Energy efficient multi-beacon guard method for periodic data gathering in time-synchronized WSN

Abstract

In periodic data-gathering, sensors can switch on the transceiver only during packet transmission to save energy. Exact clock-synchronization is challenging to achieve because of error present in synchronization protocols. Clock-disagreement increases with time in the absence of synchronization. This paper proposes an Energy-efficient Multi Beacon Guard method (EMBG) to decrease the energy consumption by minimizing the awake time of sender and receiver by periodically switching on and off the receivers during the guard-time. We determine the optimal number-of-times the receiver needs to wake up along with the wake-up intervals to collectively minimize the total energy consumption of the sender–receiver pair during transmission. This paper shows the effectiveness of EMBG in energy conservation and compares with existing approaches using ns2 simulation.

Appendix

Lemma 1

Assume the actual wake-up time of sender follows normal distribution and the guard-time is divided into $N_r$ intervals. The expected number of time the receiver wakes up for a successful packet transmission is $\frac{N_r+1}{2}$.

Proof

The receiver wakes up, sends the beacon message, and waits for round trip time (RTT) to receive data from the sender, at times $t_i, 1 \le i \le N_r$. The receiver has to wake up $i+1$ times if the sender wakes up in the interval $(t_i,t_{i+1}]$, assuming propagation delay is negligible. In other words, the expected number of times the receiver wakes up can be estimated from the probability of the sender wakes up within the time interval $(t_i,t_{i+1}]$, times the number of times the receiver wakes up. Assume, $P(X=x:t_i>x \le t_{i+1})$ denotes the probability of the sender wakes up within the $i^{th}$ interval, and for such a case the receiver needs to wake up $i+1$ times before a successful transmission. Moreover, let $P(X=x:x\le t_{1})$ denotes the probability of the sender wakes up before the first interval and for such a case the receiver sends one beacon message before a successful transmission. Hence, the expected number-of-times the receiver wakes up is,

$$\begin{aligned}&P(X=x:x\le t_{1}) \times 1 \\&\qquad + \sum \limits _{i = 1}^{{N_r-1}} {P(X = x:{t_i} > x \le {t_{i + 1}}) \times (i + 1)} \\&\quad =\frac{N_r + 1}{2}. \end{aligned}$$

$\square$

Lemma 2

Assume the actual time of the sender follows normal the distribution within the interval $[\tau _p - T_g,\tau _p + T_g]$, where $\tau _p$, $T_g$ respectively denotes the scheduled wake-up time of the sender and the half of the guard-time. The expected waiting time of a sender for a successful packet transmission is $\frac{T_g}{N_r}$.

Proof

If the area under the normal curve is equally divided into $N_{r}$ regions, then the expected waiting time of the sender is,

$$\begin{aligned} W_s&= \int _{-T_g}^{t_{1}} P(X=x)( t_{1} - x )dx \\&\quad + \sum \limits _{i = 1}^{{N_r-1}} {\int _{{t_i}}^{{t_{i + 1}}} P (X = x)({t_{i + 1}} - x)dx}. \end{aligned}$$

If $N_r$ is odd, we can rewrite the equation as,

$$\begin{aligned} W_s&=\int _{ - T_g}^{{t_1}} P (X = x)({t_1} - x)dx \nonumber \\&\quad + \int _{ {t_{{N_{r - 1}}}}}^{{t_{{N_r}}}} P (X = x)({t_1} - x)dx \nonumber \\&\quad + \sum \limits _{i = 1}^{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor - 1} {\int _{{t_i}}^{{t_{i + 1}}} P (X = x)({t_{i + 1}} - x)dx}\nonumber \\&\quad + \sum \limits _{i = 1}^{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor - 1} {\int _{{t_{{N_r} - (i + 1)}}}^{{t_{{N_r} - i}}} P (X = x)({t_{{N_r} - (i)}} - x)dx} \nonumber \\&\quad + \int _{{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor }}}^{{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor + 1}}} P (X = x)({t_{_{\left\lfloor {{N_{r/2}}} \right\rfloor + 1}}} - x)dx. \end{aligned}$$

(36)

In order to prove this lemma, we first show that $\int _{{t_i}}^{{t_{i + 1}}} P (X = x)({t_{i + 1}} - x)dx + \int _{{t_{{N_{r} - (i + 1)}}}}^{{t_{{N_r} - i}}} P (X = x)({t_{{N_r} - i}} - x)dx = \frac{1}{{{N_r}}}\left| {{t_i} - {t_{i + 1}}} \right|.$

