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Published bySebastian Daly Modified over 2 years ago

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Combinatorial Auction

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Conbinatorial auction t 1 =20 t 2 =15 t 3 =6 f(t): the set X F with the highest total value the mechanism decides the set of winners and the corresponding payments Each player wants a bundle of objects t i : value player i is willing to pay for its bundle if player i gets the bundle at price p his utility is u i =t i -p F={ X {1,…,N} : winners in X are compatible} v 1 =20 v 2 =16 v 3 =7

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Combinatorial Auction (CA) problem – single-minded case Input: n buyers, m indivisible objects each buyer i: Wants a subset S i of the objects has a value t i for S i Solution: X {1,…,n}, such that for every i,j X, with i j, S i S j = Measure (to maximize): Total value of X: i X t i

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CA game each buyer i is selfish Only buyer i knows t i (while S i is public) We want to compute a good solution w.r.t. the true values We do it by designing a mechanism Our mechanism: Asks each buyer to report its value v i Computes a solution using an output algorithm g( ٠ ) takes payments p i from buyer i using some payment function p

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More formally Type of agent buyer i: t i : value of S i Intuition: t i is the maximum value buyer i is willing to pay for S i Buyer is valuation of X F: v i (t i,X)= t i if i X, 0 otherwise SCF: a good allocation of the objects w.r.t. the true values

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How to design a truthful mechanism for the problem? Notice that: the (true) total value of a feasible X is: i v i (t i,X) the problem is utilitarian! …VCG mechanisms apply

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VCG mechanism M= : g(r): arg max x F j v j (r j,x) p i (x): for each i: p i = ji v j (r j,g(r - i )) - ji v j (r j,x) g(r) has to compute an optimal solution… …can we do that?

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Approximating CA problem within a factor better than m 1/2- is NP-hard, for any fixed >0. Theorem proof Reduction from independent set problem

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Maximum Independent Set (IS) problem Input: a graph G=(V,E) Solution: U V, such that no two verteces in U are jointed by an edge Measure: Cardinality of U Approximating IS problem within a factor better than n 1- is NP-hard, for any fixed >0. Theorem

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the reduction CA instance has a solution of total value k if and only if there is an IS of size k G=(V,E) each edge is an object each node i is a buyer with: S i : set of edges incident to i t i =1 …since m n 2 …

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How to design a truthful mechanism for the problem? Notice that: the (true) total value of a feasible X is: i v i (t i,X) the problem is utilitarian! …but a VCG mechanism is not computable in polynomial time! what can we do? …fortunately, our problem is one parameter!

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A problem is binary demand (BD) if 1. a i s type is a single parameter t i 2. a i s valuation is of the form: v i (t i,o)= t i w i (o), w i (o) {0,1} work load for a i in o when w i (o)=1 well say that a i is selected in o

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An algorithm g() for a maximization BD problem is monotone if agent a i, and for every r -i =(r 1,…,r i-1,r i+1,…,r N ), w i (g(r -i,r i )) is of the form: Definition 1 Ө i (r -i ) riri Ө i (r -i ) {+ }: threshold payment from a i is: p i (r)= Ө i (r -i )

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Our goal: to design a mechanism satisfying: 1. g( ٠ ) is monotone 2. Solution returned by g( ٠ ) is a good solution, i.e. an approximated solution 3. g( ٠ ) and p( ٠ ) computable in polynomial time

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A greedy m-approximation algorithm 1. reorder (and rename) the bids such that 2. W ; X 3. for i=1 to n do 1. if S i X= then W W {i}; X X {S i } 4. return W v 1 / |S 1 | v 2 / |S 2 | … v n / |S n |

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The algorithm g( ) is monotone Lemma proof It suffices to prove that, for any selected agent i, we have that i is still selected when it raises its bid Increasing v i can only move bidder i up in the greedy order, making it easier to win

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How much can bidder i decrease its bid before being non- selected? Computing the payments …we have to compute for each selected bidder i its threshold value

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Computing payment p i v 1 / |S 1 | … v i / |S i | … v n / |S n | Consider the greedy order without i index j Use the greedy algorithm to find the smallest index j (if any) such that: 1. j is selected 2. S j S i p i = v j |S i |/ |S j | p i = 0 if j doent exist

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Let OPT be an optimal solution for CA problem, and let W be the solution computed by the algorithm, then Lemma i W i OPT v i m i W v i proof OPT i ={j OPT : j i and S j S i } i W OPT i =OPT since it suffices to prove: j OPT i v j m v i crucial observation for greedy order we have v i |S j | i W j OPT i |S i | v j

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proof we can bound Cauchy–Schwarz inequality i W j OPT i v j j OPT i vivi |S i | |S j | j OPT i |S j | |OPT i | j OPT i |S j | |S i | m m v i |S i | m

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Cauchy–Schwarz inequality y j = |S j | x j =1 n= |OPT i | for j=1,…,|OPT i | …in our case…

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