Fighting terrorism: How to position rapid response teams?

In light of recent terrorist attacks, we introduce and study a Stackelberg game between a government and a terrorist. In this game, the government positions a number of heavily‐armed rapid response teams on a line segment (e.g., a long boulevard or shopping avenue) and then the terrorist attacks a location with the highest potential impact of an attack. This potential impact, which we call damage, is the product of the time it takes the closest rapid response team to react and the damage caused per time unit, which is modeled via a damage rate function. We prove that there exists a subgame perfect Nash equilibrium that balances the possible damage on all intervals of the line segment that result from positioning the rapid response teams. We discuss the implications for various types of damage rate functions including one mimicking a busy boulevard with various hotspot locations.


INTRODUCTION
Terrorism has claimed hundreds of innocent lives in Europe over the past years.Examples are the Bataclan attack in Paris, the van attack on Barcelona's la Rambla, and the truck incident in Nice.These attacks have also affected the economy by destruction of property, increased market uncertainty and loss of tourism resulting in monetary losses in the order of millions.
Improving protectability against terrorism is thus an important societal concern.To address this concern, governmental agencies have formulated several strategic counter terrorism agendas throughout the years (Bossong, 2021).A recent initiative in these strategic counter terrorism agendas is the deployment of rapid response teams, which we refer to as "response teams" in this article.These heavily-armed and highly-trained response teams are located at high potential attack locations and are capable to respond to attacks within minutes.It is important to select the location of these response teams carefully: on the one hand they should be positioned close enough to high risk spots (i.e., places where many people cluster), on the other hand they should also be sufficiently dispersed to cover many risk spots within reasonable time.Models and methods that are closely related to this positioning problem can be found in location theory.
The theory of locations focuses on the positioning of resources, such as supermarkets and warehouses (Eiselt et al., 2004), as well as the allocation of resources to incidents, such as ambulances and snow removal vehicles (Wang & Liu, 2019).In this theory, it is typically assumed that demand (e.g., consumers that need to go to a supermarket or patients who need an ambulance) is independent of the position of the resources.However, in our setting, demand is a result of the decision of a terrorist.For instance, a terrorist might observe potential attack locations and security measures upfront, and based on that information decides on which location(s) to attack.As a consequence, results from the field of location theory cannot be directly applied to our setting.
To account for dependent/strategic demand one can make use of game theory, which analyses and models the strategic behavior of decision makers (Osborne, 2004).Game theory has been applied to terrorism protection before, including protection of critical infrastructure (see, e.g., Bier et al., 2007;Brown et al., 2011;Hausken & Zhuang, 2011;Holzmann & Smith, 2019;Jiao & Luo, 2019;Musegaas et al., 2022;Powell, 2009;Scaparra & Church, 2008;Zhuang & Bier, 2007) and security of airports (see, e.g., Korzhyk et al., 2011;Pita et al., 2009;Shieh et al., 2012).The majority of these papers do not explicitly model the geographical positioning of protection resources.There are, however, some exceptions.The first exception in that regard is the so-called ambush game (see, e.g., Baston & Bostock, 1987;Baston & Kikuta, 2004;Hausken, 2010).In ambush games there is an intruder crossing a line between two points and a defender that needs to position a number of detectors on the line.The intruder wants to cross this line without detection, while the goal of the defender is to maximize detection.Different variants of the ambush game can be found in the literature.For example, Baston and Kikuta (2004) analyzes the game with an unknown number of intruders and Hausken (2010) analyzes the game with uncooperative detectors and cooperative detectors, where detectors can be positioned further apart if they cooperate.A paper that also explicitly models the geographical position of the resources is Berman and Gavious (2007).This paper studies a Stackelberg game where the leader first allocates a number of supportive resources (e.g., ambulances) over a network and the follower subsequently attacks one node in this network: the one that leads to the highest expected damage for the leader.This damage depends on the initial value of the nodes as well as the time to transport the closest resource to the attack node.The paper solves this problem efficiently for the case where one resource needs to be positioned somewhere in the network (i.e., on a node or arc).The paper also considers the setting where multiple resources need to be positioned somewhere on nodes only.This problem is formulated as an integer linear programming problem and is illustrated via a case study of the 20 largest cities in the US.A similar type of problem has been addressed in Meng et al. (2013), except that multiple resources, each with capacity constraints, can be used in response to an attack.The paper formulates the problem as an integer linear programming problem and applies it to a case study with 19 districts in Shanghai.
Among these exceptions, Meng et al. (2013) and Berman and Gavious (2007) are most relevant for our setting, especially when realizing that the supportive resources have a similar role as our response teams.However, both papers assume the number of potential attack and resource locations to be limited when multiple supportive resources need to positioned (around 20 nodes).This is in sharp contrast to our setting, in which attack locations and response team locations are not limited to a finite set of points.As such, we decided to model our setting by a continuous space environment rather than using the discrete network setting as suggested in Berman and Gavious (2007) and Meng et al. (2013).Continuous location problems are, however, typically notoriously hard problems (see, e.g., García & Marín, 2015).Therefore, in this article, we focus on a continuous line segment only, rather than on, for example, a 2-dimensional space environment.
