2. Combinatorial Auctions ( Bidding and Allocation) Adapted from Noam Nisan

3. What is a Combinatorial Auction? Set of Products:Set of Products: Each customer can bid: $700 for { AND } $1200 for { } OR $8 for { } $6 for { } XOR $30 for { } $3 for {ANY 3}Each customer can bid: $700 for { AND } $1200 for { } OR $8 for { } $6 for { } XOR $30 for { } $3 for {ANY 3}

4. The Model m items for sale: X = {x 1,…,x m }m items for sale: X = {x 1,…,x m } n bidders: v 1,…,v n Every v i is a valuation of subsets of X: v i : 2 X  R + v i (S ) = how much would I pay for S  Xn bidders: v 1,…,v n Every v i is a valuation of subsets of X: v i : 2 X  R + v i (S ) = how much would I pay for S  X Bids are handed in sealed envelops.Bids are handed in sealed envelops. Auctioneer allocates the items among bidders trying to maximize its revenue: Find n disjoint sets - S 1,…,S n  X s.t.  v i (S i ) is maximal.Auctioneer allocates the items among bidders trying to maximize its revenue: Find n disjoint sets - S 1,…,S n  X s.t.  v i (S i ) is maximal.

5. Issues BiddingBidding –Expressiveness –Simplicity AllocationAllocation –Hardness Results –Approximation Algorithms PaymentPayment StrategyStrategy

6. Bidding No Externalities v (S ) depends only on S.No Externalities v (S ) depends only on S. Free Disposal S  T  v (S )  v (T )Free Disposal S  T  v (S )  v (T ) Normalization v (  ) = 0Normalization v (  ) = 0

7. Bidding Let S and T be disjoint item subsets. We say S and T are:Let S and T be disjoint item subsets. We say S and T are: –Complementary, if: v (S  T ) > v (S ) + v (T ) –Substitutes, if: v (S  T ) < v (S ) + v (T )

8. Bidding - Examples The additive valuationThe additive valuation –v (S ) = |S | –No substitutabilities and no complementarities. The single item valuationThe single item valuation –Want to buy just one item. –v (S ) = 1(iff S is not empty) –All items are substitutes of each other. The K-budget valuationThe K-budget valuation –Want to buy up to K items. –v (S ) = min {K, |S |}

9. Bidding - Examples The majority valuationThe majority valuation –Want to buy most of the items. –v (S ) =0for |S | < m / 2 1otherwise Symmetric valuationSymmetric valuation –Let p 1,…,p n be non-negative numbers. v (S ) =  j=1…|S| p j –Additive: p j =1 –K-budget: p j =0for j > K 1otherwise –Majority: p j =1for j = m/2 0otherwise

10. Bidding - Examples Downward sloping symmetric valuationDownward sloping symmetric valuation –A symmetric valuation with p 1  p 2  …  p n  0 –Is viewed as the “normal” economic case.

11. Bidding - Asymmetric Examples The monochromatic valuationThe monochromatic valuation –m/2 red items, m/2 blue items. –Want to buy items of just one color. –For S having k reds and l blues, v (S ) = max{k,l } One-of-each-kind valuationOne-of-each-kind valuation –m/2 pairs of items. –Want to buy just one item from each pair. –For S having k pairs and l singletons (|S |=2k+l), v (S ) = k +l

12. Bidding Languages A simple language:A simple language: –Specify v explicitly as a 2 m vector. –Impractical - bids are too big. Language must allow:Language must allow: –To express any “reasonable” valuation with polynomial (in m) size expressions. –To be computationally easy: given v and S, compute v (S ) in polynomial time. The applet-language:The applet-language: –Specify v as a computer program. –Doesn’t allow efficient allocation algorithms.

13. Basic Bidding Languages Atomic bidsAtomic bids –v = (S,p) v (T ) =pif T  S 0otherwise –Can’t represent the additive valuation. OR bidsOR bids –v = (S 1,p 1 ) OR (S 2,p 2 ) OR … OR (S k,p k ) –If S i and S j are disjoint, v (S i  S j ) = p i + p j –Can express all bids with no substitutabilities and only them.

14. Basic Bidding Languages XOR bidsXOR bids –v = (S 1,p 1 ) XOR (S 2,p 2 ) XOR … XOR (S k,p k ) –v (S ) = max p i s.t. S  S i –Can express all bids. –The additive valuation requires 2 m atoms in XOR language but only m atoms in OR language.

