help code with lsqnonlin

greenting

please i want to know where i have made an error of this code. the code try to estimate 8 unkwons values of x for a function with lsqnonlin.

this is the code

1-

2-

3- format long

4- Rmin = [0.001 0.01 0.02 0.03 0.04 0.08 0.16 0.32 0.64 1.25 2.5 5.0];

5- Rmax = [0.01 0.02 0.03 0.04 0.08 0.16 0.32 0.64 1.25 2.5 5.0 10.0];

6- Nreel = [100 200 30 20 40 60 200 180 60 20 5 1];

7- N = 916;

8-

9- Nest = ((N.*x(1)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(2))/log(x(3))))-erf((sqrt(2)./2).*(log(Rmin./x(2)). /log(x(3)))))+(N.*x(4)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(5))./log(x(6))))-erf((sqrt(2)/2).*(log(Rmin./x(5))./log(x(6)))))+(N.*(1-x(1)-x(4))./2).*(erf((sqrt(2)/2).*(log(Rmax./x(7))./log(x(8))))-erf((sqrt(2)/2).*(log(Rmin./x(7))./log(x(8)))))-Nreel);

10-

11- % x0 = [3.,3.,3.,3.,3.,3.,3.,3.];

12- x0 = [0.11;1;1.5;0.86;1;2;1;2];

13- lb = zeros(8,1);

14- lb(:) = -inf;

15- ub = zeros(8,1);

16- ub(:) = inf;

17-

19- % options = optimoptions('lsqnonlin','Display','iter');

20- % options.Algorithm = 'levenberg-marquardt';

21- % 'trust-region-reflective'

22- options = optimoptions(@lsqnonlin,'Algorithm','levenberg-marquardt');

24- [x,resnorm,residual,exitflag,output] = lsqnonlin(@(x)Nest,x0,[],[],options)

25-

but when is run the code that's the answer

extimation

Undefined function or variable 'x'.

Error in extimation (line 9)

Nest =

((N.*x(1)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(2))/log(x(3))))-erf((sqrt(2)./2).*(log(Rmin./x(2))./log(x(3)))))+(N.*x(4)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(5))./log(x(6))))-erf((sqrt(2)/2).*(log(Rmin./x(5))./log(x(6)))))+(N.*(1-x(1)-x(4))./2).*(erf((sqrt(2)/2).*(log(Rmax./x(7))./log(x(8))))-erf((sqrt(2)/2).*(log(Rmin./x(7))./log(x(8)))))-Nreel);

>>

[Matt J]

I recommend not including line numbers when you post your code. It makes it harder for responders to copy/paste it.

Answer 1

[Bjorn Gustavsson]

At the very least you'll have to make Nest into a function - otherwise you have nothing to optimize. Perhaps something like this:

Rmin = [0.001 0.01 0.02 0.03 0.04 0.08 0.16 0.32 0.64 1.25 2.5 5.0];
Rmax = [0.01 0.02 0.03 0.04 0.08 0.16 0.32 0.64 1.25 2.5 5.0 10.0];
Nreel = [100 200 30 20 40 60 200 180 60 20 5 1];
N = 916;
% Not that this is the function-definition: (using @(x) creates a function-handle to a 
% function that depends on x and takes all other variables in the
% definition as they are at the time of function-handle-creation
Nest = @(x) ((N.*x(1)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(2))/log(x(3))))-erf((sqrt(2)./2).*(log(Rmin./x(2)).   /log(x(3)))))+(N.*x(4)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(5))./log(x(6))))-erf((sqrt(2)/2).*(log(Rmin./x(5))./log(x(6)))))+(N.*(1-x(1)-x(4))./2).*(erf((sqrt(2)/2).*(log(Rmax./x(7))./log(x(8))))-erf((sqrt(2)/2).*(log(Rmin./x(7))./log(x(8)))))-Nreel);
 
% x0 = [3.,3.,3.,3.,3.,3.,3.,3.];
x0 = [0.11;1;1.5;0.86;1;2;1;2];
lb = zeros(8,1);
lb(:) = -inf;
ub = zeros(8,1);
ub(:) = inf;

% options = optimoptions('lsqnonlin','Display','iter');
% options.Algorithm = 'levenberg-marquardt';
% 'trust-region-reflective'
options = optimoptions(@lsqnonlin,'Algorithm','levenberg-marquardt');
[x,resnorm,residual,exitflag,output] = lsqnonlin(@(x) Nest(x),x0,[],[],options);
% Above we generate a dynamic function that depends on x. This to make the
% slightly more general procedure easier to get to:
Nest2 = @(x,Rmax,Rmin,Nreel) ((N.*x(1)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(2))/log(x(3))))-erf((sqrt(2)./2).*(log(Rmin./x(2)).   /log(x(3)))))+(N.*x(4)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(5))./log(x(6))))-erf((sqrt(2)/2).*(log(Rmin./x(5))./log(x(6)))))+(N.*(1-x(1)-x(4))./2).*(erf((sqrt(2)/2).*(log(Rmax./x(7))./log(x(8))))-erf((sqrt(2)/2).*(log(Rmin./x(7))./log(x(8)))))-Nreel);
% Nest2 is now a function that takes 4 arguments. This makes it possible to
% re-use this function for multiple different Rmax, Rmin or Nreel
% parameters and fit for the correspongin optimal x:
[x,resnorm,residual,exitflag,output] = lsqnonlin(@(x) Nest2(x,Rmax,Rmin,Nreel),x0,[],[],options);

