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 ${\displaystyle U_{i}^{'}(x)=p}$ .

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.

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:

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.