Inspection Games with Local and Global Allocation Bounds

Transcription

1 Inspection Games with Local Global Allocation Bounds Yael Deutsch Boaz Golany Noam Goldberg 2 Uriel G Rothblum Faculty of Industrial Engineering Management Technion Israel Institute of Technology Haifa Israel 2 Math Computer Science Division Argonne National Laboratory 9700 South Cass Avenue Argonne Illinois Received 8 June 202; revised 28 November 202; accepted 9 December 202 DOI 0002/nav2524 Published online 6 February 203 in Wiley Online Library (wileyonlinelibrarycom Abstract This article discusses a two-player noncooperative nonzero-sum inspection game There are multiple sites that are subject to potential inspection by the first player (an inspector The second player (potentially a violator has to choose a vector of violation probabilities over the sites so that the sum of these probabilities do not exceed one An efficient method is introduced to compute all Nash equilibria parametrically in the amount of resource that is available to the inspector Sensitivity analysis reveals nonmonotonicity of the equilibrium utility of the inspector considered as a function of the amount of resource that is available to it; a phenomenon which is a variant of the well-known Braess paradox 203 Wiley Periodicals Inc Naval Research Logistics Keywords inspection games; resource allocation; computing nash equilibria INTRODUCTION Inspection games model situations where an inspector verifies that one or more agents that are subject to inspection (denoted hereafter inspectees follow some rules or regulations which were agreed on by all parties Also the inspector s verification process is defined in this agreement Typically the inspector has limited resources so this verification can only be partial A mathematical analysis of these games allows the understing of inspection processes where agents are strategic rational More importantly the analysis of these games can also help with design of efficient inspection processes Because the inspector the inspectees (that is the agents each optimize its own utility these processes should be modeled as games An extended summary of the literature on inspection games is available in Ref [5] Here we will review a few of the two person inspection games that are related to this article some of the inspection games that have recently appeared in the literature In Ref [9] Hohzaki et al considered a (single-shot multistage two-person zero-sum game Each player has a finite Correspondence to Y Deutsch (yaely@txtechnionacil Posthumous Professor Uriel G Rothblum passed away on March number of opportunities to act it does not know the history of actions taken by its opponent The inspector decides whether to inspect or not the inspectee decides whether to violate or not The game ends when the inspectee is captured or when the preplanned period expires Hohzaki et al developed dynamic programming formulation to exhaust equilibrium points on a strategy space of each player In Ref [8] Hohzaki considered a multistage two-person zero-sum stochastic game where the inspectee decides how much contrab to smuggle A closed-form equilibrium is derived for this specific case In Ref [7] Haphuriwat et al considered a two-person nonzero-sum sequential game The effects of inspection retaliation on the inspectee s decision were tested Results show that unless the inspector imposes high retaliation costs on the inspectee 00% inspection is likely to be needed deterrence with partial inspection may not be achievable in practice (even though it is possible in theory In Ref [3] Xiaoqing et al considered a two-person nonzero-sum game where the inspectee is a (potentially polluting firm The authors explored the effects of subsidies penalties other policy variables on implementation of cleaner production In Ref [5] Deutsch et al considered an inspection game between a single inspector whose inspection resource is constrained multiple independent inspectees The authors 203 Wiley Periodicals Inc

2 26 Naval Research Logistics Vol 60 (203 derived all Nash equilibria solutions for this game discussed their properties This article focuses on a special case of the game formulated in Ref [5] Specifically it considers a nonzero-sum inspection game between a single inspector a single inspectee There are multiple sites where the two players act both of them have limited resources The model of Ref [5] can be viewed as a two player game where there is a coordinator that determines the actions of the multiple independent inspectees has sufficient resources to enable violation in all of the sites Such a model can be used to describe many real-world scenarios for example an owner of multiple (potentially polluting plants who decides which of its plants will comply with the pollution policies to what extent In this article in addition to the global resource limitation the inspector has to adhere to local limitations on the amounts of resource it can allocate to the inspection of the various sites These additional local constraints make the analysis of the game more challenging Nevertheless we show how to compute all Nash equilibria solutions of this game in an efficient way Sensitivity analysis of the Nash equilibria with respect to the inspector s available resource reveals a counterintuitive phenomenon which is a variant of the Braess Paradox Ref [4] the inspector s payoff function is not monotonically increasing in its resource Two-person nonzero-sum inspection games have been modeled using different approaches in previous work In Ref [] the game has a finite number of suspicious events where the inspectee can violate The inspectee can violate only once The inspector may inspect a finite number of times it chooses which events to inspect Also the inspector can announce its strategy in advance may compensate the inspectee in case there is no violation In Ref [2] a similar model is introduced with a partition of the events to those in which the inspector can inspect others in which it is not allowed to do so In Ref [3] the authors develop methods to inspect whether containers contain suspicious objects They formulate an LP model to determine the optimal strategy of inspecting the contents of containers deciding whether they are good or bad In Ref [2] there are two models related to the model of this article The first model formulates a two-player inspection game where each one of the players picks one out of multiple sites In the second model the decisions of the inspector are different The inspector s budget is given by an integer number (greater or equal to one it has to decide how much resource to allocate to each site where there are no limitations on the amounts of resource that can be allocated to each site In this article the amount of resource available to the inspector is a real number the inspector has to adhere to site-specific bounds when allocating its resource The inspectee has to choose whether to violate the regulations or not If it decides to violate then it has to select a vector of violation probabilities over the sites such that the sum of these probabilities do not exceed the value of one Also unlike Ref [2] we are interested in computing all Nash equilibria The article is organized as follows Section 2 formulates the inspection game defines Nash equilibria Section 3 establishes the existence of Nash equilibria provides efficiently computable explicit expressions for all of them Section 4 conducts sensitivity analysis demonstrates how the inspector s amount of resource affects the Nash equilibria solutions The proofs of results in Sections 2 4 appear in the Appendix Section 5 illustrates the results through a numerical example Finally Section 6 concludes with a summary directions for future research 2 MODEL FORMULATION This article concerns an inspection game between an inspector an inspectee denoted I V respectively There is a set N { n} of sites where the inspector the inspectee may have conflicting interests The inspectee has to choose a vector of violation probabilities (whose sum do not exceed one the inspector has to determine the allocation of its inspection resources over the sites to detect the violations When the inspector has access to unlimited resources at no cost or when it has no resources at all the problem becomes trivial So the inspector is assumed to have a limited amount of inspection resource B>0 The inspector has to decide on the amounts x x n to be allocated respectively to the inspection of the sites where each such amount is limited from above So the inspector has to select a vector from { X x (x x n IR n 0 x i α i for i N i N } x i B ( where α α n B are given positive numbers that satisfy i N α i B (because otherwise the global resource limitation is redundant On the other h the inspectee has to decide on a vector of violation probabilities whose sum is constrained not to exceed one that is it can violate with certainty in no more than a single site Thus the inspectee has to determine a vector from { Y y (y y n IR n y i 0 for i N i N } y i (2

