Transcript • Chapter 28 Oligopoly extremes (pure competition and pure

• Chapter 28 Oligopoly
• It is the case that lies between two
extremes (pure competition and pure
monopoly). Often there are a number of
competitors but not so many as to regard
each of them as having a negligible effect
on price.
• Unreasonable to expect one grand model
for many behaviors observed in real life,
hence we will introduce some models.
• We will restrict to the case of two firms
or duopoly.
• The first model is the Stackelberg model,
for instance, Apple and its follower. It is a
quantity setting game and 1 is the leader
who chooses to produce y1 while after
seeing that, the follower, 2 decides to
produce y2. The total output of the market
is therefore y1+y2 and the price is p(y1+y2).
(sequential)
• Solving backwards, follower’s problem
maxy p(y1+y2)y2-c2(y2). FOC:
p(y1+y2)+p’(y1+y2)y2=c2’(y2). So basically,
from FOC, we can derive the reaction
function of 2. That is, given y1, there is an
optimal level of y2. In the following we
will work with the case where p(Y)=a-bY
and c2(y2)=0 for all y2.
• 2=(a-b(y1+y2))y2=ay2-by1y2-by22 and FOC:
a-by1-2by2=0 or y2=(a-by1)/2b (reaction
function).
2
• Two points worth mentioning. When
y1=a/b (1 has flooded the market), then
y2=0. On the other hand, when y1=0, it is
as if 2 is the monopolist, so y2=a/2b.
• Let us suppose c1(y1)=0 for all y1 and
work out the leader’s problem: maxy
p(y1+y2)y1-c1(y1). Now, we can plug in 2’s
reaction curve. So 1=(a-b(y1+(aby1)/2b))y1=(a-by1)y1/2. FOC: a/2-by1=0
So y1=a/2b.
1
• Plugging this into 2’s reaction curve, we
get y2=(a-b(a/2b))/2b=a/4b. Show this
graphically.
• Now turn to the Cournot model. Two firms
simultaneously decide output levels. We
look for the case where 1’s output is a best
response to 2’s and vice versa 2’s is a best
response to 1’s. Graphically, it is the
intersection of the two reaction curves.
• Since y2=(a-by1)/2b and symmetrically
y1=(a-by2)/2b, solving these two together,
we get y1=y2=a/3b.
• Suppose the two quantity setting firms get
together and attempt to set outputs so that
their joint profit is maximized, i.e., they
collude or form a cartel, what will the
output levels be?
• Their problem becomes maxy ,y (ab(y1+y2))(y1+y2). FOC: y1+y2=a/2b.
1
2
• Together they produce the monopoly
output.
• But there is the temptation to cheat. Look
at FOC again. Profit is (a-bY)Y
(Y=y1+y2), or PY so FOC P+(dP/dY)Y=0
or P+(dP/dY)(y1+y2)=0. From 1’s
perspective, increasing output (if 2’s
output is fixed), its profit increases by
P+(dP/dY)y1=-(dP/dY)y2>0. Hence it is a
big issue for the cartel to deter members
from cheating.
• Often some repeated interactions would
help. Consider the punishment strategies.
Each firm produces half of the monopoly
output and gets profit m. If there is any
cheating in the past, switch to Cournot
competition forever and each gets c. So
if a firm deviates, it can at best get d for
one shot and then forever it gets c. Note
that c<m<d. So a firm is comparing m
+m /r> d+c/r. As long as r<(m-c)/(dm) or future important enough, it will not
deviate.
• Lastly, we talk a bit about Bertrand
competition.
• Two firms simultaneously announce
prices pi and pj. There is a market demand
x(p). If pi<pj, then xi(pi,pj)=x(pi). If pi = pj,
then xi(pi,pj)=x(pi)/2. If pi>pj, then
xi(pi,pj)=0. Suppose both firms have
marginal cost c.
• There is a unique pure Nash equilibrium.
• Suppose pi≤pj and pi<c, then i can deviate
to pic.
• Suppose pi=c and pj>c, i can increase the
price a bit to earn positive profit.
• Suppose c<pi≤pj, j earns at most (pic)x(pi)/2, undercutting i a bit, it could
earn close to (pi-c)x(pi).
• We are left with the case where pi=pj=c.