Travelling Salesman Problem: Can the nearest neighbor algorithm be $n$ times longer than the optimal solution? This is inspired by a recent question.
Given a positive integer $n\in\mathbb{N}$, is there a setting of finitely many points and a designated "starting point" $s$ in $\mathbb{R}^2$ such that the nearest-neighbor algorithm (described below) gives a tour that is $n$ times longer than the optimal solution starting at $s$?


Starting at $s$, pick the nearest neighbor not visited so far as the next node to visit.


EDIT: If the answer is no, what is the maximum value that the ratio $r$ of "nearest neighbor trip" vs "best trip" can take?
 A: The nearest-neighbor (NN) heuristic (among others) is analyzed in this paper:

Johnson, David S., and Lyle A. McGeoch. "The traveling salesman problem: A case study in local optimization." Local search in combinatorial optimization 1 (1997): 215-310.
  (PDF download link.)

They say:



In the specific case of "random Euclidean instances"
(in contrast to "random distance matrices"),
they observe experimentally a fixed ~25% longer path length, which
would exceed any fixed $n$ for large enough nodes $N$.
In terms of both theory and experiments with random distance matrices,
the growth rate is $\log N$.
Either way, any fixed $n$ could be exceeded with sufficiently large $N$,
so the answer to the OP's question is Yes.

The cited paper is:
Rosenkrantz, Daniel J., Richard E. Stearns, and Philip M. Lewis, II. "An analysis of several heuristics for the traveling salesman problem." SIAM Journal on Computing 6.3 (1977): 563-581.
(Journal link.)

Added (in response to question from Manfred).
Here is the essence of the Rosenkrantz et al. lower bound:

          


          

(Slide from David Johnson PowerPoint Lecture.)


