sci.op-research

On Feb 27, 2:03 pm, Paul Rubin wrote:
> (P1) min x^2 s.t. x >= 3 - a; x <= 1
>
> (P2) min a^2 s.t. a <= 2*x.

Let's say (P2) is the inner problem. Since we can obtain closed form
solution for this simple example, I just do that instead of using
first order necessary condition.
We must first solve the inner problem for a and treat x as a
parameter. We get:

a*=0 if x>=0
a*=infeasible if x<0

So we only have one feasible candidate. Plug a=0 and x>=0 in the outer
problem and solve for x. You get " Min x^2 s.t. x>=3, x<=1, x>=0 "
which is infeasible.

A bit more interesting problem would be obtained if in (P2) "a<= 2*x"
is replaced by "a>= 2*x". Then you get a*=0 when x<=0 and a*=2*x when
x>0.

Pluging the first case in (P1) yields infeasible, but for the second
case we get "Min x^2 s.t. x >= 3 - 2*x; x <= 1 , x>=0" where x*=1 is
the solution (and so a*=2).

Now let me solve your problem with necessary condition. Returning to
the inner problem (P2):
2*a+u=0
a <= 2*x
u*(a-2*x)=0
u>=0

Adding these constraints to P1 we get
min x^2 s.t. x >= 3 - a; x <= 1 , 2*a+u=0 , a <= 2*x , u*(a-2*x)=0 ,
u>=0 . As expected, the feasible set is empty.

Now let's go over my modified P2:
2*a-u=0
a>=2*x
u*(2*x-a)=0
u>=0

Adding these constraints to P1 we get
min x^2 s.t. x >= 3 - a; x <= 1 , 2*a-u=0 , a >= 2*x , u*(2*x-a)=0 ,
u>=0 .
The feasible set is only a point x=1, a=2, u=4 and hence x*=1 and
a*=2, which is equal to what we obtained from closed form solution.

I hope this could clarify the problem.

Thanks

G.D.