MS&E 235
Network Analytics
Stanford University
Practice Midterm
Solution
1
Recall that an auction is truthful if and only if it is a weakly dominant strategy for every agent to report
his or her true utility/value. Suppose we run a single item auction with n ≥ 3 bidders. Each bidder has
a private value vi (1 ≤ i ≤ n) for the item, and each bids bi (1 ≤ i ≤ n). The utility gained by each
bidder is 0 if he of she does not win and vi − p if he or she does win, where p is the price paid. An auction
is truthful if the a bidder should always bid bi = vi , regardless of other bids bj . We can show that the
second price auction (where p is the second highest bid) is truthful. Show with an example that the third
price auction (where the price paid is the third highest bid) is not truthful.
Solution: Consider the example with three bidders such that v1 > v2 > v3 . Also suppose bidders 1 and
3 submit truthful bids (b1 = v1 and b3 = v3 ). If bidder 2 submits a truthful bid, he will lose the auction,
but if he submits an untruthful bid b2 > b1 , then he wins the auction and only pays a price of v3 < v2 .
Therefore, third-price auctions are not truthful.
2
In this problem we ask how much a seller can expect to receive for his object in a second-price, sealed
bid auction. Assume that all bidders have independent, private values vi that are either 0 or 1. The
probability of 0 and 1 are both 21 .
1. Suppose there are two bidders. Then there are four possible pairs of their values (v1 , v2 ): (0, 0), (1, 0), (0, 1),
and (1, 1). Each pair of values has probability 41 . Show that the seller’s expected revenue if 41 . (Assume that if there is a tie at a bid of x for the highest bid, the winner is selected at random from
among the highest bidders and the price is x.)
Solution: The expected revenue is just the expected price of the auction.
1
1
E[rev] = (0 + 0 + 0 + 1) =
4
4
2. What is the seller’s expected revenue if there are three bidders?
Solution: Now we have eight possible triplets of values (v1 , v2 , v3 ) = (0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0),(1
and (1, 1, 1) - each with probability 18 . Now we just do the same calculation as before.
1
1
E[rev] = (0 + 0 + 0 + 1 + 0 + 1 + 1 + 1) =
8
2
3. This suggests a conjecture that, as the number of bidders increases, the seller’s expected revenue
also increases. In the example we consider that the seller’s expected revenue actually converges to 1
as the number of bidders grows. Explain why this should occur. You do not need to write a proof;
an intuitive explanation is fine.
Solution: The price of the auction is the second largest value of the bidders. As the number
of bidders grows, the probability that at least two bidders have value 1 also grows - making the
expected revenue grow to 1 as well.
1
MS&E 235
Network Analytics
Stanford University
3
Suppose a search engine has two ad slots that it can sell. Slot a has a clickthrough rate of 10 and slot b
has a clickthrough rate of 5. There are three advertisers who are interested in these slots. Advertiser x
values clicks at 3 per click, advertiser y values clicks at 2 per click, and advertiser z values clicks at 1 per
click.
Compute the socially optimal allocation (which advertisers get the ad slots) and the VCG prices for it.
Give a brief explanation of your answer.
Solution: We will consider the given CTR’s as percentages. The values for the first slot are 0.3, 0.2
and 0.1 respectively for advertisers x, y and z. While those for the second slot are 0.15, 0.1 and 0.05.
Therefore, advertiser x wins slot a and advertiser b wins slot b. Now we compute the VCG prices, where
the price for a certain bidder is computed by the difference of the welfare/value of the other bidders in
the auction if that bidder wasn’t participating and the welfare/value of the other bidders in the current
auction:
px = (0.2 + 0.05) − (0.1 + 0) = 0.15
py = (0.3 + 0.05) − (0.3 + 0) = 0.05
Now we divide by CTR to get price per click. px = 1.5 per click and py = 1 per click.
4
Consider a second-price, sealed-bid auction with two bidders. Each bidder has independent value drawn
uniformly from {1, 2, . . . , 100}.
1. What is the expected revenue of this auction?
Solution: The revenue is the min of two independent random variables. Let X1 , X2 be drawn from
the given uniform distribution. First we compute the probability distribution for the revenue.
P r[rev = t] = P r[X1 = t]P r[X2 > t] + P r[X2 = t]P r[X1 > t] + P r[X1 = X2 = t]
1 100 − t
1 2
=2·
·
+
100
100
100
And expected revenue is:
E[rev] =
100
X
t=1
1 100 − t
·
+
2t ·
100
100
1
100
2
= 33.835
2. What is the expected revenue of this auction with reserve price k, for some constant k?
Solution: With a reservation price, we now break the auction into cases. If both bidders bid below
2
k, the revenue is 0, which happens with probability k−1
.
100
If one bid is greater than or equal to k and the other
is below k, the revenue of the auction is k,
k−1 100−k+1
and this happens with probability 2 · 100
.
100
If both bidders bid greater than or equal to k, then the revenue of the auction is again the minimum
bid. We may treat as we did above.
The expected revenue of the auction then is, as a function of k:
X
100
k − 1 100 − k + 1
1 100 − t
1 2
E[rev] = 2k ·
+
2t ·
·
+
100
100
100
100
100
t=k
2
MS&E 235
Network Analytics
Stanford University
3. For what value of k is the expected revenue of the auction above maximized?
Solution: We may optimize the above equation for k. Also, we can plot it as a function of k.
The max occurs at k close to 51.
3