HTH

[franklin momo]

i am sorry but i am not understand i am new in the programmation with matlab. but try this other code

Rmin = [0.001 0.01 0.02 0.03 0.04 0.08 0.16 0.32 0.64 1.25 2.5 5.0];

Rmax = [0.01 0.02 0.03 0.04 0.08 0.16 0.32 0.64 1.25 2.5 5.0 10.0];

Nreel = [100 200 30 20 40 60 200 180 60 20 5 1];

N = 916;

x0 = [0.11;1;1.5;0.86;1;2;1;2];

lb = zeros(8,1);

lb(:) = -inf;

ub = zeros(8,1);

ub(:) = inf;

options = optimoptions(@lsqnonlin,'Algorithm','levenberg-marquardt');

[x,resnorm,residual,exitflag,output] = lsqnonlin(@(x)paramter(x,N,Rmin,Rmax,Nreel),x0,[],[],options)

function [Nest1] = paramter(x,N,Rmin,Rmax,Nreel)

Nest1 = ((N.*x(1)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(2))/log(x(3))))-erf((sqrt(2)./2).*(log(Rmin./x(2))./log(x(3)))))+(N.*x(4)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(5))./log(x(6))))-erf((sqrt(2)/2).*(log(Rmin./x(5))./log(x(6)))))+(N.*(1-x(1)-x(4))./2).*(erf((sqrt(2)/2).*(log(Rmax./x(7))./log(x(8))))-erf((sqrt(2)/2).*(log(Rmin./x(7))./log(x(8)))))-Nreel);

end

and answer was

Error using erf

Input must be real and full.

Error in est>paramter (line 21)

Nest1 =

((N.*x(1)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(2))/log(x(3))))-erf((sqrt(2)./2).*(log(Rmin./x(2))./log(x(3)))))+(N.*x(4)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(5))./log(x(6))))-erf((sqrt(2)/2).*(log(Rmin./x(5))./log(x(6)))))+(N.*(1-x(1)-x(4))./2).*(erf((sqrt(2)/2).*(log(Rmax./x(7))./log(x(8))))-erf((sqrt(2)/2).*(log(Rmin./x(7))./log(x(8)))))-Nreel);

Error in est>@(x)paramter(x,N,Rmin,Rmax,Nreel)

Error in levenbergMarquardt (line 212)

trialCostFun = feval(funfcn{3},reshape(trialX,xShape),varargin{:});

Error in lsqncommon (line 174)

levenbergMarquardt(funfcn,xC,flags.verbosity,options,defaultopt,initVals.F,initVals.J, ...

Error in lsqnonlin (line 240)

lsqncommon(funfcn,xCurrent,lb,ub,options,defaultopt,caller,...

Error in est (line 16)

[x,resnorm,residual,exitflag,output] = lsqnonlin(@(x)paramter(x,N,Rmin,Rmax,Nreel),x0,[],[],options)

[Bjorn Gustavsson]

In your parameters-function you have calls to erf with with arrays of Rmax and Rmin scaled with various components of x inside calls to log. If any of those R/x(n) components become negative the log will return something complex-valued which makes erf throw and error. The way you set up your upper and lower bounds doesn't constrain the search at all since they allow all components of x to vary between -inf and inf. You'll have to think about how to properly constrain the values of x properly to avoid that.

[franklin momo]

ok thanks let me try another thing and let you know.

[franklin momo]

ok went i try with x0 = [1 1 1 1 1 1 1 1] and x0 = [1 1 1 1 1 1 1 2] it works but don't works went i use x0 = [1 1 1 1 2 1 1 2] so really the problem is coming from the limite (lb and ub). i will work those limits. thanks so muuch for your help

[Bjorn Gustavsson]

Good.

I think you "just" have to make sure that all of the log(Rmax/x(n)) type components are real for both log(Rmax/lb(n)) and log(Rmax/ub(n)).

