Too Many Input Arguments

Hello! I've checked my code multiple times, and I keep getting a "too many inputs" error. Would anyone be able to help?

function b = kinetics123(v)

k1 = v(1);

k2 = v(2);

k3 = v(3);

k4 = v(4);

k5 = v(5);

b =(k1*k3*102*sqrt(k2*100)*0.0008*(1+k1*102+sqrt(k2*100)+(k3/k4)*k1*102*(0.609/k5))^2 - (6.1*10^-5));

end

v = [0.1, 0.1, 0.1, 0.1, 0.1];

options = optimset('MaxFunEvals',1000);

q = fminsearchbnd(@kinetics123, v, [], [], options)

Error using kinetics123

Too many input arguments.

Error in fminsearchbnd>@(x,varargin)fun(xtransform(x),varargin{:}) (line 233)

intrafun = @(x, varargin) fun(xtransform(x), varargin{:});

Error in fminsearch (line 201)

fv(:,1) = funfcn(x,varargin{:})

[Madeline Alzamora]

Also when I go through debugging, the function kinetics123 reads all of the input in correctly.

Answer 1

[Star Strider]

This works (with and without the norm call) —

v = [0.1, 0.1, 0.1, 0.1, 0.1];

[B,fv] = fminsearch(@kinetics123, v)                            % Without 'norm'

 
Exiting: Maximum number of function evaluations has been exceeded
         - increase MaxFunEvals option.
         Current function value: -21938913608024237095924990053658057124452464574261387201643260652167428249316775121043664379710237162164047785759422273482999418904150474824722563045423035223507880578537363894487937679396863294586303477699253665349954558661450752149220753408.000000 

B = 1×5

1.0e+53 * -1.3634 0.3493 0.0546 0.4180 0.2841

fv = -2.1939e+241

[B,fv] = fminsearch(@(b)norm(kinetics123(b)), v)                % With 'norm'

B = 1×5

-0.0111 0.0804 0.0890 0.1668 0.1356

fv = 6.1002e-05

function b = kinetics123(v)
k1 = v(1);
k2 = v(2);
k3 = v(3);
k4 = v(4);
k5 = v(5);
b =(k1*k3*102*sqrt(k2*100)*0.0008*(1+k1*102+sqrt(k2*100)+(k3/k4)*k1*102*(0.609/k5))^2 - (6.1*10^-5));
end

Adding the norm call (finds the minimum greater than sero) produces different parameter estimates. The code runs without error either way here.

I have no idea what the code is su0pposed to do otherwise.

.

[Madeline Alzamora]

This helps a lot! I'm basically trying to do a fit for kinetic parameters and output the constants that make the system zero.

Thank you!!

[Star Strider]

My pleasure!

That’s what I thought.

Are they being fitted to data? If so, uploading-posting that data here (preferably as a text or spreadsheet file) would allow me to help fit it to the equations.

Right now, I’m not certain what the objective is, other than to minimise the norm of the function (so I would advise using the version with the norm call).

.

[Madeline Alzamora]

Hi! Here's my data :)

I straight up just used the first values when i was computing the constants because I wasn't sure how to do a for loop for fmin search. with 102 = acetone, 0.609 = isopropanol out, and (6.1*10^-5) = experimental rate. I'm also attaching the edits that I made to the code based on a personal typing error and setting a lower bound. My v values are complete guesses, and the final answers keep changing whenever I guess them. Is that supposed to happen?

You're the best!!

function b = kinetics123(v)

%I know that all of the rate constants should be positive, and k3 should

%theoretically be the smallest because it is the rate limiting step

k1 = v(1);

k2 = v(2);

k3 = v(3);

k4 = v(4);

k5 = v(5);

%broke it into steps so I could read it clearer when typing

c = k1*k3*102*sqrt(100*k2)*0.0008; %100 and 0.0008 are set values, 102 is ac in excel

f = (1+k1*102+sqrt(k2*100) + (0.609/k5))^2; %102 is ac in excel, 0.609 is iso in excel

b = c/f - (6.1*10^-5);

%though there are more values, the experimental rate doesn't change too

%much

end

v = [1, 1, 1, 1, 1];

kinetics123(v);

A = [];

b= [];

Aeq = [];

beq = [];

[B,fv] = fmincon(@kinetics123, v, A, b, Aeq, beq, [0 0 0 0 0], [])

[Star Strider]

It is difficult for me to follow that, because it is not obvious to me how to apply the data to the regression equations.

