Question: A search engine has three positions to sell. The positions receive 300, 200 and 100 clicks per day, respectively. There are three advertisers, with per-click values $10, $7 and $2. If bidder obtains position and pays per click, its proﬁt will be where is the bidder’s value per click, is the number of clicks it gets and is the price per click.
(a) Find the lowest market-clearing prices(/click) for the three positions. Edit
Answer: The minimum length chain for Θ−i does not contain a downward link followed by an upward link. Proof. Suppose it does contain such a sequence. Then, some bidder i1 moved from slot i1 to slot i2 > i1, and bidder i2 moved from slot i2 to a slot i3 < i2. An alternate solution, and thus a candidate solution for Θ−i is to have bidder i1 move from slot i1 to slot i3, have bidder i2 remain in slot i2, and keep everything else the same. (Bidder i1 can move to slot i3 since i3 < i2 and i2 is in range for bidder i1 (by the fact that i1 moved to i2 in Θ−i).) The difference between the two solutions is ci2 (vi2 − vi1 ) + ci3 (vi1 − vi2 ) = (ci3 −ci2 )(vi1 −vi2 ). We know ci3 > ci2 since i3 < i2. We also know vi1 ≥ vi2 since otherwise Θ could switch bidders i1 and i2 (note again that bidder i1 can move to slot i2, since it did so in Θ−i). Thus the difference is non-negative, and so this alternate solution to Θ−i has either greater valuation or a shorter chain.
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