How to choose one between two constraint conditions

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')

[ocean]

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

Answer 1

[Matt J]

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;

[ocean]

clear;clc;
x = optimvar('x',2,2);
y=optimvar('y',2,2)
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?
[sol,faval,exit]=solve(prob,'Solver','ga')

[ocean]

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

[Torsten]

There are 2^4 = 16 possibilities to combine the constraints

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;

Thus call "solve" 16 times and choose the solution with the lowest objective value.

[ocean]

of ocourse if these are 16 constraint you have you can do that,but if your matrix size is 10x10,you have to run 2^10 times........and actrully my real engineer's matrix is unfortunately 10x10

[Torsten]

The number of necessary runs would be 2^100, not 2^10.

2^100

ans = 1.2677e+30

Another way is described here:

https://math.stackexchange.com/questions/3990707/either-or-condition-for-equality-constraints

It requires introducing 100 binary variables.

[Matt J]

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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[ocean]

unbelivable! yes it's a very good solution for this problem

Answer 2

[Walter Roberson]

prob.Constraints.con1=x<=10 or prob.Constraints.con1=x^2<=10

You have a lower bound of 0 on x. Under the conditions, x^2<=10 is the more restrictive condition, so just use

prob.Constraints.con1=x<=sqrt(10)