Inoculation strategies for victims of viruses

Copyright (C) 2005 by Aleksandr Yampolskiy

Inoculation Strategies for Victims of Viruses and the Sum-of-Squares Partition Problem

James Aspnes, Kevin Chang, and Aleksandr Yampolskiy

(Yale University)


Outline

ØMotivationnOur ModelnNash StrategiesnOptimal StrategiesnSum-of-Squares Partition ProblemnConclusion


Question: Will you install anti-virus software?

Norton AntiVirus 2005 = $49.95

Value of your data = $350.00

Infection probability = 1/10

Expected loss = $35.00


Answer: Probably not.

Norton AntiVirus 2005 = $49.95

Value of your data = $350.00

Infection probability = 1/10

Expected loss = $35.00


This selfish behavior…n …fails to achieve the social optimum.


What if instead…n …a benevolent dictator decided which

computers install an anti-virus?

Center node must install an anti-virus

or else!


Outline

nMotivationØOur ModelnNash StrategiesnOptimal StrategiesnSum-of-Squares Partition ProblemnConclusion


Our Model

n The network is an undirected graph G = (V,E).

n Installing anti-virus software is a single round non-cooperative game.

n The players are the network nodes: V = {0,1,…,n-1}.


Our Model : Strategies

n Each node has two actions: do nothing or inoculate itself.

n Strategy profile summarizes players’ choices.

n ai = probability that node i installs anti-virus software


Our Model : Attack Model

n After the nodes choose their strategies, the adversary picks a starting point for infection uniformly at random

n Node i gets infected if it has no anti-virus software installed and if any of its neighbors become infected.


0

2

1

Our Model : Attack Model (cont.)

3

54

n Example: Only node 3 installs anti-virus software. Adversary chooses to infect node 2.


Our Model : Attack Graph

0 1

2 3

54

0

2 3

54

network graph G attack graph Ga= G - Ia

1


Our Model : Individual Costs

n Anti-virus software costs C. Expected loss from virus is L.

n Cost of strategy to node i:

n Here, pi(a) = Pr[i is infected | i does not install an anti-virus]


Our Model : Social Cost

n Social cost of is simply a sum of individual costs:


Outline

nMotivationnOur ModelØNash StrategiesnOptimal StrategiesnSum-of-Squares Partition ProblemnConclusion


Nash Strategies

n Def: Strategy profile is in Nash equilibrium if no node can improve its payoff by switching to a different strategy:

for i = 0,...,n-1 and any x 2 [0,1],

n Fact: Nash strategies do not optimize total social cost (cf. Prisoner’s Dilemma)


Nash Strategies (cont.)

Thm: There is a threshold t=Cn/L such that each node in a Nash equilibrium¨ will install an anti-virus if it would otherwise end up in

a component of expected size > t¨ will not install an anti-virus if it would end up in a

component of expected size < t.¨ is indifferent between installing and not installing

when the expected size = t.



n Corollary: Let t = Cn/L. Then a pure strategy is a Nash equilibrium if and only if¨Every component in Ga has size · t¨ Inserting any secure node j and its edges into

Ga yields a component of size ¸ t.


Nash Strategies (cont.)n Example: Let C=0.5,L=1 so that t=Cn/L=2.5.

Then is not a Nash equilibrium.

0 1

2 3

54

0

2 3

54

network graph G attack graph Ga= G - Ia

1


Nash Strategies (cont.)Thm: It is NP-hard to compute a pure Nash

equilibrium with lowest (resp., highest) cost.Proof sketch: By reduction to VERTEX COVER

(resp., INDEPENDENT DOMINATING SET).¨ Set C, L so that t=Cn/L=1.5. ¨ In a Nash equilibrium, (a) every vulnerable node

has all neighbors secure; (b) every secure node has an insecure neighbor



n If V’µ V is a minimal vertex cover, then installing software on its nodes satisfies (a) because V’ is a vertex cover and (b) because V’ is minimal.

n Conversely, if V’ are secure nodes in a Nash equilibrium, then V’ is a vertex cover by (a).



n Nash Theorem guarantees our game has a mixed Nash equilibrium.

n But does it make sense talking about pureNash equilibria?



