Download Our Inoculation Strategies for Victims of Viruses and the Sum

Our Inoculation Strategies for Victims of Viruses and the Sum-ofSquares Partition Problem*
Yafet
Farnaz
Zelalem
Spring 2017
*Aspnes, James, Kevin Chang, and Aleksandr Yampolskiy. "Inoculation strategies for victims of viruses and the sum-of-squares partition
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problem." Journal
of Computer
and SystemEmory
Sciences
72.6 (2006): 1077-1093.
Outline
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Motivation
Problem definition
Model
Nash Equilibria
Optimization
Sum of Squares problem
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Motivation
Question: is it worth to install an antivirus software?
Kaspersky 2017 = $59.99
Value of data = $250, Probability of infection = 10%
Expected loss = $25
Answer: probably not
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Problem definition (Cont.)
▶ Consider n (nodes in our graph) machines connected by a network, each
machine may or may not be protected
▶ Any virus that infects a machine also infects its unprotected neighbours
▶ Suppose that protecting a machine by installing an antivirus software costs the
owner of the machine C, but having the same machine be infected cost L (may
or may not be greater than C)
▶ Owner vs Society: If C > L no owner will install the antivirus, affecting the
whole society, selfish owner
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Benevolent Dictator
▶
The Benevolent Dictator: decides which computers install an antivirus
▶
▶
selective inoculation of a few central machines can limit the spread of the virus
Questions to be addressed:
▶ How much of an improvement a centralized solution can provide?
▶ How easy is it to find a good centralized solution
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Outline
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Motivation
Problem definition
Model
Nash
Optimization
Sum Square problem
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Model
▶ The network is an undirected graph G = (V,E).
▶ Installing anti-virus software is a one round noncooperative game.
▶ There are n player, the nodes of the graph, V =
{0,1,…,n-1}.
▶ Initially all nodes are not secure
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Model: Strategies
▶ Each node has two options: do nothing or install the
antivirus
▶ ai = probability that node i installs anti-virus software
▶ The strategies of all n players can be given as a vector
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Model: Strategies (Cont.)
▶ If ai is 0 or 1, node i adopts a pure strategy
▶ If 0<ai<1, node I adopts a mixed strategy
▶ Ia, set of nodes that have installed antivirus, secure
nodes
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Model: Attack Model
▶
After all the nodes choose their strategies, an adversary
picks a starting point for infection uniformly at random
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A node i gets infected if it has no anti-virus software
installed and if any of its neighbors become infected
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Model: Attack Model (Cont.)
▶
Example: Only node 3 installs anti-virus software.
Adversary chooses to infect node 2.
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Model: Attack Graph
▶
Attack Graph Ga: subgraph of G induced by removing
secure nodes and their edges
network graph G
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attack graph Ga= G - Ia
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Model: Individual Costs
▶
Anti-virus software costs C. Loss from infection is L.
▶
For simplicity, assume C and L are known and the same
for all nodes
Cost of mixed strategy to node i:
▶
▶
Here, pi( ) = Pr[i is infected | i does not install an
antivirus]
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Model: Social Cost
▶
Social cost of
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is simply a sum of individual costs:
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Outline
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Motivation
Problem definition
Model
Nash Equilibria
Optimization
Sum Square problem
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Strategies
▶ Two
▶
scenarios
each node decides individually
▶ Nash
▶
equilibrium
centralized plan
▶
dictator computes plan
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Nash Equilibrium
▶
A stable state of a system involving the
interaction of different participants in
which:
no participant can gain by a unilateral change
of strategy
▶ if the strategies of the others remain
unchanged.
▶
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Nash Equilibrium
▶ Assumptions
for our problem:
nodes are aware of each other’s choices
▶ nodes can easily evaluate C and L
▶ Nodes can compute Nash equilibrium in a
reasonable amount of time
▶
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Nash Strategies: Characterization
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 in the attack graph.
▶ will not install an anti-virus if it would end up in a component
of expected size < t in the attack graph.
▶ is indifferent between installing and not installing when the
expected size = t in the attack graph.
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Mixed Nash
S(i) expected size of insecure component
▶ A profile is in Nash equilibrium, iff:
▶
▶ For
all i, ai=1, S(i)>=t
▶ For all i, ai=0, S(i)<=t
▶ For all i, 0<ai<1, S(i)=t
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Pure Nash
A simplified version with no random choices
▶ A profile is in Nash equilibrium, iff:
▶
▶ Every
component in attack graph has size at most t
▶ Inserting any secure node yields a component of size at least t
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Nash Strategies
▶ Example:
▶
Then
Let C=0.5,L=1 so that t= Cn/L= 3.
is not a Nash equilibrium.
attack graph Ga= G - Ia
network graph G
0
1
0
1
2
3
2
3
4
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Nash Strategies
▶ Suppose ai=0,
node i expects that it will lie in component of size greater
than t=Cn/L in attack graph
▶ Prob[i infected] = t/n.
▶
▶ Then
its expected loss from infection is >L·(Cn/L)/n = C
and it will switch to ai = 1.
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▶
▶
It is NP-hard to compute a pure Nash equilibrium with
lowest and highest cost.
There exists a pure Nash equilibrium, which can be
achieved by a distributed, iterative process in 2n steps.
