Abstract
We are concerned with designing feedback-based adaptive network coding schemes with the aim of minimizing decoding delay in each transmission in packet-based erasure networks. We study systems where each packet brings new information to the destination regardless of its order and require the packets to be instantaneously decodable. We first formulate the decoding delay minimization problem as an integer linear program and then propose efficient algorithms for finding its optimal solution(s). We show that our problem formulation is applicable to memoryless erasures as well as Gilbert-Elliott erasures with memory. We then propose a number of heuristic algorithms with worst case linear execution complexity that can be used when an optimal solution cannot be found in a reasonable time. We verify the delay and speed performance of our techniques through numerical analysis. This analysis reveals that by taking channel memory into account in network coding decisions, one can considerably reduce decoding delays.
receivers, denoted by 
for 
. The data to be broadcast is divided into 
packets, denoted by 
for 
. Time is slotted and the source can transmit one (possibly coded) packet per slot.
connects the source to each individual receiver 
. Erasures in different links can be independent or correlated with each other. Different erasures in a single link can be independent (memoryless) or correlated with each other (with memory) over time.
can occur with a probability of 
in each packet transmission round independent of previous erasures.
to the bad state 
in link 
is 
and the probability of moving from the bad state 
to the good state 
is 
, where 
is the time slot index. Steady-state probabilities are given by 
and 
. Following [
as 
, which signifies the persistence of the channel in remaining in the same state. A small 
means a channel with little memory and a large 
means a channel with large memory.
. The coefficient vector used for linear network coding is sent in the packet header so that each destination can at some point recover the original packets. Since in this paper we are only dealing with instantaneously decodable packet transmission, it suffices to consider linear network coding over 
. That is, coded packets are formed using binary XOR of the original source packets. Thus, network coding is performed in a similar manner as in [
if it is a linear combination of source packets containing at most one source packet that 
has not decoded yet. A scheme is called instantaneously decodable if all transmissions have this property for all receivers.
in an instantaneously decodable scheme, the knowledge of receiver 
is the set consisting of all packets that the receiver has decoded so far. The receiver can therefore, compute any linear combination of the packets that it has decoded for decoding future packets.
if it only contains source packets that the receiver has decoded so far. Otherwise, the packet is innovative.
, receiver 
experiences one unit of delay if it successfully receives a packet that is either non-innovative or not instantaneously decodable. If we impose instantaneous decodability on the scheme, a delay can only occur if the received packet is not innovative.
is 
, receiving 
will cause 
to experience one unit of delay, whereas 
is innovative and instantaneously decodable, hence does not incur any delay.
uncoded, for example.
and 
receivers, there exists an offline algorithm that is both rate optimal and delay-free. For 
the authors prove that a zero-delay algorithm does not exist. By offline we mean that the algorithm needs to know future realizations of erasures in broadcast links. In contrast, an online algorithm decides on what to send in the next time slot based on the information received in the past and in the current slot. In this paper, we focus on designing online algorithms.
and consider the knowledge of all receivers, which is also available at the source because of the feedback. The state of the entire system at time index 
(in terms of packets that are still needed by the receivers) can be described by an 
binary receiver-packet incidence matrix 
with elements
are denoted by 
to 
. We assume that packets received by all receivers are removed from the receiver-packet incidence matrix. Hence, 
does not contain any all-zero columns.
receivers and 
packets. Before the transmission begins, the receiver-packet incidence matrix 
is an all-one 
matrix. If we send packet 
in the first transmission round 
and assuming that only receiver 
successfully receives it, 
will become
in the next transmission round 
and assuming that only receiver 
successfully receives it, 
will then be
. In the example above, at 
, we cannot send 
because it contains more than one packet unknown to 
.
represent a binary decision vector of length 
that determines which packets are being coded together. The transmitted packet consists of the binary XOR of the source packets for which 
. More formally, we can define the instantaneous decodability constraint for all receivers as 
, where 
represents an all-one vector of length 
and the inequality is examined on an element-by-element basis (Note that although 
is a binary or Boolean vector, 
is calculated in real domain. Hence, 
is in fact a pseudo-Boolean constraint.). This condition ensures that a transmitted coded packet contains at most one unknown source packet for each receiver. A vector 
is called infeasible if it does not satisfy the instantaneous decodability condition. In other words, 
is called infeasible if and only if there exists at least one 
for which 
in 
. A vector 
is called a solution if and only if it satisfies 
. In the rest of this paper, "
" and "
is a solution" are used interchangeably.
