# Lecture 11: Oligopoly ### 2) Bertrand S.P.S. - Strategic Price Setting

﻿$Q = 14 - P$﻿ ﻿$C_1 = 2q_1$﻿ ﻿$C_2 = 2q_2$﻿

#### a) 3 Scenarios

A: ﻿$P_1=P_2$﻿ split the marker

﻿$q_1 = 7 - \frac{1}{2}P_1$﻿ ﻿$q_2 = 7 - \frac{1}{2}$﻿

﻿$G_{1_1} = (P-AC)q$﻿ ﻿$AC = 2$﻿

﻿$G_{1_1} = G_{1_2} = (P-2)(7 - \frac{1}{2}P)$﻿

B: ﻿$P_1 > P_2$﻿ so ﻿$q_1 = 0$﻿ ﻿$G_{1_1} =0$﻿ ﻿$q_2 = 14 - P_2$﻿

﻿$G_{1_2} = (P_2-2)(14-P_2)=2G_{1_2} \ ^A$﻿ ﻿$\Rightarrow$﻿ Twice ﻿$G_1$﻿ at ﻿$A$﻿

C: ﻿$P_1 < P_2$﻿ so ﻿$q_2 = 0$﻿ ﻿$G_{1_2} =0$﻿

#### b) Strategy for firm 1

if ﻿$P_2 > AC_1 = 2$﻿ ﻿$G_{1_1^A} = 5.5$﻿

﻿$G_{1_1^B} = 0$﻿

﻿$G_{1_1} \ ^C = \checkmark$﻿ Under cut

if ﻿$P_2 < AC_1 = 2$﻿ ﻿$G_{1_1^A} =$﻿ Negative

﻿$G_{1_1^B} =$﻿ Blackout ﻿$\checkmark$﻿

﻿$G_{1_1^C} =$﻿ Double losses

### 3) Cournot SQS - Strategic quantity setting

﻿$P = 14-Q$﻿

#### a) Duoplay ﻿$C_1 = 2q_1$﻿﻿$C_2 = 2q_2$﻿﻿$Q = q_1 + q_2$﻿

Firm 1: ﻿$﻿P_1 = 14 - q_1 - q_2 = \{ 14-q_2 \} -q_1$﻿ ﻿$\Rightarrow$﻿ Intercept, depends on ﻿$f_2$﻿ output

﻿$R_1 = 14 q_1 - q_1q_2 -q_1 \ ^2$﻿

﻿$MR_1 = 14 -q_2 -2q_1$﻿

﻿$MC_1 = 2$﻿

﻿$q_1 = 6 - \frac{1}{2} q_2$﻿ ﻿$\Rightarrow$﻿ BRF, best response function

Firm 2: ﻿$P_2 = \{\ln -q_1 \} -2q_2\}$﻿

﻿$\left.\begin{matrix} MR_2 = 14-q_1-2q_2 \\ MC_2=2 \end{matrix}\right\}$﻿ ﻿$q_2 = 6 - \frac{1}{2} q_1$﻿

Equilibrium - "Stable" - Agreed on a stable outcome ﻿$\Rightarrow$﻿ No changing mind initial offer slope at ﻿$E$﻿ ﻿$q_1 = 6 - \frac{1}{2} [6 -\frac{1}{2} q_1]$﻿

﻿$=3 + \frac{1}{4} q_1$﻿ ﻿$\frac{3}{4} q_1 = 3$﻿

﻿$q_1 = 4$﻿ 1﻿$q_2 =4$﻿ ﻿$Q=8$﻿ ﻿$p=14-Q=6$﻿

#### b) Cournot triopoly

﻿$Q=14-P$﻿ ﻿$AC=2$﻿ `﻿$m=3$﻿ ﻿$P=14-Q$﻿

﻿$Q=q_1+q_2+q_3$﻿ ﻿$P= 14-q_1-q_2-q_3$﻿

Firm 1: ﻿﻿$P = (14-q_2-q_3)-q_1$﻿ ﻿$MR_1 = 14 -q_2-q_3-2q_1$﻿ ﻿$Mc=2$﻿

﻿$\left.\begin{matrix} q_1 = 6- 0.5q_2 -0.5q_3 \\ q_2 = 6- 0.5q_1 -0.5q_3 \\ q_3 = 6- 0.5q_1 -0.5q_2 \end{matrix}\right\}$﻿ ﻿$\begin{matrix} 18 = 2(q_1+q_2+q_3) \\ q_1+q_2+q_3=g \\ q_1=q_2+q3-3 \ Q=9 \ p=5 \end{matrix}$﻿

