统计代写|MATH4364 Network Analysis

Statistics-lab™可以为您提供uh.edu MATH4364 Network Analysis网络分析课程的代写代考辅导服务!

MATH4364 Network Analysis课程简介

Network Analysis has become a widely adopted method for studying the interactions between social agents, information and infrastructures. The strong demand for expertise in network analysis has been fueled by the widespread acknowledgement that everything is connected and the popularity of social networking services. This interdisciplinary course introduces students to fundamental theories, concepts, methods and applications of network analysis in a practical manner. Students learn and practice hands-on skills in collecting, analyzing and visualizing network data.

PREREQUISITES 

Understand fundamental concepts and theories from the fields of social network analysis and network science.
Apply this knowledge to solve real-world, network-centric problems.
Use basic and advanced analysis methods and tools to visualize and analyze network data.

MATH4364 Network Analysis HELP(EXAM HELP, ONLINE TUTOR)

问题 1.

THEOREM (2.2): Let $\left{E_w:\right.$ weVW $}$ be a collection of equilibration operators associated with the the following conditions.
(1′) $E_w \mathscr{F}=\mathscr{F}$ for some $\mathscr{F} \in \mathscr{L}$ implies that $\mathscr{F}$ satisfies (1.13) for this fixed $w$.
(2′) $E_w$ is a continuous mapping from $\mathscr{Z}$ to $\mathscr{Z}$.
(3′) $C\left(\overline{E_{\mathrm{w}} \mathscr{F}}\right) \leqq C(\overline{\mathscr{F}})$ for all $\mathscr{F} \in \mathscr{F}$.
(4) $C\left(E_w \mathscr{F}\right)=C(\mathscr{F})$ for some $\mathscr{F} \in \mathscr{Z}$ implies that $E_w \mathscr{F}=\mathscr{F}$.
Then any equilibration operator associated with $\mathscr{T}$ and constructed by composition of the above collection $\left{E_w: w \in \mathscr{W}\right}$ satisfies conditions $1-4$ of Theorem (2.1).

Proof: Assumption 1 follows easily from $1^{\prime}$ and the structure of an equilibration operator associated with a pair of nodes. Assumption 2 is an obvious consequence of 2 ‘. Similarly 3 follows immediately from $3^{\prime}$. Finally 4 follows by a combination of $3^{\prime}$ and $4^{\prime}$.

The above theorem reduces the problem of checking conditions 1-4 of Theorem (2.1) to the much simpler problem of checking conditions 1 ‘-4’ of Theorem (2.2).

Sometimes an equilibration operator $E$ associated with a transportation network satisfies conditions 1 and 2 of Theorem (2.1) but it does not satisfy (or at least we cannot prove that it satisfies) conditions 3 and 4 . Then of course we do not know whether $E$ induces an algorithm for the solution of $P_1[\mathscr{F}]$. Nevertheless the sequence $\left{E^n \mathscr{F}^{(0)}\right}$ may lead to the solution of problem $P_1[\mathscr{F}]$ as shown by the following theorem, the proof of which is similar to the proof of Theorem (2.1).

问题 2.

THEOREM (2.3): Suppose that an equilibration operator $E$ satisfies conditions 1,2 of Theorem (2.1). Suppose further that for some choice of $\mathscr{F}^{(0)}$ the sequence $\left{\overline{E^n \mathscr{F}(0)}\right}$ converges as $n \rightarrow \infty$. Then $\left{\overline{E^N \mathscr{F}^{(0)}}\right}$ converges to the solution $\overline{\mathscr{S}}_1$, of the problem $P_1$.

REMARK (2.1): We have seen that an equilibration operator $E$ which induces an algorithm for the solution of problem $P_1$ enables us to calculate through $(2.2)$ the unique $\bar{F}_1$ associated with a problem $P_1[\mathscr{F}]$. Then we know that $R\left[\bar{F}_1\right]$ is the set of solutions of problem $P_1$. The calculation of an element of $R\left[\bar{F}_1\right]$, given $\bar{F}_1$, amounts to finding a solution to the system (1.1), (1.4), which might be accomplished by phase 1 of the Simplex method. This requires a rather tedious calculation. However, as shown in the proof of Theorem (2.1), some elements of $R\left[\overline{\mathscr{F}}_1\right]$ can be obtained directly from the algorithm as limits of the convergent subsequences of $\left{E^{n \prime} \mathscr{F}^{(0)}\right}$. In particular, if $R\left[\overline{\mathscr{F}}_1\right]$ consists of a unique element then
$$
E^N \mathscr{F}^{(0)} \rightarrow \mathscr{F}_1, n \rightarrow \infty
$$

Textbooks


• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

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统计代写|MATH4364 Network Analysis

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