Asked by StarSign1997
on 22 Apr 2019

Hello!

I am fitting Monod equation to a data containing substrate (s), biomass (x), and ethanol (p) concentration against time. The objective is to get the parameters: 1) umax, 2) ks, 3) Yxs, and 4)Yps that will best represent the data. The differential equations are:

Here is my initial code using assumed values of the four parameters:

umax = 0.5;

ks = 6.5;

Yxs = 0.2;

Yps = 1.2;

%a(1) = x

%a(2) = s

%a(3) = p

f = @(t,a) [umax*a(1)*a(2)/(ks + a(2)); -(1/Yxs)*umax*a(1)*a(2)/(ks + a(2)); (1/Yps)*umax*a(1)*a(2)/(ks + a(2))];

xt0 = [0.0904,9.0115,0.0151];

[tspan,a] = ode45(f,[0 25],xt0);

figure

plot(tspan,a(:,1),tspan,a(:,2),tspan,a(:,3))

Here is the code for trying to fit it into the actual data (script file):

function pos = paramfun1(x,tspan)

umax = x(1);

ks = x(2);

Yxs = x(3);

Yps = x(4);

xt0 = x(5:7);

f = @(t,a) [umax*a(1)*a(2)/(ks + a(2)); -(1/Yxs)*umax*a(1)*a(2)/(ks + a(2)); (1/Yps)*umax*a(1)*a(2)/(ks + a(2))];

[~,pos] = ode45(f,tspan,xt0);

Here is my call function (in the command window):

xt0 = zeros(1,7);

xt0(1) = umax;

xt0(2) = ks;

xt0(3) = Yxs;

xt0(4) = Yps;

data =[0 3 5 8 9.5 11.5 14 16 18 20 25 27, 0.0904 0.1503 0.2407 0.3864 0.5201 0.6667 0.8159 0.9979 1.0673 1.1224 1.1512 1.2093; 0 3 5 8 9.5 11.5 14 16 18 20 25 27, 9.0115 8.8088 7.9229 7.2668 5.3347 4.911 3.5354 1.4041 0 0 0 0; 0 3 5 8 9.5 11.5 14 16 18 20 25 27, 0.0151 0.0328 0.0621 0.1259 0.2949 0.3715 0.4199 0.522 0.5345 0.6081 0.07662 0.7869];

%time =[0 3 5 8 9.5 11.5 14 16 18 20 25 27];

[pbest,exitflag,output] = lsqcurvefit(@paramfun,xt0,tspan,data);

fprintf('New parameters: %f, %f, %f, %f',pbest(1:4));

The error is function value not equal to YDATA. Btw, this code was based from an example in MATLAB. (https://www.mathworks.com/help/optim/ug/fit-differential-equation-ode.html)

My data is:

time = [0 3 5 8 9.5 11.5 14 16 18 20 25 27]

x = 0.0904 0.1503 0.2407 0.3864 0.5201 0.6667 0.8159 0.9979 1.0673 1.1224 1.1512 1.2093

s = 9.0115 8.8088 7.9229 7.2668 5.3347 4.911 3.5354 1.4041 0 0 0 0

p = 0.0151 0.0328 0.0621 0.1259 0.2949 0.3715 0.4199 0.522 0.5345 0.6081 0.07662 0.7869

Please help! I do not know how to input my data into the lsqcurvefit function.

Thanks in advance!

Answer by David Wilson
on 22 Apr 2019

Edited by David Wilson
on 22 Apr 2019

Accepted Answer

Here's my solution.

First set up the measured data, plot it, and look to see that it is all OK.

t = [0 3 5 8 9.5 11.5 14 16 18 20 25 27]'

x = [0.0904 0.1503 0.2407 0.3864 0.5201 0.6667 0.8159 0.9979 1.0673 1.1224 1.1512 1.2093]';

s = [9.0115 8.8088 7.9229 7.2668 5.3347 4.911 3.5354 1.4041 0 0 0 0]';

p = [0.0151 0.0328 0.0621 0.1259 0.2949 0.3715 0.4199 0.522 0.5345 0.6081 0.07662 0.7869]';

plot(t,[x,s,p],'o-')

No problems there. Note that the measured data is in columns.

Now, we need a function to export the predicted measurements as a function of the unknown parameters and initial conditions. My function is:

function ypred = paramfun1(p,t)

umax = p(1); ks = p(2); Yxs = p(3); Yps = p(4); % use disperse here ...

xt0 = p(5:7); % initial conditions

f = @(t,a) [umax*a(1)*a(2)/(ks + a(2)); ...

-(1/Yxs)*umax*a(1)*a(2)/(ks + a(2)); ...

(1/Yps)*umax*a(1)*a(2)/(ks + a(2))];

[~,ypred] = ode45(f,t,xt0);

end

May I suggest that you think about naming your functions sensible names. Also it is more elegant to use disperse (from the user's group) here.

