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Non-approximability of the single crane container transhipment problem

机译:单个起重机集装箱转运问题的非近似性

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The makespan of operations at container terminals is crucial for the lead time of cargo and consequently the reduction of transportation costs. Therefore, an efficient transhipment and short storage of containers are demanded. Our paper refers to the consolidation process of trains in a container transhipment terminal as well as to the intermediate storage of containers in seaports in order to accelerate the loading and unloading of the vessels. It can also be encountered in automated storage/retrieval systems. Each of these (container) storage and retrieval moves corresponds to a crane operation, carrying a load from its pickup to its drop-off position. The problem is to find a permutation of the loaded crane moves that minimises the total empty crane travel time, which is the sum of times the crane needs to get from the last drop-off point of a load to the next pickup point of a load. We address the problem as an extension of an asymmetric travelling salesman problem (ATSP), assuming that n ordered pairs of points in the two-dimensional Euclidean space need to be traversed. Each point corresponds to a crane operation carrying a load from its pickup to its drop-off position. Despite that the problem seems to be easier than the ATSP, because a simple constant factor approximation exists, which was for a long time an open question for the ATSP, we are the first to prove that there is no polynomial-time approximation algorithm with an approximation guarantee less than unless .
机译:集装箱码头的Mapspan手动对于货物的提前期并且因此降低运输成本至关重要。因此,需要高效的转运和容器的储存。我们的论文是指集装箱转运终端中列车的整合过程,以及海港的容器中间储存,以加速船舶的装载和卸载。它也可以在自动存储/检索系统中遇到。这些(容器)存储和检索动作中的每一个对应于起重机操作,从其拾取到其下降位置携带负载。问题是找到负载起重机移动的置换,这最小化了总空的起重机行程时间,这是起重机需要从负载的最后一个下降点到加载的下一个拾取点的次数。我们将问题解决作为不对称旅行推销员问题(ATSP)的延伸,假设需要遍历二维欧几里德空间中的N个点点对。每个点对应于将负载从其拾取到其下降位置的起重机操作。尽管问题似乎比ATSP更容易,所以存在简单的恒定因子近似,这是一个很长时间的ATSP的打开问题,我们是第一个证明没有多项式时间近似算法近似保证少于否则。

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