In particular, we propose a leader follower game in which the leader (e.g., government) positions response teams on a line segment.This line segment could, for instance, represent a shopping avenue or boulevard such as the Ocean drive in Miami, La Rambla in Barcelona, or the strip in Las Vegas.We assume that after the response teams are positioned, one terrorist attack takes place.This attack will occur at a location that maximizes the damage to the government.Here, a proxy for the damage is the product of the time it takes the closest response team to react and the damage that an attack causes per time unit per location.We represent this damage per time unit per location by a damage rate function over the interval of the line segment.The related damage of an attack can be seen as a composition of the amount of people that are affected and the damage to the surroundings such as buildings or statues.The leader wants to position the response teams such that the damage of the attack is minimized.Thus, the leader minimizes the maximum damage of an anticipated attack, which boils down to solving a min-max problem.We prove that there exists a subgame perfect Nash equilibrium.We call the strategy of the leader in this subgame perfect Nash equilibrium an optimal strategy of the leader.We prove that an optimal strategy of the leader balances the possible damage on all intervals of the line segment that result from positioning the response teams.We do so by decomposing the inner maximization problem into several local subproblems, where every subproblem resembles the local maximal damage on an interval.By using Berge's maximum theorem, we are able to prove that we can always reposition a single response team such that the damages of the two neighboring intervals coincide.We exploit this idea to the first two intervals and subsequently balance the local damage of the third interval with the first two intervals, while keeping them balanced as well.By iteratively using this procedure, we can ultimately prove the existence of an optimal strategy that balances the local damage of all intervals.We also discuss the implications for various types of damage rate functions.In particular, we study a setting where (i) the damage rate is more or less equal over the line segment, (ii) the damage rate increases from one side to the other, (iii) the damage rate is centered around a specific point and (iv) the damage rate mimics a boulevard with various hotspot locations.For each of these cases, we derive expressions for the optimal positioning of response teams.For the last case, the optimal positioning of response teams is also compared to two alternative positioning strategies that might be proposed by policy makers from the field.One strategy positions the response teams at the most crowded locations and the other positions the response teams such that the maximal response time to any location on the boulevard is minimized (i.e., neglecting the damage rate function).Analyzing the four types of damage rate functions leads to four managerial insights.First, if people or buildings are more clustered, response teams will also be close around them, leading to a lower total damage compared to a setting where people or buildings are more equally spread.Second, it is not always beneficial to allocate more response teams if the government wants to hedge against the worst possible outcome.Third, governments should be careful with using intuitive rules as they might perform bad under certain conditions.Finally, there exists a trade-off between minimizing damage and minimizing the maximal response time to any location on the line.
From a mathematical perspective, our article has some overlap with a p-center problem.A p-center consists of a set of p resources that minimizes the maximum distance between a demand point and a closest resource belonging to that set (Calik et al., 2015).In the classical p-center problem, the resources and the demand points are positioned only on the nodes of the graph (see, e.g., Calik & Tansel, 2013;Davidovic et al., 2011;Mladenovic et al., 2003;Tansel, 2011).Additionally, the p-center problem has multiple variants, one of them being the continuous p-center problem, where the resources can be positioned anywhere on the graph and the demand can be anywhere on the graph (see, e.g., Chandrasekaran & Tamir, 1980;Hansen et al., 1991;Tamir, 1987Tamir, , 1988)).Another variant of the p-center problem is the weighted p-center problem in which the resources can also be positioned anywhere, but the demand points are positioned on the nodes of the graph and have different weights (see, e.g., Averbakh & Berman, 1997;Burkard & Dollani, 2003;Bhattacharya & Shi, 2007;Calik & Tansel, 2013).Similar to the continuous p-center problem, we also consider a min-max problem to position resources.However, the continuous p-center problem considers distance as the only objective, while we focus on the product of distance and a damage rate function.Additionally, the weighted p-center problem considers the demand to be only on the nodes of the graph, while we consider demand to be anywhere on the line segment.Thus, our problem can be seen as a combination of the continuous p-center problem and the weighted p-center problem for a graph with only two nodes and one edge, which is outlined in Table 1.Our problem with a uniform damage rate function is the same as the continuous p-center problem with two nodes and one edge in between the nodes.More specifically, the optimal value of the objective function in Chandrasekaran and Tamir (1980) and Tamir (1987) for a graph with two nodes and one edge of length one, is the same
The rest of the article is organized as follows.Section 2 provides definitions and results which will be used frequently in several proofs of this article.Section 3 introduces the Stackelberg protection location game and Section 4 studies a best response of the follower and the leader, respectively.In Section 5, different instances of the game are evaluated including a boulevard setting.At the end of Section 5, the impact of the speed of the response teams on the damage is discussed.Finally, Section 6 concludes.

