Provisioning required reliability of wireless data communication in smart grid neighborhood area networks

Abstract

A possible candidate for implementing smart grid neighborhood area network (NAN) is wireless communications because of its advantages, such as flexibility and low installation and maintenance cost. However, one of the problems with wireless communication is its unreliability, whereas communication reliability is a basic requirement of smart grid applications. In this paper, we consider a wireless mesh network for NAN and then propose a method on the basis of hop-by-hop ARQ that achieves the required data communication reliability in wireless NAN and, at the same time, satisfies the communication latency constraint which is another requirement of smart grid applications. The proposed approach provides the optimal number of allowed retransmissions at each hop considering the loss probabilities of all hops and end-to-end delay constraint. Comparing to the typical application of ARQ in which the number of retransmissions permitted is the same at each hop, it is demonstrated that the proposed approach achieves a higher level of reliability while meeting the delay constraint. Moreover, the proposed method is evaluated by simulation, and the results are consistent with the results of the theoretical model.

Appendix

Consider an M / M / 1 queue with batch arrival, and P(k) being the probability mass function (pmf) of the size of the batch. The mean arrival rate of batches is $\lambda$, and the mean service rate is $\mu$. We also denote the mean number of packets already in the queue (including the one in service) by $\bar{N}$, and the average total waiting time (including the service time) by T.

We consider a randomly chosen packet just arriving to the queue in a batch of size k and “tag” this packet. Following the tagged packet through the queue, the packet encounters three delay sources.

• Waiting for packets already in the queue and in service. The average of the waiting time is equal to $\bar{N} /\mu$.

• Waiting for other packets of the batch that receive service before the tagged packet. The waiting time and its average are denoted by W and $\bar{W}$, respectively.

• Receiving service. The average of the service time is $1/\mu$.

Thus, the average of the total waiting time of the tagged packet is:

$$\begin{aligned} T=\bar{N} /\mu +\bar{W}+1/\mu . \end{aligned}$$

(19)

To determine the mean waiting time due to other packets in the batch (i.e., $\bar{W}$ in the above equation), we first suppose that the batch size is fixed and equal to k. As the tagged packet waits for half of ($k-1$) other packets of the batch on average, the mean waiting time for other packets of the batch is:

$$\begin{aligned} E\left[ {W|\hbox {batch size}={k}} \right]= & {} \left( {\frac{k-1}{2}} \right) \mathrm{\times mean\,service\,time}=\left( {\frac{k-1}{2}} \right) . \nonumber \\ \frac{1}{u}= & {} \frac{k-1}{2u} \end{aligned}$$

(20)

When the batch size is random, applying the total probability theorem we can write:

$$\begin{aligned} \bar{W} =\sum \nolimits _k E\left[ {W\hbox {|batch size}={k}} \right] \times {P}\left[ {\hbox {packet belongs to a batch of size}}\,k \right] . \end{aligned}$$

(21)

The probability that an individual packet belongs to a batch of size k is equal to the mass-weighted distribution of P(k) (the pmf of the batch size):

$$\begin{aligned} {P}\left[ {\hbox {packet belongs to a batch of size} \,\, k} \right] =\frac{{kP}\left( {k} \right) }{\bar{{k}}}, \end{aligned}$$

(22)

where $\bar{{k}}$ is the average of batch size. Therefore:

$$\begin{aligned} \bar{W}= & {} \sum _k \left( {\frac{k-1}{2u}} \right) \left( {\frac{{kP}\left( {k} \right) }{\bar{{k}}}} \right) =\frac{1}{2u\bar{{k}}}\sum \nolimits _k k\left( {k-1} \right) P\left( k \right) \nonumber \\= & {} \frac{\sum \nolimits _k k\left( {k-1} \right) P\left( k \right) }{2\mu \sum \nolimits _k kP\left( k \right) }. \end{aligned}$$

(23)

When batches arrive at rate $\lambda$, the total arrival rate of packets is $\bar{k} \lambda$. Thus, according to Little’s law:

$$\begin{aligned} \bar{N} =\left( {\bar{k} \lambda } \right) T. \end{aligned}$$

(24)

Replacing Eqs. (23) and (24) into Eq. (19) and solving for T yields:

$$\begin{aligned} T=\frac{\sum \nolimits _k k\left( {k+1} \right) P\left( k \right) }{2\left( {\sum \nolimits _k kP\left( k \right) } \right) \left( {\mu -\lambda \sum \nolimits _k kP\left( k \right) } \right) }=\frac{E\left[ {k^{2}} \right] +E\left[ k \right] }{2E\left[ k \right] \left( {u-\lambda E\left[ k \right] } \right) } \end{aligned}$$