Integer programming models for the routing and spectrum allocation problem

Abstract

One of the most promising solutions to deal with huge data traffic demands in large communication networks is given by flexible optical networking, in particular the flexible grid (flexgrid) technology specified in the ITU-T standard G.694.1. In this specification, the frequency spectrum of an optical fiber link is divided into narrow frequency slots. Any sequence of consecutive slots can be used as a simple channel, and such a channel can be switched in the network nodes to create a lightpath. In this kind of networks, the problem of establishing lightpaths for a set of end-to-end demands that compete for spectrum resources is called the routing and spectrum allocation problem (RSA). Due to its relevance, this problem has been intensively studied in the last couple of years. It has been shown to be NP-hard (Christodoulopoulos et al. in IEEE J Lightw Technol 29(9):1354–1366, 2011; Wang et al. in IEEE J Opt Commun Netw 4(11):906–917, 2012) and several models and formulations have been proposed, leading to different solution approaches. In this work, we explore integer programming models for RSA, analyzing their effectiveness over known instances. We resort to several modeling techniques, to find natural formulations of this problem. Since integer programming techniques are known to provide successful practical approaches for several combinatorial optimization problems, the aim of this work is to explore a similar approach for RSA.

Appendices

Appendix

A consecutive ones proposition

Fix a demand and a link used by this demand. Let $S \in \mathbb {N}$ be the amount of available slots. If $s \in \{1, \dots , S\}$ is a slot then call $X_s$ the binary variable equal to one if and only if the slot s is used by the demand. Fix the demand volume h such that $h\le S$ and consider the following constraints:

$$\begin{aligned}&{\mathop {\mathop {\sum }\limits _{s \in \{1, \dots , S\}:}}\limits _{s ~ \equiv ~ j~(mod~h)}} X_{s} ~ = ~ 1&\begin{array}{r} \forall j \in \{1, \dots , h\} \end{array} \end{aligned}$$

(52)

$$\begin{aligned}&{\mathop {\mathop {\sum }\limits _{s \in \{1, \dots , i\}:}}\limits _{s ~ \equiv ~ i~(mod~h)}} X_{s} ~ \ge ~ {\mathop {\mathop {\sum }\limits _{s \in \{1, \dots , i-1\}:}}\limits _{s +1 ~ \equiv ~ i~(mod~h)}} X_{s}&\begin{array}{r} \forall i \in \{1, \dots , S+1-h\} \end{array} \end{aligned}$$

(53)

Theorem 1

Constraints  (52) and (53) ensure that there are exactly h consecutive variables in $\{X_1, \dots , X_S\}$ taking value 1.

Proof

Using constraints (53) and a property of constraints (52) we shall prove that there are at least h consecutive variables taking value 1, and using (52) we shall prove that those h variables are the only ones taking this value.

For $i=1,\dots , s+1-h$, let $C_i$ be the constraint (53) associated to the index i, and let $\mathrm{LHS}_i$ and $\mathrm{RHS}_i$ be the left-hand and the right-hand side of $C_i$ respectively, i.e.,

$$\begin{aligned} C_i~:~\mathrm{LHS}_i ~\ge ~ \mathrm{RHS}_i. \end{aligned}$$

Since $\mathrm{LHS}_i ~ = ~ \mathrm{RHS}_{i+1}$ for $i=1,\dots ,s-h$, then $C_i$ takes the form

$$\begin{aligned} \mathrm{LHS}_{i+1} ~\ge ~ \mathrm{LHS}_i. \end{aligned}$$

(54)

Let $L_i$ be the set of indices of $\mathrm{LHS}_i$, such that $X_s$ appears in $\mathrm{LHS}_i$ if and only if $s \in L_i$. We have $i\in L_i$ but $i \not \in L_{i+1},\dots , i \not \in L_{i+h-1}$, implying that i is the largest element in $L_i$, i.e., $i = \max (L_i)$. For $q = 0, \dots , h-1$ and $i=1,\dots ,S-q$, we conclude that the only index larger than i in $L_{i+q}$ is $i+q$, i.e.,

$$\begin{aligned} \{i': i'\ge i,\;i'\in L_{i+q}\} = \{i+q\} \end{aligned}$$

(55)

Now we use these results to prove that any feasible solution that satisfies constraints (52) and (53) has at least h consecutive variables set to 1. For $i=1,\dots ,S$, we have

$$\begin{aligned} \{s \in \{1, \dots , i\}:~s ~ \equiv ~ i~(mod~h)\} \subseteq \{s \in \{1, \dots , S\}:~s ~ \equiv ~ i~(mod~h)\}. \end{aligned}$$

Thus, constraints (52) imply:

$$\begin{aligned} \mathrm{LHS}_i ~ \le 1 \end{aligned}$$

(56)

for $i=1,\dots ,S+1-h$.

Let $X^* \in \{0,1\}^{S}$ be a feasible solution, and let p be the lowest index such that $X^*_p = 1$. Let $i^*$ be the lowest index such that $X^*_p \in \mathrm{LHS}_{i^*}$. Because of (56) and since $x_p=1$ then $\mathrm{LHS}_{i^*} = 1$. Likewise, because of (54) and (56), we have $\mathrm{LHS}_k=1$ for every $k\ge i^*$. In particular,

$$\begin{aligned} \mathrm{LHS}_k = 1~ \text{ for }~ k=i^*,\dots ,i^*+h. \end{aligned}$$

By the definition of p, we have $X_{p'} = 0$ for $p'<p$. This implies that in $\mathrm{LHS}_k$ such that $i^*<k<i^*+h$ only variables with indices greater than p can take value 1; but because of (55) those are $\{X_{p+1}, \dots , X_{p+h-1}\}$.

Therefore, we have at least h consecutive variables taking value 1. Consider now the union of the sets $\{s\in \{1, \dots ,S\}:~s \equiv j~(mod~h)\}$ for $j \in \{1, \dots , h\}$. Since all their elements are disjoint, then

$$\begin{aligned}&\bigcup _{j=1}^{h}\{s\in \{1,\dots , S\}:~s \equiv j~(mod~h)\} \\&\qquad =\{s\in \{1,\dots , S\}:~s\equiv j~(mod~h),~ j \in \{1,\dots , h\}\}= \{1,\dots , S\}. \end{aligned}$$

This implies that there are at most h variables taking value 1, i.e.,

$$\begin{aligned}&\sum _{j\in \{1,\dots , h\}}{\mathop {\mathop {\sum }\limits _{s \in \{1, \dots , S\}:}}\limits _{s ~ \equiv ~ j~(mod~h)}} X_{s} ~ \le ~ h \end{aligned}$$

Since there are at least and at most h consecutive variables taking value 1, there are exactly h variables taking value 1, hence concluding the proof. $\square$