# Performance Analysis and Optimal Allocation of Layered Defense M/M/N Queueing Systems

Authors: Longyue Li; Fuxian Liu; Guangzheng Long; Huizhen Zhao; Yingying Mei
Journal: Mathematical Problems in Engineering
 (2016)

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## Abstract

One important mission of strategic defense is to develop an integrated layered Ballistic Missile Defense System (BMDS). Motivated by the queueing theory, we presented a work for the representation, modeling, performance simulation, and channels optimal allocation of the layered BMDS M/M/N queueing systems. Firstly, in order to simulate the process of defense and to study the Defense Effectiveness (DE), we modeled and simulated the M/M/N queueing system of layered BMDS. Specifically, we proposed the M/M/N/N and M/M/N/C queueing model for short defense depth and long defense depth, respectively; single target channel and multiple target channels were distinguished in each model. Secondly, we considered the problem of assigning limited target channels to incoming targets, we illustrated how to allocate channels for achieving the best DE, and we also proposed a novel and robust search algorithm for obtaining the minimum channel requirements across a set of neighborhoods. Simultaneously, we presented examples of optimal allocation problems under different constraints. Thirdly, several simulation examples verified the effectiveness of the proposed queueing models. This work may help to understand the rules of queueing process and to provide optimal configuration suggestions for defense decision-making.


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## Body

## 1. Introduction

These years, ballistic missile (BM) technology has spread to more and more countries. Nations all over the world are developing missiles capable of reaching enemies. One important mission of strategic defense is to develop an integrated layered BMDS to defend homeland, deployed forces, allies, and friends from ballistic missile attacks [1]. The BMDS is based on a multilayer defense concept and therefore contains more than one defense weapon; it will include different types of defense weapons located on land or ships used to destroy ballistic missiles [2]. Layered BMDS has two advantages: (1) interception mainly can be divided into 3 phases: boost phase, midphase, and reentry phase. Since the reentry phase is too short and it is the last chance for a shot, BMDS should not rely on a single defense weapon but on defense weapons placed at different locations forming a layered BMDS; the layered BMDS allows for more shot opportunities that will certainly increase the probability of a successful interception [3]. (2) For given affordable BMs penetration probability (or expected kill probability), cooperation between different missile defense weapons may reduce the expected resources consumption and provide an efficient way of using interceptors. The common methods used in the research on the process simulation and performance evaluation of missile defense are the mathematical programming method [4, 5], the probability calculation method [6], the system simulation method [7], the Markov method [8, 9], and so forth.Queueing theory is a mathematical theory of stochastic service system which was first proposed by Erlang [10]. Queueing systems have a wide range of applications, such as resource allocation [11], system optimization [12], and communication planning [13]. Similarly, in order to make full use of defense capabilities, queuing theory also has a lot of applications in defense weapons operation research; it can solve problems of weapons configuration or efficiency analysis [14–16]. Following are the two questions that need our attention: (1) there are many factors that affect DE, such as the number of layers, the number of defense weapons, and Single Shot Kill Probability (SSKP); these are also factors that affect the requirement of defense weapons; how are defense weapons, number of layers, BMs, SSKP, and DE interrelated and how can we understand this relationship for achieving the best allocation plan? (2) If we have deployed different types of defense weapons, then how do we deal with them?Using M/M/N queueing system to simulate the missile defense process is feasible; the reasons are as follows: (1) the Poisson process has the simplest mathematical expressions, though BMs arrival is not fully consistent with the Poisson process; it represents the most difficult scenario (worst-case scenario) for the BMDS to deal with. As long as the BMDS can deal with Poisson arrivals, it has certain adaptability to other types of arrivals distribution. (2) BMs usually have fixed and highly predictable trajectories, though some of them may have limited maneuvering potential; we think this has no influence on our research. (3) The incoming directions, firing tactics, technical characteristics, and time intervals of BM arrivals have some Poisson features; these can be viewed as customers waiting to be served by servers. (4) The targets capacity (number of target channels, servers) and shooting times for each target (service times) are limited. When BMs that arrived find that all channels are occupied (not idle) or there was little time for a shot, they will penetrate the BMDS directly. In summary, the M/M/N queueing system can be used to analyze the DE of BMDS, summarize the rules of defense, and provide suggestions for system configuration for defense decision-making. The remainder of the paper is structured as follows. Section 2 proposes the framework for M/M/N queueing model. Section 3 discusses M/M/N queueing models. Section 4 is dedicated to optimal allocation of target channels. Section 5 provides numerical examples. Section 6 includes concluding remarks and future work.


## 2. M/M/N Queueing Framework

We consider an M/M/N queueing system with BMs arrival rateλ and shooting rate μ of defense weapons. M stands for “memoryless” or “Markovian” and means that the process being represented by M comes from an exponential distribution [17].(1) Suppose that BMs arrive randomly and independently of each other to a defense weapon and that the average rate at which they arrive is given by the parameter λ [18]; that is, (1)Pnt=λtnn!e-λt,t>0,n=0,1,2,….This is known as a Poisson arrival process;Pn(t) is the probability that n BMs arrive within time t. Suppose that the time intervals between arrivals are randomly taken from the exponential distribution with parameter λ; their probability density function and distribution function are(2)fTt=λe-λt,t≥00,t<0,FTt=1-λe-λt,t≥00,t<0.The exponential distribution is memoryless, which indicates that the BMs arrivals are random.(2) Suppose that the BMs are shot in the order of their arrivals; the shooting time for a BM is also exponentially distributed at rate μ; then, its probability density function and distribution function are(3)fst=μe-μt,t≥00,t<0,Fst=1-μe-μt,t≥00,t<0,where μ=1/tsmean, where tsmean is the mean shooting time. The shooting time depends on the reaction time of the defense weapon and the time of interceptor flying from the launch point to the calculated encounter point, which is related to the technical capabilities of defense weapons. Introducing ρ=λ/μ, ρ<1 means that the queue is stable if the mean shooting time is less than the mean arrival time intervals; it can be understood as the firing density (shooting intensity) [19].(3) Suppose that the waiting times of BMs are also exponentially distributed at rate ν; then their probability density function and distribution function are(4)fwt=νe-νt,t≥00,t<0,Fwt=1-νe-νt,t≥00,t<0,where μ=1/twmean, where twmean is the mean waiting time.(4) Additionally, if there is an idle target channel when a BM arrives at the system, then the defense weapon will shoot it immediately. In this paper, we divided the queueing system into two types: (1) loss system (when BMs that arrived find that all target channels were occupied (not idle), they will penetrate the BMDS directly (leave the system without service)) and (2) mixed system (when BMs that arrived find that all target channels were occupied, BMs will not be penetrated but will wait for a limited time (depending on the time of BMs flying in the killing zone of BMDS) until a target channel becomes available). We use the term “defense depth” to distinguish between the loss system and the mixed system. “Short defense depth” is defined as the case when waiting times of BMs are shorter than shooting times of defense weapons (loss system), and the “long defense depth” is defined as the case when waiting times of BMs are longer than shooting times of defense weapons (mixed system), as shown in Figure 1.Figure 1
Long defense depth and short defense depth.
(a)
Long defense depth (b)
Short defense depth


## 3. M/M/N Queueing Models

### 3.1. Identical Weapons, Short Defense Depth, and M/M/N/N System

#### 3.1.1. Multiple Target Channels and M/M/N/N System

BMDS with short defense depth can be seen as a random loss system that does not wait. Letn be the number of target channels of BMDS; that is, the number of BMs shot by defense weapons cannot exceed n at the same time; Pf is the detection probability of BMDS radars. Possible states of the system are as follows:S0:
all target channels are idle; there is no BM in the system.S1:
1 target channel is busy; there is 1 BM in the system.⋮Sk:
k target channels are busy; there are k BMs in the system.⋮Sn:
all target channels are busy; there aren BMs in the system.Figure 2 is the state transition diagram of M/M/N/N system. In Figure 2, the arrow → indicates incoming BMs, and the arrival rate is λ. Sk means k target channels are busy, and the state is transferred to Sk+1 when a new BM arrives. Arrow ← indicates shootings of defense weapons, when the shooting process for a BM is done, and then the state is transferred from S1 to S0 with rate μ, transferred from S2 to S1 with rate 2μ,…, and transferred from Sk to Sk-1 with rate kμ. Balance equations are as follows:S0:
all target channels are idle; there is no BM in the system:(5)λS0=μS1⟹S1=λμS0=ρS0.S1:
1 target channel is busy; there is 1 BM in the system:(6)λS1=2μS2⟹S2=ρ22!S0.S2:
2 target channels are busy; there are 2 BMs in the system:(7)λS2=2μS3⟹S3=ρ33!S0.⋮Sk-1:
k-1 target channels are busy; there are k-1 BMs in the system:(8)λSk-1=kμSk⟹Sk=ρkk!S0.Sn:
n target channels are busy; there are n BMs in the system:(9)λSn-1=nμSn⟹Sn=ρnn!S0.Figure 2
State transition diagram of M/M/N/N system.Due to the fact thatPf is the probability of the BM found by the radar, we have(10)ρf=ρ·Pf=Pf·λμ=Pf·λ·tsmean.ρ f is the mean number of BMs found to arrive during the mean shooting time. Because S0+S1+S2+⋯+Sn=1, then the expression for S0 can be written in the form(11)S0=1+ρf+ρf22!+⋯+ρfkk!+⋯+ρfnn!-1=∑k=0nρfkk!-1.The optimal operation of the queueing system can be analyzed through several performance parameters, some of which follow.(1)
BM loss probabilities (penetration probabilities):(12)Plost=Sn=ρfnn!S0=ρfnn!∑k=0nρfkk!-1.(2)
Relative probabilities of BMs that will be shot:(13)Pshoot=1-Plost=1-ρfnn!∑k=0nρfkk!-1.(3)
Absolute probabilities of BMs that will be shot:(14)Pshoot′=λPshoot=λ1-ρfnn!∑k=0nρfkk!-1.(4)
Mean number of occupied target channels:(15)Z-=0·S0+1·S1+⋯+n·Sn=∑k=0nkρfkk!S0=ρfS0∑k=0nρfkk!-ρfnn!=ρf1-ρfnn!∑k=0nρfkk!-1.The mean occupancy rate of target channels is η=Z-/n.


#### 3.1.2. Single Target Channel and M/M/1/1 System

The M/M/1/1 queueing system can be viewed as a special case of Section3.1.1 for n=1; then, the expression for S0 can be rewritten in the form(16)S0=∑k=0nρfkk!-1n=1=11+ρf.Performance parameters are as follows:(1)
BM loss probabilities:(17)Plost=Sn=ρfnn!S0=ρf1+ρf.(2)
Relative probabilities of BMs that will be shot:(18)Pshoot=1-Plost=1-ρf1+ρf=11+ρf.(3)
Absolute probabilities of BMs that will be shot:(19)Pshoot′=λPshoot=λ1+ρf.The concept of DE is the product of probabilities of BM that will be shot times SSKP; that is,(20)E=Pshoot·Pk=1-Plost·Pk.


### 3.2. Identical Weapons, Long Defense Depth, and M/M/N/C System

From Section1, we know that BMDS with long defense depth can be regarded as a stochastic service system with limited waiting time, that is, the mixed queueing system. For each incoming BM, the defense weapons use idle target channels to shoot. When BMs that arrived find that all target channels were occupied, BMs will not be penetrated but will wait for a limited time until a target channel becomes available. Possible states of the system are as follows:S0:
all target channels are idle; there is no BM in the system.S1:
1 target channel is busy; there is 1 BM in the system.⋮Sk:
k target channels are busy; there are k BMs in the system.⋮Sn:
all target channels are busy; there aren BMs in the system.Sn+1:
all target channels are busy; there is 1 BM waiting.⋮Sn+s:
all target channels are busy; there ares BMs waiting.⋮Figure 3 is the state transition diagram of M/M/N/C system. In Figure 3, the arrow → indicates incoming BMs, and the arrival rate is λ. Sk means k target channels are busy, and the state is transferred from lower state to higher state when new BMs arrive. Arrow ← indicates shootings of defense weapons, when the shooting process for a BM is done, and then the state is transferred from higher state to lower state with rate kμ or nμ+sν. Balance equations are(21)S0S1⋮Sk⋮SnSn+1⋮Sn+s⋮0=-λS0+μS10=λS0-λ+μS1+2μS2⋮0=λSk-1-λ+kμSk+k+1μS2⋮0=λSn-1-λ+nμSn+nμ+νSn+10=λSn-λ+nμ+νSn+1+nμ+2νSn+2⋮0=λSn+s-1-λ+nμ+νSn+s+nμ+s+1νSn+s+1⋮.Figure 3
State transition diagram of M/M/N/C system.Then, we have the constraint that∑i=0∞Si=1 for (21). Let ρ=λ/μ be the mean number of BMs that arrived during the mean shooting time; let ϑ=ν/μ be the mean number of penetrated (leaving without being shot) BMs during the mean shooting time. Then, (21) can be written in the form(22)S0S1⋮Sk⋮SnSn+1⋮Sn+s⋮0=-ρS0+S10=ρS0-ρ+1S1+2S2⋮0=ρSk-1-ρ+kSk+k+1S2⋮0=ρSn-1-ρ+nSn+n+ϑSn+10=ρSn-ρ+n+ϑSn+1+n+2ϑSn+2⋮0=ρSn+s-1-ρ+n+ϑSn+s+n+s+1ϑSn+s+1⋮.When k≤n, we have(23)S1=λμS0=ρS0=ρ11!S0,S2=12ρ+1S1-ρS0=12ρS1+S1-ρS0=ρ2S1=ρ22!S0⋮Sn=ρnSk-1=ρnn!S0.When k>n, we have(24)Sn+1=1n+ϑρ+nSn-ρSn-1=1n+ϑρSn+nSn-ρSn-1=ρn+ϑSn=ρn+ϑρnn!S0=ρnn!ρn+ϑS0,Sn+2=1n+2ϑρ+ρ+ϑSn+1-ρSn=1n+2ϑρSn+1+ρ+ϑSn+1-ρSn=1n+2ϑρSn+1+ρSn-ρSn=ρnn!ρ2n+ϑn+2ϑS0⋮Sn+s=ρn+sϑSn+s-1=ρnn!ρs∏m=1sn+mϑS0.The constraint for (24) is(25)∑m=0nSm+∑s=1∞Sm+s=1.With direct substitution of (23) and (24) into (25), it follows that(26)1=S0∑m=0nρmm!+ρnn!∑s=1∞ρs∏m=1sn+mϑ-1,S0=∑m=0nρmm!+ρnn!∑s=1∞ρs∏m=1sn+mϑ-1-1.Then, with substitution of (26) into (24), we have steady-state probability(27)Sk=ρkk!∑m=0nρmm!+ρnn!∑s=1∞ρs∏m=1sn+mϑ-1-1,0≤k≤n,Sn+s=ρn/n!ρs∏m=1sn+mϑ-1∑m=0nρm/m!+ρn/n!∑s=1∞ρs∏m=1sn+mϑ-1,s≥1.Performance parameters are as follows:(1)
BM loss probabilities (s BMs waiting to be shot):(28)Plost=νλ∑s=1∞sSn+s=ν/μλ/μ∑s=1∞sSn+s=ϑρ∑s=1∞sSn+s=ϑρ∑s=1∞sρn/n!ρs∏m=1sn+mϑ-1∑m=0nρm/m!+ρn/n!∑s=1∞ρs∏m=1sn+mϑ-1.(2)
Relative probabilities of BMs that will be shot:(29)Pshoot=1-Plost.(3)
Absolute probabilities of BMs that will be shot:(30)Pshoot′=λPshoot.(4)
Mean number of occupied target channels:(31)Z-=∑k=1nkSk+∑s=1∞nSn+s=∑k=1nkSk+n∑s=1∞Sn+sorZ-=λPshootμ=ρPshoot.The mean occupancy rate of target channels is(32)η=ρPshootn.DE is the product of probabilities of BM that will be shot times SSKP; that is,(33)E=Pshoot·Pk=1-Plost·Pk.Similarly, the M/M/1/C queueing system can be viewed as a special case for n=1.