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Nash strategies
Nash equilibrium can be computed in time O(n3)
▶ Using DFS we can test each node once.
▶A
▶ Compute
▶
the size of each component
O(|V| +|E|)=O(n2)
▶ 2n. n2=
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O(n3)
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Price of Anarchy
▶ Price
of anarchy measures how far away a Nash
equilibrium can be from the social optimum
▶ Formally, it is the worst-case ratio between cost of Nash
equilibrium and cost of social optimum
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Price of Anarchy
▶ For
network G and costs C, L, we denote it:
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A lower bound
Consider a star graph K1,n.
Let C=L(n-1)/n so that t=n-1.
1
n-1
2
n-2
3
0
…
G = K1,n
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Price of Anarchy (cont.)
Then,
C+L(n-1)/n.
is an optimum strategy with cost
1
1
n-1
n-1
2
n-2
3
n-2
3
0
0
…
…
G = K1,n
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Ga*
Price of Anarchy (cont.)
Then,
C+L(n-1)2/n.
is an optimum strategy with cost
1
n-1
1
2
n-2
n-1
3
n-2
3
0
0
…
…
G = K1,n
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Ga*
Price of Anarchy
▶ Therefore,
▶
lower bound price of anarchy is:
(G, C, L)= L(n-1)/(2L(n-1)/n) = n/2
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Price of Anarchy
▶ The
▶
▶
upper bound price of anarchy is:
(G, C, L) <= Ln/n=n.
(G, C, L) <=Cn/C=n
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If C>L
if C<=L
Lower Bound: For a star graph K1,n,
(G, C, L) = n/2.
Upper Bound: For any graph G and any C, L,
(G, C, L)≤ n.
Price of anarchy in the game is
(G, C, L) = (n).
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Outline
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▶
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▶
Motivation
Problem definition
Model
Nash Equilibria
Optimization
Sum Square problem
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Optimization
▶
So, allowing users to selfishly choose whether or not to
install anti-virus software may be very inefficient
▶
Instead, let’s have a benevolent dictator compute and
impose a solution maximizing overall welfare
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▶
Unfortunately,
Theorem: It is NP-hard to compute an optimum strategy
Fortunately,
Theorem: a strategy with cost at most O(log1.5 n)
OPT
can be computed in polytime
▶
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Outline
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Motivation
Problem definition
Model
Nash
Optimization
Sum Square problem
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Sum of Squares Problem
Total Cost = C*(number of secure nodes)+L* Pr(infected node v) =
C |m| + (L/n)Σ Ki 2
Assumption: The number of node (m) is known
The network security problem is reduced to find the best
solution to the problem of removing m nodes to minimize
the sum of square.
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Sum of square partition problem
NP-hard
Given a graph G = (V, E), by removing a set of at most m
nodes, partition the graph into mutually disconnected
components {Hi}, such that ∑|Hi|2 is minimum.
(𝝰,𝝱)-bicriterion approximation algorithm:
The output is a node cut consisting of at most 𝝰m nodes and
partitioning graph into connectd components {Hi}, such that ∑|Hi|2 <=
𝝱 OPT
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Theorem
There exists a
polynomial time (O(log1.5n), O(1))-bicriterion approximation algorithm
for the sum-of- squares partition problem
OPT: The cost of the optimum solution for the inoculation
problem,then there exists a polynomial time approximation
algorithm that finds a solution with cost at most
O(log1.5n)OPT
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Sparsest Cut
Sparsest Cut
S
S’
C-balanced separator
Both NP-hard
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Arora-Rao-Vazirani Algorithm*
S. Arora, S. Rao, and U. Vazirani. Expander flows, geometric embeddings and graph partitioning.36th Annual ACM Symposium on the Theory of Computing,
Exists O(log 0.5 n)-approximation for the following
problem
Given graph G=(V,E), find a node cut that partitions the nodes
of G into three sets:
▶ Two disconnected subgraphs with nodes sets V1 and V2
▶ A set of removed nodes R, such that the following quantity is
maximized.
Sparsity of the Cut
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Cost-Effectiveness
Connected subgraph H with k nodes that is split into sets
V1 and V2, and R then the cut’s cost-effectiveness is:
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Cost-Effectiveness vs. Sparsity
▶
Sparsity
▶
Cost effectiveness
𝝱
𝝰
𝝰 > 2𝝱
Cost effectivenss > 2 sparsity
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Algorithm
Input: G = Graph(V,E)
M>0
Find the approximate most cos-effective cut in
each connected component
G
Partition the graph G into H1,H2 by removing
a sparse node cut.
If the sparse cut removes at most (20c log 𝑛 )
m nodes in H1 … Hk, continues, if there is not
such component halts the process.
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H1
45
H2
Algorithm
Choose the component Hj among those components considered in previous
step for which the cost-effectiveness is highest.
Set F = F U R , if |F| > (36 log 1.5 n)m nodes then halt.
Otherwise, repeat.
H1
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Runtime
▶ Algorithm
halts as soon as we augment the set of marked
nodes |F| >(36c log1.5 n)m
▶ At the beginning of each iteration, F contains at most (36c
log1.5 n)m marked nodes.
▶ Since we add at most (20c log0.5 n)m marked nodes per
iteration.
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