How can I find maximum value of this function???
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
youn chan choi
am 28 Apr. 2022
Beantwortet: David Hill
am 28 Apr. 2022
P1 = 25;
P2 = 37.5;
P3 = 20;
W1 = [5, 10];
W2 = [5, 10];
W3 = [5, 10];
L1 = [40, 60];
L2 = [25, 40];
L3 = [15, 25];
r1 = [6, 15];
r2 = [6, 15];
r3 = [6, 15];
T2 = [0.25*pi, 0.5*pi];
T3 = [0, 0.5*pi];
Fa = 1000
% Each moment arms
M1 = (P1^2+W1^2)/P1;
M2 = (P2^2+W2^2)/P2;
M3 = (P3^2+(sqrt((2*W3)^2-(P3-(L3-2*W3))^2)-W3)^2)/P3;
% Each contact Forces
PA = P3 + L2*cos(T3) + L1*cos(T2 + T3);
PB = P2 + L1*cost(T2);
F1 = (Fa*r1-F2*PB-F3*PA)/M1;
F2 = (Fa*r2-F3(P3+L2*cos(T3)))/M2;
F3 = (Fa*r3)/M3;
Favg = (F1+F2+F3)/3;
% Objective functions for the dimension optimization
d1 = F1+F2+F3;
d2 = (Favg-F1)^2+(Favg-F2)^2+(Favg-F3)^2;
fdim = (d1 + 1000/(1 + d2))
I need to find maximum value of 'fdim' when there are 8 variables have to considering.
Honestly, I have no idea of how to find it... I tried many times with several methods. But, I finally couldn't do this.
So, let me know how to find the maximum value of 'fdim'.
3 Kommentare
David Hill
am 28 Apr. 2022
PB = P2 + L1*cost(T2);
% I assume you mean:
PB = P2 + L1*cos(T2);
Akzeptierte Antwort
David Hill
am 28 Apr. 2022
This is a brute force method.
P1 = 25;
P2 = 37.5;
P3 = 20;
W1 = [5, 10];
W2 = [5, 10];
W3 = [5, 10];
L1 = [40, 60];
L2 = [25, 40];
L3 = [15, 25];
r1 = [6, 15];
r2 = [6, 15];
r3 = [6, 15];
T2 = [0.25*pi, 0.5*pi];
T3 = [0, 0.5*pi];
Fa = 1000;
[p1,p2,p3,w1,w2,w3,l1,l2,l3,R1,R2,R3,t2,t3,fa]=ndgrid(P1,P2,P3,W1,W2,W3,L1,L2,L3,r1,r2,r3,T2,T3,Fa);
M1 = (p1.^2+w1.^2)./p1;
M2 = (p2.^2+w2.^2)./p2;
M3 = (p3.^2+(sqrt((2*w3).^2-(p3-(l3-2*w3)).^2)-w3).^2)./p3;
PA = p3 + l2.*cos(t3) + l1.*cos(t2 + t3);
PB = p2 + l1.*cos(t2);
F3 = (fa.*R3)./M3;
F2 = (Fa.*R2-F3.*(p3+l2.*cos(t3)))./M2;
F1 = (fa.*R1-F2.*PB-F3.*PA)./M1;
Favg = (F1+F2+F3)/3;
d1 = F1+F2+F3;
d2 = (Favg-F1).^2+(Favg-F2).^2+(Favg-F3).^2;
fdim = (d1 + 1000./(1 + d2));
Max=max(fdim,[],'all');%did you expect your calculations to go complex?
0 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Surrogate Optimization finden Sie in Help Center und File Exchange
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!