Cheap Labor Can Be Expensive Ning Chen, Anna R. Karlin Michael König.
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Transcript of Cheap Labor Can Be Expensive Ning Chen, Anna R. Karlin Michael König.
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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:
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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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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)
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
U
V
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u0
v0
u3
v3
u3’
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uk
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Perfect Bipartite Matching Markets
u2
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u1
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1 1
1 1 1
pM = k
1
pM’ = 1 Cheap labor cost = k = O(|V|)
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ui
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ui’
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Perfect Bipartite Matching Markets
1
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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
/
S
S’
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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
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Thanks for your attention!
Questions?