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A bandwidth-sharing game is a type of resource allocation game designed to model the real-world allocation of bandwidth to many users in a network. The game is popular in game theory because the conclusions can be applied to real-life networks. The game is described as follows:

The game

• $n$ players
• each player $i$ has utility $U_{i}(x)$ for amount $x$ of bandwidth
• user $i$ pays $w_{i}$ for amount $x$ of bandwidth and receives net utility of $U_{i}(x)-w_{i}$
• the total amount of bandwidth available is $B$

We also use assumptions regarding $U_{i}(x)$

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

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

The problem

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 possible solution

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

Proof

$argmax_{x_{i}}U_{i}(x_{i})-w_{i}$

$\implies argmax_{w_{i}}U_{i}({\frac {w_{i}}{\sum _{j}w_{j}}}*B)-w_{i}$
$\implies U_{i}^{'}({\frac {w_{i}}{\sum _{j}w_{j}}}*B)({\frac {1}{\sum _{j}w_{j}}}*B-{\frac {w_{i}}{(\sum _{j}w_{j})^{2}}}*B)-1=0$
$\implies U_{i}^{'}(x_{i})({\frac {1}{p}}-{\frac {1}{p}}*{\frac {x_{i}}{B}})-1=0$
$\implies U_{i}^{'}(x_{i})(1-{\frac {x_{i}}{B}})=p$

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