### 3.3. Different Defense Weapons

Defense weapons in Sections3.1 and 3.2 are identical, and the BMDS may deploy different types of defense weapons. For different types of defense weapons, the waiting time of BMs, the detection probability of radars, and SSKP may be different. In order to model the queueing system, we choose one type of defense weapon as the reference and then substitute the reference defense (type r) for defense weapons of type i. The “replacement process” is named equivalent replacement method [20]; basic equations are(34)λr·Pshootr·Pkr·Pfr=λi·Pshooti·Pki·Pfi,nr·1+twmeanr·Pkr=ni·1+twmeani·Pki,nr=ni·1+twmeani·Pki1+twmeanr·Pkr.Subscripts(·)(r) and (·)(i) in (34) are used to distinguish defense weapons of type r from defense weapons of type i. From the above equations, we can substitute D(r) for D(i); the intensity of BMs killed by defense weapons of type i and the intensity of BMs arrivals for defense weapons of type i are(35)ρi=Di·Pfi·Pkitwtimei,λi=λρi·∑i=1Rρi-1,respectively, where R denotes the total number of types.


## 3.1. Identical Weapons, Short Defense Depth, and M/M/N/N System

### 3.1.1. Multiple Target Channels and M/M/N/N System

BMDS with short defense depth can be seen as a random loss system that does not wait. Letn be the number of target channels of BMDS; that is, the number of BMs shot by defense weapons cannot exceed n at the same time; Pf is the detection probability of BMDS radars. Possible states of the system are as follows:S0:
all target channels are idle; there is no BM in the system.S1:
1 target channel is busy; there is 1 BM in the system.⋮Sk:
k target channels are busy; there are k BMs in the system.⋮Sn:
all target channels are busy; there aren BMs in the system.Figure 2 is the state transition diagram of M/M/N/N system. In Figure 2, the arrow → indicates incoming BMs, and the arrival rate is λ. Sk means k target channels are busy, and the state is transferred to Sk+1 when a new BM arrives. Arrow ← indicates shootings of defense weapons, when the shooting process for a BM is done, and then the state is transferred from S1 to S0 with rate μ, transferred from S2 to S1 with rate 2μ,…, and transferred from Sk to Sk-1 with rate kμ. Balance equations are as follows:S0:
all target channels are idle; there is no BM in the system:(5)λS0=μS1⟹S1=λμS0=ρS0.S1:
1 target channel is busy; there is 1 BM in the system:(6)λS1=2μS2⟹S2=ρ22!S0.S2:
2 target channels are busy; there are 2 BMs in the system:(7)λS2=2μS3⟹S3=ρ33!S0.⋮Sk-1:
k-1 target channels are busy; there are k-1 BMs in the system:(8)λSk-1=kμSk⟹Sk=ρkk!S0.Sn:
n target channels are busy; there are n BMs in the system:(9)λSn-1=nμSn⟹Sn=ρnn!S0.Figure 2
State transition diagram of M/M/N/N system.Due to the fact thatPf is the probability of the BM found by the radar, we have(10)ρf=ρ·Pf=Pf·λμ=Pf·λ·tsmean.ρ f is the mean number of BMs found to arrive during the mean shooting time. Because S0+S1+S2+⋯+Sn=1, then the expression for S0 can be written in the form(11)S0=1+ρf+ρf22!+⋯+ρfkk!+⋯+ρfnn!-1=∑k=0nρfkk!-1.The optimal operation of the queueing system can be analyzed through several performance parameters, some of which follow.(1)
BM loss probabilities (penetration probabilities):(12)Plost=Sn=ρfnn!S0=ρfnn!∑k=0nρfkk!-1.(2)
Relative probabilities of BMs that will be shot:(13)Pshoot=1-Plost=1-ρfnn!∑k=0nρfkk!-1.(3)
Absolute probabilities of BMs that will be shot:(14)Pshoot′=λPshoot=λ1-ρfnn!∑k=0nρfkk!-1.(4)
Mean number of occupied target channels:(15)Z-=0·S0+1·S1+⋯+n·Sn=∑k=0nkρfkk!S0=ρfS0∑k=0nρfkk!-ρfnn!=ρf1-ρfnn!∑k=0nρfkk!-1.The mean occupancy rate of target channels is η=Z-/n.


### 3.1.2. Single Target Channel and M/M/1/1 System

The M/M/1/1 queueing system can be viewed as a special case of Section3.1.1 for n=1; then, the expression for S0 can be rewritten in the form(16)S0=∑k=0nρfkk!-1n=1=11+ρf.Performance parameters are as follows:(1)
BM loss probabilities:(17)Plost=Sn=ρfnn!S0=ρf1+ρf.(2)
Relative probabilities of BMs that will be shot:(18)Pshoot=1-Plost=1-ρf1+ρf=11+ρf.(3)
Absolute probabilities of BMs that will be shot:(19)Pshoot′=λPshoot=λ1+ρf.The concept of DE is the product of probabilities of BM that will be shot times SSKP; that is,(20)E=Pshoot·Pk=1-Plost·Pk.


## 3.1.1. Multiple Target Channels and M/M/N/N System

BMDS with short defense depth can be seen as a random loss system that does not wait. Letn be the number of target channels of BMDS; that is, the number of BMs shot by defense weapons cannot exceed n at the same time; Pf is the detection probability of BMDS radars. Possible states of the system are as follows:S0:
all target channels are idle; there is no BM in the system.S1:
1 target channel is busy; there is 1 BM in the system.⋮Sk:
k target channels are busy; there are k BMs in the system.⋮Sn:
all target channels are busy; there aren BMs in the system.Figure 2 is the state transition diagram of M/M/N/N system. In Figure 2, the arrow → indicates incoming BMs, and the arrival rate is λ. Sk means k target channels are busy, and the state is transferred to Sk+1 when a new BM arrives. Arrow ← indicates shootings of defense weapons, when the shooting process for a BM is done, and then the state is transferred from S1 to S0 with rate μ, transferred from S2 to S1 with rate 2μ,…, and transferred from Sk to Sk-1 with rate kμ. Balance equations are as follows:S0:
all target channels are idle; there is no BM in the system:(5)λS0=μS1⟹S1=λμS0=ρS0.S1:
1 target channel is busy; there is 1 BM in the system:(6)λS1=2μS2⟹S2=ρ22!S0.S2:
2 target channels are busy; there are 2 BMs in the system:(7)λS2=2μS3⟹S3=ρ33!S0.⋮Sk-1:
k-1 target channels are busy; there are k-1 BMs in the system:(8)λSk-1=kμSk⟹Sk=ρkk!S0.Sn:
n target channels are busy; there are n BMs in the system:(9)λSn-1=nμSn⟹Sn=ρnn!S0.Figure 2
State transition diagram of M/M/N/N system.Due to the fact thatPf is the probability of the BM found by the radar, we have(10)ρf=ρ·Pf=Pf·λμ=Pf·λ·tsmean.ρ f is the mean number of BMs found to arrive during the mean shooting time. Because S0+S1+S2+⋯+Sn=1, then the expression for S0 can be written in the form(11)S0=1+ρf+ρf22!+⋯+ρfkk!+⋯+ρfnn!-1=∑k=0nρfkk!-1.The optimal operation of the queueing system can be analyzed through several performance parameters, some of which follow.(1)
BM loss probabilities (penetration probabilities):(12)Plost=Sn=ρfnn!S0=ρfnn!∑k=0nρfkk!-1.(2)
Relative probabilities of BMs that will be shot:(13)Pshoot=1-Plost=1-ρfnn!∑k=0nρfkk!-1.(3)
Absolute probabilities of BMs that will be shot:(14)Pshoot′=λPshoot=λ1-ρfnn!∑k=0nρfkk!-1.(4)
Mean number of occupied target channels:(15)Z-=0·S0+1·S1+⋯+n·Sn=∑k=0nkρfkk!S0=ρfS0∑k=0nρfkk!-ρfnn!=ρf1-ρfnn!∑k=0nρfkk!-1.The mean occupancy rate of target channels is η=Z-/n.


## 3.1.2. Single Target Channel and M/M/1/1 System

The M/M/1/1 queueing system can be viewed as a special case of Section3.1.1 for n=1; then, the expression for S0 can be rewritten in the form(16)S0=∑k=0nρfkk!-1n=1=11+ρf.Performance parameters are as follows:(1)
BM loss probabilities:(17)Plost=Sn=ρfnn!S0=ρf1+ρf.(2)
Relative probabilities of BMs that will be shot:(18)Pshoot=1-Plost=1-ρf1+ρf=11+ρf.(3)
Absolute probabilities of BMs that will be shot:(19)Pshoot′=λPshoot=λ1+ρf.The concept of DE is the product of probabilities of BM that will be shot times SSKP; that is,(20)E=Pshoot·Pk=1-Plost·Pk.


## 3.2. Identical Weapons, Long Defense Depth, and M/M/N/C System

From Section1, we know that BMDS with long defense depth can be regarded as a stochastic service system with limited waiting time, that is, the mixed queueing system. For each incoming BM, the defense weapons use idle target channels to shoot. When BMs that arrived find that all target channels were occupied, BMs will not be penetrated but will wait for a limited time until a target channel becomes available. Possible states of the system are as follows:S0:
all target channels are idle; there is no BM in the system.S1:
1 target channel is busy; there is 1 BM in the system.⋮Sk:
k target channels are busy; there are k BMs in the system.⋮Sn:
all target channels are busy; there aren BMs in the system.Sn+1:
all target channels are busy; there is 1 BM waiting.⋮Sn+s:
all target channels are busy; there ares BMs waiting.⋮Figure 3 is the state transition diagram of M/M/N/C system. In Figure 3, the arrow → indicates incoming BMs, and the arrival rate is λ. Sk means k target channels are busy, and the state is transferred from lower state to higher state when new BMs arrive. Arrow ← indicates shootings of defense weapons, when the shooting process for a BM is done, and then the state is transferred from higher state to lower state with rate kμ or nμ+sν. Balance equations are(21)S0S1⋮Sk⋮SnSn+1⋮Sn+s⋮0=-λS0+μS10=λS0-λ+μS1+2μS2⋮0=λSk-1-λ+kμSk+k+1μS2⋮0=λSn-1-λ+nμSn+nμ+νSn+10=λSn-λ+nμ+νSn+1+nμ+2νSn+2⋮0=λSn+s-1-λ+nμ+νSn+s+nμ+s+1νSn+s+1⋮.Figure 3
State transition diagram of M/M/N/C system.Then, we have the constraint that∑i=0∞Si=1 for (21). Let ρ=λ/μ be the mean number of BMs that arrived during the mean shooting time; let ϑ=ν/μ be the mean number of penetrated (leaving without being shot) BMs during the mean shooting time. Then, (21) can be written in the form(22)S0S1⋮Sk⋮SnSn+1⋮Sn+s⋮0=-ρS0+S10=ρS0-ρ+1S1+2S2⋮0=ρSk-1-ρ+kSk+k+1S2⋮0=ρSn-1-ρ+nSn+n+ϑSn+10=ρSn-ρ+n+ϑSn+1+n+2ϑSn+2⋮0=ρSn+s-1-ρ+n+ϑSn+s+n+s+1ϑSn+s+1⋮.When k≤n, we have(23)S1=λμS0=ρS0=ρ11!S0,S2=12ρ+1S1-ρS0=12ρS1+S1-ρS0=ρ2S1=ρ22!S0⋮Sn=ρnSk-1=ρnn!S0.When k>n, we have(24)Sn+1=1n+ϑρ+nSn-ρSn-1=1n+ϑρSn+nSn-ρSn-1=ρn+ϑSn=ρn+ϑρnn!S0=ρnn!ρn+ϑS0,Sn+2=1n+2ϑρ+ρ+ϑSn+1-ρSn=1n+2ϑρSn+1+ρ+ϑSn+1-ρSn=1n+2ϑρSn+1+ρSn-ρSn=ρnn!ρ2n+ϑn+2ϑS0⋮Sn+s=ρn+sϑSn+s-1=ρnn!ρs∏m=1sn+mϑS0.The constraint for (24) is(25)∑m=0nSm+∑s=1∞Sm+s=1.With direct substitution of (23) and (24) into (25), it follows that(26)1=S0∑m=0nρmm!+ρnn!∑s=1∞ρs∏m=1sn+mϑ-1,S0=∑m=0nρmm!+ρnn!∑s=1∞ρs∏m=1sn+mϑ-1-1.Then, with substitution of (26) into (24), we have steady-state probability(27)Sk=ρkk!∑m=0nρmm!+ρnn!∑s=1∞ρs∏m=1sn+mϑ-1-1,0≤k≤n,Sn+s=ρn/n!ρs∏m=1sn+mϑ-1∑m=0nρm/m!+ρn/n!∑s=1∞ρs∏m=1sn+mϑ-1,s≥1.Performance parameters are as follows:(1)
BM loss probabilities (s BMs waiting to be shot):(28)Plost=νλ∑s=1∞sSn+s=ν/μλ/μ∑s=1∞sSn+s=ϑρ∑s=1∞sSn+s=ϑρ∑s=1∞sρn/n!ρs∏m=1sn+mϑ-1∑m=0nρm/m!+ρn/n!∑s=1∞ρs∏m=1sn+mϑ-1.(2)
Relative probabilities of BMs that will be shot:(29)Pshoot=1-Plost.(3)
Absolute probabilities of BMs that will be shot:(30)Pshoot′=λPshoot.(4)
Mean number of occupied target channels:(31)Z-=∑k=1nkSk+∑s=1∞nSn+s=∑k=1nkSk+n∑s=1∞Sn+sorZ-=λPshootμ=ρPshoot.The mean occupancy rate of target channels is(32)η=ρPshootn.DE is the product of probabilities of BM that will be shot times SSKP; that is,(33)E=Pshoot·Pk=1-Plost·Pk.Similarly, the M/M/1/C queueing system can be viewed as a special case for n=1.


## 3.3. Different Defense Weapons

Defense weapons in Sections3.1 and 3.2 are identical, and the BMDS may deploy different types of defense weapons. For different types of defense weapons, the waiting time of BMs, the detection probability of radars, and SSKP may be different. In order to model the queueing system, we choose one type of defense weapon as the reference and then substitute the reference defense (type r) for defense weapons of type i. The “replacement process” is named equivalent replacement method [20]; basic equations are(34)λr·Pshootr·Pkr·Pfr=λi·Pshooti·Pki·Pfi,nr·1+twmeanr·Pkr=ni·1+twmeani·Pki,nr=ni·1+twmeani·Pki1+twmeanr·Pkr.Subscripts(·)(r) and (·)(i) in (34) are used to distinguish defense weapons of type r from defense weapons of type i. From the above equations, we can substitute D(r) for D(i); the intensity of BMs killed by defense weapons of type i and the intensity of BMs arrivals for defense weapons of type i are(35)ρi=Di·Pfi·Pkitwtimei,λi=λρi·∑i=1Rρi-1,respectively, where R denotes the total number of types.


