Simon Fraser University
Econ 383 D300 Auction Theory Midterm Exam
Instructor: Songzi Du
Monday March 3, 2014, 12:30 – 2:20 PM
Spring 2014
1.
(5 points) There are four buyers { x, y, z, w } and four houses { a, b, c, d } .
The valuations are given in
Table 1 . Suppose that
p a
= 1, p b these prices market-clearing? Explain.
= 1, p c
= 2, and p d
= 2. Are
Buyer value for house a value for house b value for house c value for house d x 1 1 4 4 y z w
0
3
0
1
3
1
7
5
5
0
6
3
Table 1:
Solution:
Let us first draw the preferred-object graph given these prices:
1
House a
Buyer x 1 , 1 , 4 , 4 y 0 , 1 , 7 , 0 1 b
2 c z 3 , 3 , 5 , 6
2 d w 0 , 1 , 5 , 3
According to the graph, at these prices buyers x, y, z and w all want either house c or house d (i.e., houses { c, d } are over-demanded), thus the market cannot clear: there is no way to assign houses c and d so that all buyers get what they want. Therefore, these prices are not market-clearing prices.
1

2.
(10 points) There are five students { x, y, z, w, u } and five available dormitories
{ a, b, c, d, e } . The valuations are given in
Table 2 . Find the optimal matching, i.e., one that
maximizes social welfare.
Student value for dorm a value for dorm b value for dorm c value for dorm d value for dorm e x 3 2 2 1 0 y z
4
3
1
2
2
0
1
0
0
0 w u
1
0
2
3
2
2
0
1
2
2
Table 2:
Solution:
Let us first find a set of market-clearing prices:
0
Dorm a
Student x 3 , 2 , 2 , 1 , 0 1
Dorm a
0 b y 4 , 1 , 2 , 1 , 0 0 b
0 c z 3 , 2 , 0 , 0 , 0
0 d w 1 , 2 , 2 , 0 , 2
0 e u 0 , 3 , 2 , 1 , 2
Round 1: S = { x, y } , N ( S ) = { a } .
Student x 3 , 2 , 2 , 1 , 0 y 4 , 1 , 2 , 1 , 0
0 c z 3 , 2 , 0 , 0 , 0
0 d w 1 , 2 , 2 , 0 , 2
0 e u 0 , 3 , 2 , 1 , 2
Round 2: S = { y, z, u } , N ( S ) = { a, b } .
2

2
Dorm a
Student x 3 , 2 , 2 , 1 , 0
3
Dorm a
Student x 3 , 2 , 2 , 1 , 0
1 b y 4 , 1 , 2 , 1 , 0
2 b y 4 , 1 , 2 , 1 , 0
0 c z 3 , 2 , 0 , 0 , 0
1 c z 3 , 2 , 0 , 0 , 0
0 d w 1 , 2 , 2 , 0 , 2
0 d w 1 , 2 , 2 , 0 , 2
0 e u 0 , 3 , 2 , 1 , 2
Round 3: S = { x, y, z, w, u } , N ( S ) = { a, b, c, e } .
1 e u 0 , 3 , 2 , 1 ,
Round 3: a perfect matching exists.
2
A set of market-clearing prices is p a
= 3 , p b
= 2 , p c
= 1 , p d
= 0 , p e
= 1, and a perfect matching given by these prices is x – c , y – a , z – b , w – e , u – d . This matching maximizes social welfare, because of it is given by a set of market-clearing prices. (The social welfare of this matching is 2 + 4 + 2 + 2 + 1 = 11 .
)
3.
(10 points) There are four buyers { x, y, z, w } and four houses { a, b, c, d } . The valuations are given in
Table 3 . Find a set of market-clearing prices and a perfect matching
given by the prices.
Buyer value for house a value for house b value for house c value for house d x 3 2 3 3 y z w
1
3
3
3
1
0
2
1
2
1
1
2
Table 3:
Solution:
3

