Bandwidth-sharing game

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
The game involves $$n$$ players. Each player $$i$$ has utility $$U_i(x)$$ for $$x$$ units of bandwidth. Player $$i$$ pays $$w_i$$ for $$x$$ units of bandwidth and receives net utility of $$U_i(x)-w_i$$. The total amount of bandwidth available is $$B$$.

Regarding $$U_i(x)$$, we assume
 * $$U_i(x)\ge0;$$
 * $$U_i(x)$$ 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 $$\underset x{\operatorname{arg\,max}}\,U_i(x)-px$$. Solving for the maximum yields $$U_i^'(x)=p$$.

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.

Possible solution
A popular idea to find the price is a method called fair sharing. In this game, every player $$i$$ is asked for the 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 $$x_i=\frac{w_i B}{\sum_jw_j}$$. This method yields an effective price $$p=\frac{\sum_jw_j}{B}$$. This price can proven to be market clearing; thus, the distribution $$x_1,...,x_n$$ is optimal. The proof is as so:

Proof
We have $$\underset{x_i}{\operatorname{arg\,max}}\,U_i(x_i)-w_i$$ $$= \underset{w_i}{\operatorname{arg\,max}}\,U_i\left(\frac{w_i B}{\sum_jw_j}\right)-w_i$$. Hence,
 * $$ U^'_i\left(\frac{w_i B}{\sum_jw_j}\right)\left(\frac{B}{\sum_jw_j}-\frac{w_i B}{(\sum_jw_j)^2}\right)-1=0$$

from which we conclude
 * $$ U^'_i(x_i)\left(\frac{1}{p}-\frac{1}{p} \left( \frac{x_i}{B}\right)\right)-1=0$$

and thus $$U^'_i(x_i)\left(1-\frac{x_i}{B}\right)=p. $$

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.