[AMPL 2586] Re: Assignment and constraints

[x-22]

Hi,

Thanks for the reply. I indeed want the former -
L[0] <= M[0], L[1] <= M[1], L[2] <= M[2], L[3] <= M[3], L[4] <= M[4].

My question was about what happens subsequently when I assign L[0] to M[j], whether this L[0] will be chosen to satisfy the constraint above L[j] <= M[j], as I expect it to. [There is a reason for why I'm doing it this way.]

Thanks so much!


--- On Sat, 6/6/09, Robert Fourer <4er@...> wrote:

> From: Robert Fourer <4er@...>
> Subject: [AMPL 2583] Re: Assignment and constraints
> To: ampl@..., "'x'" <to_rent_2000@...>
> Date: Saturday, June 6, 2009, 10:42 AM
>
>
> When you write "c1{j in 0..4}: L[j] <= M[j]" AMPL sets j
> successively to the
> values 0, 1, 2, 3, 4 and generates the corresponding
> constraints L[0] <=
> M[0], L[1] <= M[1], L[2] <= M[2], L[3] <= M[3],
> L[4] <= M[4]. 
>
> Evidently you're looking for something else, so you need to
> write your
> constraint differently.  The constraint "c3{j in
> 0..4}: L[0] <= M[j]" seems
> more nearly like what you are describing, since it
> generates L[0] <= M[0],
> L[0] <= M[1], L[0] <= M[2], ...
>
> To be sure you could get a listing of all the constraints
> generated by
> giving the AMPL commands "expand c1;", "expand c3;" and the
> like.  The
> solver does not have any effect on what constraints AMPL
> generates; only
> after AMPL generates all the constraints (and the
> objective) does the solver
> try to find a solution for them.  If FilMINT is
> crashing on some problem
> then you best bet is to bring that to the attention of
> FilMINT's developers.
>
> Bob Fourer
> 4er@...
>
>
> > -----Original Message-----
> > From: ampl@...
> [mailto:ampl@...]
> On Behalf Of x
> > Sent: Thursday, June 04, 2009 1:16 PM
> > To: ampl@...
> > Subject: [AMPL 2576] Assignment and constraints
> >
> >
> >
> > Hi all,
> >
> > I have the following. I'm keeping the syntax simple
> here for ease of
> > discussion.
> >
> > var L{0..4}, M{0..4};
> > var final >= 0;
> >
> > c1{j in 0..4}: L[j] <= M[j];
> >
> > c2{j in 1..4}: L[j] = L[0];
> >
> > c{j in 0..4}: final = sum{j} (M[j] - L[j]);
> >
> > minimize{j in 0..4}: final;
> >
> > #need this explicitly?
> > #c3{j in 0..4}: L[0] <= M[j];
> >
> > This crashes FilMINT solver. Unfortunately, I don't
> get any indication of
> > what or where the problem is when the solver crashes..
> So, I tried it with
> > MINOS solver; it ignores integrality (not related to
> the variables shown
> > here) but solves the problem. However, it picks L[0]
> to have a value that
> > is not <= each M[j] which makes final negative. But
> this is not what I
> > expected.
> >
> > So, my question is this - when L[1..4] is assigned
> L[0], will L[0] value
> > be selected by the solver to be <= each of the M[j]
> per constraint c1?
> >
> > [Or, does the solver pick L[0] <= M[0], assign L[0]
> to the rest of the
> > L[j] and then crash if c1 or final_LB is violated? In
> that case, I'd have
> > to include c3 constraint but even that hasn't helped
> me with FilMINT so
> > far. However, me forcing L[0] to have a valid value
> (through a constraint)
> > helps..]
> >
> > Sincere thanks in advance for any and all
> information!
> >
>
>
>
> >
>

     

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