CN108053068A

CN108053068A - The method that mankind attacker cooperation behavior modeled and formulated corresponding Defending Policy

Info

Publication number: CN108053068A
Application number: CN201711324273.1A
Authority: CN
Inventors: 王彬入; 张渊; 仲盛
Original assignee: Nanjing University
Current assignee: Nanjing University
Priority date: 2017-12-13
Filing date: 2017-12-13
Publication date: 2018-05-18

Abstract

The present invention discloses the mankind attacker cooperation behavior modeling in a kind of security protection and the method for formulating corresponding Defending Policy, first, an adaptive human behavior model is proposed, for optimizing distribution of the tool there are two limited defence resource in the repetition SSG for the attacker that may cooperate；Secondly, devise an efficient algorithm and approximate solve the Nonlinear Nonconvex optimization problem obtained by model, final output Defending Policy；Finally, true man's simulated experiment has been carried out to test the validity of model of the present invention and method.

Description

The method that mankind attacker cooperation behavior modeled and formulated corresponding Defending Policy

Technical field

The present invention relates to the mankind attacker cooperation behavior modeling in a kind of security protection and formulate corresponding Defending Policy Method belongs to technical field of safety protection.

Background technology

The safe games of Stackelberg (SSG) are applied in a variety of safeguard protection scenes in real life, such as beautiful State's Coast Guard, federal air marshal, Los Angeles Airport etc..

Instantly it is to patrol to needing the most common defence method that target to be protected is taken in the case where defence resource is limited It patrols.Patrolman only has limited defence resource (such as human resources), so they cannot be a piece of very big in guard of same time Area every nook and cranny.In addition, attacker can be with the patrol rule of observation patrol person, and corresponding attack action is designed to keep away Exempt to be grabbed.Therefore, it is significant for defence organization's optimization Defending Policy.

In traditional single-wheel SSG, defender (leaders) arranges patrol strategy first, then attacker (follower) root Target attack is selected to the observation of Defending Policy according to it.

The main problem of traditional single-wheel SSG is the incomplete reasonability for not accounting for mankind attacker, and without abundant Utilize the passing data-optimized Defending Policy of attack.

The content of the invention

Goal of the invention：For problems of the prior art and deficiency, the present invention provides the people in a kind of security protection The method that class attacker cooperation behavior modeled and formulated corresponding Defending Policy, first, it is proposed that an adaptive human behavior Model, for optimizing distribution of the tool there are two limited defence resource in the repetition SSG for the attacker that may cooperate；Secondly, design One efficient algorithm approximate solve the Nonlinear Nonconvex optimization problem obtained by model, final output Defending Policy； Finally, true man's simulated experiment has been carried out to test the validity of model of the present invention and method.

Technical solution：A kind of side of the modeling of mankind attacker cooperation behavior and the corresponding Defending Policy of formulation in security protection Method, including following content：

(1) cooperative mechanism in SSG is repeated

Repeatedly defender and two attackers, the two attackers can choose whether to cooperate there are one cooperative games；It is anti- The person of keeping periodically arranges new Defending Policy as the leader in game, and attacker is as follower, according to it to defender Attack is made in the observation of the Defending Policy of arrangement；This game is formed more by taking turns, thus defender can periodically according to front-wheel The data that are collected into secondary change Defending Policy；Attacker independently attacks or cooperation attack；In the case where independently attacking, Whether the income that attacker attacks every time is only dependent upon the attacker this attack and succeeds.And in the case where cooperation is attacked, it attacks The income that the person of hitting attacks every time by be two people's income sums half；An extra returns ∈ is introduced simultaneously, i.e., when an attack When person has successfully carried out once cooperation attack, he will obtain extra returns ∈；

Attackers need to make two main selections in game：1) selection attack target and 2) choose whether with separately One attacker cooperates.Only when two attackers select cooperation in certain attack, a cooperative relationship just calculates foundation.

The defence resource summation M possessed by defender defends a target collection T={ 1 ..., | T | }, and T is by T₁With T₂Two are not overlapped subset and are combined into；T₁Represent the target collection that first attacker may attack, T₂Represent second attacker The target collection that may be attacked, and T₁=T-T₂；Vector x is defended, per one-dimensional element x_iDefender is represented in each target The defended probability of the defence resource namely target i of input.

Income of one attacker in target of attack i depend on 1) this target it is whether defended and 2) two attack Whether person cooperates.When an attacker selects independent attack, the income that he attacks not defended target i isAttack quilt The income of the target i of defence isRelatively, defender is in the income of the previous caseIn the income of latter situation ForThe total revenue of defender for it in two incomes by target of attack and.AndIf two A attacker agrees to cooperate, they will obtain extra returns ∈ when attacking not defended target.

(2) incomplete reasonability and adaptivity

In some illegal activities, attacker is typically considered Imperfect Rationality.For the mankind's in SSG is repeated Incomplete reasonability proposes a model.

Tendentiousness of the attacker for some target is defined first.

Define 1：One attacker is in R wheels to the tendentiousness of target iIt is defined by following equation：

WhereinIt isWithDifference,Represent the single attack that attacker r wheels carry out at target i Average yield,Represent the single attack average yield for all attacks that attacker carries out in r wheels, Q^rFor a vector, Its i-th dimension isC be a constant, Var (Q^r) represent { Q^rVariance.Therefore in fixation { Q^rDistribution in the case of, increase AddAlso can accordingly increase.First equation explanationForAverage (r=1 ..., R).

When attacker tends to target of attack iTo be just, on the contrary will be negative, and attacker be not revealed inclining for target Tropism is set to 0.

Adaptive subjectivity revenue function：

D=1/ (N are set_r- r), wherein vector ω=(ω₁, ω₂, ω₃) for parameter to be learned, N_rAlways to take turns number, r is represented For learning the round of ω.

(3) optimization problem provides

With reference to human behavior model and from the parameter that attack data learning obtains, established by equation 1 to equation 19 Generate the optimization problem of optimal Defending Policy.

D=d₁+d₂ (2)

α₁, α₂, β ∈ { 0,1 } (3)

β≤α₁ (17)

β≤α₂ (18)

α₁+α₂≤β+1 (19)

In all equatioies, Z represents a larger constant, and c represents cooperation, and nc represents uncooperative.

