Adaptive wireless communications under competition and jamming in energy constrained networks

Abstract

We propose a framed slotted Aloha-based adaptive method for robust communication between autonomous wireless nodes competing to access a channel under unknown network conditions such as adversarial disruptions. With energy as a scarce resource, we show that in order to disrupt communications, our method forces the reactive adversary to incur higher energy cost relative to a legitimate node. Consequently, the adversary depletes its energy resources and stops attacking the network. Using the proposed method, a transmitter node changes the number of selected time slots and the access probability in each selected time slot based on the number of unsuccessful transmissions of a data packet. On the receiver side, a receiver node changes the probability of listening in a time slot based on the number of unsuccessful communication attempts of a packet. We compare the proposed method with two other framed slotted Aloha-based methods in terms of average energy consumption and average time required to communicate a packet. For performance evaluation, we consider scenarios in which: (1) Multiple nodes compete to access a channel. (2) Nodes compete in the presence of adversarial attacks. (3) Nodes compete in the presence of channel errors and capture effect.

Appendices

Appendix 1: Proof of Proposition 1

Proof

When $c\le T$, a TX and its intended RX operate in mode 1, in which the TX selects a time slot randomly (with uniform distribution), transmits a data packet in the selected time slot, while its intended RX listens sequentially. It is easy to see that for N competing TX/RX pairs the probability of successful communication for each TX/RX pair in mode 1 is $(1-1/s)^{N-1}$, and the expected number of frames $E[N_F]$ required to communicate a data packet successfully is

$$\begin{aligned} E[N_F]=(1-1/s)^{1-N}. \end{aligned}$$

(1)

This means that using the proposed method the average number of frames required to successfully communicate a data packet is not more than the threshold value $T=\lceil 1/e^{\frac{N}{s}}\rceil$, as $E[N_F]=(1-1/s)^{1-N}$ is less than or equal to the selected threshold value $T=\lceil 1/e^{\frac{N}{s}}\rceil$. For M packets this value is TM.

The expected energy cost for a TX and its intended RX to communicate a data packet successfully is not more than $(S+L)$ and $(\frac{(s+1)}{2}L+ S)$ respectively. The reason is as follows: the TX transmits and listens on average once per frame, on average the RX listens in not more than $\left( \frac{(s+1)}{2}\right)$ time slots and on average it transmits ACK once per frame. Hence, using the proposed method, the expected energy cost for a TX and its intended RX to communicate M packets successfully is not more than $TM(S+L)$ units and $TM(\frac{(s+1)}{2}L+ S)$ units respectively, this proves our claim. $\square$

Appendix 2: Proof of Proposition 2

Proof

In a frame, the probability of a successful communication of a data packet in the presence of a sequential jammer ($J_s$) is

$$\begin{aligned} \begin{aligned} P[S_c\mid J_s]=&Pr\{{\text {TX }}i {\text { transmits in a time slot }}|{\text { TX}} i{\text { has transmitted once before in a time slot in that frame}}\}\\ {}&\times Pr\{{\text {RX }}i{\text { receives in that time slot}}\} \end{aligned} \end{aligned}$$

(2)

which is given by

$$\begin{aligned} \begin{aligned} P[S_c\mid J_s]= \left\{ \begin{array}{ll} 0,&{}{\text{if\,Mode 1}}\\ \sum \nolimits _{i=2}^{v_T}\left( \frac{1}{v_T}\right) ^i\left( \frac{1}{v_R}\right) \left( 1-\frac{1}{v_R}\right) ^{i-2}+\\ \left[ \sum \nolimits _{i=2}^{v_T}\left\{ \left( {\begin{array}{c} v_T \\ i \end{array}}\right) -1\right\} \left( \frac{1}{v_T}\right) ^i\left( 1-\frac{1}{v_T}\right) ^{v_T-i}\right] \left( \frac{1}{v_R}\right) \left( 1-\frac{1}{v_R}\right) ^{i-2},&{\text {otherwise}} \end{array}\right. \end{aligned} \end{aligned}$$

