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doc: add connectivity loss search analysis tex file

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\documentclass[11pt]{article}
\usepackage[utf8x]{inputenc}
\usepackage[top=2cm,bottom=2cm]{geometry}
\usepackage{forloop}
\newcounter{counter}
% maths
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{mathrsfs}
\usepackage{shadethm}
\usepackage{amsthm}
\newshadetheorem{shadeDef}{Definition}[section]
\newtheorem{remark}{Remark}[section]
\renewcommand{\P}{\mathbb{P}}
\usepackage{float}
\usepackage{booktabs}
\setlength{\parindent}{0ex}
\setlength{\parskip}{0.5em}
\begin{document}
\title{Annex 1: Probabilistic analysis of connectivity changes}
\author{Adrien Béraud, Simon Désaulniers, Guillaume Roguez}
\maketitle
\pagestyle{empty}
\begin{shadeDef}
A node flagged as \emph{``expired''} by a node $n$ is a node which has not responded to any
of $n$'s last three requests.
\end{shadeDef}
\begin{remark}
An expired node will not be contacted before 10 minutes from its expiration time.
\end{remark}
Let $N$ the DHT network, $n_0\in N$, a given node and the following probabilistic events:
\begin{itemize}
\item $A$: $\forall n \in N$ $n$ is unreachable by $n_0$, \emph{i.e.} $n_0$ lost connection
with $N$;
\item $B$: $S\subset N$, the nodes unreachable by $n_0$ with $k={|S|\over|N|}$;
\item $C$: $m \le |N|$ nodes are flagged as ``expired''.
\end{itemize}
We are interested in knowing $\P(A|C)$, \emph{i.e.} the probability of the event where $A$ occurs
prior to $C$. From the above, we immediately get
$$\left\{
\begin{array}{ll}
\P(C|A) & = 1\\
\P(A) + \P(B) & = 1
\end{array}
\right.$$
Also, the event $A|C$ can be abstracted as the urn problem of draw without replacement. Then,
$$\P(C|B) = \prod_{i=0}^m \left[k|N| - i \over |N|\right] = \prod_{i=0}^m \left[k - { i \over |N| }\right]$$
Furthermore, using Bayes' theroem we have
\begin{align*}
\P(A|C) & = { \P(C|A)\P(A) \over \P(C|A)\P(A) + \P(C|B)\P(B)}\\
& = { \P(A) \over \P(A) + \P(C|B)\P(B) }\\
& = { \P(A) \over \P(A) + \P(C|B)\left[1 - \P(A)\right] }\\
\Rightarrow \forloop{counter}{0}{\value{counter} < 5}{\qquad}\quad \P(A) & =
\P(A|C)\left[\P(A) + \P(C|B)\left(1 - \P(A)\right) \right] \\
\Rightarrow \forloop{counter}{0}{\value{counter} < 2}{\qquad}\;\:\, \P(A)\left[{ 1 \over \P(A|C)} - 1\right] & =
\P(C|B)\left(1 - \P(A)\right) \\
\end{align*}
Finally,
\begin{equation}
\label{eq:final}
\left[{\P(A) \over 1 - \P(A)}\right]\left[{ 1 \over \P(A|C)} - 1\right] =
\prod_{i=0}^m \left[k - { i \over |N| }\right]
\end{equation}
From \eqref{eq:final}, we may set a plausible configuration $\{\P(A),\P(A|C),k,|N|\}$ letting us
produce results such as in table \ref{tbl:k_1_2}, \ref{tbl:k_2_3} and \ref{tbl:k_3_4}.
\begin{table}[H]
\centering
\caption{The values for $m$ assuming $\P(A|C) \ge 0.95,\, k = {1 \over 2}$}
\label{tbl:k_1_2}
\begin{tabular}{lcccc}
\toprule
$|N| \diagdown \P(A)$ & ${1 \over 10 }$ & ${1 \over 100 }$ & ${1 \over 1000 }$ & ${1 \over 10000 }$\\
\midrule
$2^0$ & 1 & 1 & 1 & 1\\
$2^1$ & 1 & 1 & 1 & 1\\
$2^2$ & 2 & 2 & 2 & 2\\
$2^3$ & 4 & 4 & 4 & 4\\
$2^4$ & 5 & 6 & 7 & 8\\
$2^5$ & 5 & 7 & 9 & 10\\
$2^6$ & 6 & 9 & 11 & 13\\
$2^7$ & 6 & 9 & 12 & 14\\
$2^8$ & 7 & 10 & 13 & 16\\
$2^9$ & 7 & 10 & 13 & 16\\
$2^{10}$ & 7 & 10 & 13 & 17\\
\bottomrule
\end{tabular}
\end{table}
\begin{table}[H]
\centering
\caption{The values for $m$ assuming $\P(A|C) \ge 0.95,\, k = {2 \over 3}$}
\label{tbl:k_2_3}
\begin{tabular}{lcccc}
\toprule
$|N| \diagdown \P(A)$ & ${1 \over 10 }$ & ${1 \over 100 }$ & ${1 \over 1000 }$ & ${1 \over 10000 }$\\
\midrule
$2^0$ & 1 & 1 & 1 & 1\\
$2^1$ & 2 & 2 & 2 & 2\\
$2^2$ & 3 & 3 & 4 & 4\\
$2^3$ & 5 & 5 & 6 & 8\\
$2^4$ & 6 & 8 & 9 & 10\\
$2^5$ & 8 & 10 & 12 & 14\\
$2^6$ & 9 & 13 & 16 & 18\\
$2^7$ & 11 & 15 & 18 & 22\\
$2^8$ & 11 & 16 & 21 & 25\\
$2^9$ & 12 & 17 & 22 & 27\\
$2^{10}$ & 12 & 18 & 23 & 28\\
\bottomrule
\end{tabular}
\end{table}
\begin{table}[H]
\centering
\caption{The values for $m$ assuming $\P(A|C) \ge 0.95,\, k = {3 \over 4}$}
\label{tbl:k_3_4}
\begin{tabular}{lcccc}
\toprule
$|N| \diagdown \P(A)$ & ${1 \over 10 }$ & ${1 \over 100 }$ & ${1 \over 1000 }$ & ${1 \over 10000 }$\\
\midrule
$2^0$ & 1 & 1 & 1 & 1\\
$2^1$ & 2 & 2 & 2 & 2\\
$2^2$ & 3 & 3 & 3 & 3\\
$2^3$ & 5 & 6 & 6 & 6\\
$2^4$ & 7 & 9 & 10 & 11\\
$2^5$ & 10 & 12 & 14 & 16\\
$2^6$ & 12 & 16 & 19 & 22\\
$2^7$ & 14 & 19 & 23 & 27\\
$2^8$ & 15 & 21 & 27 & 32\\
$2^9$ & 16 & 23 & 30 & 36\\
$2^{10}$ & 17 & 24 & 31 & 38\\
\bottomrule
\end{tabular}
\end{table}
\end{document}
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