﻿$G_{1_1} = G_{1_2} - G_{1_3} = (5-2)(3) = 9$﻿

If Duopoly: ﻿$G_{1} = (6-2)(4)=16$﻿ As ﻿$n \uparrow , \ G_1$﻿ for each ﻿$\downarrow$﻿

Each firm looses 7, and the third earns 9. It is worth for the original to bribe the third not to enter! A monopoly would make 36. In duopoly it would be 16 so the monopoly can bribe the duopolist 16 dollars- CRTS causes this. If we had upward slopping ﻿$\mu c$﻿,there would be a built-in disadvantage to go large so we bribe.

#### c) Cournot N-opoly

﻿$P_1 = 14 - \underset{i=1}{\overset{m}{{\sum }}}$﻿ ﻿$MR_1 = 14-2q_1- \underset{i=2}{\overset{n}{{\sum }}} q_1$﻿

﻿$MC_1=2$﻿ ﻿$2 = 14-2q_1- \underset{i=2}{\overset{n}{{\sum }}} q_1$﻿

Note: ﻿$q_1=q_2=q_3=q_n$﻿ because some cost

﻿$\Rightarrow 2 = 14-2q_1- \underset{i=2}{\overset{n}{{\sum }}} q$﻿ ﻿$2 = 14-2q- \underset{i=2}{\overset{n}{{\sum }}}$﻿ ﻿$2 = 12 - 2q - q (n-1)$﻿

﻿$\Rightarrow 2 = 12 -q(n+1)$﻿ ﻿$q(n+1)=12$﻿ ﻿$q = \frac{12}{n+1}$﻿ ﻿$Q = \frac{12n}{n+1}$﻿ ﻿$\infty \rightarrow \lim _{n \rightarrow \infty} \frac{n}{n+1} = 0$﻿

﻿$\frac {\lim \ 12n}{\lim\ n+1} = 12$﻿

### 3) Stackelberg - Leader/Follower (SQS)

Follower ﻿$q_2 = 6 - \frac{1}{2} q_1$﻿ BRF2 Leader ﻿$P_1 = 14-q_1-q_2$﻿

﻿$=14-q_1$﻿

﻿$=8-\frac{1}{2} q_1$﻿

﻿$MR = 8 - q, \ MC = 2$﻿

﻿$q_1 = 6 \ q_2 = 3 \ Q=9 \ P = S$﻿

﻿$G_{1_1}=16 \ G_{1_2}$﻿﻿$=9$﻿ Taking advantage

Triopoly - Simultaneous or sequential

Do both follow one or third follows second and second follows first?

﻿$p = 14 - q_1 -q_2 -q_3$﻿ ﻿$f_38_2$﻿ follow ﻿$f_1$﻿ simult

Firm 2: ﻿$MR_2 = 12-q_1 -q_3 -2q_2$﻿ ﻿$MC_2 = 2$﻿

﻿$q_2 = 6 -0.5q_1 -0.5q_3 \ (1)$﻿

﻿$q_3 = 6 -0.5q_1 -0.5q_2 \ (2)$﻿

﻿$\Rightarrow 0.5q_2 = 6 - 0.5q_1 - q_3$﻿ ﻿$2 \times (2) - (1)$﻿

﻿$q_2 = 12-q_1 -2q_3$﻿ ﻿$0=6-0.5q_1 -0.5q_3$﻿

﻿$1.5q_3 = 6-0.5q_1$﻿

﻿$\underset{Followers}{\underbrace{q_3 = 4 -\frac{1}{3} q_1 , \ ﻿q_2=4 - \frac{1}{3} q_1}}$﻿