(3)

where $2\le v_T\le s$. For example, when in mode 1 a TX selects only one time slot in a frame and transmits in it with probability one. In this case $P[S_c\mid J_s]=0$, as the jammer by employing sequential jamming can jam this transmission with probability one. When the jammer employs the sequential jamming strategy, it sequentially listens in time slots until it detects a transmission, blocks the transmission and then waits for the next frame, so for $P[S_c\mid J_s] >0$, the TX needs to transmit more than once in a frame. In other words it can only successfully transmit in mode 2, where $v_T\ge 2$ and the TX selects two or more time slots in a frame and transmits in each of them with probability $1/v_T$. Suppose that there are $s=3$ time slots in every frame, in this case the TX and its intended RX can successfully communicate with $P[S_c\mid J_s] >0$, when $v_T=2$ or when $v_T=3$. For $v_T=2$, the TX selects two time slots in a frame and it transmits in each of them with probability $1/v_T$. The TX will be unsuccessful in the first transmission and can be successful in the second transmission if the RX listens in that time slot. In this case the probability of success obtained from Eq. (3) is: $(\frac{1}{v_T})(\frac{1}{v_T})(\frac{1}{v_R})$. For higher values of s and $v_T$, Eq (3) simply calculates the probability that in how many ways the TX and the RX can successfully communicate given that the TX selects $v_T$ time slots in a frame and transmits in each of them with probability $1/v_T$ while the RX listens in every time slot with probability $1/v_R$.

In a frame, the probability of a successfull communication of a data packet in the presence of an arbitrary adversary (when the adversary picks $s_J > 0$ out of s time slots in a frame) is given by

$$\begin{aligned} \begin{aligned} P[S_c\mid J_a]= \left\{ \begin{array}{ll} 0,&{}{\text {if }}\,s_J=s\\ \sum \nolimits _{i=1}^{v_T-s_J}\left\{ P_{s,i}\prod _{j=1}^{i-1}(1-P_{s,j})\right\} , &{}{\text {if }}\,v_T=s, 0\le s_J<s\\ \sum \nolimits _{x=0}^{{\text {min}} \{v_T, s_J\} }\left[ \frac{\left( {\begin{array}{c} v_T \\ x \end{array}}\right) \left( {\begin{array}{c} s-v_T \\ s_J-x \end{array}}\right) }{\left( {\begin{array}{c} s \\ s_J \end{array}}\right) }\left\{ \sum \nolimits _{i=1}^{v_T-x}P_{s,i}\prod _{j=1}^{i-1}(1-P_{s,j})\right\} \right] ,&{\text {otherwise}} \end{array}\right. \end{aligned} \end{aligned}$$

(4)

where $P_{s,i}$ is an element of the s-length vector $\mathbf {P}_s$. The vector $\mathbf {P}_s$ is given by $\mathbf {P}_s=[P_{s,1}, P_{s,2},\ldots , P_{s,s}]=[\frac{1}{v_T}\frac{1}{v_R}, \frac{1}{v_T}\frac{1}{v_R}, \ldots , \frac{1}{v_T}\frac{1}{v_R}].$

For example, in the frames where the adversary select $s_J=s$ slots for attack, i.e., it selects every slot in the frame, the probability of success is $P[S_c\mid J_a]=0$. When the adversary randomly selects $s_J < s$ time slots in a frame for attack, a TX and its intended RX may successfully communicate in those time slots that are not selected by the adversary. To calculate the probability of success, we need to find the following: (1) The probability that x out of $v_T$ time slots are selected by the adversary, where $x=0, 1, \ldots , v_T$. The probability of this event happening is given as: $(\frac{\left( {\begin{array}{c} v_T \\ x \end{array}}\right) \left( {\begin{array}{c} s-v_T \\ s_J-x \end{array}}\right) }{\left( {\begin{array}{c} s \\ s_J \end{array}}\right) });$ (2) In the $v_T-x$ remaining time slots which are not selected by the adversary, the probability that the TX will transmit and the RX will listen in the same slot is given as: $\sum _{i=1}^{v_T-x}P_{s,i}\prod _{j=1}^{i-1}(1-P_{s,j})$.