Formula 1 gives the expected revenus of the target of optimization problem, i.e. defender.From 2 visual target of formula by d₁And d₂Two parts It forms.We define three binary variables, wherein α in formula 3₁And α₂Attacker is represented each for the selection of cooperation, β generations The final cooperation state of table.Only when two attackers agree to cooperation, β 1, β is 0 in all other cases.This It can be ensured by the constraints in formula 17 and 19.Formula 4 and 5 ensure that the every one-dimensional all between [0,1] of defence vector x, and The total resources of defender are not more than M.In formula 6 to 11, based on α₁、α₂Two parts of defender's expected revenus are defined with β --- d₁And d₂。

With d₁Exemplified by, if (i.e. β=0) is not reached in cooperation, for d₁Constraints will be determined by formula 6 and 7.On the contrary, If (i.e. β=1), d are reached in cooperation₁Constraints will be determined by formula 8.Similarly, formula 9 to 11 is constituted to d₂Constraint.

Formula 12 to 14 respectively define attacker cooperate with it is uncooperative when expected revenus.Here two attackers are distinguished Modeling, so they have different expected revenusesWithBy formula 15 and 16, α ensure that₁And α₂It is to be received based on expectation What benefit determined.As previously mentioned, three constraintss in formula 17 to 19 are for ensuring that correctly running for cooperative mechanism, i.e., only There is the cooperation when two attackers select cooperation just to reach.

(4) optimal policy is calculated

Because above-mentioned optimization problem is complicated, this optimization problem is decomposed into 4 subproblems.By solving subproblem, and Globally optimal solution is selected in the optimal solution of subproblem, just can solve former problem, obtains defence vector x.

PROBLEM DECOMPOSITION

The angle of cooperation whether is selected from attacker, may be occurred there are four types of situation altogether.Table 3 summarizes defender at this Expected revenus (while being also the object function of subproblem) D in the case of four kinds₁、D₂、D₃And D₄。

Table 3

Expected revenus in given table 3, just can correspondingly establish 4 subproblems.Without loss of generality, one of them is only provided The definition (formula 20 to 24) of subproblem.In this subproblem, first attacker selects cooperation and second attacker selects It is uncooperative.The mark used in subproblemWithCome from the definition in previous formula 12 to 14.

Based on approximate method

Four subproblems suffer from the simple form of comparison, but the globally optimal solution for finding subproblem is still difficult , therefore piecewise linear function is introduced, constraints is loosened, it is secondary that subproblem is converted into a MIXED INTEGER quadratic constraints Planning problem (MIQCQP) solves.Such issues that can be with any MIQCQP solutions come numerical solution.

Some variable replacements are done first.Order So D₄A quadric form can be rewritten as, sees formula 26.Because introduce four A new variable to s and z according to definition, it is necessary to increase by four constraintss.Here as example is only provided for s_pPact Beam, others constraint can be provided similarly.

Formula 25 is a nonlinear restriction, therefore it is replaced with two groups and slightly loosened to obtain quadratic constraints condition.Now It illustrates how to find such two set of segmentation linear functions.

OrderIt needs to find two groups of suitably piecewise linear functionsWithSo thatHere f (x_i) on [0,1] it is a monotone decreasing convex function, because For negative.

Use x_ikDefinitionWith

WhereinIt representsThe slope of middle kth section broken line,It representsThe slope of middle kth section broken line.So In order to defineWithOnly it needs to be determined thatWith

Theorem 1：According to piecewise linear functionWithMeet

After the piecewise linear function met the requirements is determined, the constraints in formula 25 can be loosened.Specifically, will Formula 25 is replaced with following two groups of inequality.

This two groups of inequality are denoted as Cons (s_p).By s_r、z_pAnd z_rThe similar other inequality introduced are denoted as respectively Cons(s_r)、Cons(z_p) and Cons (z_r)。

Based on the modification above to subproblem, corresponding MIQCQP is established：

Cons(s_p), Cons (sr), Cons (z_p), Cons (z_r) (30)

Subproblem is rewritten with the variable newly introduced in formula 26 to 29.Formula 30 represents the quadratic constraints of generation.31 He of formula 32 be to x_tkConstraint.

MIQCQP there are many ∈-global optimization solution, come by the SCIP methods proposed using Vigerske S.et al. Solve MIQCQP.

Theorem 2：NoteForWithThe distance between, then lim_K→+∞ Dis=0.

Description of the drawings

Fig. 1 be as K=4,WithComparison diagram；

Fig. 2 be as K=10,WithComparison diagram；

Fig. 3 is the experiment interface of participant in experiment；

Fig. 4 is the earnings structure S1 and S2 used in two groups of experiments；

When Fig. 5 is K=10, using earnings structure S1 manoeuvre second wheel in " Maximin " method generate strategy and The strategy of model generation proposed by the present invention, wherein (a) is the strategy of " Maximin " method generation, (b) proposes for the present invention The strategy of model generation；

Fig. 6 is the loss comparison diagram that defender often takes turns；

Fig. 7 is the ratio chart that cooperation attack accounts for general offensive number.

Specific embodiment

With reference to specific embodiment, the present invention is furture elucidated, it should be understood that these embodiments are merely to illustrate the present invention Rather than limit the scope of the invention, after the present invention has been read, those skilled in the art are to the various equivalences of the present invention The modification of form falls within the application scope as defined in the appended claims.

A kind of method of the modeling of mankind attacker cooperation behavior and the corresponding Defending Policy of formulation in security protection, including such as Lower content：

(1) cooperative mechanism in SSG is repeated

Repeatedly defender and two attackers, the two attackers can choose whether to cooperate there are one cooperative games；One A defender or attacker can represent a tissue and not necessarily only represent personal；Defender as the leader in game, Periodically arrange new Defending Policy, and attacker does the observation of the Defending Policy of defender's arrangement according to it as follower Go out attack；This game is formed more by taking turns, so defender can periodically change according to the data that be collected into round before Defending Policy；Attacker independently attacks or cooperation attack；In the case where independently attacking, the income that attacker attacks every time is only Depending on the attacker, whether this attack succeeds.And in the case where cooperation is attacked, the income that attacker attacks every time will be The half of two people's income sums；An extra returns ∈ is introduced simultaneously, i.e., when an attacker is successfully once cooperated During attack, he will obtain extra returns ∈；

Generally speaking, attackers need to make two main selections in game：1) target of selection attack and 2) choosing It selects and whether cooperates with another attacker.Only when two attackers select cooperation in certain attack, once cooperation is closed System just calculates and establishes.

