Chapter 15 Solutions

Contributed by Anna Fairbain 2016

15.1
a.
\[\begin{align*} P & = 150 - Q\\ \pi & = (150 - Q)Q - 0\\ \pi & = 150Q - Q^2\\ \frac{\partial\pi}{\partial Q} & = 150 - 2Q\\ 0 & = 150 - 2Q\\ \end{align*}\]

Solving for Q we get the solutions:

\[\begin{align*} Q^* & = 75\\ P^* & = 150 - 75\\ P^*& = \$75\\ \pi & = 75*75 - 0\\ \pi & = \$5,625\\ \end{align*}\]

\[\begin{align*} Firm 1: \pi & = [150 - q_1 - q_2]q_1\\ \pi & = 150q_1 - q_1^2 - q_2q_1\\ \frac{\partial\pi}{\partial q_1} & = 150 - 2q_1 - q_2\\ 0 & = 150 - 2q_1 - q_2 (1)\\ \end{align*}\] \[\begin{align*} Firm 2: \pi & = [150 - q_1 - q_2]q_2\\ \pi & = 150 - q_1q_2 - q_2^2\\ \frac{\partial\pi}{\partial q_2} & = 150 - q_1 - 2q_2\\ 0 & = 150 - q_1 - 2q_2 (2)\\ \end{align*}\]

Solve the system of best response equations 1 and 2. Multiply equation 2 by -2 and add to equation 1.

\[\begin{align*} 0 & = -150 + 3q_2\\ q_2* & = 50\\ q_1* & = 50\\ Q^* & = 100\\ P^* & = \$50\\ \pi_i & = \$2,500\\ \end{align*}\]

In Bertrand Nash equilibrium, P = MC, which is zero in this case:

\[\begin{align*} P & = 0 \rightarrow Q = 150\\ q_1 & = q_2 = \frac{Q}{2} = 75\\ \pi_1 & = \pi_2 = 0\\ \end{align*}\]

15.2
a.

Monopolist

\[\begin{align*} \pi & = (a - bQ)Q - C*Q\\ \pi & = aQ - bQ^2 - CQ\\ \frac{\partial\pi}{\partial Q} & = a - 2bQ - c \\ 0 & = a - 2bQ - c\\ \end{align*}\] \[\begin{align*} Q^* & = \frac{a - c}{2b}\\ P & = a - b(\frac{a - c}{2b})\\ P & = a - \frac{(a - c)}{2}\\ P^* & = \frac{a}{2} + \frac{c}{2}\\ \pi & = (\frac{a + c}{2})(\frac{a - c}{2b}) - c(\frac{a - c}{2b})\\ \end{align*}\]

\[\begin{align*} \pi_1 & = [a - b(q_1 + q_2)]q_1 - Cq_1\\ \pi_1 & = aq_1 - bq_1^2 -bq_1q_2 - Cq_1\\ \frac{\partial \pi_1}{\partial q_1} & = a - 2bq_1 - bq_2 - C\\ 0 & = a - 2bq_1 - bq_2 - C\\ \end{align*}\] \[\begin{align*} \pi_2 & = [a - b(q_1 + q_2)]q_2 - Cq_2\\ \pi_2 & = aq_2 - bq_1q_2 - bq_2^2 - Cq_2\\ \frac{\partial \pi_2}{\partial q_2} & = a - bq_1 - 2bq_2 - C\\ 0 & = a - bq_1 - 2bq_2 - C\\ \end{align*}\] \[\begin{align*} 0 & = a - 2bq_1 - bq_2 - C\\ 0 & = -2a + 2bq_1 + 4bq_2 + 2C\\ 0 & = -a + 3bq_2 + C\\ q_2^* & = \frac{a -c}{3b}\\ \end{align*}\] \[\begin{align*} 0 & = a - bq_1 - 2b(\frac{a-c}{3b}) - c\\ 0 & = a - bq_1 - \frac{2}{3}(a - c) - c\\ 0 & = \frac{1}{3}a - bq_1 - \frac{1}{3}c\\ q_1^* & = \frac{a - c}{3b}\\ \end{align*}\] \[\begin{align*} P & = a - b\Big[\frac{2(a - c)}{3b}\Big]\\ P & = a - \frac{2}{3}(a - c)\\ P^* & = \frac{1}{3}a + \frac{2}{3}c\\ \end{align*}\] \[\begin{align*} Firm profit: \pi & = \frac{(a+2c)}{3}\frac{(a-c)}{3b} - \frac{c(a - c)}{3b}\\ Industry profit: \pi & = 2\Big[\frac{(a+2c)}{3}\frac{(a-c)}{3b} - \frac{c(a - c)}{3b}\Big]\\ Market output: Q & = q_1 + q_2 = \frac{2(a-c)}{3b}\\ \end{align*}\]

Nash equilibrium is:

\[\begin{align*} p_1 & = p_2 = C\\ C & = a - bQ\\ Q & = \frac{a - c}{b}\\ \end{align*}\]

And firms split output, so:

\[\begin{align*} q_1 & = q_2 = \frac{a -c}{2b}\\ \pi_1 & = 0 \\ \pi_2 & = 0 \\ \pi & = 0 \\ \end{align*}\]

\[\begin{align*} \pi_i & = [a - b(q_1 + q_2 + ... + q_n)]q_i - Cq_i = 0\\ \frac{\partial \pi_i}{\partial q_i} & = a - bq_1 - bq_2 - ... - bq_{i-1} - bq_{i+1} - bq_n - 2bq_i - C = 0\\ 0 & = a - \Sigma bq_i - bq_i - c\\ 0 & = a - bq_i(n + 1) - c\\ \end{align*}\]

Solving for \(q^*_i\):

\[\begin{align*} q_i^* & = \frac{a - c}{b(n + 1)}\\ Q & = (\frac{n}{n + 1})(\frac{a - c}{b})\\ Q^* & = a - \frac{n}{n + 1}(a - c)\\ \end{align*}\] \[\begin{align*} Firm \pi & = \Big[a - (\frac{n}{n+1})(a - c)\Big]\frac{(a - c)}{b(n + 1)} - \frac{c(a - c)}{b(n + 1)}\\ Industry \pi & = n\Bigg[\bigg(a - (\frac{n}{n+1})(a - c)]\bigg)\frac{(a - c}{b(n + 1)} - \frac{c(a - c)}{b(n + 1)}\Bigg]\\ \end{align*}\]

Set \(n=1\)

\[\begin{align*} P & = a - \frac{1}{2}(a - c)\\ P^* & = \frac{a - c}{2}\\ q^* & = \frac{1}{2}(\frac{a - c}{b})\\ \end{align*}\]

Set \(n=2\)

\[\begin{align*} P & = a - \frac{2}{3}(a - c)\\ P^* & = \frac{a - 2c}{3}\\ q^* & = \frac{a - c}{3b}\\ \end{align*}\]

Set \(n=\infty\)

\[\begin{align*} P & = a - 1(a - c)\\ P^* & = c\\ Q & = (\frac{n}{n+1})(\frac{a - c}{b})\\ Q^* & = 1(\frac{a - c}{b})\\ \end{align*}\]

Notice that when you set \(n = \infty\), the result gets you perfect competition.
Even more remarkable, only 2 firms gets you to competitive outcome in the Bertrand model.