Lecture 11: Oligopoly

1) Base Case - Oligopoly is about business strategy

2) Bertrand S.P.S. - Strategic Price Setting

Q=14PQ = 14 - P C1=2q1C_1 = 2q_1 C2=2q2C_2 = 2q_2

a) 3 Scenarios

A: P1=P2P_1=P_2  split the marker

q1=712P1q_1 = 7 - \frac{1}{2}P_1 q2=712q_2 = 7 - \frac{1}{2} 

G11=(PAC)qG_{1_1} = (P-AC)q AC=2AC = 2

G11=G12=(P2)(712P)G_{1_1} = G_{1_2} = (P-2)(7 - \frac{1}{2}P)

B: P1>P2P_1 > P_2  so q1=0q_1 = 0  G11=0G_{1_1} =0 q2=14P2q_2 = 14 - P_2

G12=(P22)(14P2)=2G12 AG_{1_2} = (P_2-2)(14-P_2)=2G_{1_2} \ ^A  \Rightarrow  Twice G1G_1 at AA

C: P1<P2P_1 < P_2 so q2=0q_2 = 0  G12=0G_{1_2} =0


b) Strategy for firm 1

if P2>AC1=2P_2 > AC_1 = 2  G11A=5.5G_{1_1^A} = 5.5

G11B=0G_{1_1^B} = 0

G11 C=G_{1_1} \ ^C = \checkmark Under cut

if P2<AC1=2P_2 < AC_1 = 2  G11A=G_{1_1^A} =  Negative

G11B=G_{1_1^B} =  Blackout \checkmark

G11C=G_{1_1^C} =  Double losses


3) Cournot SQS - Strategic quantity setting

P=14QP = 14-Q

a) Duoplay C1=2q1C_1 = 2q_1  C2=2q2C_2 = 2q_2 Q=q1+q2Q = q_1 + q_2

Firm 1: P1=14q1q2={14q2}q1 P_1 = 14 - q_1 - q_2 = \{ 14-q_2 \} -q_1 \Rightarrow Intercept, depends on f2f_2 output

R1=14q1q1q2q1 2R_1 = 14 q_1 - q_1q_2 -q_1 \ ^2

MR1=14q22q1MR_1 = 14 -q_2 -2q_1

MC1=2MC_1 = 2

q1=612q2q_1 = 6 - \frac{1}{2} q_2 \Rightarrow BRF, best response function

Firm 2: P2={lnq1}2q2}P_2 = \{\ln -q_1 \} -2q_2\}

MR2=14q12q2MC2=2}\left.\begin{matrix} MR_2 = 14-q_1-2q_2 \\ MC_2=2 \end{matrix}\right\} q2=612q1q_2 = 6 - \frac{1}{2} q_1

Equilibrium - "Stable" - Agreed on a stable outcome \Rightarrow No changing mind initial offer slope at EE

q1=612[612q1]q_1 = 6 - \frac{1}{2} [6 -\frac{1}{2} q_1]

=3+14q1=3 + \frac{1}{4} q_1  34q1=3\frac{3}{4} q_1 = 3

q1=4q_1 = 4  1q2=4q_2 =4 Q=8Q=8 p=14Q=6p=14-Q=6


b) Cournot triopoly

Q=14PQ=14-P AC=2AC=2 `m=3m=3 P=14QP=14-Q

Q=q1+q2+q3Q=q_1+q_2+q_3 P=14q1q2q3P= 14-q_1-q_2-q_3

Firm 1: P=(14q2q3)q1P = (14-q_2-q_3)-q_1 MR1=14q2q32q1MR_1 = 14 -q_2-q_3-2q_1 Mc=2Mc=2

q1=60.5q20.5q3q2=60.5q10.5q3q3=60.5q10.5q2} \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\} 18=2(q1+q2+q3)q1+q2+q3=gq1=q2+q33 Q=9 p=5 \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}

G11=G12G13=(52)(3)=9G_{1_1} = G_{1_2} - G_{1_3} = (5-2)(3) = 9

If Duopoly: G1=(62)(4)=16G_{1} = (6-2)(4)=16  As n, G1n \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 μc\mu c,there would be a built-in disadvantage to go large so we bribe.

c) Cournot N-opoly

P1=14mi=1P_1 = 14 - \underset{i=1}{\overset{m}{{\sum }}} MR1=142q1ni=2q1MR_1 = 14-2q_1- \underset{i=2}{\overset{n}{{\sum }}} q_1

MC1=2MC_1=2 2=142q1ni=2q12 = 14-2q_1- \underset{i=2}{\overset{n}{{\sum }}} q_1

Note: q1=q2=q3=qnq_1=q_2=q_3=q_n  because some cost

2=142q1ni=2q\Rightarrow 2 = 14-2q_1- \underset{i=2}{\overset{n}{{\sum }}} q 2=142qni=22 = 14-2q- \underset{i=2}{\overset{n}{{\sum }}}  2=122qq(n1)2 = 12 - 2q - q (n-1)