You might also get good use of setting:

dbstop if error

which makes matlab stop at the point where the error occurs. From there you can inspect the variables in the function workspaces, and also step up in the function-call to get to your Nest-function where you can check which is the offending term

[franklin momo]

greeting

i am trying to solve my problem with many input values as you know there is some values which give an error cause of the log(Rmax/x(n)) i want to know if there is function which can skip those values and permit the program to continu the resolution with ,the next value

[Bjorn Gustavsson]

That sounds like "a very dubious idea", because that would be to dynamically change/modify the function you're trying to optimize during the optimization. My suggestion is that you instead constrain the search to the region of parameters that gives sensible outputs (the way I understand your problem and the error is that you have some physics/principled based reasons that all the calls to log should give real results). To get towards that I've rewritten your function as:

function [Nest1] = paramters(x,N,Rmin,Rmax,Nreel)

Nest1 = ((N.*x(1)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(2))/log(x(3)))) - ...
                        erf((sqrt(2)./2).*(log(Rmin./x(2))./log(x(3))))) + ...
         (N.*x(4)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(5))./log(x(6)))) - ...
                        erf((sqrt(2)/2).*(log(Rmin./x(5))./log(x(6))))) + ...
         (N.*(1-x(1)-x(4))./2).*( ...
             erf((sqrt(2)/2).*(log(Rmax./x(7))./log(x(8)))) - ...
             erf((sqrt(2)/2).*(log(Rmin./x(7))./log(x(8)))))-Nreel);

end

This gives you one erf-call per line, and makes it easier to isolate the possible real-log-violators. If I get it those should be: log(Rmax./x(2)), log(x(3)), log(Rmin./x(2)), log(Rmax./x(5)), log(x(6)), log(Rmin./x(5)), log(Rmax./x(7)), log(x(8)), and log(Rmin./x(7)). Since all of Rmin and Rmax are positive this means that all of x([2 3 5 6 7 8]) has to be positive. Therefore you could (perhaps: should) enforce lower bound on those components of x. Perhaps something like:

Rmin = [0.001 0.01 0.02 0.03 0.04 0.08 0.16 0.32 0.64 1.25 2.5 5.0];
Rmax = [0.01 0.02 0.03 0.04 0.08 0.16 0.32 0.64 1.25 2.5 5.0 10.0];
Nreel = [100 200 30 20 40 60 200 180 60 20 5 1];
N = 916;
 
x0 = [0.11;1;1.5;0.86;1;2;1;2];
 

lb(:) = -inf(8,1);
ub(:) = inf(8,1);
lb([2 3 5 6 7 8]) = 0;

options = optimoptions(@lsqnonlin,'Algorithm','levenberg-marquardt');
[x,resnorm,residual,exitflag,output] = lsqnonlin(@(x)paramter(x,N,Rmin,Rmax,Nreel),...
                                                 x0,lb,ub,...
                                                 options);

Since the first and fourth elements of lb are -inf they will remain unconstrained, while the values you allow for the other components are constrained to be positive.

HTH

[Bjorn Gustavsson]

The lower bounds should be just larger than zero to be sure that we never get there (from the help:

X = lsqnonlin(FUN,X0,LB,UB) defines a set of lower and upper bounds on

the design variables, X, so that the solution is in the range LB <= X

<= UB.)

So change:

lb([2 3 5 6 7 8]) = 0;

to something like:

lb([2 3 5 6 7 8]) = eps(1);

Answer 2

[franklin momo]

greeting

thank for your reply. i know that there are input values thats will not give an answers, i just want to skip those input values and go to the next. i am try to input some thing like 1000 values and put them one by one will be easy so i write a code and want that the code skip all input values and continu running to the next one. so i will have many input values and many answers. we can [0 0 0 0 0 0 0 0] as answers for the input values that does not have answers.

below the code that i have written

i am trying to check the code that you wrote if it can help me.

% Rmin = [0.001 0.01 0.02 0.03 0.04 0.08 0.16 0.32 0.64 1.25 2.5 5.0];

% Rmax = [0.01 0.02 0.03 0.04 0.08 0.16 0.32 0.64 1.25 2.5 5.0 10.0];

Rmin = [1 10 20 30 40 80 160 320 640 1250 2500 5000];

Rmax = [10 20 30 40 80 160 320 640 1250 2500 5000 10000];

Nreel = [100 200 30 20 40 60 200 180 60 20 5 1];

