sci.op-research

On Mar 22, 4:20=A0am, Wit Jakuczun wrote:
> Dnia Sat, 22 Mar 2008 09:18:27 +0100
> Wit Jakuczun napisa=B3(a):
>
>
>
>
>
> > Dnia Fri, 21 Mar 2008 15:35:08 -0700 (PDT)
> > "xgenera...@gmail.com" napisa=B3(a):
>
> > > if x(i)=3Dxp(i)-xn(i) where xp(i) and xn(i) are two nonnegative slack
> > > variables, then to enforce exactly one of them to be zero =A0I would
> > > need to maximize
> > > |x(i)|=3Dz=3Dxp(i)+xn(i).
> > > But this is not the case for me.
> > You can also bind with them two binary variables: xpb(i) and xnb(i)
> > such that:
> > =A0 =A0xp(i) <=3D xpb(i)*UB
> > =A0 =A0xn(i) <=3D xnb(i)*UB
>
> > By adding constraint: xpb(i) + xnb(i) =3D 1 you will get exactly one
> > of them 0 without maximizing |x(i)|.
>
> Having: xpb(i) + xnb(i) <=3D 1 you could use the variables to
> estimate whether x(i) is zero.
>
> Best regards
> --
> [ Wit Jakuczun =A0 =A0 ]
> [ WLOG Solutions =A0http://www.wlogsolutions.com]- Hide quoted text -
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> - Show quoted text -

Actually, xpb(i) + xnb(i) <=3D 1 cannot be used to determine if x(i) is
zero or not. b/c x(i)=3D0 implies xp(i)=3D0 and xn(i) =3D 0, and we will
have:

xp(i) <=3D xpb(i)*UB =3D> 0 <=3D xpb(i)*UB
xn(i) <=3D xnb(i)*UB =3D> 0 <=3D xnb(i)*UB

and this implies that at least one of xnb(i) or xpb(i) can be still 1.