3 Deutsch et al Inspection Games with Allocation Bounds 27 Here y i y i 0 0 <y i < mean respectively full violation full compliance partial compliance at site i The utility functions of I V depend on (x y X Y are expressed by U I (x y i N y i (d i x i (3 As Nash equilibria with the inspector s utility function given by (3 are invariant with respect to the d i s henceforth inspector s utility function will be given by Û I (x y i N y i x i for all (x y X Y (9 U V (x y i N y i (a i x i (4 respectively where the quintuples (a i d i α i for i N are given the α i s satisfy α i d i for all i N (5 the latter expresses the assumption that no matter what amount of resource the inspector deicides to invest it will never cause its utility function as defined in (3 to be positive Also the confrontation at each site should cause nonnegative profit to the inspectee However we can assume ( we will demonstrate that when the return from violation in site i N is negative then the inspectee decides to cooperate so a similar assumption for α i a i for each i N is unnecessary The parameter a i represents the inspectee s incentive to violate in site i the parameter represents the inspectee s penalty from violating in site i if inspected the parameter d i represents the inspector s penalty for not inspecting site i the parameter represents the inspector s incentive to inspect site i A joint set of actions (x y X Y is a Nash equilibrium if no player can benefit from deviating unilaterally from its strategy that is U I (x y max x X U I (x y (6 U V (x y max y Y U V (x y (7 We say that x is a best response to y if it satisfies (6 similarly y is a best response to x if (7 is satisfied Let IR {z IR z 0} a b c d α IR n Suppose x XIfa 0 then every y Y is an inspectee s best response Otherwise if b 0 a i > 0 for some i N then the inspectee s best response is j arg max i N {a j } y j y i 0 otherwise If c T α 0 then every x X is a best response for the inspector Further if α 0 (regardless of c then x 0 is a best response for the inspector To avoid such degenerate situations it is assumed throughout that a i d i α i > 0 for all i N (8 it will express the reduced cost for the inspector rather than cost Throughout we shall consider the game defined by ((2(9(4 (5 (8 The following proposition establishes the existence of Nash equilibria for the game PROPOSITION The game has Nash equilibria The next section computes all Nash equilibria of the game 3 DETERMINING ALL NASH EQUILIBRIA This section computes all Nash Equilibria of the game defined by ( (2 (9 (4 (5 (8 For convenience it is assumed henceforth that the a i s the s are respectively distinct the sites are ordered as follows a >a 2 >>a n >a n+ 0 (0 When there is only one site the inspectee will cheat with probability if this site has a positive return will comply with probability if it has a negative return Hence to avoid the need to discuss degenerate situations we also assume that n 2 For each i N for all z IR define p i (z a i z q i (z z Under a fixed inspection strategy x X the problem that the inspectee faces is the ordinary linear knapsack problem with profit coefficients p i (xi s unit cost for each i N a total budget of a single unit Suppressing the dependence of the p i (xi s on x the inspectee s problem is expressed by max i N p i y i (a st y i (b i N y i 0 for each i N (c Further for a given violation strategy y Y q i (yi is the expected profit (reduced cost of the inspector per unit invested in inspecting site i N As > 0 yi 0 q i (yi 0 for all y Y Suppressing the dependence of

4 28 Naval Research Logistics Vol 60 (203 the q i (y i s on y the problem that the inspector faces is then given by max q i x i (2a i N st x i B (2b i N 0 x i α i for each i N (2c which is the well-known bounded knapsack problem (see Ref [0] In the following we will refer interchangeably to optimal solutions of ( (2 best responses respectively to x y Lemmas 2 record the solutions of ( (2 respectively As their proofs are stard they are omitted Throughout the empty sum is defined to be 0 LEMMA (Unbounded Knapsack ( has an optimal solution Further letting M arg max j N p j y IR n is optimal for (2 if only if y 0 i N\M y i 0 one of the following cases holds i max j N p j > 0 i M y i In this case there exists a w N such that p w max j N p j > 0 yw > 0 ii max j N p j 0 0 i M y i In this case if yw > 0 for some w N then p w max j N p j 0 Further if p j 0 for all j N then every y Y is optimal for ( iii max j N p j < 0 i M y i 0 LEMMA 2 (Bounded Knapsack (2 has an optimal solution Assume that q q 2 q n 0 let k j α j < B k j α j for some k N E {i N q i q k } Then x IR n is optimal for (2 if only if for i { k}\ex i α i for i {k + n}\ex i 0 xi i E { B i {k}\e α i if q k > 0 B i {k}\e α i if q k 0 The next corollary of Lemma 2 records simple necessary conditions for optimality in (2 COROLLARY Suppose that x is optimal for (2 for some u N q u > 0 x u <α u Then j N x j B x k 0 for each k N with q k <q u A goal of the inspector is to reduce the coefficients p i ( s to zero which would induce the inspectee to fully comply But the inspector is subject to two types of constraints The bound constraints (2c restrict the amounts of resource that can be allocated to each individual site whereas the budget constraint (2b restricts the total amount of resource that can be spent on all sites When a i α i p i (α i 0 a i (that is α i for each i N the bound constraints are redundant The following two theorems establish necessary sufficient conditions for a Nash equilibrium distinguishing between the case where this condition is satisfied strictly the case where it is not Specifically let a i ν {i N <α i } ν 0 {i N α i } ν + {i N a i >α i } then ν + implies that the bound constraints are redundant THEOREM Suppose ν + (α i a i for each i N Necessary sufficient conditions for (x y to be a Nash equilibrium along with the resulting payoffs are listed below for the four mutually exclusive collectively exhaustive cases i B> j N a j a i ( ai (xi y i ρ i if i ν 0 (ξ i 0 if i ν (3 with a i ξ i α i for i ν i ν ξ i B a i i ν 0 ρ i 0 for i ν 0 i ν 0 ρ i The corresponding payoffs are ii B j N Û I (x y i ν 0 ρ i a i (4 U V (x y 0 a j for some ζ 0 (xi y i ( ai ρ i ( ai ζ if i ν 0 if i ν (5 with ρ i ζ for i ν 0 i ν 0 ρ i + i ν ζ The corresponding payoffs are Û I (x y i ν 0 ρ i a i + i ν ζ a i (6 U V (x y 0

