Algorithms for Dynamic NFV Workload

Abstract

The dynamic NFV placement problem captures one of the main challenges facing the telecom industry following the emergence of the Network Function Virtualization (NFV) networking paradigm, that is, deciding the placement of network functions while taking into consideration the dynamic nature of networks and workloads. We model the problem as a generalization of the classic Uncapacitated Facility Location (UFL) problem, where we consider both multiple types of commodities and dynamic clients whose location changes over time.

We show that under reasonable assumptions we are able to develop a 7-approximation algorithm for the Dynamic Facility Location (DFL) problem, improving the logarithmic approximation of Eisenstat et al. [6]. We build upon this result to develop the first virtualized services placement algorithm that accounts for dynamic changes. Our tri-criteria approximation algorithms provide constant approximation factors with respect to the overall performance and size constraints, and logarithmic approximation factors with respect to capacity constraints.

Appendices

A Useful Procedures

Our model generalizes several known problems. For example, [5] showed that the generalized assignment problem (see Sect. 2.3) and the uncapacitated facility location problem (see Introduction) are special cases of the NFV problem.

Here we describe several known procedures for theses problems which we use throughout the paper.

1.1 A.1 Cover-Growing Algorithm

Here we describe a cover-growing algorithm for the uncapacitated facility location (UFL) problem. We present it together with its analysis, since we use it later and take advantage of its local properties. We assume that our input is a fractional solution to the UFL problem. The output is an integral solution, i.e., a set of open facilities such that each client is assigned to an open facility.

We can view a fractional solution to the facility location problem as inducing a probability distribution over the facilities from which a client gets service. Thus, the fractional connection cost of a client is an expectation, since it is a sum of weighted distances (where the service fractions serve as weights). A cover (or ball) around a client, having radius twice the expected distance, contains at least half of the client’s fractional service. Thus, by doubling the fractions inside the cover, the client gets all of its service from it.

Rounding Algorithm

• 1. Define a cover around each client with radius twice the expected distance.

• 2. Until all clients are satisfied:

  • (a) among all unconnected clients, find client i with minimum radius cover.

  • (b) open facility f that minimizes the installation cost in the cover.

  • (c) for every client $i'$ whose cover intersects the cover of client i (there exists a facility that serves both): connect it to facility f.

It follows from the rounding algorithm that every client is either connected to a facility in its own cover or connected to a facility in an intersecting cover (which does not have a larger radius). In this case we say that a client j is connected to a facility in its representative cover. Thus, the connection costs are at most 6 times the sum of the expected distances, and the installation costs are at most twice the fractional installation costs. In total, the approximation factor achieved is 6 for the uncapacitated facility location.

1.2 A.2 Generalized Assignment Problem

In the generalized assignment problem (GAP) we are given m machines and n jobs that need to be assigned to the machines. Job j has cost $c_{ij}$ and size $w_{ij}$ on machine i; machine i has total size $w_i$. Our goal is to assign each job to a machine, without violating machine size constraints, while minimizing the total assignment cost. Assume we are given a feasible fractional solution to GAP. In our algorithms we apply a rounding procedure to the given fractional solution due to Shmoys and Tardos (see [10]). The output of the rounding procedure is an integral solution whose cost is at most the cost of the fractional solution, and the size of every machine is violated by at most the maximum size of a job assigned fractionally to the machine, i.e., by at most a factor of two. GAP is a special case of the NFV placement problem in which all distances in the metric are set to be zero. In this case at most one copy of each function (i.e., a job) is installed, yielding a GAP instance.

1.3 A.3 Interval Selection

In the linear relaxation (presented in Sect. 3) of the DFL problem we pay for fractional changes in the assignment of a client to a facility between consecutive time steps. Eisenstat et al. [6] gave a procedure that breaks the time horizon into intervals, separately for each client, such that in each interval the fractional connection is static. The fractional change in each interval is bounded. The idea behind the procedure is to iteratively construct the intervals for each client. An interval terminates at the latest time step t in which the fractional changes that were accumulated through until t are bounded. The procedure for client j is as follows:

• 1. Set $t_0^j=1$ and $\ell =1$.

• 2. Next interval starts at the maximal t, $t\in (t_\ell ^j,T+1]$, such that $\sum _{i\in F}(\min _{\{t_\ell ^j\le u<t\}}x_{ij}^u)\ge \theta$ (where $\theta \in [0,1]$).

• 3. If $t=T+1$, all intervals are selected; otherwise, set $\ell \leftarrow \ell +1$ and select next interval.

For each interval we set the new static fractions $\hat{x}$ as follows. For each t in the $\ell$th interval,

$$\begin{aligned} \hat{x}_{ij}^t = \frac{\min _{\{t_\ell ^j\le u<t_{\ell +1}^j\}}x_{ij}^u}{\sum _{i'\in F}\min _{\{t_\ell ^j\le u<t_{\ell +1}^j\}}x_{i'j}^u}. \end{aligned}$$

It is straightforward to verify that the fractions in the solution are at most multiplied by $\frac{1}{\theta }$, since the numerator is smaller than all fractions in its interval and the denominator is at least $\frac{1}{\theta }$. If we multiply the installation fractions by $\frac{1}{\theta }$, then the solution is feasible. Next, we want to show that fractional changes in each interval are at least $1-\theta$. If so, since we only change the assignment at the end of each interval, we pay at most $\frac{1}{1-\theta }$ times the change cost. If we consider the $\ell$th interval, the total fractional change in assignment in the interval is

$$\begin{aligned} \sum _{ i\in F} \sum _{t_\ell ^j\le u\le t_{\ell +1}^j} z_{ij}^u&\ge \sum _{i\in F}(x_{ij}^{t_\ell ^j}-\min _{\{ t_\ell ^j\le u\le t_{\ell +1}^j \} }x_{ij}^u) \\&= 1 - \sum _{i\in F}\min _{\{ t_\ell ^j\le u\le t_{\ell +1}^j \} }x_{ij}^u \ge 1 - \theta . \end{aligned}$$

The first inequality follows since the change is at least the first fractional connection minus the min fractional connection.

