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Bandwidth-sharing game

The game[edit]

The game involves ${\displaystyle n}$ players. Each player ${\displaystyle i}$ has utility ${\displaystyle U_{i}($x)} for ${\displaystyle x}$ units of bandwidth. Player ${\displaystyle i}$ pays ${\displaystyle w_{i}}$ for ${\displaystyle x}$ units of bandwidth and receives net utility of ${\displaystyle U_{i}($x)-w_{i}} . The total amount of bandwidth available is ${\displaystyle B}$.

Regarding ${\displaystyle U_{i}($x)} , we assume

• ${\displaystyle U_{i}($x)\geq 0;}
• ${\displaystyle U_{i}($x)} is increasing and concave;
• ${\displaystyle U($x)} is continuous.

The game arises from trying to find a price ${\displaystyle p}$ so that every player individually optimizes their own welfare. This implies every player must individually find ${\displaystyle {\underset {x}{\operatorname {arg\,max} }}\,U_{i}($x)-px} . Solving for the maximum yields ${\displaystyle U_{i}^{'}($x)=p} .

Problem[edit]

With this maximum condition, the game then becomes a matter of finding a price that satisfies an equilibrium. Such a price is called a market clearing price.

Possible solution[edit]

A popular idea to find the price is a method called fair sharing.^[1] In this game, every player ${\displaystyle i}$ is asked for the amount they are willing to pay for the given resource denoted by ${\displaystyle w_{i}}$. The resource is then distributed in ${\displaystyle x_{i}}$ amounts by the formula ${\displaystyle x_{i}={\frac {w_{i}B}{\sum _{j}w_{j}}}}$. This method yields an effective price ${\displaystyle p={\frac {\sum _{j}w_{j}}{B}}}$. This price can proven to be market clearing; thus, the distribution ${\displaystyle x_{1},...,x_{n}}$ is optimal. The proof is as so:

Proof[edit]

We have ${\displaystyle {\underset {x_{i}}{\operatorname {arg\,max} }}\,U_{i}(x_{i})-w_{i}}$ ${\displaystyle ={\underset {w_{i}}{\operatorname {arg\,max} }}\,U_{i}\left({\frac {w_{i}B}{\sum _{j}w_{j}}}\right)-w_{i}}$. Hence,

${\displaystyle U_{i}^{'}\left({\frac {w_{i}B}{\sum _{j}w_{j}}}\right)\left({\frac {B}{\sum _{j}w_{j}}}-{\frac {w_{i}B}{(\sum _{j}w_{j})^{2}}}\right)-1=0}$

from which we conclude

${\displaystyle U_{i}^{'}(x_{i})\left({\frac {1}{p}}-{\frac {1}{p}}\left({\frac {x_{i}}{B}}\right)\right)-1=0}$

and thus ${\displaystyle U_{i}^{'}(x_{i})\left(1-{\frac {x_{i}}{B}}\right)=p.}$

Comparing this result to the equilibrium condition above, we see that when ${\displaystyle {\frac {x_{i}}{B}}}$ is very small, the two conditions equal each other and thus, the fair sharing game is almost optimal.

References[edit]