### Estimation of SIR model’s parameters

In the current study, the standard susceptible–infectious–recovered (SIR) model was adopted. In this model, we divide the total population (N) into three categories: *Susceptible (S), Infected (I) and Recovered (R).* The susceptible is the part of the total population which is vulnerable and is at risk of being infected (if previously unexposed to the pandemic disease); the infected is the fraction that has been infected (currently colonized by the pandemic disease), and the recovered is the fraction of the total population having recovered/removed (either by death or recovery). The basic structure of SIR mathematical model is described in Fig. 3 (Bagal et al. 2020).

The specific description of the SIR model is as follows:

$$\frac{{{\text{d}}S}}{{{\text{d}}t}} = - \frac{\beta }{N}{\text{SI }}$$

(1)

$$\frac{{{\text{d}}I}}{{{\text{d}}t}} = \frac{\beta }{N}SI - \gamma I$$

(2)

$$\frac{{{\text{d}}R}}{{{\text{d}}t}} = \gamma I$$

(3)

where *β* represents the effective transmission rate and *γ* represents the removal or the recovery rate. *γ* is defined as the inverse of the duration of recovery *d* (*γ* = 1/*d*). Knowing that the total population size (N = S + I + R) is time independent. The population of Algeria is estimated at 43,851,044 inhabitants in the middle of the year according to United Nations reports (Worldmeter 2020) which is the (N) value for the SIR modeling.

Initially, in the absence of infection we have I + R = 0 and S≈N. We obtain the following equation from Eq. (2):

$$\frac{{{\text{d}}I}}{{{\text{d}}t}}\sim I\left( {\beta - \gamma } \right)$$

(4)

Then, integration of Eq. (4) gives the Equation:

$$I = I_{0} e^{{\left( {\beta - \gamma } \right)t}}$$

(5)

### Determination of transmission rate (*β*) and the *m* (*m* = *β*−*γ*) value

When the infection occurs, the susceptible population is almost equal to the total population (S ≈ N). Subsequently, the number of infectious individuals *I*(*t*) first increases exponentially.

$$\frac{{{\text{d}}I}}{{{\text{d}}t}}\sim mI$$

(6)

*m* is the difference between transmission and recovery rates (*m* = *β*−*γ*.

$$I\left( t \right) = I_{0} e^{mt}$$

(7)

$${\text{In }}I = mt + {\text{In}} I_{0}$$

(8)

We can calculate the value of m from the log-plot data and for the best-line fit we can use, for example, the least squares.

In the current work, the used data file was considered from the JHU-CSSE (Johns Hopkins University Center for Systems Science and Engineering) 2019 Novel Coronavirus Visual Dashboard database (JHU-CSSE 2020). The simulation of the SIR model’s Python code (SMPC 2020) based on the dataset of Algeria was done on the Google Colab platform, and the online date calculator was used to estimate the futuristic dates (DC 2020).

### Determination of the recovery rate (*γ*)

If we suppose that *I* is a constant (*I*(*t*) =* I*_{0}), we obtain the equation:

$$\frac{{{\text{d}}R}}{{{\text{d}}t}} = \gamma I_{0}$$

(9)

By integrating Eq. (9), then it is found the following Equation:

$$R\left( t \right) = \gamma tI_{0}$$

(10)

If time of recovery*t* = T days, and \(R\left(T\right)={I}_{0}\)*,* or \(\gamma T=1\).

Therefore, we obtain the next Equation:

$$\gamma \approx \frac{1}{T}$$

(11)

when time change d*t* = *α*, we obtain the following Equation from Eq. (3):

$$\frac{{R\left( {t + \alpha } \right)}}{\alpha } = \gamma I$$

(12)

or

$$\gamma \approx \frac{{R \left( {t + 1} \right) - R\left( t \right)}}{I\left( t \right)}$$

(13)

Estimation of *I*_{max}(percentage infected persons at the peak of the epidemic) and *S*_{inf} (percentage of susceptible people remaining after the end of the epidemic):

when we divide Eq. (2) by Eq. (1), we get the equation that follows:

$${\raise0.7ex\hbox{${\left( {\frac{{{\text{d}}I}}{{{\text{d}}t}}} \right)}$} \!\mathord{\left/ {\vphantom {{\left( {\frac{{{\text{d}}I}}{{{\text{d}}t}}} \right)} {\left( {\frac{{{\text{d}}S}}{{{\text{d}}t}}} \right)}}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{${\left( {\frac{{{\text{d}}S}}{{{\text{d}}t}}} \right)}$}} = - 1\frac{\gamma }{\beta } N \frac{1}{S}$$

