# how to have function with scalar value

조회 수: 9(최근 30일)
Ali Esmaeilpour 2019년 9월 6일
답변: Catalytic 2019년 9월 6일
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
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Ali Esmaeilpour 2019년 9월 6일
consider weighted summation of those Obj1 and Obj2 and make a function named FWS and give it to fmincon.

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### 답변(1개)

Catalytic 2019년 9월 6일
"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.

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R2018b

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