Deployment Optimization Method of Multistatic Radar for Constructing Circular Barrier Coverage

To construct circular barrier coverage (CBC) with multistatic radars, a deployment optimization method based on equipartition strategy is proposed in this paper. In the method, the whole circular area is divided into several sub-circles with equal width, and each sub-circle is blanketed by a sub-CBC that is built based on the multistatic radar deployment patterns. To determine the optimal deployment patterns for each sub-CBC, the optimization conditions are firstly studied. Then, to optimize the deployment of the whole circular area, a model based on minimum deployment cost is proposed, and the proposed model is divided into two sub-models to solve the optimization issue. In the inner model, it is assumed that the width of a sub-circle is given. Based on the optimization conditions of the deployment pattern, integer linear programming (ILP) and exhaustive method (EM) are jointly adopted to determine the types and numbers of deployment patterns. Moreover, a modified formula is introduced to calculate the maximum valid number of receivers in a pattern, thus narrowing the search scope of the EM. In the outer model, the width of a sub-circle is assumed to be a variable, and the EM is adopted to determine the minimum total deployment cost and the optimal deployment patterns on each sub-circle. Moreover, the improved formula is exploited to determine the range of width for a sub-circle barrier and reduce the search scope of the EM. Finally, simulations are conducted in different conditions to verify the effectiveness of the proposed method. The simulation results indicate that the proposed method can spend less deployment cost and deploy fewer transmitters than the state-of-the-artwork.