, where 
is the nonempty set of receivers that still need source packet 
. Note that these sets can be easily determined by looking at the columns of matrix 
. The "importance" of packet 
can be, for example, taken to be the size of set 
, which is the number of receivers that still need 
.
. This is a standard problem in combinatorial optimization, usually called set packing [
with the largest total size. In the (most desirable) case when equality holds in 
for every receiver, we also speak of a set partition. This is equivalent to a zero-delay transmission.
in tailoring the optimization problem according to the application requirements or channel conditions, such as memory in erasure links.
, which signify packet importance, are all positive. If one has already found a solution such as 
with 
, then changing this solution into 
by changing 
into 
can only result in a 
strictly smaller than 
. We say that given solution 
, 
is clearly suboptimal and hence, can be discarded in an algorithm that searches for the optimal solution(s).
in a solution. Or formally, 
and 
are constrained if for any 
satisfying 
, 
(Again, note that the addition of variables takes place in real domain.). We also say that 
is constrained to 
and vice versa. It can be proven that 
and 
are constrained if and only if there exits at least one row index 
in 
for which 
.
is called the constrained set of 
and is denoted by 
. That is,
and 
are not constrained to each other (
and 
), then columns 
and 
in 
cannot have nonzero elements in the same row position. That is, for each row index 
, 
and 
.
is said to be unconstrained if 
. The set of all unconstrained variables is denoted by 
and is referred to as the unconstrained set.
is an unconstrained variable, then for each row index 
, 
for all 
(otherwise, 
and 
would become constrained).
and 
are constrained because for 
, 
. Variables 
and 
are not constrained to each other because columns 
and 
do not have a nonzero element in the same row position. Variable 
is unconstrained because no other column has a nonzero element in rows 6 or 7. In summary, 
, 
, 
, 
and 
.
are positive). In the above example, 
and 
must be set to 1 because no other variable is constrained to them (we will make this statement formal in the optimality proof of the algorithm in the appendix).
forces 
and 
to zero.
, column-wise, in descending order of the number of 
's in each column. Setting the "most important" head variable 
(with the highest 
) to 
is likely to result in the largest constrained set (because it potentially overlaps with many other variables) and hence, many variables will be resolved in the next recursion. We will refer to the approach based on finding the largest constrained set as the greedy pruning strategy and to the alterative approach as the sorted pruning search strategy.
denote the problem of size 
whose input is an 
receiver-packet incidence matrix 
and whose output is a set of solutions of the form 
of length 
which satisfy the instantaneous decodability condition 
. The algorithms can be described as shown in Algorithm 1.
) Start with the original problem of size
.
) if sorted pruning strategy is desired then
) Rearrange the variables in
in descending order of packet importance (number of
's in each column).
) end if
:
then
(since the variable is not constrained).
to
.
and the cardinality of its constrained
.
with cardinality
and also the set of unconstrained
.
by
.
and the variables in its corresponding constrained set
to 0.
accordingly.
(Note that
unconstrained variables are set to one,
and
variables constrained by
variables are resolved.).
.
accordingly.
(Note that
unconstrained variables are set to one and
, hence a total of
variables
and 
within itself.
such variables, we can start step 
above from 
instead of 
. The Matlab code for both the greedy and sorted pruning algorithms can be found at 
by 
. According to Algorithm 1, this problem is always broken into two smaller problems of size 
and 
. Therefore, one can find the number of recursions required to solve 
by recursively computing 
. The recursion stops when one reaches a problem of size 1 (only one packet to transmit) where 
.
is no longer determined by how many receivers need 
, but by the probability that 
will be successfully decoded by the receivers that need it. In computing this probability, one can use the fact that successive channel erasures in a link are usually correlated with each other and hence, their history can be used to make predictions about whether a receiver is going to experience erasure or not in the next time slot. To present the idea, we focus on the GEC model for representing channel erasures. More general memory models for erasure can also be incorporated into our framework.
of sending a packet to receiver 
as the probability of successful reception by 
in the next time slot: 
, where 
is the state of 
in the previous transmission round (Statements like "state of 
" should be interpreted as the state of the physical link 
connecting the source to 
.). The total reward or importance of sending packet 
is then
for which there is a higher chance of successful reception, because the receivers that need 
are more likely to be in good state in the next time slot. With this newly defined weight vector, one can try to solve the optimization problem given in (4) under the same instantaneous decodability condition.
