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Is there a way to solve semi infinite optimization problem?

Posted 5 years ago

Hello

Let's say your constraint contains a parameter, for example $x*cos(t)+y < 1$, it must hold for all $t\in [0;2pi]$ . Is it possible to include a constraint like this? I don't see any solution in the documentation of Minimize/Maximize:

The constraints cons can be any logical combination of:
lhs==rhs equations
lhs!=rhs inequations
lhs>rhs or lhs>=rhs inequalities
{x,y,…}?reg region specification
Exists[x,cond,expr] existential quantifiers

POSTED BY: Richard Toth
3 Replies
Posted 5 years ago

I wrote some code, which might look messy, but hope you can see through it and tell me how to make it work. Minimize outputs itself, which is becouse it doesn't have enough information to solve the problem?

n = 2;
k = Binomial[n,2];
m = n + k;
Xs = Array[x,n];
Ds = Array[d,m];
PHIs = Array[phis,n-1];
Cs = Array[c,n];
c[1]=0
c[2]=0
x[1]=d[1]*Cos[d[3]]*Cos[phis[1]]-d[2]*Sin[d[3]]*Sin[phis[1]]+c[1]
x[2]=d[1]*Sin[d[3]]*Cos[phis[1]]+d[2]*Cos[d[3]]*Sin[phis[1]]+c[2]
expr = x[1]^2-x[2]-2
objfun = -Times @@ Part[Ds,1;;n] 
Minimize[{objfun,{phis[1]>=0,phis[1]<=2*Pi,!Exists[phis[1],expr>0]}},{d[1],d[2],d[3]}]

The code supposed to solve:
min $-d_1*d_2$
s.t
$x_1^2-x_2-2\leq 0$
where $x_1=d_1*cos(d_3)*cos(\phi_1)-d_2*sin(d_3)*sin(\phi_1)+c_1$ and $x_2=d_1*sin(d_3)*cos(\phi_1)+d_2*cos(d_3)*sin(\phi_1)+c_2$
for all $\phi_1$ in the [0;2pi] interval, which can be transformed to the !Exist form according to Frank's suggestion

This problem is finding the maximal area ellipse in the $x_2\geq x_1^2-2$ region, given it's center (0,0).

POSTED BY: Richard Toth

If that expression holds over that range then there doesn't exist any point where it is violated in that range, so negating Exists might work.

POSTED BY: Frank Kampas
Posted 5 years ago

Thanks, sounds good

POSTED BY: Richard Toth
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