University:
Amherst College
Course:
ECON 301 | Advanced Microeconomics
Academic year:

2021
Views:

49

Pages:

3
Author:

Sedleckogew

Economics 58 The Cournot Model The purpose of this handout is to illustrate the Cournot model in its most elementary form. The end result is to show how the Nash concept can be used to describe an equilibrium outcome in a market with an arbitrary number of firms. Firms There are n firms. Each produces the same good and each has the same cost function Ci = mqi i = 1, n . Demand Market demand takes the linear form p = A − ∑ qi . i Profit Maximization π i = pqi − Ci = Aqi − qi2 − ∑ q j qi − mqi Using the Cournot assumption that ∂q j =0 ∂qi yields the following first order condition for a maximum (this is the Cournot “reaction function”): A − m − ∑ qj ∂π i j ≠i * = ( A − m) − 2qi − ∑ q j = 0 or qi = ∂qi 2 j ≠i j ≠i Nash-Cournot Equilibrium To find a Nash equilibrium, these n reaction functions must be solved simultaneously. That solution is easy here because we can assume symmetry: q*j = qi* for all values of j. Hence: qi* = ( A − m) − (n − 1)qi* A−m or qi* = . 2 (n + 1) Similarly: p* = A − n A n ⋅ ( A − m) = + ⋅m (n + 1) n + 1 (n + 1) π i* = ( p − m)qi = ( A − m) ( A − m) ( A − m) 2 . ⋅ = (n + 1) (n + 1) (n + 1) 2 Monopoly For n = 1 the equilibrium solutions are therefore: A−m 2 A +m p1* = 2 q1* = ⎛ A−m⎞ π =⎜ ⎟ ⎝ 2 ⎠ 2 * 1 Perfect Competition If n → ∞ we have q∞* → 0 p∞ → m π∞ → 0 Numerical Example: If: A = 100, m = 10 q1* = 45 p1* = 55 π 1* = 452 = 2025 For the case n = 2, say, we have q2* = 30 p2* = 40 π 2* = 900 Total quantity expands and price falls as the number of firms increases. First-mover Advantage A player in the Cournot game can gain advantage by moving first and relying on the opponent to follow his/her reaction function. Consider the case n = 2 and let firm 1 be A − q1 − m the first-mover. Then firm 1 knows that firm 2 will react by q2 = . Hence 2 p = A − q1 − q2 = A − q1 − [( A − q1 − m) / 2] = ( A − q1 + m) 2 and q12 ( A − m) π 1 = ( p − m)q1 = [( A − q1 − m) 2] ⋅ q1 = q1 − 2 2 The first-order condition for a maximum is: ∂π 1 A − m A−m = − q1 = 0 or q1 = . ∂q1 2 2 With A = 100, m = 10 this results in q1 = 45 q2 = 22.5 p = 32.5 π 1 = 1012.5 π 2 = 506.25 Moving first yields a big advantage. Of course, if both firms try to move first, each will A−m choose qi = and p = A − 2qi = m, π i = 0. 2

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