Less traffic with fewer roads?

German mathematician Dietrich Braess pointed towards the fact that increasing the capacity of a network can perhaps “decrease the efficiency of the journeys around it, even without increasing the number of trips made”. This is known as the Braess’s Paradox. [read more about paradoxes here]

Consider the following road network:

Example

The rate of cars entering the network at A is 1500 cars per hour. Drivers can choose between two routes: route 1 which crosses bridge a or route 2 which crosses bridge b. Let us define L and R as the number of cars that arrive at B per hour, via route 1 and 2 respectively. If we suppose that the travel time through both bridges is directly proportional to the number of cars per hour we can assume that the travel time is L/100 for bridge a and R/100 fro bridge b. The rest of route 1 or 2 has a travel time of 20 minutes.

To further analyse this information, I would like to introduce the steady-state or Nash equilibrium, which was developed by the Nobel Prize winning mathematician John Nash in order to study non-cooperative games. It imagines that:

“each driver has gone through the network many times, as is the case for someone driving every day at rush hour, and has developed a particular strategy, that is perceived as minimising travel time. Under this assumption, the travel time must be the same for all the drivers, otherwise there would be an incentive for some of the drivers to change their strategy.”

Using this, we can construct two simultaneous equations:

\[  \frac{L}{100}+20=\frac{R}{100}+20\text {.}  \]

and

\[  L+R=1500\text {.}  \]

as the number of cars per hour – L and R – must add up to the flow at A.

Hence, \[  L=R=750\text {.}  \] Therefore, the average travel time will be 27.5 minutes.

Now, we add another road (let’s call it road c), for which the travel time is 7 minutes.

Example

There are now three routes to choose from; the third route consists of driving through bridge a, road c and bridge b. The following equations can be constructed:

\[  \frac{L+C}{100}+20=\frac{R+C}{100}+20=\frac{L+C}{100}+7+\frac{R+C}{100} \text {,} \]

hence $\displaystyle \frac{R+C}{100}+20\text {,}  $ and $\displaystyle \frac{L+C}{100}+7+\frac{R+C}{100}\text {.}  $

Also \[ L+R+C=1500\text {.}  \]

**Note that C is the flow of cars on road c.

Solving these simultaneous equations gives us L = R = 200 and C = 1100, giving an average travel time of 33 minutes, which is longer than before.

This paradox is not merely a “mathematical quirk”, it can actually be of use when applied to real life situations; there have been real cases where removing roads, rather than constructing them, has improved the traffic. For example, in South Korea one of the roadways which was built in the 1960s was removed, which resulted in reduced transit times throughout the city significantly. This was due to the fact that there was a “more efficient distribution of cars across the remaining network”.

Sources: 1 | 2 | 3

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