HIV epidemic control—a model for optimal allocation of prevention and treatment resources

Abstract

With 33 million people living with human immunodeficiency virus (HIV) worldwide and 2.7 million new infections occurring annually, additional HIV prevention and treatment efforts are urgently needed. However, available resources for HIV control are limited and must be used efficiently to minimize the future spread of the epidemic. We develop a model to determine the appropriate resource allocation between expanded HIV prevention and treatment services. We create an epidemic model that incorporates multiple key populations with different transmission modes, as well as production functions that relate investment in prevention and treatment programs to changes in transmission and treatment rates. The goal is to allocate resources to minimize R ₀, the reproductive rate of infection. We first develop a single-population model and determine the optimal resource allocation between HIV prevention and treatment. We extend the analysis to multiple independent populations, with resource allocation among interventions and populations. We then include the effects of HIV transmission between key populations. We apply our model to examine HIV epidemic control in two different settings, Uganda and Russia. As part of these applications, we develop a novel approach for estimating empirical HIV program production functions. Our study provides insights into the important question of resource allocation for a country’s optimal response to its HIV epidemic and provides a practical approach for decision makers. Better decisions about allocating limited HIV resources can improve response to the epidemic and increase access to HIV prevention and treatment services for millions of people worldwide.

Appendix

Proof of full budget spending

The Lagrangian for the optimization problem in (7) is:

$$ L\left( {{x_1},{x_2},\lambda } \right)=\frac{{\gamma \beta \left( {{x_1}} \right)\nu \left( {{x_2}} \right)+\beta \left( {{x_1}} \right){d_2}}}{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]{d_2}}}+\lambda \left( {{x_1}+{x_2}-C} \right)-{\lambda_1}{x_1}-{\lambda_2}{x_2}. $$

(A1)

The following Karush-Kuhn-Tucker conditions must hold at optimality:

$$ \frac{dL }{{d{x_1}}}=\frac{{\beta \prime \left( {{x_1}} \right)\left[ {\gamma \nu \left( {{x_2}} \right)+{d_2}} \right]}}{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]{d_2}}}+\lambda -{\lambda_1}=0 $$

(A2)

$$ \frac{dL }{{d{x_2}}}=\frac{{\beta \left( {{x_1}} \right)\nu \prime \left( {{x_2}} \right)\left[ {\gamma {d_1}-{d_2}} \right]}}{{{{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]}}^2}{d_2}}}+\lambda -{\lambda_2}=0 $$

(A3)

$$ \lambda \left( {{x_1}+{x_2}-C} \right)=0 $$

(A4)

$$ {\lambda_1}{x_1}=0 $$

(A5)

$$ {\lambda_2}{x_2}=0 $$

(A6)

$$ {x_1}+{x_2}-C\leq 0 $$

(A7)

$$ \lambda, {\lambda_1},{\lambda_2}\geq 0 $$

(A8)

If β(x ₁) is monotonically decreasing, then β′ (x ₁) < 0 for all x ₁. We know that λ ₁ ≥ 0. By Eq. (A2), λ > 0, which implies that x ₁ + x ₂ − C = 0. Similarly, if ν (x ₂) is monotonically increasing, then ν′ (x ₂) > 0 for all x ₂, and $\gamma <\frac{{{d_2}}}{{{d_1}}}$ is equivalent to γd ₁ − d ₂ < 0. We know that λ ₂ ≥ 0. By Eq. (A3), λ > 0, which implies that x ₁ + x ₂ − C = 0. Hence, the optimal investment $\left( {x_1^{*},x_2^{*}} \right)$ will utilize the entire budget, C.

