I am confused by the error: "Error evaluating inline model function"

myfunc=inline('fitting(c,x)','c','x');
c0=[1.5,1.5];
c=nlinfit(x,y,myfunc,c0)

function ytotal = fitting(c,x)

xspan = x;
y0 = zeros(2,1);
[~,ye] = ode45(@eq2, xspan, y0);
ytotal = ye(:,1).*(1+0.2181) + ye(:,2) + 0.0667.*ye(:,2).*(c(1)-ye(:,1).*(1+0.2181))./(1+0.0667.*ye(:,2));
end 

function dy=eq2(x,y)
global c
%y(1)=CE,y(2)=LE
dy=zeros(2,1);
dy(1)=0.9728.*(2.5-(1 + 0.2181).*y(1)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1 + 0.2181))./(1 + 0.0667.*y(2))).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))-0.5766.*y(1);
dy(2)=0.5596.*(2.5-(1+0.2181).*y(1)-y(1)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(2)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2)))-0.5646.*y(2);
end

When I run the program, the error came out as below.

Error using nlinfit (line 207)

Error evaluating inline model function.

Error in youhuachuli1228 (line 65)

c=nlinfit(x,y,myfunc,c0)

Caused by:

Error using inlineeval (line 14)

Error in inline expression ==> fitting(c,x)

Undefined function 'fitting' for input arguments of type 'double'.

I think the problem is that the function 'fitting(c,x)' is in type of syms instead of 'double'. But I don't know how to fix it. Does anyone have any ideas?

Answer 1

[Walter Roberson]

When you use inline() the expression is evaluated in the base workspace. Any function mentioned must be resolved from the base workspace. That effectively prohibits local functions and nested functions. 'fitting' would have to have its own fitting.m file instead of defined in the same file.

The function fitting is not symbolic at all.

You also failed to initialize the global variable c.

Do not use inline() unless someone imposes using it as a job requirement (for example a homework example showing how inline breaks and so should be avoided)

global is not recommended.

https://www.mathworks.com/help/matlab/math/parameterizing-functions.html

[Jinglei]

Thank you. Then are there any other function that can replace inline() here?

[Stephen23]

"Then are there any other function that can replace inline() here? "

At the very top of the INLINE() documentation is this big yellow box (it is very hard to miss):

Did you consider following what the MATLAB documentation recommends and using an anonymous function?

https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html

[Jinglei]

Thank you. I tried it in both of the following ways,

myfunc=@(x) fitting(c,x);

c0=[1.5,1.5];

c=nlinfit(x,y,myfunc,c0)

or

c0=[1.5,1.5];

c=nlinfit(x,y,fitting(c,x),c0)

However, it still went wrong and several errors came out. Could you please show me the way to fix it?

[Stephen23]

"However, it still went wrong and several errors came out."

You did not define the x and y inputs to NLINFIT(). What values do you want to use for x and y ?

[Jinglei]

Values for x and y have been given in previous commands as below.

global c

x = [0

5

10

15

20

25

30

35

40

45

50

55

60

65

70

80

90

100

110

120

150

500

1000

5000

10000

];

y = [0

0.17917

NaN

NaN

0.44143

NaN

0.68912

NaN

NaN

NaN

0.76926

NaN

NaN

NaN

NaN

0.78383

NaN

0.7984

NaN

1.06795

0.90768

NaN

NaN

NaN

NaN

];

[Torsten]

And you think fitting NaN values will work out fine ? It won't of course.

[Jinglei]

It won't be that good result but shouldn't go wrong. I used those NaN to make more points on the line so that the line will look smoother.

[Torsten]

Your fitting function will return NaN and the solver will quit.

x = [0
    5
    20
    30
    50
    80
    100
    120
    150
    ];
y = [0
    0.17917
    0.44143
    0.68912
    0.76926
    0.78383
    0.7984
    1.06795
    0.90768
    ];
myfunc=@fitting;%inline('fitting(c,x)','c','x');
c0=[1.5,1.5];
c=nlinfit(x,y,myfunc,c0)

c = 1×2

0.1720 0.8251

%c = lsqcurvefit(myfunc,c0,x,y)
ytotal = fitting(c,x);
hold on
plot(x,ytotal)
plot(x,y,'o')
hold off

function ytotal = fitting(c,x)
xspan = x;
y0 = zeros(2,1);
[~,ye] = ode45(@(x,y)eq2(x,y,c), xspan, y0);
ytotal = ye(:,1).*(1+0.2181) + ye(:,2) + 0.0667.*ye(:,2).*(c(1)-ye(:,1).*(1+0.2181))./(1+0.0667.*ye(:,2));
end 

function dy=eq2(x,y,c)
%y(1)=CE,y(2)=LE
dy=zeros(2,1);
dy(1)=0.9728.*(2.5-(1+0.2181).*y(1)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))-0.5766.*y(1);
dy(2)=0.5596.*(2.5-(1+0.2181).*y(1)-y(1)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(2)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2)))-0.5646.*y(2);
end

[Jinglei]

Thank you for your suggestions. I tried an easier function in inline directly before with the same data and it gave the final results succesively. So I don't think that matters a lot. Do you have any other suggestions to solve the problem?

