AOM: adaptive mobile data traffic offloading for M2M networks

Abstract

With the increasing application of machine-to machine (M2M) communication through cellular networks, such as telematics, smart metering, point-of-sale terminals, and home security, more data traffice has been produced in the cellular network. Although many schemes have been proposed to reduce data traffic, they are inefficient in practical application due to poor adaption. In this paper, we focus on how to adaptively offload data traffic for cellular M2M networks. To this end, we propose an adaptive mobile data traffic offloading model (AOM). This model can decide whether to adopt opportunistic communications or communicate via cellular networks adaptively. In the AOM, we introduce traffic offloading rate (called TOR) and local resource consumption rate (called LRCR) and analyze them based on continue time Markov chain. Theory proof and extensive simulations demonstrate that our model is accurate and effective, and can adaptively offload data traffic of cellular M2M networks.

Appendices

Appendix 1: Proof of Lemma 1

We take an arbitrary C _k (t) to prove that it follows the CTMC.

We assume that the minimum time of the number of services node from i nodes to (i + 1) nodes is T _i in group k (k = 1, 2,…,K), and s _k indicates the number of seeds. From (2), we have:

$$\begin{aligned} & P\left\{ {C_{k} (0) = s_{k} } \right\} = 1 \\ & P\left\{ {C_{k} (T_{1} ) = s_{k} + 1\left| {C_{k} (0) = s_{k} } \right.} \right\} = \exp \left( { - \lambda_{s,M - S}^{\text{suc}} T_{1} } \right) = P\left\{ {C_{k} (T_{1} ) = s_{k} + 1} \right\}, \\ & P\left\{ {C_{k} \left( {T_{1} + T_{2} } \right) = s_{k} + 2\left| {C_{k} (T_{1} ) = s_{k} + 1,C_{k} (0) = s_{k} } \right.} \right\} \\ & \quad = P\left\{ {C_{k} \left( {T_{1} + T_{2} } \right) = s_{k} + 2,C_{k} (T_{1} ) = s_{k} + 1,C_{k} (0) = s_{k} } \right\}/ \\ & \quad \quad P\left\{ {C_{k} (T_{1} ) = s_{k} + 1,C_{k} (0) = s_{k} } \right\} \\ &P\left\{ {C_{k} \left( {T_{1} + T_{2} } \right) = s_{k} + 2\left| {C_{k} (T_{1} ) = s_{k} + 1 } \right.} \right\}\\ & \ldots , \\ & P\left\{ {C_{k} \left( {T_{1} + T_{2} + \cdots + T_{i} + T_{i + 1} } \right)} \right. \\ & \left.\quad = s_{k} + i + 1\left| {C_{k} \left( {T_{1} } \right) = s_{k} + i, \ldots C_{k} (0) = s_{k} }\right\} \right. \\ & \quad = P\left\{ {C_{k} \left( {T_{1} + T_{2} + \cdots + T_{i} + T_{i + 1} } \right) = s_{k} + i + 1\left| {C_{k} \left( {T_{1} } \right) = s_{k} + i} \right.} \right\} \\ \end{aligned}$$

Thus, C _k (t) is a CTMC. Because each group is mutual independence, each one group is a CTMC. So the process $\{ {\mathbb{C}}(t) :t \ge 0\}$ is a K-dimensional CTMC. □

Appendix 2: Proof of Lemma 2

According to the define of t _i , the conditional probability of D(t) = j + 1 is:

$$P\left\{ {D(t) = j + 1\left| {C(t) = s + i + 1} \right.} \right\} = \frac{N - j}{M - s - i}.$$

And

$$\begin{aligned} & P\left\{ {D(t) = j + 1\left| {C(t) = s + i + 1} \right.} \right\} \\ & \quad = P\left\{ {D\left( t \right) = j + 1\left| {C(t) = s + i + 1} \right.} \right\}/P\left\{ {C(t) = s + i + 1} \right\} \\ \end{aligned}$$

According to the formula total probability, we have:

$$\begin{aligned}&P\left\{ {D\left( {t_{i + 1} } \right) = j + 1} \right\} \\&\quad= \sum\limits_{i} {P\left\{ {D(t) = j + 1,C(t) = s + i + 1} \right\}} ,\end{aligned}$$

And (7) has been proved in Lemma 2. □

Appendix 3: Proof of Theorem 1

Assume that the minimum time of C(t) from i (i = 1, 2,…,M − s) nodes to (i + 1) nodes is T _i . Let $t_{i} = \sum\nolimits_{h = 0}^{i} {T_{h} }$; then, according to Lemma 2, the probability of T _min is as follows:

$$\begin{aligned} & P\left\{ {T_{\hbox{min} } < t_{N} } \right\} = 0, \\ & P\left\{ {T_{\hbox{min} } = t_{N} } \right\} = \frac{{P\left\{ {C(t) = s + N\left| {C\left( {t_{N - 1} } \right) = s + N - 1} \right.} \right\}}}{M - s - N + 1}, \\ & P\left\{ {T_{\hbox{min} } = t_{N + 1} } \right\} = \frac{{P\left\{ {C(t) = s + N + 1\left| {C\left( {t_{N} } \right) = s + N} \right.} \right\}}}{M - s - N}, \\ & \ldots , \\ & P\left\{ {T_{\hbox{min} } = t_{M - s - 1} } \right\} \\ &\quad= \frac{{P\left\{ {C(t) = M - s - 1\left| {C\left( {t_{M - s - 2} } \right) = M - s - 2} \right.} \right\}}}{2} \\ & P\left\{ {T_{\hbox{min} } = t_{M - s} } \right\} = P\left\{ {C(t) = M - s\left| {C\left( {t_{M - s - 1} } \right) = M - s - 1} \right.} \right\}. \\ \end{aligned}$$

