Statistical reliability for energy efficient data transport in wireless sensor networks

Abstract

Typical wireless sensor network deployments are expected to be in unattended terrains where link packet error rate may be as high as 70% and path length could be up to tens of hops. In coping with such harsh conditions, we introduce a new notion of statistical reliability to achieve a balance between data reliability and energy consumption. Under this new paradigm, the energy efficiency of a comprehensive set of statistically reliable data delivery protocols are analyzed. Based on the insight gained, we propose a hybrid system which combines the energy efficient and statistically reliable transport (eESRT) protocol with the implicit and explicit ARQ (ieARQ) protocol. This hybrid system adaptively switches between eESRT and ieARQ machanisms according to a dynamic hop threshold H_sw proposed in this work. Simulation and experiment results confirm our theoretical findings and demonstrate the advantages the hybrid system in boosting energy efficiency, reducing end to end delay, and in overcoming the “avalanche” effect.

Appendix

1.1 Proof of Proposition 1

Suppose that each sensed data has N backups. By the time independency assumption, the number of sensed data successfully delivered across h link hops having failure probabilities of p = {p _i }, X _h,p, is binomially distributed with N and success probability of $\prod_{i=0}^{h-1}\bar{p}_i.$ Thus,

$$ \beta(N,{\bf p})\mathop{=}\limits^{def} P\left(X_{h,{\bf p}} \geq 1\right) = 1- \left(1 - \prod_{i=0}^{h-1}\bar{p}_i\right)^N. $$

(10)

For statistical reliability at level β, L _ESRT(β) is the minimum integer satisfying β(N, p) ≥ β, which implies the first part of the Proposition.

The energy consumption, E _ESRT, is determined by the convolution of successful transmissions along the h hops, given that L _ESRT(β) backups for each sensed data are transmitted from the source sensor.

Let K(i) be the number of successful transmissions in link hop i = 0, ..., h − 1. Clearly, K(0) = L _ESRT(β) and by the Bayesian rule

$$ E[K(i)] = E\left[E[K(i) | K(i-1)] \right] = \bar{p}_{i-1} E[K(i-1)]. $$

Then, by recursion

$$ E[K(i)] = L_{\rm ESRT}(\beta) \prod_{k=0}^{i-1}\bar{p}_{k}, i = 1,\ldots,h-1. $$

The second part of the Proposition follows by noticing that the successful transmissions at hop i determines the number of transmissions at the next hop i + 1.$\square$

1.2 Proof of Proposition 2

A sensed data is received by the sink node successfully with probability $\prod_{i=0}^{h-1}\bar{p}_i,$ regardless of the ACK outcome. If an ACK is not received at the source within a predetermined timeout, the sensed data is retransmitted. Given a maximum number of transmissions per sensed data, N, the sensed data is delivered successfully with probability $1- \left(1- \prod_{i=0}^{h-1}\bar{p}_i\right)^N.$

For statistical reliability at level β, the minimum N, L ^e2e_ARQ (β), is given by the smallest integer satisfying

$$ 1- \left(1- \prod_{i=0}^{h-1}\bar{p}_i\right)^N \geq \beta, $$

(11)

which is resolved by L ^e2e_ARQ (β) of the Proposition.

For evaluating E ^e2e_ARQ note that unlike successful delivery, retransmissions of a sensed data is stopped only if it reaches the sink and its respective ACK reaches the source. This event occurs with probability $\prod_{i=0}^{h-1} \bar{p}_i \bar{q}_i.$ Otherwise, the source sensor retransmits the sensed data.

Let K ₂ be the number of transmissions (sensed data and ACKs) using E2E ARQ with L ^e2e_ARQ (β). Also, let X _i and Y _i be the number of transmissions of a single sensed data and an ACK made by and to node i along the route, respectively, i = 0, ..., h.