[Torsten]

Do you have any other suggestions to solve the problem?

Is there any problem left ? The code is working - although the results could be better.

[Jinglei]

No,it's not working. Several errors were found and the result couldn't be achieved. The code I ran succesfully is as follow:

myfunc=inline('c(1)+c(2).*x','c','x');

c0=[1.5,1.5];

c=nlinfit(x,y,myfunc,c0)

However, when I changed 'c(1)+c(2).*x' into 'fitting(c,x)' or the commands I stated before, errors happened again. For instance, for the following codes,

myfunc=@(x) fitting(c,x);

c0=[1.5,1.5];

c=nlinfit(x,y,myfunc,c0)

The errors are as below.

Error using nlinfit (line 213)

Error evaluating model function '@(x)fitting(c,x)'.

Error in youhuachuli1228 (line 65)

c=nlinfit(x,y,myfunc,c0)

Caused by:

Error using youhuachuli1228>@(x)fitting(c,x)

Too many input arguments.

Which makes me really confused.

[Torsten]

But as you can see from the output, the code above works. I don't know which wrong modifications you've done to it.

[Jinglei]

I am sorry that I have just seen the result you pasted. But the same code just ran wrong on my Matlab. Maybe it is caused by the edition? I am not sure. I will try to fix it. Thank you for your help. I really appreciate it.

[Torsten]

As you can see from the definition of "myfunc" and the error message you got from running your code, you haven't run the same code as pasted above.

[Jinglei]

Thank you. I tried pasting what you pasted to my Matlab and got the c values. However, the image could not achieved and errors still existed. In addition, c values are too small to get a good fitting since as you can see, the initial values I gave are (1.5, 1.5) and they fitted the data well although not the optimal. I am not sure whether it is caused by the errors.

[Torsten]

I am not sure whether it is caused by the errors.

Which errors ?

[Jinglei]

Error using nlinfit (line 213)

Error evaluating model function 'fitting'.

Error in youhuachuli1228 (line 33)

c=nlinfit(x,y,myfunc,c0)

Caused by:

Index exceeds array bounds.

[Torsten]

Since the code above works, I will no longer comment on this.

Use the code below and you won't get any syntax errors.

x = [0;5;20;30;50;80;100;120;150];
xs = 0:0.1:150;
y = [0;0.17917;0.44143;0.68912;0.76926;0.78383;0.7984;1.06795;0.90768];
myfunc=@fitting;
c0=[1.5,1.5];
ytotal0 = fitting(c0,xs);
c=nlinfit(x,y,myfunc,c0)

c = 1×2

0.1720 0.8251

xs = 0:0.1:150;
ytotal = fitting(c,xs);
%c = lsqcurvefit(myfunc,c0,x,y)
hold on
plot(x,y,'o')
plot(xs,ytotal0)
plot(xs,ytotal)
hold off

function ytotal = fitting(c,x)
xspan = x;
y0 = zeros(2,1);
[~,ye] = ode45(@(x,y)eq2(x,y,c), xspan, y0);
ytotal = ye(:,1).*(1+0.2181) + ye(:,2) + 0.0667.*ye(:,2).*(c(1)-ye(:,1).*(1+0.2181))./(1+0.0667.*ye(:,2));
end 
function dy=eq2(x,y,c)
%y(1)=CE,y(2)=LE
dy=zeros(2,1);
dy(1)=0.9728.*(2.5-(1+0.2181).*y(1)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))-0.5766.*y(1);
dy(2)=0.5596.*(2.5-(1+0.2181).*y(1)-y(1)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(2)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2)))-0.5646.*y(2);
end

[Jinglei]

I am sorry but still, the same errors happen.

Also, why does the curve look so bad? Is it because of the lack of data or the code itself?

[Walter Roberson]

@Jinglei if you copy Torsten's code from https://www.mathworks.com/matlabcentral/answers/1886712-i-am-confused-by-the-error-error-evaluating-inline-model-function#comment_2542627 exactly as-is then do you get error messages? If so what are they, and which MATLAB release are you using?