What kinetic equation (or equations) is the system modeling?

Are they differential equations or functions (analytically inetgrated differential equations)?

.

[Madeline Alzamora]

Sorry!! I'm trying to model equation "b", and it should be a function. I just combined it into a few steps because I was having a difficult time making sure that all of my parenthesis were in the same place.

The three changing variables are acetone (102 here), the change in isopropanol (0.609 here), and the total rate is the output. I derived this function by hand based on theoretical mechanisms provided in my problem statement.

[Star Strider]

Thank you!

This runs without error. I’m not certain what ‘total rate’ is (there’s no specific variable with that name), so substitute that in the ‘TotalRate’ variable I created if mine isn’t correct. I need to know what it is so that I can experiment with it

Data = readtable('https://www.mathworks.com/matlabcentral/answers/uploaded_files/808424/kinetics%20project.xlsx', 'VariableNamingRule','preserve')

Data = 400×17 table
    Acetone in    Iso out    R_Iso_exp     R_Iso_mech    Sq Diff    Var6    Acetone in_1    Iso out_1    R_Iso_exp_1    R_Iso_mech_1    Sq Diff_1    Var12    Acetone in_2    Iso out_2    R_Iso_exp_2    R_Iso_mech_2    Sq Diff_2
    __________    _______    __________    __________    _______    ____    ____________    _________    ___________    ____________    _________    _____    ____________    _________    ___________    ____________    _________

       102        0.61038     6.118e-05       NaN          NaN      NaN         102          3.5818      0.00028721         NaN            NaN        NaN         102          4.7952      0.00032042         NaN            NaN   
       104        0.60501    6.0642e-05       NaN          NaN      NaN         104          3.5881      0.00028771         NaN            NaN        NaN         104          4.8352       0.0003231         NaN            NaN   
       106         0.5992     6.006e-05       NaN          NaN      NaN         106          3.6137      0.00028977         NaN            NaN        NaN         106          4.8849      0.00032642         NaN            NaN   
       108        0.59493    5.9632e-05       NaN          NaN      NaN         108          3.6297      0.00029105         NaN            NaN        NaN         108          4.9199      0.00032876         NaN            NaN   
       110         0.5903    5.9167e-05       NaN          NaN      NaN         110          3.6417      0.00029201         NaN            NaN        NaN         110          4.9597      0.00033141         NaN            NaN   
       112        0.58668    5.8804e-05       NaN          NaN      NaN         112          3.6558      0.00029315         NaN            NaN        NaN         112          5.0024      0.00033427         NaN            NaN   
       114        0.58086    5.8222e-05       NaN          NaN      NaN         114          3.6619      0.00029363         NaN            NaN        NaN         114          5.0438      0.00033704         NaN            NaN   
       116        0.57643    5.7777e-05       NaN          NaN      NaN         116          3.6763      0.00029479         NaN            NaN        NaN         116          5.0806      0.00033949         NaN            NaN   
       118        0.57281    5.7414e-05       NaN          NaN      NaN         118          3.6839       0.0002954         NaN            NaN        NaN         118          5.1134      0.00034168         NaN            NaN   
       120        0.56897    5.7029e-05       NaN          NaN      NaN         120          3.6948      0.00029627         NaN            NaN        NaN         120           5.157       0.0003446         NaN            NaN   
       122        0.56303    5.6434e-05       NaN          NaN      NaN         122          3.7038        0.000297         NaN            NaN        NaN         122           5.178        0.000346         NaN            NaN   
       124         0.5581     5.594e-05       NaN          NaN      NaN         124          3.7137      0.00029778         NaN            NaN        NaN         124           5.214      0.00034841         NaN            NaN   
       126        0.55447    5.5576e-05       NaN          NaN      NaN         126           3.735       0.0002995         NaN            NaN        NaN         126          5.2546      0.00035112         NaN            NaN   
       128        0.55023    5.5151e-05       NaN          NaN      NaN         128          3.7462      0.00030039         NaN            NaN        NaN         128          5.2993      0.00035411         NaN            NaN   
       130        0.54616    5.4743e-05       NaN          NaN      NaN         130          3.7468      0.00030044         NaN            NaN        NaN         130          5.3328      0.00035635         NaN            NaN   
       132        0.54246    5.4372e-05       NaN          NaN      NaN         132           3.762      0.00030166         NaN            NaN        NaN         132          5.3578      0.00035801         NaN            NaN   