## 4. Optimal Allocation of Target Channels

### 4.1. Identical Defense Weapons and M/M/N/N System

#### 4.1.1. Defense Effectiveness

LetM be the number of layers; the number of target channels deployed along the ith layer is denoted by ni; the probability of BMs that will be shot by the M-layer defense with short defense depth is (36)PshootM=1-ρNN!∑k=0Nρkk!-1,where N is the total number of target channels, N=∑i=1Mni. The probabilities of BMs that will be shot by the M layer defense with short defense depth are as follows:1st layer:(37)ρ1=ρf1=λ·twmean·Pf,Pshoot1=1-ρ1n1n1!∑k=0n1ρ1kk!-1,E1=Pshoot1·Pk.2nd layer:(38)ρ2=1-E1·λ·twmean·Pf=1-E1·ρ1,Pshoot2=1-ρ2n2n2!∑k=0n2ρ2kk!-1,E2=Pshoot2·Pk.Mth layer: (39)ρM=1-EM-11-EM-2⋯1-E1λ·twmean·Pf=∏i=1M-11-Ei·ρ1=1-EM-1·ρM-1,PshootM=1-ρMnMnM!∑k=0nMρMkk!-1,EM=PshootM·Pk.DE of the whole M-layer BMDS is(40)E=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEM=E1+∑k=2MEk·∏i=1M-11-Ei.Then, we define the optimization problem as finding numbers of layers and target channels deployed along the ith layer so as to maximize DE, subject to a given set of constraints:(41)maxE=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEMs.t.ρi=1-Ei-11-Ei-2⋯1-E1λ·twmean·Pf=∏x=1i-11-Ei·ρ1=1-Ei-1·ρi-1,Pshooti=1-ρinini!∑k=0niρikk!-1,n=∑i=1Mni,Ei=Pshooti·Pk.For the nonlinear optimization problem (41), when problem size is small, we can use algebra, dynamic programming, or enumeration method to solve it. When the size of the problem is very large, an approximate solution can be obtained by using some advanced algorithms, for example, genetic, neural network, and heuristic algorithms. In order to get some potential and useful allocation rules, we analyze a scenario.Scenario 1.
Suppose that the number of target channels is 10, SSKP is 0.7, the probability that BMs will be detected by radars is 0.8,λ=5BMs/min, and tsmean=0.75min.
Tables1, 2, and 3 are the DE of two-layer, three-layer, and four-layer defense, respectively.
Letn(M)=(n1,n2,…,nM) be the allocation plan of target channels, where M is the number of layers and ni is the number of target channels deployed along the ith layer. From the results of Tables 1, 2, and 3, we noticed that plans n(2)=(6,4), n(3)=(5,3,2), and n(4)=(4,3,2,1) have the biggest DE. We also found that n(2)=(6,4)<n(3)=(5,3,2)<n(4)=(4,3,2,1), and then we have Theorem 1.Table 1
DE of two-layer defense.
1st-layer channel number 1st-layer DE 2nd-layer channel number 2nd-layer DE Total DE 9 56.00% 1 14.56% 62.41% 8 56.00% 2 25.20% 67.09% 7 54.32% 3 32.48% 69.16% 6 52.08 % 4 41.44 % 70.94 % 5 47.04% 5 44.24% 70.47% 4 35.84% 6 44.80% 64.58% 3 34.16% 7 53.76% 69.56% 2 27.44% 8 54.88% 67.26% 1 11.20% 9 56.00% 60.93%Table 2
DE of three-layer defense.
1st-channel number 1st-layer DE 2nd-channel number 2nd-layer DE 3rd-channel number 3rd-layer DE Total DE 8 55.552% 1 11.872% 1 11.872% 65.479% 7 53.256% 2 22.176% 1 11.872% 67.941% 7 53.256% 1 11.872% 2 22.176% 67.941% 6 50.344% 2 22.176% 2 22.176% 69.925% 6 50.344% 3 32.368% 1 11.872% 70.404% 6 50.344% 1 11.872% 3 32.368% 70.404% 5 46.368 % 3 32.368 % 2 22.176 % 71.771 % 5 46.368% 2 22.176% 3 32.368% 71.771% 5 46.368% 4 38.264% 1 11.872% 70.821% 5 46.368% 1 11.872% 4 38.264% 70.821% 4 38.264% 3 32.368% 3 32.368% 71.761% 4 38.264% 4 38.264% 2 22.176% 70.339% 4 38.264% 2 22.176% 4 38.264% 70.339% 4 38.264% 5 46.368% 1 11.872% 70.821% 4 38.264% 1 11.872% 5 46.368% 70.821% 3 32.368% 4 38.264% 3 32.368% 71.761% 3 32.368% 3 32.368% 4 38.264% 71.761% 3 32.368% 5 46.368% 2 22.176% 71.771% 3 32.368% 2 22.176% 5 46.368% 71.771% 3 32.368% 6 50.344% 1 11.872% 70.404% 3 32.368% 1 11.872% 6 50.344% 70.404% 2 22.176% 4 38.264% 4 38.264% 70.339% 2 22.176% 5 46.368% 3 32.368% 71.771% 2 22.176% 3 32.368% 5 46.368% 71.771% 2 22.176% 6 50.344% 2 22.176% 69.925% 2 22.176% 2 22.176% 6 50.344% 69.925% 2 22.176% 7 53.256% 1 11.872% 67.941% 2 22.176% 1 11.872% 7 53.256% 67.941% 1 11.872% 5 46.368% 4 38.264% 70.821% 1 11.872% 4 38.264% 5 46.368% 70.821% 1 11.872% 6 50.344% 3 32.368% 70.404% 1 11.872% 3 32.368% 6 50.344% 70.404% 1 11.872% 7 53.256% 2 22.176% 67.941% 1 11.872% 2 22.176% 7 53.256% 67.941% 1 11.872% 8 55.552% 1 11.872% 65.479% 1 11.872% 1 11.872% 8 55.552% 65.479%Table 3
DE of four-layer defense.
1st-channel number 1st-layer DE 2nd-channel number 2nd-layer DE 3rd-channel number 3rd-layer DE 4th-channel number 4th-layer DE Total DE 7 53.200% 1 11.872% 1 11.872% 1 11.872% 67.968% 6 50.344% 2 17.360% 1 11.872% 1 11.872% 68.129% 6 50.344% 1 11.872% 2 17.360% 1 11.872% 68.129% 5 47.600% 3 34.920% 1 11.872% 1 11.872% 73.515% 5 47.600% 2 17.360% 2 17.360% 1 11.872% 68.463% 5 47.600% 1 11.872% 3 34.920% 1 11.872% 73.515% 4 45.920 % 3 34.920 % 2 17.360 % 1 11.872 % 74.368 % 4 45.920% 3 34.920% 1 11.872% 2 17.360% 74.368% 4 45.920% 2 17.360% 1 11.872% 3 34.920% 74.368% 4 45.920% 2 17.360% 3 34.920% 1 11.872% 74.368% 4 45.920% 1 11.872% 2 17.360% 3 34.920% 74.368% 4 45.920% 1 11.872% 3 34.920% 2 17.360% 74.368% 3 34.920% 5 47.600% 1 11.872% 1 11.872% 73.515% 3 34.920% 4 45.920% 2 17.360% 1 11.872% 74.368% 3 34.920% 4 45.920% 1 11.872% 2 17.360% 74.368% 3 34.920% 3 34.920% 2 17.360% 2 17.360% 71.075% 3 34.920% 2 17.360% 3 34.920% 2 17.360% 71.075% 3 34.920% 2 17.360% 2 17.360% 3 34.920% 71.075% 3 34.920% 2 17.360% 4 45.920% 1 11.872% 74.368% 3 34.920% 2 17.360% 1 11.872% 4 45.920% 74.368% 3 34.920% 1 11.872% 4 45.920% 2 17.360% 74.368% 3 34.920% 1 11.872% 2 17.360% 4 45.920% 74.368% 3 34.920% 1 11.872% 5 47.600% 1 11.872% 73.515% 2 17.360% 6 50.344% 1 11.872% 1 11.872% 68.129% 2 17.360% 5 47.600% 2 17.360% 1 11.872% 68.463% 2 17.360% 4 45.920% 3 34.920% 3 34.920% 71.071% 2 17.360% 3 34.920% 3 34.920% 2 17.360% 71.075% 2 17.360% 3 34.920% 2 17.360% 3 34.920% 71.075% 1 11.872% 7 53.200% 1 11.872% 1 11.872% 67.968% 1 11.872% 6 50.344% 2 17.360% 1 11.872% 68.129%Theorem 1.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and N=∑i=1Mni, ni≥1. When N is constant, then max⁡E(M,n(M)) is stochastically increasing as M increases; that is,(42)max⁡EM-1,nM≤max⁡EM,nM≤max⁡EM+1,nM.Proof of Theorem1 is similar to Lemma  1 in [8]. Now, we continue to compute DE of M=5, N=10 and M=6, N=10. Tables 4 and 5 are the DE of five-layer and six-layer defense, respectively. Another useful rule is that the number of target channels deployed along the ith layer should be not less than (i+1)th layer; this rule is summarized in Theorem 2.Table 4
DE of five-layer defense.
1st number 1st DE 2nd number 2nd DE 3rd number 3rd DE 4th number 4th DE 5th number 5th DE Total DE 5 45.5% 2 23.4% 1 12.0% 1 12.0% 1 12.0% 71.5% 4 38.6% 3 30.8% 1 12.0% 1 12.0% 1 12.0% 71.0% 4 38.6% 2 23.4% 2 23.4% 1 12.0% 1 12.0% 72.0% 3 30.8% 3 30.8% 2 23.4% 1 12.0% 1 12.0% 71.6% 3 30.8 % 3 30.8 % 2 23.4 % 1 12.0 % 1 12.0 % 74.6 % 2 23.4% 2 23.4% 2 23.4% 2 23.4% 2 23.4% 73.5%Table 5
DE of six-layer defense.
1st number 1st DE 2nd number 2nd DE 3rd number 3rd DE 4th number 4th DE 5th number 5th DE 6th number 6th DE Total DE 5 45% 1 14% 1 14% 1 14% 1 14% 1 14% 74% 4 39% 2 26% 1 14% 1 14% 1 14% 1 14% 75% 3 41 % 2 26 % 2 26 % 1 14 % 1 14 % 1 14 % 79 % 2 26% 3 41% 2 26% 1 14% 1 14% 1 14% 79% 2 26% 2 26% 2 26% 2 26% 1 14% 1 14% 77%Theorem 2.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and N=∑i=1Mni, ni≥1. For allocation plan n(M)∗, the number of target channels deployed along the ith layer is decreasing as i increases; that is, for all i, ni≥ni+1. When N is constant, then one has E(M,n(M)∗)≥E(M,n(M)).Proof.
Suppose we have a plann(M)Θ=(n1,n2,…,ni∗,ni+1∗,…,nM), where ni∗=ni+1, ni+1∗=ni. In order to prove Theorem 2, we only need to prove E(M,n(M)∗)≥E(M,n(M)Θ), which can be simplified by proving E(2,n(2)∗)≥E(2,n(2)Θ), where n(2)∗=(ni,ni+1), n(2)Θ=(ni∗,ni+1∗). For function(43)fρ,n=Plost=ρnn!∑k=0nρkk!-1,when ρ is constant, f(ρ,n) decreases with the increase of n, and when n is constant, f(ρ,n) increases with the increase of ρ [2, 20]. For proof simplified by letting Pk=1 for all defense weapons, then f(ρ,ni), f(ρ,ni+1), f(ρ,ni∗), and f(ρ,ni+1∗) are(44)fρ,ni=ρnini!∑k=0niρkk!-1,fρ,ni+1=ρ∗ni+1ni+1!∑k=0ni+1ρ∗kk!-1,fρ,ni∗=ρni+1ni+1!∑k=0ni+1ρkk!-1,fρ,ni+1∗=ρΘnini!∑k=0niρΘkk!-1,respectively, where ρ∗=f(ρ,ni)·ρ, ρΘ=f(ρ,ni∗)·ρ. DE of the two-layer defense are(45)E2,n2∗=Pk·1-fρ,ni+fρ,ni·1-fρ,ni+1=1-fρ,nifρ,ni+1,E2,n2Θ=Pk·1-fρ,ni∗+fρ,ni∗·1-fρ,ni+1∗=1-fρ,ni∗fρ,ni+1∗.Since the expansion of ex is 1+x+x2/2!+⋯+xn/n!, then the approximate representations of f(ρ,ni), f(ρ,ni+1), f(ρ,ni∗), and f(ρ,ni+1∗) are(46)fρ,ni≈ρnini!eρ-1,fρ,ni+1≈ρ∗ni+1ni+1!eρ∗-1,fρ,ni∗≈ρni+1ni+1!eρ-1,fρ,ni+1∗=ρΘnini!eρΘ-1,respectively. Because ni≥ni+1, we have (47)fρ,ni∗≥fρ,ni,fρ,ni∗≥fρ,ni+1,fρ,ni≥fρ,ni+1∗.In order to prove Theorem 2, we need to prove(48)1-fρ,nifρ,ni+1≥1-fρ,ni∗fρ,ni+1∗⟸fρ,ni∗fρ,ni+1∗-fρ,nifρ,ni+1≥0⟸fρ,ni∗fρ,ni+1∗fρ,nifρ,ni+1≥1,fρ,ni∗fρ,ni+1∗fρ,nifρ,ni+1=ρni+1/ni+1!eρ-1ρΘni/ni!eρΘ-1ρni/ni!eρ-1ρ∗ni+1/ni+1!eρ∗-1=ρni+1ρΘnieρ∗ρniρ∗ni+1eρΘ=ρni+1ρΘniefρ,ni·ρρniρ∗ni+1efρ,ni∗·ρ=ρni+1fρ,ni∗·ρniefρ,ni·ρρnifρ,ni·ρni+1efρ,ni∗·ρ=fρ,ni∗niefρ,ni·ρfρ,nini+1efρ,ni∗·ρ=ni!ni+1eni+1·ρefρ,ni·ρni+1!nieni·ρefρ,ni∗·ρ=ni!ni+1ni+1!nieni+1+fρ,ni-ni-fρ,ni∗·ρ.When ni=ni+1, Theorem 2 is true. Obviously, when ni>ni+1, we have ni!ni+1≫ni+1!ni; then, E(2,n(2)∗)>E(2,n(2)Θ), and the result of Theorem 2 follows.