[Walter Roberson]

You can try a number of different starting points and hope for a better answer. If you do, then you will get ode failures, and nlinfit failures. You can get a better fit, in theory, but in practice it is still quite far from where you would hope.

x = [0;5;20;30;50;80;100;120;150];
y = [0;0.17917;0.44143;0.68912;0.76926;0.78383;0.7984;1.06795;0.90768];
myfunc=@fitting;

rng(123456);

best_r = inf;
best_c = [-inf, -inf];

for iters = 1 : 50
   c0 = [1.5 1.5] + randn(1,2)/5;
   try
   [c, r] =nlinfit(x,y,myfunc,c0);
   nr = norm(r);
   if nr < best_r
       fprintf('improved on iteration %d\n', iters);
       best_r = nr;
       best_c = c;
   end
   catch ME
       fprintf('failed iteration %d\n', iters);
   end
end

improved on iteration 1
improved on iteration 3

Warning: Failure at t=1.146514e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 4

improved on iteration 6
improved on iteration 7

Warning: Failure at t=2.115100e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (5.684342e-14) at time t.

failed iteration 8

improved on iteration 10
improved on iteration 11

Warning: Failure at t=6.933508e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 15

Warning: Failure at t=4.658234e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.136868e-13) at time t.

failed iteration 16

Warning: Failure at t=2.178430e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (5.684342e-14) at time t.

failed iteration 17

Warning: Failure at t=1.178626e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 25

Warning: Failure at t=4.211467e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.136868e-13) at time t.

failed iteration 29

Warning: Failure at t=8.001151e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 30

Warning: Failure at t=8.857406e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 31

Warning: Failure at t=8.085854e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 32

Warning: Failure at t=1.162031e+02. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.273737e-13) at time t.

failed iteration 33

Warning: Failure at t=5.427196e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 36

Warning: Failure at t=2.784805e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 39

Warning: Failure at t=2.469110e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 44

Warning: Failure at t=6.325847e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.136868e-13) at time t.

failed iteration 47

improved on iteration 48

c = best_c

c = 1×2

1.4989 -1.8552

%c = lsqcurvefit(myfunc,c0,x,y)
ytotal = fitting(c,x);
hold on
plot(x,ytotal)
plot(x,y,'o')
hold off

function ytotal = fitting(c,x)
xspan = x;
y0 = zeros(2,1);
[t,ye] = ode15s(@(x,y)eq2(x,y,c), xspan, y0);
%protect against failure of integration
if t(end) ~= xspan(end)
    ytotal = inf(numel(xspan),1);
else
    ytotal = ye(:,1).*(1+0.2181) + ye(:,2) + 0.0667.*ye(:,2).*(c(1)-ye(:,1).*(1+0.2181))./(1+0.0667.*ye(:,2));
end
end 
function dy=eq2(x,y,c)
%y(1)=CE,y(2)=LE
dy=zeros(2,1);
dy(1)=0.9728.*(2.5-(1+0.2181).*y(1)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))-0.5766.*y(1);
dy(2)=0.5596.*(2.5-(1+0.2181).*y(1)-y(1)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2))).*(c(2)-y(2)-0.0667.*y(2).*(c(1)-y(1).*(1+0.2181))./(1+0.0667.*y(2)))-0.5646.*y(2);
end

[Jinglei]

I got the line this time but errors are still there. And the result is not that good, which is worse compared with the initial values for the constant I have given. So that couldn't be correct. In addition, my matlab version is 9.4.0.813654 (R2018a). Thos errors can be seen as follows.

Error using nlinfit (line 213)

Error evaluating model function 'fitting'.

Error in youhuachuli1228 (line 33)

c=nlinfit(x,y,myfunc,c0)

Caused by:

Index exceeds array bounds.

>> fortry20230101

c =

0.1720 0.8251

>> fortry20230101

c =

0.1720 0.8251

>> fortry20230101

Error: File: fortry20230101.m Line: 6 Column: 6

Invalid text character. Check for unsupported symbol, invisible character, or pasting of non-ASCII characters.

[Walter Roberson]

Try the iterative version I just posted, as it has some error control.

[Torsten]

@Jinglei

I suspect that the ODE solver is not capable to solve some intermediate integration problem with the parameters supplied. You can't do anything else but changing your initial guesses for the parameters.

[Walter Roberson]

When ode45 or ode15s cannot meet integration tolerances, then they end the session early -- in which case the size of the output might not be the same as the size of xspan . That can potentially lead to indexing problems. The t(end) ~= xspan(end) in my code checks for the case of early ending and substitutes values that nlinfilt will recognize as being very far from the desired values.

[Jinglei]

@TorstenThank you for your explanation. Actually I used Excel before to optimize the constants, which, however, changed dramatically under different initial values. I used Matlab just to find a better result in a quicker way since Excel is relatively slow and may easily break down when processing. Here, I just can't understand that since the initial values are already better than many others especially the final chosen ones, why the program will go to a less fitted one. Under this situation, choosing better initial values would be a more tough thing.

@Walter RobersonThe matlab also told a size mismatch and gave a possible reason as the use of matrix operators (*, /, ^) instead of the corresponding elementwise operators (.*, ./, .^), which is obviously incorrect as can be seen in the code. Previously, I used ode to solve the fitting problem and got the function, which was not the optimal one since two of the parameters were not so correct based on experimental data. Therefore, I used this to find those two parameters to get the optimal curve. So I don't think this is caused by ODE. And I would like to know whether there might be any problem with nlinfit.