Proof of proposition 1

If ν (x ₂) is increasing and linear in x ₂, then equivalently, ν (x ₁) is decreasing and linear in x ₁. Thus, we can rewrite R ₀ in terms of one variable, x, assuming x ₂ = C − x ₁. ${R_0}(x)=\frac{{\gamma \beta (x)\nu (x)+\beta (x){d_2}}}{{\left[ {\nu (x)+{d_1}} \right]{d_2}}}$. Let f(x) = γβ (x) ν (x).

$$ \begin{array}{*{20}c} {f\prime (x)=\left. {\gamma \beta (x)\nu \prime (x)+\beta \prime (x)\nu (x)} \right]<0} \\ {f\prime \prime (x)=\left. {\gamma \beta (x)\nu \prime \prime (x)+2\beta \prime (x)\nu \prime (x)+\beta \prime \prime (x)\nu (x)} \right]>0} \\ \end{array} $$

because γ > 0, β(x) > 0, β′(x) < 0, β″ (x) > 0, ν(x) > 0, ν′(x) < 0, ν″(x) = 0. Thus, f(x) is decreasing and convex.

Let g (x) = f(x) + β(x)d ₂. Then

$$ \begin{array}{*{20}c} {g\prime (x)=f\prime (x)+{d_2}\beta \prime (x)<0} \hfill \\ {g\prime \prime (x)=f\prime \prime (x)+{d_2}\beta \prime \prime (x)>0} \hfill \\ \end{array} $$

because d ₂ > 0, f′ (x) < 0, f″ (x) > 0, β′ (x) < 0, β″ (x) > 0. Thus, g(x) is also decreasing and convex.

Let $h(y)=\frac{y}{{\nu (x)}}$, which is increasing in y. If g(x) is convex and thus quasi-convex, then the composite function h(g(x)) is quasi-convex. Therefore, R ₀ (x ₁,x ₂) is quasi-convex.

Proof of proposition 2

The proof follows from the Arrow-Enthoven sufficient conditions for optimality with quasi-convex objective functions [70].

Proof of proposition 3

We compare the marginal rate of substitution (MRS) for prevention $\left( {\frac{{d{R_0}}}{{d{x_1}}}} \right)$ and treatment $\left( {\frac{{d{R_0}}}{{d{x_2}}}} \right)$. As long as they are different, in order to find the optimal investment we will continue to shift the investment towards the program with the lower MRS, until we reach the budget limit C. At the optimal solution $\left( {x_1^{*},x_2^{*}} \right)$, either the two MRS are equal $\left( {\frac{{d{R_0}}}{{d{x_1}}}=\frac{{d{R_0}}}{{d{x_2}}}} \right)$ so we are indifferent between the two programs, or we have already invested all the available budget in the most favorable intervention.

• Case 1: Since we are minimizing, we invest only in prevention if

  $$ \begin{array}{*{20}c} {\frac{{d{R_0}}}{{d{x_1}}} < \frac{{d{R_0}}}{{d{x_2}}}} \hfill \\ {\frac{{\left[ {\frac{d}{{d{x_1}}}\beta \left( {{x_1}} \right)} \right]\left[ {\gamma \nu \left( {{x_2}} \right)+{d_2}} \right]}}{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]{d_2}}} < \frac{{\beta \left( {{x_1}} \right)\left[ {\frac{d}{{d{x_2}}}\nu \left( {{x_2}} \right)} \right]\left[ {\gamma {d_1}-{d_2}} \right]}}{{{{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]}}^2}{d_2}}}} \hfill \\ \end{array} $$

  (A9)

  We have $\frac{d}{{d{x_1}}}\beta \left( {{x_1}} \right)<0$ because β(x ₁) is decreasing, and $\frac{d}{{d{x_2}}}\nu \left( {{x_2}} \right)>0$ because ν(x ₂) is increasing. Also, γν (x ₂) + d ₂ > 0 and ν (x ₂) + d ₁ > 0 because ν (x ₂) > 0 and γ, d ₁, and d ₂ are positive constants. In order for (44) to always hold, γd ₁ − d ₂ > 0, or equivalently, $\gamma >\frac{{{d_2}}}{{{d_1}}}$. If γd ₁ − d ₂ < 0, or equivalently, $\gamma <\frac{{{d_2}}}{{{d_1}}}$, then the condition that must hold at x ₁ = C, x ₂ = 0 is

  $$ \frac{d}{{d{x_1}}}\beta \left( {{x_1}} \right)<\frac{{\beta \left( {{x_1}} \right)\frac{d}{{d{x_2}}}\nu \left( {{x_2}} \right)\left[ {\gamma {d_1}-{d_2}} \right]}}{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]\left[ {\gamma \nu \left( {{x_2}} \right)+{d_2}} \right]}}. $$

