Evaluating the Efficacy of Tuberculosis Management Strategies in Nigeria: A Mathematical Modelling Approach

TB is a global illness that mainly affects the lungs and is caused by Mycobacterium, which can be spread when an individual with the illness coughs, sneezes, or sings, passing through the air to susceptible individuals. TB is one of the most contagious diseases, claiming more lives every day (W​o​r​l​d​ ​H​e​a​l​t​h​ ​O​r​g​a​n​i​z​a​t​i​o​n​,​ ​2​0​1​9). In the exploration of TB management strategies, S​u​l​a​y​m​a​n​ ​&​ ​A​b​d​u​l​l​a​h​ ​(​2​0​2​2​) identified a significant distinction between the approaches to health education and the lack thereof, which divided infectious individuals into two classes: infected individuals treated at home versus those treated in hospital settings (H​u​o​ ​&​ ​Z​o​u​,​ ​2​0​1​6). In every population, at least one-third has a TB disease, which yields millions of deaths and new cases annually (C​a​s​t​i​l​l​o​-​C​h​a​v​e​z​ ​&​ ​S​o​n​g​,​ ​2​0​0​4; S​u​d​r​e​ ​e​t​ ​a​l​.​,​ ​1​9​9​2; U​l​l​a​h​ ​e​t​ ​a​l​.​,​ ​2​0​1​9). In order to identify ways to control diseases in the population, several studies have been conducted in mathematical modelling (A​t​a​n​g​a​n​a​ ​&​ ​D​o​u​n​g​m​o​ ​G​o​u​f​o​,​ ​2​0​1​4; G​o​u​f​o​ ​e​t​ ​a​l​.​,​ ​2​0​1​6; G​o​u​f​o​ ​e​t​ ​a​l​.​,​ ​2​0​1​7; L​e​o​n​ ​e​t​ ​a​l​.​,​ ​2​0​1​7; N​d​o​n​d​o​ ​e​t​ ​a​l​.​,​ ​2​0​1​6; T​c​h​e​p​m​o​ ​D​j​o​m​e​g​n​i​ ​e​t​ ​a​l​.​,​ ​2​0​1​8), which is developed and applied to the spread of diseases in order to understand the epidemiological transition phenomenon (K​a​s​e​r​e​k​a​ ​e​t​ ​a​l​.​,​ ​2​0​1​4; K​a​s​e​r​e​k​a​ ​e​t​ ​a​l​.​,​ ​2​0​1​8) at different locations.

Every country has different scenarios, but there are major surprises when it comes to Nigerians in terms of diseases. Some Nigerians don’t go to the hospital before they get cured of their illness, because maybe they have a strong immune system, lack money or don’t believe in hospital treatment. In fact, some people go to the clinic/chemist that they can afford. Chemists provide a healthcare centre where individuals get treated at a low price, but the healthcare officer at the place is not professional. Some formal doctors used to call them quack doctors (which means unprofessional) because they did not want formal education. These people are healthcare providers with professional training and are acquainted with their boss. Some individuals go to either a secondary or tertiary hospital for treatment. This study focuses on people with TB who try medication or treatment, as well as those who do not take medication or treatment or go to the hospital. How can they get cured? Some research (S​a​i​t​o​ ​&​ ​N​a​g​a​s​a​k​i​,​ ​2​0​0​8) shows that individuals with high-level melanin hormone can get cured because the hormone fights the virus to some extent. The model proposed in this study presents the transmission dynamics of TB among individuals, examining scenarios both in the presence and absence of treatment interventions. The motive behind this is because TB is among the most communicable diseases in the world. The model reveals that some individuals recover without treatment at each stage, while others become chronically infected, rendering them incurable. They have to return home for treatment, and some people got vaccinated to avoid being infected. After the treatment, all the individuals move to the susceptible group so that the process can start again and again. This work designed a model for TB, incorporating both treated and untreated scenarios. Nigeria was used as a pilot. The model was analysed using a compartmental modelling approach to study the transmission dynamics of TB at different stages of infection.

