Protection vs. false targets in series systems

Protection vs. false targets in series systems Reliability Engineering and System Safety(2009) Kjell Hausken , Gregory Levitin Advisor: Frank,Yeong-Sung Lin Presented by Jia-Ling Pan NTU IM OPLab

NTU OPLab Agenda • Introduction • The model • Scenario 1:The attacker attacks one element • Scenario 2:The attacker evenly attacks all elements • Scenario 3:The attacker evenly attacks a subset of the elements • Conclusion 1 2010/3/30 NTU OPLab

NTU OPLab Introduction • The paper analyses the optimal distribution of the defense resources between protecting the genuine system elements and deploying false elements in a series system which is destroyed when any genuine element is destroyed. • False and genuine elements cannot be distinguished by the attacker.

NTU OPLab Introduction • The probability of element destruction in the case of attack is defined as a contest function depending on the ratio of the defender’s and attacker’s effort and on a contest intensity parameter. • The dependence of the minmax defense strategy (number of false elements) and the most harmful attack strategy (number of attacked elements) on the amount of resources available to the counterparts, on the number of genuine system elements and on the contest intensity is analyzed

NTU OPLab The model

NTU OPLab The model • The system consists of N genuine elements connected in series. • Destruction of any GE causes destruction of the entire system. • Defender protects all GEs, and distributes the protection resource among the GEs evenly. • The even resource distribution appears to be optimal if the protection effort cost is the same for different GEs .

NTU OPLab The model • The attacker is not able to distinguish GEs and FEs. • Both the defender and the attacker have complete information about the structure of the game, the strategy sets, and all parameters, and are fully rational. • The defender’s one free-choice variable is the number of false targets H. • The attacker’s one free-choice variable is the number of elements to attack Q.

NTU OPLab The model • The defender’s resource is r • The resource needed to deploy one FE is x where 0<x<r . • If the defender deploys H FEs, the resource remaining for the protection is r-Hx. • The maximum number of deployed FEs is H=[r/x]. • The defender allocates its protection resource r-Hx evenly among N GEs achieving protection effort t=(r-Hx)/N per element.

NTU OPLab The model • The attacker’s resource is R • If the attacker attacks Q elements out of N+H, it achieves the per element effort TQ =R/Q. • The vulnerability of any attacked GE is determined by the ratio form of the attacker–defender contest success function. • where m ≥ 0 , ∂v/∂T > 0 and ∂v/∂t < 0.

NTU OPLab Scenario 1:The attacker attacks one element • Attacker has the ability to attack only once against any one of the N+H elements. • The probability that the attacked element is a GE is p = N/(N+H). • When one element is attacked, Q = 1 ,

NTU OPLab Scenario 1:The attacker attacks one element • Thispaperassume m = 1first because it is the most natural choice if a choice is to be made, and second because it is usually the easiest case to handle analytically. • Although this does not allow generalization to ma1,thepresenceof m in the exponent in (1) shows that, especially when t and TQ have similar sizes, the vulnerability changes moderately as m increases moderately above or below m = 1,and changes more extensively as m approaches infinity or 0.

NTU OPLab Scenario 1:The attacker attacks one element • When m = 1,differentiating V with respect to H and equating the derivative with zero gives the interior solution. • The second-order derivative is positive at the interior minimum.

NTU OPLab Scenario 1:The attacker attacks one element • Property 1. • When m = 1 ,at the interior solution the optimal number H of deployed FEs decreases in the deployment cost x, increases in both the defender’s and the attacker’s resource r and R, and in the ratio r/R, increases in the number N of GEs when R>x, and decreases in N when R<x.

NTU OPLab Scenario 1:The attacker attacks one element r/R=5.85 r/R=5.85 • Fig. 1. Optimal number H* of deployed FEs and vulnerability V* as functions of r/R for different m, where N =5 and x/R =0.5. The stepwise movement follows since H* can take only integer values

NTU OPLab Scenario 1:The attacker attacks one element • Fig. 2. System vulnerability V as a function of H for various r/R, where m = 3, N = 5, and x/R = 0.5. r/R=6, V(1)>V(12) r/R=8, V(1)<V(12)

NTU OPLab Scenario 1:The attacker attacks one element m=2 Fig. 3. (r/R)* asafunctionof m, x/R, and N.

NTU OPLab Scenario 1:The attacker attacks one element Fig. 4. Optimal H* and V* as functions of N for different values of m, where r/R = 5 and x/R = 0.5.

NTU OPLab Scenario 1:The attacker attacks one element • Fig. 5. System vulnerability V as a function of H for various N when m = 3 and0.7,where r/R = 5 and x/R = 0.5.

NTU OPLab Scenario 1:The attacker attacks one element Fig. 4. Optimal H* and V* as functions of N for different values of m, where r/R = 5 and x/R = 0.5.

NTU OPLab Scenario 2:The attacker evenly attacks all elements • Attacker attacks all Q = N+H elements evenly distributing its resource among them. • The probability that one GE survives is s=1-v. • Hence the probability that the system is destroyed is

NTU OPLab Scenario 2:The attacker evenly attacks all elements • When m = 1,differentiating V with respect to H and equating the derivative with zero gives the interior solution. • The second-order derivative is positive at the interior minimum.

