Simon Fraser University
Spring 2015
Econ 383 D100 Auction Theory Final Exam Solution
This brief solution guide does not have the explanations necessary for full marks.
1. (20 points) There are three buyers {1, 2, 3} and three houses {a, b, c}. The buyers’
values are tabulated below:
Buyer value for house a value for house b value for house c
4
3
1
1
2
2
3
2
3
2
1
0
Table 1
In parts (i)–(iv) you will calculate the allocation and payments in the VCG mechanism
given truthful reports of value.
(i) There are six matchings that assign buyers to distinct houses. Calculate the total value
(i.e., social welfare) for each of them. What is the optimal (social-welfare maximizing)
matching? This is the matching implemented by the VCG mechanism.
1
1
1
1
1
1
–
–
–
–
–
–
Matching
a, 2 – b, 3 – c
a, 2 – c, 3 – b
b, 2 – a, 3 – c
b, 2 – c, 3 – a
c, 2 – a, 3 – b
c, 2 – b, 3 – a
Total value
7
7
5
7
4
6
Solution: An optimal matching is: 1 – a, 2 – b, 3 – c. Suppose the VCG mechanism
implements this matching.
(ii) Consider buyer 1. What is the total value of others when buyer 1 is present in the
VCG mechanism? What is the total value of others when buyer 1 is not present in the
VCG mechanism? (To answer the second question calculate others’ total value in all
matchings without buyer 1, and choose the optimal one.) Finally, what is the payment
that buyer 1 makes in the VCG mechanism?
1
Matching
2 – b, 3 – c
2 – c, 3 – b
2 – a, 3 – c
2 – c, 3 – a
2 – a, 3 – b
2 – b, 3 – a
Total value without buyer 1
3
3
2
4
3
5
Solution: The total value of others when buyer 1 is present in the VCG mechanism
is 3. The total value of others when buyer 1 is not present in the VCG mechanism is
5. So buyer 1 pays 2 in the VCG mechanism.
(iii) Repeat (ii) for buyer 2.
Matching
1 – a, 3 – c
1 – a, 3 – b
1 – b, 3 – c
1 – b, 3 – a
1 – c, 3 – b
1 – c, 3 – a
Total value without buyer 2
4
5
3
5
2
3
Solution: The total value of others when buyer 2 is present in the VCG mechanism
is 4. The total value of others when buyer 2 is not present in the VCG mechanism is
5. So buyer 2 pays 1 in the VCG mechanism.
(iv) Repeat (ii) for buyer 3.
Matching
1 – a, 2 – b
1 – a, 2 – c
1 – b, 2 – a
1 – b, 2 – c
1 – c, 2 – a
1 – c, 2 – b
Total value without buyer 3
7
6
5
5
3
4
2
Solution: The total value of others when buyer 3 is present in the VCG mechanism
is 7. The total value of others when buyer 3 is not present in the VCG mechanism is
7. So buyer 3 pays 0 in the VCG mechanism.
(v) Run the ascending-price auction on the values in Table 1. (Consider only integer prices.)
For each step of the auction, write down the current prices and the preferred-object
graph, and identify the constricted set of buyers and the over-demanded set of houses.
What are the market-clearing prices and the matching from this auction?
Solution:
House
Buyer
House
Buyer
0
a
1
4, 3, 1
1
a
1
4, 3, 1
0
b
2
2, 3, 2
0
b
2
2, 3, 2
0
c
3
2, 1, 0
0
c
3
2, 1, 0
Round 1: constricted set S = {1, 3}, N (S) = {a}. Round 2: constricted set S = {1, 2, 3}, N (S) = {a, b}.
House
Buyer
2
a
1
4, 3, 1
1
b
2
2, 3, 2
0
c
3
2, 1, 0
Round 3: a perfect matching exists.
The market-clearing prices are pa = 2, pb = 1, pc = 0. A perfect matching given by the
auction is 1 − a, 2 − b, 3 − c.
2. (10 points) Suppose that there are three advertisers {1, 2, 3} and three advertisement
slots {1, 2, 3}, and each advertiser wants only one slot. Assume multiplicative valuation
vi,j = ri vj , where ri is slot i’s clickthrough rate and vj is buyer j’s revenue per click. And
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assume it is common knowledge that r1 = 8, r2 = 7, r3 = 1, v1 = 10, v2 = 6 and v3 = 1.
Consider the generalized second price (GSP) auction; let bj be buyer j’s bid on a per-click
basis.