[Torsten]

@Jinglei

I added the solution curve for c0 = [1.5 1.5] in the above code. It doesn't seem to match the experimental data better than the solution c = [0.1720 0.8251] from nlinfit. Do you agree ?

[Walter Roberson]

I have attached the exact code I worked with. I tested this in R2018a.

I was getting a fat lot of error messages that I was having trouble tracking down until I realized that I did not have the Statistics and Machine Learning toolbox installed.

The code I attach here only has minor differences from what I posted before -- differences in that it specifically turns off some warning messages. It lets the nonlinear fitting fail, but catches the event and handles it smoothly.

The eventual coefficients it comes out with are still very different from what you might hope. Perhaps a different range of initial values should be searched over.

[Jinglei]

I am so so sorry that I pasted the wrong data before and the correct data can be seen as follow,

x = [0;5;10;25;30;35;40;45;50;55;70;80;90;120];

y =[0;1.34288;1.43759;1.59058;1.59786;1.60515;1.60515;1.55415;1.54687;1.58329;1.64886;1.72171;1.72171;1.72171];

In this case, a good fitting could be achieved at c=[1.4448 1.4846].

I am really sorry for making that mistake and bother you so much @Walter Roberson@Torsten. But I really appreciate all your efforts and suggestions cause that make it possible for me to get the c values successfully although still with a lot of errors. Thank you for all your help.

One last question, for the resulting curve, I am confused why it looks like two lines instead of a smooth curve. Is this caused by my data or the function used? I asked this cause I could get a smoother curve before without using nlinfit.

[Torsten]

@Jinglei

After you get back the fit coefficients from nlinfit, change x to x=0:0.1:120 and call "fitting" with this array for x together the c array from nlinfit to get a better resolution for ytotal.

I modified the code above accordingly.

[Jinglei]

@TorstenThank you for your help and it does become smoother. But it could be better. I mean the fitting. As you can see in the following image, the curve inflection point x value is too small based on its relationship with the experimental points, which might be caused by the c0 chosen or the fitting process.

I tried different initial c0, which did little help. So I think the code could be further improved. Anyway, it can already lead to a certain result, which is much better than before. Thank you very much for your help.

[Torsten]

As you can see in the following image, the curve inflection point x value is too small based on its relationship with the experimental points, which might be caused by the c0 chosen or the fitting process.

Or your fitting model which seems the most plausible reason.

[Jinglei]

Yes, it could be although I don't want to think that way.

[Jinglei]

I used the following code finally and changed two constants into seven. I found this is much more effective than my former one. I would like to know how to make all constants larger than 0 since the result showed one constant negative. In addition, the result of running seven constants is much less fitted compared with two and I would like to know whether this is caused by some errors in the code and how to correct them.

x = [0;5;10;25;30;35;40;45;50;55;70;80;90;120];
xs = 0:0.1:120;
y = [0;1.34288;1.43759;1.59058;1.59786;1.60515;1.60515;1.55415;1.54687;1.58329;1.64886;1.72171;1.72171;1.72171];
myfunc=@fitting;
rng(123456);
best_r = inf;
best_c = [-inf, -inf, -inf, -inf, -inf, -inf, -inf];
for iters = 1 : 50
    c0 = [0.9 0.4 1.5 1 0.2 1.5 0.7] + randn(1,7)/5;
    ytotal0 = fitting(c0,xs);
    try
        [c, r] =nlinfit(x,y,myfunc,c0);
        nr = norm(r);
        if nr < best_r
            fprintf('improved on iteration %d\n', iters);
            best_r = nr;
            best_c = c;
        end
    catch ME
        fprintf('failed iteration %d\n', iters);
    end
end

c = best_c

%c = lsqcurvefit(myfunc,c0,x,y)
ytotal = fitting(c,x);
hold on
plot(x,y,'o')
plot(xs,ytotal0)
plot(xs,ytotal)
hold off

function ytotal = fitting(c,x)
xspan = x;
y0 = zeros(2,1);
[t,ye] = ode15s(@(x,y)eq2(x,y,c), xspan, y0);
%protect against failure of integration
if t(end) ~= xspan(end)
    ytotal = inf(numel(xspan),1);
else
    ytotal = ye(:,1).*(1+0.39976) + ye(:,2);
end
end 
function dy=eq2(x,y,c)
%y(1)=CE,y(2)=LE
dy=zeros(2,1);
dy(1)=c(1) .* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(3) - y(1) .* (1 + c(2))) - c(4) .* y(1);
dy(2)=c(5).* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(6) - y(2) - c(4) .* y(2) .* (c(6) - y(2))) - c(7).* y(2);
end

[Torsten]

Sorry, but it's mathematical nonsense to fit a model with 7 parameters for which you have 14 measurements.

Take a look at "overfitting" :

https://en.wikipedia.org/wiki/Overfitting

[Walter Roberson]

With a repair on plotting, and adding information about change from the original parameters, and tweaking the random divisor to have fewer rejected locations.