Table 1

In order to further describe this game, the mark being introduced into table 1.One target collection T=is defendd by M { 1 ..., | T | }, T is by T₁And T₂Two are not overlapped subset and are combined into.T₁Represent the object set that first attacker may attack It closes, T₂Represent the target collection that second attacker may attack, and T₁=T-T₂.The two subsets are non-overlapped, if becauseAttacker may attack identical target, this probability that them will be caused all to be arrested in once attacking Rise.Vector x is defended, per one-dimensional element x_iRepresent defence resource namely the target i that defender is put into each target Defended probability.

Table 2

Income of one attacker in target of attack i depend on 1) this target it is whether defended and 2) two attack Whether person cooperates.When an attacker selects independent attack, the income that he attacks not defended target i isAttack quilt The income of the target i of defence isRelatively, defender is in the income of the previous caseIn the income of latter situation ForThe total revenue of defender for it in two incomes by target of attack and.Because income and mistake that attacker successfully obtains Loss and income of the loss being subject to respectively from defender are lost, so it is contemplated that zero-sum game.Then haveWith AndIf two attackers agree to cooperate, they will obtain extra returns when attacking not defended target ∈.The income of defender and each attacker are shown in Table 2 under cooperation.In table 2, i ∈ T₁Represent first attacker's selection Target of attack, j ∈ T₂Represent the target of attack of second attacker's selection." A1 " represents first attacker, and " A2 " is represented Second attacker." S " is represented successfully, and " F " represents failure.Such as " A1S, A2F " represent first attacker success and second A attacker's failure.

(2) incomplete reasonability and adaptivity

Tendentiousness of the attacker for some target is defined first.

Generally speaking, when the average single attack income that attacker is obtained in some target i in passing round is more higher than institute There is the overall average single attack income of target,It will be higher.This is also consistent with intuition --- and attacker should be more likely to attack It hits him and has obtained the target of higher average single income.

It should be noted that it is likely to occur the feelings that some target is not all attacked in a certain even all rounds of wheel Condition, the information of these targets need to be disclosed.Because when attacker tends to target of attack iWill be just, otherwise will be it is negative, So attacker is set to 0 by us in the tendentiousness for not being revealed target.

There is above tendentious definition, we have proposed adaptive subjective revenue functions：

ASU(x_i) in there are a kind of adaptation mechanism, it by introduce d andTo realize.Here d is one to be directed to The parameter that target information is insufficient and introduces.D=1/ (N are set_r- r), wherein vector ω=(ω₁, ω₂, ω₃) it is ginseng to be learned Number, N_rAlways to take turns number, r represents the round for being used for learning ω.So will increase with the increase d of round, represent attacker and obtain Obtained more information.

In x_iWeight before multiplyButWithMultiply beforeThis is because Attacker will be according to tendentiousness in modelIts subjective revenue function is adjusted, it is necessary to ensure after parameter substitution learn, Work as tendentiousnessDuring increase, ASU (x_i) also corresponding increase.In an experiment, all ω learnt₁And ω₃0 is all no more than, All ω₂0 is all greater than, so being made that such adjustment to weight.

Finally, the model of the present invention is summarized in the case where cooperation repeats SSG.In game, two attackers can select be No cooperation.At each occurrence, they have different attack probability at target i, are denoted asWithout loss of generality, only provide belowCalculating side Method, other three probability can also be calculated similarly.

Above formula subjectivity revenue functionIt is given by：

Wherein parameterIt is with non-conjunction of the maximum Likelihood (MLE) from first attacker Make what attack data learning arrived.For ease of use, we willIt is rewritten as x_iLinear expression.

(3) optimization problem provides

With reference to human behavior model and from the parameter that attack data learning obtains, we arrive equation 19 by equation 1 Establish the optimization problem for generating optimal Defending Policy.

D=d₁+d₂ (2)

α₁, α₂, β ∈ { 0,1 } (3)

β≤α₁ (17)

β≤α₂ (18)

α₁+α₂≤β+1 (19)

In all equatioies, it is (desirable that Z represents one big constantC generations Table cooperation, nc represent uncooperative.

Formula 1 gives the expected revenus of the target of optimization problem, i.e. defender.From 2 visual target of formula by d₁And d₂Two parts It forms.We define the integer variable that three values are only 0 or 1, wherein α in formula 3₁And α₂It is each right to represent attacker In the selection of cooperation, β represents final cooperation state.Only when two attackers agree to cooperation, β 1, at any other In the case of β all be 0.This can be ensured by the constraints in formula 17 and 19.Formula 4 and 5 ensure that defence vector x per it is one-dimensional all Between [0,1], and the total resources of defender are not more than M.

In formula 6 to 11, based on α₁、α₂Two parts of defender's expected revenus are defined with β --- d₁And d₂.With d₁For Example, if (i.e. β=0) is not reached in cooperation, for d₁Constraints will be determined by formula 6 and 7.On the contrary, if cooperation is reached (i.e. β=1), d₁Constraints will be determined by formula 8.Similarly, formula 9 to 11 is constituted to d₂Constraint.

(4) optimal policy is calculated

Because above-mentioned optimization problem is complicated, this optimization problem is decomposed into 4 subproblems.By solving subproblem, and Globally optimal solution is selected in the optimal solution of subproblem, just can solve former problem, obtains defence vector x.As described in table 1, x is represented (i.e. M is defended resource preventing every time to the specific defence scheme of each target any time defended probability, x and defender Which position should be appeared in keeping) it is of equal value, therefore be just equivalent to obtain Defending Policy we can say that having solved x.

PROBLEM DECOMPOSITION

Table 3

Based on approximate method

OrderIt needs to find two groups of suitably piecewise linear functionsWithSo thatHere f (x_i) on [0,1] it is a monotone decreasing convex function, because For negative.Based on descending and convexity, we are illustrated in Fig. 1 as K=10, f (x_i) and piecewise linear function WithFunctional image on [0,1].