PRELIMINARIES
In this section, we provide definitions and results essential for several proofs in the article.Let S ⊆ R m and let Moreover, a correspondence  is continuous if it is upper hemicontinuous and lower hemicontinuous at all x ∈ S. Below, we provide a necessary and sufficient condition for a correspondence to be upper (lower) hemicontinuous at x ∈ S. In this article, it suffices to use these conditions, see Herings (1996) for an explicit definition of upper (lower) hemicontinuity.
Theorem 2.1 (Herings (1996)).Let S ⊆ R m , let T ⊆ R n , let x be an element of S, and let  ∶ S → T be a compact-valued correspondence.Then the correspondence  is upper hemicontinuous at x if and only if for every sequence (x n ) n∈N in S converging to x and every sequence (y n ) n∈N in T with y n ∈ (x n ), for all n ∈ N, converging to y ∈ R n it holds that y ∈ (x).
Theorem 2.2 (Herings (1996)).Let S ⊆ R m , let T ⊆ R n , and let x be an element of S. Then the correspondence  is lower hemicontinuous at x if and only if for every sequence (x n ) n∈N in S converging to x and for every element y of (x) there exists a sequence (y n ) n∈N in T such that y n ∈ (x n ), for all n ∈ N, and lim n→∞ y n = y.
We now provide a specific case of a correspondence that is compact-valued.
Lemma 2.3.Let S ⊆ R m and let T ⊆ R n .Additionally, let p ∶ S → R and q ∶ S → R be two continuous and bounded functions such that p(x) ≤ q(x) for all x ∈ S. Then the correspondence  ∶ S → T with (x) = [p(x), q(x)] for all x ∈ S is continuous and compact-valued, and (x) is non-empty for all x ∈ S.
Proof.See Appendix. ▪ Finally, we provide Berge's theorem which gives sufficient conditions for the continuity of a specific class of optimized functions.
Theorem 2.4 (Berge's maximum theorem (Herings (1996))).Let S ⊆ R m , let T ⊆ R n , and let  ∶ S → T be a continuous, compact-valued correspondence.Let h ∶ S × T → R be a continuous function and let the relation g ∶ S → R be defined by g(x) = max y∈(x) h(x, y), for all x ∈ S. Then g is a continuous function.

MODEL
We introduce Stackelberg protection location games (SPL games) where a government is the leader and the opponent (e.g., a terrorist) the follower.In such a game, first the leader has to position n ∈ N response teams on line segment [0, 1].We denote the location of these response teams by The second component is the continuous damage rate function f ∶ [0, 1] → R ≥0 which represents the damage of an attack per time unit.The product of these two components equals the damage of an attack and is denoted by The follower tries to maximize this damage, while the leader tries to minimize it.Formally, the leader is interested in solving the following optimization problem:1  ∶= min d∈D max a∈A D(d, a). (1) Please, note that the speed of the response teams is a constant.Consequently, it does not have an impact on the strategies of both the leader and follower and thus can be disregarded in the model.For that reason, in the remainder of this article, we assume that the speed of the response teams equals v = 1. 2 We denote a specific SPL situation by  = (n, f ) with n the number of response teams and f the damage rate function.We denote the set of all SPL situations by Θ. 0, 1], d = (0.2, 0.8), and a = 0.4.We conclude this section with an example of an SPL situation.Note that the parameters are not set to represent reality, but to keep the calculations simple and easy to follow.

ANALYSIS OF SPL GAMES
Given the sequential order in which the leader and follower make a decision in this SPL game, we start in Section 4.1 with studying a specific best response for the follower.Thereafter, in Section 4.2, we focus on a strategy of the leader within a specific subgame perfect Nash equilibrium.We refer to this as an optimal strategy of the leader or an optimal solution.

Response follower
In this section, we study the response of the follower.In doing so, let us first reconsider Example 1.
Example 2. Reconsider  of Example 1.In order to investigate an optimal response of the follower, we first visualize D(d, a) with d = (0.2, 0.8) and for a ∈ A. This is represented in Figure 2.
To identify an optimal response of the follower, we divide Figure 2 in three areas; to the left of d 1 , in between d 1 and d 2 , and to the right of d 2 .In the left area it is optimal to attack at a = 0, for which D(d, 0) = 0.2.In the middle area it is optimal to attack at a = 0.5, for which D(d, 0.5) = 0.3, and in the right area it is optimal to attack at a = 1, for which D(d, 1) = 0.2.Comparing the associated damages per area, we conclude that the damage is maximized when the follower attacks at a = 0.5 with corresponding damage D(d, 0.5) = 0.3.
In Example 2, we separated the maximization problem of the follower into several local maximization problems.This turns out to be useful in general.For every SPL situation  ∈ Θ and d ∈ D n , we separate the maximization problem of the follower, that is, the inner maximization problem in Equation ( 1), into n + 1 local maximization problems.We refer to them as local damage problems and call their optimal values the local damages.The first local damage problem, which identifies the maximal damage to the left of the first response team, is denoted by Then, we introduce n − 1 local damage problems that each identify the maximal damage between two adjacent response teams.
which is the set of feasible locations of response teams i and i + 1.Then, the local damage problem between response team i and i + 1 is given by The last local damage problem, which identifies the maximal damage to the right of the last response team, is denoted by The damage realized by the follower can be determined by taking the maximum over all local damages.That is, for every d ∈ D n , the damage realized under a best response reads We now illustrate this by means of our leading example.
Example 3. Reconsider  of Example 2. We can separate the inner maximization problem of the follower in the following three local damage problems:

Strategy leader
In this section, we focus on a leader's strategy within a subgame perfect Nash equilibrium.As already mentioned in the introduction of this article, we refer to this as an optimal strategy of the leader.We start by characterizing an optimal strategy of the leader for our leading example.
Example 4. Reconsider  of Example 3. To find an optimal strategy of the leader, we solve  in two steps.First, we fix a location d ′ 2 ∈ [0, 1] and subsequently optimize the location of d 1 .Since the local damage of ℛ(d ′ 2 ) is not affected by this decision, our subproblem reads min (3) The minimum is attained at , where the local damages of the first two local damage problems are equal to each other).Note that this is not always a unique minimum, if there exist multiple points where the minimum is attained.In Figure 3B, we illustrate the outcome of the optimization of d 1 given that d ′ 2 = 0.8.By comparing this figure with the one of Figure 3A, we see that the local damage of ℐ 1 has decreased from 0.3 to 4 15 and the local damage of ℒ has increased from 0.2 to 4 15 .Since the local damage of ℛ remains constant, we can conclude that the damage of  decreased from 0.3 to 4 15 .The second step is to optimize d ′ 2 , while keeping an optimized solution The minimum is attained at d * 2 = 3 4 , and so d * 1 = 1 4 .For this solution, we also have an equivalence between the local damages of ℐ 1 and ℛ.
The damage of an attack when d 1 and d 2 are repositioned.
Consequently, we obtain Hence, under optimal d * = ( 1 4 , 3 4 ), the local damage of each local damage problem is the same, implying that the follower is indifferent between attacking in the interval [0,  3B, we see that local damages of ℒ and ℐ 1 have been reduced simultaneously until they meet with the local damage of the ℛ, leading to a (local) damage of 1 4 for ℒ , ℐ 1 , and ℛ.
In Example 4, we illustrated the existence of an optimal strategy d * ∈ D n for which In the remainder of this section, we show that there always exists such an optimal strategy.That is, there always exists an optimal strategy for the leader for which the local damages of all local damage problems are exactly the same.We refer to such a strategy as a balanced strategy.We now explain how we prove existence and optimality of a balanced strategy.
For every  ∈ Θ and i ∈ {2, 3, … , n} we introduce a function ℒ i ∶ [0, 1] → ℛ.This function ℒ i determines the minimal damage to the left of response team i and as such can be recognized as a subproblem of  , restricted to the left of d i .Formally, function ℒ i is recursively defined as follows: with ℒ 1 = ℒ .By exploiting the recursive structure of ℒ i , our problem  can be reformulated as min Now we prove the existence of a balanced strategy and the optimality of all balanced strategies by showing the following four steps: 1.There exists a d ′ ∈ D n that solves the following system of equations: 2. If d ′ solves (7), then the minimum in ( 6) is attained at d ′ n and the minimum in ( 5) with 7), then d ′ is balanced.4. All balanced strategies have the same damage.
For the first step, we need to prove that there exists a d ′ ∈ D n that solves (7).Since ℛ(1) = ℒ n (0) = 0, it suffices to show that ℛ is continuous and non-increasing and ℒ n continuous and non-decreasing.Because ℒ n has a recursive structure, we also need to show continuity and non-decreasing/increasing properties of ℒ i for all i ∈ {1, 2, … , n} and ℐ i for all i ∈ {1, 2, … , n − 1}.
We start by showing that ℒ , ℛ and ℐ i for all i ∈ {1, 2, … , n − 1} are continuous.we do so by using Berge's maximum theorem (see Theorem 2.4).This theorem provides sufficient conditions for continuity of an optimized function and we exactly check for these properties to prove our case.
We also show the non-increasing/non-decreasing behavior of ℒ , ℛ and ℐ i for all i ∈ {1, 2, … , n − 1}.To prove these results, we each time evaluate the functions on two different response team locations.We then show that an optimal solution of the optimization problem associated to the function of one of these response team locations is a feasible solution of the optimization problem associated to the function of the other response team location.
Proof.See Appendix.▪ By combining Lemmas 4.1 and 4.2, we can show that ℒ i is continuous and non-decreasing for all i ∈ {2, 3, … , n}.For continuity, we apply Berge's maximum theorem taking into account that (i) a minimization problem can be reformulated as a maximization problem and (ii) the maximum of two continuous functions (using an inductive argument on ℒ i−1 and ℐ i−1 ) is still continuous.Non-decreasing can be shown using similar arguments as in Lemma 4.2.
Proof.See Appendix.▪ Since ℛ is continuous and non-increasing (Lemmas 4.1 and 4.2), ℒ n is continuous and non-decreasing (Lemma 4.3), and Additionally, via an inductive way, it holds that there exists a Thus, the first step of our four step plan holds and we formalize this in the upcoming lemma.
Proof.See Appendix.▪ By Lemma 4.4, we know that there exists a d ∈ D n that solves (7).As ℒ i is non-decreasing for all i ∈ {1, … , n} and ℐ i−1 is non-increasing in its first argument for all i ∈ {2, … , n}, we know that (7) leads to a minimum for ( 5) and ( 6) with 7), the minimum of ( 6) is attained at d ′ n .Thus, d ′ n is optimal.
Proof.See Appendix.▪ For the third step, we prove that if d ′ solves (7), then d ′ is balanced.We, again, make use of the fact that the minimum of ( 6) is attained at d ′ n and the minimum of (5) with ) for all i ∈ {2, 3, … , n}.By exploiting this relationship several times we show that d ′ is balanced. ▪ For the fourth step, we prove that all balanced strategies lead to the same damage.We do so by showing that it cannot be true that two balanced strategies exist that lead to different damages.
Proof.See Appendix.▪ By combining Lemmas 4.4 to 4.6, we can conclude that there always exists a balanced strategy that is optimal.Combining this with the fact that all balanced strategies lead to the same damage (Lemma 4.7), we know that all balanced strategies are optimal.Together, this results in the following main result.
Theorem 4.8.Let  = (n, f ) ∈ Θ.There exists a balanced strategy and any balanced strategy is optimal.
Proof.See Appendix.▪ We would like to stress that balancedness is a sufficient, but not a necessary condition for an optimal strategy.This is illustrated in the upcoming example.

