Re: [Help-glpk] KPI simple function
Hi Xypron, I'm sorry for disturbing you again. Following your suggestion, I created this model: var Z1; var Z2; var U; maximize obj : 0.18 * Z1 + 0.82 * Z2; #maximize obj : 0.5 * Z1 + 0.5 * Z2; s.t. c1 : 20 * Z1 + 110 * U = 83; # (0.3,0.7)-(0.85,0.6) s.t. c2 : 28 * Z1 - 8 * U = 19; # (0.85,0.6)-(0.75,0.25) s.t. c3 : 4 * Z1 + 40 * U = 13; # (0.75,0.25)-(0.25,0.3) s.t. c4 : 80 * Z1 - 10 * U = 17; # (0.25,0.3)-(0.3,0.7) s.t. c5 : 40 * Z2 - 5 * U = 9; # (0.25,0.2)-(0.3,0.6) s.t. c6 : 30 * Z2 - 110 * U = - 57; # (0.3,0.6)-(0.85,0.75) s.t. c7 : 80 * Z2 - 20 * U = 53; # (0.85,0.75)-(0.75,0.35) s.t. c8 : 12 * Z2 - 40 * U = -5; # (0.75,0.35)-(0.25,0.2) solve; printf U =%6.3f, Z1 = %6.3f, Z2 = %6.3f\n, U, Z1, Z2; end; When I try to solve this problem I get the same solution whatever is the objective function, but I need that when I change the objective function the result is different. The real problem is this: U is an action and Z1 and Z2 are two KPIs that depend on U. I'm looking for just a model to represent this in a simple way to finish my project. Sadly, I don't have the real KPI. Thanks Simone On 23/ott/09, at 19:31, xypron wrote: Hello Simone, There was a typo s.t. c4 : y =0; # (3,0)-(0,0) Best regards Xypron xypron wrote: Hello Simone, if the solution of a linear program is unique, it will always be in a vertex of the convex polyeder described by the constraints. The objective function gives the optimization direction and hence decides which vertex of the polygon is the solution. For a two dimensional problem lets think of an polygon given by the following vertices: (0,0) (1,1) (2,1) (3,0) This corresponds to the following inequalities: s.t. c1 : x - y = 0; # (0,0)-(1,1) s.t. c2 : y = 1; # (1,1)-(2,1) s.t. c3 : x + y = 3; # (2,1)-(3,0) s.t. c4 : x =0; # (3,0)-(0,0) If our optimization direction is (1,1) the objective is maximize obj : x + y; The solution is vertex (2,1) If our optimization direction is (-1,1) the objective is maximize obj: -x + y; The solution is vertex (1,1); The complete model is: var x; var y; # uncomment the appropriate objective #maximize obj : x + y; # direction (1,1); maximize obj : -x + y; # direction (-1,1); s.t. c1 : x - y = 0; # (0,0)-(1,1) s.t. c2 : y = 1; # (1,1)-(2,1) s.t. c3 : x + y = 3; # (2,1)-(3,0) s.t. c4 : x =0; # (3,0)-(0,0) solve; printf x = %6.3f, y = %6.3f\n, x, y; end; Best regards Xypron Simone Atzeni wrote: Hi all, I'm looking for two functions that could represent simple KPIs. In other world, I would like two MILP, in this way: MILP 1: MAX J = 0.5 * Z1 + 0.5 * Z2 Z1 = -AX + C Z2 = BX + D and MILP 2: MAX J = 0.32 * Z1 + 0.68 * Z2 Z1 = -AX + C Z2 = BX + D Z1 and Z2 are the values of the KPI and they depend on X. The constraints should be equal but the results (the values of Z1 and Z2) should be different changing the coefficients fo the objective function, in this case (0.5 - 0.5) for the MILP1 and (0.32 - 0.68) for the MILP 2. I can't find a good function. I need just functions where Z1 and Z2 depend on X but changing the coefficients in the objective functions change the values of Z1, Z2 and X. MILPs I'm using are the follow: MAX J = 0.5 Z.1 + 0.5 Z.2 Z.1 = 5X (0.196116135138184 Z.1 - 0.98058067569092 U.1 = 0) (the equations have been normalized) Z.2 = -3X + 4 (0.196116135138184 Z.2 + 0.115384615384615 U.1 = 0.153846153846154) and MAX J = 0.32 Z.1 + 0.68 Z.2 Z.1 = 5X Z.2 = -3X + 4 This is the picture of the two functions: Both MILPs have the same solution. Z.1 = 1 Z.2 = 0.666795 X = 0.2 In this case the weights, (0.5 - 0.5) for the MILP1 and (0.32 - 0.68) for the MILP 2, don't influence the results of the MILP. I want something in a way that the weights influence the results, so that the two MILPs have different result but they should being equal. Can someone help me? Thanks Simone ___ Help-glpk mailing list Help-glpk@gnu.org http://lists.gnu.org/mailman/listinfo/help-glpk -- View this message in context: http://www.nabble.com/KPI-simple-function-tp26025826p26030427.html Sent from the Gnu - GLPK - Help mailing list archive at Nabble.com. ___ Help-glpk mailing list Help-glpk@gnu.org http://lists.gnu.org/mailman/listinfo/help-glpk ___ Help-glpk mailing list Help-glpk@gnu.org http://lists.gnu.org/mailman/listinfo/help-glpk
[Help-glpk] KPI simple function
