Adaptive topology formation for peer-to-peer video streaming

Abstract

In this paper we propose an adaptive P2P video streaming framework to address the challenges due to bandwidth heterogeneity and peer churn on the Internet. This adaptive streaming framework consists of two major components, source rate adaptation and adaptive overlay topology formation, to maximize the video quality and fully utilize the overall peer upload capacity. In the source rate adaptation, the video server adapts the video source rate automatically based on the local measurement of peers’ download rates, so that the P2P network is not overloaded beyond its bandwidth capacity and peers are able to achieve smooth video playback. To combat bandwidth heterogeneity, we propose to construct a desirable link-level homogeneous overlay topology using a Markov chain Monte Carlo method, so that peers achieve an equal per-connection upload/download bandwidth. In this link-level homogeneous network, video flows do not encounter any bottlenecks along the delivery paths, and peers achieve high download rates to ensure smooth video playback. We also design a fully distributed algorithm to implement the dual mechanisms of the adaptive topology formation and the source rate maximization. To evaluate the performance of our streaming framework, we conduct both mathematical analysis and extensive simulations. The simulation results confirm our analysis and show that the proposed distributed algorithm is able to maximize the video playback quality with fast convergence.

Appendix

A Solution of problem J in Eqs. 1–3

Problem J is a linear programming problem. The KKT conditions [33] are both necessary and sufficient for Problem J. Let $d_{i}^{\ast}(t)$, $\forall i\in V(t)\backslash\{s\}$, be the optimal solution of Problem J. We introduce the Lagrange multiplier φ to relax the constraint of Eq. 2. φ is the shadow price if upload capacities are not fully utilized. We also introduce the Lagrange multiplier ψ _i to relax the constraint of Eq. 3. ψ _i is the shadow price if peer i does not achieve the required download rate. Then, we have

$$\begin{array}{lll} L(d_{i}(t),r(t);\varphi ,\psi _{i})&=&r(t)+\varphi \left( \underset{i\in V(t)}{\sum }C_{i}-\underset{i\in V(t)\backslash\{s\}}{\sum }d_{i}(t)\right) \\&& +\underset{i\in V(t)\backslash\{s\}}{\sum }\psi _{i}\left( d_{i}(t)-r(t)\right).\end{array}$$

We obtain the KKT conditions as follows.

$$ \underset{i\in V(t)\backslash\{s\}}{\sum }d_{i}(t)\!\leq\! \underset{i\in V(t)}{\sum }C_{i}; \ \ \ d_{i}(t)\!\geq\! r(t), \ \forall i\in V(t)\backslash\{s\}, $$

(31)

$$ \varphi \geq 0; \ \ \ \psi _{i}\geq 0, \ \forall i\in V(t)\backslash\{s\}, $$

(32)

$$\begin{array}{lll} \varphi \left( \underset{i\in V(t)}{\sum }C_{i}\!-\!\underset{i\in V(t)\backslash\{s\}}{\sum }d_{i}(t)\right) \!=\! 0;&& \ \ \ \psi _{i}\left( d_{i}(t)\!-\!r(t)\right) \!=\!0, \\&&\quad \forall i\!\in\! V(t)\backslash\{s\},\end{array}$$

(33)

$$ \frac{\partial L}{\partial d_{i}}=-\varphi +\psi _{i}=0, \ \forall i\in V(t)\backslash\{s\}, $$

(34)

$$ \frac{\partial L}{\partial r}=1-\underset{i\in V(t)\backslash\{s\}}{\sum}\psi _{i}=0. $$

(35)

By Eq. 34, we have $\varphi =\psi _{i},\forall i\in V(t)\backslash\{s\}$. If φ = 0, we have $\psi _{i}=0,\forall i\in V(t)\backslash\{s\}$; however, this conflicts with Eq. 35. Therefore, $\varphi =\psi_{i}>0,\forall i\in V(t)\backslash\{s\}$. Because φ > 0 and ψ _i  > 0, by Eq. 33, we have $\underset{i\in V(t)}{\sum}C_{i}-\underset{i\in V(t)\backslash\{s\}}{\sum }d_{i}(t)=0$ and $d_{i}(t)-r(t)=0,\forall i\in V(t)\backslash\{s\}$, respectively. Therefore,

$$\underset{i\in V(t)}{\sum}C_{i}-(N(t)-1)r(t)=0.$$

Finally, we have

$$ d_{i}^{\ast }(t)=r^{\ast }(t)=\frac{\underset{j\in V(t)}{\sum }C_{\!j}}{N(t)-1}, \ \forall i\in V(t)\backslash\{s\}. $$

(36)

B Derivation of the average out-degree of nodes

In the following, we show that if each node is of a constant in-degree m, the average out-degree of nodes is approximately m.

