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1 Sealed Bid Multi-object Auctions with Necessary Bundles and its Application to Spectrum Auctions ver. 1.0 University of Tokyo 東京大学 松井知己 Tomomi Matsui Iwate Prefectural University 岩手県立大学 渡辺隆裕 Takahiro Watanabe

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2 Multi object Auction Multi-object Auction: trading oil leases, furniture, pollution rights, airport time slots, spectrum licenses, and delivery routes, etc. Bidders’ preferences are defined on sets of objects. (combinatorial auction, simultaneous auction) Results: (1) Analysis from the point of view of game theory. (2) Apply the result to spectrum auction.

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3 Main result We introduce following assumptions; (1)each bidder has a positive reservation value only for one special subset of objects (necessary bundle) (2)admissible bid is a pair of one subset of objects and its price, Game theoretic approach: show the existence of a Nash equilibrium when bidding unit is sufficiently small Application to spectrum auction: polynomial algorithm for the problem to maximize auctioneer’s revenue explicit description of a Nash equilibrium

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4 Purpose of this talk purpose of this talk = Find friends ! Combinatorial optimization Game theory Multi-object Auction Multi-agent system searched on internet ⇒ found PRIMA2001 ⇒ submitted paper ⇒ give a talk ⇒ find friends ⇒ further work ! Communication

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5 Definitions (bidders) Game theoretic descriptions N ={1,2,…, n}: players (bidders) M = {1,2,…, m}: objects bundle: subset of objects sealed bid auction: submit bids simultaneously open bid auction (English, Japanese, Dutch,…) strategies (admissible bids) of player i: (B i, b i ) ∈ 2 M ×R + : (bundle, bidding price) The bidding price b i is the amount of money that player i is willing to pay for bundle B i.

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6 Assumption (strategies) Assumption 1 each player i submits only one pair of bundle and its price (B i, b i ) ∈ 2 M ×R + : restricted but practical combinatorial auction: each player i submits bidding prices of all the bundles f i : 2 M →R + general but impractical (2 M is a huge family)

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7 bidding unit ε : bidding unit (bidding grid) Each bidding price is a non-negative multiple of ε. Z ε ={ε ｊ ｜ｊ is a non-negative integer} strategies (admissible bids) of player i: (B i, b i ) ∈ 2 M ×Z ε profile of bids : vector of bids of all the players ((B 1,b 1 ),(B 2,b 2 ),…, (B n, b n ))=(B, b) B =(B 1, B 2,…, B n ) b = ( b 1, b 2,…, b n )

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8 Definitions (auctioneer) Given a profile of bids ((B 1,b 1 ),(B 2,b 2 ),…, (B n, b n ))=(B, b), auctioneer determines the set of winners which maximizes auctioneer’s revenue. Bundle Assignment Problem BAP(B, b) (winner determination problem) max. b T x = b 1 x 1 + b 2 x 2 + ‥＋ b n x n s. t. Ax ≦ 1, x ∈ {0,1} N. A=(a ji ) 0-1matrix {0,1} M×N a ji =1 ⇔ object j is in bundle B i

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9 Bundle assignment problem BAP(B, b) max. b T x = b 1 x 1 + b 2 x 2 + ‥＋ b n x n s. t. Ax ≦ 1, x ∈ {0,1} N. ● A=(a ji ) 0-1matrix {0,1} M×N a ji =1 ⇔ object j is in bundle B i ● x i =1 ⇔ auctioneer assigns bundle B i to player i. ● Ax ≦ 1: each object must belong to at most one player

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10 Bundle assignment problem Bundle assignment problem has many names as winner determination problem, max. weight set packing problem, max. weight independent set problem, and max. weight clique problem. theoretically hard: NP-hard practically tractable: many commercial codes solve BAP efficiently (e.g. CPLEX) [Andersson, Tenhunen and Ygge (2000)]

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11 multiple-optimal solutions If BAP has multiple-optimal solutions, then auctioneer chooses an optimal solution uniformly at random. Further work: Construct an algorithm for selecting an optimal solution of BAP uniformly at random. The problem is much harder than BAP.

