The A=1, B=1 solution is the solution that finds the most equalized set of
variables. If that is what you want, then one way is to minimize the absolute
value of the deviations around the average of the variables.
For example, instead of A, B (and to slightly generalize this), suppose you
have a set N and variables
var x{N} >= 0;
Also suppose that you want to find the solution that has as many of the x's the
same as possible. Use three more sets of variables:
var r_plus{N} >=0;
var r_neg{N} >=0;
var x_mean;
Add in the following constraints:
s.t. Calculate_Average_x: x_mean = (sum{j in N} x[j])/card(N);
s.t. EqualizeValues{j in N}: r_plus[j] - r_neg[j] = x[j] - x_mean;
Then in your objective function, add in a term that penalizes the absolute
deviation of the x's around their average:
minimize some_objective:
[your other terms in the objective function] +
99*sum{j in N} (r_plus[j] + r_neg[j]);
The "99" is a penalty - you would need to modify this penalty to see whether
the solutions you are achieving are the ones that you desire.
-Marc
From: [email protected]
[mailto:[email protected]] On Behalf Of
esma mehiaoui
Sent: Tuesday, October 09, 2012 2:53 AM
To: [email protected]
Subject: [Help-glpk] multi-objectives function
Hello everyone,
I would like to know if it is possible to maximize several objectives.
I mean, if i would like to maximize two variables (A and B) and i have three
solutions for which the sum is maximized:
A=0 and B =2
A=2 and B =0
A=1 and B =1
In the three above cases the sum A+B is equal to 2, so all solutions are the
best ones, but i would like to get the third solution (A=1 and B=1) because
this solution maximize both values of A and B (not only one of them).
.
Will someone help?
Thank you
Esma
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