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

fmincon nonlinear constraint intial pointMATLAB

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

Best Answer

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.
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