Modeling the COVID-19 / Coronavirus pandemic – 2.Simple Models
Armed with the data on the coronavirus such as the serial interval, incubation period, and the base growth rate, we are now in a position to start modeling the pandemic. Note that as the title suggests, these are simple models. Any conclusions drawn from this specific page should be taken with a grain of salt. More realistic modeling will be carried out in subsequent posts.
SIR - A very simple model
Using the above numbers, we are pretty much ready to start modeling the pandemic. We start with the simplest model that can encapsulate the exponential growth.
The simplest model for the pandemic growth is the well known SIR model, which includes the number of uninfected people $S$, the total population $N$, the number of infected and contagious individuals $I$ and the number of recovered individuals $R$. The set of ordinary differential equations (ODEs) describing the behavior is: \begin{eqnarray} {dS \over dt} &=& - \beta \left(S \over N \right) I , \\ {dI \over dt} &=& + \beta \left(S \over N \right) I - \gamma I, \\ {dR \over dt} &=& + \gamma I . \end{eqnarray} Here $\beta$ is the transmitting coefficient, which depends on the social behavior and of course some inherent characteristics of the virus. $\gamma$ is the recovery rate or the rate at which the contagious person leaves the contagious state (e.g., gets hospitalized or quarantined), in units of one over time.
This equation is nonlinear because when a large fraction of the population gets infected, $S/N$ starts decreasing, quenching the epidemic. We want (at least at first) to better understand the behavior when only a small fraction of the population is infected.
Thus, the equation of interest, assuming $S/N \approx 1$, is \begin{equation} {dI \over dt} = + \beta I - \gamma I. \end{equation} If we guess an exponential behavior (since it is a homogeneous linear ODE) of the form $X \propto \exp(r t)$ (where $X$ is any variable), we find: \begin{equation} r I = (\beta - \gamma)I ~~~\rightarrow~~~ r = \beta - \gamma. \end{equation} This immediately tells us that the infection can grow and become an epidemic if $\beta$ is larger than $\gamma$.
In fact, we can relate $r$ to the basic reproduction number $R_0$, which is the initial number of people that will be infected by an infectious individual (before any measures are taken). It is \begin{equation} R_0 = \int_0^{\infty} \beta \exp(-\gamma t) dt = {\beta \over \gamma} = {r + \gamma \over \gamma}. \end{equation} This is because the probability that an infected individual remains contagious at time $t$ is proportional to $\exp(-\gamma t)$.
If we compare our results to the nominal growth rate of 0.3 ± 0.07 day$^{-1}$ and take $\gamma$ to be the reciprocal of the serial interval, i.e., 1 / (6.6 ± 1.3) day$^{-1}$ (assuming the errors on the fit for the distribution are uncorrelated), we obtain that $R_0$ = 3.0 ± 0.6. This is the average number of infections from a contagious person. We also find $\beta$ = 0.46 ± 0.07 day$^{-1}$.
Based on this simple model, we see that in order to guarantee overcoming the pandemic growth, we need to reduce $\beta - \gamma$ and make it negative. This requires either reducing $R_0$ (i.e., $\beta$), by a factor of 3 or even 4, which is not really reasonable (effectively making the infected people less contagious) or increasing $\gamma$, which implies shortening the time that an infected person is contagious (by quarantining him), or a combination of both. Let us see how this changes if we introduce a latent period where the person is non-contagious.
Adding a non-contagious latent period
One generalization of the simplest model is to include a period when the infected person is noncontagious, namely, it is a latent period. (This isn't the clinical incubation period, which is the time until the onset of symptoms, as people can be contagious even before symptoms develop, if they develop). Thus, our model now includes the number of uninfected people $S$, the number of infected people $L$, in the "latent period", that are still noncontagious, the number $C$ of contagious infected people, and the number of recovered individuals $R$. The equations describing the behavior here will be \begin{eqnarray} {dS \over dt} &=& - \beta \left(S \over N \right) C , \\ {dL \over dt} &=& + \beta \left(S \over N \right) C - \lambda L, \\ {dC \over dt} &=& + \lambda L - \gamma C, \\ {dR \over dt} &=& + \gamma C . \end{eqnarray} Here, $\lambda$ is the rate at which infected people become contagious. Also, we again guess exponential behavior for the linear case (for which $\left(S / N \right) \rightarrow 1$), and get \begin{eqnarray} r L &=& + \beta C - \lambda L, \\ r C &=& + \lambda L - \gamma C, \\ \end{eqnarray} Because this is a homogeneous set of equations, it is an eigenvalue problem. The solution is obtained when the determinant vanishes: \begin{equation} \left | \begin{array}{c c} \lambda + r & - \beta \\ - \lambda & \gamma +r \\ \end{array} \right| = 0 \end{equation} This gives two solutions. The positive one (describing the pandemic) is: \begin{equation} r = {1\over 2} \left( -(\lambda + \gamma) + \sqrt{(\lambda - \gamma)^2 + 4 \lambda \beta } \right) \end{equation} We can invert this relation to find $\beta$ given the growth rate $r$ which we measure: \begin{equation} \beta = { (\lambda + r)(r+\gamma)\over \lambda}. \end{equation} For a very short latent period, the rate at which noncontagious become contagious, $\lambda$, is very large and we recover the equation from the previous section.
