On the Complexity of the Regenerator Cost Problem in General Networks with Traffic Grooming

Abstract

In a communication network one often needs to combine several communication requests into a path in a physical layer of the network. In these cases the cost is measured in terms of the total length of these paths or the total hardware cost of maintaining these paths. In this paper we consider a problem belonging to this general family of optimization problems. We consider the problem of minimizing the number of regenerators in optical networks with traffic grooming. In this problem we are given a network with an underlying topology of a graph G, a set of requests that correspond to paths in G and two positive integers g and d. There is a need to put a regenerator every d edges of every path, because of a degradation in the quality of the signal. Each regenerator can be shared by at most g paths, g being the grooming factor. On the one hand, we show that even in the case of d=1 the problem is APX-hard, i.e. a polynomial time approximation scheme for it does not exist (unless P=NP). On the other hand, we solve such a problem for general G and any d and g, by providing an O(logg)-approximation algorithm and thus extending previous results holding only for specific topologies and specific values of d or g.

Appendix: Proof of Lemma 3

Without loss of generality assume that the elements are numbered in the order they are covered by GreedySetCover ₂. Whenever a set S _j is chosen to the cover, we distribute its cost weight[S _j ] to the new elements it covers, by assigning a price of price(i)=eff[S _j ] to each element i added to Covered. Clearly the cost of the solution is $\sum_{i=1}^{n} \mathit{price}(i)$.

In² any iteration, the leftover sets of the optimal solution can cover the remaining elements at a cost of at most OPT. Therefore, among these sets, there must be one having cost-effectiveness of at most OPT/|A∖Covered|. In the iteration in which element i was covered, A∖Covered contained at least n−i+1 elements. Since i was covered by a set with cost-effectiveness at most ρ times the minimum, it follows that $\mathit{price}(i) \leq\rho\cdot\frac{\mathit{OPT}}{|A \setminus \mathit{Covered}|} \leq\rho\cdot\frac{\mathit{OPT}}{n - i + 1}$. Therefore the cost of the solution returned by GreedySetCover ₂ is

$$\sum_{i=1}^n \mathit{price}(i) \leq\sum _{i=1}^n \rho\cdot\frac {\mathit {OPT}}{n-i+1} = \rho \cdot\mathit{OPT} \cdot H_n. $$

In order to improve the above result so that to H _n is replaced by H _k where k is the size of the largest set, one can take an existing proof (e.g. [2]) and modify it as we did in the above proof or, alternatively, use the above result as a black box as we do in the following proof.

Proof

Let $S_{i_{1}}, S_{i_{2}}, \ldots$ be the sets chosen by GreedySetCover ₂, and for every such $S_{i_{j}}$ let $\bar{S_{i_{j}}} = S_{i_{j}} \setminus\bigcup_{k=1}^{j-1}S_{i_{k}}$, the subset of $S_{i_{j}}$ consisting of the new elements covered by it. Let also $S^{*}_{1}, S^{*}_{2}, \ldots$ be the sets of an optimal solution, and $\bar{S^{*}_{j}}$ defined similarly.

Consider an oracle that knows the sets $S^{*}_{1}, S^{*}_{2}, \ldots, S^{*}_{\ell}$ and partitions each set S _i of the input into sets S _i1,S _i2,…, such that $S_{ij} = S_{i} \cap\bar{S^{*}_{j}}$. Moreover this oracle distributes the weight of each set S _i to the sets S _ij in a manner proportional to their size, i.e. $\mathit {weight}[S_{ij}]=\mathit{weight}[S_{i}] \frac{\vert S_{ij}\vert }{\vert S_{i}\vert }$. Now consider the following possible run of GreedySetCover ₂ when its input consists of the sets S _ij . It chooses one of the sets $S_{{i_{1}}j}$, (because they have the same efficiency as $S_{i_{1}}$ which is at most ρ times the best possible), followed by all the other sets $S_{{i_{1}}j}$. Then it chooses all the sets $S_{{i_{2}}j}$, etc. … The cost of the solution obtained in this run is equal to the cost of the solution returned by GreedySetCover ₂ on the original input, i.e., $\sum\mathit{weight}[S_{{i_{k}}j}] = \sum \mathit{weight}[S_{i_{k}}]$. We can see this run as ℓ copies of GreedySetCover ₂ running in parallel each of which has as goal to cover one set $\bar {S^{*}_{j}}$. Each copy computes a $\rho\cdot H_{|\bar{S^{*}_{j}}|}$ approximated solution to the optimum, which is at most $\mathit {weight}[S^{*}_{j}]$. Summing up for all the copies, we get that the overall cost is at most

$$\sum_{j=1}^\ell\rho\cdot H_{|\bar{S^*_j}|} \cdot\mathit {weight}\bigl[S^*_j\bigr] \leq \rho\cdot H_k \sum_{j=1}^\ell\mathit{weight} \bigl[S^*_j\bigr] = \rho\cdot H_k \cdot \mathit{OPT} $$

where k is the size of the largest set in the input. □