Seminar 5 - Solutions

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Seminar 5 - Solutions
Industrial Organisation (ES30044)
Seminar Five: Oligopoly (ii)
Industrial Organisation (ES30044)
Seminar Five:
Oligopoly (ii)
1. Consider a market where there are two differentiated goods. The demand for good 1 is
given by q1 = a − bp1 + δ p2 and the demand for good 2 is given by q2 = a − bp2 + δ p1 q2
where a > 0 b > δ > 0 . The production cost of each good is zero. Suppose that both
goods are produced by the same firm (a monopolist). Compute the prices set by the
monopolist. Suppose now that a different firm produces each good and the firms choose
prices simultaneously. Compute the Bertrand-Nash equilibrium prices and confirm that
they are lower than the monopoly prices. Now assume that each good is produced by a
different firm but the firms set prices sequentially; in particular, firm 2 can observe the
price set by firm 1 before setting its own price. Compute the subgame-perfect equilibrium
price of firm 1 in this two-stage game.
Monopoly
(
)
(
π = p1q1 + p2 q2 = p1 α − β p1 + δ p2 + p2 a − bp2 + δ p1
Thus:
∂π
= a − 2bp1 + 2δ p2 = 0
∂p1
⇒
p1 =
(
1
a + 2δ p2
2b
)
∂π
= a − 2bp2 + 2δ p1 = 0
∂p2
⇒
p2 =
(
1
a + 2δ p1
2b
)
Thus:
1
)
Industrial Organisation (ES30044)
p1 =
Seminar Five: Oligopoly (ii)
⎡1
1 ⎧⎪
a + 2δ p1
⎨ a + 2δ ⎢
2b ⎪⎩
⎣ 2b
(
⎫
)⎤⎥ ⎪⎬⎪
⎦⎭
⇒
2bp1 = a +
δ
a + 2δ p1
b
(
⇒
)
(
2b2 p1 = ab + δ a + 2δ p1
)
⇒
2b2 p1 = ab + aδ + 2δ 2 p1
⇒
p1 =
a b+δ
)
2 b −δ
2
(
(
2
(
a b+δ
=
)
=
a
2 b−δ
) 2 ( b + δ )( b − δ ) (
)
Thus:
p2 =
⎡
⎤ ⎫⎪
1 ⎧⎪
a
⎥⎬
⎨ a + 2δ ⎢
2b ⎪
2
b
−
δ
⎢
⎥⎦ ⎪⎭
⎣
⎩
(
)
⇒
⎡ a
2bp2 = a + δ ⎢
⎢⎣ b − δ
⇒
(
(
)
)
(
⎤
⎥
⎥⎦
)
2b b − δ p2 = a b − δ + aδ
⇒
(
)
2 b2 − bδ p2 = ab − aδ + aδ
⇒
p2 =
(
ab
2 b − bδ
2
=
a
2 b−δ
) (
)
Thus:
p1 = p2 = p m =
a
2 b−δ
(
)
Bertrand
(
Max π 1 = p1 a − bp1 + δ p2
p1
)
2
Industrial Organisation (ES30044)
Seminar Five: Oligopoly (ii)
∂π 1
= a − 2bp1 + δ p2 = 0
∂p1
⇒
p1 =
(
1
a + δ p2
2b
)
(
Max π 2 = p2 a − bp2 + δ p1
p2
)
∂π 2
= a − 2bp2 + δ p1 = 0
∂p2
⇒
p2 =
(
1
a + δ p1
2b
)
Thus:
p1 =
⎡1
1 ⎧⎪
a − δ p1
⎨a + δ ⎢
2b ⎩⎪
⎣ 2b
(
⎫
)⎤⎥ ⎪⎬⎪
⎦⎭
⇒
⎡1
⎤
2bp1 = a + δ ⎢
a − δ p1 ⎥
⎣ 2b
⎦
⇒
(
)
4b2 p1 = 2ab + aδ − δ 2 p1
⇒
( 4b
2
⇒
p1 =
)
(
+ δ 2 p1 = a 2b + δ
(
a 2b + δ
( 4b
2
+δ
2
)=
(
)
a 2b + δ
)
) ( 2b + δ )( 2b − δ )
=
a
2b − δ
And:
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Industrial Organisation (ES30044)
⎛ a ⎞⎤
1 ⎡
⎢a + δ ⎜
⎥
2b ⎣
⎝ 2b − δ ⎟⎠ ⎦
p2 =
⇒
Seminar Five: Oligopoly (ii)
(
)
(
)
2b 2b − δ p2 = a 2b − δ + aδ
⇒
( 4b
2
)
− 2bδ p2 = 2ab − aδ + aδ
⇒
p2 =
2ab
ab
a
= 2
=
4b − 2bδ 2b − bδ 2b − δ
2
Thus
p1 = p2 = p b =
a
2b − δ
Note that p m > p b if δ > 0 vis:
pm =
a
a
a
=
>
= pb
2b − 2δ 2b − δ
2 b−δ
(
)
Stackelberg
(
Max π 2 = p1 a − bp2 + δ p1
p2
)
∂π 2
= a − 2bp2 + δ p1 = 0
∂p2
⇒
p2 =
(
1
a + δ p1
2b
)
(
Max π 1 = p1 a − bp1 + δ p2
p1
)
st
p2 =
(
1
a + δ p1
2b
)
4
Industrial Organisation (ES30044)
Seminar Five: Oligopoly (ii)
∂π 1
δ 2 p1
= a − 2bp1 + δ p2 +
=0
∂p1
2b
⇒
a2b − 2b2 p1 + 2bδ p2 + δ 2 p1 = 0
⇒
(
)
p1 4b2 − δ 2 = 2ab + 2bδ p2
⇒
⎡1
⎤
p1 4b2 − δ 2 = 2ab + 2bδ ⎢
a + δ p1 ⎥
⎣ 2b
⎦
⇒
(
)
(
)
(
)
(
(
p1 4b2 − δ 2 = 2ab + δ a + dp1
⇒
)
)
p1 4b2 − δ 2 = 2ab + aδ + δ 2 p1
⇒
(
) (
p1 4b2 − 2δ 2 = a 2b + δ
⇒
p1 =
(
a 2b + δ
( 4b
2
)
− 2δ 2
)
(
a 2b + δ
=
) 2 ( 2b
2
)
−δ2
)
And:
p2 =
(
)
⎧
⎡ a 2b + δ
1 ⎪
⎨a + δ ⎢
2b ⎪
⎢ 2 2b2 − δ 2
⎣
⎩
(
⇒
(
)
⎡ a 2b + δ
2bp2 = a + δ ⎢
⎢ 2 2b2 − δ 2
⎣
⇒
(
(
)
(
)
)
⎤ ⎫⎪
⎥⎬
⎥⎪
⎦⎭
⎤
⎥
⎥
⎦
)
(
4b 2b2 − δ 2 p2 = 2 2b2 − δ 2 a + aδ 2b + δ
)
⇒
(8b − 4bδ ) p
3
2
2
= 4ab2 − 2aδ 2 + 2abδ + aδ 2
⇒
p2 =
4ab2 − aδ 2 + 2abδ
(8b − 4bδ )
3
2
5
Industrial Organisation (ES30044)
Seminar Five: Oligopoly (ii)
2. Suppose that there are only two firms (Firm 1 and Firm 2) selling coffee. Let α i denote
the advertising level of firm i, i = 1, 2. Assume that the profits of the two firms are
affected by advertising as follows:
π 1 (α1 , α 2 ) = 4α1 + 3α1α 2 − α1α1
π 2 (α1 , α 2 ) = 2α 2 + α1α 2 − α 2α 2
(a) Calculate and draw the best-response function of each firm – i.e. for any given level
of advertising of firm j, find the profit maximising advertising level of firm i;
max π 1 (α1 , α 2 ) = 4α1 + 3α1α 2 − α1α1
α1
⇒
∂π 1
α1 , α 2 ) = 4 + 3α 2 − 2α1∗ = 0
(
∂α1
⇒
3
α1∗ = 2 + α 2
2
And:
max π 2 (α1 , α 2 ) = 2α 2 + α1α 2 − α 2α 2
α2
⇒
∂π 1
α1 , α 2 ) = 2 + α1 − 2α 2∗ = 0
(
∂α 2
⇒
1
α 2∗ = 1 + α1
2
Thus:
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Industrial Organisation (ES30044)
Seminar Five: Oligopoly (ii)
Figure 1: Advertising Best-Response Functions
(b) Infer whether the strategies are strategic complements or strategic substitutes;
The two best-response functions are upward sloping and are hence strategic
complements.
