# imaginary value from solution nonlinier least square

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im trying to fitting data from reflectance measurement with lavenberg marquardt algorithm , ive got this code LMFnlsq in file exchange but i dont understand how to use it in my case

here is the equation

R = (0.6/(a + b))*(sqrt(3*a*(a + b))+(1/c))*(exp(-sqrt(3*a*(a + b ))*c))/(c^2)

a and b is parameter that im trying to solve

R is known vector from measurement reflectance

c is known vector from distance in measurement

im using the code like this

r = [0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.6]';

data = [0.787357544 0.63199586 0.393109753 0.271890049 0.179074593 0.133498607 0.112568249 0.05274273]';

res = @(x) ((0.6./(x(1)+x(2))).*(sqrt(3.*x(1).*(x(1)+x(2)))+(1./r)).* (exp(-sqrt(3*x(1).*(x(1)+x(2)))*r))./(r.^2)) - data;

x0 = [0.1,10];

[x,ssq,cnt] = LMFnlsq(res,x0)

hold on

plot(r,data)

plot(r,res(x)+ data,'y'), grid

the problem is it sure did the iteration but the final value is in imaginary solution like this

x =

0.0528 - 0.2730i

1.1469 + 0.6248i

ssq =

0.0091

cnt =

29

anything wrong with the code?

##### 0 Comments

### Accepted Answer

Miroslav Balda
on 12 Mar 2013

Let us reformulate the problem:

data = (q + 1./r).*exp(-r*q).*(0.6/(p(1)*r.^2)),

where

q = sqrt(p(2));

The problem may be solved without any troubles by introducing a penalty function

(q<=0)*w

into vector of residuals. Thus, the main script could be

% Surpan.m Application of LMFnlsq

%%%%%%%%%%% 12. 3. 2013

% Answers: Imaginary value from solution nonlinear least square

clc

clear all

close all

%

global data r p0 w

%

data = [0.779543143 0.624771688 0.387776261 0.266986051 0.174732672 ...

0.130313689 0.110540421 0.051412959]'; % column vector

r = [0.8 0.9 1 1.1 1.2 1.3 1.4 1.6]'; % column vector

w = 10;

%

while 1

% function inp from www.mathworks.com/matlabcentral/fileexchange/9033

w = inp('weight', w); % weight of penalty function

if w<=0, break, end

p0 = rand(2,1);

[p,ssq,cnt] = LMFnlsq(@resid,ones(2,1),'Display',-100,'MaxIter',1e3);

p = p.*p0;

a = p(2)/(3*p(1))

b = p(1)-a

res = resid(p./p0);

plot(r,data,'o-', r,res(1:end-1)+data,'-or')

end

The function for evaluating residuals has the form:

function res = resid(p) % Residuals of Surpans' equations

%%%%%%%%%%%%%%%%%%%%%%%%%

global data r w p0

%

p = p.*p0; % p(1) = (a+b)

q = sqrt(p(2)); % q = sqrt(3*a*(a+b))

%

res = [(q + 1./r).*exp(-r*q).*(0.6./(p(1)*r.^2)) - data

(q>=0)*w % penalty function

];

The nonlinear fit is made by the function LMFnlsq, which may be found on

www.mathworks.com/matlabcentral/fileexchange/17534

##### 0 Comments

### More Answers (1)

Matt J
on 28 Feb 2013

##### 4 Comments

Matt J
on 28 Feb 2013

but i must solve it with lavenberg marquardt algorithm

Meaning it's an assignment? Then get clarification from the one who assigned it to you. LM is not applicable to constrained problems.

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