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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数学分析代写Mathematical Analysis代考|Numerical Simulation

I Iere first the model system (1) is simulated for the following set of parameters:
$$\begin{gathered} \Lambda=500, \mu=0.0166, \beta_{T}=0.00001, \beta_{S}=0.000014, \beta_{T S}=0.000015, \sigma=0.06, \ p=0.45, \rho=0.000013, \alpha=0.03, \zeta=0.0945, \omega=1.05, \delta=0.04, \ \nu=0.08, \rho_{S}=0.00002, \sigma_{S}=0.12, \delta_{S}=0.06, \nu_{S}=0.05, \eta=1.04, \epsilon=0.8 . \end{gathered}$$
For this set of parameters both $R_{0}^{T}$ and $R_{0}^{S}$ are greater than one. Here $R_{0}^{T}=1.214$ and $R_{0}^{S}=5.505$. The equilibria $(30120,0,0,0,0,0,0)$ where both disease and smokers are not there. This equilibrium point is unstable. The smoking-free equilibrium point $(11033,7227.4,2872,0,0,0,2068)$ and TB-free equilibrium point (5471.4, $0,0,24649,0,0,0)$ are also unstable. The non-trivial equilibrium point $(7175.6,3949.5,2557.9,2206.9,3251.3,697.13,1598.7)$ is stable. Here Fig. 2 is demonstrating the stability of non-trivial equilibrium point. Next the parameter $p$ is changed from $0.45$ to $0.75$ and all other parameters are kept as mentioned above. For this set of parameters, $R_{0}^{T}=0.5527<1$ and $R_{0}^{S}=5.505>1$. Hence $R_{0}^{S T}=5.505>1$. In this case also a stable non-trivial equilibrium point as $(6996.8,3420.8,1525.8,4422.7,5869.1,722.79,873.5)$ exists. This suggest that backward bifurcation occurs. The stability of this equilibrium point is shown in Fig. 3 .
Next, the model is simulated for different values of $\beta_{S}$ to see the impact of this parameter on the equilibrium level of different population. The parameter $p$ is taken as $0.65$ and all other parameters are as mentioned above. Here it is observed that with the increase in this parameter the number of smokers and smokers infected with TB and TB-infectives increase (Figs. 4, 5 and 6). Here for all values of $\beta_{S}, R_{0}^{T}$ is less than 1 and $R_{0}^{S}$ is greater than one. But similar results appear for the set of parameters where both $R_{0}^{S}$ and $R_{0}^{T}$ are greater than one.

## 数学代写|数学分析代写Mathematical Analysis代考|Optimal Control Problem

Here the mathematical model (1) is extended to optimal control problem. Optimal control is a method to determine time dependent control and state variables for a given dynamical system over a period of time to design suitable cost-effective strategies. Here the parameters $\nu$ (the recovery rate constant for TB-infectives) and $\sigma_{S}$ (the rate at which smokers infected with TB leave smoking) are made time dependent. These two parameters are identified keeping in mind the fact that it is possible to regulate these parameters by increasing the number of TB-detection centres and counseling centres for smokers. If these are zero then there is no effort being placed in these controls at time $t$ and if they are equal to one then maximum effort is applied. Keeping in view of the above assumptions, the optimal control model is formulated as follows:
\begin{aligned} \frac{d S}{d t} &=\Lambda-\mu S-\lambda_{S} S-\lambda_{T} S+\sigma I_{S} \ \frac{d E_{T}}{d t} &=p \lambda_{T} S-(\mu+\rho) E_{T}+\alpha E_{T S}+\zeta R_{T}-\lambda_{S} E_{T}-\omega \lambda_{T} E_{T} \ \frac{d I_{T}}{d t} &=(1-p) \lambda_{T} S-(\mu+\delta+\nu(t)) I_{T}+\omega \lambda_{T} E_{T}+\rho E_{T}+\sigma_{S}(t) I_{T S} \ \frac{d I_{S}}{d t} &=\lambda_{S} S-(\mu+\sigma) I_{S}-\lambda_{T S} I_{S} \ \frac{d E_{T S}}{d t} &=\lambda_{S}\left(E_{T}+R_{T}\right)+p \lambda_{T S} I_{S}-\omega \lambda_{T} E_{T S}-\left(\alpha+\mu+\rho_{S}\right) E_{T S} \end{aligned}

## 数学代写|数学分析代写Mathematical Analysis代考|Numerical Simulation

$$\Lambda=500, \mu=0.0166, \beta_{T}=0.00001, \beta_{S}=0.000014, \beta_{T S}=0.000015, \sigma=0.06, p=0.45, \rho=0.000013, \alpha=0.03, \zeta=0.0945, \omega=1.05, \delta=0.04, \nu$$

$(7175.6,3949.5,2557.9,2206.9,3251.3,697.13,1598.7)$ 是稳定的。这里图 2 展示了非平凡平衡点的稳定性。接下来是参数 $p$ 从改变 $0.45$ 至 $0.75$ 并且所有其他 参数都如上所述保留。对于这组参数， $R_{0}^{T}=0.5527<1$ 和 $R_{0}^{S}=5.505>1$. 因此 $R_{0}^{S T}=5.505>1$. 在这种情况下，也是一个稳定的非平凡平衡点 $(6996.8,3420.8,1525.8,4422.7,5869.1,722.79,873.5)$ 存在。这表明发生了后向分叉。该平衡点的稳定性如图3所示。

## 数学代写|数学分析代写Mathematical Analysis代考|Optimal Control Problem

$$\left.\frac{d S}{d t}=\Lambda-\mu S-\lambda_{S} S-\lambda_{T} S+\sigma I_{S} \frac{d E_{T}}{d t} \quad=p \lambda_{T} S-(\mu+\rho) E_{T}+\alpha E_{T S}+\zeta R_{T}-\lambda_{S} E_{T}-\omega \lambda_{T} E_{T} \frac{d I_{T}}{d t}=(1-p) \lambda_{T} S-(\mu+\delta+\nu(t)) I_{T}+\omega\right)$$

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师