Now we are ready to test it, and make sure our output is the same shape as the actual data.

xt0 = [0.1;9;0.1]; % read off from data plot [x0 ,S0, p0]';

p0 = [umax; ks; Yxs; Yps; xt0]

Ypred = paramfun1(p0,t); % measurement predictions

Now we are ready for the fit. I'm using your initial conditions, and we can get reasonable start values from the data.

pfit = lsqcurvefit(@paramfun1,p0,t,[x,s,p])

Now plot the fitted curve and the actual data. They should line up.

umax = pfit(1); ks = pfit(2);Yxs = pfit(3); Yps = pfit(4);

f = @(t,a) [umax*a(1)*a(2)/(ks + a(2)); -(1/Yxs)*umax*a(1)*a(2)/(ks + a(2)); (1/Yps)*umax*a(1)*a(2)/(ks + a(2))];

xt0 = pfit(5:7);

[tspan,a] = ode45(f,[0 25],xt0);

plot(tspan,a(:,1),tspan,a(:,2),tspan,a(:,3), '-', ...

t,x,'o',t,s,'+',t,p,'s')

Plot shows the regressed fit is not too bad:

Vinoj Liyanaarachchi
on 19 Oct 2019 at 4:30

StarSign1997 Can you kindly provide the final code.... It would be a great help.

I also have to model the same set of equations. Experimental data are as follows:

t = [0 2 4 6 8 10 12 14 16]';

x = [0.06 0.11 0.46 0.78 1.42 2.36 2.49 2.49 2.54]';

s = [9.54 9.33 9.52 9.06 8.05 7.27 6.03 5.80 5.51]';

p = [0.00 0.00 0.00 0.01 0.18 0.35 0.49 0.53 0.55]';

I used the code provided by David Wilson (I don't know how to disperse the intial values, So I used same code provided)

Curves were not fitted to the experimental data well.

Please help!

Thanks in advance!

StarSign1997
on 19 Oct 2019 at 13:27

Ahh yes. I tried your data and we have the same results.

Here is the script I used in the command window. The function parafun1 is the same as the one posted.

t = [0 2 4 6 8 10 12 14 16]';

x = [0.06 0.11 0.46 0.78 1.42 2.36 2.49 2.49 2.54]';

s = [9.54 9.33 9.52 9.06 8.05 7.27 6.03 5.80 5.51]';

p = [0.00 0.00 0.00 0.01 0.18 0.35 0.49 0.53 0.55]';

plot(t,[x,s,p],'o-')

umax = 0.5;

ks = 2.5;

Yxs = 5;

Yps = 1.2;

xt0 = [0.06;9.54;0.00]; % read off from data plot [x0 ,S0, p0]';

p0 = [umax; ks; Yxs; Yps; xt0]

ypred = paramfun1(p0,t); % measurement predictions

options = optimoptions ('lsqcurvefit','FiniteDifferenceStepSize',1e-4,'FiniteDifferenceType','central');

[pfit, presnorm, presidual] = lsqcurvefit(@paramfun1,p0,t,[x,s,p],[],[],options);

pfit = lsqcurvefit(@paramfun1,p0,t,[x,s,p])

umax = pfit(1); ks = pfit(2);Yxs = pfit(3); Yps = pfit(4);

xt0 = pfit(5:7);

[tspan,a] = ode45(f,[0 16],xt0);

plot(tspan,a(:,1),'-',tspan,a(:,2),'--',tspan,a(:,3),'k-.', ...

t,x,'o',t,s,'+',t,p,'s')

title('Monod Equation Simulation')

xlabel('Time')

ylabel('Concentration')

legend('Simulated Biomass','Simulated Substrate','Simulated Ethanol','Biomass (Actual)','Substrate (Actual)','Ethanol (Actual)')

Vinoj Liyanaarachchi
on 19 Oct 2019 at 16:17

Are there any techniques to improve the parameter estimation? Above shapes of the curves are not acceptable in my case (microalgae cultivation).

Thanks in advance!

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Answer by Alex Sha
on 21 Oct 2019 at 7:33

The follow results obtained from 1stOpt may be used for comparison:

Code:

Variable t,x,s,p;

ODEFunction x'=umax*s*x/(ks + s);

s'=-(1/Yxs)*umax*s*x/(ks + s);

p'=(1/Yps)*umax*s*x/(ks + s);

Data;

t = [0 2 4 6 8 10 12 14 16];

x = [0.06 0.11 0.46 0.78 1.42 2.36 2.49 2.49 2.54];

s = [9.54 9.33 9.52 9.06 8.05 7.27 6.03 5.80 5.51];

p = [0.00 0.00 0.00 0.01 0.18 0.35 0.49 0.53 0.55];

Results:

Root of Mean Square Error (RMSE): 0.250520910244571

Sum of Squared Residual: 1.50625743527444

Correlation Coef. (R): 0.981509121545543

R-Square: 0.963360155677103

Adjusted R-Square: 0.914727231437843

Determination Coef. (DC): 0.942399473562309

F-Statistic: 14.7361249635671

Parameter Best Estimate

-------------------- -------------

umax 0.117013263727751

ks -5.83263810513223

yxs 0.699511367324087

yps 5.01252941213249

Answer by Alex Sha
on 21 Oct 2019 at 8:42

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