According to the probability distribution of T _min, we can obtain the expectation of T _min:

$$\begin{aligned} E[T_{\hbox{min} } ] & = \sum\limits_{1}^{M - s} {t_{i} P\left\{ {T_{\hbox{min} } = t_{i} } \right\}} = \sum\limits_{N}^{M - s} {t_{i} P\left\{ {T_{\hbox{min} } = t_{i} } \right\}} \\ & = \sum\limits_{i - N}^{M - s} {\frac{{P\left\{ {C(t) = s + i\left| {C\left( {t_{N - 1} } \right) = s + i - 1} \right.} \right\}}}{M - s - i + 1} \cdot \frac{1}{M\psi }t_{i} } . \\ \end{aligned}$$

Let $\varphi \left( {t_{i} } \right) = \frac{{P\left\{ {C(t) = s + i\left| {C\left( {t_{N - 1} } \right) = s + i - 1} \right.} \right\}}}{M - s - i + 1};$ then, we have:

$$E[T_{\hbox{min} } ] = \sum\limits_{i = N}^{M - s} {\varphi \left( {t_{i} } \right) \cdot \frac{1}{M\psi }t_{i} }$$

Similarly, we calculate the expectation of t _i :

$$E[t_{i} ] = \sum\limits_{i} {E\left[ {T_{i} } \right]} .$$

Because T _i follows exponential distribution, according to (2), we have:

$$\begin{aligned} E\left[ {t_{i} } \right] & = \sum\limits_{i} {E\left[ {T_{i} } \right]} = \frac{1}{\psi }\sum\limits_{1}^{i} {\frac{1}{(s + i)(M - s - i)}} \\ & \approx \frac{1}{\psi M}\int_{1}^{i} {\left( {\frac{1}{s + t} + \frac{1}{M - s - t}} \right){\text{d}}t} \\ & \approx \frac{1}{\psi M}\ln \left( {\frac{M - s - 1}{M - s - i} \cdot \frac{s + i}{s + 1}} \right). \\ \end{aligned}$$

Summary, Theorem 1 has been proved. □

Appendix 4: Proof of Theorem 2

According to Theorem 1 at time t, we can obtain the probability distribution of $\hat{T}_{\hbox{min} }$ and T _min:

$$\begin{aligned} &P\left\{ {T_{\hbox{min} } = t_{N + i} } \right\} \\&\quad = \frac{{P\left\{ {C(t) = s + i - N\left| {C\left( {t_{N - 1} } \right) = s + i - N - 1} \right.} \right\}}}{M - s - N - i + 1}, \\ &P\left\{ {\hat{T}_{\hbox{min} } = t_{N + i} } \right\} \\& \quad= \frac{{P\left\{ {C(t) = s + i + N\left| {C\left( {t_{N - 1} } \right) = s + i + N - 1} \right.} \right\}}}{M - s - N - i + 1}, \\ \end{aligned}$$

i.e., $P\left\{ {T_{\hbox{min} } = t_{N + i} } \right\} = P\left\{ {\hat{T}_{\hbox{min} } = t_{N + i} } \right\}$. From (12), we have,

$$Et_{i} = \frac{1}{\psi M}\ln \left( {\frac{M - s - 1}{M - s - i} \cdot \frac{s + i}{s + 1}} \right).$$

And because $\hat{\lambda }_{\alpha ,\beta }^{\text{suc}} = \omega \lambda_{\alpha ,\beta }^{\text{suc}}$, i.e., $\hat{\psi } = \omega \psi$, the expectation of $\hat{t}_{i}$ denotes as follows:

$$\begin{aligned} E\hat{t}_{i} & = \frac{1}{{\hat{\psi }M}}\ln \left( {\frac{M - s - 1}{M - s - i} \cdot \frac{s + i}{s + 1}} \right) \\ & = \frac{1}{\omega \psi M}\ln \left( {\frac{M - s - 1}{M - s - i} \cdot \frac{s + i}{s + 1}} \right). \\ & = \omega^{ - 1} Et_{i} \\ \end{aligned}$$

In other words, $P\left\{ {\hat{T}_{\hbox{min} } < t} \right\} = P\left\{ {\omega^{ - 1} T_{\hbox{min} } < t} \right\}$; then,

$$\hat{T}_{\hbox{min} } \mathop = \limits^{d} \omega^{ - 1} T_{\hbox{min} } .$$

□