Optimizing transport frequency in multi-layered urban transportation networks for pandemic prevention

Abstract

In this paper, we show how transport policy decisions regarding vehicle scheduling frequency can affect the pandemic dynamics in urban populations. Specifically, we develop a multi-agent simulation framework to model infection dynamics in complex transportation networks. Our agents periodically commute between home and work via a combination of walking routes and public transit, and make decisions intelligently based upon their location, available routes, and expectations of public transport arrival times. Our infection scheme allows for different levels of contagiousness, as a function of where the agents interact (i.e., inside or outside). The results show that the pandemic’s scale is heavily impacted by the network’s structure, and the decision making of the agents. In particular, the progression of the pandemic greatly differs when agents primarily infect each other in a crowded urban transportation system, opposed to while walking. We also assess the effect of modifying the public transport’s running frequency on the virus spread. Lowering the running frequency can discourage agents from taking public transportation too often, especially for shorter distances. On the other hand, the low frequency contributes to more crowded streetcars or subway cars if the policy is not designed correctly, which is why such an analysis may prove valuable for finding “sweet spots” that optimize the system. The proposed approach has been validated on real-world data, and a model of the transportation network of downtown Toronto, Canada. The used framework is flexible and can be easily adjusted to model other urban environments, and additional forms of transportation (such as carpooling, ride-share and more). This general approach can be used to model contiguous disease spread in urban environments, including influenza or various COVID-19 variants.

Appendices

Proofs of theorems from Sect. 3

To prove Theorem 3.1, we first prove the following lemma.

Lemma A.1

There exists a function $\epsilon _{1} = \epsilon _{1}(n) = o(1)$, such that for any $0 \le t < \tau _{c_f}$, a.a.s. it holds that

$$\begin{aligned} \left| X_{t+1} - X_{t} - (n - X_t)\left( \sum _{i=1}^{2} \alpha _i \left( 1 - \exp \left( -\frac{b_i \lambda _i \alpha _i X_t}{n}\right) \right) \right) \right| \le \epsilon _1. \end{aligned}$$

Note that $\epsilon _1$ does not depend on t.

Proof of Lemma A.1

We shall apply a standard form of Chernoff bound [see, for example, Corollary 2.3 in Janson et al. (2011)]. Let $X \in \text {Bin}(k,p)$ be a random variable with the binomial distribution with parameters k and p, and suppose $0 < \epsilon \le 3/2$ and $\mu := \mathbb {E}X = k p$. In this case,

$$\begin{aligned} \mathbb {P}[ |X- \mu | \ge \epsilon \cdot \mu ] \le 2\exp \left( - \frac{\epsilon ^2 \cdot \mu }{3} \right) . \end{aligned}$$

(8)

Given $s \ge 0$ and $j \in \{1,2\}$, define $D_{s}^{j}$ as the set of the agents that choose the path $P_{j}$ on the commute s. Moreover, for $s \ge 0$, define $Y_{s}^j$ to be the number of the agents which become infected during commute s while taking $P_j$. Clearly, $Y_{s}^j = |D_{s}^j \cap I_{s} \cap \lnot I_{s-1}|$ for $s \ge 1$. Let us now condition on $I_t$ and $D_{t+1}^{j}$ for $0 \le t < \tau _{c_f}$ and fix agents $a,b \in [n]$ where $a \ne b$. Observe then that if $a \in [n] \setminus I_t$, $b \in I_t$, and both a and b select $P_j$ then

$$\begin{aligned} \mathbb {P}\left[ \hbox {Agent}\ a \ \hbox {is infected in commute}\ t+1 \,\textrm{by agent}\, b \, | \, I_t, D_{t+1}^{j}\right] = \beta _{j} \lambda _{j}. \end{aligned}$$

To see this, observe that if both agents select path $P_j$, then agent b infects a provided a and b interact, and their interaction is contagious. This occurs with a probability of precisely $\beta _{j} \lambda _{j}$. As the agents of $I_t \cap D_{t+1}^{j}$ infect a independently of one another, we have that

$$\begin{aligned} \mathbb {P}\left[ \hbox {Agent}\ a \ \hbox {is infected in commute}\, t+1 \, | \, I_t, D_{t+1}^{j}\right] =1 - (1 - \beta _j \lambda _j)^{|I_t \cap D_{t+1}^j|}. \end{aligned}$$

(9)