• Case 2: We invest only in treatment if $\gamma <\frac{{{d_2}}}{{{d_1}}}$ and

  $$ \begin{array}{*{20}c} {\frac{{d{R_0}}}{{d{x_1}}} > \frac{{d{R_0}}}{{d{x_2}}}} \hfill \\ {\frac{{\left[ {\frac{d}{{d{x_1}}}\beta \left( {{x_1}} \right)} \right]\left[ {\gamma \nu \left( {{x_2}} \right)+{d_2}} \right]}}{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]{d_2}}} > \frac{{\beta \left( {{x_1}} \right)\left[ {\frac{d}{{d{x_2}}}\nu \left( {{x_2}} \right)} \right]\left[ {\gamma {d_1}-{d_2}} \right]}}{{{{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]}}^2}{d_2}}}} \hfill \\ {\frac{d}{{d{x_1}}}\beta \left( {{x_1}} \right)>\frac{{\beta \left( {{x_1}} \right)\frac{d}{{d{x_2}}}\nu \left( {{x_2}} \right)\left[ {\gamma {d_1}-{d_2}} \right]}}{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]\left[ {\gamma \nu \left( {{x_2}} \right)+{d_2}} \right]}}} \hfill \\ \end{array} $$

  (A10)

  which must hold at x ₁ = 0, x ₂ = C.

Proof of proposition 4

We have shown that at optimal points the entire budget C is used, hence if $x_1^{*}={x^{*}}$ then $x_2^{*}=C-{x^{*}}$.

$$ \begin{array}{*{20}c} {\frac{{d{R_0}}}{{d{x_1}}}=\frac{{d{R_0}}}{{d{x_2}}}} \hfill \\ {\frac{{\left[ {\frac{d}{{d{x_1}}}\beta \left( {{x_1}} \right)} \right]\left[ {\gamma \nu \left( {{x_2}} \right)+{d_2}} \right]}}{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]{d_2}}}=\frac{{\beta \left( {{x_1}} \right)\left[ {\frac{d}{{d{x_2}}}\nu \left( {{x_2}} \right)} \right]\left[ {\gamma {d_1}-{d_2}} \right]}}{{{{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]}}^2}{d_2}}}} \hfill \\ {\frac{d}{{d{x_1}}}\beta \left( {{x_1}} \right)=\frac{{\beta \left( {{x_1}} \right)\frac{d}{{d{x_2}}}\nu \left( {{x_2}} \right)\left[ {\gamma {d_1}-{d_2}} \right]}}{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]\left[ {\gamma \nu \left( {{x_2}} \right)+{d_2}} \right]}}} \hfill \\ \end{array} $$

(A11)

For optimality, this condition must hold at $x_1^{*}={x^{*}},\;x_2^{*}=C-{x^{*}}$, which yields the condition described in the proposition statement.

Proof of corollary 1

Suppose γ = 0 (that is, treatment completely eliminates the chance of disease transmission), which most optimistically favors treatment over prevention. By the previous proposition, we should invest only in treatment if at (x ₁ = 0, x ₂ = C):

$$ \begin{array}{*{20}c} {\frac{{d{R_0}}}{{d{x_1}}} > \frac{{d{R_0}}}{{d{x_2}}}} \hfill \\ {\frac{{\left[ {\frac{d}{{d{x_1}}}\beta \left( {{x_1}} \right)} \right]\left[ {\gamma \nu \left( {{x_2}} \right)+{d_2}} \right]}}{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]{d_2}}} > \frac{{\beta \left( {{x_1}} \right)\left[ {\frac{d}{{d{x_2}}}\nu \left( {{x_2}} \right)} \right]\left[ {\gamma {d_1}-{d_2}} \right]}}{{{{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]}}^2}{d_2}}}} \hfill \\ {\frac{d}{{d{x_1}}}\beta \left( {{x_1}} \right)>-\frac{d}{{d{x_2}}}\nu \left( {{x_2}} \right)\frac{1}{{\nu \left( {{x_2}} \right)+{d_1}}}\beta \left( {{x_1}} \right)} \hfill \\ \end{array} $$

because γ = 0. If ν (x ₂) = kx ₂, then at the point (x ₁ = 0, x ₂ = C):

$$ \begin{array}{*{20}c} {\frac{d}{{d{x_1}}}\beta \left( {{x_1}} \right)>-k\frac{1}{{k{x_2}+{d_1}}}\beta (0)} \hfill \\ {\frac{d}{{d{x_1}}}\beta (0)>-\frac{1}{{C+\frac{{{d_1}}}{k}}}\beta (0)} \hfill \\ \end{array} $$

