auctions
v( Combinatorial ) = $500
v(
) = $700
Vincent Conitzer [e-mail]
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...