,exit]=solve(prob,x0,"Solver","ga")
Solving problem using ga.
ga stopped because the average change in the penalty function value is less than options.FunctionTolerance and 
the constraint violation is less than options.ConstraintTolerance.
sol = struct with fields:
    c: [2x2 double]
    x: [2x2 double]
    y: [2x2 double]
faval = 715.0000
exit = 
    SolverConvergedSuccessfully

Matt J 2024-5-9

编辑：Matt J 2024-5-9

在 MATLAB Online 中打开

xb=[10 6; 10 4];
yb=[15 3;15 2];
x = optimvar('x',2,2);
y=optimvar('y',2,2);
prob=optimproblem('Objective',sum(x.^2+y.^2,'all'));
prob.Constraints.con=fcn2optimexpr( @(x,y) min(xb-x,yb-y) , x,y  )<=0;
soltemp=solve(prob,'Solver','ga');
Solving problem using ga.
Optimization finished: average change in the fitness value less than options.FunctionTolerance and constraint violation is less than options.ConstraintTolerance.
[sol,faval,exit]=solve(prob,soltemp,'Solver','fmincon');
Solving problem using fmincon.

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.
sol.x,
ans = 2x2
   10.0000   -0.0000
   10.0000    0.0000
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sol.y
ans = 2x2
   -0.0000    3.0000
    0.0000    2.0000
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