﻿$P = 14-q-(4-\frac{1}{3}q_1)-(4-\frac{1}{3}q_1)$﻿

﻿$= 6 - \frac{1}{2}q_1$﻿

﻿$\left.\begin{matrix} MR_1 = 6 - \frac{2}{3}q_1 \\ MC_1 = \frac{2}{3}q_1 \end{matrix}\right\}$﻿ ﻿$q_1 = 4.5$﻿ ﻿$q_2 = q_3 - 2.5$﻿ ﻿$Q=9.5$﻿ ﻿$p=4.5$﻿

﻿$G_{1_1} =4.5(4.5) - \frac{1}{3} 4.5^2=13.5$﻿

### ﻿$G_{1_{2,3}} = 4.5(2.5)-2(2.5)=6.25$﻿

(Sequential-solve BRf3 sub into firm 2 demand to get BRf2 and sub both into leader demand)(If two don't have same C's, ﻿$q_2 \neq q_3)$﻿

### 4) Chamberlin-Monopolistic Competition Product Differentiation - Advertising to get monopoly power.﻿$P_1^1$﻿ - demand for firm one when there is only one firm. As firms enter﻿$P_1 \ ^2, P_1 \ ^3, P_1 \ ^4$﻿ - market share gets smaller until ﻿$G_1 = 0$﻿ an entry stops. Long run ﻿$E$﻿: ﻿$1)$﻿ ﻿$P=AC \Rightarrow G_1 = 0$﻿

﻿$2)$﻿ ﻿$MR=MC \Rightarrow$﻿ Maximizing = 0

﻿$P=12-n Q$﻿ ﻿$c=Q^2+6$﻿ when ﻿$n=1$﻿﻿$,$﻿ ﻿$P=12 - Q$﻿

﻿$MR = 12 -2Q$﻿ ﻿$MC = 2Q$﻿ ﻿$MC = MR$﻿ ﻿$\rightarrow 12 - 2nQ = 2Q \ (1)$﻿

﻿$(2) \ P=AC$﻿ ﻿$AC=Q + \frac{6}{Q} = 12-nQ =P$﻿

﻿$(1) 6-Q=nQ \Rightarrow n = \frac{6}{Q}-1$﻿ into ﻿$(2)$﻿ ﻿$12-(\frac{6}{Q}-1) Q =Q + \frac{6}{Q} \Rightarrow 6+Q=Q + \frac{6}{Q} \Rightarrow Q=1$﻿ & ﻿$n=5$﻿

﻿$p = 12 - 5 Q$﻿ ﻿$MR = 12 - 10Q$﻿ ﻿$p=7$﻿ ﻿$MR=2$﻿ #### b) Predatory pricing

﻿$\left.\begin{matrix} P=12-Q\\ C_1=Q_1 \ ^2\\ \underset{entrant}{C_2=Q_2 \ ^2 +8 } \end{matrix}\right\} \begin{matrix} u=1 \ \ MR = 12 -2q_1=Mc=2Q\\ Q_1=3 \ \ P_1 = 9 \ G_{1_1}\\ (critical \ that \ the \ entrant \ has \ fixed \ costs \\ you \ don't \ have \ so \ you \ can\ take \ advantage) \end{matrix}$﻿

Drop ﻿$p﻿ ﻿$G_1 \downarrow$﻿ but you can make ﻿$G_{1_2}=0$﻿ so they wont enter.

﻿$P_2 = 12 - Q_1 - Q_2$﻿ ﻿$\left.\begin{matrix} MR_2 = 12 - Q_1 - 2Q_2 \\ MC_2 = 2Q_2 \end{matrix}\right\} Q_2 = 3 - \frac{1}{3}Q_1 \ BRf_2$﻿

﻿$P_2 = 12 - Q_1 - [3-\frac{1}{3}Q_1] = 9 - \frac{2}{3} Q_1$﻿ ﻿$G_{1_2} = P_2Q_2-C_2$﻿

﻿$=(9-\frac{2}{3}Q_1)(3-\frac{1}{3} Q_1)-(3-\frac{1}{3}Q_1)^2-8$﻿

﻿$\Rightarrow G_{1_2} = 2 (3-\frac{1}{3}Q_1)^2$﻿ ﻿$2=3 - \frac{1}{3}Q_1$﻿ ﻿$3=Q_1$﻿ ﻿$Q_2=0$﻿