1. 1 Combinatorial Auctions V.S. Subrahmanian

2. Fall 2002, © V.S. Subrahmanian 2 Motivation Bidders can bid on a set of items. The auctioneer can select a set of bids which maximize his revenue. Of course, he cannot sell the same item to two different people.

3. Fall 2002, © V.S. Subrahmanian 3 Example A,B 500 C 200 A,C 450 B,C 400

4. Fall 2002, © V.S. Subrahmanian 4 Example A,B 500 C 200 A,C 450 B,C 400 A,B 500 C 200 winners

5. Fall 2002, © V.S. Subrahmanian 5 Bids Suppose I is a finite set of items for auction. A bid is a pair (S,p) where – S is a subset of I – p is a price. Example: I = {A,B,C} ({A,B},500) is a bid. ({B,C},450) is a bid.

6. Fall 2002, © V.S. Subrahmanian 6 Succinct Bids A set B of bids is succinct iff there are no two bids (s,p1),(s,p2) in B. If B is not succinct you can make it succinct pretty easily. Why ? E.g. if B = { …,({A,B},500),({A,B},480),…} you can delete the bid ({A,B},480). We henceforth assume all sets of bids are succinct.

7. Fall 2002, © V.S. Subrahmanian 7 Labelings Suppose B is a finite (succinct) set of bids. A labeling is a mapping L from B to {0,1} such that: – If L(S1,p1)=1 and L(S2,p2)=1, then S1 and S2 are disjoint.

8. Fall 2002, © V.S. Subrahmanian 8 Example labeling A,B 500 C 200 A,C 450 B,C 400 1001

9. Fall 2002, © V.S. Subrahmanian 9 Example labeling A,B 500 C 200 A,C 450 B,C 400 0100

10. Fall 2002, © V.S. Subrahmanian 10 Value of a Labeling Suppose I is a set of items and B is a succinct set of bids. The value v(L) of labeling L is defined as SUM (S,P) in B and L(S,p)=1 p. Intuitively, the value of a labeling determines the total amount of money the auctioneer takes in from the auction.

11. Fall 2002, © V.S. Subrahmanian 11 Example labeling A,B 500 C 200 A,C 450 B,C 400 1001 Value = 700

12. Fall 2002, © V.S. Subrahmanian 12 Example labeling A,B 500 C 200 A,C 450 B,C 400 0100 Value = 400

13. Fall 2002, © V.S. Subrahmanian 13 Winner determination problem (WDP) Suppose I is a set of items and B is a succinct set of bids. A labeling L is a winner iff there is no other labeling L’ such that v(L’) > v(L). WDP:Find a winner ! WDP is NP-complete.

14. Fall 2002, © V.S. Subrahmanian 14 Winner computation How do you find a winner ? Construct a tree. Each node in the tree contains a subset of B such that no two bids in B conflict. Root contains the empty set of bids. If a node N has b1 as its set of bids, and if b is a bid in B – b1 which does not conflict with any bid in b1, then N has a child labeled b1 U {b}. Thus, node N may potentially have lots of children. We would like to find a leaf node in this tree such that the value of that leaf node is maximal.

15. Fall 2002, © V.S. Subrahmanian 15 Example ({A,B},500)({C},200)({A,C},450}({B,C},400) ({},0)

16. Fall 2002, © V.S. Subrahmanian 16 Example Tree ({A,B},500)({C},200)({A,C},450}({B,C},400) ({},0) ({C},200)({A,B},500) (NOTE: label of a node is union of all labels on the path to that node).

17. Fall 2002, © V.S. Subrahmanian 17 Winner computation Clearly, such a tree is clearly HUGE. We do not want to construct this whole tree. Instead, we would like to find a winner by pruning this tree. How can we prune this tree?

18. Fall 2002, © V.S. Subrahmanian 18 Winner computation Associate with each node, two values. G(N): total revenues associated with the bids labeling the node. H(N): an overestimate of the cost of a solution reachable from that node.

19. Fall 2002, © V.S. Subrahmanian 19 Example Tree: G(N) ({A,B},500)({C},200)({A,C},450}({B,C},400) ({},0) ({C},200)({A,B},500) 500 700

20. Fall 2002, © V.S. Subrahmanian 20 How to compute H(N) Let bids(N) be the set of bids labeling N. Associate with each item i not contained in any bid in bids(N), a cost C(i). Set H(N) to be the sum of all the C(i) such that i is not contained in any bid in bids(N),

21. Fall 2002, © V.S. Subrahmanian 21 How to compute C(i) Find each item i not contained in any bid in bids(N).. For each such i, find each bid b in B-bids(N) which contains I and which does not conflict with bids(N). Let C(i) = MAX { avgcost(b) | b in B-bids(N) and b contains i and b does not conflict with bids(N) }.

22. Fall 2002, © V.S. Subrahmanian 22 Example Tree: H(N) ({A,B},500)({C},200)({A,C},450}({B,C},400) ({},0) ({C},200)({A,B},500)

23. Fall 2002, © V.S. Subrahmanian 23 Example: H(N) ({C},200) does not contain both A,B. Bids containing A: ({A,B},500), ({A,C},450). ({A,C},450) conflicts with ({C},200). Cost(A)=250. Bids containing B: ({A,B},500),({B,C},400). ({B,C},400) conflicts with ({C},200). Cost(B) = 250. H(N) = 500

24. Fall 2002, © V.S. Subrahmanian 24 Search Strategy Maintain – OPEN list (nodes not expanded yet) in descending order of G(N)+H(N). – BestSol – BestVal. If BestVal > G(N)+H(N), then prune N from search tree.

25. Fall 2002, © V.S. Subrahmanian 25 Example Tree: G(N) + H(N) ({A,B},500)({C},200)({A,C},450}({B,C},400) ({},0) ({C},200)({A,B},500) 725650700

26. Fall 2002, © V.S. Subrahmanian 26 Example Tree: G(N) + H(N) ({A,B},500)({C},200)({A,C},450}({B,C},400) ({},0) ({C},200) 725650700 BestSol BestVal = 700 As no other node has G(N)+H(N) > BestVal, STOP

27. Fall 2002, © V.S. Subrahmanian 27 Conclusions Many improvements possible. But this algorithm will product an optimal solution.