Airplane refueling problem (ARP) is a scheduling problem with an objective function of fractional form. Given a fleet of $n$ airplanes with mid-air refueling technique, each airplane has a specific fuel capacity and fuel consumption rate. The fleet starts to fly together to a same target and during the trip each airplane could instantaneously refuel to other airplanes and then be dropped out. The question is how to find the best refueling policy to make the last remaining airplane travels the farthest. We give a definition of the sequential feasible solution and construct a sequential search algorithm, whose computational complexity depends on the number of sequential feasible solutions referred to $Q_n$. By utilizing combination and recurrence ideas, we prove that the the upper bound of $Q_n$ is $2^{n-2}$. Then we focus on the worst-case and investigate the complexity of the sequential search algorithm from a dynamic perspective. Given a worst-case instance under some assumptions, we prove that there must exist an index $m$ such that when $n$ is greater than $2m$, $Q_n$ turns out to be upper bounded by $\frac{m^2}{n}C_n^m$. Here the index $m$ is a constant and could be regarded as an "inflection point": with the increasing scale of input $n$, $Q_n$ turns out to be a polynomial function of $n$. Hence, the sequential search algorithm turns out to run in polynomial time of $n$. Moreover, we build an efficient computability scheme by which we shall predict the complexity of $Q_n$ to choose a proper algorithm considering the available running time for decision makers or users.
翻译:空中加油问题( ARP) 是一个日程安排问题, 其客观功能为分数形式。 由于机队为美元, 机队为美元, 其计算复杂性取决于以美元为单位的连续可行解决方案数量, 每架飞机都有特定的燃料容量和燃料消耗率。 机队开始一起飞向同一目标, 在旅途中, 每架飞机可以瞬间再加油到其他飞机, 然后退出。 问题是如何找到最佳的加油政策, 使最后剩下的飞机旅行最远。 我们给出了序列可行性解决方案的定义, 并构建了顺序搜索算法, 其计算复杂性取决于以美元为单位的连续可行解决方案数量。 我们利用组合和重现的想法, 证明$的上限是2美元, 美元, 而每架飞机的上限是2美元。 然后我们从动态角度关注最坏的情况, 调查序列搜索算法的复杂性。 根据某些假设最坏的情况, 我们证明一定存在美元指数, 当美元比美元高时, 当我们2万美元, 美元, 美元的计算复杂性取决于 美元, 美元 以美元为美元为美元 。 美元 美元为美元 美元 。 以美元 美元 美元 正在运行一个固定 的 的 递增 。