Reactants = [Data.('Acetone in') Data.('Iso out')];
TotalRate = sum(Reactants,2);
% v = [0.1, 0.1, 0.1, 0.1, 0.1];
v = rand(5,1);

[B,fv] = fminsearch(@(b)norm(TotalRate - kinetics123(b,Reactants)), v)

 
Exiting: Maximum number of function evaluations has been exceeded
         - increase MaxFunEvals option.
         Current function value: 1002.894706 

B = 5×1

0.0007 0.1608 1.3782 0.0005 0.3635

fv = 1.0029e+03

EstTotalRate = kinetics123(B,Reactants);

EstTotalRate = 400×1

33.8815 35.2729 36.6236 38.1728 39.6947 41.3730 42.7589 44.3520 46.0861 47.8085

figure
scatter3(Reactants(:,1), Reactants(:,2), TotalRate, 50, 'p', 'filled')
hold on
plot3(Reactants(:,1), Reactants(:,2), EstTotalRate, '-r')
hold off
grid on
xlabel('CH_3CHO')
ylabel('CH_3CH_2OH')
zlabel('Output')
legend('Data', 'Fitted Regression', 'Location','best')

function b = kinetics123(v,Reactants)
k1 = v(1);
k2 = v(2);
k3 = v(3);
k4 = v(4);
k5 = v(5);
Acetn = Reactants(:,1);
IPrOH = Reactants(:,2);
b =(k1.*k3.*Acetn.*sqrt(k2.*100.)*0.0008.*(1+k1.*Acetn+sqrt(k2.*100)+(k3/k4).*k1.*Acetn.*(IPrOH/k5)).^2 - 6.1E-5);
end

.

[Madeline Alzamora]

Hi! I swapped out the total rate for R_Iso_exp because that's the experimental data that I'm trying to fit kinetics123 to (so that both of them are ~6.1*10^-5). It looks so nice!!! Is there a way to ensure that "EstTotalRate" converges to (6.1*10^-5) definitively without any k values being negative? Just to check, I swapped in fmincon and put in [0 0 0 0 0] for the lower bound, but it keeps reaching .82.

Thank you!!

Data = readtable('https://www.mathworks.com/matlabcentral/answers/uploaded_files/808424/kinetics%20project.xlsx', 'VariableNamingRule','preserve');

Reactants = [Data.('Acetone in') Data.('Iso out')];

%TotalRate = sum(Reactants,2);

TotalRate = [Data.('R_Iso_exp')];

% v = [0.1, 0.1, 0.1, 0.1, 0.1];

v = rand(5,1);

[B,fv] = fminsearch(@(b)norm(TotalRate - kinetics1234(b,Reactants)), v)

EstTotalRate = kinetics1234(B,Reactants);

%when you put in fmincon

A = [];

b= [];

Aeq = [];

beq = [];

[B,fv] = fmincon(@(b)norm(TotalRate - kinetics1234(b,Reactants)), v, A, b, Aeq, beq, [0 0 0 0 0], [])

[Star Strider]

This works, although I can’t get a good fit to those data.

Check to see that the objective funciton is coded correctly so that the output of the function is appropriate to the data. I have no idea what the reaction is or how the differential equation system was integrated, so I can’t check thiat myself.

Data = readtable('https://www.mathworks.com/matlabcentral/answers/uploaded_files/808424/kinetics%20project.xlsx', 'VariableNamingRule','preserve')

Data = 400×17 table
    Acetone in    Iso out    R_Iso_exp     R_Iso_mech    Sq Diff    Var6    Acetone in_1    Iso out_1    R_Iso_exp_1    R_Iso_mech_1    Sq Diff_1    Var12    Acetone in_2    Iso out_2    R_Iso_exp_2    R_Iso_mech_2    Sq Diff_2
    __________    _______    __________    __________    _______    ____    ____________    _________    ___________    ____________    _________    _____    ____________    _________    ___________    ____________    _________

