Reliability Analysis for a Data Flow in Event-Driven Wireless Sensor Networks

Abstract

For the purpose of designing more reliable networks, we extend the traditional reliability analysis from wired networks to wireless networks with imperfect components. This paper aims to study the reliability of a data flow in event-driven wireless sensor networks with acknowledgment-based transmission scheme. Initially, an event-driven wireless sensor network model is described in terms of limited node battery energy and shadowed fading channels. Then, in order to analyze the network reliability, wireless link reliability and node energy availability are investigated, respectively. Further the analytical expressions of the instantaneous network reliability and the mean time to failure are derived. Finally, the simulation results validate the correctness and accuracy of the analytical results.

Appendices

Appendix 1: Derivation of the Probability Distribution Law of Random Variable $S_n \left( t \right)$

The probability distribution law of random variable $S_n \left( t \right) ,\forall n\in \left\{ {0,1,\cdots ,N} \right\}$ will be deduced in this “appendix”.

Firstly the random variable $S_0 \left( t \right)$ which describes the number of packets that are sent out from the source node (Node $0)$ during $\left[ {0,t} \right]$ will be considered. It has been known that totally $M\left( t \right)$ packets are detected by the source node during $\left[ {0,t} \right]$. In this paper, we don’t concerned propagation delay, hence $M\left( t \right)$ is also equal to the number of packets that are successfully received by the first relay node during $\left[ {0,t} \right]$. It is because that according to the acknowledgment-based scheme, the sender doesn’t stop transmitting copies until an acknowledgment for a correctly received copy is received. $p_0$ is the probability that first relay node successfully receive a packet from the source node. Moreover, it has been assume that the acknowledgment information from receiver will never be lost. Thus, conditional on $M\left( t \right) =k,k\ne 0$, the process that the source node sends packets seems to be a sequence of independent Bernoulli trials. The process that the source node sends a packet is a trail and each trial having two potential outcomes that the first relay node successfully receive a packet or not called “success” and “failure”. In each trial the probability of success is $p_0$ and of failure is $1-p_0$. We are observing this sequence until a predefined number $k$ of successes has occurred. Then the total number of trials needed to get $k$ successes is $S_0 \left( t \right)$. Further conditional on $M\left( t \right) =k,k\ne 0$, assume random variable $Y_j ,\forall j\in \left\{ {1,2,\ldots ,k} \right\}$ denotes the random total number of Bernoulli trials needed to get the $j$th success with successful probability $p_0$ and failure probability $1-p_0$. Hence, it can obtained that the random variables $Y_1,Y_2 ,\ldots Y_k$ are independent Geometrically distributed variables with parameter $p_0$ and

$$\begin{aligned} \left\{ {S_0 \left( t \right) \left| {M\left( t \right) =k,k\ne 0} \right. } \right\} =\sum \limits _{i=1}^k {Y_k } \end{aligned}$$

(42)

According to Pitman [20, 21] and the definition of Negative Binomial distribution in [22], the random variable $\left\{ {S_0 \left( t \right) \left| {M\left( t \right) =k} \right. ,k\ne 0} \right\}$ follows a Negative Binomial (or Pascal) distribution with parameters $k$ and $1-p_0$, namely:

$$\begin{aligned} \left\{ {S_0 \left( t \right) \left| {M\left( t \right) =k,k\ne 0} \right. } \right\} \sim \hbox {NB}\left( {k,1-p_0 } \right) \end{aligned}$$

(43)

The probability distribution of such a random variable can be given by

$$\begin{aligned} \Pr \left\{ {S_0 \left( t \right) =i\left| {M\left( t \right) =k,k\ne 0} \right. } \right\} =\left\{ {{\begin{array}{l@{\quad }l} {\left( \begin{array}{ll} i-1\\ k-1\end{array}\right) p^{k}\left( {1-p} \right) ^{i-k}}&{} \qquad {i\ge k} \\ 0&{} \qquad {i<k} \\ \end{array} }} \right. \end{aligned}$$

(44)

where $\left( \begin{array}{l} i\\ k\end{array} \right)$ can be calculated by

$$\begin{aligned} \left( \begin{array}{l} i\\ k\end{array} \right) =\frac{i!}{k!(i-k)!} \end{aligned}$$

(45)

