I am trying to solve the following question:

'In a scenario in which there exist multiple identical firms with a large supply of products available, each firm must decide how much to provide to the market. Each firm has the same cost function, given by C(q) = 10q and market demand is given by Q = 150 - P.

Entering the market entails a fixed cost F to be incurred by each firm in the entry stage. Suppose there is one period of Cournot competition after entry.

Determine the long-run equilibrium number of firms n^0.'

How can this be determined without knowing the exact initial fixed cost? Are long-run profits under Cournot competition 0 as well (like they are under perfect competition)? Thank you!


1 Answer 1


As Giskard’s answer to the linked question says, since the firms have symmetric costs, in the long run, they all make $0$ profit.

The inverse demand function is

$Q = 150 - P \implies P(Q) = 150 - Q$

The profit function is

$\Pi_i = P(Q) q_i - TC_i(q_i)$

Substituting all the given functions,

$\Pi_i = 150 q_i - Q q_i - 10 q_i - F$

Now we find the optimal production for each firm:

By the product rule on $Q q_i$ and noting $\frac{\partial Q}{\partial q_i} = 1$,

$\frac{\partial \Pi_i}{\partial q_i} = 140 - Q - q_i = 0$

Since the firms have symmetric costs, they have the same optimal productions, which implies $ Q = n q_i$

$140 - (n+1) q_i = 0$

Isolating $q_i$

$q_i = \frac{140}{n+1}$

$Q = \frac{140 n}{n+1}$

From the inverse demand function we get

$P = \frac{10 (n+15)}{n+1}$

Substituting these into the $0$ profit condition,

$10 \cdot \frac{n+15}{n+1} \cdot \frac{140}{n+1} - 10 \cdot 35 - F = 0$

After doing some algebra we get

$\frac{F + 350}{1400} (n+1)^2 = n + 15$

$\frac{F + 350}{1400} (n+1)^2 = (n+1) + 14$

$\frac{F + 350}{1400} (n+1)^2 - (n+1) - 14 = 0$

$(F + 350)(n+1)^2 - 1400 (n+1) - 19600 = 0$

This is a quadratic equation in $n+1$ with coefficients

$a = F + 350, b = -1400, c = -19600$

Using the quadratic formula we get

$n + 1 = \frac{140(5 \pm \sqrt{F + 375})}{F+350}$

$n = \frac{140(5 \pm \sqrt{F + 375})}{F+350} - 1$

Since $\sqrt{F + 375} \geq \sqrt{375} > \sqrt{25} = 5$, the solution with the minus sign is negative so we discard it.

Therefore, the number of firms is

$n = \frac{140(5 + \sqrt{F + 375})}{F+350} - 1$

We got the number of firms as a function of the fixed cost $F$. To get an actual number, you would need to know the fixed cost value.

EDIT: Maybe you have to use the fact that in the long run, the fixed costs are $0$, so you would follow the same procedure and just get rid of $F$.

In that case, the number of firms would be:

$n = \frac{140(5+\sqrt{375})}{350} - 1 \approx 8.75$

Since there can’t be fractional firms, here I’d say the number of firms would be $8$ since it’s the highest integer value in which the firms don’t make negative economic profits.


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