Intelligent Dynamical Buffer Scheduling Mechanism for Intermittently Connected Mobile Network

Abstract

According to the store-carry-forward packet transmission method, nodes can communicate with each other in intermittently connected mobile network flexibly. As can be seen, the successful transmission of packets is assisted by multiple copies injected into the network. Therefore, the limited buffer should be utilized reasonably in this situation. In this paper, an adaptive buffer scheduling mechanism is proposed with the aid of packet transmission status estimation. According to the activity degree of node and the number of packet copies, the status of packet transmission in the network can be evaluated. Furthermore, with the estimated outcome of packet redundancy, the packets in the buffer are scheduled dynamically. Numerical results show that the activity degree can be estimated accurately, especially when the networks become larger. The number of packet copies can be proved that it follows normal distribution. Compared with other buffer scheduling mechanisms, our mechanism displays better performance, e.g., the packet delivery probability is enhanced by 21–50 %, and the latency is reduced by 15–23 %.

Appendix 1: Prove C\((T_{m})\) Follows Normal Distribution

Let $C$ denote the number of packet copies in intermittently connected mobile network, and $C_1 ,C_2 ,\ldots ,C_n$ are the samples of $C$. Based on the Skewness-Kurtosis verify method [30], $G_1 ={H_3 }/{H_2^{3/2} },G_2 ={H_4 }/{H_2^2 }$ are the samples of skewness and kurtosis respectively. $H_k (k=2,3,4)$ is the $k$-order central moment of the sample, and $A_k =\frac{1}{n}\sum \nolimits _{i=1}^n {C_i^k (k=1,2,3,4)}$ is $k$-order sample moments. As can be seen, $H_{2}, H_{3}, H_{4}$ can be calculated by the following Equations.

$$\begin{aligned}&H_2 =A_2 -A_1^2\\&H_3 =A_3 -3A_2 A_1 +2A_1^3\\&H_4 =A_4 -4A_3 A_1 +6A_2 A_1^2 -3A_1^4 ..\\ \end{aligned}$$

While the number of packet copies $C$ follows normal distribution and $n$ is large enough, we can assume approximately:

$$\begin{aligned}&G_1 \sim N\left( {0,\left. {\frac{6(n-2)}{(n+1)(n+3)}} \right) } \right. , G_2 \sim N\left( {3-\frac{6}{n+1}} \right. ,\left. {\frac{24n(n-2)(n-3)}{(n+1)^{2}(n+3)(n+5)}} \right) \\&\sigma _1 =\sqrt{\frac{6(n-2)}{(n+1)(n+3)}}, \sigma _2 =\sqrt{\frac{24n(n-2)(n-3)}{(n+1)^{2}(n+3)(n+5)}}, \mu _2 =3-\frac{6}{n+1}, U_1 ={G_1 }/{\sigma _1 },\\&U_2 =(G_2 -\mu _2 )/\sigma _2 , U_1 \sim N(0,1), U_2 \sim N(0,1). \end{aligned}$$

Let $u_{1}$ and $u_{2}$ denote the observation value of $U_{1 }$and $U_{2}$ respectively. According to the verify method, the probability that $C$ follows normal distribution is $\gamma$ if ${\vert }u_{1}{\vert }\le \mathrm{z}_{\gamma /4}$ or ${\vert }u_{2}{\vert }\le \mathrm{z}_{\gamma /4}$, where $z_{\gamma }$ is the $\gamma$ percentiles of standard normal distribution.

After a large number of measurements, the results show that the number of copies follows normal distribution. To make sure that our conclusion is reasonable, the distribution of simulation data is validated. For the case of $\mathrm{T}_{\mathrm{m}}=75$ % TTL, the data of $C(T_{m})$ is shown in Table 4.

According to the Skewness–Kurtosis method, $n=145, \sigma _{1}=0.1993, \sigma _{2}=0.3864, \mu _{2}=2.9589, H_{2}=125.6014, H_{3}=272.6525, H_{4}=42941.7521, G_{1}=0.1937, G_{2}=2.722$. The observation values are ${\vert }\mathrm{u}_{1}{\vert }=0.972$ and ${\vert }\mathrm{u}_{2}{\vert }=0.613$, and $z_{\gamma /4}=z_{0.0125}=2.24$.

For ${\vert }\mathrm{u}_{1}{\vert }=0.972<2.24,{\vert }\mathrm{u}_{2}{\vert }=0.613<2.24$, so the conclusion is that the data in Table 4 follows normal distribution, and the probability is higher than 95 % (Table 5).

Table 5 Data distribution of $C(T_{m})$ ($T_{m}=75$ % TTL)

Figure 10 shows the cumulative distribution function of $C(T_{m})$ when $\mathrm{T}_{\mathrm{m}}=75\,\%$ TTL. From the results, we can see that $C(T_{m})$ is approximate to normal distribution. For the case of $\mathrm{T}_{\mathrm{m}}=50$ % TTL and $\mathrm{T}_{\mathrm{m}}=90$ % TTL, the same conclusion can be drawn as Figs. 11 and 12 show.

Fig. 10

Cumulative distribution comparing of $C(T_{m})$ ($T_{m}=75$ % TTL)

Fig. 11

Cumulative distribution comparing of $C(T_{m})$ ($T_{m}=50$ % TTL)

Fig. 12

Cumulative distribution comparing of $C(T_{m})$ ($T_{m}=90$ % TTL)