Now, $\beta _j = b_j/n$, so in particular, $\beta _j \lambda _j = o(1)$. Thus, if $a \in [n] \setminus I_t$ and $a \in D_{t+1}^j$ then

$$\begin{aligned} \mathbb {P}\left[ \hbox {Agent}\ a \ \hbox {is infected in commute}\ t+1 \, | \, I_t, D_{t+1}^{j}\right] \sim 1- \exp \left( -\frac{b_j \lambda _j |I_t \cap D_{t+1}^j|}{n}\right) . \end{aligned}$$

Thus,

$$\begin{aligned} \mathbb {E}\left[ Y_{t+1}^j \, | \, I_t, D_{t+1}^j\right] \sim |D_{t+1}^{j} \cap \lnot I_t| \left( 1- \exp \left( -\frac{b_j \lambda _j |I_t \cap D_{t+1}^j|}{n}\right) \right) \end{aligned}$$

(10)

To simplify (10), observe that conditional on $I_t$, the random variables $|I_t \cap D_{t+1}^j|$ and $|D_{t+1}^{j} \cap \lnot I_t|$ are distributed as $\text {Bin}(|I_t|, \alpha _j)$ and $\text {Bin}( |\lnot I_t|, \alpha _j)$, respectively. Moreover, $c_{s} n \le |I_t| \le c_{f} n$, as $0 \le t \le \tau _{c_f}$, so we may apply (8) with, say, $\epsilon (n):= 1/n^{1/3} =o(1)$ to conclude that a.a.s. $\ell :=|I_t \cap D_{t+1}^j| = (1+ o(1)) \alpha _j X_{t}$ and $k:=|\lnot I_t \cap D_{t+1}^j| =(1 + o(1)) \alpha _j (n - X_t)$. Thus, a.a.s.

$$\begin{aligned} \mathbb {E}\left[ Y_{t+1}^j \, | \, I_t, D_{t+1}^j\right] \sim \alpha _j (n - X_t) \left( 1- \exp \left( -\frac{b_j \lambda _j \alpha _j X_{t}}{n}\right) \right) = \Theta (n). \end{aligned}$$

(11)

On the other hand, $Y_{t+1}^j$ conditional on $I_t$ and $D_{t+1}^j$ is distributed as a binomial $\text {Bin}(k,p)$ with parameters $k = |D_{t+1}^{j} \cap \lnot I_t|$ and $p= 1 - (1 - \beta _j \lambda _j)^{|I_t \cap D_{t+1}^j|} = 1 - (1 - \beta _j \lambda _j)^{\ell }$. Moreover, a.a.s.

$$\begin{aligned} \mu _j:= \alpha _j (n - X_t) \left( 1- \exp \left( -\frac{b_j \lambda _j \alpha _j X_{t}}{n}\right) \right) \sim k p, \end{aligned}$$

(12)

where the approximation for p follows from (9). As a result, by taking $\epsilon (n) = 1/n^{1/3}$ and applying (8), we get that a.a.s.

$$\begin{aligned} \mathbb {P}\left[ |Y_{t+1}^j - \mu _j| \ge \epsilon \mu _j \, | \, I_t, D_{t+1}^{j}\right] \le \exp \left( -\Omega \left( n^{1/3}\right) \right) . \end{aligned}$$

(Formally, one should approximate stochastically lower and upper bound $Y_{t+1}^j$ by $\text {Bin}(k_{-},p_-)$ and $\text {Bin}(k_+,p_+)$ with some deterministic functions $k_{\pm }$ and $p_{\pm }$, such that $p_+/p_- \rightarrow 1$ and $k_+/k_- \rightarrow 1$ as $n \rightarrow \infty$.) Thus, a.a.s.,

$$\begin{aligned} |Y_{t+1}^j - \mu _j| \le \epsilon \mu _j \end{aligned}$$

(13)

for each $j =1,2$. On the other hand, $X_{t+1} - X_{t} = \sum _{j=1}^{2} Y_{t+1}^j$, and

$$\begin{aligned} \mu _{1} + \mu _2 \sim (n - X_t)\left( \sum _{i=1}^{2} \alpha _i \left( 1 - \exp \left( -\frac{b_i \lambda _i \alpha _i X_t}{n}\right) \right) \right) , \end{aligned}$$

so the proof is complete. $\square$

Proof of Theorem 3.1

Let us set $A_{i}:= b_i \lambda _i \alpha _i \ge 0$ for $i =1,2$, and $A:= \max \{1, A_1, A_2\}$ for convenience. Given $0 \le t_0 < \tau _{c_f}$, our goal is to show that there exists a function $\epsilon _0 = \epsilon _0(n) = o(1)$ such that a.a.s.