$$\begin{aligned}&\int _{{t_i}}^{{t_{i + 1}}} P (X = x)({t_{i + 1}} - x)dx + \int _{{t_{{N_{r} - (i + 1)}}}}^{{t_{{N_r} - i}}} P (X = x)({t_{{N_r} - i}} - x)dx \\&\quad = \int \limits _{{t_i}}^{{t_{i + 1}}} {P(X = x)({t_{i + 1}})dx} - \frac{{ - \sigma }}{{\sqrt{2\pi } }}\left( {{e^{\frac{{ - {t_{i + 1}}^2}}{{2{\sigma ^2}}}}} - {e^{\frac{{ - {t_i}^2}}{{2{\sigma ^2}}}}}} \right) \\&\qquad + \int \limits _{{t_{{N_{r} - (i + 1)}}}}^{{t_{{N_{r} - i}}}} {P(X = x)({t_{{N_{r} - i}}})dx} - \frac{{ - \sigma }}{{\sqrt{2\pi } }}\left( {{e^{\frac{{ - {t_{{N_{r - i}}}}^2}}{{2{\sigma ^2}}}}} - {e^{\frac{{ - {t_{{N_{r} - (i + 1)}}}^2}}{{2{\sigma ^2}}}}}} \right) , \\& \text {by substituting } P(X=x)=\frac{1}{{{\sigma _p}\sqrt{2\pi } }}{e^{\frac{{ - {x^2}}}{{2\sigma _p^2}}}} \end{aligned}$$

Since the function is symmetric on both side of mean, and $\left\| { t_{i} - {t_{i+1}}} \right\| = \left\| {t_{{N_{r} - (i + 1)}}} - {t_{{N_{r} - (i)}}} \right\|$, $\left| {{t_{{N_r} - i}}} \right| = \left| {{t_i}} \right|$ for $i \le \left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor - 1$. The above equation can be rewritten as,

$$\begin{aligned}&{t_{i + 1}}\int _{{t_i}}^{{t_{i + 1}}} P (X = x)dx + {t_{{N_r} - i}}\int _{{t_{{N_{r} - (i + 1)}}}}^{{t_{{N_r} - i}}} P (X = x)dx \nonumber \\&\quad = {t_{i + 1}}\frac{1}{{{N_r}}} + {t_{{N_r} -i}}\frac{1}{{{N_r}}}, \,\,\, \text {by substituting }P(X=x)=\frac{1}{{{\sigma _p}\sqrt{2\pi } }}{e^{\frac{{ - {x^2}}}{{2\sigma _p^2}}}} \nonumber \\&\quad = \frac{1}{{{N_r}}}\left| {{t_{i + 1}} - {t_i}} \right| . \end{aligned}$$

(37)

Since the midpoint of the middle interval $[{t_{\left| {\frac{{{N_r}}}{2}} \right| }},{t_{\left| {\frac{{{N_r}}}{2}} \right| + 1}}]$ is mean of PDF, we can write $\int _{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor }}^{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor +1}} P(X=x)( t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor +1} - x )dx$ of Eq. 36 as,

$$\begin{aligned}&\int _{{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } }}^0 P (X = x)({t_{{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } + 1}})dx - \int _{{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } }}^0 P (X = x)(x)dx \\&\quad + \int _0^{{t_{{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } + 1}}} P (X = x)({t_{{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } + 1}})dx - \int _0^{{t_{{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } + 1}}} P (X = x)(x)dx\\&\quad ={t_{{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } + 1}}\int \limits _{{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor }}}^0 {P(X = x)dx} - \frac{{ - \sigma }}{{\sqrt{2\pi } }}\left( {1 -{e^{\frac{{ - {t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor }}^2}}{{2{\sigma ^2}}}}}} \right) \\&\qquad + {t_{{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } + 1}}\int \limits _0^{{t_{{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } + 1}}} {P(X = x)dx} - \frac{{ - \sigma }}{{\sqrt{2\pi } }}\left( {{e^{\frac{{ - {{({t_{{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } + 1}})}^2}}}{{2{\sigma ^2}}}}} - 1} \right) , \\& \text {by substituting }P(X=x)=\frac{1}{{{\sigma _p}\sqrt{2\pi } }}{e^{\frac{{ - {x^2}}}{{2\sigma _p^2}}}} \end{aligned}$$

As we know that the function is symmetric in both side of the mean, hence $\left| {{t_{{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } + 1}}} \right| = \left| {{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor }}} \right|$ and the above expression becomes,

$$\begin{aligned}&\int _{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor }}^{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor +1}} P(X=x)( t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor +1} - x )dx \nonumber \\&\quad = {t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor + 1}}\int _{{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor }}}^0 P (X = x)dx + {t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor + 1}}\int _0^{{t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor + 1}}} P (X = x)dx, \nonumber \\&\quad =t_{{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor } +1}\times \frac{1}{N_r}. \end{aligned}$$

(38)

Note that $t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor +1}$ can be written as $\left| {0 - {t_{\left\lfloor {\frac{{{N_r}}}{2}} \right\rfloor }}} \right|$. Similarly, $\int _{ - T_g}^{{t_1}} P (X = x)({t_1} - x)dx + \int _{ {t_{{n_{r - 1}}}}}^{{t_{{N_r}}}} P (X = x)({t_1} - x)dx = \frac{1}{{{N_r}}}\left| {{t_1} - T_g} \right|.$ Hence Eq. 36 can be written as,

$$\begin{aligned} W_s&= \frac{1}{{{N_r}}}\left| {{t_1} - T_g} \right| + \frac{1}{{{N_r}}}\left( {\sum \limits _{i = 1}^{\left\lfloor {\frac{{{N_r}}}{2} - 1} \right\rfloor } {\left| {{t_i} - {t_{i + 1}}} \right| } } \right) \\&\quad + \frac{1}{{{N_r}}}\left| {{t_{\left| {\frac{{{N_r}}}{2}} \right| }} - 0} \right| , \quad \text {by substituting Eqs.~}37, 38. \\&= \frac{T_g}{{{N_r}}}. \end{aligned}$$