Note that $\sum\limits_{i\in V(t)}O_{i}(t_{i},t)=\sum\limits_{i\in V(t)\backslash\{s\}}I_{i}(t)$ as the overall out-degree equals to the overall in-degree, the expected out-degree of nodes is

$$ \frac{1}{N(t)}\sum_{i\in V(t)}O_{i}(t_{i},t)=\frac{1}{N(t)}\sum_{i\in V(t)\backslash\{s\}}I_{i}(t). $$

Because $\sum\limits_{i\in V(t)\backslash\{s\}}I_{i}(t)=(N(t)-1)m$ as each node has in-degree m except the server s, we have

$$ \frac{1}{N(t)}\sum_{i\in V(t)}O_{i}(t_{i},t)=\frac{1}{N(t)}(N(t)-1)m. $$

When N(t) ≫ 1, we have

$$ \frac{1}{N(t)}\sum_{i\in V(t)}O_{i}(t_{i},t)\approx m. $$

(37)

C Derivation of Eq. 25

To derive $\mathcal{Z}$, substituting Eq. 23 into Eq. 21 yields

$$\begin{array}{lll} \mathcal{Z} &=&\lim\limits_{t\rightarrow\infty}\left\langle \int_{0}^{t}\lambda e^{-\mu(t-t_{i})}\frac{C_{i}^{2}}{O_{i}(t_{i},t)}dt_{i}\right\rangle \\ &=&\lim\limits_{t\rightarrow\infty}\left\langle \lambda C_{i}\int_{0}^{t} \frac{e^{-\mu(t-t_{i})}}{\sqrt{\frac{2m\mathcal{N}}{\mathcal{Z}}\left( 1-e^{-2\mu\left( t-t_{i}\right) }\right) }}dt_{i}\right\rangle . \end{array}$$

(38)

Note that the $\left\langle {\cdot}\right\rangle$ operation is taken with respect to the peer capacity C _i in Eq. 38. Therefore, we have

$$ \mathcal{Z}=\left\langle C\,\right\rangle \lim\limits_{t\rightarrow\infty}\int_{0} ^{t}\frac{\lambda}{\sqrt{\frac{2m\mathcal{N}}{\mathcal{Z}}}}\frac {e^{-\mu(t-t_{i})}}{\sqrt{1-e^{-2\mu\left( t-t_{i}\right) }}}dt_{i} . $$

(39)

By denoting τ = t − t _i , Eq. 39 becomes

$$ \mathcal{Z}=\frac{\lambda\left\langle C\,\right\rangle }{\sqrt{\frac {2m\mathcal{N}}{\mathcal{Z}}}}\lim\limits_{t\rightarrow\infty}\int_{0}^{t} \frac{e^{-\mu\tau}}{\sqrt{1-e^{-2\mu\tau}}}d\tau. $$

(40)

Then, changing the integration variable x = e ^− μτ and plugging it into Eq. 40, we have

$$ \mathcal{Z}=\frac{1}{\mu}\frac{\lambda\left\langle C\,\right\rangle } {\sqrt{\frac{2m\mathcal{N}}{\mathcal{Z}}}}\int_{0}^{1}\frac{1}{\sqrt{1-x^{2}} }dx. $$

(41)

Note that $\int_{0}^{1}\frac{1}{\sqrt{1-x^{2}}}dx=\frac{\pi}{2}$ and plugging it into Eq. 41; therefore, we have

$$ \mathcal{Z}=\frac{\pi}{2\mu}\frac{\lambda\left\langle C\,\right\rangle } {\sqrt{\frac{2m\mathcal{N}}{\mathcal{Z}}}}. $$

(42)

Plugging $\lambda=\mathcal{N}\mu$ into Eq. 42, we obtain

$$ \mathcal{Z}=\frac{\pi}{2}\frac{\left\langle C\,\right\rangle }{\sqrt {\frac{2m\mathcal{N}}{\mathcal{Z}}}}\mathcal{N}. $$

(43)

Solving Eq. 43 for $\mathcal{Z}$, we finally have

$$ \mathcal{Z}=\frac{\pi^{2}\left\langle C\,\right\rangle ^{2}}{8m}\mathcal{N}. $$

(44)

D Derivation of Eqs. 29 and 30

If the nodes reselect their parent nodes periodically, Eq. 14 becomes

$$\begin{array}{lll} \frac{dO_{i}(t_{i},t)}{dt}&=&\left( \lambda m+N(t)\mu m+N(t)m\mu ^{\prime }\right) \pi _{i}(t) \\&&-\,O_{i}(t_{i},t)\mu -O_{i}(t_{i},t)\mu ^{\prime },\end{array}$$

(45)

where the term N(t)mμ′ appears because the nodes re-select each parent at an average rate μ′; $O_{i}(t_{i},t)\mu ^{\prime }$ is the additional rate of out-degree reduction because node i has O _i (t _i , t) child nodes which are disconnected from node i at the average rate μ′ due to this reselection operation. Solving Eq. 45, we have

$$\begin{array}{lll} O_{i}(t_{i},t)&=&C_{i}\sqrt{\frac{\mathcal{N}}{\mathcal{Z}^{\prime}}\left( \frac{2\mu m+m\mu ^{\prime }}{\mu +\mu ^{\prime }}\right) \left( 1-e^{-2(\mu +\mu ^{\prime })\left( t-t_{i}\right) }\right)}, \\&& t\geq t_{i}, \end{array}$$

