Heterogeneous-cost Cournot Competition

The Set-up

Suppose that we have inverse-demand given by

[math]p=A-BQ, \mbox{where }Q = \sum_{i}^{n} q_i [/math]

and there are [math]\;n[/math] firms in the market. Firm profit is then given by:

[math]\pi_i = q_i \left ( p(Q) - c_i \right)[/math]

Solving for optimum quantities

Begin by taking a first-order condition:

[math]2q_i^* = \frac{A - c_i}{B} - \sum_{j \ne i} q_j[/math]

We now need two constraints to find the solution. First, we need to solve for [math]\sum_{j\ne i}\;q_j^*[/math].

There are [math]\;n[/math] of these first order conditions:

[math] \begin{array}{lclllllllllllll} 2q_1^* &=& \frac{A - c_1}{B}& - &(&0 &+ &q_2 &+ &\ldots &+ &q_{n-1} &+ &q_n &)\\ 2q_2^* &=& \frac{A - c_2}{B}& - &(&q_1 &+ &0 &+ &\ldots &+ &q_{n-1} &+ &q_n &)\\ \quad\vdots & & \quad\vdots &&&\vdots&&\vdots&&&&\vdots&&\vdots& \\ 2q_n^* &=& \frac{A - c_n}{B}& - &(&q_1 &+ &q_2 &+ &\ldots &+ &q_{n-1} &+ &0 &) \end{array} [/math]

As each of these conditions holds in equilibrium, their sum must all also hold:

[math]\sum_{i=1}^n 2 q_i^* = \frac{n A - \sum_{i=1}^n c_i - B (n-1) \sum_{i=1}^n q_i}{B} [/math]

Noting that [math]\;Q=Q^*[/math] in equilibrium, and rearranging gives

[math]\sum_{j \ne i}^n q_j^* = - q_i^* + \frac{n A - \sum_{i=1}^n c_i}{B (n+1)}[/math]

Now we can solve for [math]\;q_i^*[/math]. Substituting this into the original first-order condition gives:

[math]$q_i^* = \frac{\left ( A - c_i(n+1) + \sum_{i=1}^n c_i \right )}{B (n+1)}[/math]

Market clearing price and profits

Prices always come from the demand function. Firm's don't make profits by setting prices, the make profits by providing less than competitive quantities. Accordingly first find the total quantity provided:

[math]Q^* = \sum_{i=1}^n q_i^* = \frac{\left ( nA - sum_{i=1}^n c_i \right )}{B (n+1)}[/math]

And then substitute into [math]\;p=A-BQ[/math] to get:

[math]p^* & = \frac{A + \sum_{i=1}^n c_i}{n+1}[/math]

Now find firm profits by subsituting both [math]\;q_i^*[/math] and [math]\;p^*[/math] into [math]\;\pi_i = q_i \left ( p(Q) - c_i \right)[/math]:

[math]\pi_i^* &= \frac{1}{B }\left (\frac{A - c_i(n+1) + \sum_{i=1}^n c_i }{(n+1)} \right )^2[/math]