Azzera filtri
Azzera filtri

fmincon, find integer values as optimal values

5 visualizzazioni (ultimi 30 giorni)
f =@(fr)(50*fr(1)^2 + 100)/fr(1) + (175*fr(2)^2 + 150)/fr(2) + (160*fr(3)^2 + 250)/fr(3)
lb = [0,0,0];
ub = [5,5,5];
A = [];
b = [];
Aeq = [];
beq = [];
fr0 = [1,1,1];
fr = fmincon(f,fr0,A,b,Aeq,beq,lb,ub)
output:
f =
function_handle with value:
@(fr)(50*fr(1)^2+100)/fr(1)+(175*fr(2)^2+150)/fr(2)+(160*fr(3)^2+250)/fr(3)
Local minimum possible. Constraints satisfied.
fmincon stopped because the size of the current step is less than
the default value of the step size tolerance and constraints are
satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
fr =
1.4142 0.9258 1.2500
I want to find positive integer values rather than decimal values for my variables. is there any way to include this condition with fmincon? any help will be highly appreciated. thank you

Risposta accettata

Sean de Wolski
Sean de Wolski il 1 Feb 2018
Modificato: Sean de Wolski il 1 Feb 2018
fmincon is not designed to deal with integer x values. You should try ga() which has an IntCon option or patternsearch() with a round() on the input values.
Of course for a small problem like this there are only 216 unique combinations of x so you could easily brute force it.
[rr,cc,pp] = ndgrid(0:5);
v = (50*rr.^2 + 100)./rr + (175*cc.^2 + 150)./cc + (160*pp.^2 + 250)./pp;
[val, idx] = min(v(:));
[rr(idx) cc(idx) pp(idx)]
  1 Commento
Mohamed Musni
Mohamed Musni il 1 Feb 2018
thank you. I found tutorial with ga() and abled to get answer with integer values. have a nice day

Accedi per commentare.

Più risposte (0)

Community Treasure Hunt

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

Start Hunting!

Translated by