0
House a
Buyer x 3 , 2 , 3 , 3 y 1 , 3 , 2 , 1
1
House a
Buyer x 3 , 2 , 3 , 3 y 1 , 3 , 2 , 1 0 b
0 c
0 b
0 c z 3 , 1 , 1 , 1 z 3 , 1 , 1 , 1
0 d w 3 , 0 , 2 , 2
Round 1: S = { z, w } , N ( S ) = { a } .
0 d w 3 , 0 , 2 , 2
Round 2: a perfect matching exists.
A set of market-clearing prices is p a
= 1 , p b
= 0 , p c
= 0 , p d
= 0, and a perfect matching given by these prices is x – c , y – b , z – a , w – d .
4.
(10 points) Consider the following simultaneous-move game with two players. Each player simultaneously chooses a bid from the set { 1 dollar , 50 dollars, 100 dollars } . The player with the higher bid wins a prize of 100 dollars; the other player (the loser) does not receive any prize. If the two bids are the same, neither player receives the prize. Each player must pay his own bid, whether or not he wins the prize. (The loser pays too.) The payoff of each player is simply his net winnings. For example, if player 1 bids 100 and player 2 bids
50, then player 1 gets 100 − 100 = 0 and player 2 gets 0 − 50 = − 50; if player 1 bids 50 and player 2 bids 50, then each gets 0 − 50 = − 50.
(i) Write the payoff matrix of this game.
(ii) Find the pure-strategy Nash equilibria of this game.
Solution:
Part(i) —
The payoff matrix is
$1 $50
$1 − 1 , − 1 − 1 , 50
$50
$100
50
0 ,
, −
− 1
1 − 50 , − 50
$100
− 1 , 0
− 50 , 0
0 , − 50 − 100 , − 100
Part(ii) —
First, notice that this is a symmetric game: the two players have the same set of strategies,
4

and their payoffs depend on strategies played but not on who is playing them. The best response to $1 is $50, the best response to $50 is $100, and the best response to $100 is $1.
Thus, no pair of strategies is mutual best responses, hence there is no pure-strategy Nash equilibrium.
(This auction is called the “penny” auction. It illustrates the phenomenon of escalation: you want to outbid your opponent to recover your loss.)
5.
(15 points) There are 10 people in a small town. The valuation of person i (1 ≤ i ≤
10) for a new park, v i
, is either $0 or $100 and is his private information. The town decides whether or not to build the park with the following voting game. Each person votes “yes” or “no” (simultaneously) on the park, and the park will be built if and only if at least five people vote “yes.” If the park will not be built, nobody pays anything, so everyone’s payoff is zero. If the park will be built, then person i ’s payoff is v i
− P i
, where P i is the amount that he pays for the project.
(i) Consider first the following “naive” payment scheme if the park will be built: those who have voted “yes” each pays $90, and those who have voted “no” pay nothing. The truthful strategy (for any person) is to vote “yes” if one’s value is $100 and to vote “no” if one’s value is $0. Argue that the truthful strategy is not a dominant strategy in the voting game given the naive payment scheme.
(ii) Now consider a more “sophisticated” payment scheme if the park will be built: those who have voted “no” still pay nothing; if person i has voted “yes,” then he pays nothing if there are at least five other people who have voted “yes” (so the park will be built even without i ’s vote), and he pays $90 if there are only four other people who have voted “yes” (so the part will not be built without i ’s vote). Argue that the truthful strategy is a dominant strategy in the voting game given this sophisticated payment scheme.
Solution:
Part(i) —
To show that the truthful strategy is not a dominant strategy, we just need to find a situation for which the truthful strategy is not a best response. Suppose that among person
2, 3, . . . , 10, six of them vote “yes” on the park, and that person 1 has a valuation of $100 for the park; then it is best for person 1 to vote “no,” contrary to the truthful strategy (because
5

no matter what person 1 votes, the park will be built, so person 1 wants to avoid paying the
$90 associated with his “yes” vote).
Part(ii) —
To show that the truthful strategy is a dominant strategy, we must show that the truthful strategy is a best response in every situation. Because this game is symmetric, let us think from the perspective of person 1:
1. If person 1 has value $0 for the park, then clearly voting “no” is always a best response, because by voting “no” he doesn’t pay anything.
2. If person 1 has value $100 for the park and at least five other people have voted “yes,” then voting “yes” is a best response for person 1 because whatever he votes the park will be built, and he will not pay anything.
3. If person 1 has value $100 for the park and exactly four other people have voted “yes,” then person 1’s vote matters: he gets 100 − 90 = 10 from voting “yes” (park will be built, and he pays $90), and 0 from voting “no” (park will not be built, and he pays nothing); thus, voting “yes” is a best response for person 1 in this case.
4. If person 1 has value $100 for the park and less than four other people have voted
“yes,” then voting “yes” is a best response for person 1 because whatever he votes the park will not be built, and he will not pay anything.
6