#### 4.1.2. Minimum Requirements of Target Channels

The requirements of target channels necessary to achieve a demanded DE can be viewed as a dual problem of (41). Assuming that the DE is held at greater than E∗, we define the optimization problem so as to minimize the requirements of target channels, subject to a given set of constraints: (49)minn=∑i=1Mnis.t.ρi=1-Ei-11-Ei-2⋯1-E1λ·twmean·Pf=∏x=1i-11-Ei·ρ1=1-Ei-1·ρi-1,Pshooti=1-ρinini!∑k=0niρikk!-1,E∗≤E1+1-E1E2+⋯+∏i=1M-11-EiEM,Ei=Pshooti·Pk.For the nonlinear optimization problem (49), when problem size is small, we can use algebra, dynamic programming, or enumeration method to solve it. When the size of the problem is very large, an approximate solution can be obtained by using some advanced algorithms, for example, genetic and heuristic algorithms. Then, we give the definition of neighborhood [21].Definition 3.
Letn(M)=(n1,n2,…,nM)∈Ω be an allocation plan, and Ω is the feasible region of allocation plans. Suppose that n(M)∗=(n1,n2,…,ni-1,ni-1,ni+1,…,nM)∈Ω, where the number of target channels deployed along the ith layer is ni-1. Then, one says that n(M)∗ is in the neighborhood of n(M); that is,(50)n1,n2,…,ni-1,ni-1,ni+1,…,nM∈φn1,n2,…,nM.Scenario 2.
Suppose the SSKP is 0.7, the probability that BMs will be detected by radars is 0.8,λ=5BMs/min, tsmean=0.75min, and E∗=65%; then the question becomes as follows: “What is the least cost of target channels to achieve a demanded DE?”
Figure4 is the schematic of search in the neighborhood of n(3)=(5,3,2) and n(4)=(4,3,2,1). We can see the least cost is 9 channels for three-layer defense and 8 channels for four-layer defense, and allocation plans are n(3)=(4,3,2) and n(4)=(3,3,1,1), respectively. So, we give a simple algorithm in finding the least cost of DWs to achieve a demanded DE; specific search methods are as follows. A feasible initial allocation plan is very important in this algorithm. Theorem 1 proposes the basic rule of finding an initial plan, which greatly simplifies the searching process.Figure 4
Search in the neighborhood ofn(3)=(5,3,2) and n(4)=(4,3,2,1).Step 1.
Suppose the number of layers isM; the demanded DE is E∗; and give an initial allocation plan n(M)=(n1,n2,…,nM)∈Ω, where E(M,n(M))≥E∗.Step 2.
Search in the neighborhood ofn(M), φn(M), and try to find an allocation plan φ1n(M) satisfying the following: (51)EM,φ1nM=max⁡EM,φnM.Step 3.
IfEM,φ1n(M)≥E∗,then n(M)=φ1n(M); go to Step 2 and repeat.Step 4.
IfEM,φ1n(M)<E∗,go to Step 5.Step 5.
Output the allocation plann(M); ∑i=1Mni is the least cost to achieve the demanded DE.


### 4.2. Different Defense Weapons and M/M/N/N System

#### 4.2.1. Different Defense Weapons and Identical SSKPs

In this section, we consider different types of defense weapons. LetM be the number of layers; the number of target channels deployed along the ith layer will be denoted by ni, assuming that the total number of types is M and defense weapons are identical in the same layer. As in Section 3.3, subscript (·)(i) indicates defense weapons of type i; then probabilities that BMs will be shot by the M-layer defense with short defense depth are as follows:1st layer:(52)ρ11=λ·twmean1·Pf1,Pshoot1=1-ρ11n1n1!∑k=0n1ρ11kk!-1,E1=Pshoot1·Pk1.2nd layer:(53)ρ22=1-E1·λ·twmean2·Pf2,Pshoot2=1-ρ22n2n2!∑k=0n2ρ22kk!-1,E2=Pshoot2·Pk2.Mth layer: (54)ρMM=1-EM-11-EM-2⋯1-E1λ·twmeanM·PfM=∏i=1M-11-Ei·λ·twmeanM·PfM,PshootM=1-ρMMnMnM!∑k=0nMρMMkk!-1,EM=PshootM·PkM.DE of the wholeM-layer BMDS is(55)E=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEM=E1+∑k=2MEk·∏i=1M-11-Ei.For certain types and numbers of target channels, we define the optimization problem as finding which type of defense weapon should be deployed on each layer so as to maximize the ED, subject to a given set of constraints:(56)maxE=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEMs.t.ρiM=1-Ei-11-Ei-2⋯1-E1λ·twmeanM·PfM=∏x=1i-11-Ei·λ·twmeanM·PfM,Pshooti=1-ρiMnini!∑k=0niρiMkk!-1,Ei=Pshooti·PkM,n=∑i=1Mni.In order to get some potential and useful allocation rules, we analyze a scenario.Scenario 3.
Suppose that the number of layers is 3, the number of defense weapon types is 2 (types I and II),Pk(I)=Pk(II)=0.7, the probability that BMs will be detected by radars is 0.8, λ=5BMs/min, tsmean(I)=0.75min, and tsmean(II)=1min.
Table6 is the DE of three-layer defense.
It can be seen from Table6 that plan n(3)=(I,I,II) has the biggest DE. We also found that n(4)=(I,I,I,II) is the best, and then we have Theorem 4.Table 6
DE of three-layer defense.
1st-layer type 1st-layer DE 2nd-layer type 2nd-layer DE 3rd-layer type 3rd-layer DE Total DE II 43.200% II 45.440% I 55.520% 86.216% II 48.960% I 51.840% II 46.560% 86.864% II 43.200% I 55.040% I 55.040% 88.518% I 54.560% II 45.280% II 46.720% 86.752% I 56.800 % I 55.360 % II 43.040 % 89.016 %Theorem 4.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and let n1=n2=⋯=nM. Allocation plan n(M)I=(I,II) indicates that defense weapons of type I are forward-deployed, and n(M)II=(II,I) indicates that defense weapons of type II are forward-deployed. Suppose that defense weapons are identical in the same layer, and Pk(I)=Pk(II) and Pf(I)=Pf(II); if tsmean(I)≤tsmean(II), then one has E(M,n(M)I)≥E(M,n(M)II).Proof.
Suppose that we have an allocation plann(M)IΘ=(IΘ,IIΘ), where tsmean(IΘ)=tsmean(II) and tsmean(IIΘ)=tsmean(I). In order to prove Theorem 4, we only need to prove E(M,n(M)I)≥E(M,n(M)IΘ). For function(57)fρ,n=Plost=ρnn!∑k=0nρkk!-1,when ρ is constant, f(ρ,n) decreases with the increase of n, and when n is constant, f(ρ,n) increases with the increase of ρ [2, 20]. Then, f(ρ,I), f(ρ,II), f(ρ,IΘ), and f(ρ,IIΘ) are(58)fρ,I=ρInn!∑k=0nρIkk!-1,fρ,II=ρIInn!∑k=0nρIIkk!-1,fρ,IΘ=ρIΘnn!∑k=0nρIΘkk!-1,fρ,IIΘ=ρIIΘnn!∑k=0nρIIΘkk!-1,where(59)ρI=λ·tsmeanI·PfI,ρIΘ=λ·tsmeanIΘ·PfIΘ,ρII=1-EIλtsmeanIIPfII,ρIIΘ=1-EIΘλtsmeanIIΘPfIΘ,EI=1-fρ,I,EIΘ=1-fρ,IΘ.DE of the defense are(60)EM,nMI=Pk·1-fρ,I+fρ,I·1-fρ,II=1-fρ,Ifρ,II,EM,nMIΘ=Pk·1-fρ,IΘ+fρ,IΘ·1-fρ,IIΘ=1-fρ,IΘfρ,IIΘ.Since the expansion of ex is 1+x+x2/2!+⋯+xn/n!, then the approximate representations of f(ρ,I), f(ρ,II), f(ρ,IΘ), and f(ρ,IIΘ) are(61)fρ,I≈ρInn!eρI-1,fρ,II≈ρIIni+1n!eρII-1,fρ,IΘ≈ρIΘnn!eρIΘ-1,fρ,IIΘ=ρIIΘnn!eρIIΘ-1.In order to prove Theorem 4, we need to prove(62)1-fρ,Ifρ,II≥1-fρ,IΘfρ,IIΘ⟸fρ,IΘfρ,IIΘ-fρ,Ifρ,II≥0⟸fρ,IΘfρ,IIΘfρ,Ifρ,II≥1,fρ,IΘfρ,IIΘfρ,Ifρ,II=ρIΘn/n!eρIΘ-1ρIIΘn/n!eρIIΘ-1ρIn/n!eρI-1ρIIni/n!eρII-1=ρIΘnρIIΘnρInρIIneρI+ρII-ρIΘ-ρIIΘ=ρIΘρIn2en+1ρI+ρII-n+1ρIΘ-ρIIΘ=λtsmeanIIPfIIλtsmeanIPfIn2eλn+1tsmeanIPfI+1-EItsmeanIIPfII-n+1tsmeanIIPfII-1-EIItsmeanIPfI.Let tsmean(II)Pf(II)=A and tsmean(I)Pf(I)=B; then,(63)fρ,IΘfρ,IIΘfρ,Ifρ,II=ABn2eλn+1B+1-EIA-n+1A-1-EIIB.Since Pf(II)=Pf(I), then A>B, and the result of Theorem 4 follows.The termsA and B that appear in the proof of Theorem 4 could be a factor in channels allocation; we extended Theorem 4 to obtain Lemma 5.Lemma 5.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and let n1=n2=⋯=nM. Allocation plan n(M)I=(I,II) indicates that defense weapons of type I are forward-deployed, and n(M)II=(II,I) indicates that defense weapons of type II are forward-deployed. Suppose that defense weapons are identical in the same layer, and Pk(I)=Pk(II); if Pf(I)tsmean(I)≤Pf(II)tsmean(II), then one has E(M,n(M)I)≥E(M,n(M)II).


#### 4.2.2. Different Defense Weapons and Different SSKPs

In this section, we consider different defense weapons and different SSKPs of layered BMDS. It is obvious that higher SSKP dose improves the DE. How SSKP affects the target channels allocation is a complex problem; however, we can summarize some useful rules; see Scenario4.Scenario 4.
Suppose that the number of layers is 3, the number of defense weapon types is 2 (types I and II), the probability that BMs will be detected by radars is 0.8,λ=5BMs/min, tsmean(I)=tsmean(II)=0.75min, Pk(I)=0.8, and Pk(II)=0.6.
Table7 is the DE of three-layer defense. It can be seen from Table 7 that plan n(3)=(I,I,II) has the biggest DE. We also found that n(4)=(I,I,I,II) is the best, and then we have Theorem 6.Table 7
DE of three-layer defense.
1st-layer type 1st-layer DE 2nd-layer type 2nd-layer DE 3rd-layer type 3rd-layer DE Total DE II 36.720% II 41.400% I 0.594 83.930% II 42.120% I 59.040% II 0.400 84.766% II 42.600% I 55.840% I 0.506 87.468% I 54.240% II 39.600% II 0.415 85.837% I 56.640 % I 54.400 % II 0.426 88.651 %Theorem 6.
LetE(M,n(M)) be the DE of the M-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and let n1=n2=⋯=nM. Allocation plan n(M)I=(I,II) indicates that defense weapons of type I are forward-deployed, and n(M)II=(II,I) indicates that defense weapons of type II are forward-deployed. Suppose that defense weapons are identical in the same layer, and tsmean(I)=tsmean(II) and Pf(I)=Pf(II); if Pk(I)>Pk(II), then one has E(M,n(M)I)≥E(M,n(M)II).Proof.
Suppose that we have an allocation plann(M)IΘ=(IΘ,IIΘ), where Pk(IΘ)=Pk(II) and Pk(IIΘ)=Pk(I). In order to prove Theorem 4, we only need to prove E(M,n(M)I)≥E(M,n(M)IΘ).
f ( ρ , I ), f(ρ,II), f(ρ,IΘ), and f(ρ,IIΘ) are(64)fρ,I=ρInn!∑k=0nρIkk!-1,fρ,II=ρIInn!∑k=0nρIIkk!-1,fρ,IΘ=ρIΘnn!∑k=0nρIΘkk!-1,fρ,IIΘ=ρIIΘnn!∑k=0nρIIΘkk!-1,where(65)ρI=λ·tsmeanI·PfI,ρIΘ=λ·tsmeanIΘ·PfIΘ,ρII=1-EIλtsmeanIIPfII,ρIIΘ=1-EIΘλtsmeanIIΘPfIΘ,EI=1-fρ,I·PkI,EIΘ=1-fρ,IΘ·PkII.DE of the defense are(66)EM,nMI=PkI-fρ,IPkI+PkII-fρ,IIPkII-PkIPkII+fρ,IIPkIPkII+fρ,IPkIPkII-fρ,Ifρ,IIPkIPkII,EM,nMIΘ=PkIΘ-fρ,IΘPkIΘ+PkIIΘ-fρ,IIΘPkIIΘ-PkIΘPkIIΘ+fρ,IIΘPkIΘPkIIΘ+fρ,IΘPkIΘPkIIΘ-fρ,IΘfρ,IIΘPkIΘPkIIΘ.In order to prove Theorem 6, we need to prove(67)EM,nMI-EM,nMIΘ=PkIPkIIfρ,II+fρ,I-fρ,Ifρ,II-fρ,IIΘ-fρ,IΘ+fρ,IΘfρ,IIΘ-fρ,IPkI-fρ,IIPkII+fρ,IΘPkII+fρ,IIΘPkI≥0.Since Pk(I)>Pk(II), f(ρ,I)=f(ρ,IΘ), then ∗fρ,I=fρ,IΘ≥fρ,IIΘ≥fρ,II,PkIPkII1-fρ,Ifρ,IIΘ-fρ,II+fρ,IPkI-PkII+fρ,IIPkII-fρ,IIΘPkI≥0.Since the expansion of ex is 1+x+x2/2!+⋯+xn/n!, then the approximate representations of f(ρ,IΘ) and f(ρ,IIΘ) are(68)fρ,IΘ≈ρIΘnn!eρIΘ-1,fρ,IIΘ=ρIIΘnn!eρIIΘ-1,fρ,IIΘfρ,IΘ=1-1-fρ,IΘ·PkIΘneρIΘ·ρIIΘ.Let tsmean(II)Pf(II)=A and tsmean(I)Pf(I)=B; then,(69)fρ,IΘfρ,IIΘfρ,Ifρ,II=ABn2eλn+1B+1-EIA-n+1A-1-EIIB.Obviously, we have(70)1-1-fρ,IΘ·PkIΘneρIΘ·ρIIΘ⟶0.The result of Theorem 4 follows.


