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An Algorithm for Single-Item Capacitated Economic Lot Sizing with Piecewise Linear Production Costs and General Holding Costs

机译:具有分段线性生产成本和一般持有成本的单项有能力经济批量的算法

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摘要

We consider the Capacitated Economic Lot Size Problem with piecewise linear production costs and general holding costs, which is an NP-hard problem but solvable in pseudo-polynomial time. A straightforward dynamic programming approach to this problem results in an O(n 2 c\bar d\bar ) algorithm, where n is the number of periods, and d\bar and c\bar are the average demand and the average production capacity over the n periods, respectively. However, we present a dynamic programming procedure with complexity O(n 2 q\bar d\bar ), where q\bar is the average number of pieces required to represent the production cost functions. In particular, this means that problems in which the production functions consist of a fixed set-up cost plus a linear variable cost are solved in O(n 2 d\bar ) time. Hence, the running time of our algorithm is only linearly dependent on the magnitude of the data. This result also holds if extensions such as backlogging and startup costs are considered. Moreover, computational experiments indicate that the algorithm is capable of solving quite large problem instances within a reasonable amount of time. For example, the average time needed to solve test instances with 96 periods, 8 pieces in every production cost function, and average demand of 100 units is approximately 40 seconds on a SUN SPARC 5 workstation.
机译:我们考虑具有分段线性生产成本和一般持有成本的容量经济批量问题,这是一个NP问题,但可以在伪多项式时间内解决。针对该问题的直接动态编程方法导致了O(n 2 c \ bar d \ bar)算法,其中n是周期数,d \ bar和c \ bar是平均需求和平均生产能力。 n个时期。但是,我们提出了一种动态编程程序,其复杂度为O(n 2 q \ bar d \ bar),其中q \ bar是表示生产成本函数所需的平均件数。尤其是,这意味着在O(n 2 d \ bar)的时间内解决了生产功能由固定设置成本加上线性可变成本组成的问题。因此,我们算法的运行时间仅线性地取决于数据的大小。如果考虑扩展,例如积压和启动成本,则此结果也成立。此外,计算实验表明,该算法能够在合理的时间内解决相当大的问题实例。例如,在SUN SPARC 5工作站上,解决具有96个周期的测试实例所需的平均时间,每个生产成本函数需要8件,平均需求为100个单位,大约需要40秒。

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