Wednesday, I will be giving a talk at the seminar at the European Network for Business and Industrial Statistics (ENBIS), to present our model on COVID-19 pandemic control: balancing detection policy and lockdown intervention under ICU sustainability. The slides can be downloaded from here.

# Tag Archives: SIR

# Talk at CMAP on optimal Control and COVID-19

Friday, I will be giving a talk at the seminar at Ecole Polytechnique, on our model on COVID-19 pandemic control: balancing detection policy and lockdown intervention under ICU sustainability. The slides can be downloaded from here.

# COVID19 pandemic control: balancing detection policy and lockdown intervention under ICU sustainability

After almost two months, with Romuald Elie, Chi (Tran Viet Chi) and Mathieu Laurière, we finally have a draft of the paper related to our recent work, entitled “COVID-19 pandemic control: balancing detection policy and lockdown intervention under ICU sustainability” (available on HAL ArXiv and MedrXiv)

Our model is an extention of the classical SIR model, but more realistic for the COVID-19 (pronounded “cider”)

We use scenarios to see the impact on various quantities

but also optimal control to see the best strategy, when it comes to lockdown, and testing. All comments are welcome…

# Modeling pandemics (2)

When introducing the SIR model, in our initial post, we got an ordinary differential equation, but we did not really discuss stability, and periodicity. It has to do with the Jacobian matrix of the system. But first of all, we had three equations for three function, but actually\displaystyle{{\frac{dS}{dt}}+{\frac {dI}{dt}}+{\frac {dR}{dt}}=0}so it means that our problem is here simply in dimension 2. Hence\displaystyle {\begin{aligned}&X={\frac {dS}{dt}}=\mu(N-S)-{\frac {\beta IS}{N}},\\[6pt]&Y={\frac {dI}{dt}}={\frac {\beta IS}{N}}-(\mu+\gamma)I\end{aligned}}and therefore, the Jacobian of the system is\begin{pmatrix}\displaystyle{\frac{\partial X}{\partial S}}&\displaystyle{\frac{\partial X}{\partial I}}\\[9pt]\displaystyle{\frac{\partial Y}{\partial S}}&\displaystyle{\frac{\partial Y}{\partial I}}\end{pmatrix}=\begin{pmatrix}\displaystyle{-\mu-\beta\frac{I}{N}}&\displaystyle{-\beta\frac{S}{N}}\\[9pt]\displaystyle{\beta\frac{I}{N}}&\displaystyle{\beta\frac{S}{N}-(\mu+\gamma)}\end{pmatrix}We should evaluate the Jacobian at the equilibrium, i.e. S^\star=\frac{\gamma+\mu}{\beta}=\frac{1}{R_0}andI^\star=\frac{\mu(R_0-1)}{\beta}We should then look at eigenvalues of the matrix.

Our very last example was

times = seq(0, 100, by=.1) p = c(mu = 1/100, N = 1, beta = 50, gamma = 10) start_SIR = c(S=0.19, I=0.01, R = 0.8) resol = ode(y=start_SIR, t=times, func=SIR, p=p) plot(resol[,"time"],resol[,"I"],type="l",xlab="time",ylab="") |

We can compute values at the equilibrium

mu=p["mu"]; beta=p["beta"]; gamma=p["gamma"] N=1 S = (gamma + mu)/beta I = mu * (beta/(gamma + mu) - 1)/beta |

and the Jacobian matrix

J=matrix(c(-(mu + beta * I/N),-(beta * S/N), beta * I/N,beta * S/N - (mu + gamma)),2,2,byrow = TRUE) |

Now, if we look at the eigenvalues,

eigen(J)$values [1] -0.024975+0.6318831i -0.024975-0.6318831i |

or more precisely 2\pi/b where a\pm ib are the conjuguate eigenvalues

2 * pi/(Im(eigen(J)$values[1])) [1] 9.943588 |

we have a damping period of 10 time lengths (10 days, or 10 weeks), which is more or less what we’ve seen above,

The graph above was obtained using

p = c(mu = 1/100, N = 1, beta = 50, gamma = 10) start_SIR = c(S=0.19, I=0.01, R = 0.8) resol = ode(y=start_SIR, t=times, func=SIR, p=p) plot(resol[1:1e5,"time"],resol[1:1e5,"I"],type="l",xlab="time",ylab="",lwd=3,col="red") yi=resol[,"I"] dyi=diff(yi) i=which((dyi[2:length(dyi)]*dyi[1:(length(dyi)-1)])<0) t=resol[i,"time"] arrows(t[2],.008,t[4],.008,length=.1,code=3) |

If we look carefully. at the begining, the duration is (much) longer than 10 (about 13)… but it does converge towards 9.94

plot(diff(t[seq(2,40,by=2)]),type="b") abline(h=2 * pi/(Im(eigen(J)$values[1])) |

So here, theoretically, every 10 weeks (assuming that our time length is a week), we should observe an outbreak, smaller than the previous one. In practice, initially it is every 13 or 12 weeks, but the time to wait between outbreaks decreases (until it reaches 10 weeks).