(46)

where $\mathcal{Z}^{\prime}=\underset{t\rightarrow\infty}{\lim}\left\langle \sum\limits_{i\in V(t)}\frac{C_{i}^{2}}{O_{i}(t_{i},t)}\right\rangle$. As shown in Eq. 46, the convergence rate to the link-level homogeneous state increases by μ′. $\mathcal{Z}^{\prime }$ can be computed in the same manner as in Appendix C by substituting Eq. 46 into Eq. 21.

$$\begin{array}{lll} \mathcal{Z}^{\prime }\!&=&\!\lim\limits_{t\rightarrow \infty}\!\left\langle \int_{0}^{t}\!\lambda e^{-\mu (t-t_{i})}\phantom{\frac{C_{i}^{2}}{C_{i}\sqrt{\frac{\mathcal{N}}{\mathcal{Z}^{\prime }}\!\left (\frac{2\mu m+m\mu ^{\prime}}{\mu +\mu ^{\prime }}\right)}}}\right. \\&& \left. \quad\quad\quad\times\frac{C_{i}^{2}}{C_{i}\sqrt{\frac{\mathcal{N}}{\mathcal{Z}^{\prime }}\!\left (\frac{2\mu m+m\mu ^{\prime}}{\mu +\mu ^{\prime }}\right)\!\!\left( 1\!-\!e^{-2(\mu +\mu ^{\prime })\!\left( t-t_{i}\right)\!}\right)\!}}dt_{i}\!\right\rangle.\end{array}$$

(47)

Denoting $a=\frac{\mu^{\prime}}{\mu}$ and substituting it into Eq. 47, we have

$$\begin{array}{lll} \mathcal{Z}^{\prime}&=&\left\langle C\,\right\rangle \sqrt {\frac{\mathcal{Z}^{\prime}\left( 1+a\right) }{\mathcal{N}m\left( 2+a\right) }}\lim\limits_{t\rightarrow\infty}\int_{0}^{t}\lambda e^{-\mu(t-t_{i} )} \\&&\times\frac{1}{\sqrt{\left( 1-e^{-2\mu(1+a)\left( t-t_{i}\right) }\right) } }dt_{i},\\ \sqrt{\mathcal{Z}^{\prime}}&=&\left\langle C\,\right\rangle \sqrt {\frac{\mathcal{N}\left( 1+a\right) }{m\left( 2+a\right) }}\lim\limits_{t\rightarrow\infty}\int_{0}^{t}\frac{\lambda e^{-\mu(t-t_{i})}}{\mathcal{N} }\\ &&\times\frac{1}{\sqrt{1-e^{-2\mu(1+a)\left( t-t_{i}\right) }}}dt_{i}. \end{array}$$

(48)

Denoting$A=\underset{t\rightarrow\infty}{\lim}\int_{0}^{t}\frac{\lambda e^{-\mu(t-t_{i})}}{\mathcal{N}}\frac{1}{\sqrt{1-e^{-2\mu(1+a)\left( t-t_{i}\right) }}}dt_{i}$ and plugging $\lambda=\mathcal{N}\mu$, we have

$$ A=\underset{t\rightarrow\infty}{\lim}\int_{0}^{t}\mu e^{-\mu(t-t_{i})}\frac {1}{\sqrt{1-e^{-2\mu(1+a)\left( t-t_{i}\right) }}}dt_{i}. $$

Changing the integration variable $x=e^{-\mu(t-t_{i})}$, we obtain a simplified expression of A in Eq. 49.

$$ A=\int_{0}^{1}\frac{1}{\sqrt{1-x^{2(1+a)}}}dx. $$

(49)

Plugging Eq. 49 into Eq. 48, we have

$$ \mathcal{Z}^{\prime}=\frac{A^{2}\left\langle C\,\right\rangle ^{2}\left( 1+a\right) }{m\left( 2+a\right) }\mathcal{N}. $$

(50)

Substituting Eq. 50 into Eq. 46, we have

$$ O_{i}(t_{i},t)=C_{i}\frac{m\left( 2+a\right) }{\left\langle C\,\right\rangle A\left( 1+a\right) }\sqrt{\left( 1-e^{-2(\mu+\mu^{\prime})\left( t-t_{i}\right) }\right) },\text{ }t\geq t_{i}. $$

Hence, as t→ ∞, the converged capacity per out-degree value due to the proactive reselection of parent nodes is

$$ \delta^{\prime}=\lim\limits_{t\rightarrow\infty}\frac{C_{i}}{O_{i}(t_{i},t)}=\frac{\left\langle C\,\right\rangle A\left( 1+a\right) }{m\left( 2+a\right)}. $$

(51)

Note that the integration with respect to x in Eq. 49 is conducted in $\left[ 0,1\right)$. As a→ ∞, x ^2(1 + a)→0. Therefore,

$$ \lim\limits_{a\rightarrow\infty}A=\lim\limits_{a\rightarrow\infty}\int_{0}^{1}\frac{1}{\sqrt{1-x^{2(1+a)}}}dx=1. $$

(52)