(c) Find the Nash equilibrium advertising levels and calculate the firms’ Nash
equilibrium level of profits.
3
α1b = 2 + α 2b
2
⇒
3⎛
1 ⎞
α1b = 2 + ⎜ 1 + α1b ⎟
⎝
2
2 ⎠
⇒
3 3
α1b = 2 + + α1b
2 4
⇒
1 b 7
α1 =
4
2
⇒
28
α1b =
= 14
2
And:
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Industrial Organisation (ES30044)
Seminar Five: Oligopoly (ii)
1
α 2b = 1 + α1b
2
⇒
1
α 2b = 1 + (14 )
2
⇒
α 2b = 8
Thus:
π 1b (α1b , α 2b ) = 4α1b + 3α1bα 2b − α1bα 2b
⇒
π 1b (14, 8 ) = 4 ⋅14 + 3⋅14 ⋅ 8 − 14 2
⇒
π 1b (14, 8 ) = 14 2
And:
π 2b (α1b , α 2b ) = 2α 2b + α1bα 2b − α 2bα 2b
⇒
π 1b (14, 8 ) = 2 ⋅ 8 + 14 ⋅ 8 − 8 2
⇒
π 1b (14, 8 ) = 8 2
⇒
π 1b (14, 8 ) = 14 2
3. Pultney Bridge is best described by the interval [0,1]. Two fast-food restaurants serving
identical food are located at the edges of the road, so that Restaurant 1 is located at the
most left-hand side and Restaurant 2 is located at he most right hand side. Consumers are
uniformly distributed on the unit interval [0,1], where at each point on the interval lives
one consumer. Each consumer buys one meal from the restaurant in which price plus the
transportation cost is the lowest. Along Pultney Bridge, the wind blows from right to left
such that the transportation cost for a consumer who travels to the right (left) is $R ($1)
per unit of distance:
(a) Let pi denote the price of a meal at restaurant i, i = 1, 2. Assume that p1 and p2 are
given and satisfy:
0 < p1 − R < p2 < 1 + p1
Denote by xˆ the location of a consumer who is indifferent as to whether he eats at
Restaurant 1 or Restaurant 2 and calculate xˆ as a function of p1 , p2 and R;
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Industrial Organisation (ES30044)
Seminar Five: Oligopoly (ii)
The indifferent consumer, xˆ , is defined implicitly by:
xˆ ∗1 + p1 = (1 − xˆ ) ∗ R + p2
⇒
xˆ =
R + ( p2 − p1 )
1+ R
(b) Suppose that the given prices satisfy p1 = p2 . What is the minimal value of the
parameter R such that all consumers will only eat at Restaurant 1?
Substituting p1 = p2 into xˆ yields:
xˆ =
R
1+ R
Thus:
1 ⎞
⎛ R ⎞
⎛ 1 + R −1⎞
⎛
lim xˆ = lim ⎜
⎟ = lim
⎜
⎟ = lim
⎜1 −
⎟ =1
R→∞
R→∞ ⎝ 1 + R ⎠
R→∞ ⎝ 1 + R ⎠
R→∞ ⎝
1+ R⎠
Therefore, all consumers will eat in Restaurant 1 (i.e. x = 1 ) only if the transportation cost for
travelling to the east is infinite. Otherwise, some consumers will always eat at Restaurant 2.
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