TITLE: The motorcyclist's challenge
QUESTION [8 upvotes]: n walkers ${A}_{i}$ start out from X to Y simultaneously with speeds ${a}_{i}$, $i=1,2,...,n$. ${a}_{i}\neq {a}_{j}$ if $i\neq j$. At the same time, a motorcyclist M with speed $m=1$ starts out from Y to carry them (as shown in the illustration attached). The motorcyclist can pick up any person he meets. He can carry at most one person. But he's free to drop off the person he carries at any time. Walkers just walk straight forwards (from X to Y), while motorcyclist is free to drive either forwards or backwards.
The motorcylist's goal is to find a plan under which all the walkers arrive at Y simultaneously, if such a plan exists. We call such a plan a feasible plan. 
Define the configuration of the problem to be the n-tuple $({a}_{1},{a}_{2},...,{a}_{n})\in {\mathbb{R}}_{++}^{n}$. 
Define ${B}^{n}\subseteq{\mathbb{R}}_{++}^{n}$ to be the set of all feasible configurations (i.e., configurations for which a feasible plan exists).
Trivially, we know that ${B}^{1}={\mathbb{R}}_{++}$ and ${B}^{2}={\mathbb{R}}_{++}^{2}$. Also it's easy to see that ${B}^{n}\subseteq {(0,1)}^{n}$ for $n>2$. However, it is nontrivial to find ${B}^{n}$ for $n>2$. Without loss of generality, assuming ${a}_{1}<{a}_{2}<...<{a}_{n}$, I find that ${B}^{3}=\{({a}_{1},{a}_{2},{a}_{3})| 2{a}_{2}\leq{a}_{1}+1,\ {a}_{3}<1 \}$, which is neat!

When I tried to figure out ${B}^{4}$, however, I found it to be significantly harder than ${B}^{3}$. So I have two related questions here:
1.How should we tackle ${B}^{n}$? Is there some systematic method we can use to characterize ${B}^{n}$ for successive n's?   
2.If not, can we speculate what ${B}^{n}$ will look like, for example by giving some upper and lower bound on it which we are able to characterize?


EDIT: $n=4$ is EXHAUSTING, and honestly I have no idea where to start. Maybe it's just impractical to find ${B}^{n}$ for $n>3$. An interesting way to simplify the model is to assume that M can teleport himself anywhere he wants (as long as the motorcycle is not loaded). This assumption may allow a cleaner analysis and nicer answer.

REPLY [2 votes]: Well, this is not really an answer. It is motivated by fedja's comment. In the original problem we do allow motorcyclist (M) to drive walkers backward. If not, I think I can say something more:
If driving walkers backward is not allowed, the best M can do is trying to successively drive slower walkers forward enough to correct positions before the fastest walker reach Y.  Conceivably in a feasible configuration then, if he is visionary enough, M can figure out a plan in which he never goes forward unloaded (in this plan, when M placed someone in correct position, he immediately turns back to pick another walker). If this is true, then half of time M is empty (going backward), and half of time it's loaded (going forward).
Let $T$ be the total time lapse of this plan, $L$ be the distance between XY, ${t}_{i}$ be the time on the motorcycle for walker $i$, then we have:
$\sum{t}_{i}=T/2$ 
${t}_{i}+(T-{t}_{i}){a}_{i}=L\    $     for $i=1,2,...,n$ 
which is n+1 equations for n+1 unknowns. What is perhaps a little surprising about the plan is that the total time $T$ doesn't depend on whom M decides to pick first, and whom to pick later. Maybe this is not very relevant to my original questions, but I can't help remark on it.