5 Deutsch et al Inspection Games with Allocation Bounds 29 iii k a j a k j (xi y i a i δ with δ <B< k a j a k+ j for some k N k j if i k (0 0 if k<i n (7 k a j j b B j k j iv B k a j a k+ j The corresponding payoffs are Û I (x y U V (x y δ B k j (8 for some k { n } (xi y i a η i a k+ c i k if i k j (0 η if i k + (0 0 if k + <i n (9 for some 0 η k+ c k+ (< The corresponding j payoffs are Û I (x y U V (x y a k+ ( ηb k j (20 The next result complements Theorem by considering the case where ν + In this case let τ max i N {p i (α i a i α i } (note that ν + implies τ>0 μ max {i N a i τ} N { μ} ν {i N a i τ <α i } ν 0 {i N a i τ α i } (note that τ a i α i for some i N that is ν 0 For i N τ a i α i that is α i a i τ ; hence ν + {i N a i τ >α i } ν ν 0 partition N Further for each i N \ N p i (x i a i <τ THEOREM 2 Suppose ν + Necessary sufficient conditions for (x y to be a Nash equilibrium along with the resulting payoffs are listed below for the three mutually exclusive collectively exhaustive cases i B> a j τ j N ( ai τ ρ i if i ν 0 (xi y i (ξ i 0 if i ν (ξ i 0 if i N \ N with a i τ (2 ξ i α i for i ν 0 ξ i α i for i N \ N a i τ i ν 0 + i ν ξ i + i N\ N ξ i B ρ i 0 for i ν 0 i ν 0 ρ i The corresponding payoffs are Û I (x y ( ai τ ρ i (22 b i ν i 0 U V (x y i ν 0 ρ i τ a j τ ii B j N for some ζ 0 ( ai τ ρ i if i ν 0 ( ai τ (xi y i ζ if i ν \{μ} or i μ a μ >τ (0 ρ μ if i μ ν a μ τ (0 0 if i N \ N (23 with ρ i ζ for i ν 0 ρ μ ζ c μ if μ ν a μ τ The corresponding payoffs are If i μ a μ >τ Û I (x y ( ai τ ρ i i ν 0 + i ν \{μ} ( ai τ ζ U V (x y i ν 0 ρ i τ + If i μ ν a μ τ Û I (x y i ν 0 ρ i ( ai τ U V (x y i ν 0 ρ i τ + + ζ i ν \{μ} i ν \{μ} ( aμ τ b μ + ζ ρ + ζ c μ τ i ν \{μ} ζ τ + ρ μ τ (24 ( ai τ ζ (25 iii 0 <B< a j τ j N (x y is a Nash equilibrium for the model where each α i is replaced by ᾱ i max{α i a i } to which Theorem applies

6 30 Naval Research Logistics Vol 60 (203 4 SENSITIVITY OF NASH EQUILIBRIA TO BUDGET CHANGES The purpose of this section is to explore the effect of the changes in the amount B that is available to the inspector on the Nash equilibria To do so we consider imperfect information variant of the game we solved In the modified game each player knows its the opponent s possible actions payoff functions but B is an exogenous parameter which is unknown to the players We assume that the players have some beliefs about B s value but are uncertain about it We will consider the case where ν + The values B k a j a k+ j for k 0 n will be referred to as singular amounts of the inspector s resource (k n corresponding to B a j j N These singular amounts are treated distinctly in Theorem The set of nonsingular amounts is the union of disjoint open intervals I k ( k a j a k j k a j a k+ j k n; the closure of each I k will be denoted I k The case where ν + is addressed in Theorem which demonstrates (i for nonsingular amounts B < a j j N Nash equilibrium allocations of the inspector the inspectee are unique (ii for singular amounts Nash equilibrium allocations of the inspector are unique whereas the inspectee has multiple Nash equilibrium allocations finally (iii for B> a j j N Nash equilibrium allocations of both players are not unique In either case the Nash equilibrium allocations of the players are in product form that is their selection (by the inspector by the inspectee are independent Consequently the notion equilibrium strategies for the inspector for the inspectee will be used Further the sets of equilibrium strategies will be indexed by B using the notation x (B y (B respectively Formally x ( y ( express point to set mappings but we refer to them as functions on subsets of their domain on which the corresponding ranges consist of singletons In addition the strategy sets will also be indexed by B that is X(B Y(B Properties of x ( y ( are next recorded The symbol ± is used to indicate either + or LEMMA 3 Assume that ν + i For B > a j j N y (B {y Y(B y i 0 for each i ν y i 0 for each i ν 0 } x (B {x X(B a i x i α i for each i N} ii x ( is a piecewise linear continuous weakly increasing function on the interval (0 a j j N ] Further for i N xi ( is zero on (0 i ] for i k n its slope on I k is j a j a i k j which decreases in i k n so xi ( is concave on [ i j n a j j ] a j a i iii For i N y i ( is zero on (0 i j i k n yi ( equals k j which decreases in k a j a i for on I k a constant iv For i N any singular amount B let yi (B ± 0 lim ɛ 0 yi (B ± ɛ IfB k a j a k+ j for k N \{n} then yi (B y i (B 0 k y j i (B + 0 k+ j for i k y i (B 0 0 y i (B + 0 k+ j for i k + {0} if i k + 2 n (26 The results of Lemma 3 are next used to determine the equilibrium utility functions of the inspector of the inspectee The notation (Û I (B (U V (B will be used for the set of equilibrium utility payoffs of the corresponding player under B for example (Û I (B {Û I (x y (x y is a Nash equilibrium when B is the amount of available resource} As is done for the equilibrium strategies we refer to (Û I ( (U V ( as functions on subsets of their domain on which the corresponding ranges consist of singletons THEOREM 3 Assume that ν + i For B> a j j N (U V (B 0 (Û I (B 0 ii (U V ( is piecewise linear continuous decreasing convex in B on the interval (0 a j j N ] further its slope on I k is k ; these negative constants j increase in k iii For k N (Û I ( is linearly increasing in B on I k with slope k ; these positive constants decrease j in k iv For singular amount B k a j a k+ j let (Û I (B ± 0 lim ɛ 0 (Û I (B ± ɛ IfB k j a j a k+ for k N then

