Exercise 1 A monopolist has an inverse demand curve given ... - IDEA

c) Socially optimal price: p = MC ⇒ p = 20
d) Socially optimal output: 20 = 100 − 2y ⇒ y = 40
e) deadweight loss: 1(p 2 M −p C )×(y C −y M ) = 0.5(60−20)×(40−20) = 400.
The deadweight loss is 400 euros.
f) Perfect discrimination deadweight loss: 0. Remember that perfect
discrimination means that the monopolist charges a price for each consumer
and charges at the highest willingness to pay of the consumer. He produces
a level of output up to a point where the price is equal to the marginal cost
as in the competitive case, all the surplus going to the monopolist. At the
marginal point he sells to the consumer who has the lowest willingness to pay,
there is no surplus made on this consumer. In the competitive case, there is
a unique price for all the consumers. At this unique price, some consumers
are happy because they would be willing to pay more for the good. The
producer loses the surplus to the benefit of those consumers.
Exercise 3
Consider a market for a homogeneous good with two firms and demand
function:
Y = 30 − 3p (1)
where Y = y 1 + y 2 . Suppose the firms have the following cost functions:
C(y 1 ) = 3y 1
C(y 2 ) = 4y 2
a) Suppose the two firms engage in Cournot competition. Find the two
reaction functions and graph them.
b) Find the equilibrium price, the quantities produced by each firm as
well as each firm’s profits.
c) Suppose the two firms engage in Stackelberg competition, where firm 1
acts first and firm 2 acts when firm 1 quantity is known. Find the equilibrium
price, the quantities produced and each firm’s profits.
d) Suppose the two firms engage in Bertrand competition. Find the
equilibrium price of each firm, as well as quantities produced and profits.
e) Suppose the two firms form a cartel and maximize joint profits. Find
the equilibrium price and the quantities produced by each firm.
Answers
a)Y = 30 − 3p ⇒ y 1 + y 2 = 30 − 3p ⇒ p = 10 − 1(y 3 1 + y 2 )
profit function π 1 (y 1 , y2) e = p(y 1 +y2)y e 1 −C(y 1 ) = [10− 1(y 3 1 +y2)]y e 1 −3y 1
⇒ π 1 (y 1 , y2) e = 7y 1 − 1 3 y2 1 − 1y 3 1y2
e
Profit-maximizing output: ∆π 1(y 1 ,y2 e)
∆y 1
= 0 ⇒ 7 − 2y 3 1 − 1 3 ye 2 = 0
Reaction curve: y 1 = 10.5 − 0.5y2
e
2