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Sealed Bid Multi-object Auctionswith 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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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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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 revenueexplicit description of a Nash equilibrium

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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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Definitions (bidders)

Game theoretic descriptions

N ={1,2,…, n}: players (bidders)

M = {1,2,…, m}: objectsbundle: subset of objects

sealed bid auction: submit bids simultaneouslyopen bid auction (English, Japanese, Dutch,…)

strategies (admissible bids) of player i:(Bi, bi) 2∈ M×R+: (bundle, bidding price)

The bidding price bi is the amount of money

that player i is willing to pay for bundle Bi .

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Assumption (strategies)

Assumption 1

each player i submits only one pair of bundle and its price (Bi, bi) 2∈ M×R+: restricted but practical

combinatorial auction:

each player i submits bidding prices of all the bundles fi : 2

M→R+

general but impractical (2M is a huge family)

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bidding unit

ε : bidding unit (bidding grid)Each bidding price is

a non-negative multiple of ε.Zε={ε ｊ ｜ｊ is a non-negative integer}

strategies (admissible bids) of player i:(Bi, bi) 2∈ M×Zε

profile of bids : vector of bids of all the players

((B1,b1),(B2,b2),…, (Bn, bn))=(B, b)

B =(B1, B2,…, Bn )

b = ( b1, b2 ,…, bn)

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Definitions (auctioneer)

Given a profile of bids

((B1,b1),(B2,b2),…, (Bn, bn))=(B, b), auctioneer determines the set of winners wh

ich maximizes auctioneer’s revenue.Bundle Assignment Problem BAP(B, b)

(winner determination problem)

max. bTx = b1x1+ b2x2+‥ ＋ bnxn

s. t. Ax ≦1, x {0,1}∈ N.A=(aji) 0-1matrix {0,1}M×N

aji=1 ⇔ object j is in bundle Bi

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Bundle assignment problem

BAP(B, b)max. bTx = b1x1+ b2x2+‥ ＋ bnxns. t. Ax ≦1, x {0,1}∈ N.

● A=(aji) 0-1matrix {0,1}M×N

aji=1 ⇔ object j is in bundle Bi

● xi =1 ⇔ auctioneer assigns

bundle Bi to player i.● Ax ≦1: each object must belong to at mos

t one player

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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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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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Definitions (bidders)

Vi(S): Each player i has a non-negative

reservation value Vi(S) for each bundle S.

Vi : 2M → Zδ (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 : (Ti , vi )

Vi(S) = vi ⇔ (S⊇Ti)Vi(S) = 0 ⇔ (otherwise)

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Nash equilibrium

profile of bids :((B1,b1),(B2,b2),…, (Bn, bn))=(B, b)

Utility of player i : Ui (B, b)

Ui (B, b)=(Vi(Bi) ｰ bi ) Pr[player i is selected]

profile (B*, b*) is a Nash equilibrium

⇔ ∀i∈N, (∀ Bi, bi) 2∈ M×Zε,Ui (B*, b*) ≧ Ui ((B*-i, b*-i), (Bi, bi))

((B*-i, b*-i), (Bi, bi)) : profile obtained from (B*, b

*) by replacing strategy of player i with (Bi, bi)

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Main results

Theorem 2: If the bidding unit ε is sufficiently small, then Nash equilibrium exists.

size of bidding unit ε δ(≦ n2n+1)δ ： unit of reservation value, n: number of players

<proof: omitted>profile (B*, b*) is a Nash equilibrium

⇔ ∀i∈N, (∀ Bi, bi) 2∈ M×Zε,Ui (B*, b*) ≧ Ui ((B*-i, b*-i), (Bi, bi))

((B*-i, b*-i), (Bi, bi)) : profile obtained from (B*, b*)

by replacing strategy of player i with (Bi, bi)

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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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Nash equilibrium

Theorem 1: Following profile (B*, b*) is a Nash equilibrium;

questionable bidder i ： (B*i, b*i) = (Ti, vi)

rejected bidder i : (B*i, b*i) = (Ti, vi)

passed bidder i : B*i =Ti,

b*i: minimal vector in Zε

satisfying Ω (B*, b*) = Ω (T, v) (solution sets are equivalent)

(Ti, vi ): (necessary bundle, reservation value)<proof: omitted>

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Application to spectrum auctions

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Spectrum auction

Spectrum auction:

objects: spectrums (frequency channel for cellular phone) are arranged in linear order

necessary bundles (Ti): sequences of consecutive channels

channels :1 2 3 4 5 6 7 8 9 10 11

T2

T1

T4

T3

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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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longest path problem

directed graph

1 2 3 4 5 6 7 8 9 10 11

T2

T1

T4

T3

v2

v1

v3 v4

nodes: barrier of channelsarcs: ( j, j+1), necessary bundles arc weight = reservation value

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longest path problem

longest path problem

1 2 3 4 5 6 7 8 9 10 11

T2

T1

T4

T3

v2

v1

v3 v4

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longest path problem

longest path problem

1 2 3 4 5 6 7 8 9 10 11

T2

T1

T4

T3

v2

v1

v3 v4

Finding a longest path = winner determination

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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 algorithmexplicit description of a Nash equilibrium: comple

mentality slackness theorem for linear programming problems

(detail is omitted)

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conclusion

Assumption 1 (multi-object auction)

each player i submits only one pair of bundle and its price (Bi, bi) 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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END

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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(≦ n2n+1)δ ： unit of reservation value, n: number of players

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random selection

BAP(B, b) max{bTx | Ax ≦1, x {0,1}∈ N}linear relaxation max{bTx | Ax ≦1, x 0}≧dual problem min{yTx | yTA ≧b, y 0 }≧

y* ： optimal dual solution

M*={j∈M | y*Tai=bi } ai : i th column vector

Lemma: M* is the set of passed and questionable bidders.

Ordinary dynamic programming procedure ⇒ random selection of longest paths