(A12)

Equation (A12) defines the condition when it is optimal to spend only on treatment. Therefore, it is optimal to spend some amount on prevention ($x_1^{*} > 0,x_2^{*}\geq 0$) if:

$$ \frac{d}{{d{x_1}}}\beta (0)<-\frac{{\beta (0)}}{{C+\frac{{{d_1}}}{k}}} $$

(A13)

Proof of corollary 2

By the envelope theorem, if the optimal investment at γ = 0 includes some prevention, then as treatment becomes less effective at reducing infectivity ($\widehat{\gamma} > 0$), the optimal solution must not increase investment in treatment. Mathematically, as γ increases, $\frac{{d{R_0}}}{{d{x_2}}}=\frac{{v\prime \left( {{x_2}} \right)\left[ {\gamma {d_1}-{d_2}} \right]}}{{{{{\left[ {\nu \left( {{x_2}} \right)+{d_1}} \right]}}^2}{d_2}}}$ increases, so R ₀ will increase as x ₂ increases. Therefore, it cannot be optimal to have $\widehat{x}_2^{*} > x_2^{*}$.

Illustrative numerical examples of optimal allocation for different production functions and a single population

Figure 3 shows numerical examples of R ₀ for different production functions. These examples assume d ₁ = 1/10 (average life expectancy of an infected individual is 10 years), d ₂ = 1/25 (average life expectancy of a treated individual is 25 years). The budget, C = 1, is entirely spent and the optimal allocation between prevention ($x_1^{*}$) and treatment ($x_2^{*}$) is given. In Fig. 3a, $\gamma >\frac{{{d_2}}}{{{d_1}}}$, and the reduction in infectivity due to treatment is too low to invest any amount in treatment ($x_1^{*}=1,x_2^{*}=0$). In Fig. 3b, $\gamma <\frac{{{d_2}}}{{{d_1}}}$ and $\frac{d}{{d{x_1}}}\beta (0)=-\infty$, and it is optimal to invest some amount in prevention ($x_1^{*}=0.6821,x_2^{*}=0.3179$). In Fig. 3c, $\gamma <\frac{{{d_2}}}{{{d_1}}}$ and $\frac{d}{{d{x_1}}}\beta (0)=-\infty$, and it is optimal to allocate more to prevention because the treatment production function ν (x ₂) has diminishing returns ($x_1^{*}=0.8488,x_2^{*}=0.1512$). In Fig. 3d, $\gamma <\frac{{{d_2}}}{{{d_1}}}$ and it is easy to verify that $\frac{d}{{d{x_1}}}\beta (0)>\frac{{\frac{d}{{d{x_2}}}\nu (C)\left[ {\gamma {d_1}-{d_2}} \right]}}{{\left[ {\nu (C)+{d_1}} \right]\left[ {\gamma \nu (C)+{d_2}} \right]}}$. Thus, it is optimal to invest only in treatment ($x_1^{*}=0,x_2^{*}=1$).

Proof of proposition 5

The Lagrangian for the redefined problem is:

