Answer Key: Problem Set 3

February 22, 2016

Problem 1

Question a

P = a− 2Q,MR = a− 4Q.By setting MR =MC, we can easily solve

Qm =a− c4

, Pm =a+ c

2, πm =

(a− c)28

, Lm =a− ca+ c

.

Question b

Solving this symmetric game by

maxqi(a− 2qi − 2q−i − c)qi.

The f.o.c is a− 4qi − 2q−i − c = 0. Imposing qi = q−i, we can solve

qci =a− c6

, P c =a+ 2c

3, πc =

(a− c)29

, Lc =a− ca+ 2c

.

Comparing (a) and (b), we can see

Qc = 2qci > qm, P c < Pm, πc < πm, Lc < Lm.

Question c

For the Bertrand game,

QB =a− c2

, PB = c, πB = 0, LB = 0.

Question d

Firms prefer Cournot competition since they can get positive profits by choosing quantities simultaneously. Con-sumers prefer Bertrand competition since they can get more surplus. Bertrand is more effi cient since it yields ahigher quantity– more specifically, greater total surplus.

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Question e

In the n-firm Cournot game,

maxqi(a− 2qi − 2Q−i − c)qi

The f.o.c. is

a− 4qi − 2Q−i − c = 0,

where Q−i = (n− 1)qi by symmetry. So

qci =a− c2(n+ 1)

, P c =a+ nc

n+ 1, πci =

1

2

(a− cn+ 1

)2, πc =

n

2

(a− cn+ 1

)2.

Clearly, both firm profits and industry profits are decreasing in n: dπci

dn = − (a−c)2(n+1)3 < 0, dπ

c

dn = (a−c)2(1−n)2(n+1)3 < 0.

As n goes to infinity, we have:

limn→∞

P c = c and limn→∞

πc = 0.

Problem 2

Question a

Here it is useful to assume that a− 2c1 + c2 > 0 (you will see why below). Firm i solves:

maxqi

qi (a− bqi − bq−i − ci) ,

so that the FOC is:

a− 2bqi − bq−i − ci = 0,

qi =a− ci2b

− q−i2.

Since we have two firms, the equation above must hold for i = 1, 2, so that we have the following system:

q1 =a− c12b

− q22,

q2 =a− c22b

− q12,

and plugging one equation into the other we obtain:

q1 =a− c12b

− 12

(a− c22b

− q12

),

=a− 2c1 + c2

4b+q14,

=a− 2c1 + c2

3b,

2

and similarly, q2 = a−2c2+c13b . Since we assumed that c1 > c2, we have that q1 < q2. Total quantity is

Q = q1 + q2,

=a− 2c1 + c2

3b+a− 2c2 + c1

3b,

=2a− c1 − c2

3b.

The market price is

p = a− b(2a− c1 − c2

3b

)=

a+ c1 + c23

.

Question b

Firm 1 profits are

π1 = q1 (p− c1)

=a− 2c1 + c2

3b

(a+ c1 + c2

3− c1

)=

(a− 2c1 + c2)2

9b,

π2 =(a− 2c2 + c1)2

9b,

so that π2 > π1 > 0 as long as a− 2c1 + c2 > 0.

Problem 3

Question a

The firm solves

maxq{(11− q)q − (2 + q)}

The f.o.c. is

11− 2q − 1 = 0which gives

qm = 5, pm = 6

πm = (10− 5)5− 2 = 23

CSm = 5 · 5 · 12= 12.5

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Question b

Firm i solves

maxqi{(11− qi −Q−i)qi − (2 + qi)}

The f.o.c.(w.r.t. qi) is

11− 2qi −Q−i − 1 = 0Symmetry implies Q−i = (n − 1)qi,which reduces the f.o.c. to 10 − (n + 1)qi = 0. (Note that if you make this

substitution before deriving the FOC, you will not get the right solution; remember that firm i does not get tochoose the quantities of its competitors!) Therefore:

qci (n) =10

n+ 1

Qc(n) =10n

n+ 1

P c(n) =11 + n

n+ 1

πc(n) = nπi =100n

(n+ 1)2− 2n

CSc(n) =1

2

(10n

n+ 1

)2.

To see how these quantities change with the number of firms, we need to take the derivatives with respect to n:

dqcidn

= − 10

(n+ 1)2< 0 , decreasing in n

dQc(n)

dn=

10

(n+ 1)2> 0 , increasing in n

dπc(n)

dn=

100

(n+ 1)3(1− n)− 2 < 0,decreasingin n

d(CSc)

dn=

100n

(n+ 1)3> 0,increasing in n

Question c

Suppose now the number of firms is determined endogenously. If there is free-entry:

πi = 0100

(n+ 1)2− 2 = 0

This gives n ≈ 6.07. Therefore π is 0; CS is 36.86. The sum CS + π = 36.86 (plug n = 6.07 into the formula inpart (b) ).

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Question d

To find the effi cient number of firms, we need to solve

maxn(CS + π)

i.e.

maxn

{1

2

(10n

n+ 1

)2+

100n

(n+ 1)2− 2n

}The f.o.c. is

100n

(n+ 1)3+100(1− n)(n+ 1)3

− 2 = 0,

from which we can solve n ≈ 2.68. Therefore CS + π = 40.94.

Question e

The ineffi ciency of free entry is due to business-stealing effect: part of an additional entrant’s profit comes at theexpense of existing firms. When firms make entry decisions, they do not take into consideration the decrease intheir opponent’s quantities when they enter. In other words, the private benefit for an additional firm exceeds thesocial benefit.

Question f

Yes. When the fixed cost associated with entry is eliminated, the effi cient number of firms will be ∞. Since entryalways ensures a positive profit, the number of firms in the free-entry equilibrium also tends to ∞. Therefore thediscrepancy disappears when there is no fixed cost.

Problem 4

Question a

In order to derive the residual demand function, note that given that prices are the same for both products, wehave:

q1 = a− b1p+ b2pwhere according to assumptions of the problem, for every dollar increase in the price of product A, for any givenprice of product B, product A loses twenty units of sales to products outside the candidate market and ten unitsof sales to product B. This means b1 = 20 + 10 = 30. But since the same is true for product B, for every dollarincrease in the price of product B, for any given price of product A, product B loses ten units of sales to productA and therefore b2 = 10. Hence we have:

q1 = a− 30p+ 10pReplacing for q1 = 1200 and p = 100, we find that a = 3200.

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Question b

Since we are given that mc = 60, we have that the maximization problem for firm i, when firms choose prices, is:

maxpi(3200− 30pi + 10p−i) (pi − 60),

from which we get the FOC:3200− 60pi + 10p−i + 1800 = 0,

and imposing symmetry pi = p−i = p, we get:

p = 100,

q = 1200.

Question c

Suppose now that a monopolist would sell both products, choosing prices. The maximization program is thus:

maxpA,pB

(3200− 30pA + 10pB) (pA − 60) + (3200− 30pB + 10pA) (pB − 60),

and the relevant first order conditions are:

3200− 60pA + 10pB + 1800 + 10pB − 600 = 0

3200− 60pB + 10pA + 1800 + 10pA − 600 = 0.

Imposing symmetry pA = pB = p, we get4400− 40p = 0,

so that p = 110, which matches the claim.

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