As I have described before, my problem is
max min c'x (P1)
s.t. Ax <= b - Du : λ1
Ex == f - Fu : λ2
g(u) <= 0
wherein u is the outer decision variables and x is the inners.
Before, I have tried to use KKT method to render it but it seems impractical because the KKT method introduces many dual variables and constraints.
So, now I will try the dual method, i.e
max -(b - Du)*λ1+(f - Fu)*λ2 (P2)
s.t. -A* λ1+E*λ2=c
λ1>=0
g(u)<=0
So, using the kkt function, we can get that details.b = b - D*u and details.f = f - F*u.
But, I donnot know whether the sdpvars in details.b and details.f is the u of the primal problem or not.
How to give the constrints for u in the problem (P2)?
The constraints [-details.A'*details.dual + details.E'*details.dualeq == details.c 0<=details.dual<=M - M<=details.dualeq<= M g(u)<=0] seems not enough.
The u seems not correspond to the sdpvars details.b and details.f.