Boğaziçi University
Department of Economics
Fall 2016
EC 301 ECONOMICS of INDUSTRIAL ORGANIZATION
Problem Set 5 - Answer Key
1. Consider a 1 km linear city with N many consumers uniformly distributed along the city. There
are only two supermarkets, A and B, serving this city. A is located right in the middle of the
city. B is located outside of the city. The supermarkets sell the same fixed basket of goods. A
has zero marginal cost and B has a marginal cost of 1TL per basket. They pick their prices
simultaneously, pA and pB . Each consumer needs only one basket of goods. Each consumer, after
observing the price of each supermarket, chooses which supermarket to buy from. Consumers
who choose to buy from A has to travel to the supermarket then travel back home, incurring a
transportation cost of 3TL per unit distance on the way to the supermarket with no basket, and
4TL per unit distance on the way back home with the basket of goods. Consumers who choose
to buy from B, do not incur any transportation cost, because supermarket B delivers the basket
for free. Find the equilibrium prices.
Answer:
Between buying from A and B, there are two indifferent consumers, one on the left half and the
other on the right half. Let the one on the left half be located at x. Then, pA +(3+4)(1/2−x) =
pB which implies
x∗ =
1 pA − pB
+
2
7
A’s demand is DA = 2 × ( 21 − x∗ ) = 1 − 2x∗ and its profit function is
1 pA − pB
2
πA = pA DA = pA (1 − 2[ +
]) = pA (pB − pA )
2
7
7
Then its best response is pBR
A = pB /2.
B’s demand is DB = 2x∗ and its profit function is
1 pA − pB
2
) = (pB − 1)(1 + (pA − pB ))
πB = (pB − 1)2x∗ = (pB − 1)2( +
2
7
7
Then its best response is pBR
B =
PA
2
+ 94 .
Solving these two best responses give us p∗A = 3/2 and p∗B = 3.
2. Suppose that there are 3000 buyers distributed over a 1 km beach such that 1000 buyers are
distributed evenly over the west half of the beach, and 2000 buyers are distributed evenly over
the east half of the beach. Each buyer incurs a transportation cost t per km. Each buyer buys
1
only one ice-cream. There are two ice cream sellers, A and B. A is located at the west end of
the beach, and B is located at the east end. Seller A has a marginal cost given by M CA = 1
TL and seller B has a marginal cost given by M CB = 2 TL. There are no fixed costs. Sellers
are simultaneously setting their prices, PA and PB .
(a) Find the location, x
b, of the buyer who is indifferent between buying from seller A and
buying from seller B, as a function of PA , PB , and t.
PB − PA 1
Answer: x
b: PA + tb
x = PB + (1 − x
b)t ⇒ x
b=
+
2t
2
(b) Suppose that the prices are such that x
b is in the west half of the beach. Write the profit of
each seller as a function of PA , PB , and t.
Answer:
PB − PA 1
(PA − 1)
πA = 2 × 1000b
x(PA − 1) = 2000
+
2t
2
1
πB = 2000 +
−x
b 2 × 1000 (PB − 2) = (3000 − 2000b
x)(PB − 2)
2
PB − PA 1
= 3000 − 2000
+
(PB − 2)
2t 2
PB − PA
= 2000 1 −
(PB − 2)
2t
3. Consider a 1 km linear city that stretches along a hill. Consumers are uniformly distributed along
the city. There are two supermarkets, A and B, in this city. A is located 0.3 km away from the
bottom of the hill, and B is located 0.1 km away from the top of the hill. The supermarkets sell
the same fixed basket of goods with zero marginal cost. They pick their prices simultaneously,
pA and pB . Each consumer needs only one basket of goods. Each consumer, after observing
the price of each supermarket, chooses which supermarket to buy from. Going downhill with a
basket of goods costs 2TL per unit distance traveled and with no basket it is costless. Going
uphill with a basket of goods costs 5TL per unit distance traveled and with no basket it costs
3TL per unit distance traveled. Note that each consumer has to travel to the supermarket then
travel back home. Find the equilibrium prices.
Answer:
Let the consumer who is indifferent between buying from A and B be located at x. Then,
pA + 5(x − 0.3) = pB + (0.9 − x)3 + (0.9 − x)2, which implies
5(x − 0.3 − 0.9 + x) = pB − pA , which gives
x=
pB − pA + 6
pA − pB + 4
and (1 − x) =
10
10
Then, πA = pA x = pA
pB − pA + 6
10
FOC(pA ): pB − 2pA + 6 = 0 =⇒ pA =
pB + 6
2
2
And πB = pB (1 − x) = pB
pA − pB + 4
10
FOC(pB ): pA − 2pB + 4 = 0 =⇒ pB =
pA + 4
2
pA + 4
+6
16
14
2
Then, pA =
⇒ 4pA = pA + 16 =⇒ p∗A = , p∗B = .
2
3
3
4. Consider the location model we discussed, where the two firms first pick their location simultaneously and then compete in prices. Show that minimal differentiation (locating at the same
point, that is, a + b = 1) is not an equilibrium, by showing that there is a deviation.
