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Wavelength assignment and generalized interval graph coloring

机译:波长分配和广义间隔图着色

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In this paper we study wavelength assignment on an optical linesystem without wavelength conversion. Consider a set of undirected demands along the line. Each demand is carried on a wavelength and any two overlapping demands on the same fiber require distinct wavelengths. Suppose μ wavelengths are available in the system and each fiber can carry all μ wavelengths. We define ℓ(e), the load on link e, to be the smallest integer such that ℓ(e)μ is at least the number of demands passing through e. Hence, ℓ(e) is the minimum number of fibers required on e in order to support all demands.We present a polynomial-time wavelength assignment algorithm that guarantees each wavelength appears at most ℓ(e) times on each link e. (This generalizes the well-known fact that interval graphs are perfect.) In the presence of MOADMs (mesh optical add/drop multiplexers), devices that multiplex distinct wavelengths from different fibers into a newfiber, we only need to deploy ℓ(e) fibers per link. On the other hand, if each demand has to stay on a single fiber, as is the case without MOADMs, we show that some links may require more than ℓ(e) fibers. In fact, we show that it is NP-complete to decide if a set of demands can be carried on a given set of fibers, or if there exists a set of fibers with a given total length that can carry all the demands.
机译:在本文中,我们研究了没有波长转换的光线路系统上的波长分配。考虑沿线的一组无方向需求。每个需求都承载一个波长,并且同一根光纤上的任何两个重叠需求都需要不同的波长。假设系统中有μ个波长,并且每根光纤都可以承载所有μ个波长。我们将链接 e 上的负载&ell;( e )定义为最小整数,使得&ell;( e )μ等于至少通过 e 传递的需求数量。因此,&ell;( e )是满足所有需求的 e 所需的最少光纤数。我们提出了多项式时间波长分配算法,该算法可以保证每个波长每个链接e最多出现&ell;( e )次。 (这概括了众所周知的事实,即间隔图是完美的。)在存在MOADM(网状光分插复用器)的情况下,将来自不同光纤的不同波长复用到新光纤中的设备,我们只需要部署&ell;(< I> e )每个链接的光纤。另一方面,如果每个需求都必须停留在一根光纤上(如不使用MOADM的情况一样),则我们表明某些链路可能需要的光纤超过&ell;( e )光纤。实际上,我们表明,确定是否可以在给定的一组光纤上执行一组需求,或者是否存在一组可以满足所有需求的,具有给定总长度的光纤是NP完全的。

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