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Does the order of magnitude of the variables in the non linear constraints influence the solution in fmincon?

조회 수: 2 (최근 30일)
Can you please help me to figure out the problem with this optimization ? Starting from the given initial point, I obtain the message :
'Solver stopped prematurely.↵↵fmincon stopped because it exceeded the function evaluation limit,↵options.MaxFunctionEvaluations = 3000 (the default value).'
When I use the given value for f, the constraint is no longer defined in the considered point and I get the following message
'Error using barrier Nonlinear constraint function is undefined at initial point. Fmincon cannot continue.'
An interesting remark is that that value of the initial point x0=0.6*b is a local minimum for the objective function @(f)0 and with the same non linear constraints when C_Ter=5*1e+11.
Does this refer to a problem of order of magnitude of my variables?
clear all;
%%%%%%% simulation parameters
n=20
m=5;
power_BS = 20
N0=10e-10;
fb = 0.18*1e+6
n_RB = 100
bandwidth= 100*1e+6;
b_v=fb*n_RB/n;
radius_BS = 500;
b=b_v*rand(n,1);
d_sq=radius_BS*rand(n+m,1).^2;
h=exprnd(1,n+m,1)./(d_sq);
SNR=h*power_BS/N0;
C_Ter=5*1e+6;
p=[];
x0=[52.8827; 45.0967; 45.3726; 46.4245; 53.6886; 1.2579*ones(15,1)] % initial point
epsilon=(2e-4);
%objective=@(f)0
objective=@(f) obj(f,n,b,SNR);
lb=zeros(n,1);
ub=b;
%opts = optimset('Display','iter','Algorithm','interior-point', 'MaxIter', 100000, 'MaxFunEvals', 100000);
options = optimoptions('fmincon','Display','iter','Algorithm','sqp');
[f,fval,exitflag,output] = fmincon(objective,x0,[],[],[],[],lb,ub,@(f)mycon_Taylor(f,SNR,n,m,bandwidth,epsilon,C_Ter,x0,b))
p=[p,-fval];
where
function fun = obj(f,n,b1,SNR)
%UNTITLED3 Summary of this function goes here
% Detailed explanation goes here
fun=-sum((b1-f).*log(1+(SNR(1:n)./((b1-f)))))
end
and
function [c,ceq] = mycon(f,SNR,n,m,bandwidth,epsilon,C_Ter,a,b)
% Compute nonlinear inequalities at x.
c=[sum(f)- bandwidth;(1/epsilon)-sum(f(1:m)'*log(1+(SNR(n+1:n+m)./f(1:m))));
(sum(f(1:m)'*log(1+(SNR(n+1:n+m)./f(1:m))))+sum((b-a).*log(1+(SNR(1:n)./(b-a)))-(f-a).*log(1+(SNR(1:n)./(b-a)))+(SNR(1:n).*(f-a))./(b-a+SNR(1:n)))-C_Ter)];
ceq=[];
end
Actually, I deduced the initial point by solving the problem without the last non linear constraint through the following function :
function [c,ceq] = mycon(f,SNR,n,m,bandwidth,epsilon,b)
% Compute nonlinear inequalities at x.
c=[sum(f)- bandwidth;(1/epsilon)-sum(f(1:m)'*log(1+(SNR(n+1:n+m)./f(1:m))))];
ceq=[];
end

채택된 답변

Matt J
Matt J 2019년 9월 26일
편집: Matt J 2019년 9월 26일
Well, clearly the feasible set gets smaller as C_Ter gets smaller. Possibly you have made C_Ter so small that the feasible set is empty.
  댓글 수: 4
wiem abd
wiem abd 2019년 9월 29일
편집: wiem abd 2019년 9월 29일
Thank you Matt for your answer.
I chnaged the values of C_Ter as :
C_Ter=5*10^7:5*10^7:20e+07
The problem has always a feasible point. However, I get this message
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.
When I keep changing the values of the initial point with the newly found value, I reach the following state :
Error using barrier
Nonlinear constraint function is undefined at initial point. Fmincon cannot continue.
Matt J
Matt J 2019년 9월 29일
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.
This one was a good message. It probably means fmincon succeeded.

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