Cheap Labor Can Be Expensive Ning Chen, Anna R. Karlin Michael Knig - PowerPoint PPT Presentation

Cheap Labor Can Be Expensive Ning Chen, Anna R. Karlin Michael König

The Problem 5 4.5 4 2 4 4 2 4 4 s t 2 3 5 2 4 4 3 7 “Nash Equilibrium” 10 7 2

The Problem 0 6 0 1 s t 2 6 5 6 7 100 6 3

The Problem 0 5 0 5 s t 5 0 5 0 Total price: 5 4

The Problem 3 5 0 5 0 5 s t 0 0 0 0 5 0 5 0 5 Total price: 10 5

Markets • Set of agents “E” • Each agent e ∈ E has a cost c(e) and bid b(e) • Customer wants to hire a team of agents • Feasible sets “F” are teams of agents capable of getting the job done 6

Feasible Sets 4 2 4 s t 2 3 7 7

Cheap Labor Cost 5 0 0 0 0 5 5 5 5 s s s s t t t t 0 0 0 0 0 5 5 5 5 0 0 0 0 5 1000 Total price: 10 Total price: 5  Cheap Labor Cost of this market is 8

Cheap Labor Cost • Cheap labor cost for a market M: • p M := total price of M 9

Questions up to this point?

GreedyAlg 1. Find the cheapest feasible set S ∈ F with respect to costs 4 2 4 s t 2 3 7 11

GreedyAlg For each e ∈ E, initialize b(e) to c(e) 2. 4 2 4 2 4 4 s t 2 3 2 3 7 7 12

GreedyAlg For each e ∈ S: 3. - Raise b(e) until there is S’ ∈ F / such that e ∈ S’ and b(S) = b(S’) 4 2 4 2 4 4 s t 2 3 3 2 3 7 7 13

GreedyAlg 1. Find the cheapest feasible set S ∈ F with respect to costs 2. For each e ∈ E, initialize b(e) to c(e) 3. For each e ∈ S: - Raise b(e) until there is S’ ∈ F / such that e ∈ S’ and b(S) = b(S’) 4. Output the bids b and the winning set S 14

Tight sets • For any NE b with winning set S: – For any e ∈ S, there is another winning feasible / set S’ ∈ F with e ∈ S’ and b(S) = b(S’) – These feasible sets are called tight sets . 4 2 4 s t 2 3 15 7

Upper Bound • The cheap labor cost of any market is at most |S|, where S ∈ F is a feasible set with minimum total cost 0 5 s t 0 0 5 0 Here: |S| = 3 16

Proof of Upper Bound • It suffices to show: – For any market M, NE b with winning set S, for any submarket M’,best NE b’ with winning set S’ b(S) ≤ |S| ⋅ b’(S’) (we choose b and S to be computed by GreedyAlg ) 17

Proof of Upper Bound S’ Case 1: e ∈ S’ \ S S [ GreedyAlg ] • b(e) = c(e)  b(S’ \ S) = c(S’ \ S) • b(S \ S’) ≤ b(S’ \ S) [S is the winning set] • c(S’ \ S) ≤ b’(S’ \ S) [bid behavior]  b(S \ S’) ≤ b’(S’ \ S) 18

Proof of Upper Bound S’ Case 2: e ∈ S’ ∩ S S For each such e there exists a tight set S’’ ( ∈ F’) • / such that e ∈ S ’’ and b’(S’) = b’(S’’). • We claim b(e) ≤ b’(S’). Otherwise: b(S) = b(S \ S’’) + b(S ∩ S’’) > b’(S’) + b(S ∩ S’’) [reverse claim] = b’(S’’) + b(S ∩ S’’) *b’(S’) = b’(S’’)+ ≥ c(S’’) + b(S ∩ S’’) [bid behavior] ≥ c(S’’ \ S) + b(S ∩ S’’) [ GreedyAlg ] = b(S’’ \ S) + b(S ∩ S’’) = b(S’’) [contradiction: S is the winning set] 19

Proof of Upper Bound Case 1 (e ∈ S’ \ S): b(S \ S’) ≤ b ’(S’ \ S) Case 2 (e ∈ S’ ∩ S): b(e) ≤ b’(S’) Putting the cases together: b(S) = b(S \ S’) + b(S ∩ S’) ≤ b’(S’ \ S) + |S ∩ S’| ⋅ b’(S’) ≤ |S| ⋅ b’(S’) 20

Perfect Bipartite Matching Markets • Customer wants to buy edges to obtain a perfect matching in a bipartite graph U V 21

Perfect Bipartite Matching Markets u 0 u 1 u 1’ u 2 u 2’ u 3 u 3’ u k u k’ 1 1 1 ... 1 1 1 1 v 0 v 1 v 1’ v 2 v 2’ v 3 v k v 3’ v k’ p M = k p M’ = 1 Cheap labor cost = k = O(|V|) 22

Perfect Bipartite Matching Markets u 0 u i u i’ 1 v 0 v i v i’ 23

Matroid Markets • Agents and feasible sets form a matroid (E, F) (F ⊆ P (E) with a bunch of special rules) • Cheap labor cost is always 1. • Natural Occurrence: buying spanning trees 24

Path Markets • Purchasing an s-t path in a directed graph 4 2 4 s t 2 3 7 25

Path Markets • Observation: There are always at least 2 edge-disjoint paths P 1 and P 2 with b(P 1 ) = b(P 2 ) = b(P), where P is the winning path. 4 2 4 2 5 5 0 5 4 4 s s t t 0 0 2 3 0 0 2 4 5 0 5 5 7 26 7

Path Markets S’ • Proof idea: S – There always are “tight paths” (tight sets) For any e ∈ S, there is another winning feasible / set S’ ∈ F with e ∈ S’ and b(S) = b(S’) – Any prefix of a tight path is optimal (otherwise the winning path would not be winning). – The union of all tight paths only contains optimal s-t paths and is two-connected. 27

Path Markets • Proposition: – Pick the two cheapest paths by cost, P 1 and P 2 . – asdf (p G := total price of G) 4 2 4 s t 2 3 28 7

Path Markets • Now observe that for the two cheapest paths by cost, P1 and P2, c(P 1 ) + c(P 2 ) gives an upper bound for p G . • Thus, p G ≤ c(P 1 ) + c(P 2 ) ≤ 2 ⋅ max{c(P 1 ), c(P 2 )} = 2 ⋅ p G*  The cheap labor cost for path markets is at most 2. 4 2 4 s t 2 3 30 7

Path Markets • This bound is tight: 0 5 s t 0 0 5 0 31

Conclusion • Short paper stuffed with proofs • Exhaustive study of “cheap labor cost” for non-cooperative markets – General upper bound |S| – Values for common market types 32

Thanks for your attention! Questions?