Maximization of Network Survivability against Intelligent and Malicious Attacks (Cont’d)

Presented by Erion Lin

Outline

Problem DescriptionModelSolution Approach

Problem Description

Problem Description

Assume the budget allocation policy is given, we want to know the minimal attack cost for an attacker to compromise a network.

The system is survivable if there is at least one available path for each critical OD-pair.

Problem Assumptions

The survivability metric is measured as the connectivity of the given critical OD-pairs.

The attacker and the defender have complete information about the targeted network topology.

The defender’s budget allocation strategy is a given parameter.

Problem Assumptions (Cont’d)

The objective of the attacker is to minimize the total attack cost of destroying all paths between one of the critical OD-pairs.

We consider node attacks only. (No link attacks are considered). If a node is attacked, its outgoing links are not functional.

We consider malicious attacks only. (No random failures are considered.)

Model

Model Description

Given Network topology A set of critical OD-pairs Total defense budget for the defender

Model Description (Cont’d)

Objective: To minimize the total cost of an attack

Subject to: There is no available path for one of the critical

OD-pairs to communicate.

To determine: Which nodes will be attacked

Given Parameters

Decision Variables

Formulation

Objective Function

subject to ii i

yi V

Min y a

l ic y M , ii V l OUT (IP 1.1)

Link cost representation

wl l pl ll L l L

t c c

,wp P w W (IP 1.2)

w

p pl wlp P

x t

,w W l L (IP 1.3)

wl lw W l L

M t c

(IP 1.4)

Formulation (Cont.)

subject to (cont.)

0 1px or ,wp P w W (IP 1.6)

0 1iy or i V (IP 1.7)

0 1wlt or ,w W l L (IP 1.8)

1w

pp P

x

w W (IP 1.5)

Reformulation

We reformulate the problem with one assumption and one argument.

Assumption

Argument the optimality condition for the defender holds if

and only if the total budget B is fully used.

,i ia b i V

The threshold attack cost to compromise a node equals to the allocated budget on it.

Reformulation (Cont.)

Objective Function

subject to ii i

yi V

Min y b

l ic y M , ii V l OUT (IP 2.1)

Link cost representation

wl l pl ll L l L

t c c

,wp P w W (IP 2.2)

,w W l L (IP 2.3)

wl lw W l L

M t c

(IP 2.4)

w

p pl wlp P

x t

Reformulation (Cont.)

subject to (cont.)

1w

pp P

x

w W (IP 2.5)

0 1px or ,wp P w W (IP 2.6)

0 1iy or i V (IP 2.7)

0 1wlt or ,w W l L (IP 2.8)

or lc M .l L (IP 2.9)

Solution Approach

Max-Flow Min-Cut Theorem

The maximum value of the flow from a source node to a sink node t in a capacitated network equals the minimum capacity among all s-t cuts.

Therefore, we gain a byproduct of the minimum cut from the maximum flow algorithm.

Genetic Augmenting Path Algorithm

Example

Example (cont.)

Questions

How to identify an augmenting path or show that the network contains no such path?

Whether the algorithm terminates in finite number of iterations?

Labeling algorithm is a specific implementation.

Exists if the residual capacity of the arc is not zero

The Labeling Algorithm

S-T Cut

A cut is a partition of the node N into two subsets S and =N – S.

We refer to a cut as an s-t cut if .S

s S and t S

Example of an S-T Cut

Theorem

The maximum value of the flow from a source node s to a sink node t in a capacitated network equals the minimum capacity among all s-t cuts.

Proof. When the labeling algorithm terminates, it also

discovered a minimum cut.

Theorem (Cont’d)

A flow x* is a maximum flow if and only of the residual network G(x*) contains no augmenting path.

Proof. If the residual network G(x*) contains an augme

nting path, clearly the flow x* is not a maximum flow.

Node Splitting

300300

Solution Approach

Combine max-flow min-cut theorem and node splitting method.

Example

300

200

50

400

70

Example (Cont’d)

300

50

200

70

400

Infinite Capacity

-200

-200

-200

-50

-50 -50

-50

Max Flow and Min Cut： 250

Time Complexity Analysis

Labeling Algorithm :O((|N|+|L|)xn) n: number of augmentations

Consider w OD-pairs O(|W|x(|N|+|L|)xn)

Thanks for Your Listening