Re: [Help-glpk] [Fwd: Find nearest point]

[glpk xypron]

Hello Paul,

GLPK only calculates with limited precision.

Unfortunately you have chosen big_number unreasonably high. If
you change it to 10, you get the expected result.

In a big M approach you should always use the smallest possible
number for M.

The problem is recurring cf.
http://lists.gnu.org/archive/html/help-glpk/2011-04/msg00021.html

I added a paragraph in the Wikibook
http://en.wikibooks.org/wiki/GLPK/Modeling_tips#Big_M

Best regards

Xypron

> -------- Forwarded Message --------
> Subject: Find nearest point
> Date: Wed, 15 Jun 2011 10:10:25 +0200 (CEST)
> 
> Hi.
> 
> I'm currently working on a project that aims to compute the consign we
> should give to an uav in order to reach a point.
> 
> The first step has been to model the problem without obstacles. It seems
> to work (we find the straight line :) ).
> 
> Now we need to add obstacles. And in my new model, glpk seems to ignore
> some constraints. 
> 
> I tried to make a minimalist example to reproduce the "problem".
> 
> When i run this example, b[0] and b[1] both equals 0 and i wonder how c6
> is satisfied ?
> 
> I suppose that i have made a mistake. Could you explain my why ?
> 
> Thank you.
> 
> Paul.
> 
> 
> 
> 
> # trivial.dat
> data;
> param big_number := 100000;
> param xref := 5;
> param xobs := 5;
> param cobs := 1;
> end;
> 
> #trivial .mod
> param big_number;
> # reference coordinate
> param xref;
> # obstacle coordinate
> param xobs;
> # obstacle side
> param cobs;
> 
> var x;
> var xerr;
> var b{i in 1..2}, binary;
> 
> minimize f : xerr;
> 
> # xerr = abs(x - xref)
> s.t. c1 :   x - xref <= xerr;
> s.t. c2 :  -x + xref <= xerr;
> 
> # b[1] = 0 if x is to the left of xobs
> s.t. c3 : x >= xobs - cobs - big_number * (1 - b[1]);
> s.t. c4 : x <= xobs - cobs + big_number *      b[1];
> 
> # b[2] = 0 if x is to the right of xobs
> s.t. c5 : x <= xobs + cobs + big_number * (1 - b[2]);
> s.t. c6 : x >= xobs + cobs - big_number *      b[2];
> 
> # x must be at least to the right or to the left
> s.t. c7 : sum{j in 1..2} b[j] <= 1;
> 
> solve;
> 
> display xobs;
> display cobs;
> display x;
> display b;
> 
> end;


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