MOD for optimization variable
8 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Hi,
I have to optimize matrix to obtain quantities that will be divisible by certain numbers. When I use MOD on numbers, it works fine, for example MOD(10,3)=1. However, when I try to create constraint that I want my optimization variable x to be divisible by let's say 3 MOD(x,3), I get an error
Undefined function 'mod' for input arguments of type 'optim.problemdef.OptimizationVariable'.
Is there anything I can do to solve this issue?
Thanks in advance.
0 commentaires
Réponse acceptée
Matt J
le 24 Août 2019
Modifié(e) : Matt J
le 24 Août 2019
No need to use constraints. Just declare x as an OptimizationExpression,
x=3*optimvar('z','Type', 'Integer','LowerBound',0)
x =
OptimizationExpression
3*z
Now use x freely to build the objective and constraints as you did previously. E.g.,
x=3*optimvar('z','Type', 'integer','LowerBound',0);
y=optimvar('y','LowerBound',0);
prob=optimproblem;
prob.Objective=2*x+4*y;
prob.Constraints=x+y>=7;
sol=solve(prob);
The solver will solve for z, not x,
>> sol
sol =
struct with fields:
y: 1.0000
z: 2.0000
but you can easily convert the optimal z to a corresponding result for x.
>> x_optimal=evaluate(x,sol)
x_optimal =
6.0000
3 commentaires
Matt J
le 25 Août 2019
annazy's comment moved here:
This is wonderful. What I actually need is this 'z', as I optimize quantities of products which must be divisible by number of products per cartoon. However this number varies depending on products. Also, I have multiple percentage constraints on quantities (not cartoons) which is why I can't focus on cartoons only, but in the end number of cartoons is what I need.
You are both amazing people, I hope one day I will be clever enough to help other lost souls like you do help me :)
Matt J
le 25 Août 2019
I'm glad it's what you need but please Accept-click the answer if it addresses your question.
Plus de réponses (1)
Walter Roberson
le 22 Août 2019
Introduce an extra integer variable and constrain equality x-3*extra = 1
4 commentaires
Walter Roberson
le 24 Août 2019
Yes, N=optimvar('N','Type','integer') is good.
If you have upper and/or lower bound on x then you can use it to figure out upper and lower bounds on N to make the optimization more efficient.
Voir également
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!