NAME_______________________________________________________________
FE461 Practice Problem
Fourth Problem Set
Due March 27th
1. (25 points) Let the inverse demand function in an industry be P(Q) = 40 – 3Q. There are
two firms in this industry, firm 1 faces costs given by: C(q1) = 5 + 10q1,and firm 2 faces
costs given by C(q2) = 5 + 13q2.
a) (15 points) Assuming Cournot competition, find the profits for each firm. (Note: the
firms face asymmetric costs.)
TR1 = (40 – 3(q1 + q2))*q1
MR1 = 40 – 6q1 – 3q2
MC1 = 10
 40 – 6q1 – 3q2 = 10
q1 = 30 – 3q2
6
TR2 = (40 – 3(q1 + q2))*q2
MR2 = 40 – 3q1 – 6q2
MC2 = 8
 40 – 3q1 – 6q2 = 13
q2 = 27 – 3q1
6
Simultaneously solving the two equations:
1
1 27 1
27 1
q 2 and q2 =
 q1 q1 = 5  (  q1 )
2
6 2
2 6 2
27 1
q1  5 
 q1
12 4
3
q1  5  2.25
4
4
q1  2.75 * ( )  3.67
3
27 1
27 1
q2 
 q1 
 (3.67)  4.5  1.84  2.66
6 2
6 2
q1 = 5 
Q = q1 + q2 = 3.67 + 2.66 = 6.33
P = 40 – 3Q = 40 – 3*6.33 = 21
So finally 1 = P*q1 – C(q1) = 21*3.67 – (5 + 10*(3.67)) = 77.07 – 5 – 36.7 = 35.37
and 2 = P*q2 – C(q2) = 21*2.66 – (5 + 13*(2.66)) = 55.86 – 5 – 34.58 = 16.28
b) (10 points) Assuming Bertand competition, find the profits for each firm.
Firm 2 has a higher marginal cost, so firm 1 will undercut the other firm P1 =12.99 and
serve the entire market.
q2  0
P1  12.99  40  3(q1 )
27
 q1  9; q2  0
3
 1  ( P  MC )Q  FIXEDCOST  (12.99  10)9  5  22
 2  ( P  MC )Q  FIXEDCOST  (13  10)0  5  5
2. (15 points) Consider a Stackleberg game of quantity competition between two firms.
Firm 1 is the leader and firm 2 is the follower. Market demand is described by the inverse
demand function P = 1,000 – 4Q. Each firm has a constant unit cost of production equal
to 40. Solve for the Nash Equilibrium quantities for each firm.
Firm 2 chooses its quantity to maximize
TR2  Q2 1000  4Q1  4Q2 
TR2
1
1
 1000  4Q1  8Q2  40  MC  Q2  960  4Q1   120  Q1
Q2
8
2
Now, Firm 1 chooses its quantity to maximize

1 

TR1  Q1 1000  4Q1  4Q2   Q1 1000  4Q1  4120  Q1    (1000  4Q1  480  2Q1 )Q1
2 


TR1
480
 520  4Q1  40  MC1  Q1 
 120  Q2  60
Q1
4
3. (25 points) Suppose that firm 1 can choose to produce either good A or good B or both
goods or nothing. Firm 2, on the other hand, can produce only good C or nothing.
Firms’ profits correspond d to each possible scenario of goods for sale are described in
the following table:
Product Selection
Firm 1’s profit
Firm 2’s profit
A
20
0
A,B
18
0
A,B,C
2
-2
B,C
-3
-3
C
0
10
A,C
9
8
B
11
0
a) (10 points) Set up the normal form game for when the two firms move simultaneously.
What is the Nash Equilibrium (equilibria)?
Firm 2 
C
Nothing
Firm 1 
A
(9, 8)
(20, 0)
B
(-3, -3)
(11, 0)
A, B
(2, -2)
(18, 0)
Nothing
(0, 10)
(0, 0)
There is a unique Nash equilibrium, where Firm 1 chooses A and Firm 2 chooses C.
b) (5 points) Now suppose that firm 1 can commit to its product choice before firm 2. Draw
the extensive form of this game and identify the Subgame Perfect Nash equilibrium
(SPE). Compare your answer to (a) and explain.
If firm 1 commits to producing A,B, firm 2 will produce nothing.
c) (10 points) The game is like the one in (b) only now suppose firm 1 can reverse its
decision after observing firm 2’s choice and this possibility is common knowledge. Does
this affect the game? If so, explain the new outcome? If not, explain why not.
Note that A is a dominant strategy for Firm A. Therefore, even if Firm 1 can commit before
Firm 2, the answer does not change.
4. (25 points) In a Centipede game suppose there are 2 players, player 1 moves first and
player 2 moves second and each player makes only one decision. The game begins with
$1 on the table. Player 1 can either “take” the $1 or “wait”; if player 1 chooses “take” the
game ends and player 1 earns $1 and player 2 earns $0. If player 1 chooses “wait” the
money quadruples (the $1 is now worth $4) and now player 2 can decide to “keep” or
“split”. If player 2 chooses “keep” player 2 earns $4 and player 1 earns $0; if player 2
chooses “split” both players earn $2.
a) (5 points) Draw the extensive form of this game.
b) (5 points) What is the subgame perfect equilibrium?
The strategy of splitting the money is never an equilibrium since once the game reaches
the point P21, the optimal strategy for the Player 2 is to take the entire $4. Because
Player 1 knows this will be the outcome at P2, Player 1 will always choose “Grab” and
the outcome will be T3 with Player 1 getting $1 and Player 2 getting nothing.
c) (15 points) Now suppose that Centipede has 3 three moves. First player 1 moves and
chooses “keep” or “wait”; then player 2 chooses “keep” or “wait”. If player 2 now
chooses “wait” the money quadruples again (so the $4 is now worth $16). Then player 1
moves last and can decided to “keep” (player 1 earns $16 and player 2 earns $0) or
“split” (both players earn $8). Draw the extensive form game and find the subgame
perfect equilibrium.
It is clear that in the third stage Player 1 will choose to keep all the money. Thus we can
eliminate this choice from the tree and consider the new game with the final node removed
(pruned). This will give
It is now obvious that Player 2 will choose to take it all at node P21W. Thus we can
eliminate this node from the tree and replace it with the payoffs to both players when
Player 2 chooses to take it all. This will give
It is now clear that Player 1 will grab the money at the initial node and the final
payoff will be (1,0).