Segment-based packet combining: how to schedule a dense relayer cluster?

Abstract

The classical three-terminal relaying scenario can be generalized to an entire cluster of relayers, which in this paper is assumed to be densely deployed. We consider the combination of cluster-based relaying with a packet-combining technique where the user data of a packet is partitioned into contiguous segments that can be individually checked for correctness. In such a setup the question arises how to schedule the transmissions of the relayers to either minimize the average transmission costs until at least one (and then all) relayers have the full set of segments, or to maximize the per-segment diversity, i.e. the number of distinct relayers sending the same segment. We investigate different options for scheduling the relayer cluster and show that under certain assumptions, average-optimal and easily implementable schedules exist. We furthermore provide numerical evidence that the adoption of a segment-based approach gives performance benefits over the “classical” method which only considers correctness of whole packets.

Appendices

Appendix 1: Independence on transmission sequence for two transmissions

Suppose that we have three nodes with n ₁ < n, n ₂ < n and n ₃ < n randomly chosen segments (out of a total of n segments), respectively, node i has n _i segments. The segment sets of all three nodes are independent of each other. The first node remains quiet. We look at two sequences of events: in the first sequence node two transmits its n ₂ segments first and is followed by node three transmitting its n ₃ segments. In the second sequence node three transmits first and node two follows. Let X ₀ = n ₁ be the number of different segments node one has in its cache before any transmission, and let X ₁ = n ₁ + δ with n ₁ + δ ≤ n be the number of different segments node one has after the other two nodes have transmitted their data. More precisely, let

$$ T_{n_1,n_2,n_3,\delta} = \hbox{Pr}\left[X_1=n_1 + \delta\vert X_0=n_1;(n_2, n_3)\right] $$

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be the conditional probability that node one receives δ additional segments when it has already n ₁ segments and when first node two transmits, followed by node three. On the other hand, let

$$ T_{n_1,n_3,n_2,\delta} = \hbox{Pr} \left[X_1=n_1 + \delta \vert X_0=n_1;(n_3, n_2)\right] $$

be the analogous quantity for the scenario in which first node three transmits, followed by node two. We assume that all involved channels are perfect. Our goal is to show that

$$ T_{n_1,n_2,n_3,\delta} = T_{n_1,n_3,n_2,\delta} $$

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holds. We first observe that from the law of total probability and assuming independence between the transmissions of n ₂ and n ₃ we can express $T_{n_1,n_2,n_3,\delta}$ as:

$$ T_{n_1,n_2,n_3,\delta} = \sum_{\nu = {\rm max}\{0, \delta - n_3\}}^{{\rm min}\{\delta, n_2\}} T_{n_1, n_2, \nu} \cdot T_{n_1+\nu, n_3, \delta-\nu} $$

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(where $T_{n_1, n_2, \nu}$ is defined according to Eq. 3), where the summation is carried out over all possible increments after one step. We first consider the case where δ ≤ min{n ₂, n ₃} so that we get

$$ \begin{aligned} &T_{n_1,n_2,n_3,\delta} \\ &=\sum_{\nu =0}^{\delta} T_{n_1, n_2, \nu} \cdot T_{n_1+\nu, n_3, \delta-\nu} \\&= \sum_{\nu = 0}^{\delta} \frac{\left(\begin{array}{c}n_1 \\n_2- \nu\end{array}\right) {\left(\begin{array}{c}n-n_1 \\\nu\end{array}\right)}} {\left(\begin{array}{c}n \\n_2\end{array}\right)}\frac{\left(\begin{array}{c}n_1 + \nu \\ n_3 -\delta + \nu\end{array}\right){\left(\begin{array}{c}n-n_1-\nu \\\delta - \nu\end{array}\right)}}{\left(\begin{array}{c}n \\n_3\end{array}\right)}\\ &=\frac{1}{\left(\begin{array}{c}n \\n_2\end{array}\right){\left(\begin{array}{c}n \\n_3\end{array}\right)}} \\ & \left[\sum_{\nu = 0}^{\delta}{\left(\begin{array}{c}n_1 \\ n_2 -\nu\end{array}\right)}{\left(\begin{array}{c}n-n_1 \\\nu\end{array}\right)}{\left(\begin{array}{c}n_1 + \nu \\ n_3 -\delta + \nu\end{array}\right)}{\left(\begin{array}{c}n-n_1-\nu \\\delta - \nu\end{array}\right)}\right] \\ &=\frac{\left(\begin{array}{c}n-n_1 \\\delta\end{array}\right)}{\left(\begin{array}{c}n \\n_2\end{array}\right){\left(\begin{array}{c}n \\n_3\end{array}\right)}} \left[\sum_{\nu = 0}^{\delta}{\left(\begin{array}{c}n_1 \\ n_2 - \nu\end{array}\right)}{\left(\begin{array}{c}n_1 + \nu \\ n_3 - (\delta -\nu)\end{array}\right)}{\left(\begin{array}{c}\delta \\\nu\end{array}\right)}\right] \\ \end{aligned} $$