## 4.1. Identical Defense Weapons and M/M/N/N System

### 4.1.1. Defense Effectiveness

LetM be the number of layers; the number of target channels deployed along the ith layer is denoted by ni; the probability of BMs that will be shot by the M-layer defense with short defense depth is (36)PshootM=1-ρNN!∑k=0Nρkk!-1,where N is the total number of target channels, N=∑i=1Mni. The probabilities of BMs that will be shot by the M layer defense with short defense depth are as follows:1st layer:(37)ρ1=ρf1=λ·twmean·Pf,Pshoot1=1-ρ1n1n1!∑k=0n1ρ1kk!-1,E1=Pshoot1·Pk.2nd layer:(38)ρ2=1-E1·λ·twmean·Pf=1-E1·ρ1,Pshoot2=1-ρ2n2n2!∑k=0n2ρ2kk!-1,E2=Pshoot2·Pk.Mth layer: (39)ρM=1-EM-11-EM-2⋯1-E1λ·twmean·Pf=∏i=1M-11-Ei·ρ1=1-EM-1·ρM-1,PshootM=1-ρMnMnM!∑k=0nMρMkk!-1,EM=PshootM·Pk.DE of the whole M-layer BMDS is(40)E=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEM=E1+∑k=2MEk·∏i=1M-11-Ei.Then, we define the optimization problem as finding numbers of layers and target channels deployed along the ith layer so as to maximize DE, subject to a given set of constraints:(41)maxE=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEMs.t.ρi=1-Ei-11-Ei-2⋯1-E1λ·twmean·Pf=∏x=1i-11-Ei·ρ1=1-Ei-1·ρi-1,Pshooti=1-ρinini!∑k=0niρikk!-1,n=∑i=1Mni,Ei=Pshooti·Pk.For the nonlinear optimization problem (41), when problem size is small, we can use algebra, dynamic programming, or enumeration method to solve it. When the size of the problem is very large, an approximate solution can be obtained by using some advanced algorithms, for example, genetic, neural network, and heuristic algorithms. In order to get some potential and useful allocation rules, we analyze a scenario.Scenario 1.
Suppose that the number of target channels is 10, SSKP is 0.7, the probability that BMs will be detected by radars is 0.8,λ=5BMs/min, and tsmean=0.75min.
Tables1, 2, and 3 are the DE of two-layer, three-layer, and four-layer defense, respectively.
Letn(M)=(n1,n2,…,nM) be the allocation plan of target channels, where M is the number of layers and ni is the number of target channels deployed along the ith layer. From the results of Tables 1, 2, and 3, we noticed that plans n(2)=(6,4), n(3)=(5,3,2), and n(4)=(4,3,2,1) have the biggest DE. We also found that n(2)=(6,4)<n(3)=(5,3,2)<n(4)=(4,3,2,1), and then we have Theorem 1.Table 1
DE of two-layer defense.
1st-layer channel number 1st-layer DE 2nd-layer channel number 2nd-layer DE Total DE 9 56.00% 1 14.56% 62.41% 8 56.00% 2 25.20% 67.09% 7 54.32% 3 32.48% 69.16% 6 52.08 % 4 41.44 % 70.94 % 5 47.04% 5 44.24% 70.47% 4 35.84% 6 44.80% 64.58% 3 34.16% 7 53.76% 69.56% 2 27.44% 8 54.88% 67.26% 1 11.20% 9 56.00% 60.93%Table 2
DE of three-layer defense.
1st-channel number 1st-layer DE 2nd-channel number 2nd-layer DE 3rd-channel number 3rd-layer DE Total DE 8 55.552% 1 11.872% 1 11.872% 65.479% 7 53.256% 2 22.176% 1 11.872% 67.941% 7 53.256% 1 11.872% 2 22.176% 67.941% 6 50.344% 2 22.176% 2 22.176% 69.925% 6 50.344% 3 32.368% 1 11.872% 70.404% 6 50.344% 1 11.872% 3 32.368% 70.404% 5 46.368 % 3 32.368 % 2 22.176 % 71.771 % 5 46.368% 2 22.176% 3 32.368% 71.771% 5 46.368% 4 38.264% 1 11.872% 70.821% 5 46.368% 1 11.872% 4 38.264% 70.821% 4 38.264% 3 32.368% 3 32.368% 71.761% 4 38.264% 4 38.264% 2 22.176% 70.339% 4 38.264% 2 22.176% 4 38.264% 70.339% 4 38.264% 5 46.368% 1 11.872% 70.821% 4 38.264% 1 11.872% 5 46.368% 70.821% 3 32.368% 4 38.264% 3 32.368% 71.761% 3 32.368% 3 32.368% 4 38.264% 71.761% 3 32.368% 5 46.368% 2 22.176% 71.771% 3 32.368% 2 22.176% 5 46.368% 71.771% 3 32.368% 6 50.344% 1 11.872% 70.404% 3 32.368% 1 11.872% 6 50.344% 70.404% 2 22.176% 4 38.264% 4 38.264% 70.339% 2 22.176% 5 46.368% 3 32.368% 71.771% 2 22.176% 3 32.368% 5 46.368% 71.771% 2 22.176% 6 50.344% 2 22.176% 69.925% 2 22.176% 2 22.176% 6 50.344% 69.925% 2 22.176% 7 53.256% 1 11.872% 67.941% 2 22.176% 1 11.872% 7 53.256% 67.941% 1 11.872% 5 46.368% 4 38.264% 70.821% 1 11.872% 4 38.264% 5 46.368% 70.821% 1 11.872% 6 50.344% 3 32.368% 70.404% 1 11.872% 3 32.368% 6 50.344% 70.404% 1 11.872% 7 53.256% 2 22.176% 67.941% 1 11.872% 2 22.176% 7 53.256% 67.941% 1 11.872% 8 55.552% 1 11.872% 65.479% 1 11.872% 1 11.872% 8 55.552% 65.479%Table 3
DE of four-layer defense.
1st-channel number 1st-layer DE 2nd-channel number 2nd-layer DE 3rd-channel number 3rd-layer DE 4th-channel number 4th-layer DE Total DE 7 53.200% 1 11.872% 1 11.872% 1 11.872% 67.968% 6 50.344% 2 17.360% 1 11.872% 1 11.872% 68.129% 6 50.344% 1 11.872% 2 17.360% 1 11.872% 68.129% 5 47.600% 3 34.920% 1 11.872% 1 11.872% 73.515% 5 47.600% 2 17.360% 2 17.360% 1 11.872% 68.463% 5 47.600% 1 11.872% 3 34.920% 1 11.872% 73.515% 4 45.920 % 3 34.920 % 2 17.360 % 1 11.872 % 74.368 % 4 45.920% 3 34.920% 1 11.872% 2 17.360% 74.368% 4 45.920% 2 17.360% 1 11.872% 3 34.920% 74.368% 4 45.920% 2 17.360% 3 34.920% 1 11.872% 74.368% 4 45.920% 1 11.872% 2 17.360% 3 34.920% 74.368% 4 45.920% 1 11.872% 3 34.920% 2 17.360% 74.368% 3 34.920% 5 47.600% 1 11.872% 1 11.872% 73.515% 3 34.920% 4 45.920% 2 17.360% 1 11.872% 74.368% 3 34.920% 4 45.920% 1 11.872% 2 17.360% 74.368% 3 34.920% 3 34.920% 2 17.360% 2 17.360% 71.075% 3 34.920% 2 17.360% 3 34.920% 2 17.360% 71.075% 3 34.920% 2 17.360% 2 17.360% 3 34.920% 71.075% 3 34.920% 2 17.360% 4 45.920% 1 11.872% 74.368% 3 34.920% 2 17.360% 1 11.872% 4 45.920% 74.368% 3 34.920% 1 11.872% 4 45.920% 2 17.360% 74.368% 3 34.920% 1 11.872% 2 17.360% 4 45.920% 74.368% 3 34.920% 1 11.872% 5 47.600% 1 11.872% 73.515% 2 17.360% 6 50.344% 1 11.872% 1 11.872% 68.129% 2 17.360% 5 47.600% 2 17.360% 1 11.872% 68.463% 2 17.360% 4 45.920% 3 34.920% 3 34.920% 71.071% 2 17.360% 3 34.920% 3 34.920% 2 17.360% 71.075% 2 17.360% 3 34.920% 2 17.360% 3 34.920% 71.075% 1 11.872% 7 53.200% 1 11.872% 1 11.872% 67.968% 1 11.872% 6 50.344% 2 17.360% 1 11.872% 68.129%Theorem 1.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and N=∑i=1Mni, ni≥1. When N is constant, then max⁡E(M,n(M)) is stochastically increasing as M increases; that is,(42)max⁡EM-1,nM≤max⁡EM,nM≤max⁡EM+1,nM.Proof of Theorem1 is similar to Lemma  1 in [8]. Now, we continue to compute DE of M=5, N=10 and M=6, N=10. Tables 4 and 5 are the DE of five-layer and six-layer defense, respectively. Another useful rule is that the number of target channels deployed along the ith layer should be not less than (i+1)th layer; this rule is summarized in Theorem 2.Table 4
DE of five-layer defense.
1st number 1st DE 2nd number 2nd DE 3rd number 3rd DE 4th number 4th DE 5th number 5th DE Total DE 5 45.5% 2 23.4% 1 12.0% 1 12.0% 1 12.0% 71.5% 4 38.6% 3 30.8% 1 12.0% 1 12.0% 1 12.0% 71.0% 4 38.6% 2 23.4% 2 23.4% 1 12.0% 1 12.0% 72.0% 3 30.8% 3 30.8% 2 23.4% 1 12.0% 1 12.0% 71.6% 3 30.8 % 3 30.8 % 2 23.4 % 1 12.0 % 1 12.0 % 74.6 % 2 23.4% 2 23.4% 2 23.4% 2 23.4% 2 23.4% 73.5%Table 5
DE of six-layer defense.
1st number 1st DE 2nd number 2nd DE 3rd number 3rd DE 4th number 4th DE 5th number 5th DE 6th number 6th DE Total DE 5 45% 1 14% 1 14% 1 14% 1 14% 1 14% 74% 4 39% 2 26% 1 14% 1 14% 1 14% 1 14% 75% 3 41 % 2 26 % 2 26 % 1 14 % 1 14 % 1 14 % 79 % 2 26% 3 41% 2 26% 1 14% 1 14% 1 14% 79% 2 26% 2 26% 2 26% 2 26% 1 14% 1 14% 77%Theorem 2.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and N=∑i=1Mni, ni≥1. For allocation plan n(M)∗, the number of target channels deployed along the ith layer is decreasing as i increases; that is, for all i, ni≥ni+1. When N is constant, then one has E(M,n(M)∗)≥E(M,n(M)).Proof.
Suppose we have a plann(M)Θ=(n1,n2,…,ni∗,ni+1∗,…,nM), where ni∗=ni+1, ni+1∗=ni. In order to prove Theorem 2, we only need to prove E(M,n(M)∗)≥E(M,n(M)Θ), which can be simplified by proving E(2,n(2)∗)≥E(2,n(2)Θ), where n(2)∗=(ni,ni+1), n(2)Θ=(ni∗,ni+1∗). For function(43)fρ,n=Plost=ρnn!∑k=0nρkk!-1,when ρ is constant, f(ρ,n) decreases with the increase of n, and when n is constant, f(ρ,n) increases with the increase of ρ [2, 20]. For proof simplified by letting Pk=1 for all defense weapons, then f(ρ,ni), f(ρ,ni+1), f(ρ,ni∗), and f(ρ,ni+1∗) are(44)fρ,ni=ρnini!∑k=0niρkk!-1,fρ,ni+1=ρ∗ni+1ni+1!∑k=0ni+1ρ∗kk!-1,fρ,ni∗=ρni+1ni+1!∑k=0ni+1ρkk!-1,fρ,ni+1∗=ρΘnini!∑k=0niρΘkk!-1,respectively, where ρ∗=f(ρ,ni)·ρ, ρΘ=f(ρ,ni∗)·ρ. DE of the two-layer defense are(45)E2,n2∗=Pk·1-fρ,ni+fρ,ni·1-fρ,ni+1=1-fρ,nifρ,ni+1,E2,n2Θ=Pk·1-fρ,ni∗+fρ,ni∗·1-fρ,ni+1∗=1-fρ,ni∗fρ,ni+1∗.Since the expansion of ex is 1+x+x2/2!+⋯+xn/n!, then the approximate representations of f(ρ,ni), f(ρ,ni+1), f(ρ,ni∗), and f(ρ,ni+1∗) are(46)fρ,ni≈ρnini!eρ-1,fρ,ni+1≈ρ∗ni+1ni+1!eρ∗-1,fρ,ni∗≈ρni+1ni+1!eρ-1,fρ,ni+1∗=ρΘnini!eρΘ-1,respectively. Because ni≥ni+1, we have (47)fρ,ni∗≥fρ,ni,fρ,ni∗≥fρ,ni+1,fρ,ni≥fρ,ni+1∗.In order to prove Theorem 2, we need to prove(48)1-fρ,nifρ,ni+1≥1-fρ,ni∗fρ,ni+1∗⟸fρ,ni∗fρ,ni+1∗-fρ,nifρ,ni+1≥0⟸fρ,ni∗fρ,ni+1∗fρ,nifρ,ni+1≥1,fρ,ni∗fρ,ni+1∗fρ,nifρ,ni+1=ρni+1/ni+1!eρ-1ρΘni/ni!eρΘ-1ρni/ni!eρ-1ρ∗ni+1/ni+1!eρ∗-1=ρni+1ρΘnieρ∗ρniρ∗ni+1eρΘ=ρni+1ρΘniefρ,ni·ρρniρ∗ni+1efρ,ni∗·ρ=ρni+1fρ,ni∗·ρniefρ,ni·ρρnifρ,ni·ρni+1efρ,ni∗·ρ=fρ,ni∗niefρ,ni·ρfρ,nini+1efρ,ni∗·ρ=ni!ni+1eni+1·ρefρ,ni·ρni+1!nieni·ρefρ,ni∗·ρ=ni!ni+1ni+1!nieni+1+fρ,ni-ni-fρ,ni∗·ρ.When ni=ni+1, Theorem 2 is true. Obviously, when ni>ni+1, we have ni!ni+1≫ni+1!ni; then, E(2,n(2)∗)>E(2,n(2)Θ), and the result of Theorem 2 follows.


### 4.1.2. Minimum Requirements of Target Channels

The requirements of target channels necessary to achieve a demanded DE can be viewed as a dual problem of (41). Assuming that the DE is held at greater than E∗, we define the optimization problem so as to minimize the requirements of target channels, subject to a given set of constraints: (49)minn=∑i=1Mnis.t.ρi=1-Ei-11-Ei-2⋯1-E1λ·twmean·Pf=∏x=1i-11-Ei·ρ1=1-Ei-1·ρi-1,Pshooti=1-ρinini!∑k=0niρikk!-1,E∗≤E1+1-E1E2+⋯+∏i=1M-11-EiEM,Ei=Pshooti·Pk.For the nonlinear optimization problem (49), when problem size is small, we can use algebra, dynamic programming, or enumeration method to solve it. When the size of the problem is very large, an approximate solution can be obtained by using some advanced algorithms, for example, genetic and heuristic algorithms. Then, we give the definition of neighborhood [21].Definition 3.
Letn(M)=(n1,n2,…,nM)∈Ω be an allocation plan, and Ω is the feasible region of allocation plans. Suppose that n(M)∗=(n1,n2,…,ni-1,ni-1,ni+1,…,nM)∈Ω, where the number of target channels deployed along the ith layer is ni-1. Then, one says that n(M)∗ is in the neighborhood of n(M); that is,(50)n1,n2,…,ni-1,ni-1,ni+1,…,nM∈φn1,n2,…,nM.Scenario 2.
Suppose the SSKP is 0.7, the probability that BMs will be detected by radars is 0.8,λ=5BMs/min, tsmean=0.75min, and E∗=65%; then the question becomes as follows: “What is the least cost of target channels to achieve a demanded DE?”
Figure4 is the schematic of search in the neighborhood of n(3)=(5,3,2) and n(4)=(4,3,2,1). We can see the least cost is 9 channels for three-layer defense and 8 channels for four-layer defense, and allocation plans are n(3)=(4,3,2) and n(4)=(3,3,1,1), respectively. So, we give a simple algorithm in finding the least cost of DWs to achieve a demanded DE; specific search methods are as follows. A feasible initial allocation plan is very important in this algorithm. Theorem 1 proposes the basic rule of finding an initial plan, which greatly simplifies the searching process.Figure 4
Search in the neighborhood ofn(3)=(5,3,2) and n(4)=(4,3,2,1).Step 1.
Suppose the number of layers isM; the demanded DE is E∗; and give an initial allocation plan n(M)=(n1,n2,…,nM)∈Ω, where E(M,n(M))≥E∗.Step 2.
Search in the neighborhood ofn(M), φn(M), and try to find an allocation plan φ1n(M) satisfying the following: (51)EM,φ1nM=max⁡EM,φnM.Step 3.
IfEM,φ1n(M)≥E∗,then n(M)=φ1n(M); go to Step 2 and repeat.Step 4.
IfEM,φ1n(M)<E∗,go to Step 5.Step 5.
Output the allocation plann(M); ∑i=1Mni is the least cost to achieve the demanded DE.


