Consider a market, where N firms with average cost c compete in quantities. Assume linear demand P=a-bQ. Assume that N is arbitrary.

a)Find the equilibrium of the game, where these N firms compete a la Cournot.

b)Assume now that one firm (out of these N firms) acts as a “leader” by setting quantity first and the other firms move second (simultaneously) knowing the leader’s choice. Find the new equilibrium.

Here is my solution for point a)



Then profit will be:



imposing symmetry: V

Therefor p



How do I apporach point b), I'm a bit lost? Also extra question: what if all firms sequentially chose quantities?


1 Answer 1


(B) Define the set of firms as $F= \{1,2,3,....N \}$

Demand function given as $P=a -bQ$

Let the leader firm's quantity produced be $q_l $ for some $l \in F$

Since it's a extensive game (assuming perfect information is available to all) of choosing quantity and the Leader being the first mover can anticipate the choice by the rest of the firms (i.e. their best response to the leader's quantity choice)

We can try to use backward induction to arrive at Subgame Perfect Nash Equilibrium

Let's try to figure out the best response of each of the remaining firms to the leader's quantity choice in the initial period $t_0$ and to the other firm's choice (simultaneous game)

Apart from the leader, each firm $j$ tries to maximise their profit given by:-

$$\underset{q_j\geq0}{max} \: \left(a-b(q_l + q_j + \sum_{k\neq \{j, l\}} q_k)\right) q_j - cq_j$$

$ \: \: \: \forall \: k \in F - \{ q_l , q_j\}$

This yields

$BR_j (q_l,\sum_{k\neq \{j, l\}} q_k)$ = $\left\{\begin{matrix} \frac{a-bq_l - b\sum_{k \neq \{j, l\}} q_k -c}{2b}& if \: \frac{a-c}{b} \geq q_l + \sum_{k\neq \{j, l\}} q_k \\ 0& otherwise \end{matrix}\right.$

Since it is symmetric for all $j$

$ Q_{-l} = \sum_{j \neq l}q_j$

$ = \left\{\begin{matrix} \frac{N-1}{bN}(a-bq_l -c) & if \: \frac{a-c}{b} \geq q_l \\ 0 & otherwise \end{matrix}\right. $

Now, Leader takes the above information into account while taking the decision

$\underset{q_l\geq0}{max} (a - b(q_l + Q_{-i}))q_l -cq_l $

$s.t. Q_{-i} = \left\{\begin{matrix} \frac{N-1}{bN}(a-bq_l -c) & if \: \frac{a-c}{b} \geq q_l \\ 0 & otherwise \end{matrix}\right. $

This would result in SPNE outcome such that $q_l^* = \frac{a-c}{2b}$ and
$q_{i \neq l}^* = \frac{a-c}{2bN}$ for all firms other than l (leader)

  • $\begingroup$ Thanks for the solution! Could you explain how you got !Q-l value after you found BR, I assume it is (N-1)*BR, since other firms are equal and we exclude leader, but I do not come to your value of !Q-l when solving like that. $\endgroup$ Oct 17, 2023 at 19:20
  • $\begingroup$ If all firms except leader simultaneously choose the strategy $q_j = BR_j$ (and satisfying non-negativity constraint), then we can find $\sum q_j$ for all j. So, we have $ Q_{-l} = \frac{(N-1)a - (N-1)b q_l - (N-2) b Q_{-l} - (N-1)c}{2b}$ . Now we can solve for $Q_{-l}$ and get the result. $\endgroup$ Oct 18, 2023 at 2:08

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