Specially, when $M\left( t \right) =0$, then it means the source node doesn’t sensing any packets during $\left[ {0,t} \right]$ that result in the source node doesn’t transmit any packets during $\left[ {0,t} \right]$. Hence

$$\begin{aligned} \Pr \left\{ {S_0 \left( t \right) =i\left| {M\left( t \right) =k,k=0} \right. } \right\} =\left\{ {{\begin{array}{l@{\quad }l} 1&{} \qquad {i=0} \\ 0&{} \qquad {i\ne 0} \\ \end{array} }} \right. \end{aligned}$$

(46)

According to the law of total probability and using (3), (44) and (46), the probability distribution law of random variable$S_{0}\left( t \right)$ can be described by

$$\begin{aligned} \begin{array}{l@{\quad }l} \Pr \left\{ {S_0 \left( t \right) =i} \right\} &{}=\sum \limits _{k=0}^\infty {\left\{ {\Pr \left\{ {S_0 \left( t \right) =i\left| {M\left( t \right) =k} \right. } \right\} \Pr \left\{ {M\left( t \right) =k} \right\} } \right\} } \\ &{}=\left\{ {{\begin{array}{l@{\quad }l} {\sum \limits _{k=1}^i {\left( \begin{array}{l} i-1\\ k-1\end{array}\right) p_0 ^{k}\left( {1-p_0 } \right) ^{i-k}\frac{\left( {\Lambda \left( t \right) } \right) ^{k}}{k!}\exp \left( {-\Lambda \left( t \right) } \right) } }&{} \qquad {i\ge 1} \\ {\exp \left( {-\Lambda \left( t \right) } \right) }&{} \qquad {i = 0} \\ 0&{} \qquad {i<0} \\ \end{array} }} \right. \end{array} \end{aligned}$$

(47)

Further, the random variable $S_n \left( t \right) ,\forall n\in \left\{ {1,2,\ldots ,N} \right\}$ which describes the number of packets that are sent out from Node $n$ during $\left[ {0,t} \right]$ will be considered in the following. It’s obviously that the number of packets that are sent out from Node $n$ and successfully received by Node $n+1$ relay node during $\left[ {0,t} \right]$ is equal to $M\left( t \right)$. Like the derivation of $S_0 \left( t \right)$ the probability distribution law of random variable $S_{n} \left( t \right) ,\forall n\in \left\{ {1,2,\ldots ,N} \right\}$ can be described by

$$\begin{aligned} \begin{array}{l@{\quad }l} \Pr \left\{ {S_n \left( t \right) =i} \right\} &{}=\sum \limits _{k=0}^\infty {\left\{ {\Pr \left\{ {S_n \left( t \right) =i\left| {M\left( t \right) =k} \right. } \right\} \Pr \left\{ {M\left( t \right) =k} \right\} } \right\} } \\ &{}=\left\{ {{\begin{array}{l@{\quad }l} {\sum \limits _{k=1}^i {\left( \begin{array}{l} i-1\\ k-1\end{array}\right) p_n ^{k}\left( {1-p_n } \right) ^{i-k}\frac{\left( {\Lambda \left( t \right) } \right) ^{k}}{k!}\exp \left( {-\Lambda \left( t \right) } \right) } }&{} \qquad {i\ge 1} \\ {\exp \left( {-\Lambda \left( t \right) } \right) }&{} \qquad {i = 0} \\ 0&{} \qquad {i<0} \\ \end{array} }} \right. \end{array} \end{aligned}$$

(48)

According to (47) and (48), the probability distribution law of random variable $S_{n} \left( t \right) ,\forall n\in \left\{ {0,1,\ldots ,N} \right\}$ can be described by

$$\begin{aligned} \begin{array}{l@{\quad }l} \Pr \left\{ {S_n \left( t \right) =i} \right\} &{}=\sum \limits _{k=0}^\infty {\left\{ {\Pr \left\{ {S_n \left( t \right) =i\left| {M\left( t \right) =k} \right. } \right\} \Pr \left\{ {M\left( t \right) =k} \right\} } \right\} } \\ &{}=\left\{ {{\begin{array}{l@{\quad }l} {\sum \limits _{k=1}^i {\left( \begin{array}{l} i-1\\ k-1\end{array}\right) p_n ^{k}\left( {1-p_n } \right) ^{i-k}\frac{\left( {\Lambda \left( t \right) } \right) ^{k}}{k!}\exp \left( {-\Lambda \left( t \right) } \right) } }&{} \qquad {i\ge 1} \\ {\exp \left( {-\Lambda \left( t \right) } \right) }&{} \qquad {i = 0} \\ 0&{} \qquad {i<0} \\ \end{array} }} \right. \end{array} \end{aligned}$$

