lsqnonlin is only giving the initial point

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

[Matt J]

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]

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

Answer 1

[Matt J]

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       

[Günef Bozkurt]

thanks for the answer! altough, the results seem to be static. It doesnt change with my code even though z changes each turn.

x = 2×4    
   33.0000   32.7483   32.7483   32.7483
   38.0000   39.8693   39.8693   39.8693

I tried doing some things different but this is still the only x I get. Am I doing something wrong?

[Günef Bozkurt]

I think this is also giving x0 as the guess

[Matt J]

I don't understand how x became 2x4.

[Matt J]

The result I get when I run my code is

x =

   31.0000
   38.0000

[Günef Bozkurt]

I changed it a bit more and this is also what im getting now. Maybe it had to do with how i place your code into mine. But output is still not correct, it doesnt change with z. The output is related to initial point and lower boundary I think. I may be saying it wrong because it really only takes the value of the lower boundary now, but its not affected from other parameters.

[Matt J]

Well, the solution has to be at a boundary, because your z are very large compared to the distance between points in la,lo and the feasible bounded region. When I simply generate random z according to,

z=rand(4,1);

I see a reasonable amount of variation in the solutions,

>> testrun

x =

   32.0343
   40.0000

>> testrun

x =

   32.3791
   39.9854

>> testrun

x =

   32.5359
   39.4659

>> testrun

x =

   32.2146
   39.9732

[Günef Bozkurt]

I'm so sorry, I've made a silly mistake and didn't transpose the z matrix. It works fine now. Thanks a lot! I wonder what made the difference?

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

Is it this part? Would you mind explaining maybe?

Thanks again!

[Günef Bozkurt]

Oh and again being silly I also took lat and lon values directly into pythagoran equation as if that would output the distance in meters... It does not... Noting just in case if helps anyone out in the future

[Matt J]

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]

thank you so much