INSIGHTS FOR DIFFERENT DAMAGE RATE FUNCTIONS
In this section, we apply the result of Theorem 4.8 to various damage rate functions and study the position of response teams.We start with a uniform damage rate function, then discuss a linear damage rate function and subsequently discuss a triangular damage rate function.For each of them, we also study the impact on damage of an increasing number of response teams.Moreover, we also study a scenario inspired by a touristic boulevard consisting of four tourist attractions on which we position four response teams.For this last scenario, we compare the optimal positioning of response teams to two alternative positioning strategies that might be proposed by policy makers from the field.The first strategy positions response teams at the most crowded locations and the second strategy positions response teams such that the maximal distance to any location on the boulevard is minimized.Finally, in Section 5.5 we discuss the impact of the speed of the response teams v on the damage of an attack.

Uniform damage rate function
Let the damage rate function be uniform, that is, we have f (x) = c for a given c ∈ R ≥0 and for all x ∈ [0, 1].This could, for instance, represent a district where the number of people is more or less the same at every location, such as a crowded marketplace or a beach boulevard.
In Example 4, we also discussed a setting with a uniform damage rate function.There, we showed that response teams should be positioned such that the distance between an optimal attack location and the closest team is equal everywhere.This structure holds in general.That is, the distance between two subsequent response teams is constant and the distance from the first (last) team to its closest boundary is half of this distance ( 1 2n FIGURE 5 Damage under optimal d * per number of response teams. Theorem 5.1.Let  = (n, f ) ∈ Θ with f being uniform.An optimal location of the response teams is given by d * i = 2i−1 2n for all i ∈ {1, … , n}, with associated damage c 2n .
Proof.See Appendix.▪ In Figure 5, we visualize this damage for c = 1 and various numbers of response teams.Note that the marginal reduction in damage is decreasing in the number of response teams.

Linear f
Let the damage rate function be linearly increasing, that is, we have f (x) = c ⋅ x for a given c ∈ R ≥0 and for all x ∈ [0, 1].This could, for instance, represent a district where the number of people increases more or less linearly when going from the beginning to the end of the district, such as a shopping avenue with a market square at the end of the street.We would like to stress that this function is also a building block for the triangle function discussed later on.We start with a useful property for linear increasing damage rate functions.It tells us how to position the response teams to the left of an arbitrary response team.In the proof, we use that the best response of the attacker is to attack exactly in the middle of any two response teams.Using this property allows us to find a relationship between the response teams, once allocated optimally.
Proof.See Appendix.▪ By applying Lemma 5.2 to the nth response team, we can characterize an optimal position of all response teams.We do so by identifying the position of the nth response team for which the minimized damage to the left of the nth team equals the local damage to the right of the nth response team.Theorem 5.3.Let  = (n, f ) ∈ Θ with f being linearly increasing.Then, an optimal location of the response teams is given by d Proof.See Appendix.▪ Theorem 5.3 tells us that an optimal position of the response teams can be described via a square root relationship.This means that relatively more response teams are positioned towards the end of the line segment, which is natural since also more damage can be realized there.A useful insight for the government could thus be that response teams should be positioned closer to each other at places with relatively more people or buildings.
Figure 6 visualizes the damage under d * for c = 2 and various numbers of response teams.Note that the marginal reduction in damage is decreasing in the number of response teams, again.
Please, also note that the damages are smaller than the ones in Figure 5 (i.e., the uniform case) for each number of response teams. 5This is interesting, because the average damage rate is the same for both cases, namely 1. 6 This might indicate that the government is able to better protect a line segment with a linear damage rate function than a line segment with a uniform damage rate function.The intuition is that damage is more centered for the linear case, namely to the right, which can be better protected.