$$\begin{aligned} \left| \frac{X_{t_0}}{n} - \widetilde{x}_{t_0} \right| \le \epsilon _0(n). \end{aligned}$$

In order to prove this, we first prove the following implication. Let us take $0 \le t < \tau _{c_f}$, and assume that $0 \le \delta = \delta (n) \le 1$ satisfies

$$\begin{aligned} \left| \frac{X_{t}}{n} - \widetilde{x}_t \right| \le \delta , \end{aligned}$$

(14)

and

$$\begin{aligned} \left| X_{t+1} - X_{t} - (n - X_t)\left( \sum _{i=1}^{2} \alpha _i \left( 1 - \exp \left( -\frac{b_i \lambda _i \alpha _i X_t}{n}\right) \right) \right) \right| \le \delta . \end{aligned}$$

(15)

Under these assumptions, we claim that $|X_{t+1}/n -\widetilde{x}_{t+1}| \le 5 A \delta$. Observe first that by (14), $X_t/n \le \widetilde{x}_t + \delta$, so

$$\begin{aligned} \exp (-A_i X_t/n) \ge \exp (-A_i( \widetilde{x}_t + \delta )). \end{aligned}$$

Now, using the inequality $1-x \le \exp (-x)$, we get that

$$\begin{aligned} \exp (-(z_1 + z_2)) \ge \exp (-z_1) (1-z_2) \ge \exp (-z_1) - z_2 \end{aligned}$$

for $z_{1},z_{2} \ge 0$. It follows that

$$\begin{aligned} \exp (-A_i( \widetilde{x}_t + \delta )) \ge \exp (-A_{i} \widetilde{x}_t) - A \delta . \end{aligned}$$

(16)

Define $g(z):= \sum _{i=1}^{2} \alpha _i (1- \exp (- A_i z))$ for $z \in \mathbb {R}$. Observe then that g(z) is an increasing function of z and so by (14) and (16),

$$\begin{aligned} g(X_t/n) \le g(\widetilde{x}_t + \delta ) \le g(\widetilde{x}_t) + 2 A \delta . \end{aligned}$$

(17)

Now, applying (15),

$$\begin{aligned} \frac{X_{t+1}}{n}&\le \frac{X_t}{n} + \left( 1 - \frac{X_{t}}{n} \right) \left( \sum _{i=1}^{2} \alpha _i \left( 1 -\exp \left( -A_i X_t/n\right) \right) \right) + \delta \\&\le \widetilde{x}_t + \delta + (1 - \widetilde{x}_t + \delta ) g(X_t/n) + \delta \\&\le \widetilde{x}_t + (1 - \widetilde{x}_t) g(X_t/n) + \delta + \delta \\&\le \widetilde{x}_t + (1 - \widetilde{x}_t) g(\widetilde{x}_t) + 2 A \delta + 2 \delta + \delta \\&\le \widetilde{x}_t + (1 - \widetilde{x}_t) g(\widetilde{x}_t) + 5 A \delta \end{aligned}$$

where the latter inequalities follow since $g(z) \le 1$ for $z \ge 0$, and by (14) and (17). Yet, $\widetilde{x}_{t+1}: = \widetilde{x}_t + (1 - \widetilde{x}_t) g(\widetilde{x}_t)$, so

$$\begin{aligned} \frac{X_{t+1}}{n} \le \widetilde{x}_{t+1} + 5 A\delta . \end{aligned}$$

An analogous argument shows that $X_{t+1}/n \ge \widetilde{x}_{t+1} - 5 A \delta$. Thus, assuming (14) and (15) for $\delta \le 1$, it follows that

$$\begin{aligned} \left| \frac{X_{t+1}}{n} - \widetilde{x}_{t+1}\right| \le 5 A \delta . \end{aligned}$$

In order to complete the argument, take $\delta (n) = \epsilon _{3}(n)$, where $\epsilon _{3} = o(1)$ is from Lemma A.1. Clearly, we may assume that $5^{t_0} A \delta \le 1$ for n sufficiently large. Now, $X_{0}/n = \widetilde{x}_0$ by assumption. Moreover, since $t_0$ is constant, we may apply Lemma A.1 to each $0 \le t \le t_0$ to ensure that (15) holds a.a.s. for all $0 \le t \le t_0$ simultaneously. Thus, proceeding by induction, we get that a.a.s.

$$\begin{aligned} \left| \frac{X_{t_0}}{n} - \widetilde{x}_{t_0}\right| \le 5^{t_0} A \delta . \end{aligned}$$

