Nonlinear regression with two variables

Hi, Im not really adept at programming but I need to fit a non linear regression model : y=a*(T-c)*(1-exp(b*(T-d)))(1-10^(e-pH)) where I have the values for y, T and pH. I used The curve fitting tool with nonlinearleastsquaremethod and a trust region algorithm, to fit a simpler version of the model where I just had the y and T.

Can someone please tell me how to solve this by uing either the curve fitting tool or writting a script?

Many thanks in advance.

[Ameer Hamza]

Can you show us the code to solve the simpler version of this model, and what are the issues extending it to the full model?

[Dipon Sarkar]

I fitted the simple model with the curve fitting app with the 'custom equation' option. This is the code that I generated from it:

%% Fit: 'untitled fit 1'.
[xData, yData] = prepareCurveData( mid35x, mid35y );

% Set up fittype and options.
ft = fittype( 'a*(x-c)*(1-exp(b*(x-d)))', 'independent', 'x', 'dependent', 'y' );
opts = fitoptions( 'Method', 'NonlinearLeastSquares' );
opts.Display = 'Off';
opts.StartPoint = [0.260107478852342 0.290941052369803 0.995531610700984 0.800330575352401];

% Fit model to data.
[fitresult, gof] = fit( xData, yData, ft, opts );

[Ameer Hamza]

What happen when you apply it to the complex model?

[Dipon Sarkar]

The simple model just has growth as the dependant variable and temperature as the independant. There is an added variable of pH in the comple model which I am unable to code for...

[Ameer Hamza]

Dipon, what is the value of 'e' in your model?

[Dipon Sarkar]

thats the regression coefficient that I would like to know from fitting the data. In the model its caled 'pHmin', which is the theoritical minimum pH required for the growth

Answer 1

[Alex Sha]

Hi, Dipon, post your y, T and pH data please, if possible.

[Dipon Sarkar]

y  T  pH 
 
  0.1157  5  5.5 
 
  0.1233  5  5.57 
 
  0.1865  12  5.56 
 
  0.2520  12  5.61 
 
  0.2519  15  5.53 
 
  0.2814  15  5.42 
 
  0.3680  18  5.37 
 
  0.3186  18  5.56 
 
  0.4549  22  5.42 
 
  0.4757  25  5.45 
 
  0.4559  25  5.49 
 
  0.4241  30  5.57 
 
  0.4697  30  5.46 
 
  0.4953  35  5.50 
 
  0.5767  35  5.42 
 
  0.4163  40  5.58 
 
  0.4279  40  5.53 

[Alex Sha]

Hi, Dipon, your original function is: y=a*(T-c)*(1-exp(b*(T-d)))(1-10^(e-pH)) , does you miss a operator symbol of "*"?, the function should be: y=a*(T-c)*(1-exp(b*(T-d)))*(1-10^(e-pH)), if right, the result should be:

Root of Mean Square Error (RMSE): 0.0327506128198954
Sum of Squared Residual: 0.0182342448813378
Correlation Coef. (R): 0.971783944382366
R-Square: 0.94436403455935

Parameter	Best Estimate
----------	-------------
a	-5.8541753677852E-13
b	0.0540580580588969
c	52.4138416996001
d	-5.49253844700665
e	15.2843009735593

[Dipon Sarkar]

Thanks a lot for taking the time to do this. Sorry I missed the operator symbol.

Could you please share the code for this?

In terms of fitting and the value of e and c determined, it doesnt really look like the data is fitting the model well, whereas the simple model fitted the data quite well.

Thanks a lot,

Dipon

[Dipon Sarkar]

Thanks a lot Alex. Could you please share the code?

[Alex Sha]

Hi, Dipon, rather than Matlab, the results above are obtained by using a package 1stOpt, easy for using, and with the ability of global optimization, no need to guess the initial start-values of each parameter, the codes look like:

Parameter a,b,c,d,e;
Variable y,T,pH;
Function y=a*(T-c)*(1-exp(b*(T-d)))*(1-10^(e-pH));
Data;
//y  T  pH
0.1157	5	5.5
0.1233	5	5.57
0.1865	12	5.56
0.2520	12	5.61
0.2519	15	5.53
0.2814	15	5.42
0.3680	18	5.37
0.3186	18	5.56
0.4549	22	5.42
0.4757	25	5.45
0.4559	25	5.49
0.4241	30	5.57
0.4697	30	5.46
0.4953	35	5.50
0.5767	35	5.42
0.4163	40	5.58
0.4279	40	5.53

a little more better result:

Root of Mean Square Error (RMSE): 0.0327506128180404
Sum of Squared Residual: 0.0182342448792722
Correlation Coef. (R): 0.971783944265729
R-Square: 0.944364034332657

Parameter	Best Estimate
----------	-------------
a	-3.90156564244422E-13
b	0.0540580605547428
c	52.4138409784489
d	-5.49253907297291
e	15.4605277403042

You may try Matlab, however, with free start-values of each parameter, it will be very hard or impossible to obtain results better than above.

Answer 2

[Ameer Hamza]

The problem you have posted has several local optimal solutions. Alex posted a solution using 1stOpt. You can also do something similar using global optimization toolbox in MATLAB. However, due to the existance of several local optima, which solution is applicable to your problem depends on the bounds on the parameters. Try the following code. I have also posted a comparison with the parameters posted by Alex for comparison; however, for further narrowing down the solution, you need to specify the bound on each parameter.

X = ...
 [0.1157  5  5.5 
  0.1233  5  5.57 
  0.1865  12  5.56 
  0.2520  12  5.61 
  0.2519  15  5.53 
  0.2814  15  5.42 
  0.3680  18  5.37 
  0.3186  18  5.56 
  0.4549  22  5.42 
  0.4757  25  5.45 
  0.4559  25  5.49 
  0.4241  30  5.57 
  0.4697  30  5.46 
  0.4953  35  5.50 
  0.5767  35  5.42 
  0.4163  40  5.58 
  0.4279  40  5.53];

y = X(:,1);
T = X(:,2);
pH = X(:,3);

Xdata = [T pH];
Ydata = y;

y = @(param, xdata) param(1).*(xdata(:,1)-param(3)).*(1-exp(param(2).*(xdata(:,1)-param(4)))).*(1-10.^(param(5)-xdata(:,2)));

problem = createOptimProblem('lsqcurvefit','x0',rand(1,5),...
    'objective', y, 'xdata', Xdata, 'ydata', Ydata);

ms = MultiStart;
solMs = run(ms,problem,50); %% <<<---- solution by MATLAB function

sol_1st_opt = [-5.8541753677852E-13 0.0540580580588969 52.4138416996001 -5.49253844700665 15.2843009735593];

plot3(T, pH, Ydata, 'b+-', ...
    T, pH, y(solMs, Xdata), 'r+--', ...
    T, pH, y(sol_1st_opt, Xdata), 'm+-.', ...
    'LineWidth', 2)

xlabel('T');
ylabel('pH');
zlabel('y');

legend({'Original Data', 'MultiStart MATLAB', '1stOpt'},...
    'FontSize', 14, 'Location', 'best')