I tweaked the code so that it will not offer negative initial coordinates. The nonlinear fitting routine does not offer constraints, so you would need to switch to something like fmincon() to prevent values from going negative; I work around it here by rejecting negative results.

I did not bother to add back in the supression of the warning messages. I figure that you must have a reason for wanting to see them, since you adopted my try/catch but not my warning off code.

x = [0;5;10;25;30;35;40;45;50;55;70;80;90;120];
xs = 0:0.1:120;
y = [0;1.34288;1.43759;1.59058;1.59786;1.60515;1.60515;1.55415;1.54687;1.58329;1.64886;1.72171;1.72171;1.72171];
myfunc=@fitting;
rng(123456);
best_r = inf;
best_c = [-inf, -inf, -inf, -inf, -inf, -inf, -inf];
c0_orig = [0.9 0.4 1.5 1 0.2 1.5 0.7];
for iters = 1 : 100
    c0 = max(0, c0_orig + randn(1,7)/20);
    ytotal0 = fitting(c0,xs);
    try
        [c, r] =nlinfit(x,y,myfunc,c0);
        if any(c < 0)
            fprintf('rejected negative output on iteration %d\n', iters);
            continue;
        end
        nr = norm(r);
        if nr < best_r
            fprintf('improved on iteration %d\n', iters);
            best_r = nr;
            best_c = c;
        end
    catch ME
        fprintf('failed iteration %d\n', iters);
    end
end

improved on iteration 1

Warning: Failure at t=5.214845e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 3

Warning: Failure at t=1.138046e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 4

Warning: Failure at t=3.027034e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 8

rejected negative output on iteration 9

Warning: Failure at t=5.172382e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 10

Warning: Failure at t=3.196325e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 11

rejected negative output on iteration 12
rejected negative output on iteration 15
rejected negative output on iteration 16

Warning: Failure at t=5.034794e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 22

Warning: Failure at t=1.404348e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 24

Warning: Failure at t=1.382413e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 25

Warning: Failure at t=1.858841e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 27

Warning: Failure at t=5.805013e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 28

rejected negative output on iteration 29

Warning: Failure at t=3.031519e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 30

Warning: Failure at t=2.133830e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 31

Warning: Failure at t=1.104143e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 32

Warning: Failure at t=4.857090e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 33

Warning: Failure at t=4.029846e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 34

Warning: Failure at t=5.848573e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 35

Warning: Failure at t=1.051578e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 37

rejected negative output on iteration 38

Warning: Failure at t=2.364404e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 39

Warning: Failure at t=4.498947e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 40

rejected negative output on iteration 41

Warning: Failure at t=2.689470e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 42

Warning: Failure at t=2.677579e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 44

Warning: Failure at t=6.993425e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 46

Warning: Failure at t=2.940197e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 48

rejected negative output on iteration 49

Warning: Failure at t=8.152841e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 50

Warning: Failure at t=2.034104e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 51

Warning: Failure at t=1.137900e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 52

rejected negative output on iteration 53

Warning: Failure at t=1.011267e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 54

Warning: Failure at t=2.787820e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 55

Warning: Failure at t=3.488307e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 56

rejected negative output on iteration 57

Warning: Failure at t=9.999214e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 58

Warning: Failure at t=1.551725e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 61

Warning: Failure at t=3.421249e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 63

rejected negative output on iteration 64
rejected negative output on iteration 65

improved on iteration 66

rejected negative output on iteration 67

Warning: Failure at t=2.873194e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 70

rejected negative output on iteration 71

Warning: The Jacobian at the solution is ill-conditioned, and some model parameters may not be estimated well (they are not identifiable). Use caution in making predictions.

Warning: Failure at t=5.456363e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 74

Warning: Failure at t=6.623573e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 75

Warning: Failure at t=3.539704e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 77

Warning: Failure at t=3.977617e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 78

Warning: Failure at t=8.294844e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.776357e-15) at time t.

failed iteration 79

Warning: Failure at t=6.023717e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 81

Warning: Failure at t=1.648890e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.

failed iteration 82

Warning: Failure at t=2.914583e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 83

rejected negative output on iteration 84
rejected negative output on iteration 85

Warning: Failure at t=4.582848e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (1.421085e-14) at time t.

failed iteration 87

Warning: Failure at t=4.580019e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t.

failed iteration 89

rejected negative output on iteration 92

Warning: Failure at t=2.528501e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 93

rejected negative output on iteration 95

Warning: Failure at t=2.098721e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (7.105427e-15) at time t.

failed iteration 97

Warning: Failure at t=1.579961e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 98

Warning: Failure at t=1.076871e+01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (2.842171e-14) at time t.

