advice on least square method

[Carrico, Paul]

Dear all,

 

The current email is intend as (an) advice(s) request regarding the
least square method calculations.

 

I'm currently using the function herebellow to calculate the
coefficients of a 2nd order polynomial using thousands of experimental
data (furthermore the function is inserted in a loop => thousands of
loop = billiards of calculations).

 

So the main time cost is these calculation ...

 

 

I'm working to improve the code to be faster and in parallele I would
like to know if these a more relevant instruction in Scilab that
performe the calculation faster (maybe more accurate) ?

 

I had a look in the leastsq instruction ... but the later is strongly
influenced by the initial parameters guess X0 (isn't it).

 

 

Any advice is welcome

 

Regards / have a good WE

 

Paul

 

PS : A give a basic exampel in attached

PS 2 : additional question (but i'm a newby : calculations to be
parallelized ???)

 

 

 

###################################################"

 

 

##########################################################

function R_square = fct_MMC(p,experience)

 

// Nota :

// p = polynomial degree

// experience = experimental data

 

 

// step 1 : "linear" least square method

// matrix size
[n,c] = size(experience);

 

// calculation of  S
S = zeros(2*p,1);
for k = 1 : (2*p)
 for i = 1 : n
  S(k) = S(k) + experience(i,1)^k;
 end
end

 

// calculation of W
W = zeros(p+1,1);
for k = 1 : (p+1)
 for i = 1 : n
  W(k) = W(k) + experience(i,2)*experience(i,1)^(k-1);
 end
end

 

// calculation of A = S matrix
A = zeros(p+1,p+1);
for i = 1 : (p+1)
 for j = 1 : (p+1)
  if (i == 1 & j == 1) then
   A(1,1) = n;
  else
   A(i,j) = S(i+j-2);
  end
 end
end

 


// calculation of  B = W matrix
B = zeros(p+1,1);
for j = 1 : (p+1)
 B(j) = W(j);
end

 


// calculation of the coefficients
if (det(A) == 0) then
 printf('
########################################################################
###############\n')
 printf(' ####                                              Singular
matrix => no solution
####\n')
 printf('
########################################################################
###############\n')

 

else
 //C = inv(A)*B;

C = lsq(A,B);   // a bit faster than the previous calculation / same
results
 // calculation of R_square
 // formula : R_square = 1 - ( sum(F_fit - F_measured)^2 / sum(F_mean -
F_measured )
 F_fit = zeros(n,1);
 R_square = 0;
 F_mean = 0;
 
 for i = 1 : n
  // Calcul de F_fit
   for k = 1 : (p+1)
    F_fit(i) = F_fit(i) + C(k)*experience(i,1)^(k-1) ;
   end
 end

 

 calculation of  F_mean
 F_mean = mean(experience(:,2));

 

 // SSE = sum (F_fit - F_mean)
 // SSTE = sum (F_mean - F_measured)
 SSE = 0;
 SST = 0;



 for i = 1 : n
  SSE = SSE + (F_fit(i) - experience(i,2))^2 ;
  SST = SST + (F_mean - experience(i,2))^2 ;
 end

 

 R_square = 1 - (SSE / SST);

 

 

end

 

clear A
clear B
clear S
clear W
clear SSE
clear SST

 


endfunction

 

 


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