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

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

Jan 12, 2024 133 likes •454 views

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

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Cheap Labor Can Be Expensive

Ning Chen, Anna R. Karlin Michael König

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2

4 4.5 4 2 7 2 5

The Problem

s t 7 2 3 4 2 4 10 3 5 4 4 4

“Nash Equilibrium”

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3

The Problem

s t 2 6 1 5 6 6 6 7 100

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4

The Problem

s t 5 5 5 5 Total price: 5

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5

The Problem

s t 5 5 5 5 5 5 Total price: 10 3

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6

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
  • f getting the job done
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7

Feasible Sets

s t 7 2 3 4 2 4

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Total price: 10 Total price: 5 s t 5 5 s t 5 s t 5 5 s t 5 5

Cheap Labor Cost

 Cheap Labor Cost of this market is 5 5 1000 5

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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

s t 7 2 3 4 2 4

  • 1. Find the cheapest feasible set S ∈ F with

respect to costs

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GreedyAlg

s t 7 2 3 4 2 4

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

2 7 3 2 4 4

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3

GreedyAlg

s t 7 2 3 4 2 4

3. For each e ∈ S:

  • Raise b(e) until there is S’ ∈ F

such that e ∈ S’ and b(S) = b(S’)

2 7 3 2 4 4

/

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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.

s t 7 2 3 4 2 4

/

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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

s t 5 5

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 Bound

Case 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’ v3’ uk vk uk’ vk’

Perfect Bipartite Matching Markets

u2 v2 u2’ v2’

...

u1 v1 u1’ v1’

1 1 1

1 1 1

pM = k

1

pM’ = 1 Cheap labor cost = k = O(|V|)

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ui vi ui’ vi’ u0 v0

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

s t 7 2 3 4 2 4

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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.

s t 7 2 3 4 2 4 2 7 4 2 4 4 s t 5 5 5 5 5 5

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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)

s t 7 2 3 4 2 4

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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(P1), c(P2)} = 2⋅pG*

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

s t 7 2 3 4 2 4

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Path Markets

  • This bound is tight:

s t 5 5

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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?