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where in the last equation we have used the easily verifiable identity:

$$ {\left(\begin{array}{c}n-n_1 \\ \nu\end{array}\right)}{\left(\begin{array}{c}n - n_1 - \nu \\ \delta - \nu\end{array}\right)} = {\left(\begin{array}{c}n-n_1\\ \delta\end{array}\right)}{\left(\begin{array}{c}\delta \\ \nu\end{array}\right)} $$

To show $T_{n_1,n_2,n_3,\delta} =T_{n_1,n_3,n_2,\delta}$ it therefore suffices to show that

$$ \begin{aligned} 0 &= \sum_{\nu = 0}^{\delta} {\left(\begin{array}{c}\delta \\ \nu\end{array}\right)} {\left(\begin{array}{c}n_1 \\ n_2 - \nu\end{array}\right)}{\left(\begin{array}{c}n_1 + \nu \\ n_3 - (\delta - \nu)\end{array}\right)}\\ & -\sum_{\nu = 0}^{\delta} {\left(\begin{array}{c}\delta \\ \nu\end{array}\right)} {\left(\begin{array}{c}n_1 \\ n_3 - \nu\end{array}\right)} {\left(\begin{array}{c}n_1 + \nu \\ n_2 - (\delta -\nu)\end{array}\right)} \end{aligned} $$

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holds. To show this, we use the identity

$$ {\left(\begin{array}{c}n \\ k - m\end{array}\right)} = \sum_{j=0}^m {\left(\begin{array}{c}n+m-j \\ k\end{array}\right)} {\left(\begin{array}{c}m \\ j\end{array}\right)} (-1)^j $$

which is easily seen by induction for m ≤ k. If we use this identity in Eq. 19 we need to consider the expression:

$$ \begin{aligned} & \left[ \sum_{\nu=0}^\delta {\left(\begin{array}{c}\delta \\ \nu\end{array}\right)}\left( \sum_{j_1=0}^\nu {\left(\begin{array}{c}n_1 + \nu - j_1 \\ n_2\end{array}\right)}{\left(\begin{array}{c}\nu \\ j_1\end{array}\right)} (-1)^{j_1} \right) \right. \cdot\\ & \left. \left( \sum_{j_2=0}^{\delta - \nu} {\left(\begin{array}{c}n_1 + \delta - j_2 \\ n_3\end{array}\right)}{\left(\begin{array}{c}\delta - \nu \\ j_2\end{array}\right)}(-1)^{j_2} \right) \right] \\ &- \left[\sum_{\nu=0}^\delta {\left(\begin{array}{c}\delta \\ \nu\end{array}\right)}\left( \sum_{j_3=0}^\nu {\left(\begin{array}{c}n_1 + \nu - j_3 \\ n_3\end{array}\right)}{\left(\begin{array}{c}\nu \\ j_3\end{array}\right)} (-1)^{j_3} \right) \right. \cdot\\ & \left. \left( \sum_{j_4=0}^{\delta - \nu} {\left(\begin{array}{c}n_1 + \delta - j_4 \\ n_2\end{array}\right)}{\left(\begin{array}{c}\delta - \nu \\ j_4\end{array}\right)}(-1)^{j_4} \right) \right] \end{aligned} $$

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This expression can be rewritten as a sum of the form

$$ \sum_{x,y=0}^\delta a_{x,y} {\left(\begin{array}{c}n_1 + x \\ n_2\end{array}\right)} {\left(\begin{array}{c}n_1 + y \\ n_3\end{array}\right)} $$

with to-be-determined coefficients a _x,y . We will show that a _x,y  = 0 holds for all $x, y \in \{0, 1, {\ldots}, \delta\}$. From inspecting Eq. 20 it is clear that for x < y only the upper sum contributes to a _x,y , for x > y only the lower sum and for x = y both sums contribute to a _x,y . Because of symmetry, it suffices to consider the cases x < y and x = y:

Case 1: We assume x = y. In this case it can be seen that j ₁ = j ₃ = 0 and j ₂ = j ₄ = δ − x and we get

$$ \begin{aligned} a_{x,x} &= {\left(\begin{array}{c}n_1 + x \\ n_2\end{array}\right)} {\left(\begin{array}{c}n_1 + x \\ n_3\end{array}\right)} \cdot \\ & \left[{\left(\begin{array}{c}\delta \\ x\end{array}\right)} {\left(\begin{array}{c}x \\ 0\end{array}\right)} (-1)^0 {\left(\begin{array}{c}\delta - x \\ \delta -x \end{array}\right)} (-1)^{\delta -x} \right. \\ & - \left[ {\left(\begin{array}{c}\delta \\ x\end{array}\right)} {\left(\begin{array}{c}x\\ 0\end{array}\right)} (-1)^0 {\left(\begin{array}{c}\delta - x \\ \delta -x \end{array}\right)} (-1)^{\delta -x} \right] \\ &= 0 \\ \end{aligned} $$