$$ \begin{array}{*{20}c} {L\left( {{x_i},{C_i},{x_j},{C_j},\lambda } \right)=\frac{{{\alpha_i}{\gamma_i}{\beta_i}\left( {{C_i}-{x_i}} \right){\nu_i}\left( {{x_i}} \right)+{\beta_i}\left( {{C_i}-{x_i}} \right){d_{i2 }}}}{{\left[ {{\nu_i}\left( {{x_i}} \right)+{d_{i1 }}} \right]{d_{i2 }}}}} \hfill \\ {+\frac{{{\alpha_j}{\gamma_j}{\beta_j}\left( {{C_j}-{x_j}} \right){\nu_j}\left( {{x_j}} \right)+{\beta_j}\left( {{C_j}-{x_j}} \right){d_{j2 }}}}{{\left[ {{\nu_j}\left( {{x_j}} \right)+{d_{j1 }}} \right]{d_{j2 }}}}} \hfill \\ {+\lambda \left( {{C_i}+{C_j}-C} \right)+{\lambda_i}\left( {{x_i}-{C_i}} \right)+{\lambda_j}\left( {{x_j}-{C_j}} \right).} \hfill \\ \end{array} $$

(A14)

The following Karush-Kuhn-Tucker conditions must hold at optimality:

$$ \frac{dL }{{d{x_i}}}=\frac{{-{\alpha_i}\beta_i^{\prime}\left( {{C_i}-{x_i}} \right)\left[ {{\gamma_i}{\nu_i}\left( {{x_i}} \right)+{d_{i2 }}} \right]}}{{\left[ {{\nu_i}\left( {{x_i}} \right)+{d_{i1 }}} \right]{d_{i2 }}}}+{\beta_i}\left( {{C_i}-{x_i}} \right)\frac{{{\alpha_i}\nu_i^{\prime}\left( {{x_i}} \right)\left[ {{\gamma_i}{d_{i1 }}-{d_{i2 }}} \right]}}{{{{{\left[ {{\nu_i}\left( {{x_i}} \right)+{d_{i1 }}} \right]}}^2}{d_{i2 }}}}+{\lambda_i}=0 $$

(A15)

$$ \frac{dL }{{d{x_j}}}=\frac{{-{\alpha_j}\beta_j^{\prime}\left( {{C_j}-{x_j}} \right)\left[ {{\gamma_j}{\nu_j}\left( {{x_j}} \right)+{d_{j2 }}} \right]}}{{\left[ {{\nu_j}\left( {{x_j}} \right)+{d_{j1 }}} \right]{d_{j2 }}}}+{\beta_j}\left( {{C_j}-{x_j}} \right)\frac{{{\alpha_j}\nu_j^{\prime}\left( {{x_j}} \right)\left[ {{\gamma_j}{d_{j1 }}-{d_{j2 }}} \right]}}{{{{{\left[ {{\nu_j}\left( {{x_j}} \right)+{d_{j1 }}} \right]}}^2}{d_{j2 }}}}+{\lambda_j}=0 $$

(A16)

$$ \frac{dL }{{d{C_i}}}=\frac{{{\alpha_i}\beta_i^{\prime}\left( {{C_i}-{x_i}} \right)\left[ {{\gamma_i}{\nu_i}\left( {{x_i}} \right)+{d_{12 }}} \right]}}{{\left[ {{\nu_i}\left( {{x_i}} \right)+{d_{i1 }}} \right]{d_{12 }}}}+\lambda -{\lambda_i}=0 $$

(A17)

$$ \frac{dL }{{d{C_j}}}=\frac{{{\alpha_j}\beta_j^{\prime}\left( {{C_j}-{x_j}} \right)\left[ {{\gamma_j}{\nu_j}\left( {{x_j}} \right)+{d_{j2 }}} \right]}}{{\left[ {{\nu_j}\left( {{x_j}} \right)+{d_{j1 }}} \right]{d_{j2 }}}}+\lambda -{\lambda_j}=0 $$

(A18)

$$ \lambda \left( {{C_i}+{C_j}-C} \right)=0 $$

(A19)

$$ {\lambda_i}\left( {{x_i}-{C_i}} \right)=0 $$

(A20)

$$ {\lambda_j}\left( {{x_j}-{C_j}} \right)=0 $$

(A21)

$$ {C_i}+{C_j}-C\leq 0 $$

(A22)

$$ {x_i}-{C_i}\leq 0 $$

(A23)

$$ {x_j}-{C_j}\leq 0 $$

(A24)

$$ \lambda, {\lambda_i},{\lambda_j}\geq 0 $$

(A25)

