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

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wiem abd 2019-9-26

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https://ww2.mathworks.cn/matlabcentral/answers/482311-does-the-order-of-magnitude-of-the-variables-in-the-non-linear-constraints-influence-the-solution-in

评论： Matt J 2019-9-29

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

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Answer 1

Matt J 2019-9-26

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/482311-does-the-order-of-magnitude-of-the-variables-in-the-non-linear-constraints-influence-the-solution-in#answer_393644

编辑：Matt J 2019-9-26

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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Matt J 2019-9-27

编辑：Matt J 2019-9-27

在 MATLAB Online 中打开

I have no way of knowing for what values C_Ter will have a solution, but clearly you cannot make it arbitrarily small. For C_Ter=-inf, you will definitely have an empty feasible set. So, you will need to have a good reason to trust that any particular reduction in C_Ter from 5e+11 will be feasible.

One thing you can try is to solve your @(f)0 problem for a series of gradually descending C_Ter values:

x0=0.6*b; %Feasible for C_Ter=5e+11
for C_Ter=5*logspace(11,6,100)
 options = optimoptions('fmincon','Display','iter','Algorithm','sqp');
 [x0,~,exitflag] = fmincon(@(f) 0 ,x0,[],[],[],[],lb,ub,@(f)mycon_Taylor(f,SNR,n,m,bandwidth,epsilon,C_Ter,x0,b))
 
 if exitflag<0
     
     disp(['No solution found for C_Ter='  num2str(C_Ter)]); break
 end
 
end

Each time the solution becomes the x0 of the next for-loop iteration, so that you incrementally track the boundary of the shrinking feasible set. If the loop makes it all the way to the last iteration, you will have found a feasible x0 for C_Ter=5*1e+6. If not, there is a chance that there is no feasible point for that C_Ter.

wiem abd 2019-9-29

编辑：wiem abd 2019-9-29

在 MATLAB Online 中打开

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 2019-9-29

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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