lsqnonlin is only giving the initial point

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Günef Bozkurt 2020-8-9

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/577198-lsqnonlin-is-only-giving-the-initial-point

评论： Günef Bozkurt 2020-8-11

I'm trying to find location of a target node using nonlinear least squares algorithm. The answer it gives is off, it only gives the initial point, and I can't figure out why. I tried mutliple things and thats why if the parts of code looks messy. I'd appreciate any help

Heres my code:

syms x y z k p lo la
P = k-sqrt((k-lo).^2+(p-la).^2)
ref_umram = txsite('Name','UMRAM',...
       'Latitude',39.873297, ...
       'Longitude',32.745859, ...
       'TransmitterFrequency', 2.5e9);
ref_ee = txsite('Name','EE',...
       'Latitude',39.872124, ...
       'Longitude',32.750538);
ref_mayfest = txsite('Name','Mayfest',...
       'Latitude',39.869953, ...
       'Longitude',32.753181);
ref_bilka = txsite('Name','Bilka',...
       'Latitude',39.864596, ...
       'Longitude',32.747785);              %4 reference nodes
   
%reference_nodes={'ref_umram','ref_ee','ref_mayfest','ref_bilka'};
ref_bilka2 = rxsite('Name','Bilka2',...
       'Latitude',39.864595, ...
       'Longitude',32.747785);
target_lat = randi([39864596 39873297])/1000000
target_lon = randi([32745859 32753181])/1000000
target = rxsite('Name','target',...
       'Latitude',target_lat, ...
       'Longitude',target_lon);
   
pm = propagationModel('freespace');
n_pathloss = 2; %path loss exponent given for freespace
d0 = distance(ref_bilka2,ref_bilka);            %short reference distance
rss0 = sigstrength(ref_bilka2,ref_bilka,pm);    %reference rss
noise=randn(1,4);   %gaussian zero mean random variable vector
rss_umram = sigstrength(target,ref_umram,pm);
d_umram= 10^((rss0-rss_umram-noise(1))/(10*n_pathloss))*d0;
rss_ee = sigstrength(target,ref_ee,pm);
d_ee= 10^((rss0-rss_ee-noise(2))/(10*n_pathloss))*d0;
rss_mayfest = sigstrength(target,ref_mayfest,pm);
d_mayfest= 10^((rss0-rss_mayfest-noise(3))/(10*n_pathloss))*d0;
rss_bilka = sigstrength(target,ref_bilka,pm);
d_bilka= 10^((rss0-rss_bilka-noise(4))/(10*n_pathloss))*d0;    %distances to reference nodes calculated
global z
z = [d_umram d_ee d_mayfest d_bilka]
z
k=[x y];
kk=dpure(k)
% lb=[31 38];
% ub=[33 40];
x0=[32 39];
options=optimoptions("lsqnonlin",'MaxIterations',1500,'FunctionTolerance',0)
[x,resnorm,residual,exitflag,output] = lsqnonlin(@dpure,x0)
function P = dpure(x)
global z
la=[39.873297 39.872124 39.869953 39.864596];
lo=[32.745859 32.750538 32.753181 32.747785];%ref node locations
P = z-sqrt((x(1)-lo).^2+(x(2)-la).^2);
end

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Matt J 2020-8-9

Some of your code requires special Toolboxes to run. It would be easier for us if you would just attach a .mat file containing the data vector z, which would make all but the last 3 lines of your code unnecessary for us.

Günef Bozkurt 2020-8-9

lsqnonlinm.m

thanks for your feedback. z normally changes each turn but heres a file with a sample.

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Answer 1

Matt J 2020-8-9

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/577198-lsqnonlin-is-only-giving-the-initial-point#answer_477178

编辑：Matt J 2020-8-9

在 MATLAB Online 中打开

I don't know if the solution is what you were expecting, but this gives exitflag 1,

la=[39.873297 39.872124 39.869953 39.864596].';
lo=[32.745859 32.750538 32.753181 32.747785].';%ref node locations
z = 10^3*[0.317139933557511   0.213426852450956   0.554382676693071   1.090925354925889].';
M=diff([eye(4);1 0 0 0]);
A=M*lo;
B=M*la;
C=M*(lo.^2+la.^2-z.^2)/2;
x0=[A,B]\C;  %The unconstrained solution, with no bounds
lb=[31 38];
ub=[33 40];
options=optimoptions('lsqnonlin','Display','iter');
[x,resnorm,residual,exitflag,output] = ...
    lsqnonlin(  @(x) z-sqrt(  (x(1)-lo).^2+(x(2)-la).^2 ),  x0,lb,ub,options)

as well as quite good convergence with respect to the first optimality measure:

                                         Norm of      First-order 
 Iteration  Func-count     f(x)          step          optimality
     0          3      1.6386e+06                      1.64e+03
     1          6     1.63248e+06        1.41421        0.00759      
     2          9     1.63248e+06     0.00232396       3.37e-11       

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Matt J 2020-8-9

编辑：Matt J 2020-8-10

在 MATLAB Online 中打开

Is it this part? Would you mind explaining maybe?

If you were to remove the bounds lb,ub the problem has a closed-form analytical solution, and this solution might be a good initial guess x0 for the constrained problem. This is because if you subtract any two of your equations from each other, you obtain a linear equation:

     (x1^2-2*x1*la(k)+la(k)^2)+(x2^2-2*x1*lo(k)+lo(k)^2) = z(k)^2
     (x1^2-2*x1*la(m)+la(m)^2)+(x2^2-2*x1*lo(m)+lo(m)^2) = z(m)^2
  -
_______________________________________________________________________________
   -2*(la(k)-la(m))*x1  -2*(lo(k)-lo(m))*x2 =(z(k)^2-la(k)^2-lo(k)^2) - (z(m)^2-la(m)^2-lo(m)^2) 

The set of equations generated this way can be solved by basic linear algebra.

Günef Bozkurt 2020-8-11