It will be seen that [0,1] is equally divided into K (K=4 here) part from Fig. 1, per portion by a new variable x_ik(k=1,2 ..., K) is represented.Obviously,And Use x_ikDefinitionWith

WhereinIt representsThe slope of middle kth section broken line,It representsThe slope of middle kth section broken line.So In order to defineWithOnly it needs to be determined thatWithIt will calculate WithStep Suddenly it is summarized as algorithm 1.

Above-mentioned algorithm 1 is specifically described as：

1) it is directly substituted into firstIt calculatesValue.

2) slope of kth section (k=1 ..., K) is then calculated.

3) Δ is made_{I, k}ForOnMaximum, calculate Δ_{I, k}The value and basis of (k=1 ..., K) Δ_{I, k}It calculates

4) Δ is taken_{I, 0}=Δ_{I, 1},

5) basisIt calculates

6) exportWith

Theorem 1：The piecewise linear function generated according to algorithm 1WithMeet

It proves：BecauseIt is [0, a 1] Convex Functions, it is clear that

Below based onIt provesGiven k ∈ 1, 2 ..., K }, orderHave

AndNotice g (x_i) andAll it isOn linear function.Therefore,

On the other hand, becauseIt can be seen that

Thus obtainSo

After the piecewise linear function met the requirements is determined with algorithm 1, the constraints in formula 25 can be loosened.Tool Body, the following two groups of inequality of formula 25 are replaced.

Cons(s_p), Cons (s_r), Cons (z_p), Cons (z_r) (30)

Subproblem is rewritten with the variable newly introduced in formula 26 to 29.Formula 30 represent the method for the present invention generation it is secondary about Beam.Formula 31 and 32 is to x_tkConstraint.

Theorem 2：NoteForWithThe distance between, then lim_K→+∞ Dis=0.

It proves：It is based onWithDefinition andIt is clear toOrder Thenφ (x) existsOn maximum be NoteThen

NoteIt obtains

Δ_{I, 1}=wlnw-w+1

Because

It obtains

Theorem 2 shows that the method for the present invention is capable of providing and constraints arbitrary extent is loosened, this is because two set of segmentation The distance between linear function can be arbitrarily small.Fig. 2 is illustrated as K=10,、f(x_i) andWhat can be reached connects Short range degree.Therefore, after a suitably K has been selected, the method for the present invention can obtain the approximate solution of four subproblems.

(5) test

In order to estimate the parameter in cooperation human behavior model and test its validity, tested, and analyze reality Test result.

Test general view

In an experiment, participant needs to play the part of the attacker for being carried out wildcat operation by Conservation areas at one piece, carries out manoeuvre. The panel of manoeuvre such as Fig. 3.

Protected area has been divided into 2 pieces of regions, and each region is divided into as 9 fritters, and each fritter represents a target. Manoeuvre needs 2 participants to simultaneously participate in, they are referred to as Alice and Bob.Alice and Bob can be with before manoeuvre starts Observe 18 all targets, income at each target when the defence probability of defender, attacker's independence success attack and Punishment during independent attack failure can all show them.As auxiliary, a temperature figure is provided to aid in participant's observation every The defence intensity of defender at a target.The color of target is deeper in figure, its defended probability is higher.In addition, also it is two A participant respectively provides a form, when form summarizes cooperation attack can obtainable income and its generation probability.Alice The target of half can only be all attacked with Bob, i.e. Alice can only attack 9, left side target, and Bob can only attack 9 targets in the right. Two participants need 1) from oneself one target of selection and can 2) to decide whether and another one participant closes in firing area Make.After two participants press attack (Attack) key, this attack terminates, and the result of attack can be illustrated in manoeuvre panel Portion, two respective remaining sums of participant can also be updated according to result.

Experiment setting

Participants, which need to play, amounts to 5 wheel manoeuvres, often takes turns manoeuvre and is made of several offices.In the first round, due to not yet Any attack data are collected into, we use Gholami, the Defending Policy for the method generation first round that S.et al. are proposed, with Under we term it " Maximin " methods.For the validity of evaluation method, the strategy for respectively generating " Maximin " method It is applied to the strategy of model proposed by the present invention generation in 5 wheel manoeuvres.Two groups of manoeuvres are provided with altogether, they have different Earnings structure S1 and S2, every group of manoeuvre tool correspond to the strategy that two methods generate respectively there are two 5 wheel manoeuvres.

The two kinds of earnings structure such as Fig. 4 (receipts of independence success attack at each target of digitized representation used in an experiment Benefit).Punishment at each target is set to -1.The extra returns ∈ of co-participant is set to 1.Total defence resource of defender M is set to 9, it means that the same time only has 9 targets can be defended.

Interpretation of result

In cooperation SSG, our main target is the cooperation destroyed between attacker.When Fig. 5 illustrates K=10, adopt It is generated with the tactful and proposed by the present invention model that " Maximin " method generates in the second wheel of the manoeuvre of earnings structure S1 Strategy (percentage that probability is defended at each target of digitized representation).As seen from Figure 5, the defence intensity that two attackers face has Significant difference.

For the incomplete reasonability of mankind attacker, we have collected data and are drilled according to often wheel (except last wheel) It practises data and has estimated model parameter.Here we listed in table 4 and table 5 from attack behavior learning to parameter.

Table 4

Table 5

In order to show validity of the model in terms of defender's average loss is reduced of the present invention, carry out in figure 6 pair Than.As seen from Figure 6, except the first round, the strategy that our model obtains in two kinds of earnings structures can provide in each round The significantly low defender loss than " Maximin " method.In addition, model of the invention is in the second wheel horse back initially applied With regard to defender's loss can be reduced.

Finally calculate cooperative rate, i.e., cooperation attack proportion in one wheel.Fig. 7 shows the experiment on two kinds of earnings structures As a result.When using the model of the present invention, the cooperative rate of the second wheel person of beginning participating in is decreased obviously and is kept low.With Upper result is consistent with the result in Fig. 6, and further illustrates the method for the present invention better than " Maximin " method.

