The city of Amsterdam has a fairly well-functioning tram system.
About two dozen different tram lines crisscross the city, making sure one
can get almost anywhere from almost anywhere without having to walk too
much, in a reasonable amount of time-especially in comparsion with the time
it would take to navigate the the city by car. Furthermore, these
trams run fairly often, normally at about once every 8 minutes during
most of the day and evening.

Given that I work in one part of the city and work in another, and I'm too lazy to go by bicycle and too cheap
to buy a car that will also cost 100% more due to tax, 8 dollar per
gallon gas and 30 dollar a day for parking, I am using the tram. This
means I have to wait. Sometimes, it rains. Often, there is wind.
This gives one pause to think. The most common thought is: "If these
trams run every 8 minutes, I should wait on average 4 minutes. Well,
I usually have to wait a whole lot longer than that-at least, that's what it
feels like.

Now, I'm a bit of a mathematician, and the combination of
boredom, wind, rain, prompted me to think about this problem. All
the potentially hot chick wore scarves or
burquas, so there were no distractions: I was going to solve this
problem

The first thing we can state is that in the part of Amsterdam where I
travel, trams don't face many traffic lights, so that
couldn't really be it. Furthermore, it is easily observable is that
trams spend more time loading, or rather, cramming, passengers in
their malodorous innards than actually riding. Furthermore, the flow of
passengers to a tram stop could be modeled with a Poisson distribution,
i.e. people come randomly, but in the long run, the statistical noise
becomes smaller compared to the average flow. This prompted me to come up
with the following model.

The time it takes a tram to travel to the next stop consists of a
fixed fraction *F*, and a time needed to cram passengers in; we shall
denote this with *C*. Now, we can assume *C* to be proportional
(with constant *K*) to the amount of passengers waiting, which we shall
denote with *P*. This *P* can be thought of, initially, as a
constant *p'* times the time difference between this tram and the
previous tram, or * -t*_{0} + t_{i}, with *I* the
initial time difference between the previous tram (0) and this one (1), and
*t* the total traveling time up to this point. Note that trams can't
overtake, so the time difference is floored at 0. In this system, the tram
will need *t*_{i}^{i+1} -t_{i}^{i} =
F+Kp'(I-t_{0}^{i} +t_{i}^{i}) until
the next stop. I'm using ^{i} here to denote that the time
for the next leg (*i+1*) of the trip depends on the time previous leg
of the trip (*i*). We can just use this formula over and over for the
entire trip. Now, there is no *a priori* reason to assume either tram 0
or tram 1 is faster, so let's assume they are equally fast. This means our
tram will need *F + Kp'I* to travel until the next stop. This doesn't
help us one bit yet.

In order to solve this, we will need to employ chaos theory. This
sounds extremely complex (and I'm sure it can be), but I'm not a real
mathematician, and I can't access Google on a tram stop. So, we are going
to do this the simple way. Assume there is a little perturbation for
tram 1; perhaps there is one extra passenger. Perhaps I'm blocking the
doors because I'm dreaming. As a consequence, tram 1 will be late.
Now, *(t*_{0}^{i} +t_{1}^{i}) will be
a positive number. The tram will be even more late on leg ^{i+1}.
More people will get on the next stop, because of this increase in
*Kp'(I-t*_{0}^{i+1} +t_{i}^{i+1} ).
This again leads to a further increase in *t*^{i+2} until the
tram gets too crowded to accept more people.

In the tram behind it, however, *(-t*_{1}^{i}
+t_{2}^{i}) becomes ever smaller, and the tram moves
faster, picking up ever less passengers; the exact opposite of what is
happening with the first tram! In an extreme case, which will occur on a
sufficiently long track (and trust me, the lines in Amsterdam are long
enough for this to happen somewhere halfway or even earlier), there is a
severely delayed, overcrowded tram leading at least one or possibly
two nearly empty trams

We can now observe another great phenomenon: the tragedy of the
commons. You see, passagers can see that tram 1 is overcrowded and tram 2
and 3 are nearly empty. If they were to let tram 1 pass and just get in tram
2 and 3, tram 1, determining the speed, would travel faster and everyone
would arrive earlier. However, if I'm the only one doing this, all I can be
sure of is that I'll be slightly later (because tram 2 travels a bit behind
tram 1). This means I won't do this, as it only helps when many people do
it.

The net result is that while on average, trams might be reasonably
empty, and travel with not too much delay, this isn't true for an
individual tram. In particular, some trams will be massively overcrowded
and delayed, and some will be on time and empty. The fun part is that you
are of course, by virtue of the larger number of passenger, much more likely
to end up in the crawling sauna in front than in the empty limousine in
the back. So, while 50% or so of trams may be on time, 50% of passengers
won't, as all the delayed trams are full and the non-delayed trams are
nearly empty.

While doing some Internet searches for this theory, I immediately found
something at the Wikipedia. It is claimed there that this
theory is that "The existence of classical pairwise bunching has not been
borne out by vehicle tracking systems data." Uhuh. So, not only I'm not the
first person to think of this, I'm also wrong.

For this whole bunching thing to work, it's important that the loading
time of the passengers is large compared to other random factors such as
speed, traffic lights, traffic, etc. For the Amsterdam
trams, this is certainly the case; there are few random factors (no traffic
lights, trams have right of way, even if they don't have it). I suspect that for buses,
these random factors are much larger, and the time spent picking up people
is comparatively less. Furthermore, the Amsterdam metro system doesn't
appear to have this bunching. There, during rush hour, all cars
are filled to the brim with people waiting on the platform; all metros
are equally crowded, so an extra influx of passengers will just have to
wait.

While I have tossed around something resembling math, this isn't an
actual scientific theory - I haven't measured it, it probably can't easily
be reproduced in a lab, stuff like that. Hell, most of it is probably my
boredom talking, the sluggish rides in full trams etch more
firmly in my memory that the normal rides. Still, if it were true, I think
it's a simple and accessible demonstration of chaos theory and the
tragedy of the commons. The astute reader will also have noticed that all
this thinking hasn't brought me one iota closer to a shorter commute.
Typical.

## Background material

- http://en.wikipedia.org/wiki/Bus_bunching
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