, which can be appropriately determined based on the application. For example, instead of allocating the same weight to a packet needed by a subset of receivers, one can allocate different weights to the same packet (looking column-wise at 
) depending on the priorities or demands of each user. In the map update example described in the Introduction, different emergency units can adaptively flag to the base station different parts of the map as more or less important depending on their distance from a certain disaster zone. The task of the base station is then to send a packet combination that satisfies the largest total priority. One can also combine user-dependent packet weights with the channel state prediction outcomes in a GEC. One possibility is to multiply the probabilities 
by the receiver priority. It could then turn out that although a receiver is more likely to be in erasure in the next transmission round, it may be served because of a high priority request.
. We then rearrange the columns of the matrix 
in descending order of 
(starting from the packet with the highest weight). Note that this is different from the sorted pruning version of the Algorithm 1, in which the columns of 
were sorted in descending order of 
to potentially result in large constrained sets. We then set the head variable 
and find its corresponding constrained set 
to resolve 
variables that are to be set to zero. We then solve the smaller problem of size 
and continue until the problem cannot be further reduced. One main difference between Heuristic 1 and Algorithm 1 is that at each recursion, the head variable is only set to one; the other possibility of 
is not pursued at all. In a sense, this heuristic algorithm finds greedy solutions to the problem at each recursion by serving the highest priority packet. In this heuristic algorithm, all 
unconstrained variables are naturally set to 1 in the course of the algorithm. The computational complexity of this method is at worst proportional to 
, which can happen when there is no constraint between packets.
based on the number of receivers that need a packet. It only reverts to sorting the unresolved variables based on the vector 
in the termination process. Secondly, if the maximum number of recursions is set to one, Algorithm 1 just performs the termination process and becomes identical to Heuristic 1.
denote the index of receivers that still need at least one packet and 
denote such receivers. The achieved throughput at time slot 
is defined as 
, where 
is the found solution and 
is an appropriate function of receivers' needs. For memoryless erasures 
and for GEC's 
(refer to Section 4 and (7)).) is higher than a desired value, there is no need to proceed any further. Otherwise, we can gradually increase the number of recursions by an appropriate step size. This heuristic stops when it either reaches the maximum allowed recursions or when increasing the number of recursions does not result in a noticeable improvement in the throughput.
packets. End-to-end decoding delay of a receiver is the sum of decoding delays for the receiver in each transmission step. In the following, when we say "the delay performance of method X", we are referring to the delay performance of the end-to-end transmission, where method X is applied at each step.
when referring to link erasure probabilities.
packets to 
to 
receivers. Channel erasures are memoryless and occur with a high probability of 
independently in every link. The median of delay is computed across all receivers and is, in fact, also the median across many stochastic runs of the algorithms. The first curve from below shows the delay obtained from Algorithm 1 (Throughout the numerical evaluations, we used the sorted pruning version of Algorithm 1.). The middle curve is the delay obtained by performing Heuristic 1. The top curve shows a reproduction of delay results reported in [
packets to 
to 
receivers. Channel erasures are memoryless and occur with a high probability of 
independently in every link. Algorithm 1, Heuristic 1 and random Heuristic [
packets to 
to 
receivers. Algorithm 1 is compared with Heuristics 1–3. Both Heuristics 2 and 3 perform very closely to Algorithm 1. The maximum number of recursions for both Heuristic 2 and 3 is set to 100.
. For example, a mean delay of 10 slots for 
packets signifies on average 
overhead, which is the price for guaranteeing instantaneous decodability. In other words, one measure of throughput is 
, where 
is the mean delay across all receivers. An example is shown in Figure 
receivers in the system, Algorithm 1, Heuristics 2, and 3 ensure an average throughput loss of 
.
. A natural question that arises is whether systematic selection of a solution with a particular property is better than others in the presence of erasures in the channel. Our experiments verify that indeed some secondary post processing on the solutions can improve the end-to-end delay. In particular, we compare two post processing techniques: 
selecting a solution which involves minimum amount of coding (lowest number of 1's in the solution vector 
) and 
selecting a solution with maximum amount of coding (highest number of 1's in the solution vector 
). Figure 
packets to be transmitted to 
receivers and at the beginning of the third transmission round, matrix 
is given as follows
or packet 
by itself, where the former involves coding and latter is uncoded. Now let us assume that we select the maximum coding strategy and send 
. If in the third transmission round only 
successfully receives, 
will become
. If in the fourth transmission round only 
successfully receives, 
will become
in the third transmission round, we can always find innovative and instantaneously decodable packets for all three receivers in the future regardless of erasures in the channel. In summary, solutions with less coding tend to cause less constrains on the problem in the future. 