Triangular f
Let the damage rate function be triangular, that is, we have: 5 This can also be concluded from the damage expressions: for all n ∈ N + . 6We define the average damage rate as For both uniform and linear increasing, we obtain 1.
for a given c ∈ R ≥0 and  ∈ [0, 1], which is the top of the triangle.This function could, for instance, represent a boulevard with one specific touristic attraction at location .An example of this damage rate function (with c = 1 and  = 1 3 ) is given in Figure 7.
To analyse the triangular case, we use that the triangular damage rate function is described by two linear functions.From the previous section, we already learned how to position response teams optimally if the damage rate function is linearly increasing.Similarly, we can also identify an optimal position of the response teams for a linear decreasing function.We do so by using our results of the linear increasing case and some relabeling of d and transformation on x.
We consider a damage rate function f to be linearly decreasing if f (x) = c⋅(1−x) for a given c ∈ R ≥0 and for all x ∈ [0, 1].We now present two results for this damage rate function.
Corollary 5.5.Let  = (n, f ) ∈ Θ with f being linearly decreasing.Then, an optimal location of the response teams is given by d By using Corollary 5.4 and Lemma 5.2, we can identify an optimal location of the response teams for a very specific instance of the triangular damage rate function.Namely the one where, if the response teams are positioned such that there is balancedness, exactly one of the response teams is positioned at the top (i.e., at ).In total, there exist n of such instances, the one where the first response team is on the top, the second one is on the top, and so on.
We identify the n instances (i.e., we describe ) as follows.We first assume that response team k ∈ {1, 2, … , n} is positioned at the top.Then, we know how to optimally position the response teams to the left of k (using Lemma 5.2), as well as how to optimally position the response teams to the right of k (using Corollary 5.4).Consequently, we can also identify the local damage to the left of response team k as well to the right of k.Subsequently, if the location of  is such that the local damages are equal to each other, we have found an optimal solution.It turns out that  = k n+1 .Theorem 5.6.Let  = (n, f ) ∈ Θ with f being triangular.If  = k n+1 for some k ∈ {1, … , n}, then an optimal location of the response teams is given by ( 8), with associated damage c⋅k 4⋅(n+1) 2 .
Proof.See Appendix.▪ From Theorem 5.6, it can be noted that relatively more response teams are positioned towards the top of the triangle, which is natural since more damage can be realized there.Similar to the managerial insight in Section 5.2, we can again conclude that a useful insight for the government could be to position response teams closer to each other at places with relatively more people or buildings.
Figure 8 shows the damage per number of response teams when  = 1 2 and c = 4, so when the average damage rate equals 1.In this case, the damage can be described by 1 2(n+1) as we take k =  ⋅(n+1), which is according to the if condition in Theorem 5.6.
Consequently, also in this case the marginal reduction in damage is decreasing in the number of response teams, which can be seen in Figure 8.Note that only the damage for an odd number of response teams is included in Figure 8 as Theorem 5.6 can only be applied to these cases.When we compare the damages with the ones of Figures 5 and 6, we see that the triangular damage rate function is the best performing one. 7This might indicate that the government is able to more efficiently protect a line segment with a triangular damage rate function than a uniform or linear damage 7 This can also be concluded from the damage expressions: Damage rate function consisting of two symmetric triangles with optimal locations for n = 2 and n = 3.rate function.The intuition is that damage is even more centered, namely to the top (at ), which can be protected better.
In the following example, we illustrate that damage is not always strictly decreasing in the number of response teams.Hence, another useful insight for the government is that adding an extra response team to a line segment will not always strictly decrease damage.

A boulevard scenario
In this subsection, we study a typical boulevard scenario that might arise in practice.Inspiration for this example stems from the strip in Las Vegas and the Ramblas in Barcelona.We first show how to optimally allocate the response teams for this boulevard scenario.Thereafter, we investigate the impact of different positioning strategies that might be proposed by policy makers from the field.
We consider a boulevard scenario, consisting of four different touristic attractions.The first attraction is located at the beginning of the boulevard, where a well-known musician is playing songs, surrounded with some audience.A little further on the boulevard, a historical fountain is located, attracting many tourists making pictures.At the end of the boulevard, there is a famous casino hotel, where also many people hold to make pictures.Next to the hotel, there is a high-quality restaurant.Because of its delicious food, many people are queuing and waiting to be served.Between these touristic attractions, there is a constant flow of pedestrians.Finally, the government has four response teams that should be positioned on this boulevard.Formally, we consider a  = (n, f ) ∈ Θ with n = 4 and f given by Equation (9).In Figure 10, the damage rate function is visualized.
FIGURE 10 The boulevard setting.
If we position the response teams according to an optimal strategy, the result is d = ( 7 30 , 1 3 , 3 5 , 4 5 ) . Moreover, the optimal attack locations are given by a ∈ { 1 10 , 17 60 , 7 15 , 11 15 , 9 10 } .It can be verified that for positioning d, we have ) ⋅ 1 10 = 1 75 .In practice, the government might decide to position response teams based on intuitive rules, such as: position response teams at the most crowded locations, or position response teams such that the maximal response time to any location on the boulevard is minimized.We remark that this latter rule is in line with equally distributing resources in Shan and Zhuang (2013a).We will now investigate how these strategies from practice perform against the optimal strategy.
If the government positions the response teams at the most crowded locations, the result is ) . In this situation, it is optimal for the terrorist to attack at a = 1 10 , with the associated damage being equal to D = ) . 8 Note that in this case, the maximal response time is 1 8 .Moreover, the optimal attack location is a = 3 4 , with the associated damage being In summary, damage increases with 50% if response teams are positioned at the most crowded locations and damage increases with more than 134% if response teams are positioned such that the maximal response time is minimized.Hence, these examples demonstrate that intuitive rules from practice might perform substantially worse compared to the optimal strategy.

Impact of the speed of the response teams
While the optimal location of the response teams and the attack is independent of v, the damage of an attack certainly is impacted by v.By studying the model formulation in Section 3, we learn that the damage is inversely proportional to v. Consequently, in order to determine the damage for an arbitrary setting, we need to multiply the expressions of Theorems 5.1, 5.3, and 5.6 with 1 v .This result implies that speed not only has a decreasing effect on the damage, but also that it has a marginal decreasing impact on this damage reduction.