N = 916;

a1=0.3; a11=0.34;

b1=8.5; b11=11.5;

c1=1.1; c11=1.2;

a2=0.59; a22=0.69;

b2=119; b22=121;

c2=1.1; c22=1.9;

b3=2258; b33=2261;

c3=1; c33=1.2;

a=(a1+a11)/2; b=(b1+b11)/2; c=(c1+c11)/2; d=(a2+a22)/2; e=(b2+b22)/2;

f=(c2+c22)/2; g=(b3+b33)/2;h=(c3+c33)/2;

x1(1)=a1; r1(1)=b1; t1(1)=c1; x2(1)=a2; r2(1)=b2; t2(1)=c2; r3(1)=b3; t3(1)=c3;

for i=1:2

x1(i+1)=x1(i)+a;

r1(i+1)=r1(i)+b;

t1(i+1)=t1(i)+c;

x2(i+1)=x2(i)+d;

r2(i+1)=r2(i)+e;

t2(i+1)=t2(i)+f;

r3(i+1)=r3(i)+g;

t3(i+1)=t3(i)+h;

end

for h=1:3

for i=1:3

for j=1:3

for k=1:3

for l=1:3

for m=1:3

for n=1:3

for o=1:3

x0 = [x1(o) r1(n) t1(m) x2(l) r2(k) t2(j) r3(i) t3(h)]

lb = [0.2 8 1 0.55 117 1 2257 1];

ub = [0.4 12 1.5 0.7 122 2 2262 1.5];

options = optimoptions(@lsqnonlin,'Algorithm','levenberg-marquardt');

[x,resnorm,residual,exitflag,output] = lsqnonlin(@(x)paramter(x,N,Rmin,Rmax,Nreel),x0,[],[],options)

% x3(i) = 1-x1(o)-x2(i)

end

end

end

end

end

end

end

end

function [Nest1] = paramter(x,N,Rmin,Rmax,Nreel)

Nest1 = ((N.*x(1)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(2))/log(x(3))))-erf((sqrt(2)./2).*(log(Rmin./x(2))./log10(x(3)))))+(N.*x(4)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(5))./log(x(6))))-erf((sqrt(2)/2).*(log(Rmin./x(5))./log(x(6)))))+(N.*(1-x(1)-x(4))./2).*(erf((sqrt(2)/2).*(log(Rmax./x(7))./log(x(8))))-erf((sqrt(2)/2).*(log(Rmin./x(7))./log(x(8)))))-Nreel);

end

and the coomiplation give

>> ext4

x0 =

1.0e+03 *

0.0003 0.0085 0.0011 0.0006 0.1190 0.0011 2.2580 0.0010

Local minimum possible.

lsqnonlin stopped because the final change in the sum of squares relative to

its initial value is less than the default value of the function tolerance.

<stopping criteria details>

x =

1.0e+03 *

0.0002 0.0109 0.0010 0.0007 0.3017 0.0023 2.2580 0.0010

resnorm =

1.6480e+04

residual =

Columns 1 through 11

-99.9898 21.9651 -28.7112 -17.0779 -9.3213 52.7751 9.8833 18.1139 33.1513 21.9651 -1.9510

Column 12

-0.8086

exitflag =

3

output =

struct with fields:

iterations: 20

funcCount: 190

stepsize: 0.0097

cgiterations: []

firstorderopt: 3.5155

algorithm: 'levenberg-marquardt'

message: 'Local minimum possible.↵↵lsqnonlin stopped because the final change in the sum of squares relative to ↵its initial value is less than the default value of the function tolerance.↵↵Stopping criteria details:↵↵Optimization stopped because the relative sum of squares (r) is changing↵by less than options.FunctionTolerance = 1.000000e-06.↵↵Optimization Metric Options↵relative change r = 3.73e-07 FunctionTolerance = 1e-06 (default)'

x0 =

1.0e+03 *

0.0006 0.0085 0.0011 0.0006 0.1190 0.0011 2.2580 0.0010

Error using erf

Input must be real and full.

Error in ext4>paramter (line 64)

Nest1 =

((N.*x(1)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(2))/log(x(3))))-erf((sqrt(2)./2).*(log(Rmin./x(2))./log10(x(3)))))+(N.*x(4)./2).*(erf((sqrt(2)/2).*(log(Rmax./x(5))./log(x(6))))-erf((sqrt(2)/2).*(log(Rmin./x(5))./log(x(6)))))+(N.*(1-x(1)-x(4))./2).*(erf((sqrt(2)/2).*(log(Rmax./x(7))./log(x(8))))-erf((sqrt(2)/2).*(log(Rmin./x(7))./log(x(8)))))-Nreel);

Error in ext4>@(x)paramter(x,N,Rmin,Rmax,Nreel)

Error in levenbergMarquardt (line 212)

trialCostFun = feval(funfcn{3},reshape(trialX,xShape),varargin{:});

Error in lsqncommon (line 174)

levenbergMarquardt(funfcn,xC,flags.verbosity,options,defaultopt,initVals.F,initVals.J, ...

Error in lsqnonlin (line 240)

lsqncommon(funfcn,xCurrent,lb,ub,options,defaultopt,caller,...

Error in ext4 (line 50)

[x,resnorm,residual,exitflag,output] =

lsqnonlin(@(x)paramter(x,N,Rmin,Rmax,Nreel),x0,[],[],options)

>>

we can see the code start running and stop when the input value do not give an anwers. i wwant to skip those input values by give them the answers [0 0 0 0 0 0 0 0] and continu to the next value