(i) Construct a Nash equilibrium of the GSP auction that is revenue-equivalent to the
VCG mechanism, and verify that no bidder has an incentive to deviate (i.e., it is indeed a
Nash equilibrium).
Solution: The VCG prices are P3VCG = 0, P2VCG = 1×(7−1) = 6, P1VCG = 1×(7−1)+
6×(8−7) = 12. So one Nash equilibrium in the GSP auction is: b1 = P1VCG /r1 +1 = 5/2, b2 =
P1VCG /r1 = 12/8 = 3/2, and b3 = P2VCG /r2 = 6/7. Given these bids, buyer 1 gets slot 1 and
has no incentive to get slot 2 by bidding 6/7 < b1 < 3/2 because 8×(10−3/2) > 7×(10−6/7),
and no incentive to get slot 3 by bidding b1 < 6/7 because 8 × (10 − 3/2) > 1 × (10 − 0).
(Add the verifications for buyer 2 and 3.)
(ii) Is b1 = 10, b2 = 4 and b3 = 1 a Nash equilibrium of the GSP auction? Why or why
not?
Solution: Given these bids, buyer 1 gets slot 1 but has an incentive to get slot 2 instead
by bidding 1 < b1 < 4 because 8 × (10 − 4) < 7 × (10 − 1). So these bids do not form a Nash
equilibrium.
3. (5 points) Consider a third-price auction of a single object (the highest bidder gets
the object and pays the third highest bid; the others do not pay). Give an example (values
and bids) that shows that truthful bidding is not a dominant strategy in this auction, and
explain your example.
Solution: Suppose there are three bidders with bids b1 = 10, b2 = 5, and b3 = 0, and
bidder 2 has a value of v2 = 5. Given these bids, bidder 2 has no incentive to bid truthfully
at b2 = 5 since by bidding b2 > 10 he wins the object, pays 0 and gets a net payoff of 5,
which is larger than his payoff of 0 from truthful bidding.
4. (10 points) Consider a second-price auction of a single object. Suppose that bidders
bid truthfully in the auction.
(i) There is a reserve price r between 0 and 1, and a single bidder whose value is randomly
drawn from the uniform distribution on [0, 1]. Calculate the expected revenue from the
auction as a function of r. What is the reserve price r that maximizes the revenue, and
what is this maximum revenue?
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Solution: Let Rev(r) be the expected revenue as a function of r. Clearly, Rev(r) =
(1 − r)r. The first order condition is Rev0 (r) = 1 − 2r = 0, so the revenue is maximized
at r = 1/2, and the maximum revenue is Rev(1/2) = 1/4.
(ii) Now suppose there is no reserve price (r = 0). There are two bidders whose values are
randomly and independently drawn from the uniform distribution on [0, 1]. Calculate
the expected revenue from the auction. Is it higher than or lower than the maximum
revenue from part (i)?
Solution: The expected revenue here is E[min(v1 , v2 )] = 1/3, which is larger than that
from part (i).
5. (10 points) There are 20 people in a small town. The valuation of person i (where
1 ≤ i ≤ 20) for a new park, vi , 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 is built if and only if at least 10
people vote “yes.” If the park is not built, nobody pays, so everyone’s payoff is zero. If the
park is built, then person i’s payoff is vi − Pi , where Pi is the amount that he pays for the
project.
(i) Consider first the following “naive” payment scheme if the park is 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.
Solution: 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, . . . , 20, eleven 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 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).
(ii) Now consider a more “sophisticated” payment scheme if the park is built: those who
have voted “no” still pay nothing; if person i has voted “yes,” then he pays nothing if
there are at least 10 other people who have also voted “yes” (so the park is built even
without i’s vote), and he pays $90 if there are exactly 9 other people who have also
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voted “yes” (so the park is not built without i’s vote). Argue that the truthful strategy
is a dominant strategy in the voting game given this sophisticated payment scheme.
Solution: 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:
(a) 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.
(b) If person 1 has value $100 for the park and at least ten 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 pays nothing
(c) If person 1 has value $100 for the park and exactly nine 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.
(d) If person 1 has value $100 for the park and less than nine 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 pays nothing.
(iii) Can you think of any problem with the sophisticated payment scheme?
Solution: One problem with the sophisticated payment scheme is low revenue: unless
there are exactly ten people who want the park (value = $100), this payment scheme
generates zero revenue!
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