CONCLUSIONS AND FURTHER RESEARCH
In light of recent terrorist attacks, we study a Stackelberg game between a government and a terrorist.The government positions response teams on a line segment and thereafter the terrorist decides where to attack.We model the situation by a min-max problem in which the terrorist maximizes the damage and the government minimizes this maximum damage.Our model thus reflects a situation where the terrorist observes attack locations and security measures upfront, and based on that information decides where to attack.At the same time, it also reflects a situation where the government is not sure about the information the terrorist has obtained, or the motives of the terrorist, and therefore wants to hedge against the worst possible outcome.
One of the main contributions is that a balanced strategy is optimal for the government.Such a strategy balances the possible damage on all intervals that result from positioning the rapid response teams.A balanced strategy makes the attacker indifferent: it is equally interesting to attack either to the left of the response teams, to the right of the response teams or between any two consecutive response teams.From our results (i.e., specific damage rate functions), we obtained four managerial insights.First, we learn that if people or buildings are more clustered, response teams will also be close around them, leading to a lower total damage compared to a setting with equal spread.Second, it is not always beneficial to allocate more response teams if the government wants to hedge against the worst possible outcome.Third, the government should be careful with using intuitive rules as they might perform bad under certain conditions.Finally, there exists a trade-off between minimizing damage and minimizing the maximal response time to any location on the line.
The results in this study are established under some assumptions.First, we assume that the terrorist observes the actions of the government and, given this observation, chooses the best location to attack.This can be seen as a rational or strategic terrorist.In the literature on terrorism protection, one can also find studies that assume a terrorist to be irrational or non-strategic (see Golany et al., 2009;Hao et al., 2009;Shan & Zhuang, 2013b).A natural future research direction could be to include irrational or non-strategic behavior.Second, in our model we assume that the government is only interested in minimizing the damage, but in practice the government might consider multiple objectives at the same time (see, Bertsimas et al., 2011;Keeney, 1980;Shan & Zhuang, 2013a).Including such additional objectives to our problem could also be a natural research direction.Third, we assume that the response teams can be positioned on a line segment only.A natural generalization would be to change the one-dimensional line segment to a two-dimensional space.
Here the response teams can not only be located on a busy shopping avenue or boulevard, but also on a more complex street network.Such a situation could be modeled by a damage rate function ranging over a two-dimensional space where the response teams can travel through the space according to Manhattan distance.Here Manhattan distance is the absolute difference of the horizontal location coefficient of the response team plus the absolute difference of the vertical location coefficient of the response team.For such a setting it is not immediately clear what a balanced strategy would be and it is likely that not all balanced strategies lead to the same damage, which severely complicates the analysis.Finally, we also assume that the terrorist will only attack at one location.In practice, we have seen that terrorists might plan multiple attacks, closely after each other.As such it is interesting to generalize our model by analyzing the possibility of multiple attacks by the terrorist and the possibility of relocating and re-assigning the available response teams by the government.
Proof.First we prove that (x) is compact and non-empty for all x ∈ S. Observe that p(x) ∈ (x) for all x ∈ S. Hence, (x) is non-empty for all x ∈ S. Additionally, as p and q are bounded and (x) is defined by a closed interval for all x ∈ S, (x) is closed and bounded and thus compact for all x ∈ S. Consequently,  is compact-valued.
Second, we prove that  is upper hemicontinuous.Let x ∈ S and let (x n ) n∈N be a sequence in S converging to x.Let (y n ) n∈N be a sequence in T with y n ∈ (x n ) for all n ∈ N, converging to y ∈ R. We know by continuity of p and q that lim n→∞ p(x n ) = p(x) and lim n→∞ q(x n ) = q(x).Since lim n→∞ y n = y and y n ∈ [p(x n ), q(x n )] for all n ∈ N, we know that p(x) ≤ y ≤ q(x) and so y ∈ (x).Note that  is compact-valued by the first part of this proof.By Theorem 2.1, it proves that  is upper hemicontinuous at x and so it proves that  is upper hemicontinuous in general.
Third, we prove that  is lower hemicontinuous.Let x ∈ S and let (x n ) n∈N be a sequence in S converging to x.Let y ∈ (x).We know that p(x) ≤ y ≤ q(x).Thus, there exists an  ∈ [0, 1] such that y =  ⋅ p(x) + (1 − ) ⋅ q(x).We set As p(x n ) ≤ q(x n ) we know that y n ≤ q(x n ) and y n ≥ p(x n ) for all n ∈ N. Consequently, y n ∈ (x n ) for all n ∈ N.Moreover, we know that lim n→∞ p(x n ) = p(x) and lim n→∞ q(x n ) = q(x).Thus, y n =  ⋅ p(x n ) + (1 − ) ⋅ q(x n ) and lim n→∞ ⋅p(x n )+(1−)⋅q(x n ) = ⋅p(x)+(1−)⋅ q(x) = y.Consequently there exists a sequence, namely (y n ) n∈N in T such that y n ∈ (x n ) for all n ∈ N, and lim n→∞ y n = y.By Theorem 2.2, it proves that  is lower hemicontinuous at x and so it proves that  is lower hemicontinuous in general.As  is upper hemicontinuous and lower hemicontinuous,  is continuous.▪ Proof of Lemma 4.1 Proof.First, we focus on ℒ , then on ℐ i for all i ∈ {1, 2, … , n − 1} and finally on ℛ.
Note that g is continuous as f is continuous and the product of two continuous functions remains continuous.Since p and q are continuous and bounded functions and p(d 1 ) ≤ q(d 1 ) for all d 1 ∈ [0, 1], by Lemma 2.3 it is true that  is continuous and compact-valued.Moreover, g is continuous and thus by Theorem 2.4 it is true that . Note that g is a continuous function as the product as well as the minimum of two continuous functions remains continuous.Since p and q are continuous and bounded functions and , by Lemma 2.3 is is true that  is continuous and compact-valued.Moreover, g is continuous and thus by Theorem 2.4 it is true that ℐ Note that g is continuous as f is continuous and the product of two continuous functions remains continuous.Since p and q are continuous and bounded functions and p(d n ) ≤ q(d n ) for all d n ∈ [0, 1], by Lemma 2.3 it is true that  is continuous and compact-valued.Moreover, g is continuous and thus by Theorem 2.4 it is true that ℛ Proof.First we focus on ℒ , then on ℛ and finally on ).The first inequality holds as d ′ 1 > d 1 .The second inequality holds as we know that a We derive the following: The first inequality holds as d i < d ′ i .The second inequality holds as we know that a We derive the following: Proof.We prove this lemma by induction.First, we show continuity.By Lemma 4.1 we know that ℒ 1 is continuous.Take a i ∈ {2, … , n − 1} and suppose that ℒ i is continuous.Note that for all d i+1 ∈ [0, 1] the following holds:  i,i+1 .Note that ℒ i is continuous, ℐ i is continuous by Lemma 4.1 and the maximum of two continuous functions is continuous.As a continuous function multiplied by −1 is still continuous, it can be concluded that g is continuous.Since p and q are continuous and bounded functions and p(d i+1 ) ≤ q(d i+1 ) for all d i+1 ∈ [0, 1], by Lemma 2.3 it is true that  is upper and lower hemicontinuous and thus continuous and that  is compact-valued.Since  is a continuous, compact-valued correspondence and g is continuous, by Theorem 1.1 it is true that ℒ i+1 given by − max d i ∈[0,d i+1 ] g(d i , d i+1 ) is continuous.We conclude by the principle of induction that ℒ i is continuous for all i ∈ {1, … , n}.
Second, we prove by induction that ℒ i is non-decreasing for all i ∈ {1, … , n}.As ℒ 1 = ℒ by definition and ℒ is non-decreasing by Lemma 4.2, we conclude that ℒ 1 is non-decreasing.Now let i ∈ {1, … , n − 1} and assume that ℒ i is non-decreasing.We prove that ℒ i+1 is also non-decreasing.By definition, we know that ℒ i+1 (d i+1 ) = min The second inequality holds as d * i > d i+1 and we have assumed that ℒ i is non-decreasing.The second equality holds as we know that ℒ i (d i+1 ) ≥ 0 and ℐ i (d i+1 , d i+1 ) = 0.The third inequality holds as d i+1 is a feasible solution to the optimization problem min Thus, we have proven that ℒ i+1 is non-decreasing.Consequently, by induction we have shown that ℒ i is non-decreasing for all i ∈ {1, … , n}. ▪

