The capacitated vehicle routing problem (CVRP) is one of the most extensively studied problems in combinatorial optimization. In this problem, we are given a depot and a set of customers, each with a demand, embedded in a metric space. The objective is to find a set of tours, each starting and ending at the depot, operated by the capacititated vehicle at the depot to serve all customers, such that all customers are served, and the total travel cost is minimized. We consider the unplittable variant, where the demand of each customer must be served entirely by a single tour. Let $α$ denote the current best-known approximation ratio for the metric traveling salesman problem. The previous best approximation ratio was $α+1+\ln 2+δ<3.1932$ for a small constant $δ>0$ (Friggstad et al., Math. Oper. Res. 2025), which can be further improved by a small constant using the result of Blauth, Traub, and Vygen (Math. Program. 2023). In this paper, we propose two improved approximation algorithms. The first algorithm focuses on the case of fixed vehicle capacity and achieves an approximation ratio of $α+1+\ln\bigl(2-\frac{1}{2}y_0\bigr)<3.0897$, where $y_0>0.39312$ is the unique root of $\ln\bigl(2-\frac{1}{2}y\bigr)=\frac{3}{2}y$. The second algorithm considers general vehicle capacity and achieves an approximation ratio of $α+1+y_1+\ln\left(2-2y_1\right)+δ<3.1759$ for a small constant $δ>0$, where $y_1>0.17458$ is the unique root of $\frac{1}{2} y_1+ 6 (1-y_1)\bigl(1-e^{-\frac{1}{2} y_1}\bigr) =\ln\left(2-2y_1\right)$. Both approximations can be further improved by a small constant using the result of Blauth, Traub, and Vygen (Math. Program. 2023).

翻译：容量车辆路径问题（CVRP）是组合优化领域研究最为广泛的问题之一。在该问题中，给定一个仓库和一组客户（每个客户具有特定需求），它们嵌入在一个度量空间中。目标是找到一组从仓库出发并返回仓库的路径，由仓库处具有容量限制的车辆执行，以服务所有客户，使得所有客户均被服务且总行驶成本最小。我们考虑不可分割变体，其中每个客户的需求必须由单条路径完整服务。令 $α$ 表示度量旅行商问题当前已知的最佳近似比。先前的最佳近似比为 $α+1+\ln 2+δ<3.1932$（其中 $δ>0$ 为小常数，Friggstad 等人，Math. Oper. Res. 2025），该结果可利用 Blauth、Traub 和 Vygen 的成果（Math. Program. 2023）进一步改进一个小常数。本文提出了两种改进的近似算法。第一种算法针对固定车辆容量的情况，实现了 $α+1+\ln\bigl(2-\frac{1}{2}y_0\bigr)<3.0897$ 的近似比，其中 $y_0>0.39312$ 是方程 $\ln\bigl(2-\frac{1}{2}y\bigr)=\frac{3}{2}y$ 的唯一根。第二种算法考虑一般车辆容量，实现了 $α+1+y_1+\ln\left(2-2y_1\right)+δ<3.1759$ 的近似比（$δ>0$ 为小常数），其中 $y_1>0.17458$ 是方程 $\frac{1}{2} y_1+ 6 (1-y_1)\bigl(1-e^{-\frac{1}{2} y_1}\bigr) =\ln\left(2-2y_1\right)$ 的唯一根。两种近似结果均可利用 Blauth、Traub 和 Vygen 的成果（Math. Program. 2023）进一步改进一个小常数。