Intlinprog function - problem

2 vues (au cours des 30 derniers jours)
Michal Malec
Michal Malec le 13 Sep 2020
Modifié(e) : Matt J le 13 Sep 2020
I want to solve a math problem that give me a minimal value of a function. The function of the objective is:
min 64x1+40x2
The invariants are:
3x1+5x2150
8X1+5X2300
0X1+1X220
I wrote the code in Matlab:
f=-[64;40]; intcon = [1;2]; A=[3,5;8,5;0,1]; B=[150;300;20];
[x,fval] = intlinprog(f,intcon,A,B,[],[],lb)
When I run this code, I get the results:
x =
30.0000
12.0000
fval =
-2.4000e+03
And this results are ok, but when I changes the value of the objective function from 64 to 64.01, I get the strongly changed values of X:
x =
35.0000
4.0000
fval =
-2.4003e+03
I would like to know why the value of x varies so much.

Réponse acceptée

Matt J
Matt J le 13 Sep 2020
Modifié(e) : Matt J le 13 Sep 2020
The optimal solution of a linear program can be discontinuous as a function of the problem data, particular when you have integer constraints. Here is an example where it is easier to understand how integer constraints can induce this behavior.
>> LB=[0,0]; x=intlinprog([1,1],1:2,[],[],[],[],LB).'
x =
0 0
One sees here that the optimum is achieved at the boundaries specified by LB. Clearly therefore, if I increase the lower bounds by even a small amount, the optimal x(i) must jump to the next integer:
>> x=intlinprog([1,1],1:2,[],[],[],[],LB+0.00002).'
x =
1 1
  1 commentaire
Matt J
Matt J le 13 Sep 2020
Modifié(e) : Matt J le 13 Sep 2020
Even in non-integer problems, the solution can jump strongly with a small perturbation in the problem data. Consider this example, where the feasible region is the unit square.
f=[0,1];
delta_f=1e-10*[1,0];%A very small perturbation
x=intlinprog(f+delta_f,[],[],[],[],[],[0,0],[1,1]).', %optimum=[0,0]
x=intlinprog(f-delta_f,[],[],[],[],[],[0,0],[1,1]).', %optimum=[1,0]
Despite the small difference by 2*delta_f in the objective coefficients, the optimal x(1) solution changes a lot. This happens because the objective, when viewed as a 2D line, is very nearly parallel to one of the faces of the unit square, so a small tipping in its slope will cause the optimum to occur at a different vertex.

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Surrogate Optimization dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by