The M.A.E Congestion Problem is a simulation project that explores traffic congestion as a collective action problem. The project focuses on the scenario where students commute to school, choosing between two transportation options: bus or car. This choice, when made by multiple individuals, leads to traffic issues due to limited road space.
Below are our different stages of our project presentations.
Click Here to View Our First Presentation Click Here to View Our Second Presentation Click Here to View Our Third PresentationThe code in this "in-progress" repository is our 3rd iteration of our project.
Make sure to have python working on your device and run the Testing_updatedAlgorithm.py to run our simulations!
When students travel to campus they have the option to take a car or bus. Depending on their decision, it can contribute to road congestion because of the limit road space available. In our game, these are the contraints:
- N players
- 2 strategies: car or bus
- Limited amount of road space
- Set number of buses
- Players can switch strategy depending on delay
- Current: we are using the repetition mechanism
- It simulates days as rounds
- We make it a possibility for players to switch strategy depending on the level of congestion on the road (delay function is high)
-
$$D_{i}(car)$$ – delay when the player takes a car -
$$D_{i}(bus)$$ – delay when the player takes the bus -
$$S_{c}$$ – size of car -
$$S_{b}$$ – space taken by busses -
$$C$$ – number of cars (other than the player) -
$$R$$ – available road space -
$$B$$ – delay caused by taking bus -
$$L$$ – large constant -
$$t$$ – threshold for traffic becoming too congested -
$$c$$ – constant for scaling x in F
Nash Equilibrium:
-
$$B > F\left(\sum_{i=1}^{C}\left(\frac{S_{c}}{R}\right) +\frac{S_{c}}{R}\right) - F\left(\sum_{i=1}^{C}\frac{S_{c}}{R}\right)$$ - when all players are traveling by car
- when
$$\sum_{i=1}^{C}\frac{S_{c}}{R} \geq t$$
-
$$B < F\left(\sum_{i=1}^{C}\left(\frac{S_{c}}{R}\right) +\frac{S_{c}}{R}\right) - F\left(\sum_{i=1}^{C}\frac{S_{c}}{R}\right)$$ - when all players are traveling by bus
-
$$B = F\left(\sum_{i=1}^{C}\left(\frac{S_{c}}{R}\right) +\frac{S_{c}}{R}\right) - F\left(\sum_{i=1}^{C}\frac{S_{c}}{R}\right)$$ - nash for all player strategy combinations, except for when
$$\sum_{i=1}^{C}\frac{S_{c}}{R} = t$$
- nash for all player strategy combinations, except for when
Pareto:
-
$$B > F\left(\sum_{i=1}^{C}\left(\frac{S_{c}}{R}\right) +\frac{S_{c}}{R}\right) - F\left(\sum_{i=1}^{C}\frac{S_{c}}{R}\right)$$ - all strategy combinations are pareto except for when
$$\sum_{i=1}^{C}\frac{S_{c}}{R} = t$$ or all players are in cars
- all strategy combinations are pareto except for when
-
$$B \leq F\left(\sum_{i=1}^{C}\left(\frac{S_{c}}{R}\right) +\frac{S_{c}}{R}\right) - F\left(\sum_{i=1}^{C}\frac{S_{c}}{R}\right)$$ - when everyone takes the bus
- Bus Constant: 10 (5 buses and each takes 2 spaces)
- Threshold of Car Multiplier: 0.4 (40% or higher of road taken up, car delay grows)
- Car Delay Multiplier: 2.5 (for cars when above threshold)
- Chance to Switch if Allowed: 0.2 (chance to make the switches gradual over the rounds)
- Congestion is affected by player choices:
- As more players take cars, more of the road is taken up, resulting in higher delays for all players
- As more players take the bus, less of the road is taken up, resulting in lower delays for all players
- With the repetition mechanism, players had the chance to change their strategy day by day; some things we noticed:
- When traffic becomes unbearable, players are more likely to switch to bus in an attempt to aleviate some traffic
- When traffic is lighter, players are much less likely to switch
- After a certain number of rounds, if the congestion gets below a certain level, players stop switching entirely
- In cases where there is less space on the road, increasing the number of rounds makes it more likely to reach an equilibrium
- In cases where there is a lot of excess space on the road, equilibrium is reached quicker and the number of rounds does not affect it as much
Example:
One-off game:
5th and 10th rounds of repeated game (with same starting values as one-off):