<p>As Giskard’s answer to the linked question says, since the firms have symmetric costs, in the long run, they all make <span class="math-container">$0$</span> profit.</p>
<p>The inverse demand function is</p>
<p><span class="math-container">$Q = 150 - P \implies P(Q) = 150 - Q$</span></p>
<p>The profit function is</p>
<p><span class="math-container">$\Pi_i = P(Q) q_i - TC_i(q_i)$</span></p>
<p>Substituting all the given functions,</p>
<p><span class="math-container">$\Pi_i = 150 q_i - Q q_i - 10 q_i - F$</span></p>
<p>Now we find the optimal production for each firm:</p>
<p>By the product rule on <span class="math-container">$Q q_i$</span> and noting <span class="math-container">$\frac{\partial Q}{\partial q_i} = 1$</span>,</p>
<p><span class="math-container">$\frac{\partial \Pi_i}{\partial q_i} = 140 - Q - q_i = 0$</span></p>
<p>Since the firms have symmetric costs, they have the same optimal productions, which implies <span class="math-container">$ Q = n q_i$</span></p>
<p><span class="math-container">$140 - (n+1) q_i = 0$</span></p>
<p>Isolating <span class="math-container">$q_i$</span></p>
<p><span class="math-container">$q_i = \frac{140}{n+1}$</span></p>
<p><span class="math-container">$Q = \frac{140 n}{n+1}$</span></p>
<p>From the inverse demand function we get</p>
<p><span class="math-container">$P = \frac{10 (n+15)}{n+1}$</span></p>
<p>Substituting these into the <span class="math-container">$0$</span> profit condition,</p>
<p><span class="math-container">$10 \cdot \frac{n+15}{n+1} \cdot \frac{140}{n+1} - 10 \cdot 35 - F = 0$</span></p>
<p>After doing some algebra we get</p>
<p><span class="math-container">$\frac{F + 350}{1400} (n+1)^2 = n + 15$</span></p>
<p><span class="math-container">$\frac{F + 350}{1400} (n+1)^2 = (n+1) + 14$</span></p>
<p><span class="math-container">$\frac{F + 350}{1400} (n+1)^2 - (n+1) - 14 = 0$</span></p>
<p><span class="math-container">$(F + 350)(n+1)^2 - 1400 (n+1) - 19600 = 0$</span></p>
<p>This is a quadratic equation in <span class="math-container">$n+1$</span> with coefficients</p>
<p><span class="math-container">$a = F + 350, b = -1400, c = -19600$</span></p>
<p>Using the quadratic formula we get</p>
<p><span class="math-container">$n + 1 = \frac{140(5 \pm \sqrt{F + 375})}{F+350}$</span></p>
<p><span class="math-container">$n = \frac{140(5 \pm \sqrt{F + 375})}{F+350} - 1$</span></p>
<p>Since <span class="math-container">$\sqrt{F + 375} \geq \sqrt{375} &gt; \sqrt{25} = 5$</span>, the solution with the minus sign is negative so we discard it.</p>
<p>Therefore, the number of firms is</p>
<p><span class="math-container">$n = \frac{140(5 + \sqrt{F + 375})}{F+350} - 1$</span></p>
<p>We got the number of firms as a function of the fixed cost <span class="math-container">$F$</span>. To get an actual number, you would need to know the fixed cost value.</p>
<p>EDIT: Maybe you have to use the fact that in the long run, the fixed costs are <span class="math-container">$0$</span>, so you would follow the same procedure and just get rid of <span class="math-container">$F$</span>.</p>
<p>In that case, the number of firms would be:</p>
<p><span class="math-container">$n = \frac{140(5+\sqrt{375})}{350} - 1 \approx 8.75$</span></p>
<p>Since there can’t be fractional firms, here I’d say the number of firms would be <span class="math-container">$8$</span> since it’s the highest integer value in which the firms don’t make negative economic profits.</p>

CC BY-SA 4.0
2023-04-13T16:20:45.350Z
2023-04-13T16:59:28.953Z
2023-04-13T16:59:28.953Z
Nicolas Torres
<p>I am trying to solve the following question:</p>
<p>'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.</p>
<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.</p>
<p><em>Determine the long-run equilibrium number of firms n^0.</em>'</p>
<p>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!</p>

CC BY-SA 4.0
2023-04-13T13:55:30.493Z
Does this answer your question? [Long Run Equilibrium of Oligopolies](https://economics.stackexchange.com/questions/13958/long-run-equilibrium-of-oligopolies)
Giskard
<microeconomics><cournot>
How to calculate the long-run equilibrium of firms in a Cournot competition?
Lucas
# How to calculate the long-run equilibrium of firms in a Cournot competition?

(Asked by: username_0 on 2023-04-13)

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!

**Tags:** microeconomics, cournot

> **Comments:**
> **username_1**: Does this answer your question? [Long Run Equilibrium of Oligopolies](https://economics.stackexchange.com/questions/13958/long-run-equilibrium-of-oligopolies)  

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## 1 answers

### Answer by username_2 on 2023-04-13. Score: -1

As username_1’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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