Peer-to-peer error recovery for wireless video broadcasting

Abstract

Wireless video broadcasting has experienced much growth in recent years. In video broadcasting, packet loss is inevitable due to dynamic channel condition. To address this, we study peer-to-peer (P2P) error recovery. In our system, a mobile station (MS) may generate some parity packets based on its received source packets and share them by broadcasting to its neighbors via a secondary channel (e.g., Wi-Fi or Bluetooth). With parity packets from its neighbors, an MS can effectively repair its lost packets locally. An important problem is to minimize the total number of parity packets generated while achieving a certain residual loss rate at the MSs. We first formulate the problem as a linear program which can be solved efficiently as the optimal performance of the system. We then propose a novel and fully distributed algorithm based on only local information at clients. Simulation results show that our distributed solution achieves high recovery efficiency and fast convergence. It generates very low recovery traffic and high video quality. Its performance is very close to the optimal solution based on centralized approach with complete network information.

Appendices

Appendix A: Linear network coding and its decoding in BOPPER

Given a group of source packets, a parity packet is a linear combination of these source packets with certain coefficients. Suppose that K source packets in a coding window are M ₁,…,M _k . A peer may select a set of coding coefficients $[{c_{1}^{j}},\ldots \ldots ,{c_{k}^{j}}]$ for parity packet j. It then generates a parity packet as

$$ \left\{x_{j} = \sum\limits_{i=1}^{K}{c_{i}^{j}}\cdot M_{i}\right\}, \text{ for} j\in [1,h]. $$

(14)

Note that a coefficient ${c_{i}^{j}}$ could be 0. In other words, if a source packet in the coding window is missing, we can simply set its coefficient to 0. This is a fundamental difference from FEC and fountain codes. Similarly, the peer can select another set of coefficients for another parity packet. If all peers share a common pseudo random number generator, only the random seed used to produce a corresponding coefficient series need to be embedded in the header of the parity packet. Therefore, a parity packet is of the same size as a single source packet, excluding a few bytes for the seed.

For decoding, suppose a peer has lost d source packets in a coding window. When it receives h parity packets, it needs to solve (14).

Here the unknowns are the lost source packets M _i . Clearly, this is a linear system with h equations and d unknowns. We can use Gauss-Jordan elimination to solve the system. We need h≥d to have a chance of recovering all unknowns. Note that h≥d is not a sufficient condition, as some of the coefficient combinations might be linearly dependent.

We use a simple example to illustrate the progressive decoding. Suppose there are 3 source packets in a coding window, which are denoted as M ₁, M ₂ and M ₃. For simplicity, we use a numerical value to represent the value of a source packet or a parity packet. Suppose error peer j only correctly received M ₁ in the coding window. It knows that M ₁=10, but not the values of M ₂ and M ₃. Suppose the first parity packet arriving at j has the value 1220, with a coefficient set (5, 18, 27). It indicates that

$$ 5M_{1} + 18M_{2} + 27M_{3} = 1220. $$

(15)

The equations can be written in the augmented matrix form as

$$\begin{array}{@{}rcl@{}} \left[\begin{array}{ccccc} 1 & 0 & 0 &\mid&10\\[-4pt] &&&\mid&\\[-4pt] 5 & 18 & 27&\mid&1220 \end{array}\right]. \end{array} $$

(16)

We use R ₁ and R ₂ to represent row 1 and row 2 in the matrix, respectively. We will use elementary row operations to transform this matrix into reduced echelon form. Let $R_{2}\leftarrow (R_{2} + (-5)R_{1})/18$. We have

$$\begin{array}{@{}rcl@{}} \left[\begin{array}{ccccc} 1 & 0 & 0 &\mid&10\\[-4pt] &&&\mid&\\[-4pt] 0 & 1 & 1.5&\mid&65 \end{array}\right]. \end{array} $$

(17)

Later, the second parity packet arrives at j with a value 570 and a coefficient set (34, 4, 5). It indicates that

$$ 34M_{1} + 4M_{2} + 5M_{3} = 570. $$

(18)

Hence, the matrix can now be written as

$$\begin{array}{@{}rcl@{}} \left[\begin{array}{ccccc} 1 & 0 & 0 &\mid&10\\[-4pt] & & &\mid&\\[-4pt] 0&1&1.5&\mid&65\\[-4pt] & & &\mid&\\[-4pt] 34 & 4 & 5&\mid&570 \end{array}\right]. \end{array} $$

(19)

Let $R_{3}\leftarrow R_{3} + (-34)R_{1} + (-4)R_{2}$. We have

$$\begin{array}{@{}rcl@{}} \left[\begin{array}{ccccc} 1 & 0 & 0 &\mid&10\\[-4pt] & & &\mid&\\[-4pt] 0&1&1.5&\mid&65\\[-4pt] && &\mid&\\[-4pt] 0 & 0 & -1&\mid&-30 \end{array}\right]. \end{array} $$

(20)

Let $R_{2}\leftarrow R_{2} + 1.5R_{3}$, and $R_{3}\leftarrow (-1)R_{3}$. We have

$$\begin{array}{@{}rcl@{}} \left[\begin{array}{ccccc} 1 & 0 & 0 &\mid&10\\[-4pt] && &\mid&\\[-4pt] 0&1&0&\mid&20\\[-4pt] && &\mid&\\[-4pt] 0 & 0 & 1&\mid&30 \end{array}\right]. \end{array} $$

(21)

This reduced echelon form matrix indicates that

$$ M_{1} = 10,\ M_{2} = 20\ \text{and}\ M_{3} = 30. $$

(22)

Therefore, with two linearly independent parity packets, error peer j could recover its lost source packets M ₂ and M ₃.