Claims

1. a kind of method of the modeling of mankind attacker cooperation behavior and the corresponding Defending Policy of formulation in security protection, feature exist In including following content：

(1) cooperative mechanism in SSG is repeated

Game is formed more by taking turns, and attackers need to make two selections in game：1) target of selection attack and 2) selection Whether cooperate with another attacker.Only when two attackers select cooperation in certain attack, a cooperative relationship It just calculates and establishes；In the case where cooperation is attacked, income that attacker attacks every time by be two people's income sums half；It introduces simultaneously One extra returns ∈, i.e., when an attacker has successfully carried out once cooperation attack, he will obtain extra returns ∈；

(2) incomplete reasonability and adaptivity

Incomplete reasonability for the mankind in SSG is repeated proposes a model；

Tendentiousness of the attacker for some target is defined first.

<mrow> <msubsup> <mi>I</mi> <mi>i</mi> <mi>R</mi> </msubsup> <mo>=</mo> <mfrac> <mrow> <msubsup> <mo>&Sigma;</mo> <mrow> <mi>r</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>R</mi> </msubsup> <msubsup> <mi>N</mi> <mi>i</mi> <mi>r</mi> </msubsup> </mrow> <mi>R</mi> </mfrac> </mrow>

WhereinIt isWithDifference,Represent being averaged for the single attack that attacker r wheels carry out at target i Income,Represent the single attack average yield for all attacks that attacker carries out in r wheels, Q^rFor a vector, I is tieed upC be a constant, Var (Q^r) represent { Q^rVariance；

When attacker tends to target of attack iTo be just, on the contrary will be negative, and attacker be not revealed the tendentiousness of target It is set to 0.

Adaptive subjectivity revenue function：

D=1/ (N are set_r- r), wherein vector ω=(ω₁,ω₂,ω₃) for parameter to be learned, N_rAlways to take turns number, r representatives are used for Learn the round of ω；

(3) optimization problem provides

With reference to human behavior model and from the parameter that attack data learning obtains, the optimization for generating optimal Defending Policy is established Problem；

(4) optimal policy is calculated

Optimization problem is decomposed into 4 subproblems.By solving subproblem, and global optimum is selected in the optimal solution of subproblem Solution, just can solve former problem, obtain defence vector x.

2. the modeling of mankind attacker cooperation behavior and the corresponding Defending Policy of formulation in a kind of security protection as described in power requires 1 Method, which is characterized in that repeat cooperative game there are one defender and two attackers, the two attackers can select be No cooperation；Defender periodically arranges new Defending Policy as the leader in game, and attacker is as follower, according to It makes attack to the observation of the Defending Policy of defender's arrangement；This game is formed more by taking turns, so defender can be regular Defending Policy is changed according to the data being collected into round before；Attacker independently attacks or cooperation attack；Independently attacking In the case of hitting, whether income that attacker attacks every time is only dependent upon the attacker this attack and succeeds.

3. the modeling of mankind attacker cooperation behavior and the corresponding Defending Policy of formulation in a kind of security protection as described in power requires 1 Method, which is characterized in that the defence resource summation M possessed by defender come defend a target collection T=1 ..., | T |, T is by T₁And T₂Two are not overlapped subset and are combined into；T₁Represent the target collection that first attacker may attack, T₂Represent the The target collection that two attackers may attack, and T₁=T-T₂；Vector x is defended, per one-dimensional element x_iDefender is represented to exist The defence resource of each target input namely the defended probability of target i.

4. the modeling of mankind attacker cooperation behavior and the corresponding Defending Policy of formulation in a kind of security protection as described in power requires 1 Method, which is characterized in that income of the attacker in target of attack i depend on 1) this target it is whether defended and 2) whether two attackers cooperate；When an attacker selects independent attack, the income that he attacks not defended target i isThe income for attacking defended target i isRelatively, defender is in the income of the previous caseIn latter feelings The income of condition isThe total revenue of defender for it in two incomes by target of attack and；AndIf two attackers agree to cooperate, they will obtain extra returns ∈ when attacking not defended target.

5. the modeling of mankind attacker cooperation behavior and the corresponding Defending Policy of formulation in a kind of security protection as described in power requires 1 Method, which is characterized in that with reference to human behavior model and from the obtained parameter of attack data learning, arrived by equation 1 Equation 19 establishes the optimization problem for generating optimal Defending Policy.

D=d₁+d₂ (2)

α₁, α₂, β ∈ { 0,1 } (3)

