Parallel systems under two sequential attacks Gregory Levitin , Kjell Hausken

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Parallel systems under two sequential attacks

Gregory Levitin, Kjell Hausken

Advisor: Frank,Yeong-Sung Lin碩一冠廷

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1.Introduction 2.The attack model 2.1. Even resource distribution between two attacks 2.2. Uneven resource distribution between two attacks 2.3. Uneven resource distribution between two attacks and between elements 3.General model of the optimal attack 4.Defender’s minmax strategy 5.Conclusions

Agenda

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Agenda

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An attacker tries to maximize the system vulnerability.

The attacker distributes its constrained resource optimally across two attacks.

The attacker can choose the number of elements to be attacked in the first attack.

The attacker observes which elements are destroyed and not destroyed in the first attack, and applies its remaining resource into attacking the remaining elements in the second attack.

Introduction

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We consider a 1-out-of-N system which means that all elements have to be destroyed to ensure a non-functioning system.

The defender distributes its constrained resource between deploying redundant elements and protecting them against the attack.

Introduction

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Introduction

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Agenda

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The vulnerability of an element that is attacked is determined by a contest between the defender and the attacker.

Contest success function ：

The attack model

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Agenda

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Two identical separated parallel elements. (N=2) The total attacker’s resource equals the total defender’s resource: r=R. The defender allocates the same resource r/2 to protection of each element. If the attacker attacks several elements, it distributes its resource evenly among the elements

2.1 Even resource distribution between two attacks

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First scenario： the attacker uses all its resources in the single attack.

1.The element vulnerability is

2.The probability that both elements are destroyed is v2=0.25.


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First scenario

R

Defender AttackerN =2

Entire defender’s resource=r Entire attacker’s resource=R

t=r/2

t=r/2

T=R/2

T=R/2

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Second scenario： the attacker distributes its resources evenly between two attacks.

1.The element vulnerability is 2. Three possible outcomes of attack A. Two elements are destroyed of first attack with probability w2.


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Second scenario(A)Defender Attacke

rN =2Entire defender’s resource=r Entire attacker’s resource=R

t=r/2

t=r/2

T=R/4

T=R/4

R/2

R/2

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2. Three possible outcomes of attack B. One element is destroyed of first attack with probability 2(1-w)w. One element is destroyed of second attack with probability v.(Attacker attacks the remaining single element with all its remaining resource R/2) Both attack with probability 2(1-w)w‧v=(1-w)w


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Second scenario(B)Defender Attacke


t=r/2

t=r/2

T=R/4

T=R/4

R/2

R/2T=R/2

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2. Three possible outcomes of attack C. Two elements are not destroyed of first attack with probability (1-w)2.

Two elements are destroyed of second attack with probability w2. Both attack with probability (1-w)2w2.


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Second scenario(C)Defender Attacke


t=r/2

t=r/2

T=R/4

T=R/4

R/2

R/2T=R/4

T=R/4

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Since three possible outcomes of attack are mutually exclusive scenarios, the overall probability of system destruction in a double attack is

The double attack with even resource distribution is beneficial if the system vulnerability in double attack exceeds this probability in single attack:


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2.1 Even resource distribution between two attacks It can be seen that the double attack with

even resource distribution is beneficial for m<=1.82.

m=1.82

r/R=1

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Agenda

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The attacker allocates a part xR of its resource in the first attack, and the remaining part (1-x)R in the second attack. (0< x <=1)

The element vulnerability in the first attack is

If one element is destroyed in the first attack, the remaining attacker’s resource per element is (1-x)R.

2.2 Uneven resource distribution between two attacks

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ScenarioDefender Attacke

rEntire attacker’s resource=R

t=r/2

t=r/2

T=xR/2

T=xR/2

xR

(1-x)R

N =2Entire defender’s resource=r

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If both elements survive the first attack ,the remaining attacker’s resource per element is(1-x)R/2.

The overall system vulnerability


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2.2 Uneven resource distribution between two attacks m 較小時，平均分配資源至兩次攻擊

m 較大時，集中資源至某一次攻擊m 小

m 大

r=R

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for r/R=0.5, m*= 3.06; for r/R=1, m* =1.87;and for r/R=2, m*=1.6

When m>m* the double attackcannot provide greater systemvulnerability than single attackfor any attacker’s resource distribution x. (ex. r/R=1)

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presents m* (the maximal value of m when double attack remains beneficial)as a function of r/R.

Double attack

Single attack

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Agenda

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The attacker attacks only one out of two elements.

The first attack allocating the resource xR to one element.


2.3 Uneven resource distribution between two attacks and between elements

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ScenarioDefender Attacke

rEntire attacker’s resource=R

t=r/2

t=r/2

T=xR

T=(1-X)R

xR

(1-x)R

N =2Entire defender’s resource=r

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The overall system vulnerability is

ref:

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m 小，攻擊者會頃向選擇所有 elements

elements m 大，攻擊者會頃向選擇部分 elements

r/R=1

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Agenda

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The attacker chooses x and the number Q of elements to attack in the first attack optimally.

The attacker can distribute its resource unevenly across the two attacks, and evenly across those elements it chooses to attack in each of the two attacks.


General model of the optimal attack

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The probability that exactly j (0<=j<=Q) elements are destroyed by the first attack is

The probability of system destruction is


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The attacker seeks for x and Q that maximize V(x,Q).


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1. If m is high, the attacker prefers to attack the partial elements. If m is low, the attacker prefers to attack all elements.2. A highly intensive contest and constant N the choice of x plays no important role.

N=4

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1. If m is high, the attacker prefers to attack the partial elements. If m is low, the attacker prefers to attack all elements.2. The attacker’s effort decreases as N increases , but decreasing x* when Q* is constant means that the attacker increases its per element effort in the second attack

r/R=1

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if the attacker can choose 1. how many elements to attack in the first attack, 0<=Q<=N, 2. how to distribute its effort between the two attacks, 0<x<=1 then a single attack is never preferable

Proposition

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Concentrating all its resource on a single attack

Hypothetically, it achieves the same per element effort xR/Q=R/N in the first attack (x=Q/N)

Proof

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which is always greater than V=v(R/N,r/N)N achieved in the single attack.

Proof

j=Q 時

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Agenda

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The defender choose the number of elements N and distribute its resource r between deploying N elements and protecting these N elements.

Observe that 1<=N<=└r/y

┘, where

└r/y

┘ is

the greatest integer that does not exceed r/y.

The resource remaining for protection is r-Ny

Defender’s minmax strategy

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The optimal values x*, Q*, and N* are determined by the following enumerative minmax procedure.


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y/R=0.2

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r/R=2

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Agenda

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The attacker can decide whether to concentrate its limited resource on a single attack or distribute it among two attacks.

The defender distributes its limited resource among deploying redundant elements and protecting them against attacks.

The defender chooses the strategy that minimizes the maximal system vulnerability that the attacker can achieve using its optimal strategy.

Conclusions

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The presented model uses the contest intensity parameter m that cannot be exactly evaluated in practice.

Two ways of handling the uncertainty of the contest intensity can be outlined ：

A. m can be defined as a fuzzy variable and fuzzy logic model can be studied. B. the range of possible variation of m takes the values that are most favorable for the attacker.

Conclusions

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Thanks for your listening.

Parallel systems under two sequential attacks Gregory Levitin , Kjell Hausken

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