Re: [Help-glpk] function

[Mate Hegyhati]

Hi Zvonko,

maybe something like this?

R >= 0
R <= P - PMIN*X
R <= PMAX*X - P

This should be enough, if R is not represented anywhere else, and it
should be as much as possible in the objective.

Otherwise I think You need an additional binary variable, Y,
representing whether P is closer to PMIN or to PMAX.

P <= (PMIN + PMAX) / 2 + (1-Y) * (PMAX-PMIN)/2
P >= (PMIN + PMAX) / 2 - (1-Y) * (PMAX-PMIN)/2

And then You can add these to the above constraints:

R >= P - PMIN - Y * PMAX
R >= PMAX - P - (1-Y)*PMAX

Hope this is what You are looking for, best regards,

Mate

On 08/17/2012 10:02 AM, Zvonko Bregar wrote:
> Hi,
> 
> My name is Zvonko Bregar.
> 
> Please help me with the following problem.
> 
> There is a real variable P having disjunct definition area {0} + [PMIN, PMAX].
> There is also a binary variable X helping in this definition.
> If X = 0 then P = 0
> And if X = 1 then PMIN < P < PMAX.
> So variables X an P work simultaneously.
> This should be a rather standard construction.
> 
> The question is to construct another variable, say R, from X an P
> Such that
> If X = P = 0 then also R = 0
> Else R = min {P-PMIN,PMAX-P}
> 
> In other words, R measures the minimal distance of P to its boundaries PMIN 
> and PMAX.
> But if P = 0 then R must be set to 0 too.
> 
> For example, if PMIN = 200 and PMAX = 400, than the required function should 
> be something like the "upside down" absolute value function:
> 
> R =  - | P - 300 | + 100  if 200 < P < 400
> 
> But also R = 0 if P = 0.
> 
> I have been trying this for quite some time but didn't succeed. It looks 
> simple and perhaps there is a simple solution? Or perhaps this cannot be done 
> at all?
> 
> Thank you
> 
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