Let $f_j$ represent a frame in which both $v_T>1$ and $v_R>1$ when the adversary employ sequential jamming attack, or let $f_j$ represent a frame where $J_s<s$ when the adversary employ arbitrary jamming attack. Let $P_j$ be the maximum probability that the packet will not be successfully communicated in a frame $f_j$. Note that for the considered scenario under sequential jamming attack, or for a given $s_J$ under arbitrary jamming attack, maximum probability of unsuccessful communication $P_j$ occurs when both $v_T=s$ and $v_R=s$. As with increasing $v_T$ the TX in each of the $v_T$ selected time slots, transmits with probability 1 / s, whereas with increasing value of $v_R$, the RX decreases the probability of listening in a time slot. Hence, $v_T=v_R=s$ corresponds to the situation in which the probability of successful communication is minimum in a frame. This probability can be calculated using Eq. 3 for sequential jamming scenario and using Eq. 4 for arbitrary jamming scenario. The probability of not being successfully communicated in n of such frames is less than or equal to $(P_j)^{n}$, in 2n frames is less than or equal to $P_j^{2n}$, etc. Since $P_j<1$, these probabilities tend to zero. Hence, $\lim _{n \rightarrow \infty } P_j^n=0$, which proves our claim. $\square$

Appendix 3: Proof of Proposition 3

Proof

In the presence of a sequential reactive adversary, for each data packet, the probability of successful communication of a packet is zero in the first T frames where $c\le T$. When $c>T$, it is easy to see (from Eq. 3) that for the proposed method, the probability that a data packet is successfully communicated in any frame is minimum when $v_T$ and $v_R$ reach their maximum value of s. As with increasing $v_T$ a TX in each of the $v_T$ selected time slots, transmits with probability 1 / s, whereas with increasing value of $v_R$, an RX decreases the probability of listening in a time slot. Hence, $v_T=v_R=s$ corresponds to the situation in which the probability of successful communication is minimum in a frame. Due to this reason, $P[S_c \mid J_s]$ in a frame is at least $P[S_c \mid J_s, v_T=v_R=s]$ and hence $E[N_F\mid J_s]<M(\frac{1}{P[S_c\mid J_s, v_T=v_R=s]}+T)$.

In the presence of an arbitrary reactive jammer, for a given $s_J$ time slots used for attack in a frame, the conditional probability of successful communication is at least $P[S_c \mid J_a, v_T=v_R=s]= \frac{1}{s^2}\sum _{i=1}^{s-s_J}(1-\frac{1}{s^2})^{(i-1)}$. The conditional probability of success $P[S_c \mid J_a, v_T=v_R=s]$ corresponds to the worst case situation when $v_T$ and $v_R$ reach their maximum value of s. Hence for successful communication of the M data packets, $E[N_F\mid J_a] <\frac{M}{P[S_c \mid J_a, v_T=v_R=s]}$. $\square$

Appendix 4: Proof of Proposition 4

Proof

Using the proposed method, a TX transmits and listens once per frame in expectation. Therefore, the average energy cost of the TX to successfully communicate M packets is $ME[N_F](S+L)$, where $E[N_F]$ is the average number of frames to successfully communicate a packet.

The arbitrary jammer that selects at least half of the time slots in a frame for attack has expected energy cost of at least $(\frac{s}{2})L+S$ per frame. Hence, the expected cost to communicate M packets successfully for the TX is at least $ME[N_F](\frac{s}{2}-1)L$ less than the jammer.

