Multi-layer Virtual Transport Network Design

Abstract

Service overlay networks and network virtualization enable multiple overlay/virtual networks to run over a common physical network infrastructure. They are widely used to overcome deficiencies of the Internet (e.g., resiliency, security and QoS guarantees). However, most overlay/virtual networks are used for routing/tunneling purposes, and not for providing scoped transport flows (involving all mechanisms such as error and flow control, resource allocation, etc.), which can allow better network resource allocation and utilization. Most importantly, the design of overlay/virtual networks is mostly single-layered, and lacks dynamic scope management, which is important for application and network management. In response to these limitations, we propose a multi-layer approach to virtual transport network (VTN) design. This design is a key part of VTN-based network management, where network management is done via managing various VTNs over different scopes (i.e., ranges of operation). We explain the details of the multi-layer VTN design problem as well as our design algorithms, and focus on leveraging the VTN structure to partition the network into smaller scopes for better network performance. Our simulation and experimental results show that our multi-layer approach to VTN design can achieve better performance compared to the traditional single-layer design used for overlay/virtual networks.

Appendix: Proof of Equation 6

Equation 6 can be derived using an absorbing Markov chain. As shown in Fig.27, each circle denotes a possible state of the current packet, where $S_0$ is the initial state where a packet is to be sent by the sender, and $S_H$ is the absorbing (final) state when the packet is received by the receiver. For any intermediate state, it has a probability of $1-P$ of transitioning to the following state if the packet does not get lost; if the packet gets lost (with probability P), it goes back to the initial state ($S_0$). This absorbing Markov chain has H transient states ($S_0$ to $S_{H-1}$), and 1 absorbing state ($S_H$), so the expected number of total transmissions for all hosts along the path to successfully deliver one packet is the same as the expected number of steps from the initial state $S_0$ to the absorbing state $S_H$.

Fig. 27

An absorbing Markov chain for delivering one packet over a TCP connection of H hops, where each circle denotes a possible state. Assume loss rate on each link is P

Generally, for an absorbing Markov chain with transition matrix P, assume it has t transient states and r absorbing state, then

$$\begin{aligned} P= \begin{bmatrix} Q&R\\ 0&I_r\\ \end{bmatrix} \end{aligned}$$

where Q is a t-by-t matrix and I is the r-by-r identity matrix. The fundamental matrix of an absorbing Markov chain is $N = ( I - Q)^{-1}$, and the expected number of step from the initial state to the absorbing states is $t = N c$, where c is a column vector all of whose entries are 1 [36].

For the absorbing Markov chain in Fig. 27, $t= H$ and $r = 1$, so its transition matrix is

$$\begin{aligned} P[ (H + 1) \times (H+ 1)]= \begin{bmatrix} P&1-P&0&0&..&0&0\\ P&0&1-P&0&\ldots&0&0\\ P&0&0&1-P&\ldots&0&0 \\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots \\ P&0&0&0&\ldots&0&1-P\\ 0&0&0&0&\ldots&0&1 \\ \end{bmatrix} \end{aligned}$$

where

$$\begin{aligned} Q[ H \times H]= \begin{bmatrix} P&1-P&0&0&..&0 \\ P&0&1-P&0&\ldots&0 \\ P&0&0&1-P&\ldots&0 \\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots \\ P&0&0&0&\ldots&1-P \\ P&0&0&0&\ldots&0 \\ \end{bmatrix} \end{aligned}$$

Then its fundamental matrix is

$$\begin{aligned} N = ( I - Q)^{-1}= \begin{bmatrix} 1-P&P-1&0&0&..&0 \\ -P&1&P-1&0&\ldots&0 \\ -P&0&1&P-1&\ldots&0 \\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots \\ -P&0&0&0&\ldots&P-1 \\ -P&0&0&0&\ldots&1 \\ \end{bmatrix}^{-1} \end{aligned}$$

$$\begin{aligned} = \begin{bmatrix} (1-P)^{-H}&(1-P)^{1-H}&\ldots&(1-P)^{-2}&(1-P)^{-1} \\ \vdots&\vdots&\ddots&\vdots&\vdots \\ \end{bmatrix} \end{aligned}$$

So the expected number of steps from $S_0$ to $S_H$ is $t = N c = (1-P)^{-1} + (1-P)^{-2} + .. + (1-P)^{-H}$ = $\displaystyle { { ({ {1}\over {1-P} }) ^H - 1 }\over {P} }$

Then we prove that the expected number of total transmissions for all hosts along the path to successfully deliver one packet is $E_{tcp} = \displaystyle { { ({ {1}\over {1-P} }) ^H - 1 }\over {P} }$.