7 Deutsch et al Inspection Games with Allocation Bounds 3 (Û I (B (Û I (B 0 (Û I (B + 0 k j ( a j a k+ k j k j ( a j a k+ k+ j (27 Lemma 3 part (ii Theorem 3 parts (iii (iv demonstrate a counterintuitive phenomenon while the x i s are monotonically increasing continuous in B (Û I ( is not In particular (Û I ( increases on each (open interval I k of nonsingular amounts then it drops at the singular amounts that is (Û I ( is not monotonically increasing in B At the following interval I k+ the slope of (Û I ( is still positive but lower This nonmonotonicity effect might be considered as a variant of the Braess paradox Ref [4] adding more resources to a network where the agents are strategic rational does not necessarily increase their payoffs Here the resources are added to the inspector affect negatively only its payoff This variant of the Braess paradox occurs from a similar reason as the original one noncooperative Nash equilibria are not necessarily Pareto efficient Ref [6] In these singular amounts the inspectee is indifferent between violating in sites k ( k<n violating in sites k + while the inspector is determined to inspect only in sites k Hence when the inspectee decides to violate in the larger set of sites ( k + the inspector s payoff decreases This observation means that for every selection of parameters for this game there is a set of budget-values beyond which there are intervals of discontinuity in the budgets allocated to the inspector In other words the inspector s true payoff function can be depicted as a step function for example there is an interval of discontinuity in budgets allocated for the inspector at [5 75] in the (following Figure 3 That is if the inspector is offered a budget within this interval than it would prefer to stay with a budget of 5 This phenomenon is further demonstrated in the forthcoming numerical examples 5 NUMERICAL EXAMPLES EXAMPLE Consider an example with the data given in Table (as B is an exogenous parameter it is not given as part of the game s data In this example there are seven sites Also α i a i for each i N soν + This case is addressed in Theorem In particular the inspector has no local limitations [see (2c] it can inspect the sites such that their p i (x i s will have a value of 0 Nash equilibria solutions of this problem are depicted in Figures 4 parameterically in B Wenext offer some interpretations for these figures Figures 2 represent equilibrium values of x i s y i s as a function of B respectively The inspector starts Table Data for example i a i α i inspecting site continues to increase its allocation until p (x satisfies p (x a b x a 2 Under this allocation the inspectee violates with certainty in site that is it invests y (as this site is the most beneficial to it complies with certainty in sites 2-7 that is it invests 0 in them At B 5 sites 2 are equally beneficial for the inspectee it switches to a mode of partial violation at them with 066 y 0 y y i 0 for i 3 7 When 5 <B<5 the extra resource B 5 of the inspector is allocated to sites 2 such that p (x p 2(x2 a 3 The inspectee partially violates in sites 2 with y 066 y complies with certainty at the other sites with yi 0 for i 3 7 When B 5 sites 2 3 are equally beneficial for the inspectee hence it partially violates at these sites with 046 y y y 3 03 yi 0 for i 4 7 When 5 <B<2 the extra resource B 5 of the inspector is allocated to sites 2 3 in the same way so on until the point where B 22 At this point all sites are inspected such that their p i (xi s have a value of 0 so the inspectee complies with certainty Figure 3 plots Û I (x y as a function of B The equilibrium utility of the inspector is expressed as a point-to-set function of the amount of resource B which exhibits drops at the points where the p i (xi s of the inspected sites have the same value as that of the next uninspected site that is at singular amounts of resource This point to set map is linear increasing between jumps with decreasing slopes in progressing intervals Figure 4 expresses U V (x y as a function of B The equilibrium utility of the inspectee is piecewise linear decreasing EXAMPLE 2 Consider an example with the same data as in Table except for α 4 α 5 which now are changed into 8 7 respectively Hence α 4 < a 4 b 4 α 5 < a 5 b 5 so ν + The case where ν + is addressed in Theorem 2 In this example τ max i N p i (α i 0 μ {max i N a i τ} 5 implying that N { 5} Hence the maximum amount of resource the inspector may use is a i τ i N 55 The inspector the inspectee act in a similar way to that explained in Example However the

8 32 Naval Research Logistics Vol 60 (203 Figure I s equilibrium strategies as a function of B when ν + [Color figure can be viewed in the online issue which is available at wileyonlinelibrarycom] inspector cannot inspect beyond the point where the p i (x i s of sites 5 have the same positive value of 0 so the inspectee is not deterred from violation The inspector s equilibrium values the inspectee s equilibrium values are not plotted here because they are almost identical to the relevant figures in Example with Figure 2 V s equilibrium strategies as a function of B when ν + [Color figure can be viewed in the online issue which is available at wileyonlinelibrarycom]

9 Deutsch et al Inspection Games with Allocation Bounds 33 Figure 3 I s equilibrium payoffs as a function of B when ν + the difference that the graphs are cut in the budget value of 55 The inspector s payoff function the inspectee s payoff function are depicted in Figures 5 6 parameterically in B 6 CONCLUDING REMARKS This article considers a (one stage noncooperative nonzero-sum inspection game between an inspector an inspectee who exhibit conflicting interests Both players have Figure 4 V s equilibrium payoffs as a function of B when ν +