Figure 1 presents a four-compartment model. In the model, the S population is considered as susceptible people who have not contacted TB with a birth rate of µ(1-p)N, where (1-p) is the unvaccinated individual rate with a death rate of µ. The S group can contract TB with a rate of to the latent (L) stage, in which the L individual can return to S with a rate of γ without going to the hospital or taking drugs. It can be noted that the L individual can become infectious (I) with the contact rate of F_LI, and the transmission rate of δ can be revised to α with a death rate of µ. The L individual can also be treated (T) without becoming I if he can be detected early with a rate of p. An I individual can also directly become T at a rate of . The T individual can also become I at a contact rate of ϵ and a transmission rate of F_TI if he is in the chronic stage. In this case, he cannot be T because the drugs do not work for him any more, or the disease is not cured. During hospitalization or treatment, the individual can also become L (not able to get I due to the drugs taken in the hospital), with a death rate of µ. Before leaving the hospital, it is assumed that all the T individuals are vaccinated at the rate of µP. Then all the survival treatment moves to S at a rate of q. After the vaccine has expired, the individual can be I again.

The number of S individuals increases and decreases as a result of natural death and latent infection at rates of μ(1-p), r₃, q, γ with respect to the total population (N), I, T and μ, β, respectively. Therefore, the change rates of susceptible populations are given as follows:

The S populations are exposed and get I at the rate of β with a transmission rate of F_SL and decreased as a result of death due to L at a rate of µ; others also contribute to the population from T and I.

Similarly, the I population is cured after being T at a rate of r₂ and dead at a rate of μ.

Finally, the T population is generated via the recovery of I individuals in the L stage and vaccinated individuals with a rate of µ and a vaccination rate of P. It can be seen from the equation that some individuals also return to both I (due to no response to treatment) and L (due to being partially cured to the extent that he/she cannot infect the population).

Therefore, based on the above descriptions and assumptions, the TB model, incorporating both treated and untreated scenarios, culminates in the derivation of a set of non-linear differential equations.

Equilibrium changes in $\frac{d S}{d t}=\frac{d L}{d t}=\frac{d I}{d t}=\frac{d T}{d t}=0$ yields:

At the initial stage, it is assumed that nothing is recorded, i.e., L = 0, I = 0, T = 0, and then it yields:

$\mu$S=$\mu$ (1-p) N+r₃*0+q*0+$\gamma$*0-$\beta$SF_SL $\frac{0}{N}$

$S=(1-P) N$

The TB-free equilibrium is E=(S, L, I, T)=((1-P)N, 0, 0, 0).

The eigenvalue is $A-\lambda I=0$, which means that the function $\mathrm{F}(t)=0$ and $A=\mathrm{DF}(t)$.

$A-\lambda I=\left[\begin{array}{cccc}-\beta F_{S L} \frac{I}{N}-\mu-\lambda & \gamma & r_3+\beta S F_{S L} \frac{1}{N} & q \\ \beta F_{S L} \frac{I}{N} & -\delta F_{L I}-\gamma-p-\mu-\lambda & \beta S F_{S L} \frac{1}{N}+C^* F_{T L} T \frac{1}{N}+\alpha & C^* F_{T L} \frac{I}{N} \\ 0 & \delta F_{L I} & -\alpha-r_3-r_2-\mu-\lambda & \epsilon F_{T I} \\ 0 & p & r_2-C^* F_{T L} T \frac{1}{N} & -q-C^* F_{T L} \frac{I}{N}-\epsilon F_{T I}-\mu-\lambda\end{array}\right]=0$

The characteristic equation of the function A−λI=0 is as follows:

$\begin{gathered}-\beta F_{S L} \frac{I}{N}-\mu-\lambda\left[\begin{array}{ccc}-\delta F_{L I}-\gamma-p-\mu-\lambda & \beta S F_{S L} \frac{1}{N}+C^* F_{T L} T \frac{1}{N}+\alpha & C^* F_{T L} \frac{I}{N} \\ \delta F_{L I} & -\alpha-r_3-r_2-\mu-\lambda & \epsilon F_{T I} \\ p & r_2-C F_{T L} T \frac{1}{N} & -q-C^* F_{T L} \frac{I}{N}-\epsilon F_{T I}-\mu-\lambda\end{array}\right] \\ -\gamma\left[\begin{array}{ccc}\beta F_{S L} \frac{I}{N} & \beta S F_{S L} \frac{1}{N}+C^* F_{T L} T \frac{1}{N}+\alpha & C^* F_{T L} \frac{I}{N} \\ 0 & -\alpha-r_3-r_2-\mu-\lambda & \epsilon F_{T I} \\ 0 & r_2-C^* F_{T L} T \frac{1}{N} & -q-C^* F_{T L} \frac{I}{N}-\epsilon F_{T I}-\mu-\lambda\end{array}\right] \\ r_3+\beta S F_{S L} \frac{1}{N}\left[\begin{array}{ccc}\beta F_{S L} \frac{I}{N} & -\delta F_{L I}-\gamma-p-\mu-\lambda & C F_{T L} \frac{I}{N} \\ 0 & \delta F_{L I} & \epsilon F_{T I} \\ 0 & p & -q-C^* F_{T L} \frac{I}{N}-\epsilon F_{T I}-\mu-\lambda\end{array}\right]+ \\ -q\left[\begin{array}{ccc}\beta F_{S L} \frac{I}{N} & -\delta F_{L I}-\gamma-p-\mu-\lambda & \beta S F_{S L} \frac{1}{N}+C^* F_{T L} T \frac{1}{N}+\alpha \\ 0 & \delta F_{L I} & -\alpha-r_3-r_2-\mu-\lambda \\ 0 & p & r_2-C^* F_{T L} T \frac{1}{N}\end{array}\right]=0\end{gathered}$

When the TB-free is substituted with A−λI, it yields:

$A-\lambda I=\left[\begin{array}{cccc}-\mu-\lambda & \gamma & r_3 & q \\ 0 & -\delta F_{L I}-\gamma-p-\mu-\lambda & \alpha & 0 \\ 0 & \delta F_{L I} & -\alpha-r_3-r_2-\mu-\lambda & \epsilon F_{T I} \\ 0 & p & r_2 & -q-\epsilon F_{T I}-\mu-\lambda\end{array}\right]=0$

The characteristic equation of the function A−λI=0 is as follows:

$\begin{array}{r}-\mu-\lambda\left[\begin{array}{ccc}-\delta F_{L I}-\gamma-p-\mu-\lambda & \alpha & 0 \\ \delta F_{L I} & -\alpha-r_3-r_2-\mu-\lambda & \epsilon F_{T I} \\ p & r_2 & -q-\epsilon F_{T I}-\mu-\lambda\end{array}\right]-\gamma\left[\begin{array}{ccc}0 & \alpha & 0 \\ 0 & -\alpha-r_3-r_2-\mu-\lambda & \epsilon F_{T I} \\ 0 & r_2 & -q-\epsilon F_{T I}-\mu-\lambda\end{array}\right] \\ +r_3\left[\begin{array}{ccc}0 & -\delta F_{L I}-\gamma-p-\mu-\lambda & 0 \\ 0 & \delta F_{T I} & \epsilon F_{T I} \\ 0 & p & -q-\epsilon F_{T I}-\mu-\lambda\end{array}\right]-q\left[\begin{array}{ccc}0 & -\delta F_{L I}-\gamma-p-\mu-\lambda & \alpha \\ 0 & \delta F_{L I} & -\alpha-r_3-r_2-\mu-\lambda \\ 0 & p & r_2\end{array}\right]=0\end{array}$

This also yields:

$-\mu-\lambda\left[\left(-\delta F_{L I}-\gamma-p-\mu-\lambda\right)\left[\left(-\alpha-r_3-r_2-\mu-\lambda\right)\left(-q-\epsilon F_{T I}-\mu-\lambda\right)-r_2 \epsilon F_{T I}\right]-\alpha\left[\delta F_{L I}\left(-q-\epsilon F_{T I}-\mu-\lambda\right)-p \epsilon F_{I I}\right]\right]=0$

Utilizing the parameter values in Table 1 yields the following results:

λ = −0.08, −0.291, −0.2903, −1.2874

To find the reproductive number, only two stages are considered, i.e., the L and I stages. The equations are as follows:

Eq. (12) at equilibrium yields:

$0=\beta F_{S L} S \frac{I}{N}+C^* F_{T L} T \frac{I}{N}+\alpha I-\delta F_L L-\gamma L-p L-\mu L$

This is equal to:

Eq. (13) at equilibrium yields:

$0=\delta F_{L I} L+\epsilon F_{T I} T-\alpha I-r_3 I-r_2 I-\mu I$

This is equal to:

$\delta F_{L I} L=I\left(\alpha+r_3+r_2+\mu\right)-\epsilon F_{T I} T$

To find the reproductive number, the following is equated:

$\mathbf{V}=\left[\begin{array}{cc}\frac{\partial\left[L\left(\delta F_{L I}+\gamma+p+\mu\right)-C^* F_{T L} T \frac{I}{N}-\alpha I\right]}{\partial L} & \frac{\partial\left[L\left(\delta F_{L I}+\gamma+p+\mu\right)-C^* F_{T L} T \frac{I}{N}-\alpha I\right]}{\partial I} \\ \frac{\partial\left[\left(\alpha+r_3+r_2+\mu\right)-\epsilon F_{T I} T\right]}{\partial L} & \frac{\partial\left[I\left(\alpha+r_3+r_2+\mu\right)-\epsilon F_{T I T} T\right]}{\partial I}\end{array}\right]=\left[\begin{array}{cc}\delta F_{L I}+\gamma+p+\mu & -C^* F_{T L} T \frac{1}{N}-\alpha \\ 0 & \alpha+r_3+r_2+\mu\end{array}\right]$

$\mathbf{V}^{-1}=\frac{1}{\left(\delta F_{L I}+\gamma+p+\mu\right)\left(\alpha+r_3+r_2+\mu\right)}\left[\begin{array}{cc}\alpha+r_3+r_2+\mu & C^* F_{T L} T \frac{1}{N}+\alpha \\ 0 & \delta F_{L I}+\gamma+p+\mu\end{array}\right]$

Assuming that there is no interaction at the beginning, the equation can be expressed as follows, according to Georges (K​a​m​a​n​g​a​,​ ​2​0​2​0):

$\mathbf{F V}^{-1}=\frac{1}{\left(\delta F_{L I}+\gamma+p+\mu\right)\left(\alpha+r_3+r_2+\mu\right)}\left[\begin{array}{cc}\alpha+r_3+r_2+\mu & C^* F_{T L} T \frac{1}{N}+\alpha \\ 0 & \delta F_{L I}+\gamma+p+\mu\end{array}\right]\left[\begin{array}{cc}0 & \beta F_{S L} \frac{1}{N} \\ \delta F_{L I} & 0\end{array}\right]$

$\mathbf{F V}^{-1}=\frac{1}{\left(\delta F_{L I}+\gamma+p+\mu\right)\left(\alpha+r_3+r_2+\mu\right)}\left[\begin{array}{cc}\left(C^* F_{T L} T \frac{1}{N}+\alpha\right) \delta F_{L I} & \left(\alpha+r_3+r_2+\mu\right) \beta F_{S L} \frac{1}{N} \\ \left(\delta F_{L I}+\gamma+p+\mu\right) \delta F_{L I} & 0\end{array}\right]$

$\mathbf{F V}^{-1}=\left[\begin{array}{cc}\frac{\left(C^* F_{T L} T \frac{1}{N}+\alpha\right) \delta F_{L I}}{\left(\delta F_{L I}+\gamma+p+\mu\right)\left(\alpha+r_3+r_2+\mu\right)} & \frac{\left(\alpha+r_3+r_2+\mu\right) \beta F_{S L} \frac{1}{N}}{\left(\delta F_{L I}+\gamma+p+\mu\right)\left(\alpha+r_3+r_2+\mu\right)} \\ \frac{\left(\delta F_{L I}+\gamma+p+\mu\right) \delta F_{L I}}{\left(\delta F_{L I}+\gamma+p+\mu\right)\left(\alpha+r_3+r_2+\mu\right)} & 0\end{array}\right]=\left[\begin{array}{cc}A & B \\ C & D\end{array}\right]$