• Condition 1: If β _i (C _i  − x _i ) is monotonically decreasing, then $\beta_i^{\prime}\left( {{C_i}-{x_i}} \right)<0$ for all x _i . By Eq. (A17), λ − λ _i  > 0, thus λ > λ _i . But λ _i  ≥ 0, hence λ > 0, which from Eq. (A19) implies C _i  + C _j  − C = 0. Similarly, if β _j (C _j  − x _j ) is monotonically decreasing, then $\beta_j^{\prime}\left( {{C_j}-{x_j}} \right)<0$ for all x _j . By (A18), λ − λ _j  > 0, which implies that C _i  + C _j  − C = 0. If ν (x ₂) is monotonically increasing, then ν′ (x ₂) > 0 for all x ₂, and $\gamma <\frac{{{d_2}}}{{{d_1}}}$ is equivalent to γd ₁ − d ₂ < 0. By (A3), λ > 0, which implies that x ₁ + x ₂ − C = 0. Hence, the optimal investment $\left( {x_1^{*},x_2^{*}} \right)$ will utilize the entire budget, C.

• Condition 2: From (A15) and (A17), $\lambda =-{\beta_i}\left( {{C_i}-{x_i}} \right)\frac{{{\alpha_i}\nu_i^{\prime}\left( {{x_i}} \right)\left( {{\gamma_i}{d_{i1 }}-{d_{i2 }}} \right)}}{{{{{\left[ {{\nu_i}\left( {{x_i}} \right)+{d_{i1 }}} \right]}}^2}{d_{i2 }}}}$. If ν _i (x _i ) is monotonically increasing and ${\gamma_i} < \frac{{{d_{i2 }}}}{{d{i_1}}}$, then $\nu_i^{\prime}\left( {{x_i}} \right)>0$ and γ _i d _i1 − d _i2 < 0, hence λ > 0, which from (A19) implies C _i  + C _j  − C = 0. Similarly, from (A16) and (A18), $\lambda =-{\beta_j}\left( {{C_j}-{x_j}} \right)\frac{{{\alpha_j}\nu_j^{\prime}\left( {{x_j}} \right)\left( {{\gamma_j}{d_{j1 }}-{d_{j2 }}} \right)}}{{{{{\left[ {{\nu_j}\left( {{x_j}} \right)+{d_{j1 }}} \right]}}^2}{d_{j2 }}}}$. If ν _j (x _j ) is monotonically increasing and ${\gamma_j} < \frac{{{d_{j2 }}}}{{{d_{j1 }}}}$, then $\nu_j^{\prime}\left( {{x_j}} \right)>0$ and γ _i d _i1 − d _i2 < 0, then λ > 0, which from (A19) implies C _i  + C _j  − C = 0.

Derivation of R ₀ for one population

For the given system, (1) – (4), we have:

$$ \begin{array}{*{20}c} {z=\left[ {I,T,S} \right]} \\ {A(z)=\left[ {\beta \left( {I+\gamma T} \right)\frac{S}{N},0,0} \right]} \\ {\Im (z)=\left[ {\nu I+{d_1}I,-\nu I+{d_2}T,-bN+\beta \left( {I+\gamma T} \right)\frac{S}{N}+bS} \right]} \\ \end{array} $$

The infected compartments are I and T, so the number of infected compartments is m = 2. Defining F and V as above, we obtain the following expression for the next generation matrix FV ⁻¹:

$$ \begin{array}{*{20}c} {F=\left[ {\begin{array}{*{20}c} \beta \hfill & {\beta \gamma } \hfill \\ 0 \hfill & 0 \hfill \\ \end{array}} \right],V=\left[ {\begin{array}{*{20}c} {\nu +{d_1}} \hfill & 0 \hfill \\ {-\nu } \hfill & {{d_2}} \hfill \\ \end{array}} \right]} \hfill \\ {F{V^{-1 }}=\left[ {\begin{array}{*{20}c} {\frac{{\gamma \beta \nu +\beta {d_2}}}{{\left( {\nu +{d_1}} \right){d_2}}}} & {\frac{{\gamma \beta }}{{{d_2}}}} \\ 0 & 0 \\ \end{array}} \right]} \hfill \\ \end{array} $$

The eigenvalues are ${R_0}=\frac{{\gamma \beta \nu +\beta {d_2}}}{{\left( {\nu +{d_1}} \right){d_2}}}$ and 0.

