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

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.

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

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

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.

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)

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 )

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

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

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)]

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.

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 )

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 )

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 )

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).

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 )

17 Application to spectrum auctions

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 : T2T2 T1T1 T4T4 T3T3

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.

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

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

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

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)

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)

25 END

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

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