packets along with the number of recursions required in Heuristics 1, 2, and 3. Algorithm 1 shows three modes of behavior: low, medium, and high number of recursions. When the number of receivers is larger than 
, Algorithm 1 finds the optimal solution very quickly and the number of recursions is very close to the number of packets 
. However, when the number of receivers is lower, the constraints that each receiver imposes on the network coding decisions cannot limit the search space enough and hence, a large number of combinations have to be tested. Obviously, Heuristic 1 has the lowest number of recursions. Compared to Heuristic 2 with 100 fixed recursions, dynamic Heuristic 3 can almost halve the number of recursions with negligible effect on delay performance (see Figure 
, 
, and 
are considered. The complexity remains linear with the number of packets for well-sized receiver populations (30 and 40 receivers). This is in agreement with observations in Figure 
), we see a sudden growth in complexity, in terms of number of recursions, when 
packets. In such situations, truncating the number of recursion to be linear with the number of packets (Heuristic 2) is a good alternative.
is plotted for three different erasure probabilities 
, 
, and 
, which are still high erasure probabilities. The number of receivers is fixed to 
. The delay performance of Heuristics 1 and 3 are shown. A few observations are made. Firstly, as expected, the delay (both absolute and normalized measures) decreases as the erasure probability decreases. Secondly, the difference in the delay performance between Heuristics 1 and 3 decreases as the erasure probability decreases. This trend has also been observed for other number of receivers. Moreover, the difference between heuristics and Algorithm 1 decreases with erasure probability, which is not shown here for clarity of figure. Finally, the normalized delay decreases as the number of packets increases. We noted, however, that the absolute delay may increase or decrease depending on the number of receivers in the system. We attribute possible decrease in the normalized delay to the fact that when there are more packets to transmit, the transmitter has more options to choose from and hence, encounters delays less often in a normalized sense.
for this particular 
(this is not always the case).
packets to transmit, does it make sense to divide them into subblocks of smaller sizes or does it make sense to treat them as one single block of packets? The short answer is to include all "order-insensitive" packets in making transmission decisions and only break the packets into subblocks when the assumption of order insensitivity between subblocks breaks down. In the extreme case, an infinite number of order-insensitive packets provides an infinite pool of packets to choose from that can satisfy the demands of all receivers and are instantaneously decodable. Figure 
packets is chosen as the reference for comparison. We can see that although the delay of transmitting 
packets, 
, can be larger than that of transmitting 
packets 
, the delay does not increase by a factor of 
. That is 
and one does not benefit from breaking 
packets into 
subblocks of size 
packets each. By treating 
subblocks of size 
as one block of size 
, we add more degrees of freedom in making decisions.
packets in Heuristic 1, 
, is taken as the reference, we can see that the delay of including 
packets in transmission is less than 
. The same observation applies to the delay of Algorithm 1. In general, it is recommended to include all "order-insensitive" packets in making transmission decisions and only break the packets into subblocks when the assumption of order insensitivity between subblocks breaks down.
packets and 
receivers. The GEC parameters for all links are identical with 
. The horizontal axis shows the memory content 
.The first curve from above shows the performance of Algorithm 1 when the transmitter does not take channel conditions into account in making coding decisions. In other words, 
is used in Algorithm 1 as if the channel states were memoryless. For relatively large memory contents, this method results in the largest mean delay. The next curve shows the delay performance of Heuristic 1. The next two curves, which are almost indistinguishable, show the performance of Algorithm 1 which takes channel states into account (using (7)) and Heuristic 2 with 100 recursions. The last curve shows the best delay that can be achieved by occasionally violating the instantaneous decodability rule for one receiver in favor of the other two receivers that are predicted to be in good state in the next transmission round. More details can be found in [
to 
receivers. Both the mean delay and mean delay plus one standard deviation of delay (across 1000 stochastic runs of the transmission) are shown. As expected, the delay increases as the number of receivers increases. Comparing the delay's standard deviation with its mean, we observe that when the number of receivers is 3–5, the delay is relatively more variant than when the number of receivers is 10–15. For example, for 
and 
, the ratio of standard deviation to mean delay is around 
, whereas for 
and 
this ratio reduce to only 
. One should keep these variations in mind when designing the transmission system.