Derivation of R ₀ for two interacting populations case

We rewrite the differential equations for the two interacting populations, (16)–(23), to include the parameter r = N _i /N _j :

$$ \frac{{d{S_i}}}{dt }={b_i}{N_i}-\left( {{\beta_{ii }}{I_i}+{\gamma_i}{\beta_{ii }}{T_i}+r{\beta_{ji }}{I_j}+r{\gamma_j}{\beta_{ji }}{T_j}} \right){{{{S_i}}} \left/ {{{N_i}-{b_i}{S_i}}} \right.} $$

(A26)

$$ \frac{{d{I_i}}}{dt }=\left( {{\beta_{ii }}{I_i}+{\gamma_i}{\beta_{ii }}{T_i}+r{\beta_{ji }}{I_j}+r{\gamma_j}{\beta_{ji }}{T_j}} \right){{{{S_i}}} \left/ {{{N_i}-{\nu_i}{I_i}-{d_{i1 }}{I_i}}} \right.} $$

(A27)

$$ \frac{{d{T_i}}}{dt }={\nu_i}{I_i}-{d_{i2 }}{T_i} $$

(A28)

$$ {N_i}={S_i}+{I_i}+{T_i} $$

(A29)

$$ \frac{{d{S_j}}}{dt }={b_j}{N_j}-\left( {{\beta_{jj }}{I_j}+{\gamma_j}{\beta_{jj }}{T_j}+{1 \left/ {r* } \right.}{\beta_{ij }}{I_i}+{1 \left/ {r* } \right.}{\gamma_i}{\beta_{ij }}{T_i}} \right){{{{S_j}}} \left/ {{{N_j}-{b_j}{S_j}}} \right.} $$

(A30)

$$ \frac{{d{S_j}}}{dt }=\left( {{\beta_{jj }}{I_j}+{\gamma_j}{\beta_{jj }}{I_i}+{1 \left/ {r* } \right.}{\beta_{ij }}{I_j}+{1 \left/ {{r*{\gamma_i}{\beta_{ij }}{T_i}}} \right.}} \right){{{{S_j}}} \left/ {{{N_j}-{\nu_j}{I_j}-{d_{j1 }}{I_j}}} \right.} $$

(A31)

$$ \frac{{d{T_j}}}{dt }={\nu_j}{I_j}-{d_{j2 }}{T_j} $$

(A32)

$$ {N_j}={S_j}+{I_j}+{T_j} $$

(A33)

Let z be the vector representing the number of individuals in each compartment, A _k (z) the rate of appearance of new infections in compartment z _k , and ℑ _k (z) the rate of change in compartment z _k by all other means. For the given system:

$$ \begin{array}{*{20}c} {z=\left[ {{I_i},{T_i},{I_j},{T_j},{S_i},{S_j}} \right]} \\ {A(z)=\left[ {\left( {{\beta_{ii }}{I_i}+{\gamma_i}{\beta_{ii }}{T_i}+r{\beta_{ji }}{I_j}+r{\gamma_j}{\beta_{ji }}{T_j}} \right){{{{S_i}}} \left/ {{{N_i},0,}} \right.}} \right.} \\ \end{array} $$

(A34)

$$ \left. {\left( {{\beta_{jj }}{I_j}+{\gamma_j}{\beta_{jj }}{T_j}+{1 \left/ {r* } \right.}{\beta_{ij }}{I_i}+{1 \left/ {r* } \right.}{\gamma_i}{\beta_{ij }}{T_i}} \right){{{{S_j}}} \left/ {{{N_j},0,0,0}} \right.}} \right] $$