<mrow> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> <mo>&cup;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> </mrow> </munder> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>&le;</mo> <mi>M</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>&Element;</mo> <mo>&lsqb;</mo> <mn>0</mn> <mo>,</mo> <mn>1</mn> <mo>&rsqb;</mo> <mo>,</mo> <mo>&ForAll;</mo> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> <mo>&cup;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msub> <mi>d</mi> <mn>1</mn> </msub> <mo>-</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msubsup> <mi>q</mi> <mi>i</mi> <mrow> <mi>n</mi> <mi>c</mi> </mrow> </msubsup> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>i</mi> <mi>d</mi> </msubsup> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>+</mo> <msubsup> <mi>P</mi> <mi>i</mi> <mi>d</mi> </msubsup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mo>&le;</mo> <msub> <mi>&alpha;</mi> <mn>1</mn> </msub> <mi>Z</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msub> <mi>d</mi> <mn>1</mn> </msub> <mo>-</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msubsup> <mi>q</mi> <mi>i</mi> <mi>c</mi> </msubsup> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>i</mi> <mi>d</mi> </msubsup> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>+</mo> <msubsup> <mi>P</mi> <mi>i</mi> <mi>d</mi> </msubsup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mo>&le;</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>&alpha;</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mi>Z</mi> <mo>+</mo> <mi>&beta;</mi> <mi>Z</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>7</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msub> <mi>d</mi> <mn>1</mn> </msub> <mo>-</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msubsup> <mi>q</mi> <mi>i</mi> <mi>c</mi> </msubsup> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>i</mi> <mi>d</mi> </msubsup> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>+</mo> <mrow> <mo>(</mo> <msubsup> <mi>P</mi> <mi>i</mi> <mi>d</mi> </msubsup> <mo>-</mo> <mo>&Element;</mo> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mo>&le;</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>&alpha;</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mi>Z</mi> <mo>+</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>&beta;</mi> <mo>)</mo> </mrow> <mi>Z</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msub> <mi>d</mi> <mn>2</mn> </msub> <mo>-</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>j</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> </mrow> </munder> <msubsup> <mi>q</mi> <mi>j</mi> <mrow> <mi>n</mi> <mi>c</mi> </mrow> </msubsup> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>j</mi> <mi>d</mi> </msubsup> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>+</mo> <msubsup> <mi>P</mi> <mi>j</mi> <mi>d</mi> </msubsup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mo>&le;</mo> <msub> <mi>&alpha;</mi> <mn>2</mn> </msub> <mi>Z</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msub> <mi>d</mi> <mn>2</mn> </msub> <mo>-</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>j</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> </mrow> </munder> <msubsup> <mi>q</mi> <mi>j</mi> <mi>c</mi> </msubsup> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>j</mi> <mi>d</mi> </msubsup> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>+</mo> <msubsup> <mi>P</mi> <mi>j</mi> <mi>d</mi> </msubsup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mo>&le;</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>&alpha;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mi>Z</mi> <mo>+</mo> <mi>&beta;</mi> <mi>Z</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msub> <mi>d</mi> <mn>2</mn> </msub> <mo>-</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>j</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> </mrow> </munder> <msubsup> <mi>q</mi> <mi>j</mi> <mi>c</mi> </msubsup> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>j</mi> <mi>d</mi> </msubsup> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>+</mo> <mrow> <mo>(</mo> <msubsup> <mi>P</mi> <mi>j</mi> <mi>d</mi> </msubsup> <mo>-</mo> <mo>&Element;</mo> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mo>&le;</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>&alpha;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mi>Z</mi> <mo>+</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>&beta;</mi> <mo>)</mo> </mrow> <mi>Z</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <mtable> <mtr> <mtd> <mrow> <msubsup> <mi>U</mi> <mi>a</mi> <mi>c</mi> </msubsup> <mo>=</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mo>{</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msubsup> <mi>q</mi> <mi>i</mi> <mi>c</mi> </msubsup> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>R</mi> <mi>i</mi> <mi>a</mi> </msubsup> <mo>+</mo> <mo>&Element;</mo> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>P</mi> <mi>i</mi> <mi>a</mi> </msubsup> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>&rsqb;</mo> <mo>+</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <munder> <mo>&Sigma;</mo> <mrow> <mi>j</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> </mrow> </munder> <msubsup> <mi>q</mi> <mi>j</mi> <mi>c</mi> </msubsup> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>R</mi> <mi>j</mi> <mi>a</mi> </msubsup> <mo>+</mo> <mo>&Element;</mo> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>P</mi> <mi>j</mi> <mi>a</mi> </msubsup> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>&rsqb;</mo> <mo>}</mo> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msubsup> <mi>U</mi> <mrow> <mi>a</mi> <mn>1</mn> </mrow> <mrow> <mi>n</mi> <mi>c</mi> </mrow> </msubsup> <mo>=</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msubsup> <mi>q</mi> <mi>i</mi> <mrow> <mi>n</mi> <mi>c</mi> </mrow> </msubsup> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>i</mi> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>P</mi> <mi>i</mi> <mi>a</mi> </msubsup> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>&rsqb;</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>13</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msubsup> <mi>U</mi> <mrow> <mi>a</mi> <mn>2</mn> </mrow> <mrow> <mi>n</mi> <mi>c</mi> </mrow> </msubsup> <mo>=</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>j</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> </mrow> </munder> <msubsup> <mi>q</mi> <mi>j</mi> <mrow> <mi>n</mi> <mi>c</mi> </mrow> </msubsup> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>j</mi> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>P</mi> <mi>j</mi> <mi>a</mi> </msubsup> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>&rsqb;</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>14</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <mo>-</mo> <msub> <mi>&alpha;</mi> <mn>1</mn> </msub> <mi>Z</mi> <mo>&le;</mo> <msubsup> <mi>U</mi> <mrow> <mi>a</mi> <mn>1</mn> </mrow> <mrow> <mi>n</mi> <mi>c</mi> </mrow> </msubsup> <mo>-</mo> <msubsup> <mi>U</mi> <mi>a</mi> <mi>c</mi> </msubsup> <mo>&le;</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>&alpha;</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mi>Z</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>15</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <mo>-</mo> <msub> <mi>&alpha;</mi> <mn>2</mn> </msub> <mi>Z</mi> <mo>&le;</mo> <msubsup> <mi>U</mi> <mrow> <mi>a</mi> <mn>2</mn> </mrow> <mrow> <mi>n</mi> <mi>c</mi> </mrow> </msubsup> <mo>-</mo> <msubsup> <mi>U</mi> <mi>a</mi> <mi>c</mi> </msubsup> <mo>&le;</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>&alpha;</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mi>Z</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>16</mn> <mo>)</mo> </mrow> </mrow>

β≤α₁ (17)

β≤α₂ (18)

α₁+α₂≤β+1 (19)

In all equatioies, Z represents a larger constant, and c represents cooperation, and nc represents uncooperative；

Formula 1 gives the expected revenus of the target of optimization problem, i.e. defender；From 2 visual target of formula by d₁And d₂Two parts structure Into；Three binary variables defined in formula 3, wherein α₁And α₂Attacker is represented each for the selection of cooperation, and β is represented most Whole cooperation state；Only when two attackers agree to cooperation, β 1, β is 0 in all other cases；This is by formula 17 Ensure with the constraints in 19；Formula 4 and 5 ensure that the every one-dimensional all between [0,1] of defence vector x, and defender's is total Resource is not more than M；

In formula 6 to 11, based on α₁、α₂Two parts of defender's expected revenus are defined with β --- d₁And d₂；

Formula 12 to 14 respectively define attacker cooperate with it is uncooperative when expected revenus；Here two attackers are built respectively Mould, so they have different expected revenusesWithBy formula 15 and 16, α ensure that₁And α₂It is based on expected revenus It determines；As previously mentioned, three constraintss in formula 17 to 19 are for ensuring that correctly running for cooperative mechanism, that is, only have Cooperation is just reached when two attackers select cooperation.

6. the modeling of mankind attacker cooperation behavior and the corresponding Defending Policy of formulation in a kind of security protection as described in power requires 1 Method, which is characterized in that optimization problem is decomposed into 4 subproblems；By solving subproblem, and in the optimal solution of subproblem Middle selection globally optimal solution just can solve former problem, obtain defence vector x.

