0 个投票

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{:})

1 个评论
显示 -1更早的评论 隐藏 -1更早的评论

Madeline Alzamora
Madeline Alzamora 2021-11-21
Also when I go through debugging, the function kinetics123 reads all of the input in correctly.

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Star Strider
Star Strider 2021-11-21
Ran in:

1 个投票

Ran in:
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.
.

10 个评论
显示 8更早的评论 隐藏 8更早的评论

Madeline Alzamora
Madeline Alzamora 2021-11-21
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
Star Strider 2021-11-21
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
Madeline Alzamora 2021-11-21
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
Star Strider 2021-11-21
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
Madeline Alzamora 2021-11-21
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
Star Strider 2021-11-21
Ran in:
Ran in:
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
Madeline Alzamora 2021-11-21
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
Star Strider 2021-11-21
Ran in:
Ran in:
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
Madeline Alzamora 2021-11-21
!!! 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
Star Strider 2021-11-21
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Interactive Control and Callbacks 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by