(A35)

$$ \Im (z)=\left[ {{\nu_i}{I_i}+{d_{i1 }}{I_i},-{\nu_i}{I_i}+{d_{i2 }}{T_i},{\nu_j}{I_j}} \right.+{d_{j1 }}{I_j},{\nu_j}{I_j}+{d_{j2 }}{T_j}, $$

$$ -{b_i}{N_i}+\left( {{\beta_{ii }}{I_i}+{\gamma_i}{\beta_{ii }}{T_i}+r{\beta_{ji }}{I_j}+r{\gamma_j}{\beta_{ji }}{T_j}} \right){{{{S_i}}} \left/ {{{N_i}+{b_i}{S_i},}} \right.} $$

$$ \left. {-{b_j}{N_j}+\left( {{\beta_{jj }}{I_j}+{\gamma_j}{\beta_{jj }}{T_j}+{1 \left/ {r* } \right.}{\beta_{ij }}{I_i}+{1 \left/ {{r*{\gamma_i}{\beta_{ij }}{T_i}}} \right.}} \right){{{{S_j}}} \left/ {{{N_j}+{b_j}{S_j}}} \right.}} \right] $$

(A36)

The infected compartments are I _i , T _i , I _j , and T _j , so the number of infected compartments is m = 4. Defining F and V as below, we obtain the following expression for the next generation matrix FV ⁻¹:

$$ \begin{array}{*{20}c} {F=\left[ {\frac{{\partial {A_k}}}{{\partial {z_l}}}\left( {{z_0}} \right)} \right],V=\left[ {\frac{{\partial {\Im_k}}}{{\partial {z_l}}}\left( {{z_0}} \right)} \right],\mathrm{with}1\leq {\it k}, {\it l} \leq {\it m}} \\ {F=\left[ {\begin{array}{*{20}c} {{\beta_i}} & {{\gamma_i}{\beta_{ii }}} & {r{\beta_{ji }}} & {r{\gamma_j}{\beta_{ji }}} \\ 0 & 0 & 0 & 0 \\ {{\beta_j}} & {{\gamma_j}{\beta_{jj }}} & {{1 \left/ {{r*{\beta_{ij }}}} \right.}} & {{1 \left/ {{r*{\gamma_i}{\beta_{ij }}}} \right.}} \\ 0 & 0 & 0 & 0 \\ \end{array}} \right],V=\left[ {\begin{array}{*{20}c} {{\nu_i}+{d_{i1 }}} & 0 & 0 & 0 \\ {-{\nu_i}} & {{d_{12 }}} & 0 & 0 \\ 0 & 0 & {{\nu_j}+{d_{j1 }}} & 0 \\ 0 & 0 & {-{\nu_j}} & {{d_{j2 }}} \\ \end{array}} \right]} \\ \end{array} $$

(A37)

$$ F{V^{-1 }}=\left[ {\begin{array}{*{20}c} {\frac{{{\beta_{ii }}\left( {{d_{i2 }}+{\gamma_i}{\nu_i}} \right)}}{{{d_{i2 }}\left( {{\nu_i}+{d_{i1 }}} \right)}}} & {\frac{{{\gamma_i}{\beta_{ii }}}}{{{d_{i2 }}}}} & {\frac{{{\beta_{ji }}r\left( {{d_{j2 }}+{\gamma_j}\nu } \right)}}{{{d_{j2 }}\left( {{\nu_j}+{d_{j1 }}} \right)}}} & {\frac{{r{\gamma_j}{\beta_{ji }}}}{{{d_{j2 }}}}} \\ 0 & 0 & 0 & 0 \\ {\frac{{{\beta_{jj }}\left( {{d_{i2 }}+{\gamma_j}{\nu_i}} \right)}}{{{d_{i2 }}\left( {{\nu_i}+{d_{i1 }}} \right)}}} & {\frac{{{\gamma_j}{\beta_{jj }}}}{{{d_{i2 }}}}} & {\frac{{{\beta_{ij }}\left( {{d_{j2 }}+{\gamma_i}{\nu_j}} \right)}}{{{d_{j2 }}r\left( {{\nu_j}+{d_{j1 }}} \right)}}} & {\frac{{{\beta_{ij }}{\gamma_i}}}{{r{d_{j2 }}}}} \\ 0 & 0 & 0 & 0 \\ \end{array}} \right] $$

(A38)

The (k, l) entry of FV ⁻¹ gives the expected number of new infections caused in compartment k by an infected individual in compartment l. R ₀ is defined as the spectral radius of FV ⁻¹ and is as given in Eq. (25).