receivers and 
packets. The figure confirms our earlier finding that selecting the maximum amount of coding among the optimal solutions provided by Algorithm 1 can result in larger end-to-end delays. We also note that serving the maximum number of receivers can have an adverse effect on the delay in GEC's. To explain this, consider an example where there are 
left packets to be transmitted to 
receivers. Packet 1 is needed by 
to 
and packet 2 is needed by 
and 
. Since both packets are needed by 
, we can either send packet 1 or 2, but not both. Now assume that 
to 
are all predicted to be in good state with probability 
and 
is predicted to be in good state with probability 
, so that 
according to (7). Therefore, transmission of either packet seems to be equally optimal. However, one can easily verify that the probability of at least one receiver among 
to 
receiving packet 1 is only 
, whereas the probability of either 
or 
receiving packet 2 is 
. Therefore, it makes sense to satisfy only two receivers, one of which has a high priority due its good channel conditions.
-step decodability and investigate the delay-throughput tradeoff.
as explained below.The algorithm generates a binary tree. Each node represents a problem of size 
and 
, and branches into two subproblems of size 
and 
. The former subproblem is a result of setting 
and the latter a result of setting 
. A leaf is reached when we need to solve 
. Without loss of generality let us assume that the variable to be examined is the first variable (
) which is followed by 
variables (
to 
) that are constrained to 
, 
variables (
to 
) that are constrained but not to 
, and finally 
unconstrained variables 
to 
. This can be easily accomplished by rearranging the columns of 
.For 
, it is clear that the only optimal solution to 
is 
which is returned by the algorithm. Hence, the minimal structure of the algorithm returns the optimal solution and our claim is true for 
. The induction hypothesis is that the two subproblems 
and 
have only discarded infeasible vectors and some suboptimal solutions. We need to prove that the same statement applies to the parent problem 
.We first look at the left branch where 
. According to the construction of the algorithm, any solution such as 
of length 
provided by the left branch 
is appended by the parent problem 
to form
is set to one, variables constrained to 
are set to zero and all unconstrained variables are set to one. We first prove that 
is indeed a solution and then show that changing any element of 
results in either an infeasible or a clearly suboptimal 
. We use Definitions 6–8.(i)For
as a weighted sum of columns of 
. That is, 
, where 
is a submatrix of 
of size 
, which is input to 
, and according to the induction hypothesis 
. But since no variable in 
is constrained to 
, no column in 
and 
can have ones in the same row position. Therefore, 
.(ii)Since 
to 
are unconstrained, no column 
to 
can have ones in the same row position. Hence, 
.(iii)Using similar arguments, we can assert that no column in 
or 
can have ones in the same row positions as 
to 
do. Therefore, 
and 
is a solution.(iv)We now argue that variables 
to 
cannot be anything other than zero. This directly follows from the fact that 
is constrained with 
for 
and hence, in any given solution they cannot be simultaneously one.(v)Since we have already found a solution 
where the first and last 
variables are one, we know that any other solution such as 
with one or more zeros in these positions becomes suboptimal and can be discarded. That is, 
due to positiveness of elements of 
.(vi)Finally, according to induction hypothesis, we know that 
cannot be changed into anything other than what 
provides without making it either infeasible or suboptimal. In summary, for each solution 
provided by the left branch 
, the constructed vector
. According to the construction of the algorithm, a given solution such as 
of length 
provided by the right branch 
is appended by the parent problem 
to form
this is indeed a solution. We then show that changing any element of 
can only result in an infeasible vector, a clearly suboptimal solution, or a duplicate solution already provided by the left branch and hence, can be discarded. We use Definitions 6–8.(i)We write 
as 
, where 
is a submatrix of 
of size 
, which is input to 
, and according to the induction hypothesis 
. Similar to the arguments for the left branch, we can assert that no column 
to 
corresponding to unconstrained variables can have ones in the same row position. Hence, 
. Furthermore, that no column in 
can have ones in the same row positions as 
to 
. Therefore, 
and 
is a solution.(ii)Since we have already found a solution 
where the last 
variables are one, we know that any other solution such as 
with one or more zeros in these positions becomes suboptimal and can be discarded.(iii)Finally, we show that any vector of the form
. If 
for any 
, then we have already shown in the analysis of the left branch that
and 
are constrained to each other. If none of 
to 
are one, then 
will be of the form
. But, 
has to be a solution of 
. Hence, considering vectors of the form
provided by the right branch 
, the constructed vector