       102        0.61038     6.118e-05       NaN          NaN      NaN         102          3.5818      0.00028721         NaN            NaN        NaN         102          4.7952      0.00032042         NaN            NaN   
       104        0.60501    6.0642e-05       NaN          NaN      NaN         104          3.5881      0.00028771         NaN            NaN        NaN         104          4.8352       0.0003231         NaN            NaN   
       106         0.5992     6.006e-05       NaN          NaN      NaN         106          3.6137      0.00028977         NaN            NaN        NaN         106          4.8849      0.00032642         NaN            NaN   
       108        0.59493    5.9632e-05       NaN          NaN      NaN         108          3.6297      0.00029105         NaN            NaN        NaN         108          4.9199      0.00032876         NaN            NaN   
       110         0.5903    5.9167e-05       NaN          NaN      NaN         110          3.6417      0.00029201         NaN            NaN        NaN         110          4.9597      0.00033141         NaN            NaN   
       112        0.58668    5.8804e-05       NaN          NaN      NaN         112          3.6558      0.00029315         NaN            NaN        NaN         112          5.0024      0.00033427         NaN            NaN   
       114        0.58086    5.8222e-05       NaN          NaN      NaN         114          3.6619      0.00029363         NaN            NaN        NaN         114          5.0438      0.00033704         NaN            NaN   
       116        0.57643    5.7777e-05       NaN          NaN      NaN         116          3.6763      0.00029479         NaN            NaN        NaN         116          5.0806      0.00033949         NaN            NaN   
       118        0.57281    5.7414e-05       NaN          NaN      NaN         118          3.6839       0.0002954         NaN            NaN        NaN         118          5.1134      0.00034168         NaN            NaN   
       120        0.56897    5.7029e-05       NaN          NaN      NaN         120          3.6948      0.00029627         NaN            NaN        NaN         120           5.157       0.0003446         NaN            NaN   
       122        0.56303    5.6434e-05       NaN          NaN      NaN         122          3.7038        0.000297         NaN            NaN        NaN         122           5.178        0.000346         NaN            NaN   
       124         0.5581     5.594e-05       NaN          NaN      NaN         124          3.7137      0.00029778         NaN            NaN        NaN         124           5.214      0.00034841         NaN            NaN   
       126        0.55447    5.5576e-05       NaN          NaN      NaN         126           3.735       0.0002995         NaN            NaN        NaN         126          5.2546      0.00035112         NaN            NaN   
       128        0.55023    5.5151e-05       NaN          NaN      NaN         128          3.7462      0.00030039         NaN            NaN        NaN         128          5.2993      0.00035411         NaN            NaN   
       130        0.54616    5.4743e-05       NaN          NaN      NaN         130          3.7468      0.00030044         NaN            NaN        NaN         130          5.3328      0.00035635         NaN            NaN   
       132        0.54246    5.4372e-05       NaN          NaN      NaN         132           3.762      0.00030166         NaN            NaN        NaN         132          5.3578      0.00035801         NaN            NaN   

Reactants = [Data.('Acetone in') Data.('Iso out')];
TotalRate = Data.('R_Iso_exp');
% v = [0.1, 0.1, 0.1, 0.1, 0.1];
v = rand(5,1)*100;

[B,fv] = fminsearch(@(b)norm(TotalRate - kinetics123(b,Reactants)), v)

B = 5×1

0.0000 102.1531 20.5651 83.6655 33.3665

fv = 9.7131e-04

EstTotalRate = kinetics123(B,Reactants);

figure
scatter3(Reactants(:,1), Reactants(:,2), TotalRate, 50, 'p', 'filled')
hold on
plot3(Reactants(:,1), Reactants(:,2), EstTotalRate, '-r')
hold off
grid on
xlabel('CH_3CHO')
ylabel('CH_3CH_2OH')
zlabel('Output')
legend('Data', 'Fitted Regression', 'Location','best')

function b = kinetics123(v,Reactants)
k1 = v(1);
k2 = v(2);
k3 = v(3);
k4 = v(4);
k5 = v(5);
Acetn = Reactants(:,1);
IPrOH = Reactants(:,2);
b =(k1.*k3.*Acetn.*sqrt(k2.*100.)*0.0008.*(1+k1.*Acetn+sqrt(k2.*100)+(k3/k4).*k1.*Acetn.*(IPrOH/k5)).^2 - 6.1E-5);
end

.

[Madeline Alzamora]

!!! I fixed a factor in my equation and played around a little bit in the structure and it's working! The esttotalrate starts off at the appropriate value, and my B values are stabilizing.

I can't thank you enough for helping me! I've been stuck on this for three days and it was driving me insane.

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.