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### cgtmd_combinatorial_auctions

Course: CPS 296, Fall 2009
School: Duke
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Word Count: 1369

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auctions v( Combinatorial ) = \$500 v( ) = \$700 Vincent Conitzer conitzer@cs.duke.edu Complementarity and substitutability How valuable one item is to a bidder may depend on whether the bidder possesses another item Items a and b are complementary if v({a, b}) &gt; v({a}) + v({b}) E.g. Items a and b are substitutes if v({a, b}) &lt; v({a}) + v({b}) E.g. Suppose your valuation function is v( ) =...

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auctions v( Combinatorial ) = \$500 v( ) = \$700 Vincent Conitzer conitzer@cs.duke.edu Complementarity and substitutability How valuable one item is to a bidder may depend on whether the bidder possesses another item Items a and b are complementary if v({a, b}) > v({a}) + v({b}) E.g. Items a and b are substitutes if v({a, b}) < v({a}) + v({b}) E.g. Suppose your valuation function is v( ) = \$200, v( ) = \$100, v( ) = \$500 Now suppose that there are two (say, Vickrey) auctions, the first one for and the second one for What should you bid in the first auction (for )? If you bid \$200, you may lose to a bidder who bids \$250, only to find out that you could have won for \$200 If you bid anything higher, you may pay more than \$200, only to find out that sells for \$1000 Inefficiency of sequential auctions Combinatorial auctions Simultaneously for sale: bid 1 , , v( bid 2 ) = \$500 ) = \$700 bid 3 v( v( ) = \$300 used in truckload transportation, industrial procurement, radio spectrum allocation, Exponentially many bundles In general, in a combinatorial auction with set of items I (|I| = m) for sale, a bidder could have a different valuation for every subset S of I Implicit assumption: no externalities (bidder does not care what the other bidders win) Must a bidder communicate 2m values? Impractical Also difficult for the bidder to evaluate every bundle Could require vi() = 0 Does not help much Could require: if S is a superset of S, v(S) v(S) (free disposal) Does not help in terms of number of values Bidding languages Bidding language = a language for expressing valuation functions A good bidding language allows bidders to concisely express natural valuation functions Example: the OR bidding language [Rothkopf et al. 98, DeMartini et al. 99] Bundle-value pairs are ORed together, auctioneer may accept any number of these pairs (assuming no overlap in items) E.g. ({a}, 3) OR ({b, c}, 4) OR ({c, d}, 4) implies A value of 3 for {a} A value of 4 for {b, c, d} A value of 7 for {a, b, c} Can we express the valuation function v({a, b}) = v({a}) = v{b} = 1 using the OR bidding language? OR language is good for expressing complementarity, bad for expressing substitutability If we use XOR instead of OR, that means that only one of the bundle-value pairs can be accepted Can express any valuation function (simply XOR together all bundles) E.g. ({a}, 3) XOR ({b, c}, 4) XOR ({c, d}, 4) implies A value of 3 for {a} A value of 4 for {b, c, d} A value of 4 for {a, b, c} XORs Sometimes not very concise E.g. suppose that for any S, v(S) = s in Sv({s}) How can this be expressed in the OR language? What about the XOR language? Can also combine ORs and XORs to get benefits of both [Nisan 00, Sandholm 02] E.g. (({a}, 3) XOR ({b, c}, 4)) OR ({c, d}, 4) implies A value of 4 for {a, b, c} A value of 4 for {b, c, d} A value of 7 for {a, c, d} The winner determination problem (WDP) Allocate a subset Si of I to each bidder i to maximize ivi(Si) (under the constraint that for ij, Si Sj = ) This is assuming free disposal, i.e. not everything needs to be allocated Complexity of the winner determination problem depends on the bidding language WDP and bidding languages Single-minded bidders bid on only one bundle Valuation is x for any subset including that bundle, 0 otherwise If we can solve the WDP for single-minded bidders, we can also solve it for the OR language Simply pretend that each bundle-value pair comes from a different bidder We can even use the same algorithm when XORs are added, using the following trick: For bundle-value pairs that are XORed together, add a dummy item to them [Fujishima et al 99, Nisan 00] E.g. ({a}, 3) XOR ({b, c}, 4) becomes ({a, dummy1}, 3) OR ({b, c, dummy1}, 4) So, we will focus on single-minded bids Dynamic programming approach to WDP [Rothkopf et al. 98] For every subset S of I, compute w(S) = the maximum total value that can be obtained when only allocating items in S Then, w(S) = max {maxi vi(S), maxS: S is a subset of S, and there exists a bid on S w(S) + w(S \ S)} Runs in O(n3m) time The winner determination problem as a weighted independent set problem Each (single-minded) bid is a vertex Draw an edge between two vertices if they share an item bid 2 v( bid 3 ) = \$700 bid 1 v( ) = \$300 v( ) = \$500 Optimal allocation = maximum weight independent set Can model each weighted independent set instance as a CA winner determination problem (1 item per edge (or clique)) But weighted independent set cannot be approximated to k = n1- unless NP = ZPP [Hstad 96] [Sandholm 02] noted that this inapproximability applies to the WDP xb equals 1 if bid b is accepted, 0 if it is not maximize b vbxb subject to for each item j, b: j in b xb 1 An integer program formulation If each xb can take any value in [0, 1], we say that bids can be partially accepted In this case, this is a linear program that can be solved in polynomial time This requires that each item can be divided into fractions if a bidder gets a fraction f of each of the items in his bundle, then this is worth the same fraction f of his value vb for the bundle Under certain conditions, the optimal solution to the Bids with few items [Rothkopf et al. 98] If each bid is on a bundle of at most two items, then the winner determination problem can be solved in polynomial time as a maximum weighted matching problem 3-item example: Value of highest bid on {A, B} Value of highest bid on {B} item A Value of highest bid on {A} Value of highest bid on {A, C} item B Bs dummy Value of highest bid on {B, C} As dummy item C Value of highest bid on {C} Cs dummy If each bid is on a bundle of three items, then the winner determination problem is NP-hard again (can reduce from exact-cover-by-3-sets problem) Bids on connected sets of items in a tree Suppose items are organized in a tree item B item A item C item D E.g. {A, B, C, G}, but not {A, B, G} item E item F item G item H Suppose each bid is on a connected set of items Then the WDP can be solved in polynomial time using dynamic programming [Sandholm & Suri 03] Tree does not need to be given: can be constructed from the bids in polynomial time if it exists [Conitzer, Derryberry, Sandholm 04] More generally, WDP can also be solved in polynomial time for gra...

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