What is the name of this logistic variant of TSP?
I have a logistic problem that can be seen as a variant of $\text{TSP}$. It is so natural, I'm sure it has been studied in Operations research or something similar. Here's one way of looking at the problem.
I have $P$ warehouses on the Cartesian plane. There's a path from a warehouse to every other warehouse and the distance metric used is the Euclidean distance. In addition, there are $n$ different items. Each item $1 \leq i \leq n$ can be present in any number of warehouses. We have a collector and we are given a starting point $s$ for it, say the origin $(0,0)$. The collector is given an order, so a list of items. Here, we can assume that the list only contains distinct items and only one of each. We must determine the shortest tour starting at $s$ visiting some number of warehouses so that the we pick up every item on the order.
Here's a visualization of a randomly generated instance with $P = 35$. Warehouses are represented with circles. 180
Red ones contain item $1$, blue ones item $2$ and green ones item $3$. Given some starting point $s$ and the order ($1,2,3$), we must pick one red, one blue and one green warehouse so the order can be completed. By accident, there are no multi-colored warehouses in this example so they all contain exactly one item. This particular instance is a case of set-TSP.
I can show that the problem is indeed $\mathcal{NP}$-hard. Consider an instance where each item $i$ is located in a different warehouse $P_i$. The order is such that it contains every item. Now we must visit every warehouse $P_i$ and find the shortest tour doing so. This is equivalent of solving an instance of $\text{TSP}$.
Being so obviously useful at least in the context of logistic, routing and planning, I'm sure this has been studied before. I have two questions:
1. What is the name of the problem?
2. How well can one hope to approximate the problem (assuming $\mathcal{P} \neq \mathcal{NP}$)?
I'm quite happy with the name and/or reference(s) to the problem. Maybe the answer to the second point follows easily or I can find out that myself. 835