Hi all,I'm looking for two functions that could represent simple KPIs.In other world, I would like two MILP, in this way:MILP 1:MAX J = 0.5 * Z1 + 0.5 * Z2Z1 = -AX + CZ2 = BX + DandMILP 2:MAX J = 0.32 * Z1 + 0.68 * Z2Z1 = -AX + CZ2 = BX + DZ1 and Z2 are the values of the KPI and they depend on X. The constraints should be equal but the results (the values of Z1 and Z2) should be different changing the coefficients fo the objective function, in this case (0.5 - 0.5) for the MILP1 and (0.32 - 0.68) for the MILP 2.I can't find a good function. I need just functions where Z1 and Z2 depend on X but changing the coefficients in the objective functions change the values of Z1, Z2 and X.MILPs I'm using are the follow:MAX J = 0.5 Z.1 + 0.5 Z.2Z.1 = 5X (0.196116135138184 Z.1 - 0.98058067569092 U.1 = 0) (the equations have been normalized)Z.2 = -3X + 4 (0.196116135138184 Z.2 + 0.115384615384615 U.1 = 0.153846153846154)andMAX J = 0.32 Z.1 + 0.68 Z.2Z.1 = 5XZ.2 = -3X + 4This is the picture of the two functions:Both MILPs have the same solution.Z.1 = 1Z.2 =0.666795X = 0.2In this casethe weights,(0.5 - 0.5) for the MILP1 and (0.32 - 0.68) for the MILP 2, don't influence the results of the MILP. I want something in a way that the weights influence the results, so that the two MILPs have different result but they should being equal.Can someone help me?ThanksSimone___ Help-glpk mailing list Help-glpk@gnu.org http://lists.gnu.org/mailman/listinfo/help-glpk
Re: [Help-glpk] KPI simple function
Hello Simone, if the solution of a linear program is unique, it will always be in a vertex of the convex polyeder described by the constraints. The objective function gives the optimization direction and hence decides which vertex of the polygon is the solution. For a two dimensional problem lets think of an polygon given by the following vertices: (0,0) (1,1) (2,1) (3,0) This corresponds to the following inequalities: s.t. c1 : x - y = 0; # (0,0)-(1,1) s.t. c2 : y = 1; # (1,1)-(2,1) s.t. c3 : x + y = 3; # (2,1)-(3,0) s.t. c4 : x =0; # (3,0)-(0,0) If our optimization direction is (1,1) the objective is maximize obj : x + y; The solution is vertex (2,1) If our optimization direction is (-1,1) the objective is maximize obj: -x + y; The solution is vertex (1,1); The complete model is: var x; var y; # uncomment the appropriate objective #maximize obj : x + y; # direction (1,1); maximize obj : -x + y; # direction (-1,1); s.t. c1 : x - y = 0; # (0,0)-(1,1) s.t. c2 : y = 1; # (1,1)-(2,1) s.t. c3 : x + y = 3; # (2,1)-(3,0) s.t. c4 : x =0; # (3,0)-(0,0) solve; printf x = %6.3f, y = %6.3f\n, x, y; end; Best regards Xypron Simone Atzeni wrote: Hi all, I'm looking for two functions that could represent simple KPIs. In other world, I would like two MILP, in this way: MILP 1: MAX J = 0.5 * Z1 + 0.5 * Z2 Z1 = -AX + C Z2 = BX + D and MILP 2: MAX J = 0.32 * Z1 + 0.68 * Z2 Z1 = -AX + C Z2 = BX + D Z1 and Z2 are the values of the KPI and they depend on X. The constraints should be equal but the results (the values of Z1 and Z2) should be different changing the coefficients fo the objective function, in this case (0.5 - 0.5) for the MILP1 and (0.32 - 0.68) for the MILP 2. I can't find a good function. I need just functions where Z1 and Z2 depend on X but changing the coefficients in the objective functions change the values of Z1, Z2 and X. MILPs I'm using are the follow: MAX J = 0.5 Z.1 + 0.5 Z.2 Z.1 = 5X (0.196116135138184 Z.1 - 0.98058067569092 U.1 = 0) (the equations have been normalized) Z.2 = -3X + 4 (0.196116135138184 Z.2 + 0.115384615384615 U.1 = 0.153846153846154) and MAX J = 0.32 Z.1 + 0.68 Z.2 Z.1 = 5X Z.2 = -3X + 4 This is the picture of the two functions: Both MILPs have the same solution. Z.1 = 1 Z.2 = 0.666795 X = 0.2 In this case the weights, (0.5 - 0.5) for the MILP1 and (0.32 - 0.68) for the MILP 2, don't influence the results of the MILP. I want something in a way that the weights influence the results, so that the two MILPs have different result but they should being equal. Can someone help me? Thanks Simone ___ Help-glpk mailing list Help-glpk@gnu.org http://lists.gnu.org/mailman/listinfo/help-glpk -- View this message in context: http://www.nabble.com/KPI-simple-function-tp26025826p26030380.html Sent from the Gnu - GLPK - Help mailing list archive at Nabble.com. ___ Help-glpk mailing list Help-glpk@gnu.org http://lists.gnu.org/mailman/listinfo/help-glpk
