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ECON 815: Economic Analysis for Business
Solutions for Problem Set 4
Professor D. Weisman
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1. P  Q   20  Q, where Q  q1  q2 .
Firm 1’s cost function: C  q1   2q1
Firm 2’s cost functions: C  q2   4q2
P
20
4
2
MC(Firm 2)
MC(Firm 1)
Q
Firm 1 puts on the market a level of output corresponding to the competitive outcome
with respect to firm 2’s cost function.
This implies that 4  20  q1  q1  Q L  16, where the superscript “L” denotes “limit.”
Also, P 16  20 16  4. Hence, P L  4, Q L  16.
2. P  20  Q and C  qi   2qi , i  1, 2 . (Cournot Duopoly)
a) The cost functions are identical for the duopolists. Compute reaction function for
firm 1.
(1) P   20  q2   q1
(2) MR1   20  q2   2q1  2  MC1
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(3) q1  R1  q2   9  q2 , and by symmetry
2
3
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(4) q2  R2  q1   9  q1
2
b) Solve for equilibrium values:
Solve (4) for q1
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(5) q2  9  q1  q1  18  2q2 and set equal to the expression in (3)
2
1
(6) 18  2q2  9  q2
2
(7) 9 
8
9
3
q2 
2
18

6

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 Equal Output Level Verification
1
*
*
q1  R1  q2   9   6   6 

2
q2* 
(10) Q*  q1*  q2*  6  6  12
(11) P  Q*   P 12   20  12  8
(12)    P  C  Q  8  212  72
1 3 6 ; 2  3 6 ;

7 2
1
2
c) Monopoly Outcome
(13) P  20  Q  MR  20  2Q .
(14) 20  2Q  2  Q  9 and P  20  9  11
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Alternatively, R1  q2   9  q2 and R1  0   9
2
4
P
20
Monopoly
P
CSm = (1/2)(9)(20-11)=40.5
20
Cournot
CSco = (1/2)(12)(20-8)=72
11
8
9
20
Q
12
20
Q
Consumers are better off under Cournot competition relative to monopoly by
CS  72  40.5  31.5
3. In a collusive outcome, the firms cooperate in choosing output levels, but the
collusive outcome is unsustainable in equilibrium.
a) The best the firms can do is to achieve the monopoly outcome. If each firm puts 4.5
units of output on the market, total output is 9 and the corresponding price is 11. Each
firm realizes profit of 40.5, so total industry profits are at the monopoly level.
b) If firm 1 believes that firm 2 will adhere to the collusive outcome and put 4.5 units of
output on the market, its best response (from problem 2) is
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q1  R1  4.5   9  (4.5)  6.75. Under these conditions Q = 4.5 + 6.75 = 11.25 and the
2
corresponding price is P = 8.75. It follows that firm 1’s profit is 1 = (8.75 – 2)(6.75)
= 45.5625 > 40.5 and firm 2’s profit is  2 = (8.75 – 2)(4.5) = 30.375 < 40.5. Hence,
firm 1 gains relative to the collusive monopoly outcome and firm 2 loses relative to the
collusive monopoly outcome.
c) If each firm cheats on the collusive agreement, believing the other firm will adhere to
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the agreement, q1  R1  4.5   9  (4.5)  6.75  q2  R2  4.5  . Hence, total output on
2
the market is Q = 6.75 + 6.75 = 13.5 and P = 20 – 13.5 = 6.5. Hence 1 =  2 = (6.5
– 2)(6.75) = 30.375 < 36 (Nash Equilibrium profit level).
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Cheat
Firm 2
Collude
Firm 1
Cheat
Collude
Firm 1
30.375, 30.375
45.5625, 30.375
30.375, 45.5625
40.5, 40.5
Note: The collusive outcome is unsustainable because each firm can increase its profits
by cheating on the collusive agreement. Hence, unless these firms sign enforceable
agreements, which are patently unlawful under the antitrust laws, the collusive outcome
is not sustainable.
4. Bertrand setting(i.e. firms choose prices) for homogeneous products.
In equilibrium, firm 2 sets a price, P2  10 (actually, slightly below 10). At this price
firm 1 will not supply any output because doing so will result in a loss. Hence, the Nash
Equilibrium Price is P  10 . The profits for each firm are as follows:
1  0
 2   P  4  Q  10  4  90   540
Note: Q 10  100 10  90
Bertrand Heterogeneous Goods.
5. Firm 1: (1) Q1  20  4 P1  2 P2
Firm 2 : (2) Q2  16  2 P2  1P1
A. Profit functions:
(3) 1  Q1  P1  c1 
ci  MCi , i  1 , 2
(4) 2  Q2  P2  c2 
B. Profit-maximizing conditions:
1 Q1

(5)
 P1  c1   Q1  0
P1
P1
(6)
 2 Q2

 P2  c2   Q2  0
P2
P2
Note:
Q1
Q
 4; 2  2
P1
P2
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(5’)
1
 4  P1  2   20  4 P1  2 P2  0
P1
(5”)
1
 28  8 P1  2 P2  0
P1
(7) P1 
28  2 P2 14  P2

8
4
(6’)
 2
 2  P2  2   16  2 P2  P1  0
P2
(6”)
 2
 20  4 P2  P1  0
P2
(8) P2 
20  P1
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C. Reaction functions
(9) P1  P2  
14  P2
 P2  4 P1  14
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(10) P2  P1  
20  P1
4
P2
Firm 1
20
Firm 2
6.2667
5
0
-14
5.0667
P1
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Solve (9) and (10) simultaneously
20  P1
4
(12) 16 P1  56  20  P1
(11) 4 P1  14 
(13) 15P1  76
(14) P1 
76
 5.0667
15
 76 
(15) P2   
 15 
76
15  376  6.2667
4
60
20 
D. Nash Equilibrium:
P1*  5.0667 1*  12.2666  5.0667  2   37.618
P2*  6.2667
 *2  8.5333  6.2667  2   36.409
Q1*  20  4  5.0667   2  6.2667   12.2666
Q2*  16  2  6.2667   1 5.0667   8.5333