Answer: If a + b = 1 then PA = PB = c. This is because
a−b
b−a
PA∗ = c + t(1 − a − b)(1 +
) and PB∗ = c + t(1 − a − b)(1 +
).
3
3
Then, with a + b = 1 both firms get zero profit. However, given the other firm sticks to this
strategy a firm would deviate to some other location so that a + b < 1, hence get a price
strictly larger than marginal cost, c. This would bring positive profit, thus would be a profitable
deviation. Thus a + b = 1 would not be equilibrium.
5. Consider the circular city model we discussed in class. Recall that the market is circular and
there is free entry. Suppose that the transportation cost to travel a distance x is T C(x) = 0.5x2 .
All firms have a marginal cost equal to c = 2TL and a fixed entry cost f = 100TL.
(a) Find the equilibrium price
1
Answer: Pi + x2 = P +
2
for a given the number of firms, n.
1 1
1
1
( − x)2 ⇒ x = (P − Pi )n +
⇒ 2x = (P − Pi )2n +
2 n
2n
n
1
max(Pi − c)2x = (Pi − c) (P − Pi )2n +
Pi
n
1
FOC: (P − Pi )2n + = 2n(Pi − 2)
n
1
⇒ 2P n + = 4Pi n − 4n
n
1
⇒ 2n(P + 2) + = 4Pi n
n
P +2
1
⇒ Pi =
+ 2
2
4n
Using symmetry (Pi = P ):
Pi + 2
1
1
Pi =
+ 2 =⇒ Pi∗ = 2 + 2
2
4n
2n
t
1
t
Note that when T C = tx we have P ∗ = c + (2 + ). When T C = tx2 you get c + 2 (2 +
n
2n
n
1
)
2n2
(b) Find the equilibrium number of firms.
Answer: For the equilibrium number of firm; need πi = 0, that is (p − c)2x = f
1
So, (p − c) = f
n
3
1
1
1
− 2) =
⇒ n∗ = 2
2n2
n
16
Warning: In the problem f = 100 is too high. Change it to f = 1/16
(2 +
(c) Find the socially optimal number of firms. Compare it to the one you found in (b) and
comment.
Answer:
Z
1/2n
min nf + t(2n
n
x2 dx)
0
n
n
n
1
1 1
1
3 1/2n
)⇒
=
+ 2n( x3 +n
+
3
16 2
16
3 8n
16 24n2
0
1
1 1
FOC:
= 2 3 ⇒ n3 = 4/3
16
24 n
nsoc.optimal = 1 < n∗ = 2
6. Nispetiye avenue is best described as the interval [0, 1]. Two kebab places serving identical
iskender are located at the edges of the avenue. Antep Iskender (A) is located at the southwest
end of the avenue, and Bursa Iskender (B) is located at the northeast end of the avenue. Both
A and B have a marginal cost equal to c > 0. Iskender consumers are uniformly distributed on
the avenue (on the interval [0, 1]), where at each point on the interval lives one consumer. Each
consumer buys one iskender from the restaurant in which the price plus the transportation cost
is the lowest. On Nispetiye Avenue, however, there is a constant wind (Lodos) blowing from the
northeast end toward the southwest end, hence the transportation cost for consumer who travels
to the northeast is w TL per unit of distance, and only 1TL for a consumer who travels to the
southwest, where w > 1. Transportation costs are linear. Let pi denote the price of an iskender
at restaurant i, where i = A, B. Let x
b (the distance to A) be the location of the consumer who
is indifferent to whether he/she eats at Antep Iskender or Bursa Iskender.
(a) Find x
b as a function of pA , pB and w.
Answer: pA + x = pB + (1 − x)w ⇒ x = pB − pA + w − xw ⇒ x
b=
pB − pA + w
1+w
(b) How does x
b change with respect to w? Provide some intuition.
Answer: As w increases (stronger lodos), x
b increases, that is the indifferent consumer gets
closer to B. This is because with a higher w, it’s more costly to travel to B, hence relatively
more buyers will prefer buying from A. Thus x
b moves toward B, and x
b increases.
(c) Find the equilibrium prices, p∗A and p∗B , as a function of w and c.
Answer:
max(pA − c) − x
b = (PA − c)(
pA
⇒
pB − pA + w
)
1+w
pB − pA + w
1
pB + w + c
=
(pA − c) ⇒ pA =
1+w
1+w
2
max(pB − c) − x
b = (PB − c)(1 −
pB
4
pB − pA + w
1 − pB + pA
) = (pB − c)(
)
1+w
1+w
1 − pB + pA
1
pA + 1 + c
=
(pB − c) ⇒ pB =
1+w
1+w
2
Solving pA and pB together, we get
3p∗
(p∗ + 1 + c)
1 3c
2p∗A = A
+w+c⇒ A = +
+w
2
2
2
2
1 + 2w + 3c
2 + w + 3c
=⇒ p∗A =
and p∗B =
3
3
(d) Suppose w = 2 and c = 1. Find x
b. Is it closer to A or B? Why?
⇒
Answer: For w = 2 and c = 1, p∗A = 8/3 p∗B = 7/3
1 + 2w
5
1
(1 − w)/3 + w
=
= > ⇒ closer to pB
Then x
b=
1+w
3(1 + w)
9
2
5