MATLAB Answers

I've got a problem with optimization and using a variable for exponent

13 views (last 30 days)
Minsu Seo
Minsu Seo on 26 Feb 2020
Commented: Minsu Seo on 28 Feb 2020
Here is my matlab codeR
x = optimvar('x','type','integer','LowerBound',0,'UpperBound',48);
y = optimvar('y','type','integer','LowerBound',0,'UpperBound',48);
obj = fcn2optimexpr(@objfunx,x,y);
prob = optimproblem('Objective',obj);
con1 = x <= 48; prob.Constraints.constr = con1;
con2 = y <= 48; prob.Constraints.constr = con2;
con3 = (1-((0.2.^x))) + ((0.2.^x)).*(1-((0.3.^y))) >= 0.99;
prob.Constraints.constr = con3;
x0.x = 0; x0.y = 0;
show(prob)
function f = objfunx(x,y)
f =(60).*(x) + ((0.2.^x)).*(y).*(20);
end
and I got this error message
Error using optim.internal.problemdef.Power
Exponent must be a finite real numeric scalar.
Error in .^
Error in Test (line 8)
I want to get solution x, y
how can I solve this problem? I'm going to appreciate if you give me whole code

Accepted Answer

Alan Weiss
Alan Weiss on 27 Feb 2020
Edited: Alan Weiss on 27 Feb 2020
Optimization Toolbox™ does not support general nonlinear integer programming. Use ga or surrogateopt for nonlinear integer programming.
Or, for this problem, you could do exhaustive search; there are fewer than 2500 = 50^2 possibilities.
Alan Weiss
MATLAB mathematical toolbox documentation
  1 Comment
Minsu Seo
Minsu Seo on 28 Feb 2020
Thanks Alan Weiss, I already solved this problem with exhaustive search you mentioned.
I just wondered if I could solve that problem using some internal function that the MATLAB provides.
Your answer is really helpful for me and thank you again.

Sign in to comment.

More Answers (0)

Products


Release

R2019b

Community Treasure Hunt

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

Start Hunting!

Translated by