Hello Xiaomi, in linear programming, the objective function is linear and the solution space is bounded by a convex polyhedron. Hence you cannot formulate your problem as an LP with x, y as variables.
Of cause reformulating the problem with binaries takes more effort then writing down the solution algorithm for this problem with GLPK. param xmin := 2; param xmax := 6; param xrange := 10; param ymin := 1; param ymax := 6; param yrange := 10; printf "x = %f\n", if (xmin <= xrange - xmax ) then xmin else xmax; printf "y = %f\n", if (ymin <= yrange - ymax ) then ymin else ymax; end; If you are not interested in the unique values of x, y you can of cause introduce new variables: u, >= min( xmin, xrange - xmax ) := min( x, xrange - x ); v, >= min( ymin, yrange - ymax ) := min( y, yrange - y ); This allows you to define an LP giving you the correct objective value. Best regards Xypron -------- Original-Nachricht -------- > Datum: Fri, 01 Oct 2010 18:58:55 -0400 > Betreff: [Help-glpk] Can this problem be solved in LP without binary > condition? > The problem is: Given a total area, let's say, from (0,0) to (10,10); > given an area for one point(x,y) that movable, let's say 2<=x<=6, > 1<=y<=6. Find the closest position of (x,y) to the boundary of the total > area. In the example, the solution should be (x,1), x can be any in the > constraint. > > Can this problem be solved in LP without binary condition? > Thanks. > > _______________________________________________ > Help-glpk mailing list > [email protected] > http://lists.gnu.org/mailman/listinfo/help-glpk -- GRATIS: Spider-Man 1-3 sowie 300 weitere Videos! Jetzt freischalten! http://portal.gmx.net/de/go/maxdome _______________________________________________ Help-glpk mailing list [email protected] http://lists.gnu.org/mailman/listinfo/help-glpk
