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