As delineated by Georges (K​a​m​a​n​g​a​,​ ​2​0​2​0), R_O can be calculated as follows:

$R_o=\frac{-B \pm \sqrt{B^2-4 A C}}{2 A}$

$R_O$ is the biggest eigenvalue of matrix $F V^{-1}$, calculated according to:

$R_o=\frac{-\frac{\left(\alpha+r_3+r_2+\mu\right) \beta F_{S L} \frac{1}{N}}{\left(\delta F_{L I}+\gamma+p+\mu\right)\left(\alpha+r_3+r_2+\mu\right)} \pm \sqrt{\left.\frac{\left(\alpha+r_3+r_2+\mu\right) \beta F_{S L} \frac{1}{N}}{\left(\delta F_{L I}+\gamma+p+\mu\right)\left(\alpha+r_3+r_2+\mu\right)}\right]^2-4 \frac{\left(C^* F_{T L} T \frac{1}{N}+\alpha\right) \delta F_{L I}}{\left(\delta F_{L I}+\gamma+p+\mu\right)\left(\alpha+r_3+r_2+\mu\right)} \frac{\left(\delta F_{L I}+\gamma+p+\mu\right) \delta F_{L I}}{\left(\delta F_{L I}+\gamma+p+\mu\right)\left(\alpha+r_3+r_2+\mu\right)}}}{2 \frac{\left(C^* F_{I L} T \frac{1}{N}+\alpha\right) \delta F_{L I}}{\left(\delta F_{L I}+\gamma+p+\mu\right)\left(\alpha+r_3+r_2+\mu\right)}}$

Ro=$\frac{-\frac{(0.099+0.15+0.8+0.08) \text{*} 21 \text{*}1 \cdot \frac{1}{200,000,000}}{(0.2 \text{*} 1+0.001+0.2+0.08)(0.099+0.15+0.8+0.08)}\pm \sqrt{\left[\frac{(0.099+0.15+0.8+0.08) \cdot 21 \cdot 1 \cdot \frac{1}{200,000,000}}{(0.2 \text{*} 1+0.001+p+0.08)(0.099+0.15+0.8+0.08)}\right]^2-4 \cdot \frac{\left(0.7 \text{*} 1 \text{*} 7,500,000 \cdot \frac{1}{200,000,000}+0.099\right) \cdot 0.2 \text{*} 1 \text{*}}{(0.2 \text{*} 1+0.001+0.01+0.08)(0.099+0.15+0.8+0.08)} \cdot \frac{(0.2 \text{*} 1+0.001+0.01+0.08) \cdot 0.2 \text{*} 1}{(0.2 \text{*} 1+0.001+0.01+0.08)(0.099+0.15+0.8+0.08)}}}{2 \cdot \frac{\left(0.7 \text{*} 1 \text{*} 7,500,000 \cdot \frac{1}{200,000,000}+0.099\right) \cdot 0.2 \text{*} 1}{(0.2 \text{*} 1+0.001+0.01+0.08)(0.099+0.15+0.8+0.08)}}$

$R_o = \frac{-0.0000002183 \pm \sqrt{\left(\frac{0.0000002183}{0.325}\right)^2 - 4 \left(\frac{0.1253}{0.3285}\right) \left(\frac{0.0582}{0.3285}\right)}}{2 \left(\frac{0.1253}{0.3285}\right)}$

$R_o=\frac{-0.0000002183 \pm \sqrt{0-0.27033632}}{0.7629}$

$R_o=0.3544 i$

Table 1 and Table 2 describe the parameters of the model.