## 4.1.1. Defense Effectiveness

LetM be the number of layers; the number of target channels deployed along the ith layer is denoted by ni; the probability of BMs that will be shot by the M-layer defense with short defense depth is (36)PshootM=1-ρNN!∑k=0Nρkk!-1,where N is the total number of target channels, N=∑i=1Mni. The probabilities of BMs that will be shot by the M layer defense with short defense depth are as follows:1st layer:(37)ρ1=ρf1=λ·twmean·Pf,Pshoot1=1-ρ1n1n1!∑k=0n1ρ1kk!-1,E1=Pshoot1·Pk.2nd layer:(38)ρ2=1-E1·λ·twmean·Pf=1-E1·ρ1,Pshoot2=1-ρ2n2n2!∑k=0n2ρ2kk!-1,E2=Pshoot2·Pk.Mth layer: (39)ρM=1-EM-11-EM-2⋯1-E1λ·twmean·Pf=∏i=1M-11-Ei·ρ1=1-EM-1·ρM-1,PshootM=1-ρMnMnM!∑k=0nMρMkk!-1,EM=PshootM·Pk.DE of the whole M-layer BMDS is(40)E=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEM=E1+∑k=2MEk·∏i=1M-11-Ei.Then, we define the optimization problem as finding numbers of layers and target channels deployed along the ith layer so as to maximize DE, subject to a given set of constraints:(41)maxE=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEMs.t.ρi=1-Ei-11-Ei-2⋯1-E1λ·twmean·Pf=∏x=1i-11-Ei·ρ1=1-Ei-1·ρi-1,Pshooti=1-ρinini!∑k=0niρikk!-1,n=∑i=1Mni,Ei=Pshooti·Pk.For the nonlinear optimization problem (41), when problem size is small, we can use algebra, dynamic programming, or enumeration method to solve it. When the size of the problem is very large, an approximate solution can be obtained by using some advanced algorithms, for example, genetic, neural network, and heuristic algorithms. In order to get some potential and useful allocation rules, we analyze a scenario.Scenario 1.
Suppose that the number of target channels is 10, SSKP is 0.7, the probability that BMs will be detected by radars is 0.8,λ=5BMs/min, and tsmean=0.75min.
Tables1, 2, and 3 are the DE of two-layer, three-layer, and four-layer defense, respectively.
Letn(M)=(n1,n2,…,nM) be the allocation plan of target channels, where M is the number of layers and ni is the number of target channels deployed along the ith layer. From the results of Tables 1, 2, and 3, we noticed that plans n(2)=(6,4), n(3)=(5,3,2), and n(4)=(4,3,2,1) have the biggest DE. We also found that n(2)=(6,4)<n(3)=(5,3,2)<n(4)=(4,3,2,1), and then we have Theorem 1.Table 1
DE of two-layer defense.
1st-layer channel number 1st-layer DE 2nd-layer channel number 2nd-layer DE Total DE 9 56.00% 1 14.56% 62.41% 8 56.00% 2 25.20% 67.09% 7 54.32% 3 32.48% 69.16% 6 52.08 % 4 41.44 % 70.94 % 5 47.04% 5 44.24% 70.47% 4 35.84% 6 44.80% 64.58% 3 34.16% 7 53.76% 69.56% 2 27.44% 8 54.88% 67.26% 1 11.20% 9 56.00% 60.93%Table 2
DE of three-layer defense.
1st-channel number 1st-layer DE 2nd-channel number 2nd-layer DE 3rd-channel number 3rd-layer DE Total DE 8 55.552% 1 11.872% 1 11.872% 65.479% 7 53.256% 2 22.176% 1 11.872% 67.941% 7 53.256% 1 11.872% 2 22.176% 67.941% 6 50.344% 2 22.176% 2 22.176% 69.925% 6 50.344% 3 32.368% 1 11.872% 70.404% 6 50.344% 1 11.872% 3 32.368% 70.404% 5 46.368 % 3 32.368 % 2 22.176 % 71.771 % 5 46.368% 2 22.176% 3 32.368% 71.771% 5 46.368% 4 38.264% 1 11.872% 70.821% 5 46.368% 1 11.872% 4 38.264% 70.821% 4 38.264% 3 32.368% 3 32.368% 71.761% 4 38.264% 4 38.264% 2 22.176% 70.339% 4 38.264% 2 22.176% 4 38.264% 70.339% 4 38.264% 5 46.368% 1 11.872% 70.821% 4 38.264% 1 11.872% 5 46.368% 70.821% 3 32.368% 4 38.264% 3 32.368% 71.761% 3 32.368% 3 32.368% 4 38.264% 71.761% 3 32.368% 5 46.368% 2 22.176% 71.771% 3 32.368% 2 22.176% 5 46.368% 71.771% 3 32.368% 6 50.344% 1 11.872% 70.404% 3 32.368% 1 11.872% 6 50.344% 70.404% 2 22.176% 4 38.264% 4 38.264% 70.339% 2 22.176% 5 46.368% 3 32.368% 71.771% 2 22.176% 3 32.368% 5 46.368% 71.771% 2 22.176% 6 50.344% 2 22.176% 69.925% 2 22.176% 2 22.176% 6 50.344% 69.925% 2 22.176% 7 53.256% 1 11.872% 67.941% 2 22.176% 1 11.872% 7 53.256% 67.941% 1 11.872% 5 46.368% 4 38.264% 70.821% 1 11.872% 4 38.264% 5 46.368% 70.821% 1 11.872% 6 50.344% 3 32.368% 70.404% 1 11.872% 3 32.368% 6 50.344% 70.404% 1 11.872% 7 53.256% 2 22.176% 67.941% 1 11.872% 2 22.176% 7 53.256% 67.941% 1 11.872% 8 55.552% 1 11.872% 65.479% 1 11.872% 1 11.872% 8 55.552% 65.479%Table 3
DE of four-layer defense.
1st-channel number 1st-layer DE 2nd-channel number 2nd-layer DE 3rd-channel number 3rd-layer DE 4th-channel number 4th-layer DE Total DE 7 53.200% 1 11.872% 1 11.872% 1 11.872% 67.968% 6 50.344% 2 17.360% 1 11.872% 1 11.872% 68.129% 6 50.344% 1 11.872% 2 17.360% 1 11.872% 68.129% 5 47.600% 3 34.920% 1 11.872% 1 11.872% 73.515% 5 47.600% 2 17.360% 2 17.360% 1 11.872% 68.463% 5 47.600% 1 11.872% 3 34.920% 1 11.872% 73.515% 4 45.920 % 3 34.920 % 2 17.360 % 1 11.872 % 74.368 % 4 45.920% 3 34.920% 1 11.872% 2 17.360% 74.368% 4 45.920% 2 17.360% 1 11.872% 3 34.920% 74.368% 4 45.920% 2 17.360% 3 34.920% 1 11.872% 74.368% 4 45.920% 1 11.872% 2 17.360% 3 34.920% 74.368% 4 45.920% 1 11.872% 3 34.920% 2 17.360% 74.368% 3 34.920% 5 47.600% 1 11.872% 1 11.872% 73.515% 3 34.920% 4 45.920% 2 17.360% 1 11.872% 74.368% 3 34.920% 4 45.920% 1 11.872% 2 17.360% 74.368% 3 34.920% 3 34.920% 2 17.360% 2 17.360% 71.075% 3 34.920% 2 17.360% 3 34.920% 2 17.360% 71.075% 3 34.920% 2 17.360% 2 17.360% 3 34.920% 71.075% 3 34.920% 2 17.360% 4 45.920% 1 11.872% 74.368% 3 34.920% 2 17.360% 1 11.872% 4 45.920% 74.368% 3 34.920% 1 11.872% 4 45.920% 2 17.360% 74.368% 3 34.920% 1 11.872% 2 17.360% 4 45.920% 74.368% 3 34.920% 1 11.872% 5 47.600% 1 11.872% 73.515% 2 17.360% 6 50.344% 1 11.872% 1 11.872% 68.129% 2 17.360% 5 47.600% 2 17.360% 1 11.872% 68.463% 2 17.360% 4 45.920% 3 34.920% 3 34.920% 71.071% 2 17.360% 3 34.920% 3 34.920% 2 17.360% 71.075% 2 17.360% 3 34.920% 2 17.360% 3 34.920% 71.075% 1 11.872% 7 53.200% 1 11.872% 1 11.872% 67.968% 1 11.872% 6 50.344% 2 17.360% 1 11.872% 68.129%Theorem 1.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and N=∑i=1Mni, ni≥1. When N is constant, then max⁡E(M,n(M)) is stochastically increasing as M increases; that is,(42)max⁡EM-1,nM≤max⁡EM,nM≤max⁡EM+1,nM.Proof of Theorem1 is similar to Lemma  1 in [8]. Now, we continue to compute DE of M=5, N=10 and M=6, N=10. Tables 4 and 5 are the DE of five-layer and six-layer defense, respectively. Another useful rule is that the number of target channels deployed along the ith layer should be not less than (i+1)th layer; this rule is summarized in Theorem 2.Table 4
DE of five-layer defense.
1st number 1st DE 2nd number 2nd DE 3rd number 3rd DE 4th number 4th DE 5th number 5th DE Total DE 5 45.5% 2 23.4% 1 12.0% 1 12.0% 1 12.0% 71.5% 4 38.6% 3 30.8% 1 12.0% 1 12.0% 1 12.0% 71.0% 4 38.6% 2 23.4% 2 23.4% 1 12.0% 1 12.0% 72.0% 3 30.8% 3 30.8% 2 23.4% 1 12.0% 1 12.0% 71.6% 3 30.8 % 3 30.8 % 2 23.4 % 1 12.0 % 1 12.0 % 74.6 % 2 23.4% 2 23.4% 2 23.4% 2 23.4% 2 23.4% 73.5%Table 5
DE of six-layer defense.
1st number 1st DE 2nd number 2nd DE 3rd number 3rd DE 4th number 4th DE 5th number 5th DE 6th number 6th DE Total DE 5 45% 1 14% 1 14% 1 14% 1 14% 1 14% 74% 4 39% 2 26% 1 14% 1 14% 1 14% 1 14% 75% 3 41 % 2 26 % 2 26 % 1 14 % 1 14 % 1 14 % 79 % 2 26% 3 41% 2 26% 1 14% 1 14% 1 14% 79% 2 26% 2 26% 2 26% 2 26% 1 14% 1 14% 77%Theorem 2.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and N=∑i=1Mni, ni≥1. For allocation plan n(M)∗, the number of target channels deployed along the ith layer is decreasing as i increases; that is, for all i, ni≥ni+1. When N is constant, then one has E(M,n(M)∗)≥E(M,n(M)).Proof.
Suppose we have a plann(M)Θ=(n1,n2,…,ni∗,ni+1∗,…,nM), where ni∗=ni+1, ni+1∗=ni. In order to prove Theorem 2, we only need to prove E(M,n(M)∗)≥E(M,n(M)Θ), which can be simplified by proving E(2,n(2)∗)≥E(2,n(2)Θ), where n(2)∗=(ni,ni+1), n(2)Θ=(ni∗,ni+1∗). For function(43)fρ,n=Plost=ρnn!∑k=0nρkk!-1,when ρ is constant, f(ρ,n) decreases with the increase of n, and when n is constant, f(ρ,n) increases with the increase of ρ [2, 20]. For proof simplified by letting Pk=1 for all defense weapons, then f(ρ,ni), f(ρ,ni+1), f(ρ,ni∗), and f(ρ,ni+1∗) are(44)fρ,ni=ρnini!∑k=0niρkk!-1,fρ,ni+1=ρ∗ni+1ni+1!∑k=0ni+1ρ∗kk!-1,fρ,ni∗=ρni+1ni+1!∑k=0ni+1ρkk!-1,fρ,ni+1∗=ρΘnini!∑k=0niρΘkk!-1,respectively, where ρ∗=f(ρ,ni)·ρ, ρΘ=f(ρ,ni∗)·ρ. DE of the two-layer defense are(45)E2,n2∗=Pk·1-fρ,ni+fρ,ni·1-fρ,ni+1=1-fρ,nifρ,ni+1,E2,n2Θ=Pk·1-fρ,ni∗+fρ,ni∗·1-fρ,ni+1∗=1-fρ,ni∗fρ,ni+1∗.Since the expansion of ex is 1+x+x2/2!+⋯+xn/n!, then the approximate representations of f(ρ,ni), f(ρ,ni+1), f(ρ,ni∗), and f(ρ,ni+1∗) are(46)fρ,ni≈ρnini!eρ-1,fρ,ni+1≈ρ∗ni+1ni+1!eρ∗-1,fρ,ni∗≈ρni+1ni+1!eρ-1,fρ,ni+1∗=ρΘnini!eρΘ-1,respectively. Because ni≥ni+1, we have (47)fρ,ni∗≥fρ,ni,fρ,ni∗≥fρ,ni+1,fρ,ni≥fρ,ni+1∗.In order to prove Theorem 2, we need to prove(48)1-fρ,nifρ,ni+1≥1-fρ,ni∗fρ,ni+1∗⟸fρ,ni∗fρ,ni+1∗-fρ,nifρ,ni+1≥0⟸fρ,ni∗fρ,ni+1∗fρ,nifρ,ni+1≥1,fρ,ni∗fρ,ni+1∗fρ,nifρ,ni+1=ρni+1/ni+1!eρ-1ρΘni/ni!eρΘ-1ρni/ni!eρ-1ρ∗ni+1/ni+1!eρ∗-1=ρni+1ρΘnieρ∗ρniρ∗ni+1eρΘ=ρni+1ρΘniefρ,ni·ρρniρ∗ni+1efρ,ni∗·ρ=ρni+1fρ,ni∗·ρniefρ,ni·ρρnifρ,ni·ρni+1efρ,ni∗·ρ=fρ,ni∗niefρ,ni·ρfρ,nini+1efρ,ni∗·ρ=ni!ni+1eni+1·ρefρ,ni·ρni+1!nieni·ρefρ,ni∗·ρ=ni!ni+1ni+1!nieni+1+fρ,ni-ni-fρ,ni∗·ρ.When ni=ni+1, Theorem 2 is true. Obviously, when ni>ni+1, we have ni!ni+1≫ni+1!ni; then, E(2,n(2)∗)>E(2,n(2)Θ), and the result of Theorem 2 follows.


## 4.1.2. Minimum Requirements of Target Channels

The requirements of target channels necessary to achieve a demanded DE can be viewed as a dual problem of (41). Assuming that the DE is held at greater than E∗, we define the optimization problem so as to minimize the requirements of target channels, subject to a given set of constraints: (49)minn=∑i=1Mnis.t.ρi=1-Ei-11-Ei-2⋯1-E1λ·twmean·Pf=∏x=1i-11-Ei·ρ1=1-Ei-1·ρi-1,Pshooti=1-ρinini!∑k=0niρikk!-1,E∗≤E1+1-E1E2+⋯+∏i=1M-11-EiEM,Ei=Pshooti·Pk.For the nonlinear optimization problem (49), when problem size is small, we can use algebra, dynamic programming, or enumeration method to solve it. When the size of the problem is very large, an approximate solution can be obtained by using some advanced algorithms, for example, genetic and heuristic algorithms. Then, we give the definition of neighborhood [21].Definition 3.
Letn(M)=(n1,n2,…,nM)∈Ω be an allocation plan, and Ω is the feasible region of allocation plans. Suppose that n(M)∗=(n1,n2,…,ni-1,ni-1,ni+1,…,nM)∈Ω, where the number of target channels deployed along the ith layer is ni-1. Then, one says that n(M)∗ is in the neighborhood of n(M); that is,(50)n1,n2,…,ni-1,ni-1,ni+1,…,nM∈φn1,n2,…,nM.Scenario 2.
Suppose the SSKP is 0.7, the probability that BMs will be detected by radars is 0.8,λ=5BMs/min, tsmean=0.75min, and E∗=65%; then the question becomes as follows: “What is the least cost of target channels to achieve a demanded DE?”
Figure4 is the schematic of search in the neighborhood of n(3)=(5,3,2) and n(4)=(4,3,2,1). We can see the least cost is 9 channels for three-layer defense and 8 channels for four-layer defense, and allocation plans are n(3)=(4,3,2) and n(4)=(3,3,1,1), respectively. So, we give a simple algorithm in finding the least cost of DWs to achieve a demanded DE; specific search methods are as follows. A feasible initial allocation plan is very important in this algorithm. Theorem 1 proposes the basic rule of finding an initial plan, which greatly simplifies the searching process.Figure 4
Search in the neighborhood ofn(3)=(5,3,2) and n(4)=(4,3,2,1).Step 1.
Suppose the number of layers isM; the demanded DE is E∗; and give an initial allocation plan n(M)=(n1,n2,…,nM)∈Ω, where E(M,n(M))≥E∗.Step 2.
Search in the neighborhood ofn(M), φn(M), and try to find an allocation plan φ1n(M) satisfying the following: (51)EM,φ1nM=max⁡EM,φnM.Step 3.
IfEM,φ1n(M)≥E∗,then n(M)=φ1n(M); go to Step 2 and repeat.Step 4.
IfEM,φ1n(M)<E∗,go to Step 5.Step 5.
Output the allocation plann(M); ∑i=1Mni is the least cost to achieve the demanded DE.