15. OR-of-XOR v = u 1 OR u 2 OR … OR u k each u i is a XOR bid.v = u 1 OR u 2 OR … OR u k each u i is a XOR bid. The bidder is willing to obtain any number of u-s for the sum of their prices.The bidder is willing to obtain any number of u-s for the sum of their prices. Downward sloping bid: [({x 1 },p 1 ) XOR … XOR ({x m },p 1 )] OR [({x 1 },p 2 ) XOR … XOR ({x m },p 2 )] OR … [({x 1 },p m ) XOR … XOR ({x m },p m )]Downward sloping bid: [({x 1 },p 1 ) XOR … XOR ({x m },p 1 )] OR [({x 1 },p 2 ) XOR … XOR ({x m },p 2 )] OR … [({x 1 },p m ) XOR … XOR ({x m },p m )]

16. OR-of-XOR Theorem: The monochromatic valuation requires an  (2 m/2 ) size OR-of-XOR expression.Theorem: The monochromatic valuation requires an  (2 m/2 ) size OR-of-XOR expression. Proof:Proof: –W.L.O.G. every (S,p) is monochromatic. –p = |S | –Can’t have a blue atom in one clause and a red atom in another. –All atoms must be in one XOR clause. –The additive valuation on m/2 red items requires XOR of 2 m/2 atoms.

17. XOR-of-OR v = u 1 XOR u 2 XOR … XOR u k each u i is an OR bid.v = u 1 XOR u 2 XOR … XOR u k each u i is an OR bid. The bidder is willing to obtain the maximal u.The bidder is willing to obtain the maximal u. Monochromatic bid: (OR over all reds) XOR (OR over all blues)Monochromatic bid: (OR over all reds) XOR (OR over all blues) Theorem: Fix K =  m/2. The K-budget valuation requires an  (2 m 1/4 ) size XOR-of-OR expression.Theorem: Fix K =  m/2. The K-budget valuation requires an  (2 m 1/4 ) size XOR-of-OR expression.

18. OR/XOR Formulae Definition: Let v and u be valuations. Then,Definition: Let v and u be valuations. Then, –(v XOR u)(S) = max{v(S),u(S)} –(v OR u)(S) = max{v(R)+u(T)|R  T= , R  T=S} Stronger than OR-of-XOR  XOR-of-OR. E.g.: v =(monochromatic on m/2 items) OR (K-budget on m/2 items)Stronger than OR-of-XOR  XOR-of-OR. E.g.: v =(monochromatic on m/2 items) OR (K-budget on m/2 items)

19. OR Bids with Phantom Items OR* bids (Fujishima et al.): Each bidder submits an OR bid whose atoms (S,p) may introduce new (phantom) items.OR* bids (Fujishima et al.): Each bidder submits an OR bid whose atoms (S,p) may introduce new (phantom) items. Phantom items are used to express constraints, e.g.: (S 1,p 1 ) XOR (S 2,p 2 ) = (S 1  {g},p 1 ) OR (S 2  {g},p 2 )Phantom items are used to express constraints, e.g.: (S 1,p 1 ) XOR (S 2,p 2 ) = (S 1  {g},p 1 ) OR (S 2  {g},p 2 ) Theorem: Any OR/XOR formula of size s can be rewritten as an OR* formula of size s and O(s 2 ) phantom items.Theorem: Any OR/XOR formula of size s can be rewritten as an OR* formula of size s and O(s 2 ) phantom items.

20. The OR* Language Theorem: The majority valuation requires at least ( m m/2 ) atoms in the OR* language.Theorem: The majority valuation requires at least ( m m/2 ) atoms in the OR* language. Proof:Proof: –(S,p) with p > 0 must have at least m/2 real items. –Every subset of m/2 real items must appear as an atom (possibly with phantom items). Open problem: Is OR* strictly stronger than OR/XOR?Open problem: Is OR* strictly stronger than OR/XOR? OR* can express externalities.OR* can express externalities.

21. Bidding and Computability Definition 1: A bidding language is polynomially interpretable if there exists a polynomial algorithm receiving a bid b in the language and a subset S as input, and outputs b (S ).Definition 1: A bidding language is polynomially interpretable if there exists a polynomial algorithm receiving a bid b in the language and a subset S as input, and outputs b (S ). Only Atomic and XOR are polynomially interpretable.Only Atomic and XOR are polynomially interpretable. Definition 2: A bidding language is polynomially verifiable if there exists an NP algorithm receiving a bid b in the language, a subset S and a proof w of a lower bound on b (S ).Definition 2: A bidding language is polynomially verifiable if there exists an NP algorithm receiving a bid b in the language, a subset S and a proof w of a lower bound on b (S ).

