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@...
>>>
http://lists.gnu.org/mailman/listinfo/help-glpk>>>
>>>
>>
>>
>
> --
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http://www.nabble.com/KPI-simple-function-tp26025826p26030427.html> Sent from the Gnu - GLPK - Help mailing list archive at Nabble.com.
>
>
>
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