2=12q(n+1)\Rightarrow 2 = 12 -q(n+1)  q(n+1)=12q(n+1)=12 q=12n+1q = \frac{12}{n+1} Q=12nn+1Q = \frac{12n}{n+1}


limnnn+1=0\infty \rightarrow \lim _{n \rightarrow \infty} \frac{n}{n+1} = 0 

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

3) Stackelberg - Leader/Follower (SQS)

Leader knows BRFf

Follower q2=612q1q_2 = 6 - \frac{1}{2} q_1  BRF2 Leader P1=14q1q2P_1 = 14-q_1-q_2

=14q1=14-q_1

=812q1=8-\frac{1}{2} q_1

MR=8q, MC=2MR = 8 - q, \ MC = 2

q1=6 q2=3 Q=9 P=Sq_1 = 6 \ q_2 = 3 \ Q=9 \ P = S

G11=16 G12G_{1_1}=16 \ G_{1_2}=9=9 Taking advantage

Triopoly - Simultaneous or sequential

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

p=14q1q2q3p = 14 - q_1 -q_2 -q_3  f382f_38_2  follow f1f_1 simult

Firm 2: MR2=12q1q32q2MR_2 = 12-q_1 -q_3 -2q_2  MC2=2MC_2 = 2

q2=60.5q10.5q3 (1)q_2 = 6 -0.5q_1 -0.5q_3 \ (1)

q3=60.5q10.5q2 (2)q_3 = 6 -0.5q_1 -0.5q_2 \ (2)

0.5q2=60.5q1q3\Rightarrow 0.5q_2 = 6 - 0.5q_1 - q_3  2×(2)(1)2 \times (2) - (1)

q2=12q12q3q_2 = 12-q_1 -2q_3  0=60.5q10.5q30=6-0.5q_1 -0.5q_3

1.5q3=60.5q11.5q_3 = 6-0.5q_1

q3=413q1, q2=413q1Followers\underset{Followers}{\underbrace{q_3 = 4 -\frac{1}{3} q_1 , \ q_2=4 - \frac{1}{3} q_1}}

P=14q(413q1)(413q1)P = 14-q-(4-\frac{1}{3}q_1)-(4-\frac{1}{3}q_1)

=612q1= 6 - \frac{1}{2}q_1

MR1=623q1MC1=23q1} \left.\begin{matrix} MR_1 = 6 - \frac{2}{3}q_1 \\ MC_1 = \frac{2}{3}q_1 \end{matrix}\right\} q1=4.5q_1 = 4.5  q2=q32.5q_2 = q_3 - 2.5  Q=9.5Q=9.5 p=4.5p=4.5 

G11=4.5(4.5)134.52=13.5G_{1_1} =4.5(4.5) - \frac{1}{3} 4.5^2=13.5

G12,3=4.5(2.5)2(2.5)=6.25G_{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, q2q3)q_2 \neq q_3)


4) Chamberlin-Monopolistic Competition

Product Differentiation - Advertising to get monopoly power.P11 P_1^1 - demand for firm one when there is only one firm. As firms enterP1 2,P1 3,P1 4 P_1 \ ^2, P_1 \ ^3, P_1 \ ^4 - market share gets smaller until G1=0G_1 = 0 an entry stops.

Long run EE: 1)1) P=ACG1=0P=AC \Rightarrow G_1 = 0

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

P=12nQP=12-n Q c=Q2+6c=Q^2+6 when n=1n=1 ,, P=12QP=12 - Q

MR=122QMR = 12 -2Q  MC=2QMC = 2Q MC=MRMC = MR  122nQ=2Q (1)\rightarrow 12 - 2nQ = 2Q \ (1)

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

(1)6Q=nQn=6Q1(1) 6-Q=nQ \Rightarrow n = \frac{6}{Q}-1 into (2)(2)  12(6Q1)Q=Q+6Q6+Q=Q+6QQ=112-(\frac{6}{Q}-1) Q =Q + \frac{6}{Q} \Rightarrow 6+Q=Q + \frac{6}{Q} \Rightarrow Q=1 & n=5n=5

p=125Qp = 12 - 5 Q  MR=1210QMR = 12 - 10Q p=7p=7 MR=2MR=2


b) Predatory pricing

P=12QC1=Q1 2C2=Q2 2+8entrant}u=1  MR=122q1=Mc=2QQ1=3  P1=9 G11(critical that the entrant has fixed costsyou dont have so you can take advantage)\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<qp<q G1 G_1 \downarrow but you can make G12=0G_{1_2}=0 so they wont enter.

P2=12Q1Q2P_2 = 12 - Q_1 - Q_2 MR2=12Q12Q2MC2=2Q2}Q2=313Q1 BRf2 \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

P2=12Q1[313Q1]=923Q1P_2 = 12 - Q_1 - [3-\frac{1}{3}Q_1] = 9 - \frac{2}{3} Q_1 G12=P2Q2C2G_{1_2} = P_2Q_2-C_2

=(923Q1)(313Q1)(313Q1)28=(9-\frac{2}{3}Q_1)(3-\frac{1}{3} Q_1)-(3-\frac{1}{3}Q_1)^2-8

G12=2(313Q1)2\Rightarrow G_{1_2} = 2 (3-\frac{1}{3}Q_1)^2 2=313Q12=3 - \frac{1}{3}Q_1 3=Q13=Q_1 Q2=0Q_2=0


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