10 34 Naval Research Logistics Vol 60 (203 Figure 5 I s equilibrium payoffs as a function of B when ν + limited resources available for their actions there are multiple sites where they can act The inspectee has to choose a vector of violation probabilities over the sites (whose sum do not exceed one the inspector has to determine the allocation of its inspection resources over the sites to detect the violations In addition to its global resource limitation the inspector has local restrictions on the amounts of resource it can allocate to the sites All Nash equilibria solutions are derived for this game In some cases the structure of our game yields an interesting phenomenon where the inspector s utility decreases when the amount of resource available to it increases Such cases Figure 6 V s equilibrium payoffs as a function of B when ν +

11 Deutsch et al Inspection Games with Allocation Bounds 35 were demonstrated in Sections 4 5 which also provide explanation as to what cause them In a follow-up work which we have already begun the game modeled here will be further extended to situations in which the inspectee can violate with certainty in more than one site Further investigating whether the counterintuitive phenomenon we observed here is still present in a repeated game of this model can also be of interest as future work APPENDIX A PROOF OF PROPOSITION As Û I (x y U V (x y are bilinear in (x y X Y are nonempty convex compact the existence of Nash equilibria for the game follows from a classic result of Rosen Ref [] APPENDIX B PROOF OF THEOREM Sufficiency i B> a j j N Assume that (x y satisfies (3 with corresponding ξ i s ρ i s Then clearly x X y Y Also p i (xi a i xi 0 for each i N implying that a best response of the inspectee to x is any vector y Y satisfying y i 0 for i N with p i (xi <0 (Lemma (ii As {i N p i(xi <0} {i N xi > a i }{i N a i <xi α i } ν {i N yi 0} y is such a (best response On the other h if yi > 0 then (by the definition of ν 0 i ν 0 xi a i α i So xi α i whenever q i (yi yi > 0 By Lemma 2 x is a best response of the inspector to y ii B a j j N Assume that ζ 0 (x y satisfies (5 with corresponding ρ i s Then clearly x X (xi a i α i for each i N j N x j a j j N B y Y Next observe that p i (xi a i xi 0 for each i N implying (by Lemma (ii that any vector y Y is a best response of the inspectee to x In particular y is such a response On the other h q i (yi yi ζ 0 for i ν q i (yi yi ρ i ζ for i ν 0 By Lemma 2 any x X with i N x i B x i α i for each i with q i (yi >ζis a best response of the inspector to y As yi q i (yi >ζimplies that i ν 0 xi a i α i x is such a response iii k a j a k j <B< k a j a k+ j for some k N kj a j Assume that (x y b B satisfies (7 with δ j kj Then bj clearly y Y Also as k a j a k j <B< k a j a k+ j δ kj a j B kj > δ kj a j k j a j a k+ kj a k+ 0 kj a j B kj a j k j kj < kj kj a j k a j a k j kj a k a j a k For i k it then follows that α i a i xi a i δ > a i a k xi 0 (the last inequality by (0 The definition of δ also assures that j N x j x X Next observe that > a i δ k j a j δ p i (xi a i xi { δ for i k a i a k+ <δ for i k + n B So implying (by Lemma (i that a best response of the inspectee to x is any vector y Y satisfying y i 0 for i k + n kj y j In particular y is such a response On the other h q i (y i y i { ki ci for i k 0 for i k + n implying (by Lemma 2 that a best response of the inspector to y is any vector x X satisfying x i 0 for i k + n kj x j B In particular x is such a response iv B k a j a k+ j for some k { n } Assume that (x y satisfies (9 with 0 η k+ c k+ j As k < n (that is c k+ is defined clearly k+ c k+ < j For each i k xi a i a k+ > 0 (by (0 α i a i > a i a k+ xi Also j N x j k a j a k+ j B So x X Next as 0 η k+ c k+ j < it follows that yi 0 for each i N Further u N y u η + k η u c kj u cj So y Y Next observe that { p i (xi a i xi ak+ for i k + a i <a k+ for i k + 2 n implying (by Lemma (i that a best response of the inspectee to x is any vector y Y satisfying y i 0 for i k + 2 n k+ j y j In particular y is such a response On the other h η ki for i k q i (yi yi ci c k+ η for i k + 0 for i k + 2 n k+ c k+ j As 0 η +c kj it follows that k+ cj 0 η + c k+ η k j ; so 0 q k+ (yk+ c k+η η kj yi q i (yi q i(yi η kj > 0 for cj cj i k Thus (by Lemma 2 any vector x X with x i 0 for i k + n k i x i B is a best response of the inspector to y In particular x is such a response Necessity Suppose that (x y is a Nash equilibrium The following five observations will be used to prove the necessity of cases (i (iv

