Combinatorial Auctions: A Survey

1. Combinatorial Auctions: A Survey Sven de Vries & Rakesh Vohra (2000)

2. Contents • Introduction • CAP • Decentralized Methods

3. Introduction(1) • Complimentarities between different assets • Bidders have preferences not just for particular items but for sets of bundels of items • Traveling to LA • (restaurants and hotels for the intermediate cities, car) or (airline ticket, taxi) • Auctions where bidders submit bids on combinations : recently been aroused • Jackson(1976),Caplice(1996),Rothkopf(1998),Fujishima(1999),Sandholm(1999) • Increases in computing power

4. Introduction(2) • Tools • ‘SBIDS’ by SAITECH-INC • ‘OptiBid’ by Logistics.com • Combinatorial Auction Problem (CAP) • Selecting the winning set of bids. • Can be formulated as an Integer Program

5. Introduction • CAP • Decentralized Methods

6. CAP • CAP • SPP • Solvable Instances of SPP • Exact Methods • Approximate Methods

7. CAP(1) CAP (Combinatorial Auction Problem) -Selecting the winning set of bids- Difficulty Resolution • Each bidder must submit a bid for every subset of objects he is interested in • How to transmit this bidding function in a succinct way to the auctioneer • To restrict the kinds of combinations that bidders may bid on • How to decide which collection of bids to accept - Solving CAP

8. CAP(2) CAP (Combinatorial Auction Problem) -Selecting the winning set of bids- Difficulty Resolution • Each bidder must submit a bid for every subset of objects he is interested in • How to transmit this bidding function in a succinct way to the auctioneer • To restrict the kinds of combinations that bidders may bid on • How to decide which collection of bids to accept - Solving CAP

9. CAP(3) • Notations • N : the set of bidders • M : the set of m distinct objects • S : subset of M • bj(S) : the bid that agent j in N has announced he is willing to pay for S

10. CAP(4) • CAP formula :

11. CAP(4) • CAP formula : • x(S) = 1 : the highest bid on the set S is to be accepted 0 : no bid on the set S are accepted

12. CAP(4) • CAP formula : • : no object in M is assigned to more than one bidder

13. CAP(4) • CAP formula : • Call this formulation CAP1

14. CAP(5) • Superadditive : • for all j∈N and A,B⊂M such that • CAP1 correctly models CAP when the bid functions bj are all superadditive • The goods complement each other. • When goods are substitutes, CAP1 is incorrect. • Why ? • Superadditive formula doesn’t hold for some j,A,B. • An optimal solution to CAP1 may assign A,B to bidder j and incorrectly record a revenue of bj(A)+bj(B) rather than

15. CAP(6) • How to obviate this difficulty ? • Through the introduction of dummy good g • bj(A) => bj(A∪{g}) bj(B) => bj(B∪{g}) bj(A∪B) remains the same M => M∪{g} • If A is assigned to j, then B cannot be assigned to j. • Through the formula CAP2

16. CAP(7) • CAP2 formulation • CAP1 formulation

17. CAP(8) • CAP2 formulation No bidder receives more than one subset

18. CAP(9) • CAP2 formulation Overlapping sets of goods are never assigned

19. CAP(10) • Assumption of CAP1,CAP2 • There is at most one copy of each object. • Extending the formulation • The case when there are multiple copies of the same object and each bidder wants at most one copy of each object : • The right hand sides of the contraints in CAP1, CAP2 take on values larger than 1. • The case when there are multiple copies and the bidder may want more than one copy of the same object : • Multi-unit combinatorial auctions (Leyton-Brown 2000)

20. CAP • CAP • SPP • Solvable Instances of SPP • Exact Methods • Approximate Methods

21. SPP(1) • Set Packing Problem • Given a ground set M of elements and a collection V of subsets with non-negative weights, find the largest weight collection of subsets that are pairwise disjoint.