(14)

or

$$\frac{{{\text{d}}I}}{{{\text{d}}S}} = - 1 + \frac{\gamma }{\beta } N\frac{1}{S}$$

(15)

If we integrate both sides, we obtain the Equation:

$$I = - S + \frac{\gamma }{\beta } N\log S + C$$

(16)

*C* is a constant.

Knowing that at the beginning of the infection, S≈N and the number of infected (I) is extremely low, this mean that At: \(t=0, I\sim 0\) and \(S\sim N\)

Therefore, if we substitute these values in Eq. (16), we get:

$$0 = - N + \frac{\gamma }{\beta }N\;In\;N + C$$

(17)

or

$$C = {\text{N}}\left( {1 + \frac{{\upgamma }}{{\upbeta }}{\text{InN}}} \right)$$

(18)

From Eq. (18), we take the value of *C* and we change it into Eq. (16). We will obtain Eq. (19):

$$I = N - S + \frac{\gamma }{\beta } N {\text{In}}\frac{S}{N}$$

(19)

Equation (19) is valid for all times. In general, the number of infected persons (*I*) increases exponentially from the beginning of the infection to reach a peak, before gradually shrinking to zero. We will estimate the percentage infected persons at the peak of COVID-19 (*I*_{max}) and remained susceptible people after the end of the epidemic.

Here, we use Eq. (19) to resolve Eq. (2) related to infection rate.

For simplification, we suppose \(\mathrm{S}=\mathrm{Ns},\mathrm{ I}=\mathrm{Ni},\) \(\mathrm{S}=\mathrm{Ns},\mathrm{I}=\mathrm{Ni},\) and \(\mathrm{R}=Nr\) \(\mathrm{S}=\mathrm{Ns}\) \(\mathrm{S}=\mathrm{Ns}\).

where* s* is the fraction of total susceptible population.

*i* is the fraction of total infected population.

*r* is the fraction of total recovered/removed population.

Therefore, following equations are obtained:

$$\frac{{{\text{d}}t}}{{{\text{d}}t}} = \beta is - \gamma i = i\left( {\beta s - \gamma } \right)$$

(20)

and

$$i = 1 - s + \frac{\gamma }{\beta }{\text{In}} S$$

(21)

At the peak of the infection when d \(\mathrm{di}/\mathrm{dt}=0\) \(\mathrm{di}/\mathrm{dt}=0\) \(\mathrm{di}/\mathrm{dt}=0\) \(\mathrm{di}/\mathrm{dt}=0\), *s* is calculated by the equation that follows:

$$s = \frac{\gamma }{\beta }$$

(22)

if we substitute the value of s from Eq. (22) into Eq. (21), we get Eq. (23):

$$i_{{\max}} = 1 + \frac{\gamma }{\beta }\left( {{\text{In}}\frac{\gamma }{\beta } - 1} \right)$$

(23)

We, now, need to find out \({S}_{inf}=\underset{t\to \infty }{\mathrm{lim}}S(t)\) the percentage of remaining susceptible individuals after the end of the infection. It is noted that at the end of infection t tends to infinity (∞), and thus, i = 0. We can rewrite Eq. (21) as follows:

$$1 - s + \frac{\gamma }{\beta }{\text{In}} s = 0$$

(24)

We can solve Eq. (24) numerically to obtain the value of s.

### Determination of the basic reproduction number (R_{0})

*R*_{0} or the basic reproduction number is defined as the average number of people infected by a single individual. Mathematically, it represents the ratio of transmission and recovery rates.

$$R_{0} = \frac{\beta }{\gamma }$$

(25)

To calculate *R*_{0}, we used the following Equations:

$$i_{{\max}} = 1 - \frac{1}{{R_{0} }}\left( {1 - {\text{In}}\frac{1}{{R_{0} }}} \right)$$

(26)

and

$$R_{0} = 1 - \frac{{{\text{In}}S_{{\inf}} }}{{S_{{\inf}} - 1}}$$

(27)