12 36 Naval Research Logistics Vol 60 (203 A If y 0 then (by Lemma (ii Lemma (iii for all i N p i (xi 0 that is x i a i B If yu > 0 for u N then (by Lemma (i Lemma (ii p u (xu 0 that is x u au C If xi < a i for some i N then p i (xi >0 (by Lemma (i {j N y j > 0} arg max j N p j (x j {j N p j (x j >0} { j N 0 x j < a j yj j N In particular for some w N yw > 0 x w < aw D If B< a j j N then {i N x i > 0} {i N y i > 0} (28 xi B (29 i N Indeed as j N x j B< a j j N xk < a k b k for some k N It next follows from (C that for some w N xw < aw α w yw > 0 the latter implying that q w(yw c wyw > 0 As x w <α w q w (yw >0(29 follows from Corollary Further if y u 0 then q u (yu c uyu 0 <q w(yw Corollary implies that xu 0 So x u > 0 implies y u > 0 verifying (28 E If B< a j j N u<i yi > 0 then yu > 0 Indeed assume that yu 0 Then q i(yi yi > 0 c u yu q u(yu (by Lemma (i Lemma (ii p u (xu p i(xi ; in particular a u xu p u(xu p i(xi a i < a u (the last inequality by (0 implying that xu > 0 As q i(yi >q u(yu xu > 0 Lemma 2 implies that x i α i a i therefore p i (xi a i xi 0 As yi > 0 Lemma (ii implies that for each j N a j xj p j (xj p i(xi 0 that is xj a j therefore a j j N j N x j B< a j j N a contradiction a j i B> j N It will be shown that (3 is satisfied with corresponding ξ i s ρ i s As (x y X Y it suffices to show that yi 0 for i ν xi a i for i N (the latter would imply that for i ν 0 a i xi α i a i assuring that xi a i Assume first that yu > 0 for some u ν we will establish a contradiction The assumption yu > 0 implies that q u(yu c uyu > 0 by (B xu au < α u Using Corollary it then follows that j N x j B > a j j N Consequently for some v N \{u} xv > av implying that p v (xv a v xv < 0 (by Lemma (i yv 0 But q u(yu >0 c vyv q v(yv xv > av > 0 xu <α u contradict Corollary Next assume that xu < au α u for some u N we will establish another contradiction As xu < au (C implies that for some w N yw > 0 x w < aw α w Replacing u by w in the above arguments leading to a contradiction under the assumption that yu > 0 xu au <α u apply So (x y satisfies (3 } a j ii B j N It will be shown that (5 is satisfied for some ζ 0 If y 0 then (A implies that xi a i for each i N As B j N x j a j j N B it follows that xi a i for each i N So (5 holds with ζ 0 Next assume that y 0 We claim that xi a i for each i N Indeed if xi < a i for some i N then (C implies that for some w N yw > 0 xw < aw α w in particular q w (yw c wyw > 0 It now follows from Corollary that j N x j B a j j N ;asxw < aw it further follows that xv > av for some v N \{w} implying that p v (xv a v xv < 0 [by Lemma (i] y v 0 But q w (yw >0 c vyv q v(yv x v > av > 0 xw <α w contradict Corollary The contradiction proves that xi a i for each i N AsB j N x j a j j N B it follows that xi a i for each i N that is x satisfies (5 In particular xi a i α i for each i ν 0 xi a i <α i for each i ν By Lemma 2 q i (yi yi is a nonnegative constant for i ν say it equals ζ q i (yi yi ζ for i ν 0 so y satisfies (5 iii k a j a k j <B< k a j a k+ j for some k N kj a j b B j kj bj It will be shown that (7 is satisfied with δ Evidently a j j N k a j a k+ j >B Hence (D (E apply assuring that (28 (29 hold for some α N { α} {j N y j > 0} arg max p j (xj (30 j N [the inclusion following from Lemma (i or Lemma (ii y 0 following from (28-(29] We will prove that α k Let δ max j N p j (xj By (30 (28 (29 for i α u > α yi > 0 yu a i xi p i (xi δ p α+(xα+ a α+ xu 0 αj xj j N x a j B Thus i δ xi a i a α+ for each i α therefore α a j a α+ j α j xj B > k a j a k j As k a j a k j is strictly increasing in k n it follows that α>k that is α k To establish a contradiction assume that α k + Then for i k δ a i xi p i (xi p k+(xk+ a k+ b k+ xk+ a k+ implying that xi a i a k+ ; so k a j a k+ j k j xj α j xj B< kj a j a k+ a contradiction As α k we have that xi a i δ for i k xu yu 0 for u > k; so x satisfies (7 Further kj a j δ kj a j b B j kj > bj k j x j j N x j B implying that δ kj a j b k a j a k+ j j kj a k+ > 0 Also for bj i k a i xi δ p k (xk a k implying that xi a i δ a i a k > 0 [the last inequality follows from (0] Hence k a j a k j <B j N x j x k + k a j δ j assuring that δ<a k xk > 0 It follows that 0 <x i a i δ < a i α i for i k xu 0 for u>k So Lemma 2 implies that for i k yi q i (yi η max j N q j (yj assuring that yi η Further as x a δ b < a b α (C implies that j N y j k η j ; hence η proof that y satisfies (7 kj cj completing the

13 Deutsch et al Inspection Games with Allocation Bounds 37 iv B k a j a k+ j for some k { n } It will be shown that (9 is satisfied Evidently B kj a j a k+ < a j j N B k a j a k+ j > kj a j a k k a j a k j AsB< a j j N the arguments of case (iii imply that for some α N (28 (30 hold Again let δ max j N p j (xj Arguments used to prove part (ii imply that 0 xi α k As k j a j δ α j a i δ a j δ for i α xu 0 for u>α α k xj B j j a j a k+ it follows that a k+ δ Thus for j k + 2 p j (xj a j < a k+ δ therefore by (30 (28 xj y j 0 in particular α k+ So α {k k+}asa j a k+ 0 for j k+ it follows that α a j δ j B k a j a k+ j α a j a k+ j Thus δ a k+ > 0 xi a i δ a i a k+ > 0 for i k xk+ B k j xj B k a j a k+ j 0 So we established that x satisfies (9 For i k 0<xi a i a k+ < a i α i for u>k+ u > α which implies yu 0 [by (28] x u 0 Further xk+ 0 With θ max j N {q j (yj yj } it now follows from Lemma 2 that y i q i (y i θ q k+(y k+ c k+y k+ for i k As x a a k+ b < a b it follows that p (x a b x > 0 therefore by Lemma (i j N y j k+ j y j Consequently η y k+ k j y j k j θ implying η kj that θ for i k yi θ cj η yk+ 0 is trite η yk+ θ c k+ η c k+ kj η c k+ + kj k η c k+ + j Hence y satisfies (9 with 0 η kj k+ c k+ j η kj cj Finally APPENDIX C PROOF OF THEOREM 2 Sufficiency i B> a j τ j N Assume that (x y satisfies (2 with corresponding ξ i s ρ i s Then clearly x X y Y Also τ if i ν 0 p i (xi a i ξ i τ a i ξ i a i <τ if i ν if i N \ N implying that max j N p j (xj τ > 0 A best response of the inspectee to x is then any vector y Y satisfying y 0 j N y i y i > 0 only if p i (xi τ that is x i a i τ [Lemma (i]; in particular y is such a response On the other h if yi > 0 then i ν 0 implying that xi a i τ α i So xi α i whenever q i (yi yi > 0 By Lemma 2 x is a best response of the inspector to y ii B a j τ j N Assume that ζ 0 (x y satisfies (23 with corresponding ρ i s Then clearly x X y Y Next observe that p i (xi a i xi τ>0for each i N p i (xi a i <τ for each i N \ N implying [by Lemma (i] that a best response of the inspectee to x is any vector y Y satisfying y i 0 for i N y i 0 for i N \ N j N y j In particular y is such a response On the other h j N y j assures that max j N {q j (yj yj } > 0 q i (yi yi ρ i ζ if i ν 0 ζ if i ν \{μ} or i μ a μ >τ c μ ρ μ ζ if i μ ν a μ τ 0 if i N \ N By Lemma 2 any vector x X with i N x i B x i α i for each i with q i (yi >ζ0 x i α i for each i with q i (yi ζ x i 0 for each i with q i (yi <ζis a best response of the inspector to y For the first of two cases assume that a μ >τin this case q i (yi >ζimplies i ν 0 for which xi a i τ α i q i (yi <ζimplies i N \ N for which xi 0; consequently x is a best response of the inspector to y Alternatively assume that a μ τ (which implies p μ (α μ <τ that is μ ν In this case q i (yi >ζimplies i ν 0 for which xi a i τ α i further q i (yi <ζimplies i N \ N or i μ in either case xi 0 So again x is a best response of the inspector to y iii 0 <B< a j τ j N Consider the model where each α i is replaced by ᾱ i max{α i a i } α i while all other data elements remain unchanged The model with the modified data will be referred to as the bar model (to be distinguished from the original model that we discuss The strategy set of the inspector in the bar model is X {x IR n 0 x i ᾱ i for i N j N x j B} X Also for each i N p i (ᾱ i a i ᾱ i 0 implying that {i N a i > ᾱ i } consequently Theorem applies to the bar model Assume that (x y is a Nash equilibrium of the bar model As 0 <B< a j τ j N < a j j N Theorem implies that (x y satisfies either of the conditions given in (iii or (iv of that theorem with a corresponding k ( these conditions do not depend on the α i s As X X to show that (x y is a Nash equilibrium it suffices to show that x X Assume first that (x y satisfies condition (iii of Theorem with k N that is (x y satisfies (7 with δ a i δ kj a j b B j kj bj To prove that x X it suffices to show that i k The definition of δ assures that B k a j δ j α i for