When the adversary employs sequential jamming attacks, we first show the cost incurred in the frames where $c \le T$. The expected cost for the sequential adversary in these frames is $\left( \frac{(s+1)}{2}L+S\right)$. This is due to the reason that in each frame the TX selects a time slot and transmits in it, while the adversary sequentially listens in time slots until it detects a transmission and blocks it.

In the frames where $c>T$, the TX instead of selecting a single time slot it changes the number of selected time slots $\nu _T$, where $1< \nu _T\le s$. The expected cost per frame for the sequential adversary in this case is given by

$$\begin{aligned} E[C_{J_s}]= \sum _{L=1}^{s}(L+S)a_L + (1-\sum _{L=1}^sa_L)L, \end{aligned}$$

(5)

where

$$\begin{aligned} a_L= & {} \sum _{k=1}^{v_T} W^kT_L^k,\\ W^k= & {} \left( 1-\frac{1}{v_T}\right) ^{k-1}\frac{1}{v_T} \end{aligned}$$

and

$$\begin{aligned} T_L^k=\left\{ \begin{array}{ll} \frac{\left( {\begin{array}{c} L-1 \\ k-1\end{array}}\right) \left( {\begin{array}{c}s-L \\ v_T-k\end{array}}\right) }{\left( {\begin{array}{c}s\\ v_T\end{array}}\right) },&{} {\text {if }} k \le L\le s-v_T+k \\ 0, &{} {\text {otherwise.}} \end{array}\right. \end{aligned}$$

It can be calculated using Eq. 5 that for any $\nu _T> 1$, the average cost of sequential jammer is greater than or equal to $\left( \frac{(s+1)}{2}L+S\right)$. The reason is as follows: For $\nu _T>1$, when the TX selects $\nu _T$ time slots and transmits in each of these time slots with probability $1/\nu _T$, the adversary now needs to listen (on average) in more than $\left( \frac{s+1}{2}\right)$ time slots to detect the transmission. Moreover, there is also now a possibility that the TX does not transmit in any time slots, in this case, the adversary incurs the maximum cost of sL in that frame. Hence the cost to adversary is at least $\left( \frac{(s+1)}{2}L+S\right)$. Due to this reason, the expected cost to communicate M packets successfully for the TX is at least $ME[N_F](\frac{s}{2}-1)L$ less than the sequential jammer.

For the RX, when $c\le T$ then $v_R=1$ and the expected energy cost to the RX is $E[C_{RX}\mid c\le T] =\left( \frac{(s+1)}{2}L+S\right)$, i.e., the same cost as incurred by the adversary. However, when $c>T$, the RX instead of listening with probability one in a time slot, listens with probability $1/\nu _R$, where $1< \nu _R\le s$. Hence the cost of the RX per frame decreases with increasing $v_R$. When $v_R=2$, the expected cost of the RX in a frame is $\left( \frac{s}{2}L+S\right)$ and so on. Finally, when $v_R=s$, the expected cost for the RX in each of the remaining frames is $(S+L)$. It can be seen that the costs for the RX (when $v_R>1$) are less than the costs incurred by the adversary. For $v_R>2$, the total expected energy cost to the RX is given as

$$\begin{aligned} E[C_{RX}\mid v_R>2]=\sum _{v=3}^s\frac{s}{v}L+(s-2)S. \end{aligned}$$

(6)

Clearly, the expected cost in the frames where $v_R>2$ is upper bounded by $(s-2)(S+L\ln s)$, i.e.,

$$\begin{aligned} E[C_{RX}\mid v_R>2]< (s-2)(S+L\ln s) \end{aligned}$$

whereas the expected cost of the jammer in these frames is at least $(s-2)(\frac{s}{2}L+S)$, as explained above. Due to this reason, the expected cost to communicate M packets successfully for the RX is at least $ME[N_F](\frac{s}{2}- \ln s)L$ less than the adversary, which proves our claim. $\square$