PROBLEM DECOMPOSITION

The angle of cooperation whether is selected from attacker, may be occurred there are four types of situation altogether；Following table summarize defender this four Expected revenus D in the case of kind₁、D₂、D₃And D₄；

Table 3

Expected revenus in given table, just can correspondingly establish 4 subproblems.

7. the modeling of mankind attacker cooperation behavior and the corresponding Defending Policy of formulation in a kind of security protection as described in power requires 6 Method, which is characterized in that find the globally optimal solution of subproblem based on approximate method, introduce piecewise linear function, will about Beam condition is loosened, and subproblem is converted into a MIXED INTEGER quadratically constrained quadratic programming problem (MIQCQP) to solve；

Variable replacement is done first；Order That D₄A quadric form can be rewritten as, sees formula 26；Because four new variables are introduced, it is necessary to according to definition to s and z Increase by four constraintss；Here only provide for s_pConstraint, others constraint can similarly provide；

<mrow> <msub> <mi>s</mi> <mi>p</mi> </msub> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msup> <mi>e</mi> <mrow> <msubsup> <mi>ASU</mi> <mn>1</mn> <mi>c</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> </mrow> </msup> <mo>-</mo> <msup> <mi>e</mi> <mrow> <msubsup> <mi>ASU</mi> <mn>1</mn> <mi>c</mi> </msubsup> <mrow> <mo>(</mo> <msub> <mi>x</mi> <mi>p</mi> </msub> <mo>)</mo> </mrow> </mrow> </msup> <mo>=</mo> <mn>0</mn> <mo>,</mo> <mo>&ForAll;</mo> <mi>p</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>25</mn> <mo>)</mo> </mrow> </mrow>

Formula 25 is a nonlinear restriction, therefore it is replaced with two groups and slightly loosened to obtain quadratic constraints condition；Present explanation How such two set of segmentation linear functions are found.

OrderIt needs to find two groups of suitably piecewise linear functionsWithSo thatHere f (x_i) on [0,1] it is a monotone decreasing convex function, because For negative；

AndUse x_ikDefinitionWith

<mrow> <msubsup> <mi>l</mi> <mi>i</mi> <mn>1</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <mn>1</mn> <mo>+</mo> <munderover> <mo>&Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>K</mi> </munderover> <msubsup> <mi>v</mi> <mrow> <mi>i</mi> <mi>k</mi> </mrow> <mrow> <mi>u</mi> <mi>p</mi> </mrow> </msubsup> <msub> <mi>x</mi> <mrow> <mi>i</mi> <mi>k</mi> </mrow> </msub> </mrow>

<mrow> <msubsup> <mi>l</mi> <mi>i</mi> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mi>l</mi> <mi>i</mi> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> <mo>+</mo> <munderover> <mo>&Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>K</mi> </munderover> <msubsup> <mi>v</mi> <mrow> <mi>i</mi> <mi>k</mi> </mrow> <mrow> <mi>l</mi> <mi>o</mi> <mi>w</mi> </mrow> </msubsup> <msub> <mi>x</mi> <mrow> <mi>i</mi> <mi>k</mi> </mrow> </msub> </mrow>

WhereinIt representsThe slope of middle kth section broken line,It representsThe slope of middle kth section broken line；So in order to DefinitionWithOnly it needs to be determined thatWith

Theorem 1：According to piecewise linear functionWithMeetIt is full determining After the piecewise linear function required enough, the constraints in formula 25 is loosened；Specifically, formula 25 is replaced with following two groups of inequality It changes；

<mrow> <mi>C</mi> <mi>o</mi> <mi>n</mi> <mi>s</mi> <mrow> <mo>(</mo> <msub> <mi>s</mi> <mi>p</mi> </msub> <mo>)</mo> </mrow> <mo>:</mo> <mfenced open = "{" close = ""> <mtable> <mtr> <mtd> <mrow> <msub> <mi>s</mi> <mi>p</mi> </msub> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msup> <mi>e</mi> <msubsup> <mi>b</mi> <mrow> <mn>1</mn> <mi>t</mi> </mrow> <mi>c</mi> </msubsup> </msup> <msubsup> <mi>l</mi> <mi>i</mi> <mn>1</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <msup> <mi>e</mi> <msubsup> <mi>b</mi> <mrow> <mn>1</mn> <mi>p</mi> </mrow> <mi>c</mi> </msubsup> </msup> <msubsup> <mi>l</mi> <mi>p</mi> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>x</mi> <mi>p</mi> </msub> <mo>)</mo> </mrow> <mo>&GreaterEqual;</mo> <mn>0</mn> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mo>&ForAll;</mo> <mi>p</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <msub> <mi>s</mi> <mi>p</mi> </msub> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msup> <mi>e</mi> <msubsup> <mi>b</mi> <mrow> <mn>1</mn> <mi>t</mi> </mrow> <mi>c</mi> </msubsup> </msup> <msubsup> <mi>l</mi> <mi>i</mi> <mn>2</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <msup> <mi>e</mi> <msubsup> <mi>b</mi> <mrow> <mn>1</mn> <mi>p</mi> </mrow> <mi>c</mi> </msubsup> </msup> <msubsup> <mi>l</mi> <mi>p</mi> <mn>1</mn> </msubsup> <mrow> <mo>(</mo> <msub> <mi>x</mi> <mi>p</mi> </msub> <mo>)</mo> </mrow> <mo>&le;</mo> <mn>0</mn> <mo>,</mo> </mrow> </mtd> <mtd> <mrow> <mo>&ForAll;</mo> <mi>p</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </mtd> </mtr> </mtable> </mfenced> </mrow>

This two groups of inequality are denoted as Cons (s_p)；By s_r、z_pAnd z_rThe similar other inequality introduced are denoted as Cons respectively (s_r)、Cons(z_p) and Cons (z_r)；

