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Mechanism Design without Money

Lecture 4

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Price of Anarchy simplest example

• G is given• Route 1 unit from A to B, through AB,AXB • OPT– route ½ on AB and ½ on AXB (check!)• NASH – route 1 on AXB• Ratio is 4/3

BA

X0

1

x

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Price of Anarchy: general functions

• G is given• Route 1 unit from A to B, through AB,AXB • OPT – choose y to minimize yC (y) + (1 – y)C (1)• Nash – route 1 on AXB• Ratio is • Also called the Pigou Bound

BA

X0

c(1)

c(x)

𝑐 (1)𝑦 ∙𝑐 (𝑦 )+(1−𝑦 ) ∙𝑐(1)

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Can things be worse?

• No!• For a set of cpu functions C, define

• In our examples before r was 1• Theorem: For any routing game G, if the cost per

unit functions come from C, the Price of Anarchy is at most a(C)

𝑎 (𝐶 )=𝑚𝑎𝑥𝑐𝑝𝑢∈𝐶, 𝑦 ,𝑟

𝑟 ∙𝑐𝑝𝑢(𝑟 )𝑦 ∙𝑐𝑝𝑢 (𝑦 )+(𝑟 − 𝑦 )∙𝑐𝑝𝑢(𝑟 )

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Another property of Nash flows

• Theorem: Let f be a Nash flow. For any other flow f* which routes the same amount, we have

• Note: The cost per unit of every edge is constant, and we just want to route the flow.

∑𝑒

𝑓 𝑒𝑐𝑝𝑢𝑒( 𝑓 𝑒)≤∑𝑒

𝑓 ∗𝑒❑𝑐𝑝𝑢𝑒( 𝑓 𝑒)

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Proof

• Define

Note that

• Therefore, we need to prove that H(f*,f) ≥ H(f,f)

• This follows by using that if a path has any flow in it in a Nash flow, its cost is minimal

𝐻 ( 𝑓 ∗ , 𝑓 )=∑𝑃

𝑓 ∗𝑃𝑐𝑝𝑢𝑃( 𝑓 )

=

∑𝑒

𝑓 𝑒𝑐𝑝𝑢𝑒( 𝑓 𝑒)≤∑𝑒

𝑓 ∗𝑒❑𝑐𝑝𝑢𝑒( 𝑓 𝑒)

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Proof that a(C) is the bound on PoA

Let C be the set of functions on the edges. Let f* be the optimal solution, and f be the Nash.

And in particular setting r=fe and y=f*e

𝑐𝑜𝑠𝑡 𝑜𝑓 𝑓𝑙𝑜𝑤 𝑓 ∗=¿=

≥𝑐𝑜𝑠𝑡𝑜𝑓 𝑓𝑙𝑜𝑤 𝑓

𝑎(𝐶)

𝑎 (𝐶 )=𝑚𝑎𝑥𝑐𝑝𝑢∈𝐶, 𝑦 ,𝑟

𝑟 ∙𝑐𝑝𝑢(𝑟 )𝑦 ∙𝑐𝑝𝑢 (𝑦 )+(𝑟 − 𝑦 )∙𝑐𝑝𝑢(𝑟 )

- )

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How big can a(C) be?

• Theorem: If C is a set of affine functions, a(C) is at most 4/3

• Proof: Do this at home. Hint: compute the derivative. You should get x = r/2

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Fighting selfishness

• Braess paradox shows that selfish agents can improve their situation if an edge is removed from the graph

• Given a graph, which edges should be removed?

• We are looking at

𝐶𝑜𝑠𝑡𝑜𝑓 𝑤𝑜𝑟𝑠𝑡 𝑁𝑎𝑠h 𝑓𝑜𝑟 𝐺𝑐𝑜𝑠𝑡 𝑜𝑓 𝑤𝑜𝑟𝑠𝑡 𝑁𝑎𝑠h 𝑓𝑜𝑟 𝑎𝑠𝑢𝑏𝑔𝑟𝑎𝑝h𝑜𝑓 𝐺

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Fighting selfishness is (computationally) hard

• Problem: Which edges should I cut to improve the worst Nash?

• It is trivial to get a 4/3 approximation – just cut nothing. The worst Nash for a subgraph is always worse than OPT for the original graph, and the PoA with linear cost functions is 4/3

• Thm: It is NP hard to approximate better than 4/3

𝐶𝑜𝑠𝑡𝑜𝑓 𝑤𝑜𝑟𝑠𝑡 𝑁𝑎𝑠h 𝑓𝑜𝑟 𝐺𝑐𝑜𝑠𝑡 𝑜𝑓 𝑤𝑜𝑟𝑠𝑡 𝑁𝑎𝑠h 𝑓𝑜𝑟 𝑎𝑠𝑢𝑏𝑔𝑟𝑎𝑝h𝑜𝑓 𝐺

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Proof• Reduction from 2DPP: Given a graph G, two sources s1,s2

and two targets t1,t2 are there two vertex disjoint paths s1t1 and s2t2

• If there are no two disjoint paths you will always have a path s2t1

s

t2

s1

s2

t1

t

G

x

x1

1

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What happens for non linear cost functions?

• We said price of anarchy can grow, but what about fighting selfishness?

• Thm: There exists a graph with n vertices and non linear cost functions, such that removing edges improves the worst Nash by a factor of n/2

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The Graph

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Bad Nash flow

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After edge removal

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Atomic flows

• Multiple equilibria (remember the examples)• Sometimes there is no pure equilibrium• Weaker bounds, different techniques

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No pure Nash

• P1 routes 1 unit from s to t

• P2 routes 2 units from s to t

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Price of anarchy example

• Not all paths in the equilibrium have the same cost• In the example: PoA of 5/2• This is the worst case for

affine functions if allplayers have the sameamount of flow

• We will prove a weaker boundwhen players control different amounts of flow

U V

W

0

0

x

x xx

s1,s2

t2,t3s4

t1,s3 ,t4

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Atomic flow for affine functions

• Edge e has CPU aex + be

• Player i sends ri

• Let f be Nash, f* be OPT. We have for every player i:

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Summing over the players:

𝑟 𝑖 ∑𝑒∈ 𝑃 𝑖

𝑎𝑒 𝑓 𝑒+𝑏𝑒≤𝑟 𝑖 ∑𝑒∈ 𝑃∗

𝑖❑

𝑎𝑒 ( 𝑓 𝑒+𝑟 𝑖 )+𝑏𝑒

𝑐𝑜𝑠𝑡 ( 𝑓 )≤𝑐𝑜𝑠𝑡 ( 𝑓 ∗)+∑𝑒

𝑎𝑒 𝑓 𝑒 𝑓 𝑒∗

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Manipulations

• You get

• Solving x2-3x+10 gives (3 + 51/2)/2

𝐶 ( 𝑓 )𝐶 ( 𝑓 ∗)

−1≤ √ 𝐶 ( 𝑓 )𝐶 ( 𝑓 ∗)

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Questions?

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Extra Slides

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Chicken

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Road example

A B

1 hour

1 hour

N minutes

N minutes

• 50 people want to get from A to B• There are two roads, each one has two segments. One takes

an hour, and the other one takes the number of people on it

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Nash in road example

• In the Nash equilibrium, 25 people would take each route, for a travel time of 85 minutes

A B

1 hour

1 hour

N minutes

N minutes

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Braess’ paradox

• Now suppose someone adds an extra road which takes no time at all. Travel time goes to 100 minutes

A B

1 hour

1 hour

N minutes

N minutes

Free