Team:TU Munich/Modeling/Yeast Growth
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== Model Selection == | == Model Selection == | ||
- | As none of our measurement for yeast growth included the stationary phase, fitting a | + | As none of our measurement for yeast growth included the stationary phase, fitting a Gompertz curve, |
[[File:Eqn9950.jpg]] | [[File:Eqn9950.jpg]] | ||
- | would result in unidentifiabilities. As the | + | would most likely result in unidentifiabilities as the model is too complex. As the data was normalized to the value at t=0, we fitted it an non-scaled exponential curve. |
[[File:Eqn9950.jpg]] | [[File:Eqn9950.jpg]] | ||
- | This means that only one parameter remains, the inverse | + | This means that only one parameter remains, the inverse doubling time, scaled by a factor of ln(2). The normalization ensures that values between measurements are comparable. |
+ | |||
== Code == | == Code == | ||
Latest revision as of 14:30, 23 October 2012
Yeast Growth Model
Model Selection
As none of our measurement for yeast growth included the stationary phase, fitting a Gompertz curve,
would most likely result in unidentifiabilities as the model is too complex. As the data was normalized to the value at t=0, we fitted it an non-scaled exponential curve.
This means that only one parameter remains, the inverse doubling time, scaled by a factor of ln(2). The normalization ensures that values between measurements are comparable.
Code
data1=[1,1.3625,2.875,5.25]; t=[0,80,270,450]; fun = @(k,data) sum(((exp(k(1)*t)-data)/k(2)).^2); [k, RN] = fminsearch(@(k) fun(k,data1) - length(t)*log(1/(sqrt(2*pi*k(2)))),[0.005 1]); kc=0.01; acc=0; nacc=0; n=10000; sample=zeros(n,1); prob=zeros(n,1); lh=zeros(n,1); accepted=zeros(n,1); kprev = k(1); s=1; while (s<n+1) if(mod(s,500)==0) (s)/n; display(['Acceptance Rate: ' num2str(acc/(nacc+acc)*100) '%']) end try % the step is sampled from a multivariate normal distribution and % scaled with kc kcur = kprev + kc*mvnrnd(0,0.001); sample(s,:) = kcur; % calculate the a-posteriori probability of the new sample prob(s) = -fun([kcur k(2)],data1); if s>1 % check whether we reject the sample or not. mind that the % probability is log-scaled if log(rand(1)) < +prob(s)-prob(s-1) % update current sample kprev = kcur; % count accepted samples acc=acc+1; % save current acceptance rate for post-processing lh(s)=acc/(nacc+acc)*100; s=s+1; else % count not accepted samples nacc=nacc+1; end else % we need to do something else for the first sample. if log(rand(1)) < prob(s)+RN kprev = kcur; acc=acc+1; %waitbar((k-1)/n) lh(s)=acc/(nacc+acc)*100; s=s+1; else nacc=nacc+1; end end catch ME disp(ME) end end disp(['Doubling Time: ' num2str(mean(log(2)./sample)) ' STD: ' num2str(std(log(2)./sample))])