NTU OPLab Scenario 2:The attacker evenly attacks all elements • Proposition. • The optimal number of false elements H is independent of the contest intensity m when the attacker attacks all elements evenly.

NTU OPLab Scenario 2:The attacker evenly attacks all elements • Property2. • When m = 1,the defender always deploys fewer FEs when the attacker attacks all elements evenly compared with attacking only one element.

NTU OPLab Scenario 2:The attacker evenly attacks all elements • Fig. 6. Optimal number of deployed FEs H*, and V*, as functions of r/R, for different m, where N = 5 and x/R = 0.5.

NTU OPLab Scenario 2:The attacker evenly attacks all elements Fig.1 attacker attacks one element Fig.6 attacker evenly attacks all elements

NTU OPLab Scenario 2:The attacker evenly attacks all elements Fig. 7. Optimal H* and V* as functions of N for different m, where r/R = 5 and x/R = 0.5.

NTU OPLab Scenario 3:The attacker evenly attacks a subset of the elements • Attacker can choose the number of attacked elements that maximizes the system vulnerability. • Analyze a two-period minmax game. The defender chooses H in the first period to minimize the maximum vulnerability V that the attacker can inflict in the second period by choosing Q.

NTU OPLab Scenario 3:The attacker evenly attacks a subset of the elements • For any given Q, the number of attacked GEs q among N+H elements can vary from max{0,Q-H} to min{Q,N}. • The probability that exactly q GEs are attacked is

NTU OPLab Scenario 3:The attacker evenly attacks a subset of the elements • If exactly q GEs are attacked, the destruction probability of the series system is V = 1-sq = 1-(1-v)q, when Q targets are attacked, the total destruction probability of the system is

NTU OPLab Scenario 3:The attacker evenly attacks a subset of the elements • The solution of the minmax game(optimal values of Q* and H* and corresponding value of the expected vulnerability V*) is obtained by the following enumerative procedure: 1. Assign V* = 1; 2. For each H = 0,…, [r/x] 2.1. Assign Vmax = 0; 2.2. For each Q = 1,…,N+H 2.2.1. Determine V(Q,H) according to(7); 2.2.2. if Vmax<V(Q,H) assign Vmax = V(Q,H) and Qopt = Q; 2.3. if Vmax< V* assign V* = Vmax, Q* = Qopt, H* = H.

NTU OPLab Scenario 3:The attacker evenly attacks a subset of the elements Fig. 8. Optimal Q*, H*, and V* as functions of r/R for different m, where N = 5 and x/R = 0.5.

Fig.4 attacker attacks one element NTU OPLab Scenario 3:The attacker evenly attacks a subset of the elements Fig. 9. Optimal Q*, H*, and V* as functions of N for different m, where r/R = 5 and x/R = 0.5.

NTU OPLab Scenario 3:The attacker evenly attacks a subset of the elements • Fig. 9. Optimal Q*, H*, and V* as functions of N for different m, where r/R = 5 and x/R = 0.5. Fig.9attacker attacks a subset of the elements Fig.4 attacker attacks one element

NTU OPLab Scenario 3:The attacker evenly attacks a subset of the elements Fig. 10. Best response functions Q*(H) and H*(Q) for various m, assuming N = 5, r/ R = 5, and x/R = 0.5.

NTU OPLab Conclusion • Series systemcan be destroyed by an attack if any one of the elements is destroyed. • To reduce the system vulnerability, the defender protects the genuine elements and deploys false elements. • The optimal number of false targets in general depends on the resources available to the attacker and the defender, on the false target cost, on the contest intensity and on the number of the genuine elements.

NTU OPLab Conclusion • When the attacker attacks only one element, as the attacker’s resource increases, it becomes more important for the defender to deploy more false elements to divert the attacker. • With lowcontest intensity, efforts have modest impact on the outcome of protection and attack, and the defender deploys many false elements to decrease the probability that a genuine element is attacked.

NTU OPLab Conclusion • With high contest intensity, the attacker easily gets contest success since it attacks only one element, while the defender protects all genuine elements. • Defender deploys a maximum number of false elements when not too resourceful. • As the defender becomes more resourceful, it abruptly decreases the deployment of false elements.

NTU OPLab Conclusion • When the attacker attacks all elements, the defender always deploys fewer FEs. • With low contest intensity, this results in high vulnerability since efforts have low impact, and the attacker is not diverted to too many FEs. • With high contest intensity, the vulnerability is low when the defender is sufficiently resourceful, The optimal number of FEs does not depend of the m and strictly decreases with the number of the GEs.

NTU OPLab Conclusion • When the attacker can choose how many elements to attack, • low contest intensity induces the attacker to attack all elements since contest success is more egalitarian regardless of effort. • with high contest intensity effort matters more and the attacker attacks overall fewer elements as the defender’s resource increases.

NTU OPLab Conclusion • The presented model applies a contest intensity parameter m that cannot be exactly evaluated in practice. • Two ways of handling the uncertainty of the contest intensity can be outlined: • m can be defined as a fuzzy variable and fuzzy logic model can be studied. • The range of possible variation of m can be determined and the ‘‘worst-case’’ defense strategy can be obtained under the assumption that m takes the values that are most favorable for the attacker.

NTU OPLab Thanks for your listening!

Protection vs. false targets in series systems