We can also see that we still obtain $r=0$ for $\lambda = \beta$. However, for other values of $\beta$ we get $|r(\gamma)| < |r(\lambda \rightarrow \infty)|$. This is because the latent period slows things down, without affecting the overall behavior of the system. Once a person becomes contagious it is a race between the infection rate $\beta C$ and the recovery rate $\gamma C$. For this reason, the basic reproduction number $R_0$ is still \begin{equation} R_0 = \int_0^{\infty} \beta \exp(-\gamma t) dt = {\beta \over \gamma}. \end{equation} If we consider the serial interval distribution we derived in the background data post, we see that taking $1/\lambda \sim 2 \pm 1$ day is reasonable. If we now take $\lambda + \gamma = 1/(6.6\pm 1.3)$ day$^{-1}$, we get \begin{eqnarray} \beta &=& 0.75 \pm 0.22 \mathrm{~day}^{-1}\\ R_0 & = & 4.6 \pm 1.6. \end{eqnarray} Namely, we obtain a higher basic reproduction number. This is because the introduction of a latent period (of order 2 days) implies that for the same infection rate and recovery rate, the overall growth rate is slower. In order to compensate for it, the infection rate and basic reproduction numbers have to be higher in order to have the same growth rate $r$. In the next post we will consider having a distribution of infection coefficient $\beta$, and in the subsequent, we will also calculate the infection with a more appropriate time dependent infection rate.
Adding Quarantining
The next step is to add the effects of quarantining of sick people. If we want to stay within the framework of the linear equations, the easiest way to incorporate quarantining is to add an additional rate $\kappa$ which describes the rate at which an infectious person is quarantined. In fact, this number can be different in the latent period (when the person hasn't developed symptoms) and in the contagious period, when he could have. Thus, we introduce $\kappa_{L}$, $\kappa_{C}$ and now consider the equations: \begin{eqnarray} {dL \over dt} &=& - \lambda L - \kappa_{L} L + \beta C , \\ {dC \over dt} &=& + \lambda L - \gamma C - \kappa_{C} C. \end{eqnarray} If we now guess ${L,C \propto \exp(r t)}$, we again find ourselves with an eigenvalue problem, of which the solution is: \begin{eqnarray} r&=&{1\over 2}\left(-(\lambda + \kappa_{L}) - (\gamma +\kappa_{C})\right) \\ \nonumber && + \sqrt{\left(+(\lambda+\kappa_{L}) - (\gamma +\kappa_{C})\right)^2 + 4 \lambda \alpha}. \end{eqnarray} This gives $r=0$ for \begin{equation} \beta_{crit} = {(\lambda + \kappa_{L}) (\gamma + \kappa_{C}) \over \lambda}. \end{equation} If for example, we cannot detect people in the latent phase ($\kappa_L = 0$), and it takes 2 days to discover that people might be infected with the coronavirus, then $\kappa_C = 1/2$ day$^{-1}$. We also have $1/\lambda = 2 \pm 1$ day and $1/\lambda + 1/\gamma = 6.6 \pm 1.3 $ days which leads to $\beta_{crit} = 0.664 \pm 0.036$. However the value of $\beta$ without social distancing and other such measures is $\beta \approx 0.75$ (in the simple model with a latent / contagious period). In other words, quarantining 2 days after a person becomes infectious, which is 4 days after he is infected is barely sufficient to increase the $\beta_{crit}$ to the base value, and probably not enough to stop the pandemic without additional means (e.g. social distancing). We will return to this calculation once we have a better description of the $\beta$, allowing it to be a function of time since the infection.