failed iteration 100

c = best_c

c = 1×7

0.7129 0.3339 1.9853 0.8968 0.0325 0.9857 0.0033

change_from_original = c - c0_orig

change_from_original = 1×7

-0.1871 -0.0661 0.4853 -0.1032 -0.1675 -0.5143 -0.6967

%c = lsqcurvefit(myfunc,c0,x,y)
ytotal = fitting(c,x);
hold on
plot(x,y,'o', 'displayname', 'original')
plot(xs, ytotal0, 'displayname', 'ytotal0')
plot(x, ytotal, 'displayname', 'ytotal')
hold off
xlim auto
ylim auto
legend show

function ytotal = fitting(c,x)
xspan = x;
y0 = zeros(2,1);
[t,ye] = ode15s(@(x,y)eq2(x,y,c), xspan, y0);
%protect against failure of integration
if t(end) ~= xspan(end)
    ytotal = inf(numel(xspan),1);
else
    ytotal = ye(:,1).*(1+0.39976) + ye(:,2);
end
end 
function dy=eq2(x,y,c)
%y(1)=CE,y(2)=LE
dy=zeros(2,1);
dy(1)=c(1) .* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(3) - y(1) .* (1 + c(2))) - c(4) .* y(1);
dy(2)=c(5).* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(6) - y(2) - c(4) .* y(2) .* (c(6) - y(2))) - c(7).* y(2);
end

[Jinglei]

@Walter RobersonThank you for your code and it does help. The problem is that the curve between the first and second points is so straight and the connection between that with the folloing ones is not so smooth. Previously, I added some points like (1,NaN), (2,NaN) to make this area be smoother but this doesn't work in this situation maybe due to the error caused by NaN. So I would like to know how to solve this problem.

[Torsten]

Previously, I added some points like (1,NaN), (2,NaN) to make this area be smoother but this doesn't work in this situation maybe due to the error caused by NaN. So I would like to know how to solve this problem.

Add values for x after fitting and plot the solution function with the extended xspan as already shown above.

c=nlinfit(x,y,myfunc,c0)
xs = 0:0.1:150;
ytotal = fitting(c,xs);
plot(xs,ytotal)

[Jinglei]

@TorstenIt works well this time. Thank you for all your help.

I have got two other questions. Firstly, how could I calculate the R2 of fitting in this case?In addition, is it possible to add several ys in this code to get five curves directly?I maen, making y a matrix with five rows. I wrote it as the following way, which, however, only showed the points without any curve.

myfunc=@fitting;
rng(123456);
best_r = inf;
best_c = [-inf, -inf, -inf, -inf, -inf, -inf, -inf];
c0_orig = [0.9 0.4 1.5 1 0.2 1.5 0.7];
for iters = 1 : 100
    c0 = max(0, c0_orig + randn(1,7)/20);
    ytotal0 = fitting(c0,xs);
    try
        [c, r] =nlinfit(x,y(:,1),y(:,2),y(:,3),y(:,4),y(:,5),myfunc,c0);
        if any(c < 0)
            fprintf('rejected negative output on iteration %d\n', iters);
            continue;
        end
        nr = norm(r);
        if nr < best_r
            fprintf('improved on iteration %d\n', iters);
            best_r = nr;
            best_c = c;
        end
    catch ME
        fprintf('failed iteration %d\n', iters);
    end
end

Unrecognized function or variable 'xs'.

c = best_c

change_from_original = c - c0_orig

%c = lsqcurvefit(myfunc,c0,x,y)
ytotal = fitting(c,x);
%for adding points
hold on
plot(x,y,'o', 'displayname', 'original')


hold off

xlim auto
ylim auto
legend show

%for adding points
hold on
xs = 0:0.1:150;
ytotal = fitting(c,xs);
plot(xs,ytotal)
hold off

[Jinglei]

I editted the code again to make it work for matrix, which, however, still didn't result in five curves successfully. Does anyone know what is wrong with my code?

x = [0 0 0 0 0;5 5 5 5 5; 10 10 10 10 10;
    15 15 15 15 15; 20 20 20 20 20;
    25 25 25 25 25; 30 30 30 30 30; 
    35 35 35 35 35; 40 40 40 40 40;
    45 45 45 45 45; 50 50 50 50 50;
    55 55 55 55 55; 60 60 60 60 60;
    70 70 70 70 70; 80 80 80 80 80;
    90 90 90 90 90;120 120 120 120];

Error using vertcat
Dimensions of arrays being concatenated are not consistent.