22. SPP(2) • Set Packing Problem • Given a ground set M of elements and a collection V of subsets with non-negative weights, find the largest weight collection of subsets that are pairwise disjoint. • Notation • x(j) = 1 if the j-th set in V with weight c(j) is selected 0 otherwise • a(i,j) = 1 if the j-th set in V contains element i∈M 0 otherwise

23. SPP(3) • Notation • x(j) = 1 if the j-th set in V with weight c(j) is selected 0 otherwise • a(i,j) = 1 if the j-th set in V contains element i∈M 0 otherwise • SPP Formulation

24. SPP(3) • Notation • x(j) = 1 if the j-th set in V with weight c(j) is selected 0 otherwise • a(i,j) = 1 if the j-th set in V contains element i∈M 0 otherwise • SPP Formulation • CAP Formulation

25. SPP(3) • Notation • x(j) = 1 if the j-th set in V with weight c(j) is selected 0 otherwise • a(i,j) = 1 if the j-th set in V contains element i∈M 0 otherwise • SPP Formulation • CAP Formulation

26. SPP(4) Other related Prolems Set Partitioning Problem (SPA) Set Covering Problem (SCP)

27. SPP(5) Set Partitioning Problem (SPA) • Bidders are sellers (rather than buyers). • Trucking companies bidding for the opportunity to ship goods from a particular warehouse to retail outlet.

28. SPP(6) Set Covering Problem (SCP) • Auction problems in procurement rather than selling terms. • Scheduling of crews for railways.

29. Complexity of SPP • No polynomial time algorithm for SPP is known. • Any algorithm for the CAP that uses directly the bids for the sets, must scan the bids and the number of such bids could be exponential in |M|. • |M| : the number of variables => |V| : the number of solutions to check = 2|M| • SPP : NP-hard (NP-complete) • Effective solution procedures for CAP • The number of distinct bids is not large • Be structured in computationally useful ways.

30. CAP • CAP • SPP • Solvable Instances of SPP • Exact Methods • Approximate Methods

31. Solvable Instances of SPP • Total Unimodularity • Balanced Matrices • Perfect Matrices • Graph Theoretic Methods • Using Preferences

32. Solvable Instances of SPP • Usual way in which instances SPP can be solved by a polynomial algorithm • When the extreme points of the polyhedron are all integral, i.e. 0-1. • In these cases, we can simply drop the integrality requirement from the SPP and solve it as a linear program • A polyhedron with all integral extreme points is called integral.

33. Total Unimodularity(TU) (1) • A matrix is TU if the determinant of every square submatrix is 0,1 or –1. • A : TU  At : TU • If A={a(i,j)}i∈M,j∈V is TU, then all extreme point of the polyhedron P(A) are integral. • There is a polynomial time algorithm to decide whether a matrix is TU.

34. Total Unimodularity(TU) (2) • Theorem 2.1) Let B be a matrix each of whose entries is 0,1 or -1. Suppose each subset S of columns of B can be divided into two sets L and R such that then B is TU. The converse is also true. • Theorem 2.2) All 0-1 matrices with the consecutive ones property are TU. • A 0-1 matrix has the consecutive ones property if the non-zero entries in each column occur consecutively.

35. Total Unimodularity(TU) (3) • For example, • Objects to be auctioned : parcels of land along a shore line • Shore line is important : it imposes a linear order on the parcels • Make a restriction to bid only contiguous parcels • The most interesting combinations would be contiguous, in the bidders eyes. • Two computational consequences. • Number of distinct bids would be limited by a polynomial in the number of objects. • The constraint matrix A of the CAP would have the consecutive ones property in the columns.

36. Balanced Matrices(1) • A 0-1 matrix B is balanced if it has no square submatrix of odd order with exactly two 1’s in each row and column. • Theorem 2.3) Let B be a balanced 0-1 matrix. Then the following linear program : has an integral optimal solution whenever the c(j)’s are integral.

37. Balanced Matrices(2) • For example, • Consider a tree T with a distance function d. • v : vertex of T • N(v,r) : set of all vertices in T that are within distance r of v. • The vertices represent parcels of land connected by a read network with no cycles. • Bidders bid for subsets of parcels which is to be of the form N(v,r). • Row of the constraint matrix : for each vertex Column : for each set of the form N(v,r) • This constraint matrix is balanced.

38. Perfect Matrices • If the contraints matrix A can be identified with the vertex-clique adjacency matrix of what is known as a perfect graph, then SPP can be solved in polynomial time. • A simple graph G is perfect if, for every induced subgraph H of G, the number of vertices in a maximum clique is • , the chromatic number of H, is the minumum k for which H is k-colorable.

