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 -t0 + ti, 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 tii+1 -tii = F+Kp'(I-t0i +tii) 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, (t0i +t1i) 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-t0i+1 +tii+1 ). This again leads to a further increase in ti+2 until the tram gets too crowded to accept more people.

In the tram behind it, however, (-t1i +t2i) 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

1. http://en.wikipedia.org/wiki/Bus_bunching
2. <\ol>

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