xs = 0:0.01:120;
y = [0	0	0	0	0;
    1.342848859	1.466695386	1.495835746	1.524976105	1.189861972;
    1.437555027	1.371989218	0.956739097	1.510405925	1.007734726;
    1.175291792	0.993164546	0.701760952	1.561401554	1.415699757;
    1.699818261	1.510405925	1.590541914	1.481265566	1.437555027;
    1.590541914	1.554116464	1.539546285	1.561401554	1.459410296;
    1.597827003	1.590541914	1.539546285	1.561401554	1.444840117;
    1.605112093	1.590541914	1.437555027	1.510405925	1.473980476;
    1.605112093	1.663392812	1.605112093	1.575971734	1.473980476;
    1.554116464	1.554116464	1.539546285	1.612397183	1.401129577;
    1.546831374	1.590541914	1.561401554	1.561401554	1.481265566;
    1.583256824	1.568686644	1.626967363	0.053387955	1.408414667;
    1.437555027	1.605112093	1.626967363	1.648822632	1.444840117;
    1.648822632	1.648822632	1.707103351	1.561401554	1.546831374;
    1.721673531	1.707103351	1.619682273	1.634252453	1.648822632;
    1.721673531	1.707103351	1.707103351	1.692533171	1.532261195;
    1.721673531	1.707103351	1.677962992	1.546831374	1.619682273
    ];
myfunc=@fitting;
rng(123456);
best_r = inf;
best_c = [-inf, -inf, -inf, -inf, -inf, -inf, -inf;
    -inf, -inf, -inf, -inf, -inf, -inf, -inf;
    -inf, -inf, -inf, -inf, -inf, -inf, -inf;
    -inf, -inf, -inf, -inf, -inf, -inf, -inf;
    -inf, -inf, -inf, -inf, -inf, -inf, -inf];
c0_orig = [0.9 0.4 1.5 1 0.2 1.5 0.7;
    0.9 0.4 1.5 1 0.2 1.5 0.7;
    0.9 0.4 1.5 1 0.2 1.5 0.7;
    0.9 0.4 1.5 1 0.2 1.5 0.7;
    0.9 0.4 1.5 1 0.2 1.5 0.7];
for iters = 1 : 100
    c0 = max(0, c0_orig + randn(5,7)/20);
    ytotal0 = fitting(c0,xs);
    try
        [c, r] =nlinfit(x,y,myfunc,c0);
        if any(c < 0)
            fprintf('rejected negative output on iteration %d\n', iters);
            continue;
        end
        nr = norm(r);
        if nr < best_r
            fprintf('improved on iteration %d\n', iters);
            best_r = nr;
            best_c = c;
        end
    catch ME
        fprintf('failed iteration %d\n', iters);
    end
end

c = best_c

change_from_original = c - c0_orig

%c = lsqcurvefit(myfunc,c0,x,y)
ytotal = fitting(c,x);
%for adding points
hold on
plot(x,y,'o', 'displayname', 'original')


hold off

xlim auto
ylim auto
legend show

%for adding points
hold on
xs = 0:0.1:150;
ytotal = fitting(c,xs);
plot(xs,ytotal)
hold off

function ytotal = fitting(c,x)
xspan = x;
y0 = zeros(2,1);
[t,ye] = ode15s(@(x,y)eq2(x,y,c), xspan, y0);
%protect against failure of integration
if t(end) ~= xspan(end)
    ytotal = inf(numel(xspan),1);
else
    ytotal = ye(:,1).*(1+0.39976) + ye(:,2);
end
end 
function dy=eq2(x,y,c)
%y(1)=CE,y(2)=LE
dy=zeros(2,1);
dy(1)=c(1) .* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(3) - y(1) .* (1 + c(2))) - c(4) .* y(1);
dy(2)=c(5).* (2.5 - (1 + c(2)) .* y(1) - y(2)) .* (c(6) - y(2) - c(4) .* y(2) .* (c(6) - y(2))) - c(7).* y(2);
end

[Walter Roberson]

90 90 90 90 90;120 120 120 120];

You only have four 120's but 5 of all other numbers.

[Jinglei]

Thank you. I corrected it. But this is not the big problem since the code still could not get the any result and thsoe c obtained were all -inf.

c =

-Inf -Inf -Inf -Inf -Inf -Inf -Inf

-Inf -Inf -Inf -Inf -Inf -Inf -Inf

-Inf -Inf -Inf -Inf -Inf -Inf -Inf

-Inf -Inf -Inf -Inf -Inf -Inf -Inf

-Inf -Inf -Inf -Inf -Inf -Inf -Inf

change_from_original =

-Inf -Inf -Inf -Inf -Inf -Inf -Inf

-Inf -Inf -Inf -Inf -Inf -Inf -Inf

-Inf -Inf -Inf -Inf -Inf -Inf -Inf

-Inf -Inf -Inf -Inf -Inf -Inf -Inf

-Inf -Inf -Inf -Inf -Inf -Inf -Inf

[Walter Roberson]

You cannot do that.

nlinfit() requires that the x and y values and initial model parameters must be vectors. You can call

nlinfit(x(:),y(:),myfunc,c0(:))

and then have myfunc make appropriate adjustments.

But inside myfunc you take the input and use it as tspan values for the ode. The ode*() functions require that the tspan consist of strictly monotonic values -- strictly increasing or strictly decreasing. Which your x(:) would not have, as your x has a lot of rows with duplicate columns.

