Braess's paradox

Consider a road network as shown in the adjacent diagram on which 4000 drivers wish to travel from point Start to End. The travel time in minutes on the Start–A road is the number of travelers (T) divided by 100, and on Start–B is a constant 45 minutes (likewise with the roads across from them). If the dashed road does not exist (so the traffic network has 4 roads in total), the time needed to drive Start–A–End route with ${\displaystyle a}$ drivers would be ${\displaystyle {\tfrac {a}{100}}+45}$. The time needed to drive the Start–B–End route with ${\displaystyle b}$ drivers would be ${\displaystyle {\tfrac {b}{100}}+45}$. As there are 4000 drivers, the fact that ${\displaystyle a+b=4000}$ can be used to derive the fact that ${\displaystyle a=b=2000}$ when the system is at equilibrium. Therefore, each route takes ${\displaystyle {\tfrac {2000}{100}}+45=65}$ minutes. If either route took less time, it would not be a Nash equilibrium: a rational driver would switch from the longer route to the shorter route.

Now suppose the dashed line A–B is a road with an extremely short travel time of approximately 0 minutes. Suppose that the road is opened and one driver tries Start–A–B–End. To his surprise he finds that his time is ${\displaystyle {\tfrac {2000}{100}}+{\tfrac {2001}{100}}=40.01}$ minutes, a saving of almost 25 minutes. Soon, more of the 4000 drivers are trying this new route. The time taken rises from 40.01 and keeps climbing. When the number of drivers trying the new route reaches 2500, with 1500 still in the Start–B–End route, their time will be ${\displaystyle {\tfrac {2500}{100}}+{\tfrac {4000}{100}}=65}$ minutes, which is no improvement over the original route. Meanwhile, those 1500 drivers have been slowed to ${\displaystyle 45+{\tfrac {4000}{100}}=85}$ minutes, a 20-minute increase. They are obliged to switch to the new route via A too, so it now takes ${\displaystyle {\tfrac {4000}{100}}+{\tfrac {4000}{100}}=80}$ minutes. Nobody has any incentive to travel A-End or Start-B because any driver trying them will take 85 minutes. Thus, the opening of the cross route triggers an irreversible change to it by everyone, costing everyone 80 minutes instead of the original 65. If every driver were to agree not to use the A–B path, or if that route were closed, every driver would benefit by a 15-minute reduction in travel time.

Generalized routing problem

The routing problem introduced in the Braess's paradox can be generalized to many different flows traversing the same graph at the same time.

Definition (Generalized flow). Let ${\displaystyle G=(V,E)}$, ${\displaystyle L}$ and ${\displaystyle w}$ be as defined above, and suppose that we want to route the quantities ${\displaystyle R=\{r_{1},r_{2},\dots ,r_{k},\;|\;r_{i}>0\}}$ through each distinct pair of nodes in ${\displaystyle \Gamma =\{(s_{1},t_{1}),(s_{2},t_{2}),\dots ,(s_{k},t_{k})\}\subseteq (V\times V)}$. A flow ${\displaystyle f_{\Gamma ,R}}$ is defined as an assignment ${\displaystyle p\mapsto \Re }$ of a real, nonnegative number to each path ${\displaystyle p}$ going from ${\displaystyle s_{i}}$ to ${\displaystyle t_{i}}$ ${\displaystyle \in \Gamma }$, with the constraint that

${\displaystyle \sum _{p:\,s_{i}\rightarrow t_{i}}{f_{p}}=r_{i}\;\;\forall (s_{i},t_{i})\in \Gamma .}$

The flow traversing a specific edge of ${\displaystyle G}$ is defined as

${\displaystyle f_{e,\Gamma ,R}=\sum _{p:\,e\in p}{f_{p}}.}$

For succinctness, we write ${\displaystyle f_{e}}$ when ${\displaystyle \Gamma ,R}$ are clear from context.

Definition (Nash-equilibrium flow). A flow ${\displaystyle f_{\Gamma ,R}}$ is a Nash-equilibrium flow iff ${\displaystyle \forall (s_{i},t_{i})\in \Gamma }$ and ${\displaystyle \forall p,q}$ from ${\displaystyle s_{i}}$ to ${\displaystyle t_{i}}$

${\displaystyle f_{p}>0\Rightarrow \sum _{e\in p}{l_{e}(f_{e})}\leq \sum _{e\in q}{l_{e}(f_{e})}.}$

This definition is closely related to what we said about the support of mixed-strategy Nash equilibria in normal-form games.

Definition (Conditional welfare of a flow). Let ${\displaystyle f_{\Gamma ,R}}$ and ${\displaystyle f_{\Gamma ,R}^{*}}$ be two flows in ${\displaystyle G}$ associated with the same sets ${\displaystyle \Gamma }$ and ${\displaystyle R}$. In what follows, we will drop the subscript to make the notation clearer. Assume to fix the latencies induced by ${\displaystyle f}$ on the graph: the conditional welfare of ${\displaystyle f^{*}}$ with respect to ${\displaystyle f}$ is defined as

${\displaystyle w^{f}(f^{*})=\sum _{e\in E}{f_{e}^{*}\cdot l_{e}(f_{e})}}$

Fact 1. Given a Nash-equilibrium flow ${\displaystyle f}$ and any other flow ${\displaystyle f^{*}}$, ${\displaystyle w(f)=w^{f}(f)\leq w^{f}(f^{*})}$.

