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Industrial Organization

Problem Set #2 Solution

Universidade Nova de LisboaFaculty of Economics

Fall 2008

Instructor: Vasco SantosGrader: David Henriques

1 Dominant Firm

(6 points)

P = 20 2QD () QD = 10 12

P

MgCA = 9

T Ci = qi (qi + 11) = q2i + 11qi; i = B;C:

MgCi = 2qi + 11; i = B;C:

Price takers: P = M gCi()

P = 2qi + 11()

qi =1

2P

11

2; i = B;C:

Qfringe (P) = 2

1

2P 11

2

= P 11 () P = Qfringe + 11 (1 point)

Firm A's residual demand,

QA =

10 12P P 11

10 12P [P 11] P > 11 () QA =

10 12P P 1121 32P P > 11

()

() P =

20 2QA QA 4:514 23QA QA < 4:5

:

Firm A's marginal revenue,

M gRA =

20 4QA QA 4:514 43QA QA < 4:5

:

Solving A's problem using the usual optimality condition:

M gRA = M gCA () 14 43

QA = 9 () QA = 3:75 < 4:5 (3 points)

and the equilibrium price is P = 14 23 3:75 = 11:5: (1 point)

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Given the price dened by the dominant rm, the competitive fringe in equilibrium produces

Qfringe (11:5) = 11:5 11 = 0:5: (1 point)

The total demand in equilibrium is given by,

QD (11:5) = Qfringe + Q

A = 0:5 + 3:75 = 4:25:

2 Monopolistic competition

a) (2 points)For any number of rms, n; NOVA's problem is,

maxq

= 90 +20

n 4q q

q2

414:05

By the f.o.c.:

90 +20

n 8q 2q = 0 () q = 9 + 2

nand p = 90 +

20

n 4

9 +

2

n

= 54 +

12

n: (1)

Checking the s.o.c.:

@2

@q2= 10 < 0. Thus, by concavity of ; the solution from (1) is a maximum.

Computing NOVA's prot we get,

(n) =

54 +

12

n

9 +

2

n

9 +2

n

2 414:05

=

9 +

2

n

45 +

10

n

414:05

= 405 +180

n+

20

n2 414:05 (2)

Therefore, in the short run, for n = 4,

q

sr = 9 +

2

4 = 9:5 (0.75 point)

psr = 54 +12

4= 57 (0.75 points)

sr = 405 +180

4+

20

42 414:05 = 37: 2 > 0: (0.5 points)

Since the prot is positive, it is expected that the number of rms operating in the marketshould increase till the zero prot condition is met.

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b) (4 points)Equating (2) to zero,

(n) = 0 () 405 +180

n +

20

n2 414:05 = 0; Solution is : fn = 0:1105g ; fn = 20:0gSince n must be non-negative, the only possible solution is nLR = 20 rms. (1 point)

By q and p derived in (1) and nLR = 20 it is obtained,

qLR = 9 +2

20= 9:1 and pLR = 54 +

12

20= 54:6 (1 point)

If F increases, the impact in the number of rms in the long run is quantied by,

@n

@F(20; 414:05) = @=@F

@=@n= 1

180n2

40n3

(20;414:05)

= n3

180n + 40

(20;414:05)

= 20091

= 2:1978 < 0 (0.5 points)

where the rst equality is assured by the Implicit Function Theorem and

(n; F) = 405 +180

n+

20

n2 F = 0 in the long run.

Alternatively, for those who are not familiar with the theorem, computing explicitly,

(n; F) = 405 +180

n

+20

n2

F = 0

()(405

F) n2 + 180n + 20 = 0

() n = 180p

1802 80 (405 F)2(405 F) .

Note: The solution with a plus before the root is ruled out since n is negative in that case.

Computing the derivative of n in order to F,

@n

@F=

1

4F2 3240F + 656 1008p

5F 360

+2

(2F 810)q

15F

;

applying the derivative when F = 414:05,

@n

@F(414:05) =

8p

5 (414:05) 3604 414:052 3240 414:05 + 656 100+

2

(2 414:05 810)q

15 414:05

= 2: 1978:

By equations in (1) it's easy to show that

@q

@F=

@q

@n

@n

@F= 2

n2:

@n

@F,

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where the rst equality holds according to the chain rule and @n@F

is known from the previous

computation. Hence, at the point n = 20, @q

@F= 2202 (2: 1978) = 1: 0989 102 > 0: (0.5

points)

Still using equations from (1),

@p

@F=

@p

@n

@n

@F= 12

n2@n

@F:

Finding the derivative at the point where n = 20, we get @p

@F= 12202 (2: 1978) = 6:

5934 102 > 0: (0.5 points)

Intuitive conclusions. (0.5 points)In the long run all rms are operating with zero prot. Thus, if the xed cost increases,

the prot becomes negative. Firms start leaving the market till the zero prot condition ismet again, so the number of rms in the LR equilibrium is decreasing with F. The rms thatstill operate observe an expansion of the faced demand function which allows for an increase

in revenue through higher prices and quantity sold. After all adjustments, the revenue increaseshould be exactly equal to the xed-cost increase.

c) (3 points)The short run consumer surplus (CSsr),

CSsr = n:

90 + 20

npsr

qsr

2= 4:

90 + 204 57

9:5

2= 722; (1.5 points)

i.e., consumers gain a surplus of 7224 = 180: 5 from each one of the products (and there are4).

The long run consumer surplus (CSLR),

CSLR = n:

90 + 20

npLR

qLR

2= 20:

90 + 2020 54:6

9:1

2= 3312:4; (1.5 points)

i.e., consumers gain a surplus of 3312:420 = 165: 62 from each one of the products (and thereare 20).

Notice that although consumers extract a lower surplus from each product (variety) in thelong-run than in the short run equilibrium, the total generated surplus by the 20 rms exceedsby far the one generated by 4 rms.

d) (2 points)Dening the Average Total Cost function (AT C),

AT C =T C

q= q+

414:05

q:

Deriving the minimum of the AT C function,

minq

AT C = q+414:05

q;

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by the f.o.c.

1 414:05q2

= 0 () qopt =p

414:05 = 20:348, the negative solution is ruled out. (1.5 points)

Checking the s.o.c.

@2AT C

@q2=

414:05 2q3

=414:05 2

20:3483= 9: 8292102 > 0, hence AT C reaches a minimum at qopt = 20:348.

However, qsr = 9:5 < 20:348, and q

LR = 9:1 < 20:348. Concluding, rms operating in thismarket are producing non-optimaly either in the SR as in the LR. (0.5 points)

e) (3 points)Suppose there are n = n0 rms in the market, then the Lerner index expression is given by

L =P c

P=

P(n0) 2q(n0)P (n0)

=

54 + 12n0

29 + 2n054 + 12

n0

=36 + 8n054 + 12

n0

=36n0 + 8

54n0 + 12=

36n0 + 8

(36n0 + 8)32

=2

3: (2 points)

P cP

=2

3() P = 3c,

i.e., the price is always 3 times the marginal cost independently of the number of rms in themarket and obviously,

limn!1

36 + 8n0

54 + 12n0

=2

3:

Intuition for the result. Each time a rm enters the market it forces the incumbents toreduce the price and quantity produced and consequently also their marginal costs (given by 2q).

Taking into account product dierentiation and the fact that price and marginal cost areproportional for any n, the Lerner index remains constant which is equivalent to stating thatrms' market power (mark-up) is independent of the number of competitors. (1 point)

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