# Modeling pandemics (1)

The most popular model to model epidemics is the so-called SIR model – or Kermack-McKendrick. Consider a population of size N, and assume that S is the number of susceptible, I the number of infectious, and R for the number recovered (or immune) individuals, \displaystyle {\begin{aligned}&{\frac {dS}{dt}}=-{\frac {\beta IS}{N}},\\[6pt]&{\frac {dI}{dt}}={\frac {\beta IS}{N}}-\gamma I,\\[6pt]&{\frac {dR}{dt}}=\gamma I,\end{aligned}}so that \displaystyle{{\frac{dS}{dt}}+{\frac {dI}{dt}}+{\frac {dR}{dt}}=0}which implies that S+I+R=N. In order to be more realistic, consider some (constant) birth rate \mu, so that the model becomes\displaystyle {\begin{aligned}&{\frac {dS}{dt}}=\mu(N-S)-{\frac {\beta IS}{N}},\\[6pt]&{\frac {dI}{dt}}={\frac {\beta IS}{N}}-(\gamma+\mu) I,\\[6pt]&{\frac {dR}{dt}}=\gamma I-\mu R,\end{aligned}}Note, in this model, that people get sick (infected) but they do not die, they recover. So here, we can model chickenpox, for instance, not SARS.

The dynamics of the infectious class depends on the following ratio:\displaystyle{R_{0}={\frac {\beta }{\gamma +\mu}}} which is the so-called basic reproduction number (or reproductive ratio). The effective reproductive ratio is R_0S/N, and the turnover of the epidemic happens exactly when R_0S/N=1, or when the fraction of remaining susceptibles is R_0^{-1}. As shown in Directly transmitted infectious diseases:Control by vaccination, if S/N<R_0^{-1} the disease (the number of people infected) will start to decrease.

Want to see it ? Start with

mu = 0 beta = 2 gamma = 1/2 |

for the parameters. Here, R_0=4. We also need starting values

epsilon = .001 N = 1 S = 1-epsilon I = epsilon R = 0 |

Then use the ordinary differential equation solver, in R. The idea is to say that \boldsymbol{Z}=(S,I,R) and we have the gradient \frac{\partial \boldsymbol{Z}}{\partial t} = SIR(\boldsymbol{Z})where SIR is function of the various parameters. Hence, set

p = c(mu = 0, N = 1, beta = 2, gamma = 1/2) start_SIR = c(S = 1-epsilon, I = epsilon, R = 0) |

The we must define the time, and the function that returns the gradient,

times = seq(0, 10, by = .1) SIR = function(t,Z,p){ S=Z[1]; I=Z[2]; R=Z[3]; N=S+I+R mu=p["mu"]; beta=p["beta"]; gamma=p["gamma"] dS=mu*(N-S)-beta*S*I/N dI=beta*S*I/N-(mu+gamma)*I dR=gamma*I-mu*R dZ=c(dS,dI,dR) return(list(dZ))} |

To solve this problem use

library(deSolve) resol = ode(y=start_SIR, times=times, func=SIR, parms=p) |

We can visualize the dynamics below

par(mfrow=c(1,2)) t=resol[,"time"] plot(t,resol[,"S"],type="l",xlab="time",ylab="") lines(t,resol[,"I"],col="red") lines(t,resol[,"R"],col="blue") plot(t,t*0+1,type="l",xlab="time",ylab="",ylim=0:1) polygon(c(t,rev(t)),c(resol[,"R"],rep(0,nrow(resol))),col="blue") polygon(c(t,rev(t)),c(resol[,"R"]+resol[,"I"],rev(resol[,"R"])),col="red") |

We can actually also visualize the effective reproductive number is R_0S/N, where

R0=p["beta"]/(p["gamma"]+p["mu"]) |

The effective reproductive number is on the left, and as we mentioned above, when we reach 1, we actually reach the maximum of the infected,

plot(t,resol[,"S"]*R0,type="l",xlab="time",ylab="") abline(h=1,lty=2,col="red") abline(v=max(t[resol[,"S"]*R0>=1]),col="darkgreen") points(max(t[resol[,"S"]*R0>=1]),1,pch=19) plot(t,resol[,"S"],type="l",xlab="time",ylab="",col="grey") lines(t,resol[,"I"],col="red",lwd=3) lines(t,resol[,"R"],col="light blue") abline(v=max(t[resol[,"S"]*R0>=1]),col="darkgreen") points(max(t[resol[,"S"]*R0>=1]),max(resol[,"I"]),pch=19) |

And when adding a \mu parameter, we can obtain some interesting dynamics on the number of infected,

times = seq(0, 100, by=.1) p = c(mu = 1/100, N = 1, beta = 50, gamma = 10) start_SIR = c(S=0.19, I=0.01, R = 0.8) resol = ode(y=start_SIR, t=times, func=SIR, p=p) plot(resol[,"time"],resol[,"I"],type="l",xlab="time",ylab="") |