In project management, a project is typically described as an activity-on-edge network (AOE network), where each activity / job is represented as an edge of some network $N$ (which is a DAG). Some jobs must be finished before others can be started, as described by the topology structure of $N$. It is known that job $j_i$ in normal speed would require $b_i$ days to be finished after it is started. Given the network $N$ with the associated edge lengths $b_1,\ldots,b_m$, the duration of the project is determined, which equals the length of the critical path (namely, the longest path) of $N$. To speed up the project (i.e. reduce the duration), the manager can crash a few jobs (namely, reduce the length of the corresponding edges) by investing extra resources into that job. However, the time for completing $j_i$ has a lower bound due to technological limits -- it requires at least $a_i$ days to be completed. Moreover, it is expensive to buy resources. Given $N$ and an integer $k\geq 1$, the $k$-crashing problem asks the minimum amount of resources required to speed up the project by $k$ days. We show a simple and efficient algorithm with an approximation ratio $\frac{1}{1}+\ldots+\frac{1}{k}$ for this problem. We also study a related problem called $k$-LIS, in which we are given a sequence $\omega$ of numbers and we aim to find $k$ disjoint increasing subsequence of $\omega$ with the largest total length. We show a $(1-\frac{1}{e})$-approximation algorithm which is simple and efficient.

翻译：摘要：在项目管理中，项目通常被描述为边活动网络（AOE网络），其中每个活动/作业表示为某个网络 $N$（一个DAG）中的一条边。某些作业必须完成后，其他作业才能开始，这由 $N$ 的拓扑结构描述。已知作业 $j_i$ 以正常速度进行时，在开始后需要 $b_i$ 天完成。给定网络 $N$ 及其关联的边长度 $b_1,\ldots,b_m$，项目工期得以确定，其等于 $N$ 的关键路径（即最长路径）长度。为加速项目（即缩短工期），管理者可通过向某些作业投入额外资源来压缩这些作业（即缩短对应边的长度）。然而，由于技术限制，完成 $j_i$ 的时间存在下界——至少需要 $a_i$ 天才能完成。此外，购买资源成本高昂。给定 $N$ 和一个整数 $k\geq 1$，$k$ 压缩问题要求将项目工期缩短 $k$ 天所需的最小资源量。我们针对该问题展示了一个简单且高效的算法，其近似比为 $\frac{1}{1}+\ldots+\frac{1}{k}$。我们还研究了一个相关问题，称为 $k$ 最长递增子序列问题（$k$-LIS），其中给定一个数字序列 $\omega$，目标是在 $\omega$ 中找出总长度最大的 $k$ 个互不相交的递增子序列。我们展示了一个简单且高效的 $(1-\frac{1}{e})$ 近似算法。