Final Point Equals Initial Point when trying to optimize

2022 7 11
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更新时间：2022 7 12
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The initial guesses for A0 are equal to what has been calculated for A. How can i trouble shoot?
I tried changing the initial values, changing the tolerance, changing functions... non seems to be working.

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Torsten
Torsten 2022-7-11
编辑：Torsten 2022-7-11
Translating your code, you use as time interval for integration [299:1307.99] and insert the actual time T during integration in the expressions exp(-Ei/(R*T)). This means that the temperature of your system rises from 299 to 1307.99 K linearly with time.
Is this really what you intend to do ?
And don't name your array of experimental values "exp". It can easily come into conflict with the MATLAB exponential function exp(x) for e^x.
Rosalind
Rosalind 2022-7-11
Yes, temperature linearly increases. I changed the name from exp to experiment but still have the same problem.

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Torsten
Torsten 2022-7-11
编辑：Torsten 2022-7-11
Ran in:

0 个投票

Ran in:
Here is a code where A does not equal A0:
A0=1e-3*[1.5*(10^7),1*(10^6),1*(10^6)]
A0 = 1×3
15000 1000 1000
sum(fit(A0).^2)
ans = 0.0945
options = optimset('TolFun',1e-7);
[A,resnorm,residual,exitflag,output,lambda,Jacobian] = lsqnonlin(@fit,A0)
Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
A = 1×3
1.0e+03 * 2.2446 1.3592 0.3087
resnorm = 0.0404
residual = 4×1
0.1025 0.0994 0.0999 0.1001
exitflag = 1
output = struct with fields:
firstorderopt: 2.9048e-07 iterations: 17 funcCount: 72 cgiterations: 0 algorithm: 'trust-region-reflective' stepsize: 393.3516 message: 'Local minimum found.↵↵Optimization completed because the size of the gradient is less than↵the value of the optimality tolerance.↵↵<stopping criteria details>↵↵Optimization completed: The first-order optimality measure, 2.904751e-07,↵is less than options.OptimalityTolerance = 1.000000e-06.'
lambda = struct with fields:
lower: [3×1 double] upper: [3×1 double]
Jacobian =
1.0e-03 * (1,1) -0.0235 (2,1) 0.0452 (3,1) -0.0126 (4,1) -0.0091 (1,2) -0.0467 (2,2) -0.0218 (3,2) 0.0873 (4,2) -0.0188 (1,3) -0.1528 (2,3) -0.0754 (3,3) -0.0899 (4,3) 0.3181
sum(fit(A).^2)
ans = 0.0404
function res = fit(A)
m0 = [0.5 0 0 0 ]';
T = [299,1307.99];
[T,m] = ode45(@(t,m) myfunction(t,m,A), T, m0);
experiment=[0,0.02589,0.0540,0.01821];
res = [m(end,1)-experiment(1);m(end,2)-experiment(2);...
m(end,3)-experiment(3);m(end,4)-experiment(4)];
end
function dmdt = myfunction(T,m,A)
R = 8.314*(10^(-3));
E1 = 140;
E2 = 133;
E3 = 121;
dmdt = zeros(4,1);
dmdt(1) = -m(1)*(A(1)*exp(-E1/(R*T))+A(2)*exp(-E2/(R*T))+A(3)*exp(-E3/(R*T)));
dmdt(2) = m(1)*A(1)*exp(-E1/(R*T));
dmdt(3) = A(2)*exp(-E2/(R*T))* m(1);
dmdt(4) = A(3)*exp(-E3/(R*T))* m(1);
end