22. Allocation Bids are given in OR*.Bids are given in OR*. Auctioneer can treat them as one OR* bid: {B i } i=1..n, where B i =(S i,p i ) is an atomic bid.Auctioneer can treat them as one OR* bid: {B i } i=1..n, where B i =(S i,p i ) is an atomic bid. Algorithmically – no difference between real and phantom items.Algorithmically – no difference between real and phantom items.

23. SPP - Hardness SPP – Set Packing ProblemSPP – Set Packing Problem Is equivalent to Max-Clique and Max- Independent-Set with weighted vertices.Is equivalent to Max-Clique and Max- Independent-Set with weighted vertices. Is approximable within O(n/log 2 n)Is approximable within O(n/log 2 n) Not approximable within n 1/2-  for any  >0.Not approximable within n 1/2-  for any  >0. Not approximable within n 1-  for any  >0, unless NP=ZPP.Not approximable within n 1-  for any  >0, unless NP=ZPP.

24. Integer Programming Formalization of the allocation problem: Maximize:  i=1…n x i p i Subject to:  i|j  S j x i  1, for each j=1…m x i  {0,1}, for each i=1…nFormalization of the allocation problem: Maximize:  i=1…n x i p i Subject to:  i|j  S j x i  1, for each j=1…m x i  {0,1}, for each i=1…n Relaxation to linear programming: x i  0, for each i=1…nRelaxation to linear programming: x i  0, for each i=1…n

25. Fractional Auctions Example: communication lines for sale B 1 = ({TA-Paris, Paris-NY, P}, 10) B 2 = ({TA-London, London-NY, P}, 10) B 1 is 1/3 winning and B 2 is 2/3 winning. Can use B 1 for a 1/3 of its bandwidth, and B 2 – for 2/3 of its bandwidth.Example: communication lines for sale B 1 = ({TA-Paris, Paris-NY, P}, 10) B 2 = ({TA-London, London-NY, P}, 10) B 1 is 1/3 winning and B 2 is 2/3 winning. Can use B 1 for a 1/3 of its bandwidth, and B 2 – for 2/3 of its bandwidth.

26. Single Item Prices An allocation x 1,…,x n is supported by single item prices y 1,…,y n if:An allocation x 1,…,x n is supported by single item prices y 1,…,y n if: –For every winning bid (x i =1), p i   j  S i y j –For every losing bid (x i =0), p i   j  S i y j The allocation is exactly supported if for every winning bid, p i =  j  S i y jThe allocation is exactly supported if for every winning bid, p i =  j  S i y j If every item belongs to some winning bid, it is a full allocation.If every item belongs to some winning bid, it is a full allocation.

27. Single Item Prices An auction admits single item prices if it has a full allocation supported by single item prices.An auction admits single item prices if it has a full allocation supported by single item prices. Theorem: An auction admits single item prices  The linear program produces {0,1} solutions. Then the supporting prices are the solutions to the dual linear program.Theorem: An auction admits single item prices  The linear program produces {0,1} solutions. Then the supporting prices are the solutions to the dual linear program.

28. Single Item Prices - Example Bidder #1:Bidder #1: Bidder #2:Bidder #2: ({A},5) XOR ({B},6)({A,P},5) OR ({B,P},6) ({B},3) Bidder #1 wins A for 5$. Bidder #2 wins B for 3$.Bidder #1 wins A for 5$. Bidder #2 wins B for 3$. Supporting prices: A = 2$, B = 3$, P = 3$Supporting prices: A = 2$, B = 3$, P = 3$

29. Cases Where LP Relaxation Is Optimal Linear Order Bids The items can be linearly ordered, G = {g 1,…,g m }, such that all bids are for sub-ranges, S = {g k,…,g l }.Linear Order Bids The items can be linearly ordered, G = {g 1,…,g m }, such that all bids are for sub-ranges, S = {g k,…,g l }. –Hierarchical Bids All sets form a nested hierarchy.  Every two bids S, T are either disjoint or one contains the other. OR-of-XORs of Singleton BidsOR-of-XORs of Singleton Bids –Single Item Bids –Downward Sloping Symmetric Bids