OpenFlow Compatible Key-Based Routing Protocol: Adapting SDN Networks to Content/Service-Centric Paradigm

Abstract

The host-to-host/content/service communication instead of the host-to-host communication offered by traditional Internet Protocol (IP) routing solutions has been demanded in the last few years. Nowadays, getting this type of communication directly at network level is an increasing demand in the framework of new networking scenarios, such as Internet of Things and data center scenarios. Inspired by Key-Based Routing (KBR) solutions which, in conjunction with Distributed Hash Tables, have offered a way of providing content-sharing solutions in overlay networks on the top of the Internet for years now, we propose OFC-KBR (OpenFlow Compatible Key-Based Routing) solution. OFC-KBR is a key-based routing solution directly implemented at network layer that makes use of the potential of Software Defined Networking. In this solution, end-points are identified by virtual identifiers. These virtual identifiers are obtained from a descriptive textual name, whose format is not fixed and can be defined depending on the requirements of the service that is going to use the proposed OFC-KBR solution. OFC-KBR is totally compatible with the current OpenFlow standard and can co-exist with other L2/L3 protocols. The proposal has been implemented and evaluated by simulation considering real topologies.

Appendix A

1.1 Mathematical Model

Let us suppose a Chord network of size $2^L$, and composed of N nodes. Without loss of generality, it is assumed that the vid assigned to one of the nodes is 0. We are interesting in calculating the probability distribution of the number of ranges obtained from the finger table of a node, assuming that the rest of the $N-1$ nodes are randomly distributed in the vid space $[0,2^L-1]$.

By definition, the finger table of a node is made up of L entries and the predecessor. The information of the finger table can be used to obtain $L+1$ ranges and the corresponding actions. The entry i associates the range $[2^i,2^{i+1}[$ with first node clockwise $2^i$ in the virtual space. If more than one consecutive entries of the finger table point to the same node, these ranges can be merged into a single range. Under these assumptions, we are going to obtain the probability that the number of different ranges obtained for a node is k, where $2\le k \le L+1$ when the total number of nodes in the network is N.

From now on, the interval $[2^i,2^{i+1}[$ defined from the finger table of node 0 (the reference node) before merging will be called the i-th interval. If the first node clockwise $2^i$ (that is, finger table value associated to this entry) is not within the ith interval, we define this i-th the interval as an empty interval. As can be seen, an i-th interval is an empty interval if any of the existing N nodes are located within the associated interval. On the other hand, an i-th interval is a non-empty interval if one or more than one of the N nodes are located within the interval. Thus, all the consecutive empty intervals will be merged with the following non-empty interval. As a conclusion, if there are $k-1$ non-empty intervals and the rest are empty intervals, then, the finger table will produce k different ranges. As shown in Fig. 3, the range ]predecessor, n] that points to the own node will always be present.

Therefore, there will be k different ranges if the distribution of the N nodes in the virtual space allows node 0 to have $k-1$ non-empty intervals and $l-(k-1)$ empty intervals. However, due to the fact that each i-th interval has its own capacity $(2^i)$, not all the $k-1$ combinations of $k-1$ non-empty intervals are valid. In these not valid cases, the total capacity is less than $N-1$, and it is not possible to distribute the $N-1$ nodes among them.

As an example, consider $l=6$, $N=10$, and $k=3$. There are 15 different combinations of 2 non-empty intervals and 4 empty intervals, but some of them are not valid. For example, if the non-empty intervals are the 0-th and 1-st intervals, the capacity is $2^0+2^1=3$, not enough to locate $N-1=9$ nodes. The same happens with the combinations 0, 2, 0, 3, 0, 4, 0, 5, and 1, 2. In the end, only 9 out of 15 combinations have enough capacity to locate N-1 nodes: 1, 3, 1, 4, 1, 5, 2, 3, 2, 4, 2, 5, 3, 4, 3, 5, and 4, 5.

Now, let us consider a valid combination of intervals of the above example (2, 3). Both intervals have the capacity to locate up to $2^2+2^3=12$ nodes. Therefore, there are 4 possibilities to distribute the $N-1=9$ nodes within the intervals: $n_2=1$, $n_3=8$, $n_2=2$, $n_3=7$, $n_2=3$, $n_3=6$ and $n_2=4$, $n_3=5$.

Consequently, among all the valid combinations of intervals, it should also be considered those that meet the following equation:

$$\begin{aligned} \varGamma \in {i,j,\ldots ,k} | n_i+n_j+\cdots +n_k=N-1 \end{aligned}$$

(3)

where $n_i$ is the number of nodes in the interval i-th. Thus, the probability that the finger table of a node causes k different ranges is:

$$\begin{aligned} P(k)=\sum _\varGamma P(n_1)P(n_2)P(n_{k-1}) \end{aligned}$$

(4)

where $P(n_i)$ is the probability that $n_i$ of the total number of nodes $N-1$ are located in the interval i-th (with capacity to hold up to $l_i$ nodes).

Finally, the probability can be easily obtained using combinatorial numbers as follows:

$$\begin{aligned} P(k)=\sum _\varGamma \frac{\left( {\begin{array}{c}l_1\\ n_1\end{array}}\right) \left( {\begin{array}{c}l_2\\ \cdots \end{array}}\right) \left( {\begin{array}{c}l_{k-1}\\ n_{k-1}\end{array}}\right) }{\left( {\begin{array}{c}2^l-1\\ N-1\end{array}}\right) } \end{aligned}$$

(5)

1.2 Validation

The mathematical model has been validated using a simulation tool programmed in C language. The simulator considers the reference node (node 0) and randomly selects other $N-1$ nodes in a virtual of $2^L-1$ values ($]0,2^L-1]$). Then, the simulator calculates the number of different ranges associated to node 0 (the reference node) accordingly to the procedure proposed in OFC-KBR solution, taking into account the existing nodes. This simulation is repeated $10^6$ times in order to reach steady state probabilities.

Fig. 12

Virtual space=$\left[ 0,2^6-1 \right]$ and N = 6, 12, 19, 25 and 32, corresponding to 10%, 20%, 30%, 40% and 50%. a Probability distribution of the number of ranges obtained by simulation. b Probability distribution of the number of ranges obtained using the mathematical model

Figure 12a shows the probability distribution of the number of ranges associated to a node, obtained by simulation, when $L=6$ (that is, a virtual space = $[0,2^6-1]$, and the number of nodes is $N=6,12,19,25$ and 32 , which corresponds to 10%, 20%, 30%, 40% and 50% of the capacity of the network.

As can be seen, as the number of nodes (N) increases, the probability distribution moves to the right. As the number of nodes increases, the probability that the ranges obtained directly from the entries of the finger table can be merged is smaller, because the entries point to different nodes. In particular, in the present study, the mean number of ranges are 3.8, 4.8, 5.4, 5.8 and 6.2 for $N=6,12,19,25$ and 32, respectively.

On the other hand, Fig. 12b shows the probability distribution of the number of ranges associated to a node, obtained by the mathematical model. The exact match between both figures clearly validates the model.