B Horizon Length Dependent Approximation

Assuming there is a feasible fractional solution to the (IGLC) program, we split the time horizon into two independent parts which we color separately, achieving an $O(\log T)$ approximation. Similarly to the previous algorithm, we exploit the fact that each clique can be legally colored.

Time-Split Algorithm

• 1. Let t be the middle point in T.

• 2. Color the intervals in clique I(t).

• 3. Continue recursively on both halves of T. At the ith iteration use the ith copy of the set of colors.

Lemma 5

Time-Split Algorithm provides $O(\log T)$ approximation for the IGLC problem.

Proof

Each clique can be seen as a matching of intervals to colors. We are guaranteed by the fractional solution (to IGLC in Sect. 2.2) that the number of colors available for any set of $\ell$ intersecting intervals is at least $\ell$. Therefore, by Hall’s theorem, there is a perfect matching in each clique from intervals to colors. Next, at each step we color the middle point with a new set of colors. After removing all colored intervals we are left with two ranges that do not share any intervals thus, we can use the next copy of the set of colors at both of them. Each time the we color a clique we create two ranges that are at most half the size of the original range, so it is obvious we will not need more than $O(\log T)$ copies of the colors to color all intervals.

C Experiments

We devote this section to test the uncapacitated dynamic NFV algorithm. To this end we consider a subnet of the physical network of Cogent, a tier 1 ISP, which offers us a realistic facilities’ deployment (using its publicly available data center locations). We choose ten data centers placed in Europe and defined facilities at their location. Next, we added a hundred clients in random positions, and for each client defined a random walk, taking a step at a random direction of random size. In addition, each client was assigned a random commodity vector describing which commodities, out of a list of five different commodities, it requires. Finally, a random size was assigned to each facility, together with a size and cost for each commodity (at each facility).

In order to evaluate the algorithm, we compare it with previous solutions for the problem, the uncapacitated (static) NFV from [5]. Since the static algorithm does not optimize over the time horizon of the dynamic problem, we define two different variants for using it. The first one uses the clients’ position over the time horizon to find the average position of each client. Using this average position we obtain a single time step instance. The second one runs the static algorithm at each time step separately, paying change costs accordingly. These two options define the two extreme options of a fully static solution, in which we do not allow any changes, and a fully dynamic one, which does not integrate between solutions to avoid overpaying for change costs and opening costs.

Fig. 2.

Performance ratio of the dynamic UNFV algorithm compared to the static UNFV with respect to the ratio between the average connection cost and the change cost. Top presents the full graph while the bottom zooms in on the smaller values.

In Fig. 2 shows the cost percentage of the two versions of the static algorithm compared to the cost of the dynamic algorithm. It can be seen that the intuition for the performance of the algorithm is correct, that is, for small change costs (relative to the connection cost), ignoring it results in higher costs, and for high change costs, it does not necessarily come with a cost. Still, giving consideration to the dynamic nature of the problem does give advantage to the dynamic algorithm, and for high change costs, the performance of the static algorithm and the dynamic algorithm, converge to one another.

Fig. 3.

Performance ratio of the dynamic UNFV algorithm compared to the static UNFV and the fractional solution with respect to theta.

Fig. 4.

Performance ratio of the dynamic UNFV algorithm compared to the static UNFV with respect to the facility size.

In Sect. 2.3 we discussed the interval selection procedure. For each client, we found the time steps so that in between substantial fractional changes in the assignment were accumulated, and we split the time horizon into intervals at these time steps. The question of at which point the “right amount” of fractional change has accumulated arises. We ran the experiments with a small change cost (0.04 of the average connection cost) and a large change cost (which equals the average connection cost). As seen in Fig. 3, this value, denoted by $\theta$, may have significant impact on the performance of the algorithm. If we choose a value too big, we may induce too many assignment changes as we break the time horizon into too many intervals. And for small values of $\theta$, we may end up with a static solution which may lose the advantage of the dynamic algorithm. Another evidence for this can be found in the comparison with the fractional solution. For values of $\theta$ in the range [0.4, 0.6], the ratio between the fractional solution and the algorithm’s solution is bigger. Usually the optimal solution and the fractional solution are not close. This leads us to assume that in practice, the ratio between the algorithm’s solution and the optimal solution is better than 2.

Lastly, in Fig. 4 we can see the performance ratio of the dynamic algorithm as a function of the expected number of commodities that can be installed in each facility. The performance of the algorithm peaks as the size constraint loosens. The lower performance ratio may be the result of tight size constraint which creates a hard problem without room for much improvement. On the other hand, when the sizes of the facilities are very large, we may install each commodity at several facilities to allow more assignment changes.