# Error Not enough input arguments using Fmincon with nlcon and linear constraints

2 Ansichten (letzte 30 Tage)
ikram shahzad am 1 Nov. 2018
Kommentiert: ikram shahzad am 4 Nov. 2018
I am trying to minimize a function say fun using fmincon, it has 18 elements, having linear and non-linear constraints. linear constraints are applied on solution s, s(1) to s(7). and 2 nonlcon constraints are applied on solution s(8) to s(18).optimization is good for linear constraints but shows error using nonlcon function(@test3), not enough input arguments.
C_test_43.csv, file is required to run this code, it is attached with.
% code
clc
clear
time=50;
C=M(1:time);
Cmax=max(C);
T=10;
n=time/T;
X=zeros(1,n);
X(1,1)=8;
x=zeros(1,time);
x(1,1)=8;
E=1+(25-1).*rand(1,time);
batcharg=1;
Ve=(25-(2*batcharg))/Cmax;
Vt=(25-(2*batcharg))/Cmax;
V=(25-(2*batcharg))/Cmax;
m=11;
fr = 0 + (16.8*10^9-0).*rand(m,time);
fmax=max(fr);
fmax=max(fmax);
p = 0 + (2*10^-3-0).*rand(m,time);
pmax=max(p);
pmax=max(pmax);
Q=rand(1,time);
QD=rand(1,time);
a=0.1;
bb=0.025;
k=7.8*10^-21;
L=6400;
Hm=-50 + (-75-50).*rand(m,time);
phi=0.1;
w=1*10^6;
N=-174;
tau=1;
DL=tau.*fr./L;
D=w.*tau.*log2(1+(Hm.*p)/N.*w);
Td=0+(20-0).*rand(1,time);
vlb=zeros(1,m+7);
vub=zeros(1,m+7);
num=m+7;
energy_ub=1500;
f=zeros(1,num);
% code
fun=@(s)f;
for t=1:time
n=ceil(t/T);
freq=fr(1:m,t);
f(1,1)=X(n);
f(1,2)=x(t);
f(1,3)=V.*C(n);
f(1,4)=Q(t);
f(1,5)=QD(t);
f(1,6)=Ve;
f(1,7)=Vt;
for u=8:m+7
f(1,u)=min(DL(u-7,t)+D(u-7,t),D(u-7,t)+D(u-7,t));
end
i=1;
j=m+7;
Aeq=ones(i,j);
for j=8:m+7
Aeq(i,j)=-1;
end
beq=1;
for j=8:m+7
vlb(j)=0;
vub(j)=fmax;
end
vlb(1)=-batcharg;
vub(1)=batcharg;
vlb(2)=-1;
vub(2)=1;
vub(3)=energy_ub;
vlb(4)=-1;
vub(4)=1;
vlb(5)=-1;
vub(5)=1;
vlb(6)=0;
vub(6)=1000;
vlb(7)=0;
vub(7)=20;
if X(n)>-V*C(n)
vub(1)=1;
else
vlb(1)=-1;
end
if x(t)>-V*C(t)
vub(2)=1;
else
vlb(2)=-1;
end
if Q(t)>-Ve*E(t)
vub(4)=1;
else
vlb(4)=0;
end
if QD(t)>-Vt*Td(t)
vub(5)=1;
else
vlb(5)=0;
end
nonlcon = @test3;
[s,fval]=fmincon(fun,freq,[],[],Aeq,beq,vlb,vub,nonlcon(tau,L,m));
% function for 2 non linear constraints
function [c,ceq] = test3(s,tau,L,m)
for n=8:m+7
ceq(1) =tau.*(s(n)/L);
ceq(2)=w(n).*tau.*log2(1+(Hm.*p)/N.*w(n));
end
c = [];
end
##### 1 Kommentar-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden
Walter Roberson am 1 Nov. 2018
Search for Paramaterizing Functions

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### Antworten (1)

Nicolas Schmit am 2 Nov. 2018
There are several problems in your code
• You must pass the optimization variable as a argument to the nonlinear constraints.
• The size of Aeq, Beq, vlb and vub do not match the size of the freq vector.
• The cost function must return a scalar.
I suggest you have a look at the examples in the documentation of fmincon.
##### 5 Kommentare3 ältere Kommentare anzeigen3 ältere Kommentare ausblenden
Walter Roberson am 3 Nov. 2018
You still have
fun=@(s)f;
which tells it to ignore the input s and to always return the content of f, a constant vector. Are you sure that is what you want to do? Is your purpose to try to find a point that violates the constraints the least -- that you do not actually care about the return value of myfun and only want to try to figure out where some points are that (nearly) satisfy the constraints?
Perhaps you want something like
fun = @(s) f*s;
??
ikram shahzad am 4 Nov. 2018
Thanks Thanks to Walter Roberson. my simulation got successful by defining fun=@(s)X(n)*s(1)+x(t)*s(2)+V*C(n)*s(3)+Q(t)*s(4)+QD(t)*s(5)+Ve*s(6)+Vt*s(7)+f(1,u)*s(u); Thank you very much.

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