Proof (By contradiction). Assume that ${\displaystyle w^{f}(f^{*})<w^{f}(f)}$. By definition,

${\displaystyle \sum _{i=1}^{k}\sum _{p:s_{i}\rightarrow t_{i}}f_{p}^{*}\cdot \sum _{e\in p}l_{e}(f_{e})<\sum _{i=1}^{k}\sum _{p:s_{i}\rightarrow t_{i}}f_{p}\cdot \sum _{e\in p}l_{e}(f_{e})}$.

Since ${\displaystyle f}$ and ${\displaystyle f^{*}}$ are associated with the same sets ${\displaystyle \Gamma ,R}$, we know that

${\displaystyle \sum _{p:s_{i}\rightarrow t_{i}}f_{p}=\sum _{p:s_{i}\rightarrow t_{i}}f_{p}^{*}=r_{i}\;\;\forall i.}$

Therefore, there must be a pair ${\displaystyle (s_{i},t_{i})}$ and two paths ${\displaystyle p,q}$ from ${\displaystyle s_{i}}$ to ${\displaystyle t_{i}}$ such that ${\displaystyle f_{p}^{*}>f_{p}}$, ${\displaystyle f_{q}^{*}<f_{q}}$, and

${\displaystyle \sum _{e\in p}l_{e}(f_{e})<\sum _{e\in q}l_{e}(f_{e}).}$

In other words, the flow ${\displaystyle f^{*}}$ can achieve a lower welfare than ${\displaystyle f}$ only if there are two paths from ${\displaystyle s_{i}}$ to ${\displaystyle t_{i}}$ having different costs, and if ${\displaystyle f^{*}}$ reroutes some flow of ${\displaystyle f}$ from the higher-cost path to the lower-cost path. This situation is clearly incompatible with the assumption that ${\displaystyle f}$ is a Nash-equilibrium flow. Q.E.D.

Note that Fact 1 does not assume any particular structure on the set ${\displaystyle L}$.

Fact 2. Given any two real numbers ${\displaystyle x}$ and ${\displaystyle y}$, ${\displaystyle x\cdot y\leq x^{2}+y^{2}/4}$.

Proof. This is another way to express the true inequality ${\displaystyle (x-y/2)^{2}\geq 0}$. Q.E.D.

Theorem. The pure PoA of any generalized routing problem ${\displaystyle (G,L)}$ with linear latencies is ${\displaystyle \leq 4/3}$.

Proof. Note that this theorem is equivalent to saying that for each Nash-equilibrium flow ${\displaystyle f}$, ${\displaystyle w(f)\leq (4/3)\cdot \min _{f^{*}}\{w(f^{*})\}}$, where ${\displaystyle f^{*}}$ is any other flow. By definition,

${\displaystyle w^{f}(f^{*})=\sum _{e\in E}f_{e}^{*}(a_{e}\cdot f_{e}+b_{e})}$

${\displaystyle =\sum _{e}(a_{e}f_{e}f_{e}^{*})+\sum _{e\in E}f_{e}^{*}b_{e}.}$

By using Fact 2, we have that

${\displaystyle w^{f}(f^{*})\leq \sum _{e\in E}\left(a_{e}\cdot \left((f_{e}^{*})^{2}+(f_{e})^{2}/4\right)\right)+\sum _{e\in E}f_{e}^{*}\cdot b_{e}}$

${\displaystyle =\left(\sum _{e\in E}a_{e}(f_{e}^{*})^{2}+f_{e}^{*}b_{e}\right)+\sum _{e\in E}a_{e}(f_{e})^{2}/4}$

${\displaystyle \leq w(f^{*})+{\frac {w(f)}{4}},}$

since

${\displaystyle (1/4)\cdot w(f)=(1/4)\cdot \sum _{e\in E}f_{e}(a_{e}f_{e}+b_{e})}$

${\displaystyle =(1/4)\cdot \sum _{e\in E}(f_{e})^{2}+\underbrace {(1/4)\cdot \sum _{e\in E}f_{e}b_{e}} _{\geq 0}.}$

We can conclude that ${\displaystyle w^{f}(f^{*})\leq w(f^{*})+w(f)/4}$, and prove the thesis using Fact 1. Q.E.D.

Note that in the proof we have made extensive use of the assumption that the functions in ${\displaystyle L}$ are linear. Actually, a more general fact holds.

Theorem. Given a generalized routing problem with graph ${\displaystyle G}$ and polynomial latency functions of degree ${\displaystyle d}$ with nonnegative coefficients, the pure PoA is ${\displaystyle \leq d+1}$.

Note that the PoA can grow with ${\displaystyle d}$. Consider the example shown in the following figure, where we assume unit flow: the Nash-equilibrium flows have social welfare 1; however, the best welfare is achieved when ${\displaystyle x=1-1/{\sqrt {d+1}}}$, in which case

${\displaystyle w=\left(1-{\frac {1}{\sqrt {d+1}}}\right)^{d}\cdot \left(1-{\frac {1}{\sqrt {d+1}}}\right)+1\cdot {\frac {1}{\sqrt {d+1}}}}$

${\displaystyle =\left(\left(1-{\frac {1}{\sqrt {d+1}}}\right)^{\sqrt {d+1}}\right)^{\sqrt {d+1}}+{\frac {1}{\sqrt {d+1}}}}$

${\displaystyle \leq e^{-{\sqrt {d+1}}}+{\frac {1}{\sqrt {d+1}}}.}$

This quantity tends to zero when ${\displaystyle d}$ tends to infinity.