Parking in the city

We focus on the spacing distribution between cars parked parallel to the curb somewhere in the city center. We will assume that the street is long enough to enable a parallel parking of many cars. Moreover we assume that there are no driveways or side streets in the segment of interest and that the street is free of any kind of marked parking lots or park meters. So the drivers are free to park the car anywhere provided they find an empty space to do it. Finally we assume that many cars a cruising for parking. So there are not free parking lots and a car can park only when another car leaves the street.

The standard way to describe random parking is the continuous version of the random sequential adsorption model known also as the “random car parking problem” -see [1], [2] for review. In this model it is assumed that all cars have the same length l 0 and park on randomly chosen places. Once parked the cars do not leave the street. This model leads to predictions that can be easily verified. First of all it gives a relation between the mean bumper -to -bumper distance D and the car length: D ∼ 0.337 l 0 . Further the probability density Q(D) of the car distances D behaves like [3], [4], [5], [6] Q

To test this results real parking data were collected recently in the center of London [7]. The average distance between the parked cars was 152 cm which fit perfectly with the relation D ∼ 0.337 l 0 for l 0 = 450 cm. The gap density (1) was however fully incompatible with the observed facts. The relation (1) gives Q(D) → ∞ for D → 0 . The data from London showed however that in reality Q(D) → 0 as D → 0. Rawal and Rogers [7] used therefore an amended version of the model with a car re-positioning process (describing the car manoeuvering) to fit the data. Later Abul-Magd [8] pointed out that the car spacings observed in London can be well described by the Gaussian Unitary Ensemble of random matrices [9], [10] if the parked cars are regarded as particles of a one dimensional interacting gas. In both cases the authors assumed that Q(D) ∼ D 2 for small D. However the origin of the particular car re-positioning or car interaction remained unclear.

Our aim here is to show that the car parking can be understood as a simple Markov process. The main difference to the previous approach is that we enable the parked cars to leave the street vacating the space for a new car to parks there. The distance distribution is obtained as a steady solution of the repeated car parking and car leaving process.

To derive the equations we describe a car parking on a roundabout junction (we know that it is not allowed to park there, but it simplifies the consideration). The idea is simple: Assume that all cars have the same length l 0 and park on a roundabout junction with a circumference L. The length L is chosen in interval 3l 0 < L < 4l 0 ) so that maximally 3 cars can park there. These cars define three spacings D 1 , D 2 , D 3 with

The parking process goes as follows: one randomly chosen car leaves the street and the two adjoining lots merge into a single one. In the second step a new car parks into the empty space and splits it into two smaller lots. Such fragmentation and coagulation processes were discussed intensively since they apply for instance to the computer memory allocationsee [11] for review. Let the car leaves and the neighboring spacings -say the spacings D 1 , D 2 -merge into a single lot

When a new car parks to D it splits it into D1 , D2 :

where a ∈ (0, 1) is a random variable with a probability density P (a). It describes the parking preference of the driver. We assume that all drivers have identical preferences, i.e. identical P (a). (The meaning of the variable a is straightforward. For a = 0 the car parks immediately in front of the car delimiting the parking lot from the left without leaving any empty space (very unworthy way to park). For a = 1/2 it parks directly to the center of the lot and for a = 1 it stops exactly behind the car on the right.) Combining ( 3) and ( 4) lead to

and the car length l 0 drops out. Taking the first of these equations and using the relation (2) we finally obtain

The pair D 1 , D 2 is in no way particular. The same works when dealing with an arbitrary pair of D k , D l ; k = l; k, l = 1, 2, 3. The last step is now easy. Hypothesize that after many steps a steady state is reached. Then the distributions of D k , k = 1, 2, 3 are identical and the distances D k become copies of one random variable D. The equation ( 6) becomes

where D n denotes the variable D at time n and a n are independent identically distributed random variables with a common probability density P (a). So D n , n = 1, 2, 3, … represents a simple Markov chain. We are interested in the solution of (7) in the limit n → ∞. One can show [12] that the limiting variable D ∞ solves the equation D ∞ a(L -3l 0 -D ∞ ).

The symbol means that left and right hand sides of the equation have identical probability distri

…(Full text truncated)…

This content is AI-processed based on ArXiv data.