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Rosalind
Rosalind 2022-7-11
I need to have the T written as T = 299 + (0.25*(1:1900)-0.25); to be able to change the timing and so on. when I put that instead of T = [299,1307.99]; I get the same A as A0.
Even when I use the code above and only change the initial mass i get the same A and A0.
Torsten
Torsten 2022-7-11
编辑：Torsten 2022-7-11
I need to have the T written as T = 299 + (0.25*(1:1900)-0.25); to be able to change the timing and so on.
Whether you write T = 299 + (0.25*(1:1900)-0.25) or T = [299,773.75] only changes the number of outputs from ODE45. Since you only need the last output for time = 773.75, it does not make a difference. Especially it doesn't make any changes in "timing and so on".
Note that the T in the list of inputs to myfunction is not the vector T = 299 + (0.25*(1:1900)-0.25), but only the actual integration time (thus a scalar between 299 and 773.75).
Even when I use the code above and only change the initial mass i get the same A and A0.
The MATLAB code works - it's up to you now to produce senseful results. I mentionned the problems with your data base.
Rosalind
Rosalind 2022-7-12
编辑：Rosalind 2022-7-12
the thing is that in order ro have the mass balance correct i should have put m0 = [0.098 0 0 0 ] instead of m0 = [0.5 0 0 0 ]. this was my mistake in the first place.And when i do change it I have the equal A0 and A, which is not correct.
Torsten
Torsten 2022-7-12
Ran in:
Ran in:
A0=1e-4*[1.5*(10^7),1*(10^6),1*(10^6)]
A0 = 1×3
1500 100 100
sum(fit(A0).^2)
ans = 0.0059
options = optimset('TolFun',1e-12,'TolX',1e-12);
[A,resnorm,residual,exitflag,output,lambda,Jacobian] = lsqnonlin(@fit,A0)
Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
A = 1×3
1.0e+03 * 1.8842 1.1707 0.3470
resnorm = 0.0013
residual = 4×1
0.0225 -0.0044 -0.0270 0.0088
exitflag = 1
output = struct with fields:
firstorderopt: 7.4758e-07 iterations: 7 funcCount: 32 cgiterations: 0 algorithm: 'trust-region-reflective' stepsize: 639.9974 message: 'Local minimum found.↵↵Optimization completed because the size of the gradient is less than↵the value of the optimality tolerance.↵↵<stopping criteria details>↵↵Optimization completed: The first-order optimality measure, 7.475833e-07,↵is less than options.OptimalityTolerance = 1.000000e-06.'
lambda = struct with fields:
lower: [3×1 double] upper: [3×1 double]
Jacobian =
1.0e-04 * (1,1) -0.0511 (2,1) 0.0952 (3,1) -0.0227 (4,1) -0.0214 (1,2) -0.1018 (2,2) -0.0382 (3,2) 0.1841 (4,2) -0.0441 (1,3) -0.3331 (2,3) -0.1322 (3,3) -0.1615 (4,3) 0.6267
sum(fit(A).^2)
ans = 0.0013
function res = fit(A)
m0 = [0.098 0 0 0 ]';
T = [299,1307.99];
options = odeset('RelTol',1e-8,'AbsTol',1e-8);
[T,m] = ode45(@(t,m) myfunction(t,m,A), T, m0, options);
experiment=[0,0.02589,0.0540,0.01821];
res = [m(end,1)-experiment(1);m(end,2)-experiment(2);...
m(end,3)-experiment(3);m(end,4)-experiment(4)];
end
function dmdt = myfunction(T,m,A)
R = 8.314*(10^(-3));
E1 = 140;
E2 = 133;
E3 = 121;
dmdt = zeros(4,1);
dmdt(1) = -m(1)*(A(1)*exp(-E1/(R*T))+A(2)*exp(-E2/(R*T))+A(3)*exp(-E3/(R*T)));
dmdt(2) = m(1)*A(1)*exp(-E1/(R*T));
dmdt(3) = A(2)*exp(-E2/(R*T))* m(1);
dmdt(4) = A(3)*exp(-E3/(R*T))* m(1);
end
Rosalind
Rosalind 2022-7-12
Thank you Torsten. it worked.
Can you please tell me what to be careful about (or consider) if I want to expand my ODE equations and have more parameters to estimate?
Torsten
Torsten 2022-7-12
Can you please tell me what to be careful about (or consider) if I want to expand my ODE equations and have more parameters to estimate?
Try to get more experimental data (especially not only for the end time, but for the complete course of the process).

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