Stackelberg Equilibrium

In a Stackelberg model, a firm enter the market first. Its quantity anticipates the quantity that the other firm will produce.

Example

The demand for coffee beans follows the inverse demand `P = 1400 - 2 Q_Z - 2 Q_Y`.

Zach chooses his quantity `Q_Z` first, then Yann follows.

Yann's Best Response Function

When Yann enters the market, he chooses the best response to `Q_Z`.

Yann's revenue is `R\left(Q_Y\right) = PQ = \left(1400 - 2 Q_Z - 2 Q_Y\right) Q_Y`.

Yann's marginal revenue is `MR\left(Q_Y\right) = 1400 - 2Q_Z - 4Q_Y`.

Yann's marginal cost is `MC\left(Q_Y\right) = 200`.

Yann maximizes profit when

$$ \begin{align*} MC\left(Q_Y\right) &= MR\left(Q_Y\right) \\ 200 &= 1400 - 2Q_Z - 4Q_Y \\ 4Q_Y &= 1200 - 2Q_Z \\ Q_Y &= 300 - \frac{Q_Z}{2} \end{align*} $$

Zach's profit maximization problem

Zach's maximizes `\Pi_Z \left(Q_Z \right) = \left(1400 - 2 Q_Z - 2 Q_Y\right) Q_Z -200 Q_Z`

Plugging `Q_Y = 300 - \frac{Q_Z}{2}` gives

$$ \begin{align*} \Pi_Z \left( Q_Z \right) &= \left(1400 - 2 Q_Z - 2 \left( 300 - \frac{Q_Z}{2} \right) \right) Q_Z - 200 Q_Z \\ &= \left(1400 - 2 Q_Z - 600 + Q_Z \right) Q_Z - 200 Q_Z \\ &= \left( 800 - Q_Z \right) Q_Z - 200 Q_Z \\ \end{align*} $$

Zach's profit is composed of a revenue `R \left( Q_Z \right) = \left( 800 - Q_Z \right) Q_Z` and a cost `C \left( Q_Z \right) = 200 Q_Z`.

When Zach maximizes profit

$$ \begin{align*} MC \left( Q_Z \right) &= MR \left( Q_Z \right) \\ 200 &= 800 - 2 Q_Z \\ 2 Q_Z &= 600 \\ Q_Z &= 300 \end{align*} $$

Zach produces 300 coffee beans. Yann produces `Q_Y = 300 - \frac{300}{2} = 150` coffee beans.

Question

The inverse demand on the market for coffee is `P = 210 - 5 ( Q_Z + Q_Y )`.

Zach faces marginal costs equal to `60`.

Yann faces marginal costs equal to `150`.

What is the Stackelberg Equilibrium?

In equilibrium, `Q_Z = 24.0` and `Q_Y = -6.0`.

Yann's marginal revenue is $$ \begin{align*} MR (Q_Y) \begin{align*} MR ( Q_Y ) &= 210 - 5 Q_Z - 2 \times 5 Q_Y \\ &= 210 - 5 Q_Z - 10 Q_Y \end{align*} \end{align*} $$ Yann's marginal cost is $$ \begin{align*} MC (Q_Y) = 150 \end{align*} $$ Yann's best response solves \begin{align*} MC ( Q_Y ) &= MR ( Q_Y ) \\ 150 &= 210 - 5 Q_Y - 10 Q_Z \\ 10 Q_Y &= 210 - 150 - 5 Q_Z \\ Q_Y &= \frac{210 - 150}{10} - \frac{Q_Z}{2} \\ Q_Y &= 6.0 - \frac{Q_Z}{2} \end{align*}

$$ \begin{align*} &\Pi_Z = \left( 210 - 5 ( Q_Z + Q_Y ) \right) Q_Z - 60 Q_Z \\ &= \left( 210 - 5 Q_Z - 5 \left( 6.0 - \frac{Q_Z}{2} \right) \right) Q_Z - 60 Q_Z \\ &= \left( 210 - 5 Q_Z - 30.0 + 5 \frac{ Q_Z }{ 2 } \right) Q_Z - 60 Q_Z \\ &= \left( 180.0 - 5 \frac{ Q_Z }{ 2 } \right) Q_Z - 60 Q_Z \end{align*} $$

Maximizing Zach's profit give

$$ \begin{align*} MC_Z &= MR_Z \\ 60 &= 180.0 - 5 Q_Z \\ 5 Q_Z &= 180.0 - 60 \\ 5 Q_Z &= 120.0 \\ Q_Z &= \frac{ 120.0 }{ 5 } \\ Q_Z &= 24.0 \end{align*} $$

Plug `Q_Z = 24.0` into Yann's Best Response Function:

$$ \begin{align*} Q_{ Y } &= 6.0 - \frac{Q_Z}{2} \\ &= 6.0 - \frac{ 24.0 }{2} \\ &= 6.0 - 12.0 \\ &= -6.0 \end{align*} $$
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