Since $t_{0}$ is a constant, $5^{t_0} A \delta = o(1)$, so the argument is complete. $\square$

Proof of Corollary 3.2

Let us first fix $c:= \widetilde{x}_{t_{c_f}+3}$, which clearly satisfies $c > c_f$. By applying Theorem 3.1, it is easy to show that $\tau _{c} > t_{c_f} + 1$. Thus, we may apply Theorem 3.1 at the values from $\{t_{c_f} -1, t_{c_f}, t_{c_f} +1\}$. Specifically, there exists $\epsilon _1 = o(1)$ such that a.a.s.,

$$\begin{aligned} \left| \frac{X_k}{n} - \widetilde{x}_k \right| \le \epsilon _{1}(n) \end{aligned}$$

(18)

for each $k \in \{t_{c_f} -1, t_{c_f}, t_{c_f} +1\}$.

Now, for $k = t_{c_f} -1$, (18) implies that a.a.s.

$$\begin{aligned} \frac{X_{t_{c_f} -1}}{n} \le \widetilde{x}_{t_{c_f} - 1} + \epsilon _1. \end{aligned}$$

On the other hand, $\widetilde{x}_{t_{c_f} - 1} < c_f$. Thus, for all n sufficiently large, $\epsilon _1(n) < c_f - \widetilde{x}_{t_{c_f} -1}$. It follows that a.a.s.

$$\begin{aligned} \frac{X_{t_{c_f} -1}}{n} < c_f, \end{aligned}$$

and so $\tau _{c_f} \ge t_{c_f}$.

Suppose now that $\widetilde{x}_{t_{c_f}} > c_f$ (part (a)). Using (18) for $k = t_{c_f}$, we get that a.a.s.,

$$\begin{aligned} \frac{X_{t_{c_f}}}{n} \ge \widetilde{x}_{t_{c_f}} - \epsilon _1. \end{aligned}$$

Since $\widetilde{x}_{t_{c_f}} > c_f$, for all n sufficiently large, $\epsilon _1(n) < \widetilde{x}_{t_{c_f}} - c_f$. It follows that a.a.s.

$$\begin{aligned} \frac{X_{t_{c_f}}}{n} > c_f, \end{aligned}$$

and so $\tau _{f} \le t_{c_f}$. The same argument can be applied for part (b) but with $X_{t_{c_f+1}}$ instead of $X_{t_{c_f}}$. $\square$

Approximation for the number of infected agents

In this section, we derive an approximate but closed formula for the number of infected agents after a given number of iterations. First, let us note that $\exp (-x) = 1 - x + O(x^2)$, and so we may approximate (3) as follows:

$$\begin{aligned} \widetilde{x}_{t+1} - \widetilde{x}_t= & {} (1 - \widetilde{x}_t) \left( \alpha _1 (1 -\exp \left( - b_1 \lambda _{1} \alpha _{1} \widetilde{x}_t\right) ) + \alpha _{2} (1 -\exp \left( -b_{2} \lambda _{2} \alpha _{2} \widetilde{x}_t \right) \right) ) \\\approx & {} (1 - \widetilde{x}_t) \widetilde{x}_t \left( b_1 \lambda _{1} \alpha _{1}^2 + b_{2} \lambda _{2} \alpha _{2}^2 \right) \\= & {} A (1 - \widetilde{x}_t) \widetilde{x}_t, \qquad \qquad \qquad \qquad \qquad \text {where} \ A:= b_1 \lambda _{1} \alpha _{1}^2 + b_{2} \lambda _{2} \alpha _{2}^2. \end{aligned}$$

On Fig. 5 (right), we compare this approximation with the actual values for Scenario 1. Since the base contagion probabilities are typically very small, the exponent is also small and so the approximation $\exp (-x) \approx 1 - x$ is relatively good.

Next, we approximate the difference equation by the differential equation:

$$\begin{aligned} x'(t) = A (1-x(t))x(t) \end{aligned}$$

with the initial condition $x(0)=c_s$. We get that

$$\begin{aligned} \widetilde{x}_{t} \approx x(t):= \frac{1}{1 + \exp (-At) (1/c_s-1)}. \end{aligned}$$

As expected, this approximation is working well as depicted on Fig. 5 (left). By solving $x(t)=c_f$, it follows that

$$\begin{aligned} t_f \approx \frac{1}{A} \ln \left( \frac{c_f}{1-c_f} \cdot \frac{1-c_s}{c_s} \right) . \end{aligned}$$

List of symbols used in the paper

See Table 4.

Table 4 List of symbols used in the paper