SIR - A very simple model
Using the above numbers, we are pretty much ready to start modeling the pandemic. We start with the simplest model that can encapsulate the exponential growth.
The simplest model for the pandemic growth is the well known SIR model, which includes the number of uninfected people $S$, the total population $N$, the number of infected and contagious individuals $I$ and the number of recovered individuals $R$. The set of ordinary differential equations (ODEs) describing the behavior is: \begin{eqnarray} {dS \over dt} &=& - \beta \left(S \over N \right) I , \\ {dI \over dt} &=& + \beta \left(S \over N \right) I - \gamma I, \\ {dR \over dt} &=& + \gamma I . \end{eqnarray} Here $\beta$ is the transmitting coefficient, which depends on the social behavior and of course some inherent characteristics of the virus. $\gamma$ is the recovery rate or the rate at which the contagious person leaves the contagious state (e.g., gets hospitalized or quarantined), in units of one over time.
This equation is nonlinear because when a large fraction of the population gets infected, $S/N$ starts decreasing, quenching the epidemic. We want (at least at first) to better understand the behavior when only a small fraction of the population is infected.
Thus, the equation of interest, assuming $S/N \approx 1$, is \begin{equation} {dI \over dt} = + \beta I - \gamma I. \end{equation} If we guess an exponential behavior (since it is a homogeneous linear ODE) of the form $X \propto \exp(r t)$ (where $X$ is any variable), we find: \begin{equation} r I = (\beta - \gamma)I ~~~\rightarrow~~~ r = \beta - \gamma. \end{equation} This immediately tells us that the infection can grow and become an epidemic if $\beta$ is larger than $\gamma$.
In fact, we can relate $r$ to the basic reproduction number $R_0$, which is the initial number of people that will be infected by an infectious individual (before any measures are taken). It is \begin{equation} R_0 = \int_0^{\infty} \beta \exp(-\gamma t) dt = {\beta \over \gamma} = {r + \gamma \over \gamma}. \end{equation} This is because the probability that an infected individual remains contagious at time $t$ is proportional to $\exp(-\gamma t)$.
If we compare our results to the nominal growth rate of 0.3 ± 0.07 day$^{-1}$ and take $\gamma$ to be the reciprocal of the serial interval, i.e., 1 / (6.6 ± 1.3) day$^{-1}$ (assuming the errors on the fit for the distribution are uncorrelated), we obtain that $R_0$ = 3.0 ± 0.6. This is the average number of infections from a contagious person. We also find $\beta$ = 0.46 ± 0.07 day$^{-1}$.
Based on this simple model, we see that in order to guarantee overcoming the pandemic growth, we need to reduce $\beta - \gamma$ and make it negative. This requires either reducing $R_0$ (i.e., $\beta$), by a factor of 3 or even 4, which is not really reasonable (effectively making the infected people less contagious) or increasing $\gamma$, which implies shortening the time that an infected person is contagious (by quarantining him), or a combination of both. Let us see how this changes if we introduce a latent period where the person is non-contagious.
Adding a non-contagious latent period
One generalization of the simplest model is to include a period when the infected person is noncontagious, namely, it is a latent period. (This isn't the clinical incubation period, which is the time until the onset of symptoms, as people can be contagious even before symptoms develop, if they develop). Thus, our model now includes the number of uninfected people $S$, the number of infected people $L$, in the "latent period", that are still noncontagious, the number $C$ of contagious infected people, and the number of recovered individuals $R$. The equations describing the behavior here will be \begin{eqnarray} {dS \over dt} &=& - \beta \left(S \over N \right) C , \\ {dL \over dt} &=& + \beta \left(S \over N \right) C - \lambda L, \\ {dC \over dt} &=& + \lambda L - \gamma C, \\ {dR \over dt} &=& + \gamma C . \end{eqnarray} Here, $\lambda$ is the rate at which infected people become contagious. Also, we again guess exponential behavior for the linear case (for which $\left(S / N \right) \rightarrow 1$), and get \begin{eqnarray} r L &=& + \beta C - \lambda L, \\ r C &=& + \lambda L - \gamma C, \\ \end{eqnarray} Because this is a homogeneous set of equations, it is an eigenvalue problem. The solution is obtained when the determinant vanishes: \begin{equation} \left | \begin{array}{c c} \lambda + r & - \beta \\ - \lambda & \gamma +r \\ \end{array} \right| = 0 \end{equation} This gives two solutions. The positive one (describing the pandemic) is: \begin{equation} r = {1\over 2} \left( -(\lambda + \gamma) + \sqrt{(\lambda - \gamma)^2 + 4 \lambda \beta } \right) \end{equation} We can invert this relation to find $\beta$ given the growth rate $r$ which we measure: \begin{equation} \beta = { (\lambda + r)(r+\gamma)\over \lambda}. \end{equation} For a very short latent period, the rate at which noncontagious become contagious, $\lambda$, is very large and we recover the equation from the previous section.