FIGURE 6
FIGURE 6Damage under optimal d * per number of response teams.

FIGURE 8
FIGURE 8Damage under optimal d * per number of response teams.
the government positions the response teams such that the maximal response time to any location on the boulevard is minimized, we obtain d = ( The first inequality holds as d i+1 < d ′ i+1 .The second inequality holds as we know that a* ∈ [d i , d i+1 ].Consequently, a * is a feasible solution to the optimization problem max a∈[d i ,d ′ i+1 ] f (a) ⋅ min{a − d i , d ′ i+1 − a}.Hence, ℐ 1 (d i , ⋅) is non-decreasing.As by definition it holds that ℐ i (d i , d i+1 ) = ℐ 1 (d i , d i+1 ) for all i ∈ {1, … , n− 1}, it follows that ℐ i (d i , ⋅) is non-decreasing for all i ∈ {1, … , i − 1}.▪Proof of Lemma 4.3

TABLE 1
The p-center problem.
The speed of these response teams is assumed to be constant and is denoted by v ∈ R >0 .Once the leader has positioned the response teams at d ∈ D n , the follower selects an attack location on the interval denoted by a ∈ A, where and ℛ(0.8) = 0.2.So, the damage under a best response equals max a∈A D((0.2, 0.8), a) = max{0.2,0.3, 0.2} = 0.3.Note that this formalizes what has been established in Example 2.
The first inequality holds as d ′ n < d n .The second inequality holds as we know that a * ∈ [d n , 1].Consequently, a * is a feasible solution to the optimization problem max a∈[d ). that max {ℒ i (d * i ), ℐ i (d * i , d i+1 )} is well-defined as d * i ∈ [0, d i+1 ].The first inequality holds as Lemma 4.2 states that ℐ i (d i , ⋅) is non-decreasing.The second inequality holds as we know that d * i is a feasible solution to the optimization problem mind i ∈[0,d i+1 ] max {ℒ i (d i ), ℐ i (d i , d i+1 )}.