14 38 Naval Research Logistics Vol 60 (203 the sufficiency proof of Theorem assures that a k+ <δ<a k So k j a j a k k j B< a j a k < μ j k j a j τ < a j δ μ j a j a μ+ (3 which implies that k <μ that is k μ Ifk μ then (3 implies that δ>τ Alternatively if k<μ then δ>a k+ a μ τ In either case for i k a i δ < a i τ α i Next assume that (x y satisfies condition (iv of Theorem with k { n } that is (x y satisfies (9 To prove that x X it suffices to show that a i a k+ α i for i k Evidently k a j a k+ j B< μ a j τ j < μ a j a μ+ j implying that k < μ a k+ a μ τ for i k a i a k+ a i τ α i Necessity Suppose (x y is a Nash equilibrium it will be shown that it satisfies the corresponding conditions Recall that ν 0 Let t ν 0 Then p t (x t p t (α t τ > 0 Lemma (i implies that j N y j in particular y 0 Further observation (B the last conclusion of observation (C of the proof of Theorem are modified to B If yu > 0 for u N then (by Lemma (i p u (xu τ that is xu au τ C If xi < a i τ for some i N (that is p i (xi >τ then for some w N yw > 0 p w(xw >τ that is x w < aw τ α w in particular a w p w (xw >τ assures w N Observations (A (E of the necessity proof of Theorem will be used a i j τ j N <B It will be shown that (2 is satisfied with corresponding ξ i s ρ i s As (x y X Y it suffices to show that yi 0 for i (N \ N ν j N y j x i a i τ for i N (the latter would imply that for i ν 0 a i τ xi α i a i τ therefore xi a i τ Consider u N \ NAsp u (xu a u <τ p t (xt u / arg max j N p j (x j by Lemma (i or Lemma (ii yu 0 Next assume that y u > 0 for some u ν we will establish a contradiction The assumption yu > 0 implies that q u (yu c uyu > 0 by (B x u au τ <α u It now follows from Corollary that j N x j B> a j τ j N therefore for some v N \{u} xv > av τ 0 by (B yv 0or for some j N \ N xj > 0 as we showed y j 0 But q u (yu >0 c vyv q v(yv or q u(yu >0 yj q j (yj xv > 0orx j > 0 x u <α u contradict Corollary Next assume that xu < au τ α u for some u N we will establish another contradiction As xu < au τ (C implies that for some w N yw > 0 x w < aw τ α w Replacing u by w in the above arguments leading to a contradiction under the assumption that yu > 0 x u au <α u apply So (x y satisfies (2 a j τ ii B j N It will be shown that for some ζ 0 (23 is satisfied with corresponding ρ i s We first prove that for each i N xi a i τ Assume that this is not the case for some i N xi < a i τ we will establish a contradiction By (C for some w N yw > 0 implying that q w(yw c wyw > 0 x w < aw τ α w ; thus by Corollary j N x j B a j τ j N As xw < aw τ it follows that for some v N \{w} either v N \ N xv > 0 or v N xv > av τ > 0 implying that p v (xv <τ Hence in either case y v 0 q v(yv c vyv 0 But q w (yw c wyw > 0 q v(yv x w < α w xv > 0 contradict Corollary So indeed xi a i τ for each i N Consequently B j N x j j N x j a j τ j N B implying that xi 0 for j N \ N xi a i τ for each i N that is x satisfies (23 (if μ ν a μ τ then aμ τ b μ 0 It follows that xi 0 for i N \ N xi a i τ α i for each i ν 0 xi a i τ <α i for each i ν xi > 0 for i ν \{μ} xμ > 0ifa μ >τ otherwise xμ 0 By Lemma 2 q i(yi yi is a nonnegative constant for i ν \{μ} say it equals ζ further q μ (yμ c μyμ ζ if a μ > τ q μ (yμ c μyμ ζ if a μ τ q i (yi yi ζ for i ν 0 Finally for i N \ N p i (xi a i <τ (B implies that yi 0 So y satisfies (23 iii 0 <B< a j τ j N Consider the bar model introduced under (iii in the sufficiency proof with the corresponding definitions of the ᾱ i s X let (x y be a Nash equilibrium of the original model To show that (x y is a Nash equilibrium of the bar model it is sufficient to prove that for each x X \ X Û I ( x y Û I (x y By (B by Lemma (i if yi > 0 for i N then p i (xi τ that is x i a i τ further as p j (xj <τfor j N \ N yj 0 for j N \ N Consider x X \X Then there >α v x v >α v av τ imply- are indexes v N such that av ing that p v ( x v <τ so yv 0 As 0 <B< j N a j τ j N there exists a w N such that x w < aw τ p w ( x w >τ By (C for some r N yr > 0 x r < a j τ < so ar τ b r But x v > av τ yv 0 x r < Corollary In particular x X \ X cannot be a part of any Nash equilibrium so Û I ( x y Û I (x y ar τ b r y r > 0 contradict APPENDIX D PROOF OF LEMMA 3 Part (i is immediate from Theorem (i As there are finitely many singular amounts which include a j j N the set of nonsingular amounts in (0 a j j N ] is the union of disjoint open intervals to which Theorem (iii applies In particular (7 shows that on any open interval of nonsingular amounts each yi ( is the corresponding constant proving (iii Also each xi ( is linear weakly decreasing in δasδ is linear decreasing in B it follows that xi ( is linear weakly increasing in B further the slopes of xi are available from (7 are as stated in the lemma as they are decreasing the concavity of xi ( on [ i a j a i j n a j j ] follows To complete the proof of (ii it remains to establish continuity of xi ( at