The expected number of transmissions per each measurement is given by

$$ E_{\rm ARQ}^{e2e} = E[K_2] = \sum_{i=0}^{h-1} \left(E[X_i] + E[Y_{i}]\right). $$

(12)

Note that X ₀ is a truncated geometrically distributed r.v. with a success probability of $\prod_{i=0}^{h-1} \bar{p}_i \bar{q}_i$ taking values in the set {1, ..., L ^e2e_ARQ (β)}. Its expected value is given by:

$$ \begin{aligned} E[X_{0}] &= L_{\rm ARQ}^{e2e}(\beta)\left(1-\prod_{i=0}^{h-1} \bar{p}_i \bar{q}_i \right)^{L_{\rm ARQ}^{e2e}(\beta)-1} \\ &+ {\sum_{k=1}^{L_{\rm ARQ}^{e2e}(\beta) -1}} k \left(\prod_{i=0}^{h-1} \bar{p}_i \bar{q}_i \right) \left(1-\prod_{i=0}^{h-1} \bar{p}_i \bar{q}_i \right)^{k-1}\\ &= {\frac{1- \left(1-\prod_{i=0}^{h-1} \bar{p}_i \bar{q}_i \right)^{L_{\rm ARQ}^{e2e}(\beta)}}{\prod_{i=0}^{h-1} \bar{p}_i \bar{q}_i}}. \end{aligned} $$

(13)

The expected number of sensed data packets from any realization of X ₀ successfully forwarded to node 1 is $\bar{p}_0.$ Thus, $E[X_1] = \bar{p}_0\,E[X_0].$ Similarly for every subsequent node along the forward route to the sink. At every subsequent hop i, the expected number of transmissions is reduced by a factor of $\bar{p}_{i-1}.$ In the backward route of the ACK, a similar expected reduction occurs. The second part of the Proposition follows from (12) and (13). $\square$

1.3 Proof of Proposition 3

The derivation of L _ARQ(i, β) is similar to the previous derivations, with the difference that here we further restrict the requirement for L _ARQ(i, β) to satisfy $(1 - p_i^{L_{\rm ARQ}(i,\beta)}) \geq \beta^{1/h}.$

Each hop component of E ₃ is derived as a special case of SW E2E ARQ with h = 1. $\square$

1.4 Proof of Proposition 4

As with explicit SW HBH ARQ (HBH eACK ARQ), a sensed data is forwarded to the next hop if it has been successfully received regardless of the implicit ACK outcome. Therefor, L _oiARQ(i, β) = L _ARQ(i, β) for every i.

For E _oiARQ, note that a forward packet from node i successfully received by the next node i + 1, may not be overheard by node i triggering a retransmission. Such events are accounted for by the spatial dependency defined above by r _i  = Pr[ success at i − 1 |success at i + 1] = Pr[success at i + 1 |success at i − 1].

Let X _i be the number of transmissions made by node i, 0 ≤ i ≤ h, for a single sensed data packet. For i = 0, the source node transmits until the sensed data and its forwarding transmission are both received at node i = 1 and i = 0, respectively, but no more than L _oiARQ(i, β). The probability of this event is $\bar{p}_0 \bar{q}_0$ and by the truncated geometric distribution its expected value is given by

$$ E[X_0] = {\frac{1-(1- \bar{p}_0 \bar{q}_0)^{L_{\rm oiARQ}(0,\beta)}}{\bar{p}_0 \bar{q}_0}}. $$

(14)

For 0 < i < h, assuming proper setting of the timeouts, the transmitter node, i, transmits until the sensed data is successfully received by both, node i − 1 and node i + 1, as well as the forwarding by node i + 1 is ‘overheard’ by node i, but no more than L _oiARQ(i, β). The probability of this event is $\bar{p}_i \bar{q}_i r_i$ and by the truncated geometric distribution its expected value is given by

$$ E[X_i] = {\frac{1-(1- \bar{p}_i \bar{q}_i r_i)^{L_{\rm oiARQ}(i,\beta)}} {\bar{p}_i \bar{q}_i r_i}}. $$

(15)

The sink node, i = h, needs to transmit an explicit ACK. As with SW HBH ARQ, the expected number of these ACKs is

$$ E[X_h] = {\frac{1 - (1- \bar{p}_{h-1} \bar{q}_{h-1})^{L_{\rm oiARQ}({h-1},\beta)}}{\bar{p}_{h-1} \bar{q}_{h-1}}} \times \bar{p}_i. $$

(16)

Combining (14–16) yields the second part of the Proposition. $\square$