Cheap Labor Can Be Expensive

Ning Chen, Anna R. Karlin

Michael König

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The Problem

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“Nash Equilibrium”

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The Problem

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The Problem

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Total price: 5

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The Problem

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Total price: 10

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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

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Feasible Sets

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Total price: 10Total price: 5

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Cheap Labor Cost

Cheap Labor Cost of this market is

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Cheap Labor Cost

• pM := total price of M

• Cheap labor cost for a market M:

Questions up to this point?

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GreedyAlg

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1. Find the cheapest feasible set S F with ∈respect to costs

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GreedyAlg

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2. For each e E, initialize b(e) to c(e)∈

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GreedyAlg

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3. For each e S:∈- Raise b(e) until there is S’ F∈

such that e S’ and b(S) = b(S’)∈

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/

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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/

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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.

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Upper Bound

• The cheap labor cost of any market is at most |S|, where S F is a feasible set with ∈minimum total cost

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Here: |S| = 3

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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)

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Proof of Upper Bound

Case 1: e S’ \ S∈• b(e) = c(e) [GreedyAlg] 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)

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S

S’Proof of Upper BoundCase 2: e 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’’)= b(S’’ \ S) + b(S ∩ S’’) [GreedyAlg]= b(S’’) [contradiction: S is the winning set]

/

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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’)

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Perfect Bipartite Matching Markets

• Customer wants to buy edges to obtain a perfect matching in a bipartite graph

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Perfect Bipartite Matching Markets

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pM’ = 1 Cheap labor cost = k = O(|V|)

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Perfect Bipartite Matching Markets

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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

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Path Markets

• Purchasing an s-t path in a directed graph

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Path Markets

• Observation: There are always at least 2 edge-disjoint paths P1 and P2 with b(P1) = b(P2) = b(P), where P is the winning path.

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• Proof idea:– 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.

Path Markets

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Path Markets

• Proposition:– Pick the two cheapest paths by cost, P1 and P2.– asdf

(pG := total price of G)

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Path Markets

• Now observe that for the two cheapest paths by cost, P1 and P2, c(P1) + c(P2) gives an upper bound for pG.

• Thus, pG ≤ c(P1) + c(P2) ≤ 2 max{c(P⋅ 1), c(P2)} = 2 p⋅ G*

The cheap labor cost for path markets is at most 2.

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Path Markets

• This bound is tight:

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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

Thanks for your attention!

Questions?