Figure 2 shows the pictorial flow of each compartment model. The red line shows how the number of S individuals is decreasing due to the virus. The line rapidly decreases from the second year until the fifth year. The blue line shows the rapid increase of L individuals due to the higher contact rate. The number of I individuals is very high to the extent that it almost meets S individuals. The green line shows the I individuals and how the disease spreads to the population. The last line, which is purple, shows T people, whose number is greater than that of all the remaining compartments. The subgraph (a) of Figure 3 shows the comparison between the S and L individuals which meet at equilibrium approximately 3.5 years later. This shows that the virus really has a large impact due to its higher reproductive number. The number of L individuals continues to grow till it reaches the population size. Therefore, a lot of people may contract the disease, known or unknown. The subgraph (b) of Figure 3 displays the comparison between the S and T individuals. The more S decreases, the more T follows. It also shows that the government shows clear concern for the disease, but the population is affecting disease control once more. The subgraph (a) of Figure 4 shows the comparison between the I and T individuals. The vaccine and treatment yield good results, with the vaccinated T population being higher than the I population, which is awesome. The subgraph (b) of Figure 4 also shows the difference between the L and T individuals. Totally, 143 T individuals are also higher than the L individuals.

The subgraph (a) of Figure 5 shows the number of deaths per compartment. It can be seen that the S has the highest number of deaths, maybe due to the population in the compartment. The number of S decreases during 0-2 years, increases rapidly from around 2.2-4 years, which is the extremely high point of the S, and finally decreases to almost zero. The number of I individuals also decreases at some point in the green line. At some point, the deaths of the S and I individuals are almost equal, which shows that the disease is killing people and individuals also die naturally, with the deaths reaching 4*108. The L individuals are represented by the blue line. The number also decreases drastically, and the death rate in the treatment also decreases rapidly, indicating that the intervention is effective. The subgraph (b) of Figure 5 shows how people are recovering from the disease. The treatment rate of recovery is very high compared to other compartments, followed by S individuals who recover within the period of 1.5 to 5.2 years before becoming stable after 5 years. The numbers of I and L individuals are also decreasing due to their willingness to receive the treatment and the assumption that they are leaving the compartment for medication. The subgraph (a) of Figure 6 shows the number of I individuals recovered to S without treatment. This shows that many I individuals have no symptoms and recover without awareness, but the virus has a higher value of risk in the fourth year. The yellow line represents the increase in S, while the purple line represents the decrease in L. Both compartments become stable after six years. The subgraph (b) of Figure 6 shows both the movement of how I people get cured without treatment and the fact that the number is higher than that of those moving to an I compartment. The negative value shows the release of individuals from their compartment, and the positive value shows their absorption in the compartment.

Figure 7 illustrates the number of patients discharged from the hospital due to their chronic condition or lack of response to medication. Therefore, the patients need to return home and enter the I compartment. As displayed in the figure, the blue line, which is T, decreases after the fourth year and increases after the second year of endemic. However, the red line, which is I, rapidly increases from zero year to the fourth, decreases after the fourth year till the sixth year, and continues to increase afterwards. This shows that the disease is still circulating among the masses and that the chronic disease has not been eradicated. The subgraph (a) of Figure 8 shows how the I compartments are losing individuals to the T ones. The numbers are equal in terms of the axis, and the negative value of I shows the number of people moving to T compartments for effective treatment. The disease is at equilibrium at five-and-a-half years. The subgraph (b) of Figure 8 shows the number of individuals vaccinated. The T has the highest vaccinated population, which is almost 7 million people, followed by the I population of more than 4 million, the L population of almost 30 million, and finally the S individuals. As shown in the figure, the number of S individuals is negative and converges to zero, which means people are moving from one compartment to another. The subgraph (a) and (b) of Figure 9 display the death rate for each compartment when the death rate parameter increases to 4.9. Itcan be seen that even if the death rate increases, the population dies in each compartment becomes stable in less than a year.

In this study, a mathematical model is developed for TB transmission by examining scenarios both in the presence and absence of treatment interventions. It can be found that the treated population is greater than the untreated population. Vaccinated and unvaccinated individuals are also compared. It can be found that vaccinated individuals are much higher than individuals in other compartments. Calculations reveal that the disease-free equilibrium and the basic reproductive number are less than one, indicating that the disease is under control. Or, it is unstable, indicating that the disease is not under control. In conclusion, the presented simulation shows that vaccination is able to prevent the disease from spreading.