We study a primitive vehicle routing-type problem in which a fleet of $n$unit speed robots start from a point within a non-obtuse triangle $\Delta$, where $n \in \{1,2,3\}$. The goal is to design robots' trajectories so as to visit all edges of the triangle with the smallest visitation time makespan. We begin our study by introducing a framework for subdividing $\Delta$into regions with respect to the type of optimal trajectory that each point $P$ admits, pertaining to the order that edges are visited and to how the cost of the minimum makespan $R_n(P)$ is determined, for $n\in \{1,2,3\}$. These subdivisions are the starting points for our main result, which is to study makespan trade-offs with respect to the size of the fleet. In particular, we define $ R_{n,m} (\Delta)= \max_{P \in \Delta} R_n(P)/R_m(P)$, and we prove that, over all non-obtuse triangles $\Delta$: (i) $R_{1,3}(\Delta)$ ranges from $\sqrt{10}$ to $4$, (ii) $R_{2,3}(\Delta)$ ranges from $\sqrt{2}$ to $2$, and (iii) $R_{1,2}(\Delta)$ ranges from $5/2$ to $3$. In every case, we pinpoint the starting points within every triangle $\Delta$ that maximize $R_{n,m} (\Delta)$, as well as we identify the triangles that determine all $\inf_\Delta R_{n,m}(\Delta)$ and $\sup_\Delta R_{n,m}(\Delta)$ over the set of non-obtuse triangles.

翻译：我们研究一个基础的车辆路径类问题，其中 $n$ 个单位速度的机器人从非钝角三角形 $\Delta$ 内部一点出发，其中 $n \in \{1,2,3\}$。目标是设计机器人的轨迹，使得访问三角形所有边所需的完工时间（makespan）最小。我们首先引入一个框架，将 $\Delta$ 划分为若干区域，每个区域内的点 $P$ 对应特定类型的最优轨迹，这涉及边的访问顺序以及最小完工时间成本 $R_n(P)$ 的确定方式（$n\in \{1,2,3\}$）。这些划分是我们主要结果的起点，即研究不同规模机器人车队之间的完工时间权衡。具体而言，我们定义 $ R_{n,m} (\Delta)= \max_{P \in \Delta} R_n(P)/R_m(P)$，并证明在所有非钝角三角形 $\Delta$ 上：(i) $R_{1,3}(\Delta)$ 的取值范围为 $\sqrt{10}$ 到 $4$，(ii) $R_{2,3}(\Delta)$ 的取值范围为 $\sqrt{2}$ 到 $2$，(iii) $R_{1,2}(\Delta)$ 的取值范围为 $5/2$ 到 $3$。在每种情况下，我们精确指出了每个三角形 $\Delta$ 中使 $R_{n,m} (\Delta)$ 最大化的起始点位置，并确定了在非钝角三角形集合上决定所有 $\inf_\Delta R_{n,m}(\Delta)$ 和 $\sup_\Delta R_{n,m}(\Delta)$ 的三角形。