Re: [Help-glpk] KPI simple function

[xypron]

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
>> address@hidden
>> http://lists.gnu.org/mailman/listinfo/help-glpk
>> 
>> 
> 
> 

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