If what you are trying to do is run a number of different independent fittings simultaneously, hoping that will be more efficient than running them independently, then you are going about that the wrong way. Yes, it is possible to get ode45() or ode15s() to effectively run multiple ODEs simulatenously, but (A) you need to build the code the right way; and (B) it is not at all certain to be more efficient; and (C ) it is questionable whether it would be of sufficient accuracy. @Jan in a recent post made the explicit claim that the numeric accuracy gets worse when you are running multiple not-coupled ODE simulatenously.

If you must make the attempt to run several simultaneous not-coupled ODE, then set up your equations using the Symbolic Toolbox, and then follow the work flow in the first example of odeFunction to build the anonymous function that would be passed to ode15s() or ode45(). (Make sure to build the ytotal equation too, so that you get the proper simultaneous totals.)

But really... just run them independently, the extra work to run them simultaneously is not worth the effort.

[Jinglei]

@Walter Roberson Thank you for your explanation. I understood what you said but I would like to get five curves in one figure. Is there any method to achieve that goal by running the fitting independently?

[Jinglei]

I just run the program with one group of my data, which, however, stopped in the middle. Those warnings and things were just like what were shown when I run successfully so I am confused of what happened for this especially considring the better data compared than others.

failed iteration 63

Warning: Failure at t=1.385053e+00. Unable to meet integration tolerances without reducing the step size below

the smallest value allowed (3.552714e-15) at time t.

> In ode15s (line 730)

In xian2mu1rawdatafor5000rat50du>fitting (line 89)

In nlinfit>@(b,x)w.*model(b,x) (line 283)

In nlinfit>LMfit (line 592)

In nlinfit (line 284)

In xian2mu1rawdatafor5000rat50du (line 47)

failed iteration 68

rejected negative output on iteration 70

Warning: Failure at t=1.187736e+01. Unable to meet integration tolerances without reducing the step size below

the smallest value allowed (2.842171e-14) at time t.

> In ode15s (line 730)

In xian2mu1rawdatafor5000rat50du>fitting (line 89)

In nlinfit>@(b,x)w.*model(b,x) (line 283)

In nlinfit>LMfit (line 592)

In nlinfit (line 284)

In xian2mu1rawdatafor5000rat50du (line 47)

failed iteration 73

Those above are part of the shown things in my matlab.

[Walter Roberson]

The code I posted in https://www.mathworks.com/matlabcentral/answers/1886712-i-am-confused-by-the-error-error-evaluating-inline-model-function#comment_2542987 specifically disables those ode15s messages.

It appears to me that if you were to loop the code using columns of your x and y, that the resulting code would already create multiple lines in the same figure, as it already has hold on and hold off all over the place.

[Jinglei]

@Walter Roberson Actually, I changed the code and only retain hold on and hold off with the points and the final line between and so there was only one curve in the final image. In addition, I still would like to ask about my former question about the problem when I run the code with another group of data.

hold on

xs = 0:0.1:150;

ytotal = fitting(c,xs);

plot(xs,ytotal)

hold off

[Jinglei]

I have got another question about the former code, which I gave as attached. I found that I could get a good curve when I used semicolon when giving the initial values for C while a matrix of 8*8 as double would be achieved as the final constants, which I couldn't understand. On the other hand, when I tried comma in the c0 matrix, no curve could be achieved. I would like to know how to fix that. In addition, when I used some other groups of data to run the code, no curve could be achieved neither. Since the data is relatively good for those situations and even worse data could lead to successful result, I am really confused about what is wrong. Could anyone offer me some help?

[Walter Roberson]

When you talk about semicolon when giving hte initial values for C are you referring to

c0_orig = [0.0019,0.0004,0.0001,1.5,0.0012,0.0011,1.5,0.0011];

If so then notice you have

    c0 = max(0, c0_orig + randn(1,8)/20);

if c0_orig were a column vector then you would get implicit expansion when you added the row vector of randn() values, which is why there would be an 8 x 8 output.

[Jinglei]

Got it and thank you. The thing is when I used a column vector, I can get a great fitting while nothing could be achieved when I used any row vector of the resulting matrix. I don't know what is wrong with the process.

I have found that each time when I got a certain group of constants and tried them as initial values, different result could be achieved. I know that may be caused by the randn process, which gains different values each time. But in this case, how can I get the optimal constants to get the best fitting result? Is there any method to improve the code?

[Walter Roberson]

Do you have any constraints on the inputs? Other than that they must be non-negative?

[Jinglei]

No,they are all the same as before except for the equations and initial values for the constants.

[Walter Roberson]

Then in that case, there are a couple of different ways you could improve the code:

• use ga or particleswarm or other similar evolutionary algorithm that tries a wide variety of different positions, instead of looping trying different random positions
• use MultiStart instead of looping trying different random positions
• do a more detailed mathematical analysis of your equations in order to figure out boundary conditions so that you can generate random starting positions that will be valid. This might be difficult as it would require analyzing the ways in which your ode can fail.

[Jinglei]

Got it and thank you for your help. I will give them a try and get better results.