Appendix B: An illustrative example of error recovery in BOPPER

In this section, we illustrate with numerical examples to show how BOPPER achieves error recovery. Suppose each coding window consists of 5 source packets, i.e., K = 5. In a window, the server transmits five source packets {M ₁,M ₂,M ₃,M ₄,M ₅}.

We first consider the full recovery case. Suppose MS P ₁ has received packets {M ₃,M ₄,M ₅} while its neighbors P ₂ and P ₃ received packets {M ₂,M ₃,M ₄,M ₅} and {M ₁,M ₃,M ₄,M ₅}, respectively. So now we have

$$ \left\{ \begin{array}{l} \mathcal{C}_{1} = \{M_{3}, M_{4}, M_{5}\},\\ \mathcal{C}_{2} = \{M_{2}, M_{3}, M_{4}, M_{5}\},\\ \mathcal{C}_{3} = \{M_{1}, M_{3}, M_{4}, M_{5}\}; \end{array} \right. $$

(23)

and

$$ \left\{ \begin{array}{l} {\mathcal{L}}_{1} = \{M_{1}, M_{2}\},\\ {\mathcal{L}}_{2} = \{M_{1}\},\\ {\mathcal{L}}_{3} = \{M_{2}\}. \end{array} \right. $$

(24)

Assume that this is the very first coding window. We have $\hat \theta _{i} = 0, \bar \theta _{i} = 0, f_{i} = 0, \forall i$.

Suppose MS P ₁ broadcasts its NAK with L ₁ and f ₁ to the neighbors P ₂ and P ₃. When P ₂ receives the NAK from P ₁, it finds that its own lost packets is only a subset of the lost packets at P ₁, i.e., ${\mathcal {L}}_{2}\subseteq {\mathcal {L}}_{1}$. Hence, P ₂ suppresses its own NAK broadcasting. Similarly, P ₃ also suppresses its NAK broadcasting.

Meanwhile, P ₂ finds that $\mathcal {C}_{2} \cap {\mathcal {L}}_{1} \neq \emptyset$. It means that P ₂ holds some source packets which are required by P ₁. P ₂ hence selects $n_{\max }$ sets of random coefficients to generate parity packets from its C ₂ (since f ₁=0) and broadcasts them. Similarly, P ₃ also generates $n_{\max }$ number of parity packets and broadcasts them for P ₁ (hopefully P ₂ could overhear them). After P ₁ receives both of P ₂’s and P ₃’s parity packets, P ₁ can use Gauss-Jordan elimination to solve the linear system to recover M ₁ and M ₂. Meanwhile P ₂ and P ₃ can also recover lost packets from each other’s parity packet.

After parity recovery, each peer i then updates f _i . Take P ₁ for example. For M ₁ it receives $n_{\max }$ number of parities from P ₂; for M ₂ it receives the same number of parities from P ₃. We also have $|{\mathcal {L}}_{1}| = 2$. Suppose we set $n_{\max } = 2$, α=0.5 and 𝜃 ₁=1.1. We have

$$ \hat \theta_{1} = \min \left(\left(\frac{2}{2}+0\right), \left(0+\frac{2}{2}\right)\right) = 1, $$

(25)

$$ \bar \theta_{1} = 0.5 \times 0 + (1 - 0.5) \times 1 = 0.5, $$

(26)

and hence

$$ f_{1} = \frac{\bar \theta_{1}}{\theta_{1}} = \frac{0.5}{1.1}. $$

(27)

In the next round, if P ₁ needs to send a NAK, the above f ₁ value will be sent together.

We next consider the partial recovery case. Suppose now we have

$$ \left\{ \begin{array}{l} \mathcal{C}_{1} = \{M_{3}, M_{4}, M_{5}\},\\ \mathcal{C}_{2} = \{M_{2}, M_{3}, M_{4}, M_{5}\},\\ \mathcal{C}_{3} = \{M_{4}, M_{5}\}; \end{array} \right. $$

(28)

and correspondingly,

$$ \left\{ \begin{array}{l} {\mathcal{L}}_{1} = \{M_{1}, M_{2}\},\\ {\mathcal{L}}_{2} = \{M_{1}\},\\ {\mathcal{L}}_{3} = \{M_{1}, M_{2}, M_{3}\}. \end{array} \right. $$

(29)

P ₃ broadcasts its NAK with ${\mathcal {L}}_{3}$ to P ₁ and P ₂, both of which suppress their NAKs after received P ₃’s NAK. P ₁ and P ₂ then generate independent parity packets for P ₃. M ₂ and M ₃ can then be recovered from P ₁’s and P ₂’s parities. However none of the three MSs is able to recover M ₁. Hence the residual loss after parity recovery is

$$ \left\{ \begin{array}{l} {\mathcal{L}}_{1} = \{M_{1}\},\\ {\mathcal{L}}_{2} = \{M_{1}\},\\ {\mathcal{L}}_{3} = \{M_{1}\}, \end{array} \right. $$

(30)

which is clearly better than the loss before cooperative recovery.