15 Deutsch et al Inspection Games with Allocation Bounds 39 singular values For this purpose express δ as a function of B [that is write δ(b] note that (7 assures that k lim δ ɛ 0 j k lim δ ɛ 0 j a j a k+ a j a k ɛ + ɛ kj a j a k+ kj a j [ kj ] a j a k+ kj [ kj ] a j a k kj a k The continuity of each xi ( in B at singular amounts in now immediate from (7 (9 Finally to prove (iv observe from (9 of Theorem (iv that for k N \{n} singular amount B k a j a k+ j the range of yi (B depends (linearly on the parameter η Using (9 for i k the extreme value η 0 yields kj cj yi (B the alternative extreme value η k+ c k+ j k+ c k+ j c kj i yi (B 0 as the highest value of k+ c k+ j ( k+ c k+ j ( kj kj c k+ ( k+ c k+ j (c kj i yi (B + 0 yields k+ j as the lowest value of y i (B; so y i (B [y i (B 0 ci kj yi (B + 0 cj ] ci k+ Further again using (9 0 y j c k+ (B 0 is j the lowest value of y k+ (B while c k+ kj cj is the highest; so y k+ (B [yi (B 0 0 y i (B + 0 ci k+ ] Finally (9 assures that j c j yi (B {0} for i k + 2 n APPENDIX E PROOF OF THEOREM 3 By Lemma 3(i y (B {y Y(B y i 0 for each i ν y i 0 for each i ν 0 } for B > j N therefore (9 (4 assure that a j Û I (x y U V (x y given by (4 can now be rewritten as (Û I (B 0 (U V (B 0 Consider k N B I k By Lemma 3 (U V (B ki [ ai xi (B] yi (B (Û I (B k i xi (By i (B with each xi ( linear y i ( constant on I k implying that (U V ( (Û I ( are linear on I k The explicit expressions for the slopes of the the xi s the value of the y i s on I k imply that the slopes of (U V ( (Û I ( are respectively k ( i k i kj kj kj kj kj kj ; ki kj kj ; ki kj kj the first set of constants are negative increasing in k whereas the second set is positive decreasing in k This completes the proof of (iii To complete the proof of (ii it remains to show that for any singular amount B (U V (B is a singleton that (U V ( is continuous at B So let B be a singular amount Lemma 3(ii assures that x (B is a singleton [to be denoted x (B] therefore for all y y (B (U V (x (B y equals max y Y (U V (x (B ywhich is independent of y So (U V (B contains a single value Further Lemma 3(iv assures that this value equals the left right limits of (U V ( at B completing the proof of (ii Finally to prove (iv consider singular amount B k a j a k+ j where k N It then follows from (9 parts (ii (iv of Lemma 3 that (20 can be written as (Û I (B 0 k (Û I (B j k j ( aj a k+ ( aj a k+ ( kj aj a k+ kj ku c u k+ u c u kj ( aj a k+ ( kj aj a k+ kj (Û I (B + 0 cj ACKNOWLEDGMENTS k+ j (32 (33 ( kj aj a k+ k+ j This research was partially supported by the Daniel Rose Technion-Yale Initiative for Research on Homel Security Counter-Terrorism REFERENCES [] JB Rosen Existence uniqueness of equilibrium points for concave N-person games Econometrica 33 ( [2] R Avenhaus JM Canty Compliance quantified An introduction to data verification Cambridge University Press Cambridge CB2 RP UK 996 [3] E Boros L Fedzhora PB Kantor K Saeger P Stroud A large-scale linear programming model for finding optimal

16 40 Naval Research Logistics Vol 60 (203 container inspection strategies Nav Res Log 56 ( [4] D Braess Ueber ein Paradoxen der Verkehrsplanung Unternehmensforschung 2 ( [5] Y Deutsch B Golany UG Rothblum Determining all Nash equilibria in a (bi-linear inspection game Eur J Oper Res 25 ( [6] P Dubey Inefficiency of Nash Equilibria Math Oper Res (986 8 [7] N Haphuriwat VM Bier HH Willis Deterring the smuggling of nuclear weapons in container freight through detection retaliation Dec Anal 8 ( [8] R Hohzaki An inspection with smuggler s decision on the amount of contrab J Oper Res Soc Japan 54 ( [9] R Hohzaki R Maehara A single-shot game of multiperiod inspection Eur J Oper Res 207 ( [0] H Kellerer U Pferschy D Pisinger Knapsack Problems Springer Heidelberg ISBN [] M Maschler A price leadership method for solving the inspector s non-constant-sum game Nav Res Log Q 3 ( [2] M Maschler The inspector s non-constant-sum game Its dependence on a system of detectors Nav Res Log Q 4 ( [3] D Xiaoqing L Chaolin L Ji W Jia H Wantao A game-theoretic analysis of implementation of cleaner production policies in the Chinese electroplating industry Resources Conservation Recycling 54 (