(49)

Appendix 2: Proof of Proposition 1

Proof

Firstly, the relationship between the random variable $M\left( t \right)$ and the random variable $S_n \left( t \right) ,\forall n\in \left\{ {0,1,\ldots ,N} \right\}$ will be discussed. It has been known that $M\left( t \right)$ denotes the number of packets that are detected by the source node during $\left[ {0,t} \right]$. If doesn’t concern propagation delay, $M\left( t \right)$ is also equal to the number of packets that are successfully received by each relay node during $\left[ {0,t} \right]$. It is because that according to the acknowledgment-based scheme, the sender doesn’t stop transmitting copies until an acknowledgment for a correctly received copy is received. $S_n \left( t \right)$ denotes the number of packets that are sent out from Node $n,\forall n\in \left\{ {0,1,\ldots ,N} \right\}$ during $\left[ {0,t} \right]$. $p_n ,\forall n\in \left\{ {0,1,\ldots ,N} \right\}$ is the probability that Node $n+1$ successful receive a packet from Node $n$. Moreover, it has been assume that the acknowledgment information from receiver will never be lost. Thus, we can obtain

$$\begin{aligned} S_n \left( t \right) \cdot p_n \approx M\left( t \right) \end{aligned}$$

(50)

Substituting (50) into (17), the energy availability of source node at time $t$ can be measured as

$$\begin{aligned} \begin{array}{l@{\quad }l} \Pr \left\{ {\hbox {A}_0 } \right\} &{}=\Pr \left\{ {S_{0} \left( t \right) \le \frac{\left( {E_0^{\mathrm{init}} -\alpha P_{0}^{\mathrm{s}} t-E_{\mathrm{th}} } \right) r}{\left( {2P_0^{\mathrm{e}} +P_{0}^{\mathrm{t}} } \right) L}} \right\} \\ &{}\approx \Pr \left\{ {M\left( t \right) \le \frac{\left( {E_0^{\mathrm{init}} -\alpha P_{0}^{\mathrm{s}} t-E_{\mathrm{th}} } \right) rp_0 }{\left( {2P_0^{\mathrm{e}} +P_{0}^{\mathrm{t}} } \right) L}} \right\} \\ &{}= \left\{ {{\begin{array}{l@{\quad }l} {\sum \limits _{k=0}^{M_0^{\prime } } {\frac{\left( {\Lambda \left( t \right) } \right) ^{k}}{k!}\exp \left( {-\Lambda \left( t \right) } \right) } }&{} {M_0^{\prime } \ge 0} \\ 0&{} {M_0^{\prime } <0} \\ \end{array} }} \right. \\ \end{array} \end{aligned}$$

(51)

where

$$\begin{aligned} M_0^{\prime } =\left\lfloor {\frac{\left( {E_0^{\mathrm{init}} -\alpha P_{0}^{\mathrm{s}} t-E_{\mathrm{th}} } \right) rp_0 }{\left( {2P_0^{\mathrm{e}} +P_{0}^{\mathrm{t}} } \right) L}} \right\rfloor \end{aligned}$$

(52)

Substituting (50) into (28), the energy availability of the $n\hbox {th},\forall n\in \left\{ {1,2,\ldots ,N} \right\}$ relay node at time $t$ can be measured as

$$\begin{aligned} \begin{array}{l@{\quad }l} \Pr \left\{ {\hbox {A}_n } \right\} &{}=\Pr \left\{ {S_n \left( t \right) \le \frac{\left( {E_n^{\mathrm{init}} -E_{\mathrm{th}} } \right) r}{\left( {4P_n^{\mathrm{e}} +P_n^{\mathrm{t}} } \right) L}} \right\} \\ &{}\approx \Pr \left\{ {M\left( t \right) \le \frac{\left( {E_n^{\mathrm{init}} -E_{\mathrm{th}} } \right) rp_n }{\left( {4P_n^{\mathrm{e}} +P_n^{\mathrm{t}} } \right) L}} \right\} \\ &{}= \left\{ {{\begin{array}{l@{\quad }l} {\sum \limits _{k=0}^{M_n^{\prime } } {\frac{\left( {\Lambda \left( t \right) } \right) ^{k}}{k!}\exp \left( {-\Lambda \left( t \right) } \right) } }&{} {M_n^{\prime } \ge 0} \\ 0&{} {M_n^{\prime } <0} \\ \end{array} }} \right. \\ \end{array} \end{aligned}$$