## 4.2. Different Defense Weapons and M/M/N/N System

### 4.2.1. Different Defense Weapons and Identical SSKPs

In this section, we consider different types of defense weapons. LetM be the number of layers; the number of target channels deployed along the ith layer will be denoted by ni, assuming that the total number of types is M and defense weapons are identical in the same layer. As in Section 3.3, subscript (·)(i) indicates defense weapons of type i; then probabilities that BMs will be shot by the M-layer defense with short defense depth are as follows:1st layer:(52)ρ11=λ·twmean1·Pf1,Pshoot1=1-ρ11n1n1!∑k=0n1ρ11kk!-1,E1=Pshoot1·Pk1.2nd layer:(53)ρ22=1-E1·λ·twmean2·Pf2,Pshoot2=1-ρ22n2n2!∑k=0n2ρ22kk!-1,E2=Pshoot2·Pk2.Mth layer: (54)ρMM=1-EM-11-EM-2⋯1-E1λ·twmeanM·PfM=∏i=1M-11-Ei·λ·twmeanM·PfM,PshootM=1-ρMMnMnM!∑k=0nMρMMkk!-1,EM=PshootM·PkM.DE of the wholeM-layer BMDS is(55)E=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEM=E1+∑k=2MEk·∏i=1M-11-Ei.For certain types and numbers of target channels, we define the optimization problem as finding which type of defense weapon should be deployed on each layer so as to maximize the ED, subject to a given set of constraints:(56)maxE=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEMs.t.ρiM=1-Ei-11-Ei-2⋯1-E1λ·twmeanM·PfM=∏x=1i-11-Ei·λ·twmeanM·PfM,Pshooti=1-ρiMnini!∑k=0niρiMkk!-1,Ei=Pshooti·PkM,n=∑i=1Mni.In order to get some potential and useful allocation rules, we analyze a scenario.Scenario 3.
Suppose that the number of layers is 3, the number of defense weapon types is 2 (types I and II),Pk(I)=Pk(II)=0.7, the probability that BMs will be detected by radars is 0.8, λ=5BMs/min, tsmean(I)=0.75min, and tsmean(II)=1min.
Table6 is the DE of three-layer defense.
It can be seen from Table6 that plan n(3)=(I,I,II) has the biggest DE. We also found that n(4)=(I,I,I,II) is the best, and then we have Theorem 4.Table 6
DE of three-layer defense.
1st-layer type 1st-layer DE 2nd-layer type 2nd-layer DE 3rd-layer type 3rd-layer DE Total DE II 43.200% II 45.440% I 55.520% 86.216% II 48.960% I 51.840% II 46.560% 86.864% II 43.200% I 55.040% I 55.040% 88.518% I 54.560% II 45.280% II 46.720% 86.752% I 56.800 % I 55.360 % II 43.040 % 89.016 %Theorem 4.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and let n1=n2=⋯=nM. Allocation plan n(M)I=(I,II) indicates that defense weapons of type I are forward-deployed, and n(M)II=(II,I) indicates that defense weapons of type II are forward-deployed. Suppose that defense weapons are identical in the same layer, and Pk(I)=Pk(II) and Pf(I)=Pf(II); if tsmean(I)≤tsmean(II), then one has E(M,n(M)I)≥E(M,n(M)II).Proof.
Suppose that we have an allocation plann(M)IΘ=(IΘ,IIΘ), where tsmean(IΘ)=tsmean(II) and tsmean(IIΘ)=tsmean(I). In order to prove Theorem 4, we only need to prove E(M,n(M)I)≥E(M,n(M)IΘ). For function(57)fρ,n=Plost=ρnn!∑k=0nρkk!-1,when ρ is constant, f(ρ,n) decreases with the increase of n, and when n is constant, f(ρ,n) increases with the increase of ρ [2, 20]. Then, f(ρ,I), f(ρ,II), f(ρ,IΘ), and f(ρ,IIΘ) are(58)fρ,I=ρInn!∑k=0nρIkk!-1,fρ,II=ρIInn!∑k=0nρIIkk!-1,fρ,IΘ=ρIΘnn!∑k=0nρIΘkk!-1,fρ,IIΘ=ρIIΘnn!∑k=0nρIIΘkk!-1,where(59)ρI=λ·tsmeanI·PfI,ρIΘ=λ·tsmeanIΘ·PfIΘ,ρII=1-EIλtsmeanIIPfII,ρIIΘ=1-EIΘλtsmeanIIΘPfIΘ,EI=1-fρ,I,EIΘ=1-fρ,IΘ.DE of the defense are(60)EM,nMI=Pk·1-fρ,I+fρ,I·1-fρ,II=1-fρ,Ifρ,II,EM,nMIΘ=Pk·1-fρ,IΘ+fρ,IΘ·1-fρ,IIΘ=1-fρ,IΘfρ,IIΘ.Since the expansion of ex is 1+x+x2/2!+⋯+xn/n!, then the approximate representations of f(ρ,I), f(ρ,II), f(ρ,IΘ), and f(ρ,IIΘ) are(61)fρ,I≈ρInn!eρI-1,fρ,II≈ρIIni+1n!eρII-1,fρ,IΘ≈ρIΘnn!eρIΘ-1,fρ,IIΘ=ρIIΘnn!eρIIΘ-1.In order to prove Theorem 4, we need to prove(62)1-fρ,Ifρ,II≥1-fρ,IΘfρ,IIΘ⟸fρ,IΘfρ,IIΘ-fρ,Ifρ,II≥0⟸fρ,IΘfρ,IIΘfρ,Ifρ,II≥1,fρ,IΘfρ,IIΘfρ,Ifρ,II=ρIΘn/n!eρIΘ-1ρIIΘn/n!eρIIΘ-1ρIn/n!eρI-1ρIIni/n!eρII-1=ρIΘnρIIΘnρInρIIneρI+ρII-ρIΘ-ρIIΘ=ρIΘρIn2en+1ρI+ρII-n+1ρIΘ-ρIIΘ=λtsmeanIIPfIIλtsmeanIPfIn2eλn+1tsmeanIPfI+1-EItsmeanIIPfII-n+1tsmeanIIPfII-1-EIItsmeanIPfI.Let tsmean(II)Pf(II)=A and tsmean(I)Pf(I)=B; then,(63)fρ,IΘfρ,IIΘfρ,Ifρ,II=ABn2eλn+1B+1-EIA-n+1A-1-EIIB.Since Pf(II)=Pf(I), then A>B, and the result of Theorem 4 follows.The termsA and B that appear in the proof of Theorem 4 could be a factor in channels allocation; we extended Theorem 4 to obtain Lemma 5.Lemma 5.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and let n1=n2=⋯=nM. Allocation plan n(M)I=(I,II) indicates that defense weapons of type I are forward-deployed, and n(M)II=(II,I) indicates that defense weapons of type II are forward-deployed. Suppose that defense weapons are identical in the same layer, and Pk(I)=Pk(II); if Pf(I)tsmean(I)≤Pf(II)tsmean(II), then one has E(M,n(M)I)≥E(M,n(M)II).


### 4.2.2. Different Defense Weapons and Different SSKPs

In this section, we consider different defense weapons and different SSKPs of layered BMDS. It is obvious that higher SSKP dose improves the DE. How SSKP affects the target channels allocation is a complex problem; however, we can summarize some useful rules; see Scenario4.Scenario 4.
Suppose that the number of layers is 3, the number of defense weapon types is 2 (types I and II), the probability that BMs will be detected by radars is 0.8,λ=5BMs/min, tsmean(I)=tsmean(II)=0.75min, Pk(I)=0.8, and Pk(II)=0.6.
Table7 is the DE of three-layer defense. It can be seen from Table 7 that plan n(3)=(I,I,II) has the biggest DE. We also found that n(4)=(I,I,I,II) is the best, and then we have Theorem 6.Table 7
DE of three-layer defense.
1st-layer type 1st-layer DE 2nd-layer type 2nd-layer DE 3rd-layer type 3rd-layer DE Total DE II 36.720% II 41.400% I 0.594 83.930% II 42.120% I 59.040% II 0.400 84.766% II 42.600% I 55.840% I 0.506 87.468% I 54.240% II 39.600% II 0.415 85.837% I 56.640 % I 54.400 % II 0.426 88.651 %Theorem 6.
LetE(M,n(M)) be the DE of the M-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and let n1=n2=⋯=nM. Allocation plan n(M)I=(I,II) indicates that defense weapons of type I are forward-deployed, and n(M)II=(II,I) indicates that defense weapons of type II are forward-deployed. Suppose that defense weapons are identical in the same layer, and tsmean(I)=tsmean(II) and Pf(I)=Pf(II); if Pk(I)>Pk(II), then one has E(M,n(M)I)≥E(M,n(M)II).Proof.
Suppose that we have an allocation plann(M)IΘ=(IΘ,IIΘ), where Pk(IΘ)=Pk(II) and Pk(IIΘ)=Pk(I). In order to prove Theorem 4, we only need to prove E(M,n(M)I)≥E(M,n(M)IΘ).
f ( ρ , I ), f(ρ,II), f(ρ,IΘ), and f(ρ,IIΘ) are(64)fρ,I=ρInn!∑k=0nρIkk!-1,fρ,II=ρIInn!∑k=0nρIIkk!-1,fρ,IΘ=ρIΘnn!∑k=0nρIΘkk!-1,fρ,IIΘ=ρIIΘnn!∑k=0nρIIΘkk!-1,where(65)ρI=λ·tsmeanI·PfI,ρIΘ=λ·tsmeanIΘ·PfIΘ,ρII=1-EIλtsmeanIIPfII,ρIIΘ=1-EIΘλtsmeanIIΘPfIΘ,EI=1-fρ,I·PkI,EIΘ=1-fρ,IΘ·PkII.DE of the defense are(66)EM,nMI=PkI-fρ,IPkI+PkII-fρ,IIPkII-PkIPkII+fρ,IIPkIPkII+fρ,IPkIPkII-fρ,Ifρ,IIPkIPkII,EM,nMIΘ=PkIΘ-fρ,IΘPkIΘ+PkIIΘ-fρ,IIΘPkIIΘ-PkIΘPkIIΘ+fρ,IIΘPkIΘPkIIΘ+fρ,IΘPkIΘPkIIΘ-fρ,IΘfρ,IIΘPkIΘPkIIΘ.In order to prove Theorem 6, we need to prove(67)EM,nMI-EM,nMIΘ=PkIPkIIfρ,II+fρ,I-fρ,Ifρ,II-fρ,IIΘ-fρ,IΘ+fρ,IΘfρ,IIΘ-fρ,IPkI-fρ,IIPkII+fρ,IΘPkII+fρ,IIΘPkI≥0.Since Pk(I)>Pk(II), f(ρ,I)=f(ρ,IΘ), then ∗fρ,I=fρ,IΘ≥fρ,IIΘ≥fρ,II,PkIPkII1-fρ,Ifρ,IIΘ-fρ,II+fρ,IPkI-PkII+fρ,IIPkII-fρ,IIΘPkI≥0.Since the expansion of ex is 1+x+x2/2!+⋯+xn/n!, then the approximate representations of f(ρ,IΘ) and f(ρ,IIΘ) are(68)fρ,IΘ≈ρIΘnn!eρIΘ-1,fρ,IIΘ=ρIIΘnn!eρIIΘ-1,fρ,IIΘfρ,IΘ=1-1-fρ,IΘ·PkIΘneρIΘ·ρIIΘ.Let tsmean(II)Pf(II)=A and tsmean(I)Pf(I)=B; then,(69)fρ,IΘfρ,IIΘfρ,Ifρ,II=ABn2eλn+1B+1-EIA-n+1A-1-EIIB.Obviously, we have(70)1-1-fρ,IΘ·PkIΘneρIΘ·ρIIΘ⟶0.The result of Theorem 4 follows.