39. Graph Theoretic Methods • There are situations where P(A) is not integral yet the SPP can be solved in polynomial time because the contraint matrix A admits a graph theoretic interpretation in terms of an easy problem. • When each column of the matrix A contains at most two 1’s. => maximum weight matching problem (can be solved in polynomial time) • At most two 1’s per row of A => NP-hard • When A has the circular ones property. • A 0-1 has the circular ones property if the non-zero entries in each column (row) are consecutive • First and last entries in each column (row) are treated consecutive • Note the resemblance to the consecutive ones property

40. Graph Theoretic Methods • There are situations where P(A) is not integral yet the SPP can be solved in polynomial time because the contraint matrix A admits a graph theoretic interpretation in terms of an easy problem. • When each column of the matrix A contains at most two 1’s. => maximum weight matching problem (can be solved in polynomial time) • At most two 1’s per row of A => NP-hard • When A has the circular ones property. => A can be identified with the vertex-clique adjacency matrix of a circular arc graph. => maximum weight independent set problem for a circular arc graph. (can be solved in poly time)

41. Using Preferences(1) • Restrictions in the preference orderings of the bidders • Suppose that bidders come in two types • Type one : bj(.) = g1(.) • Type two : bj(.) = g2(.) where g1 andg2 are non-decreasing integer valued supermodular functions The dual of CAP2 is :

42. Using Preferences(1) • Restrictions in the preference orderings of the bidders • Suppose that bidders come in two types • Type one : bj(.) = g1(.) • Type two : bj(.) = g2(.) where g1 andg2 are non-decreasing integer valued supermodular functions The dual of CAP2 is : This Problem is an instance of the polymatroid intersection problem. (polynomially solvable)

43. Using Preferences(1) • Restrictions in the preference orderings of the bidders • Suppose that bidders come in two types • Type one : bj(.) = g1(.) • Type two : bj(.) = g2(.) where g1 andg2 are non-decreasing integer valued supermodular functions • Using the method to solve problems with three or more types of bidders is not possible. • It is known in those cases that the dual problem above admits fractional extreme points. • The problem of finding an in integer optimal solution for the intersection of three or more polymatroids is NP-hard.

44. Using Preferences(2) • Restrictions in the preference orderings of the bidders • When each of the bj(.) have the gross substitutes property, CAP2 reduces to a sequence of matroid partition problems, each of which can be solved in polynomial time.

45. CAP • CAP • SPP • Solvable Instances of SPP • Exact Methods • Approximate Methods

46. Exact Methods(1) • The upper bound on the optimal solution value is obtained by solving a relaxation of the optimization problem. • Replace the given problem by one with a larger feasible region that is more easily solved. • Lagrangean relaxation • Will be discussed later • Linear programming relaxation • Only the integrality constraints are relaxed

47. Exact Methods(2) • Exact methods • Branch and bound • Cutting planes • Hybrid called branch and cut

48. Exact Methods(2) • Exact methods • Branch and bound • At each stage, after solving the LP, a fractional variable xj is selected and two subproblems are set up, one where xj=1 and the other where xj=0. (Branch) • Solve the LP relaxation of the two subproblems. • From each subproblem with a nonintegral solution we branch again to generate two subproblems and so on. • By comparing the LP bound across nodes in different branches of the tree, one can prune some branches in advance. (Bound) • Cutting planes • Hybrid called branch and cut

49. Exact Methods(3) • Exact methods • Branch and bound • Cutting planes • Find linear inequalities (cuts) that are violated by a solution of a given relaxation but are satisfied by all feasible zero-one solution. • If one adds enough cuts, one is left with integral extreme points. • Hybrid called branch and cut

50. Exact Methods(4) • Exact methods • Branch and bound • Cutting planes • Hybrid called branch and cut • Works like branch and bound, but tightens the bounds in every node of the tree by adding cuts. • Since even small instances of the CAP1 may involve a huge number of columns (bids), this method needs to be augmented with another method known as column generation. (It works by generating a column when needed rather than all at once.)