(53)

where

$$\begin{aligned} M_n^{\prime } =\left\lfloor {\frac{\left( {E_n^{\mathrm{init}} -E_{\mathrm{th}} } \right) rp_n }{\left( {4P_n^{\mathrm{e}} +P_n^{\mathrm{t}} } \right) L}} \right\rfloor \end{aligned}$$

(54)

Using the upper incomplete gamma function defined in Haight [23], (51) and (53) the energy availability of Node $n,\forall n\in \left\{ {0,1,\ldots ,N} \right\}$ can be further modified into

$$\begin{aligned} \Pr \left\{ {\hbox {A}_n } \right\} =\left\{ {{\begin{array}{l@{\quad }l} {\frac{1}{M_n^{\prime } !}\Gamma \left( {M_n^{\prime } +1,\Lambda \left( t \right) } \right) }&{} {M_n^{\prime } \ge 0} \\ 0&{} {M_n^{\prime } <0} \\ \end{array} }} \right. ,\quad \forall n\in \left\{ {0,1,\ldots ,N} \right\} \end{aligned}$$

(55)

In (55), the function $\Gamma \left( {\mu ,x} \right)$ is the upper incomplete gamma function which is defined as

$$\begin{aligned} \Gamma \left( {\mu ,x} \right) =\int \limits _x^\infty {e^{-\tau }\tau ^{\mu -1}\hbox {d}\tau } \end{aligned}$$

(56)

It can be obviously find the expressions of energy availability of source node and relay node in (51) and (53) are the CDF of a Poisson distribution with mean $\Lambda \left( t \right)$ and variance $\Lambda \left( t \right)$ at $M_n^{\prime }$. According to the [24], for sufficiently large values of $\Lambda \left( t \right)$, (say $\Lambda \left( t \right) >1000)$, the normal distribution with mean $\Lambda \left( t \right)$ and variance $\Lambda \left( t \right)$ (standard deviation $\sqrt{\Lambda \left( t \right) })$, is an excellent approximation to the Poisson distribution. If $\Lambda \left( t \right)$ is greater than about 10, then the normal distribution is a good approximation if an appropriate continuity correction is performed, i.e., $\Pr \left\{ {X\le x} \right\}$, where (lower-case) $k$ is a non-negative integer, is replaced by $\Pr \left\{ {X\le x+0.5} \right\}$. Therefore, the energy availability of Node $n,\forall n\in \left\{ {0,1,\ldots ,N} \right\}$ at time $t$ can further be expressed approximately as:

$$\begin{aligned} \begin{array}{l@{\quad }l} \Pr \left\{ {\hbox {A}_n } \right\} &{}\approx \left\{ {{\begin{array}{l@{\quad }l} {\Phi \left( {\frac{M_n^{\prime } +0.5-\Lambda \left( t \right) }{\sqrt{\Lambda \left( t \right) }}} \right) }&{} {M_n^{\prime } \ge 0} \\ 0&{} {M_n^{\prime } <0} \\ \end{array} }} \right. \\ &{}=\left\{ {{\begin{array}{l@{\quad }l} {\frac{1}{2}\left[ {1+\hbox {erf}\left( {\frac{M_n^{\prime } +0.5-\Lambda \left( t \right) }{\sqrt{2\Lambda \left( t \right) }}} \right) } \right] }&{} {M_n^{\prime } \ge 0} \\ 0&{} {M_n^{\prime } <0} \\ \end{array} }} \right. \\ \end{array} \end{aligned}$$

(57)

where

$$\begin{aligned} M_n^{\prime } =\left\{ {{\begin{array}{l@{\quad }l} {\left\lfloor {\frac{\left( {E_0^{\mathrm{init}} -\alpha P_{0}^{\mathrm{s}} t-E_{\mathrm{th}} } \right) rp_0 }{\left( {2P_0^{\mathrm{e}} +P_{0}^{\mathrm{t}} } \right) L}} \right\rfloor }&{} {n=0} \\ {\left\lfloor {\frac{\left( {E_n^{\mathrm{init}} -E_{\mathrm{th}}} \right) rp_n }{\left( {4P_n^{\mathrm{e}} +P_n^{\mathrm{t}} } \right) L}} \right\rfloor }&{} {\forall n\in \left\{ {1,2,\ldots ,N} \right\} } \\ \end{array} }} \right. \end{aligned}$$

(58)

Thus, Proposition 1 is proved. $\square$