Case 2: We assume x < y. In this case we get:

$$ \begin{aligned} &a_{x,y}\\ &= {\left(\begin{array}{c}n_1+x \\ n_2\end{array}\right)}{\left(\begin{array}{c}n_1+y \\ n_3\end{array}\right)} \\ &\left[ \sum_{\nu=x}^y {\left(\begin{array}{c}\delta \\ \nu\end{array}\right)}{\left(\begin{array}{c} \nu \\ \nu - x\end{array}\right)} (-1)^{\nu - x} {\left(\begin{array}{c}\delta - \nu \\ \delta - y\end{array}\right)} (-1)^{\delta - y} \right] \\ &= {\left(\begin{array}{c}n_1+x \\ n_2\end{array}\right)}{\left(\begin{array}{c} n_1+y \\ n_3\end{array}\right)} (-1)^{\delta - x - y} \\ & \left[ \sum_{\nu=x}^y {\left(\begin{array}{c}\delta \\ \nu \end{array}\right)} {\left(\begin{array}{c}\nu \\ x\end{array}\right)} {\left(\begin{array}{c}\delta - \nu \\ \delta - y\end{array}\right)} (-1)^{\nu} \right] \\ &= {\left(\begin{array}{c}n_1+x \\ n_2\end{array}\right)}{\left(\begin{array}{c}n_1+y \\ n_3\end{array}\right)} (-1)^{\delta - x - y} \\ & \frac{\delta!}{x! (\delta - y)! (y-x)!} \left[ \sum_{\nu=x}^y {\left(\begin{array}{c}y-x \\ \nu - x\end{array}\right)} (-1)^{\nu} \right] \\ &= C \cdot \sum_{\nu=0}^{y-x} {\left(\begin{array}{c}y-x \\ \nu\end{array}\right)} (-1)^\nu = 0 \\ \end{aligned} $$

We finally have to extend this proof to the cases where n ₂ < δ or n ₃ < δ holds (compare Eq. 17). However, from inspecting Eq. 18 it can be seen that it is not harmful to let ν run from 0 to δ, since ${\left(\begin{array}{c}n \\ k\end{array}\right)}$ becomes zero for k < 0.⁷ We can therefore in all cases adopt Eq. 18.

Appendix 2: Designated collector case: $U_{i} \circ U_{j} = U_{j} \circ U_{i}$

To establish U _i (U _j (p)) = U _j (U _i (p)) for the designated collector case, we consider for given initial state $p(\cdot)$ the distribution of the number of segments in node R after node i’s and node j’s transmissions. After expanding (see Eq. 4), we get:

$$ \begin{aligned} U_i(U_j(p))(m) = \frac{1}{(1-\beta(U_j(p),i))\cdot(1-\beta(p,j))} \cdot \sum_{\nu=0}^m \sum_{\mu=0}^\nu p(\mu) \cdot T_{\nu,n_i,m-\nu}\cdot T_{\mu,n_j,\nu-\mu} \end{aligned} $$

We consider the two factors of this expression separately. For the double sum term, one can establish after simple algebraic rearrangement the identity:

$$ \begin{aligned} {\sum_{\nu=0}^m \sum_{\mu=0}^\nu p(\mu) \cdot T_{\nu,n_i,m-\nu}\cdot T_{\mu,n_j,\nu-\mu}} =\sum_{\nu=0}^m p(\nu) \cdot T_{\nu,n_j,n_i,m-\nu} =:S(m,n_j,n_i) \end{aligned} $$

(see Eq. 15 for the definition of $T_{\nu,n_{j},n_{i},m-\nu}$). An auxiliary result (Eq. 16 in "Appendix 1") establishes that S(m, n _j , n _i ) = S(m, n _i , n _j ) holds. For the other factor we consider $(1-\beta(U_j(p),i))\cdot(1-\beta(p,j))$. Expanding β(U _j (p), i) we get:

$$ \begin{aligned} {\beta(U_j(p),i)} = \sum_{\nu=0}^{n-1} U_j(p)(\nu) \cdot T_{\nu,n_i,n-\nu} = \frac{1}{1-\beta(p,j)}\sum_{\nu=0}^{n-1}\sum_{\mu=0}^\nu p(\mu) \cdot T_{\nu,n_i,n-\nu} \cdot T_{\mu,n_j,\nu-\mu} = \frac{1}{1-\beta(p,j)} \left(S(n,n_j,n_i) - \sum_{\mu=0}^n p(\mu) \cdot T_{n,n_i,0} \cdot T_{\mu,n_j,n-\mu}\right) = \frac{1}{1-\beta(p,j)}\left(S(n,n_j,n_i) - \beta(p,j)\right) \end{aligned} $$

where in the first equation the fact that the sum over ν runs only to n − 1 is due to U _j (p)(n) = 0 after a failed transmission of j (compare Eq. 4), and in the second-last line the fact $T_{n,n_{i},0} = 1$ has been used. With this result we get:

$$ \begin{aligned} (1-\beta(U_j(p),i))\cdot(1-\beta(p,j)) = 1 - \beta(p,j) - (S(n,n_j,n_i) - \beta(p,j)) = 1 - S(n,n_j,n_i) \end{aligned} $$

which, by again referring to Eq. 16 in "Appendix 1"), is insensitive to exchanging i and j. This proves the claim.