Yes, it does!

Thm: If at each step some node with suboptimal strategy switches its strategy, the system converges to a pure Nash equilibrium in · 2n steps.


Price of Anarchy [KP99]n Price of anarchy measures how far away a

Nash equilibrium can be from the social optimum

n Formally, it is the worst-case ratio between cost of Nash equilibrium and cost of social optimum

n For network G and costs C, L, we denote it:


Price of Anarchy (cont.)Lower Bound: For a star graph K1,n,

ρ(G, C, L) = n/2.Upper Bound: For any graph G and any C, L,

ρ(G, C, L)· n.

Thm: Price of anarchy in our game is ρ(G, C, L) = Θ(n).


Price of Anarchy (cont.)Proof for lower bound:Consider a star graph K1,n. Let C=L(n-1)/n so that t=Cn/L=n-1.

G = K1,n

0

n-11

2

3n-2

…


Price of Anarchy (cont.)

Then, is an optimum strategy with cost C+L(n-1)/n.

G = K1,n

0

n-11

2

3n-2

…

Ga*

0

n-11

2

3n-2

…



And is worst-cost Nash with cost C+L(n-1)2/n.

G = K1,n

0

n-11

2

3n-2

…

Ga*

0

n-11

2

3n-2

…



n Therefore,

n Proof for upper bound uses similar ideas.


Outline

nMotivationnOur ModelnNash StrategiesØOptimal StrategiesnSum-of-Squares Partition ProblemnConclusion


Optimal Strategies

n So, allowing users to selfishly choose whether or not to install anti-virus software may be very inefficient

n Instead, let’s have a benevolent dictatorcompute and impose a solution maximizing overall welfare


Optimal Strategies (cont.)

n We can show:Thm: Let t=Cn/L. If is an optimum strategy, then every component in Ga has size · max(1, (t+1)/2).

n Unfortunately,Thm: It is NP-hard to compute an optimal strategy.


Optimum Strategies (cont.)n Naturally, we consider approximating the

solution.

0 1

2 3

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0 1

2 3

54

network graph G attack graph Ga=G - Ia

k1=2

k2=2

secure nodes

Ia


Optimum Strategies (cont.)

n For pure strategy , we have:

we concentrate on this part


Outline

nMotivationnOur ModelnNash StrategiesnOptimal StrategiesØSum-of-Squares Partition ProblemnConclusion


Sum-of-Squares Partition

n We guess that there are m=|Ia| secure nodes.

n Problem: By removing a set of at most m · n nodes, partition the graph into components H1, …, Hk such that ∑i |Hi|2 is minimum.


Sum-of-Squares Partition (cont.)

Thm: We can find a set of O(log2 n)¢m nodes whose removal partitions the graph into components H1,…,Hk such that ∑i |Hi|2 · O(1)¢OPT.

Proof sketch: We use the Leighton-Rao sparse cut algorithm [LR99]. The approach is similar to greedy log n approximation algorithm for set cover. We repeatedly remove the node cut that gives the best per-node benefit.


Outline

nMotivationnOur ModelnNash StrategiesnOptimal StrategiesnSum-of-Squares Partition ProblemØConclusion


Conclusionn We proposed a simple game for modeling

containment of viruses in a network.n Nash equilibria of our game have a simple

characterization.n We showed that, in the worst case, they can be

far off from the optimal solution.n However, a near-optimal deployment of anti-

virus software can be computed by reduction to the sum-of-squares partition problem.


Open Problems

n Introduce a discount (or taxation) mechanism into the system.

n Suppose nodes can lie about their level of security (or about who their neighbors are). How do we make truth-telling a dominant strategy?

n Consider a “smart” adversary who targets the biggest graph component.

n How do we evaluate what C and L are?n Is there an algorithm for the sum-of-squares partition

problem with a better approximation ratio?


Acknowledgments

Joan Feigenbaum, Hong Jiang, and YangRichard Yang


Thank you!

Inoculation strategies for victims of viruses

Technology

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