We can also see that we still obtain $r=0$ for $\lambda = \beta$. However, for other values of $\beta$ we get $|r(\gamma)| < |r(\lambda \rightarrow \infty)|$. This is because the latent period slows things down, without affecting the overall behavior of the system. Once a person becomes contagious it is a race between the infection rate $\beta C$ and the recovery rate $\gamma C$. For this reason, the basic reproduction number $R_0$ is still \begin{equation} R_0 = \int_0^{\infty} \beta \exp(-\gamma t) dt = {\beta \over \gamma}. \end{equation} If we consider the serial interval distribution we derived in the background data post, we see that taking $1/\lambda \sim 2 \pm 1$ day is reasonable. If we now take $\lambda + \gamma = 1/(6.6\pm 1.3)$ day$^{-1}$, we get \begin{eqnarray} \beta &=& 0.75 \pm 0.22 \mathrm{~day}^{-1}\\ R_0 & = & 4.6 \pm 1.6. \end{eqnarray} Namely, we obtain a higher basic reproduction number. This is because the introduction of a latent period (of order 2 days) implies that for the same infection rate and recovery rate, the overall growth rate is slower. In order to compensate for it, the infection rate and basic reproduction numbers have to be higher in order to have the same growth rate $r$. In the next post we will consider having a distribution of infection coefficient $\beta$, and in the subsequent, we will also calculate the infection with a more appropriate time dependent infection rate.
Adding Quarantining
The next step is to add the effects of quarantining of sick people. If we want to stay within the framework of the linear equations, the easiest way to incorporate quarantining is to add an additional rate $\kappa$ which describes the rate at which an infectious person is quarantined. In fact, this number can be different in the latent period (when the person hasn't developed symptoms) and in the contagious period, when he could have. Thus, we introduce $\kappa_{L}$, $\kappa_{C}$ and now consider the equations: \begin{eqnarray} {dL \over dt} &=& - \lambda L - \kappa_{L} L + \beta C , \\ {dC \over dt} &=& + \lambda L - \gamma C - \kappa_{C} C. \end{eqnarray} If we now guess ${L,C \propto \exp(r t)}$, we again find ourselves with an eigenvalue problem, of which the solution is: \begin{eqnarray} r&=&{1\over 2}\left(-(\lambda + \kappa_{L}) - (\gamma +\kappa_{C})\right) \\ \nonumber && + \sqrt{\left(+(\lambda+\kappa_{L}) - (\gamma +\kappa_{C})\right)^2 + 4 \lambda \alpha}. \end{eqnarray} This gives $r=0$ for \begin{equation} \beta_{crit} = {(\lambda + \kappa_{L}) (\gamma + \kappa_{C}) \over \lambda}. \end{equation} If for example, we cannot detect people in the latent phase ($\kappa_L = 0$), and it takes 2 days to discover that people might be infected with the coronavirus, then $\kappa_C = 1/2$ day$^{-1}$. We also have $1/\lambda = 2 \pm 1$ day and $1/\lambda + 1/\gamma = 6.6 \pm 1.3 $ days which leads to $\beta_{crit} = 0.664 \pm 0.036$. However the value of $\beta$ without social distancing and other such measures is $\beta \approx 0.75$ (in the simple model with a latent / contagious period). In other words, quarantining 2 days after a person becomes infectious, which is 4 days after he is infected is barely sufficient to increase the $\beta_{crit}$ to the base value, and probably not enough to stop the pandemic without additional means (e.g. social distancing). We will return to this calculation once we have a better description of the $\beta$, allowing it to be a function of time since the infection.
- Background data
- Simple Modeling (this page)
- Effects of a population with a variable infection rate
- Modeling with at time variable infection rate
- Numerical Model (coming soon!)
- Discussion and Conclusions (coming soon!)