6.4 Cournot Model, Revisited

Recall the Cournot model from assignment 5.10 on oligopoly.

Question 1) Suppose market demand is given by \(P = 80 - Q\) and there are two identical firms with marginal cost \(MC = 20\).

  1. Show that firm 1’s revenue function is \(TR = (80 - Q_1 - Q_2) Q_1\).

  2. Firm 1’s reaction function is their profit maximizing \(Q_1\) given the quantity firm 2 chooses to produce. To derive this, set firm 1’s marginal revenue equal to their marginal cost. Solve for \(Q_1\) to get the reaction function.

  3. Firm 2 is symmetrical in every way to firm 1, so \(Q_2 = Q_1\). Use the equation from part b along with this equation to solve for the two unknowns, \(Q_1\) and \(Q_2\).

  4. Calculate the market price and each firm’s profit.

Question 2) Solve the Cournot model from question 1 for 3, 4, and 5 identical firms. You should find that as the number of firms increase, price and profit fall.

Question 3) What is the price under perfect competition?

Question 4) How many firms need to exist in the oligopoly for the price to be driven under $30?

Infinite Geometric Series

In the final couple assignments in this unit, you will be asked to solve a number of infinite geometric series by hand. Here is some practice doing that.

An infinite geometric series is a sum like this:

\[a + ar + ar^2 + ar^3 + ar^4 + ...\]

It has:

  • a first term \(a\)
  • a common ratio \(r\) (each term is multiplied by \(r\) to get the next)

When \(|r| < 1\), the series converges. To solve for the value it converges to, here’s a trick:

  • Let \(S\) be the value the series converges to: \(S = a + ar + ar^2 + ar^3 + ar^4 + ...\)
  • Notice that if you multiply \(S\) by \(a\), you get the same thing as if you subtract \(S\) by \(r\). So \(S - a = rS\). This equation lets you solve for \(S\) in terms of \(a\) and \(r\).

For example, take \(100 + .9 (100) + .9^2 (100) + .9^3 (100) + ...\)

  • The common ratio is 0.9, which is less than 1, so we know this series converges.
  • Let \(S = 100 + .9 (100) + .9^2 (100) + .9^3 (100) + ...\)
  • Notice that \(S - 100\) is the same thing as \(.9 S\).
  • Solve for S:

\[\begin{align} S - 100 &= .9 S\\ .1 S &= 100\\ S &= \frac{100}{.1}\\ S &= 100 \times 10\\ S &= 1000 \end{align}\]

Question 5)

Solve: \(100 + .8 (100) + .8^2 (100) + .8^3 (100) + ...\)

Question 6)

Solve: \(5 + .9 (5) + .9^2 (5) + .9^3 (5) + ...\)

Question 7)

Solve: \(0 + .9 (5) + .9^2 (5) + .9^3 (5) + ...\)

Question 8)

Solve: \(2 + .9 (2) + .9^2 (5) + .9^3 (5) + ...\)