## 4.2.1. Different Defense Weapons and Identical SSKPs

In this section, we consider different types of defense weapons. LetM be the number of layers; the number of target channels deployed along the ith layer will be denoted by ni, assuming that the total number of types is M and defense weapons are identical in the same layer. As in Section 3.3, subscript (·)(i) indicates defense weapons of type i; then probabilities that BMs will be shot by the M-layer defense with short defense depth are as follows:1st layer:(52)ρ11=λ·twmean1·Pf1,Pshoot1=1-ρ11n1n1!∑k=0n1ρ11kk!-1,E1=Pshoot1·Pk1.2nd layer:(53)ρ22=1-E1·λ·twmean2·Pf2,Pshoot2=1-ρ22n2n2!∑k=0n2ρ22kk!-1,E2=Pshoot2·Pk2.Mth layer: (54)ρMM=1-EM-11-EM-2⋯1-E1λ·twmeanM·PfM=∏i=1M-11-Ei·λ·twmeanM·PfM,PshootM=1-ρMMnMnM!∑k=0nMρMMkk!-1,EM=PshootM·PkM.DE of the wholeM-layer BMDS is(55)E=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEM=E1+∑k=2MEk·∏i=1M-11-Ei.For certain types and numbers of target channels, we define the optimization problem as finding which type of defense weapon should be deployed on each layer so as to maximize the ED, subject to a given set of constraints:(56)maxE=E1+1-E1E2+1-E11-E2E3⋯+∏i=1M-11-EiEMs.t.ρiM=1-Ei-11-Ei-2⋯1-E1λ·twmeanM·PfM=∏x=1i-11-Ei·λ·twmeanM·PfM,Pshooti=1-ρiMnini!∑k=0niρiMkk!-1,Ei=Pshooti·PkM,n=∑i=1Mni.In order to get some potential and useful allocation rules, we analyze a scenario.Scenario 3.
Suppose that the number of layers is 3, the number of defense weapon types is 2 (types I and II),Pk(I)=Pk(II)=0.7, the probability that BMs will be detected by radars is 0.8, λ=5BMs/min, tsmean(I)=0.75min, and tsmean(II)=1min.
Table6 is the DE of three-layer defense.
It can be seen from Table6 that plan n(3)=(I,I,II) has the biggest DE. We also found that n(4)=(I,I,I,II) is the best, and then we have Theorem 4.Table 6
DE of three-layer defense.
1st-layer type 1st-layer DE 2nd-layer type 2nd-layer DE 3rd-layer type 3rd-layer DE Total DE II 43.200% II 45.440% I 55.520% 86.216% II 48.960% I 51.840% II 46.560% 86.864% II 43.200% I 55.040% I 55.040% 88.518% I 54.560% II 45.280% II 46.720% 86.752% I 56.800 % I 55.360 % II 43.040 % 89.016 %Theorem 4.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and let n1=n2=⋯=nM. Allocation plan n(M)I=(I,II) indicates that defense weapons of type I are forward-deployed, and n(M)II=(II,I) indicates that defense weapons of type II are forward-deployed. Suppose that defense weapons are identical in the same layer, and Pk(I)=Pk(II) and Pf(I)=Pf(II); if tsmean(I)≤tsmean(II), then one has E(M,n(M)I)≥E(M,n(M)II).Proof.
Suppose that we have an allocation plann(M)IΘ=(IΘ,IIΘ), where tsmean(IΘ)=tsmean(II) and tsmean(IIΘ)=tsmean(I). In order to prove Theorem 4, we only need to prove E(M,n(M)I)≥E(M,n(M)IΘ). For function(57)fρ,n=Plost=ρnn!∑k=0nρkk!-1,when ρ is constant, f(ρ,n) decreases with the increase of n, and when n is constant, f(ρ,n) increases with the increase of ρ [2, 20]. Then, f(ρ,I), f(ρ,II), f(ρ,IΘ), and f(ρ,IIΘ) are(58)fρ,I=ρInn!∑k=0nρIkk!-1,fρ,II=ρIInn!∑k=0nρIIkk!-1,fρ,IΘ=ρIΘnn!∑k=0nρIΘkk!-1,fρ,IIΘ=ρIIΘnn!∑k=0nρIIΘkk!-1,where(59)ρI=λ·tsmeanI·PfI,ρIΘ=λ·tsmeanIΘ·PfIΘ,ρII=1-EIλtsmeanIIPfII,ρIIΘ=1-EIΘλtsmeanIIΘPfIΘ,EI=1-fρ,I,EIΘ=1-fρ,IΘ.DE of the defense are(60)EM,nMI=Pk·1-fρ,I+fρ,I·1-fρ,II=1-fρ,Ifρ,II,EM,nMIΘ=Pk·1-fρ,IΘ+fρ,IΘ·1-fρ,IIΘ=1-fρ,IΘfρ,IIΘ.Since the expansion of ex is 1+x+x2/2!+⋯+xn/n!, then the approximate representations of f(ρ,I), f(ρ,II), f(ρ,IΘ), and f(ρ,IIΘ) are(61)fρ,I≈ρInn!eρI-1,fρ,II≈ρIIni+1n!eρII-1,fρ,IΘ≈ρIΘnn!eρIΘ-1,fρ,IIΘ=ρIIΘnn!eρIIΘ-1.In order to prove Theorem 4, we need to prove(62)1-fρ,Ifρ,II≥1-fρ,IΘfρ,IIΘ⟸fρ,IΘfρ,IIΘ-fρ,Ifρ,II≥0⟸fρ,IΘfρ,IIΘfρ,Ifρ,II≥1,fρ,IΘfρ,IIΘfρ,Ifρ,II=ρIΘn/n!eρIΘ-1ρIIΘn/n!eρIIΘ-1ρIn/n!eρI-1ρIIni/n!eρII-1=ρIΘnρIIΘnρInρIIneρI+ρII-ρIΘ-ρIIΘ=ρIΘρIn2en+1ρI+ρII-n+1ρIΘ-ρIIΘ=λtsmeanIIPfIIλtsmeanIPfIn2eλn+1tsmeanIPfI+1-EItsmeanIIPfII-n+1tsmeanIIPfII-1-EIItsmeanIPfI.Let tsmean(II)Pf(II)=A and tsmean(I)Pf(I)=B; then,(63)fρ,IΘfρ,IIΘfρ,Ifρ,II=ABn2eλn+1B+1-EIA-n+1A-1-EIIB.Since Pf(II)=Pf(I), then A>B, and the result of Theorem 4 follows.The termsA and B that appear in the proof of Theorem 4 could be a factor in channels allocation; we extended Theorem 4 to obtain Lemma 5.Lemma 5.
LetE(M,n(M)) be the DE of theM-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and let n1=n2=⋯=nM. Allocation plan n(M)I=(I,II) indicates that defense weapons of type I are forward-deployed, and n(M)II=(II,I) indicates that defense weapons of type II are forward-deployed. Suppose that defense weapons are identical in the same layer, and Pk(I)=Pk(II); if Pf(I)tsmean(I)≤Pf(II)tsmean(II), then one has E(M,n(M)I)≥E(M,n(M)II).


## 4.2.2. Different Defense Weapons and Different SSKPs

In this section, we consider different defense weapons and different SSKPs of layered BMDS. It is obvious that higher SSKP dose improves the DE. How SSKP affects the target channels allocation is a complex problem; however, we can summarize some useful rules; see Scenario4.Scenario 4.
Suppose that the number of layers is 3, the number of defense weapon types is 2 (types I and II), the probability that BMs will be detected by radars is 0.8,λ=5BMs/min, tsmean(I)=tsmean(II)=0.75min, Pk(I)=0.8, and Pk(II)=0.6.
Table7 is the DE of three-layer defense. It can be seen from Table 7 that plan n(3)=(I,I,II) has the biggest DE. We also found that n(4)=(I,I,I,II) is the best, and then we have Theorem 6.Table 7
DE of three-layer defense.
1st-layer type 1st-layer DE 2nd-layer type 2nd-layer DE 3rd-layer type 3rd-layer DE Total DE II 36.720% II 41.400% I 0.594 83.930% II 42.120% I 59.040% II 0.400 84.766% II 42.600% I 55.840% I 0.506 87.468% I 54.240% II 39.600% II 0.415 85.837% I 56.640 % I 54.400 % II 0.426 88.651 %Theorem 6.
LetE(M,n(M)) be the DE of the M-layer BMDS; n(M)=(n1,n2,…,nM) is the allocation plan, ni is the number of target channels deployed along the ith layer, and let n1=n2=⋯=nM. Allocation plan n(M)I=(I,II) indicates that defense weapons of type I are forward-deployed, and n(M)II=(II,I) indicates that defense weapons of type II are forward-deployed. Suppose that defense weapons are identical in the same layer, and tsmean(I)=tsmean(II) and Pf(I)=Pf(II); if Pk(I)>Pk(II), then one has E(M,n(M)I)≥E(M,n(M)II).Proof.
Suppose that we have an allocation plann(M)IΘ=(IΘ,IIΘ), where Pk(IΘ)=Pk(II) and Pk(IIΘ)=Pk(I). In order to prove Theorem 4, we only need to prove E(M,n(M)I)≥E(M,n(M)IΘ).
f ( ρ , I ), f(ρ,II), f(ρ,IΘ), and f(ρ,IIΘ) are(64)fρ,I=ρInn!∑k=0nρIkk!-1,fρ,II=ρIInn!∑k=0nρIIkk!-1,fρ,IΘ=ρIΘnn!∑k=0nρIΘkk!-1,fρ,IIΘ=ρIIΘnn!∑k=0nρIIΘkk!-1,where(65)ρI=λ·tsmeanI·PfI,ρIΘ=λ·tsmeanIΘ·PfIΘ,ρII=1-EIλtsmeanIIPfII,ρIIΘ=1-EIΘλtsmeanIIΘPfIΘ,EI=1-fρ,I·PkI,EIΘ=1-fρ,IΘ·PkII.DE of the defense are(66)EM,nMI=PkI-fρ,IPkI+PkII-fρ,IIPkII-PkIPkII+fρ,IIPkIPkII+fρ,IPkIPkII-fρ,Ifρ,IIPkIPkII,EM,nMIΘ=PkIΘ-fρ,IΘPkIΘ+PkIIΘ-fρ,IIΘPkIIΘ-PkIΘPkIIΘ+fρ,IIΘPkIΘPkIIΘ+fρ,IΘPkIΘPkIIΘ-fρ,IΘfρ,IIΘPkIΘPkIIΘ.In order to prove Theorem 6, we need to prove(67)EM,nMI-EM,nMIΘ=PkIPkIIfρ,II+fρ,I-fρ,Ifρ,II-fρ,IIΘ-fρ,IΘ+fρ,IΘfρ,IIΘ-fρ,IPkI-fρ,IIPkII+fρ,IΘPkII+fρ,IIΘPkI≥0.Since Pk(I)>Pk(II), f(ρ,I)=f(ρ,IΘ), then ∗fρ,I=fρ,IΘ≥fρ,IIΘ≥fρ,II,PkIPkII1-fρ,Ifρ,IIΘ-fρ,II+fρ,IPkI-PkII+fρ,IIPkII-fρ,IIΘPkI≥0.Since the expansion of ex is 1+x+x2/2!+⋯+xn/n!, then the approximate representations of f(ρ,IΘ) and f(ρ,IIΘ) are(68)fρ,IΘ≈ρIΘnn!eρIΘ-1,fρ,IIΘ=ρIIΘnn!eρIIΘ-1,fρ,IIΘfρ,IΘ=1-1-fρ,IΘ·PkIΘneρIΘ·ρIIΘ.Let tsmean(II)Pf(II)=A and tsmean(I)Pf(I)=B; then,(69)fρ,IΘfρ,IIΘfρ,Ifρ,II=ABn2eλn+1B+1-EIA-n+1A-1-EIIB.Obviously, we have(70)1-1-fρ,IΘ·PkIΘneρIΘ·ρIIΘ⟶0.The result of Theorem 4 follows.


## 5. Numerical Examples

In this section, we use numerical examples to generate some insights into the performance of the proposed queuing models in Section3. Firstly, using the formula in Section 3.1.2, we draw the relationship of the mean shooting time and intensity of BM arrivals (see Figure 5). Then, we consider using the formula in Section 3.1.1 and calculate the loss probability of the M/M/N/N system. We draw the relationship of the BM loss probability and offered density of shootings with N=1, 2, 3, 4, 5, 6, 8, 12, 16, 20, 30, 40, and 50 (see Figure 6). From Figures 5 and 6, we can see that the probability of BM loss increases with the increasing mean shooting time and intensity of BM arrivals.Figure 5
Performance parameters as functions of the mean shooting time and the intensity of BM arrivals (M/M/1/1 system).
(a) (b) (c)Figure 6
The probability of BM loss as functions of the number of target channels and density of shooting.We set two scenarios: scenario 1 (the number of total BMs is 30, the number of target channels is 3, SSKP is 0.7, the probability that BMs will be detected by radars is 0.8,λ=3 BMs/min, and tsmean=1min (μ=1)) and scenario 2 (the number of total BMs is 50, the number of target channels is 8, SSKP is 0.8, the probability that BMs will be detected by radars is 0.9, λ=5 BMs/min, and tsmean=1min (μ=1)). We will firstly simulate the M/M/N/N queueing system and use Matlab to get the figure of the performance of the two scenarios (see Figure 7). Then, we will secondly simulate the M/M/N/C queueing system and use Matlab to get the figure of the performance of the two scenarios (see Figure 8). Table 8 shows the results of operating parameters of the two queueing systems.Table 8
Results of operating parameters of the two queueing systems.
Parameters M/M/N/N M/M/N/C Scenario1 Scenario2 Scenario1 Scenario2 Mean occupancy rate of target channels 22.6% 10.6% 64.5% 1.7% Mean number of waiting BMs 0.000 0.000 1.847 0.000 Mean waiting time of BMs 0.000 0.000 0.801 0.000 Mean number of BMs in system 1.732 5.018 4.326 3.830 Number of penetrated BMs 7 3 0 0 Probability of BM penetration 23.3% 6.0% 0% 0% Total DE 42.95% 67.68% 56% 72%Figure 7
Figures of the performance of the M/M/N/N queueing system.
(a)
BM arrival and finish time of scenario 1 and scenario 2 (b)
BM waiting time and weapons shooting time of scenario 1 and scenario 2 (c)
Queue length of scenario 1 and scenario 2 (d)
Total number of BMs in queue and last BM that arrived of scenario 1 and scenario 2Figure 8
Figures of the performance of the M/M/N/C queueing system.
(a)
BM arrival time of scenario 1 and scenario 2 (b)
Shooting time for each BM of scenario 1 and scenario 2 (c)
Waiting time of each BM of scenario 1 and scenario 2 (d)
Queue length of scenario 1 and scenario 2 (e)
Finish time of shooting of scenario 1 and scenario 2In order to explore the changes in the relationship between the performance of systems and different factors, we adjust the parameters in scenario 1, that is, (a) increase the number of arriving BMs, (b) increase the intensity of arriving BMs, (c) increase the mean shooting time for each BM, and (d) reduce the number of target channels. Figure9 is the queue length as functions of the number of BMs, and Table 9 shows the results of operating parameters of the adjusted queueing system.Table 9
The results of operating parameters of the adjusted queueing system.
Parameters Before adjustment Number of BMs, 3-fold increase Intensity of BMs arriving, 2-fold increase Mean shooting time, 2-fold increase Target channels, 1/3 reduction Mean occupancy rate of target channels 22.6% 27.7% 40.9% 43.9% 63.9% Mean number of waiting BMs 0.000 0.000 0.000 0.000 0.000 Mean waiting time of BMs 0.000 0.000 0.000 0.000 0.000 Mean number of BMs in system 1.732 1.887 2.059 2.084 1.491 Number of penetrated BMs 7 45 15 19 21 Probability of BM penetration 23.3% 30%28.7% increase 50%114% increase 63.3%171% increase 70%200% increase Total DE 42.95% 39.2%8.7% reduction 28%34.8% reduction 20.6%50.0% reduction 16.8%60.9% reductionFigure 9
The queue length as functions of the number of BMs.
(a)
3-fold increase in the number of arriving BMs (b)
2-fold increase in the intensity of arriving BMs (c)
2-fold increase in the mean shooting time for each BM (d)
1/3 reduction in the number of target channelsThrough adjustment of the system configuration, we can observe and summarize the queueing system running condition. This can be useful for decision-making of BMDS operation control and adjusting system configuration.


## 6. Concluding Remarks and Future Work

In this paper, we presented a theory for the representation, modeling, and performance simulation of the layered BMDS M/M/N queueing systems. In addition to the queueing model, four simple rules have been developed for use on the complex channel allocation problems. The main aim of this work is to study a stochastic missile defense process close to the Poisson process and to find allocation rules that maximize DE or minimize the cost to achieve a required DE. In Section4.1.2, we also proposed a novel and robust search algorithm for obtaining the minimum requirements across a set of neighborhoods. Numerical examples using models in Section 2 revealed some guidelines for adjusting queueing parameters.This work may help to understand the rules of queueing process and to provide optimal configuration suggestions for defense decision-making. The work can easily be expanded and continued to handle a broader array of queueing scenarios. Several areas for potentially valuable future research have emerged from this work; we suggest the following areas of further research [22–26]. (1) We proposed some approximation in our computations; an important question is the discussion of the accuracy of BM arrivals distribution. The Poisson arrival process is not the only fitting model for the provided queueing model in our paper. We will consider Bernoulli or Markov BM arrival process in our future research. This also includes relaxing the assumptions of exponential shooting times and allowing waiting times to vary by BM and defense weapon. (2) For the convenience of calculation and verification, the 2 scenarios in Section 5 are simple and need further analysis in the follow-up work. (3) One such suggestion is an exploration into expanding the allocation rules presented in Section 4. (4) When the size of (41)–(56) is very large, an approximate solution can be obtained by using some advanced algorithms, for example, genetic, neural network, and heuristic algorithms.



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