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- #Let Bo be the baseline risk for time "To"
- #The probability of not getting infected is exp(-kt)
- #
- #Therefore:
- #Bo=exp(-kTo) -> k=-ln(1-Bo)/To
- #0.01 was suggested as a low baseline risk in: https://www.thelancet.com/action/showPdf?pii=S0140-6736%2820%2931142-9
- #Bo (Chu) =0.01
- #
- #Some (misguided people?) suggested 2 covid infections a year might not be so bad.
- #So let's have the mean time to infection half a year.
- #
- #The mean of a binomial distribution is N*p, let N be days. So over a year
- #N*p=2 -> p=2/365 (the probability of infection per day), To=2 day
- #Therefore
- Bo=2/365; #Baselinke risk (Bo) over time (To)
- To=1;
- PFm=0.95; #is the Mask protection factor
- m=(1-PFm)
- #Calculate the decay constants
- #Not that in matlab/octave log(x) is the natural log of f
- ko=-log(1-Bo)/To;
- km=-log(1-Bo*m)/To;
- #Let's consider what halps if we only wear the mask part time for the year
- Tf_max=(365/2);
- Tf_values=Tf_max/5:Tf_max/5:Tf_max;
- y=zeros(length(Tf_values),101);
- clf
- hold on
- #Commented code to consider other time frames
- #for I=1:length(Tf_values)
- I=5 #This is the index for the full time frame (half a year)
- Tf=Tf_values(I);
- t=[0 Tf/100:Tf/100:Tf];
- y(I,:)=exp(-ko*(Tf-t)).*exp(-km*(t));
- x=t/Tf;
- plot(x,y(I,:))
- Tf #The end of the time interal
- y(I,100) #The probability of not getting infected at t=Tf w/100% mask use
- y(I,1) #The probability of not getting infected with zero mask use
- #end
- hold off
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