Similarly, it can be proved for $N_r$ is even. $\square$

Lemma 3

The expression for the expected energy consumption in the multi-beacon approach given in Eq. 9 is convex with respect to $N_r$ and the optimal value occurs at

$$\begin{aligned} \hat{N_r}= \sqrt{\frac{2P_{idle} \times T_g}{E_{sw}+E_{txbcn}+ P_{idle}T_{rtt}}} \end{aligned}$$

(39)

where $N_r,P_{idle},T_g,E_{sw},$ $E_{txbcn},T_{rtt}$ denote the number-of-times the receiver wakes up, power required for idle listening, guard time, transition energy required from sleep to awake state, energy required to transmit a beacon and average round trip time respectively.

Proof

The total expected energy consumption in multi-beacon approach ($E_{mb}$) is

$$\begin{aligned} E_{mb}&= ( E_{sw}+E_{txbcn}+ P_{idle}T_{rtt} )\frac{\hat{N_{r}}+1}{2}+ E_{rxdata} \nonumber \\&\quad + E_{txack} + E_{sw} + P_{idle}{\frac{T_g}{\hat{N_{r}}}} + E_{rxbcn} \nonumber \\&\quad +E_{txdata} + E_{rxack}. \end{aligned}$$

(40)

The function derived in Eq. 40 is convex because it is second order differentiable, $\frac{\mathrm {d} }{\mathrm {d} \hat{N_r}}(\frac{\mathrm {d} }{\mathrm {d} \hat{N_r}} {E_{mb}}(\hat{N_r})) = P_{idle}*\frac{1}{\hat{N_r}^3} > 0$ since $P_{idle},\hat{N_r} > 0$. We can find the optimal value of $E_{mb}$ occurs at,

$$\begin{aligned} \hat{N_r}= \sqrt{\frac{2P_{idle}(T_g)}{E_{sw}+E_{txbcn}+ P_{idle}T_{rtt}}}, \end{aligned}$$

(41)

from the equation $\frac{\mathrm {d} }{\mathrm {d} \hat{N_r}} ({E_{mb}}(\hat{N_r}))=0$. $\square$

Lemma 4

The expected energy consumption in guard-time approach is

$$\begin{aligned} E_{T_g}&= (E_{sw} + E_{txdata} + E_{rxack}) \nonumber \\&\quad + ( E_{sw}+P_{idle}\times T_g + E_{rxdata} +E_{txack}). \end{aligned}$$

(42)

Proof

In guard time approach, the receiver is awake during the whole guard-time and the sender sends the data packet as soon as it wakes up. As mentioned earlier wake up times of the sender follows the normal distribution within the guard time. If x denotes the instant when the sender wakes up, then the amount of time the receiver waits is $T_g+x$, where $-T_g \le x \le T_g$. Hence, the expected amount of time the receiver waits is

$$\begin{aligned} W_{r\_T_g}&= \int _{-T_g}^{T_g} P(X=x)( T_g+ x )dx, \\&= T_g \int \limits _{ - T_g}^{T_g} {\left( {\frac{1}{{\sigma \sqrt{2\pi } }}{e^{\frac{{ - {x^2}}}{{2{\sigma ^2}}}}}} \right) } \,dx + \int \limits _{ - T_g}^{T_g} {\left( {\frac{1}{{\sigma \sqrt{2\pi } }}{e^{\frac{{ - {x^2}}}{{2{\sigma ^2}}}}}} \right) } \,\,x\,\,dx, \\& \text {by substituting } P(X=x)=\frac{1}{{{\sigma _p}\sqrt{2\pi } }}{e^{\frac{{ - {x^2}}}{{2\sigma _p^2}}}} \\&= T_g. \end{aligned}$$

The receiver sends an acknowledgment after receiving the data packet from the sender and switches off the transceiver. Hence, the expected energy consumption of the receiver using the notations given in Table 1 is given as follows: $E_{r\_T_g} = E_{sw}+P_{idle}\times {(W_{r-T_g})} + E_{rxdata} +E_{txack}.$ The energy consumption of sender is $E_{s\_T_g} = E_{sw} + E_{txdata} + E_{rxack}$. Hence, the total expected energy consumption is

$$\begin{aligned} E_{T_g}&= (E_{sw} + E_{txdata} + E_{rxack}) \nonumber \\&\quad + ( E_{sw}+P_{idle}{(W_{r\_T_g})} + E_{rxdata} +E_{txack} ). \end{aligned}$$

(43)

$\square$