<mrow> <munder> <mi>max</mi> <mrow> <mi>x</mi> <mo>,</mo> <mi>s</mi> <mo>,</mo> <mi>z</mi> </mrow> </munder> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msub> <mi>s</mi> <mi>i</mi> </msub> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>i</mi> <mi>d</mi> </msubsup> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>+</mo> <msubsup> <mi>P</mi> <mi>i</mi> <mi>d</mi> </msubsup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mo>+</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>j</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> </mrow> </munder> <msub> <mi>z</mi> <mi>j</mi> </msub> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>j</mi> <mi>d</mi> </msubsup> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>+</mo> <msubsup> <mi>P</mi> <mi>j</mi> <mi>d</mi> </msubsup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> <mo>&rsqb;</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>26</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> <mo>&cup;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> </mrow> </munder> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>&le;</mo> <mi>M</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>27</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <mtable> <mtr> <mtd> <mrow> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msub> <mi>z</mi> <mi>i</mi> </msub> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>i</mi> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>P</mi> <mi>i</mi> <mi>a</mi> </msubsup> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>&rsqb;</mo> <mo>&le;</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mo>{</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msub> <mi>s</mi> <mi>i</mi> </msub> <mo>&lsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>R</mi> <mi>i</mi> <mi>a</mi> </msubsup> <mo>+</mo> <mo>&Element;</mo> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>P</mi> <mi>i</mi> <mi>a</mi> </msubsup> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>&rsqb;</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>j</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> </mrow> </munder> <msub> <mi>s</mi> <mi>j</mi> </msub> <mo>&lsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>R</mi> <mi>j</mi> <mi>a</mi> </msubsup> <mo>+</mo> <mo>&Element;</mo> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>P</mi> <mi>j</mi> <mi>a</mi> </msubsup> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>&rsqb;</mo> <mo>}</mo> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>28</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <mtable> <mtr> <mtd> <mrow> <munder> <mo>&Sigma;</mo> <mrow> <mi>j</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> </mrow> </munder> <msub> <mi>z</mi> <mi>j</mi> </msub> <mo>&lsqb;</mo> <msubsup> <mi>R</mi> <mi>j</mi> <mi>a</mi> </msubsup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>P</mi> <mi>j</mi> <mi>a</mi> </msubsup> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>&rsqb;</mo> <mo>&GreaterEqual;</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mo>{</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>i</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> </mrow> </munder> <msub> <mi>s</mi> <mi>i</mi> </msub> <mo>&lsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>R</mi> <mi>i</mi> <mi>a</mi> </msubsup> <mo>+</mo> <mo>&Element;</mo> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>P</mi> <mi>i</mi> <mi>a</mi> </msubsup> <msub> <mi>x</mi> <mi>i</mi> </msub> <mo>&rsqb;</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <munder> <mo>&Sigma;</mo> <mrow> <mi>j</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> </mrow> </munder> <msub> <mi>s</mi> <mi>j</mi> </msub> <mo>&lsqb;</mo> <mrow> <mo>(</mo> <msubsup> <mi>R</mi> <mi>j</mi> <mi>a</mi> </msubsup> <mo>+</mo> <mo>&Element;</mo> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>P</mi> <mi>j</mi> <mi>a</mi> </msubsup> <msub> <mi>x</mi> <mi>j</mi> </msub> <mo>&rsqb;</mo> <mo>}</mo> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>29</mn> <mo>)</mo> </mrow> </mrow>

Cons(s_p), Cons (s_r), Cons (z_p), Cons (z_r) (30)

<mrow> <msub> <mi>x</mi> <mi>t</mi> </msub> <mo>=</mo> <munderover> <mo>&Sigma;</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>K</mi> </munderover> <msub> <mi>x</mi> <mrow> <mi>t</mi> <mi>k</mi> </mrow> </msub> <mo>,</mo> <mo>&ForAll;</mo> <mi>t</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> <mo>&cup;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>31</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <mn>0</mn> <mo>&le;</mo> <msub> <mi>x</mi> <mrow> <mi>t</mi> <mi>k</mi> </mrow> </msub> <mo>&le;</mo> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <mo>,</mo> <mo>&ForAll;</mo> <mi>t</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> <mo>&cup;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> <mo>,</mo> <mi>k</mi> <mo>=</mo> <mn>1</mn> <mo>,</mo> <mn>...</mn> <mo>,</mo> <mi>K</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>32</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msub> <mi>y</mi> <mrow> <mi>t</mi> <mi>k</mi> </mrow> </msub> <mo>&Element;</mo> <mo>{</mo> <mn>0</mn> <mo>,</mo> <mn>1</mn> <mo>}</mo> <mo>,</mo> <mo>&ForAll;</mo> <mi>t</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> <mo>&cup;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> <mo>,</mo> <mi>k</mi> <mo>=</mo> <mn>1</mn> <mo>,</mo> <mn>...</mn> <mo>,</mo> <mi>K</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>33</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msub> <mi>y</mi> <mrow> <mi>t</mi> <mi>k</mi> </mrow> </msub> <mfrac> <mn>1</mn> <mi>K</mi> </mfrac> <mo>&le;</mo> <msub> <mi>x</mi> <mrow> <mi>t</mi> <mi>k</mi> </mrow> </msub> <mo>,</mo> <mo>&ForAll;</mo> <mi>t</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> <mo>&cup;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> <mo>,</mo> <mi>k</mi> <mo>=</mo> <mn>1</mn> <mo>,</mo> <mn>...</mn> <mo>,</mo> <mi>K</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>34</mn> <mo>)</mo> </mrow> </mrow>

<mrow> <msub> <mi>x</mi> <mrow> <mi>t</mi> <mo>,</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> </msub> <mo>&le;</mo> <msub> <mi>y</mi> <mrow> <mi>t</mi> <mi>k</mi> </mrow> </msub> <mo>,</mo> <mo>&ForAll;</mo> <mi>t</mi> <mo>&Element;</mo> <msub> <mi>T</mi> <mn>1</mn> </msub> <mo>&cup;</mo> <msub> <mi>T</mi> <mn>2</mn> </msub> <mo>,</mo> <mi>k</mi> <mo>=</mo> <mn>1</mn> <mo>,</mo> <mn>...</mn> <mo>,</mo> <mi>K</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>35</mn> <mo>)</mo> </mrow> </mrow>

Subproblem is rewritten with the variable newly introduced in formula 26 to 29；Formula 30 represents the quadratic constraints of generation；Formula 31 and 32 is To x_tkConstraint；

MIQCQP is solved using the SCIP methods of Vigerske S.et al. propositions；

Theorem 2：NoteForWithThe distance between, then lim_K→+∞Dis= 0。