how to have function with scalar value
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Greetings people! I have a code and 2 objective functions when I want to use fmincon it says function has to be scalar
here is my code:
function z = MyCost(x)
z1 = Obj1(x);
z2 = Obj2(x);
z = [z1;z2];
end
I know that my problem is because of above code z=[z1;z2] but I dont know how to have z which consist of z1 and z2 in a way to have scalar value of z
clc;
clear;
close all;
%% Problem Definition
nVar=3;
VarSize=[1 nVar];
VarMin=10^(-3);
VarMax=10^2;
%% Weighted-Sum Approach
N=60;
w1=linspace(0,1,N);
w2=1-w1;
for i=1:N
FWS=@(x) w1(i)*Obj1(x)+w2(i)*Obj2(x);
x0=unifrnd(VarMin,VarMax,VarSize);
A=[];
b=[];
LB=10^(-3);
UB=10^2;
options=optimoptions('fmincon','TolFun',10^(-12),'TolCon',10^(-10),'MaxFunEvals',4*10^3);
sol(i).Position=fmincon(FWS,x0,A,b,[],[],LB,UB,@NLC,options);
sol(i).Cost=MyCost(sol(i).Position);
end
Costs=[sol.Cost];
figure;
plot(Costs(1,:),Costs(2,:),'.');
function [C, Ceq]=NLC(x)
n=2;
C=[((30767*((1000*x(1:n))/29011 - 50500/29011)^2)/100000 - (37571*x(1:n))/2901100 + (33539*((1000*x(1:n))/29011 - 50500/29011)^3)/100000 - (8057*((1000*x(1:n))/29011 - 50500/29011)^4)/12500 + (7961*((1000*x(1:n))/29011 - 50500/29011)^5)/25000 + (1278950236579183*((1000*x(1:n))/29011 - 50500/29011)^6)/18014398509481984 - (1433*((1000*x(1:n))/29011 - 50500/29011)^7)/10000 + (360206905396347*((1000*x(1:n))/29011 - 50500/29011)^8)/9007199254740992 + (6604510839140323*((1000*x(1:n))/29011 - 50500/29011)^9)/576460752303423488 - (3107872853893447*((1000*x(1:n))/29011 - 50500/29011)^10)/576460752303423488 + 37931111499167682390657/52261571515858183782400)^2 + ((787049070879267875*x(1:n))/1045231430317163675648 + (27059*((1000*x(1:n))/29011 - 50500/29011)^2)/50000 - (75971*((1000*x(1:n))/29011 - 50500/29011)^3)/100000 + (19587*((1000*x(1:n))/29011 - 50500/29011)^4)/100000 + (3329*((1000*x(1:n))/29011 - 50500/29011)^5)/10000 - (14281*((1000*x(1:n))/29011 - 50500/29011)^6)/50000 + (634413072308427*((1000*x(1:n))/29011 - 50500/29011)^7)/18014398509481984 + (2918764904100309*((1000*x(1:n))/29011 - 50500/29011)^8)/72057594037927936 - (3668884458035139*((1000*x(1:n))/29011 - 50500/29011)^9)/288230376151711744 + (1831068088942187*((1000*x(1:n))/29011 - 50500/29011)^10)/295147905179352825856 - 284748233055908747705/2090462860634327351296)^2-x(n+1:end)
-x(n+1:end)];
Ceq=[];
end
function z = Obj2(x)
yref = 0;
p1 = -0.0053913;
p2 = 0.011457;
p3 = 0.039991;
p4 = -0.1433;
p5 = 0.070996;
p6 = 0.31844;
p7 = -0.64456;
p8 = 0.33539;
p9 = 0.30767;
p10 = -0.37571;
p11 = 0.071788;
mu = 50.5;
sigma = 29.011;
s = [10^6 10^4 10^2];
S = diag(s);
n=2;
g = (x(1:n)-mu)/sigma;
z = sum((yref-(p1*g.^10+p2*g.^9+p3*g.^8+p4*g.^7+p5*g.^6+p6*g.^5+p7*g.^4+p8*g.^3+p9*g.^2+p10*g+p11)).^2)+x(n+1:end)'*S*x(n+1:end);
end
function z = Obj1(x)
yref = 0;
p1 = -6.2039e-06;
p2 = 0.012729;
p3 = -0.040506;
p4 = -0.035217;
p5 = 0.28562;
p6 = -0.3329;
p7 = -0.19587;
p8 = 0.75971;
p9 = -0.54118;
p10 = -0.021845;
p11 = 0.098187;
mu = 50.5;
sigma = 29.011;
s = [10^6 10^4 10^2];
S = diag(s);
n=2;
g = (x(1:n)-mu)/sigma;
z = sum((yref-(p1*g.^10+p2*g.^9+p3*g.^8+p4*g.^7+p5*g.^6+p6*g.^5+p7*g.^4+p8*g.^3+p9*g.^2+p10*g+p11)).^2)+x(n+1:end)'*S*x(n+1:end);
end
2 commentaires
Stephen23
le 6 Sep 2019
Modifié(e) : Stephen23
le 6 Sep 2019
"but I dont know how to have z which consist of z1 and z2 in a way to have scalar value of z"
Only you can decide that, depending on how those functions are related to each other, e.g. their relative weighting or something similar.
Consider: what should happen to the output value z if z1 increases by 1 and z2 decreases by 1: should the output z increase or decrease or stay the same?
Some ideas: sum, weighed sum, subtraction, mean, weighted mean, RMS, etc.
Réponses (1)
Catalytic
le 6 Sep 2019
"I have a code and 2 objective functions"
If you really do want to treat this as a multi-objective problem, you should probably use gamultiobj or paretosearch instead of fmincon.
0 commentaires
Voir également
Catégories
En savoir plus sur Solver Outputs and Iterative Display 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!