Nash equilibrium for n players game

There is a question that I am trying to solve but I am not sure about my approach and is hoping I could get some help. Here is the question:

There are $n$ companies sharing a water reservoir, let's say the water pipe has a maximum transfer capacity 1 unit. Each company $i$ can receive $x_{i}$ units of water where $x_{i} \in [0, 1]$. Each company wants to take as much water as possible, but the water quality gets worse with the total water taken from the pipe. If the total water taken by all companies, $\sum_{i=1}^{n} x_{i}$ exceeds the maximum transfer capacity of 1 then the utility of each company is 0. The utility for company i is given by

$$u_i = \begin{cases} e^{x_i}\prod_{j=1}^{n} e^{{-x_i}{x_j}} -1, & \text{if } \sum_{i=1}^{n} x_{i} < 1, \\ 0, & \text{if } \sum_{i=1}^{n} x_{i} \geq 1. \end{cases}$$

We want to find the nash equilibrium and the social welfare of the equilibrium.

My idea of the equilibrium is the case where all players are at capacity of 1, so there is no incentive of any players to decrease, hence that is a Nash equilibrium. I am not sure if this is the right approach.

• What do you mean by "all players are at capacity of $1$"? – joriki Sep 27 '15 at 21:55
• Basically each company uses the maximum transfer capacity of 1, therefore all of their utilities are at 0. – gametheorybeginner Sep 27 '15 at 21:58
• The question is misleadingly phrased in that it talks about "the" Nash equilibrium. There is more than one Nash equilibrium. You found one, but I suspect that the question aims at another, more interesting one. – joriki Sep 27 '15 at 22:00
• That's what I'm thinking, there is another part of the question that says the optimal social welfare is arbitrarily larger than the social welfare at equilibrium. – gametheorybeginner Sep 27 '15 at 22:03
• What do you mean by "arbitrarily larger"? – joriki Sep 27 '15 at 22:04

You found one Nash equilibrium. There is an entire class of similar Nash equilibria: Any strategy profile with $\sum_ix_i-x_j\ge1$ for all $j$ is a Nash equilibrium, since no company can individually reduce the consumption below capacity.

However, there is also a more interesting equilibrium with total consumption below capacity. Setting the derivative of the exponent of $u_i$ with respect to $x_i$ to $0$ yields

$$\frac{\partial}{\partial x_i}\left(x_i-\sum_jx_ix_j\right)=1-\sum_jx_j-x_i=0\;,$$

or $x_i=1-\sum_jx_j$, and then summing over $i$ yields $\sum_ix_i=n\left(1-\sum_jx_j\right)$, with solution $\sum_ix_i=\frac n{n+1}$ and thus $x_i=\frac1{n+1}$. The social welfare in this case is

$$n\left(\sum_i\mathrm e^{\frac1{n+1}-\sum_j\frac1{(n+1)^2}}-1\right)=n\left(\mathrm e^{\frac1{(n+1)^2}}-1\right)\approx\frac n{(n+1)^2}\;.$$

The maximum social welfare is achieved if a single company consumes with $x_i=\frac12$, and is given by

$$\mathrm e^\frac14-1\approx0.284\;.$$