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12 Definitions (bidders) V i (S): Each player i has a non-negative reservation value V i (S) for each bundle S. V i : 2 M → Z δ (non-negative multiple of δ) Assumption 2: Each bidder has a positive reservation value only for one special bundle. ⇒ necessary bundle necessary (bundle, price) of player i : ( T i, v i ) V i (S) = v i ⇔ ( S ⊇ T i ) V i (S) = 0 ⇔ ( otherwise )

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13 Nash equilibrium profile of bids : ((B 1,b 1 ),(B 2,b 2 ),…, (B n, b n ))=(B, b) Utility of player i : U i (B, b) U i (B, b)=(V i ( B i ) ｰ b i ) Pr[player i is selected] profile (B*, b*) is a Nash equilibrium ⇔ ∀ i ∈ N, ∀ (B i, b i ) ∈ 2 M ×Z ε, U i (B*, b*) ≧ U i ((B* -i, b* -i ), (B i, b i )) ((B* -i, b* -i ), (B i, b i )) : profile obtained from (B*, b*) by replacing strategy of player i with (B i, b i )

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14 Main results Theorem 2: If the bidding unit ε is sufficiently small, then Nash equilibrium exists. size of bidding unit ε ≦ δ(n2 n +1) δ ： unit of reservation value, n: number of players profile (B*, b*) is a Nash equilibrium ⇔ ∀ i ∈ N, ∀ (B i, b i ) ∈ 2 M ×Z ε, U i (B*, b*) ≧ U i ((B* -i, b* -i ), (B i, b i )) ((B* -i, b* -i ), (B i, b i )) : profile obtained from (B*, b*) by replacing strategy of player i with (B i, b i )

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15 description of a Nash equilibrium Classify the bidders Passed bidders: bidders contained in every optimal solution of BAP(B, b). Questionable bidders: bidders contained in not all but at least one optimal solution of BAP(B, b). Rejected bidders: bidders never appearing in any optimal solutions of BAP(B, b). Optimal solution set: Ω (B, b): set of all the optimal solutions of BAP(B, b).

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16 Nash equilibrium Theorem 1: Following profile (B*, b*) is a Nash equilibrium; questionable bidder i ： (B* i, b* i ) = (T i, v i ) rejected bidder i : (B* i, b* i ) = (T i, v i ) passed bidder i : B* i =T i, b* i : minimal vector in Z ε satisfying Ω (B*, b*) = Ω (T, v) (solution sets are equivalent) (T i, v i ): ( necessary bundle, reservation value )

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17 Application to spectrum auctions

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18 Spectrum auction Spectrum auction: objects: spectrums (frequency channel for cellular phone) are arranged in linear order necessary bundles (T i ): sequences of consecutive channels channels : 1234567891011 T2T2 T1T1 T4T4 T3T3

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19 Longest path problem BAP corresponding to spectrum auction satisfies the conditions; (1)coefficient matrix A is totally unimodular, (2)liner relaxation of BAP has an integer valued optimal solution, (3)equivalent to longest path problem.

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20 longest path problem directed graph 1234567891011 T2T2 T1T1 T4T4 T3T3 v2v2 v1v1 v3v3 v4v4 nodes: barrier of channels arcs: ( j, j+1), necessary bundles arc weight = reservation value

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21 longest path problem 1234567891011 T2T2 T1T1 T4T4 T3T3 v2v2 v1v1 v3v3 v4v4

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22 longest path problem 1234567891011 T2T2 T1T1 T4T4 T3T3 v2v2 v1v1 v3v3 v4v4 Finding a longest path = winner determination

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23 random selection Generally, solving BAP and random selection from multiple-optimal solutions are hard. Spectrum auction: bundle assignment ⇒ longest path problem random selection ⇒ random path generation Key idea: BAP(B, b) ⇒ linear relaxation ⇒ dual problem bundle assignment: dynamic programming random selection: path counting algorithm explicit description of a Nash equilibrium: complementality slackness theorem for linear programming problems (detail is omitted)

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24 conclusion Assumption 1 (multi-object auction) each player i submits only one pair of bundle and its price (B i, b i ) ∈ 2 M ×R + Assumption 2: Each bidder has a positive reservation value only for one special bundle, called necessary bundle. Theorem 2: If the bidding unit ε is sufficiently small, then Nash equilibrium exists. Theorem 1 : (Characterization of a Nash equilibrium)

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25 END

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26 Main results Theorem 2: If the bidding unit ε is sufficiently small, then (pure strategy) Nash equilibrium exists. mixed strategy: Nash showed that every strategic form n-persons game with finite number of strategies has a mixed strategy Nash equilibrium. size of bidding unit ε ≦ δ(n2 n +1) δ ： unit of reservation value, n: number of players

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27 random selection BAP(B, b) max { b T x | Ax ≦ 1, x ∈ {0,1} N } linear relaxation max{b T x | Ax ≦ 1, x ≧ 0} dual problem min{y T x | y T A ≧ b, y ≧ 0 } y* ： optimal dual solution M*={j ∈ M | y* T a i =b i } a i : i th column vector Lemma: M* is the set of passed and questionable bidders. Ordinary dynamic programming procedure ⇒ random selection of longest paths

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