# Cournot Oligopoly with N Firms

I recently took ECON 325 with Basyl Golovedstkyy (SFU) for my summer semester. It’s a fairly comprehensive course focusing on Industrial Organization modelling how firms can behave when they have market power but aren’t completely in a monopolistic setting.

One of the problems I had an issue on personally on the final exam was this model of a Cournot oligopoly with a variable amount (n) of firms. Being so frustrated with myself, I decided to go over it again slowly now that my semester is over because I know that it certainly will pop up in the future.

If you’re looking for a clear explanation of the model, look no further! Now, let’s begin.

# A Cournot-Style Oligopoly

## Model Setup

Let’s keep things fairly general.

Market inverse demand curve:

$P\quad =\quad A\quad -\quad BQ$

Cost function:

$P\quad=\quad{Cq }_{ i }$

We know in a Cournot model that at equilibrium, all firms will produce exactly the same amount ${ q }_{ i }$. Therefore, the sum of all those individual yet outputs will equal ${ Q }^{ * }$:

${ Q }^{ * }\quad=\quad\sum _{i = 1 }^{ N }{ { q }_{ i } }\quad=\quad({q}_{1}\quad+\quad{q}_{2}\quad+\quad{q}_{3}\quad+\quad\cdots\quad+\quad{q}_{N})$

Armed with this knowledge, let’s expand our inverse demand function by substituting $Q$ with $({q}_{1}\quad+\quad{q}_{2}\quad+\quad{q}_{3}\quad+\quad\cdots\quad+\quad{q}_{N})$:

$P\quad =\quad A\quad -\quad B({q}_{1}\quad+\quad{q}_{2}\quad+\quad{q}_{3}\quad+\quad\cdots\quad+\quad{q}_{N})$

Let’s factor out one ${q}_{i}$ since we’ll be solving for one.

Hmm, ${q}_{1}$ seems like a good choice.

$P\quad =\quad A\quad -\quad B({ q }_{ 1 })\quad-\quad B({ q }_{ 2 }\quad +\quad { q }_{ 3 }\quad +\quad { q }_{ 4 }\quad +\quad \cdots \quad +\quad { q }_{ N })$

Now, to simplify a little, let us pause and consider $({ q }_{ 2 }\quad +\quad { q }_{ 3 }\quad +\quad { q }_{ 4 }\quad +\quad \cdots \quad +\quad { q }_{ N })$. What exactly is this?

It’s actually ${Q}_{N}$ with one of the taken out. This means we can write that as:

$({ q }_{ 2 }\quad +\quad { q }_{ 3 }\quad +\quad { q }_{ 4 }\quad +\quad \cdots \quad +\quad { q }_{ N })\quad =\quad \sum _{i = 1 }^{ N }{ { q }_{ i } }\quad -\quad { q }_{ 1 }\quad =\quad Q\quad -\quad { q }_{ 1 }\quad=\quad { Q }_{ n-1 }$

This allows us to rewrite and greatly simplify our demand function:

$P\quad =\quad A\quad -\quad B{ q }_{ 1 }\quad -\quad B{ Q }_{ n-1 }$

## Marginal Revenue & Costs

To get ${q}_{1}^{*}$ we must equate marginal revenue to marginal cost (${MR}_{1} = {MC}_{1}$) and solve for ${q}_{1}$.

Let’s find our marginal revenues and costs: the derivatives of our demand and cost functions with respect to ${q}_{1}$.

We know that marginal cost is simply the derivative of the cost function we are given with respect to the ${q}_{i}$ we need.

${MC}_{1}\quad =\quad \frac { \delta C{q}_{1} }{ \delta {q}_{1} }\quad=\quad C$ ${ MR }_{ 1 }\quad =\quad \frac { \delta PQ }{ \delta { q }_{ 1 } } \quad =\quad \frac { \delta (A\quad -\quad B{ q }_{ 1 }\quad -\quad B{ Q }_{ n-1 }){ q }_{ 1 } }{ \delta { q }_{ 1 } } \quad =\quad\frac { \delta (A{ q }_{ 1 }\quad -\quad B{ q }^{ 2 }_{ 1 }\quad -\quad B{ q }_{ 1 }{ Q }_{ n-1 }) }{ \delta { q }_{ 1 } }$

Taking the derivative and simplifying further (we can also use our knowledge that marginal revenue is simply the inverse demand with double the slope for the ${q}_{i}$ in question):

${ MR }_{ 1 }\quad =\quad A\quad -\quad 2B{q}_{1}\quad -\quad B{ Q }_{ n-1 }$

## Profit Maximization

Now that we have both marginal cost and revenue, we can now profit maximize and find ${q}^{*}_{1}$.

${ MR }_{ 1 }\quad =\quad {MC}_{1}$ $A\quad -\quad 2B{q}_{1}\quad -\quad B{ Q }_{ n-1 }\quad =\quad C$

We know ${ Q }_{ n-1 } = Q - { q }_{ 1 }$ so let’s substitute it in our equation above:

$A\quad -\quad 2B{q}_{1}\quad -\quad B(Q - { q }_{ 1 })\quad =\quad C$ $A\quad -\quad 2B{ q }_{ 1 }\quad -\quad BQ\quad+\quad B{ q }_{ 1 }\quad =\quad C$ $A\quad -\quad B{ q }_{ 1 }\quad -\quad BQ\quad =\quad C$

Solving for ${q}_{1}$:

${ q }^{ * }_{ 1 }\quad =\frac { \quad A\quad -\quad B Q \quad -\quad C }{ B }$

Now, we know that $\sum _{ i=1 }^{ N }{ {q}_{i}} = n{q}_{i}$. Let’s substitute this for $Q$.

${ q }^{ * }_{ 1 }\quad =\frac { \quad A\quad -\quad nB{q}_{i} \quad -\quad C }{ B }$ ${ q }^{ * }_{ 1 }\quad =\quad \frac { A }{ B } \quad -\quad n{ q }_{ i }\quad -\quad \frac { C }{ B }$ ${ q }^{ * }_{ 1 }\quad +\quad n{ q }_{ i }\quad =\quad \frac { A }{ B } \quad -\quad \frac { C }{ B }$ $(1\quad +\quad n){ q }^{ * }_{ 1 }\quad =\quad \frac { A }{ B } \quad -\quad \frac { C }{ B }$ ${ q }^{ * }_{ 1 }\quad =\quad (\frac { A }{ B } \quad -\quad \frac { C }{ B } )(\frac { 1 }{ (1\quad +\quad n) })$ ${ q }^{ * }_{ 1 }\quad =\quad (\frac { A\quad-\quad C}{B} )(\frac { 1 }{ (1\quad +\quad n) } )$

And again, since we know $n{ q }^{ * }_{ 1 }=Q$:

${Q}^{*}\quad =\quad (n)(\frac { A\quad -\quad C }{ B } )(\frac { 1 }{ (1\quad +\quad n) } )$

Now that we have total market output at equilibrium, we can deduce what equilibrium price will be in the market.

${P}^{*}\quad =\quad A\quad -\quad B((n)(\frac { A\quad -\quad C }{ B } )(\frac { 1 }{ (1\quad +\quad n) } ))$

From hereon in, you can go ahead and find the revenues or loses, equilibrium profits for each firm or the total profit the industry makes.