Introduction
In the last decades, the world witnessed the increasing popularity of wireless sensor networks (WSNs) because of their robust fault tolerance, high energy efficiency, strong data processing capability, and advanced wireless communication technology [1][2][3][4].
Barrier coverage is one of the essential applications of WSNs. It can be exploited to monitor the targets traversing through the boundary of an FoI [5]. To build an effective barrier coverage, the coverage area of the sensors should construct a continuous barrier covering the FoI. Compared with point coverage [6] and area coverage [7], barrier coverage obtains a balance between the number of sensors and the coverage area. Therefore, barrier coverage is efficient in monitoring intruders.
Most previous works on barrier coverage are based on the disk coverage model [8] or the sector coverage model [9]. The disk coverage model indicates that the coverage area is a circle with the sensor being located at the center, and the radius of the circle is the monitor range. While in the sector coverage model, the coverage area is a sector of the circle.
Thirdly, a two-stage optimization algorithm with an equipartition strategy is proposed to solve these two sub-optimization models. In the condition of minimum deployment cost, the optimal deployment is derived by considering the transmitter-receiver cost ratio. The simulation results indicate that the proposed method needs less deployment cost and requires fewer transmitters than the works in the literature.
The rest of this paper is organized as follows. Section 2 reviews the related work. Section 3 introduces the concepts and details of the optimization issue. Section 4 proposes the solution for the optimization issue. Section 5 conducts the simulations to validate the performance of the proposed method. Section 6 concludes the paper.
In the beginning, the works on the disk coverage model were reviewed. In [8], the notion of weak and strong barrier coverage was introduced, and a critical condition was derived for constructing the weak barrier coverage. In [22], for the barrier coverage composed of a set of wireless sensors with adjustable ranges, the constant-approximation algorithms were designed to assign the ranges with the minimum cost, and the constant depends on the covering radius related to the power of sensors. In [23], crossed barrier coverage was built to detect the intrusion behavior in both horizontal and vertical directions. Meanwhile, the branch-and-bound algorithm and multi-round shortest path algorithm are exploited to determine the maximum number of crossed barriers. Given a set of mobile sensors distributed on the plane and a set of targets located on a line, an exact algorithm was introduced to minimize the maximum movement of the sensors the work and cover the targets with minimum energy consumption [24]. In [25], the inspection robot was integrated into WSNs to form an efficient dynamic barrier coverage for monitoring the transmission lines. Moreover, a discrete heuristic method based on game theory was proposed to solve this issue.
Considering the works based on the sector coverage model, in [9], sensors with arbitrarily adjustable directions were exploited to construct the strong barrier coverage. To determine the appropriate directions of sensors in the one-dimension scenario, a polynomialtime algorithm was adopted to achieve strong barrier coverage with a minimum number of sensors. Meanwhile, in the two-dimension scenario, the directional carrier graph was introduced to solve the issue. In [26], the researchers surveyed the concepts and properties of the sector coverage model, and they analyzed the sensing characters and behaviors of directional sensors. Besides, each category of the sector coverage model was summarized in terms of issue description, assumption, application, solution, and performance. The researchers in [27] addressed the barrier gap problem of weak and strong barrier coverage and deployed the directional sensors with a line deployment strategy. Besides, they proposed a gap-finding algorithm to determine sub-barriers and locate barrier gaps, and they fixed the gaps through two gap-mending algorithms. In [28], the sector coverage model was applied in a three-dimension scenario, and a realistic resolution criterion of the camera sensor was proposed to capture the face characters of the invader. Based on this criterion, a feasible deployment strategy was studied for barrier coverage. In [29], the virtual target-barrier construction method was adopted to construct a closed circular belt around a target. The closed virtual barrier curve was exploited to simplify the optimization problem, and then a distributed particle swarm optimization algorithm was designed to deploy an approximate optimal coverage on the virtual barrier curve.
There is increasing attention on the Cassini coverage model. In [30], given the numbers of transmitters and receivers, an optimal linear barrier coverage was constructed. However, this barrier coverage does not have a requirement of minimum width. This means that in some parts, the barrier is only covered by some points, and a high-speed intruder can cross the barrier without detection. The authors in [31] stated that it is optimal to place multistatic radars on the shortest deployment line connecting the left and right boundaries of the region. Based on these achievements, an optimization deployment method was proposed to build barrier coverage on the rectangular area [32]. Specifically, the method divides the whole rectangular area into multiple sub-rectangles with equal width. Then, it sets the minimum width for a sub-rectangle. Finally, several identical deployment patterns are exploited to build barrier coverage for each sub-rectangle.
Besides, some works focus on constructing an optimal barrier coverage on a circular area, which is a challenging issue due to the circular shape. As for this shape, the width and radius of a sub-circular barrier should be considered. An optimal model was introduced to build one CBC [20]. Practically, one circular barrier is inadequate to cover a vast circular area, and the width of a circular barrier is not set with a threshold. A model with a threshold of width was mentioned in [33], and the whole circular area was divided into several subcircular barriers with the same width. Each sub-circular barrier was composed of several identical deployment patterns, and only a varied one was exploited to cover the remaining part. Besides, EM was adopted to determine the optimal results. However, the composition of a sub-circular barrier and the searching scope of EM can be further improved. Inspirited by [33], our work makes the following improvements: (1) several novel conditions are introduced to determine the types of deployment patterns for an optimal sub-circular barrier; (2) ILP is adopted to obtain the corresponding number of deployment patterns; (3) an improved formula is introduced to calculate the maximum valid number of receivers in a deployment pattern; (4) a novel formula is adopted to calculate the upper threshold of width to narrow the searching scope of sub-circular barrier width.

Conceptions of Multistatic Radar Coverage
In this paper, the deployment pattern of multistatic radar is composed of a transmitter T and several receivers R i , (i = 1, 2, . . . , m), and it can be denoted as F m = (T, R 1 , . . . , R m ). Thus, the multistatic radar can be regarded as a combination of multiple bistatic radars that share a transmitter. In terms of coverage area, it is desirable to use the bistatic radar to illustrate the coverage properties of the multistatic radar.
Taking F 1 as an example, the signal-to-noise ratio (SNR) at point A, i.e., SNR(A), can be computed by the radar equation of coherent pulse bistatic radar [16], shown as follows: where K b is assumed to be a constant determined by the working parameters of the bistatic radar [32]. Let d T (A) and d R 1 (A) represent the distance from point A to the transmitter and the receiver, respectively.
Let γ represent the minimum SNR that is the threshold for target detection. If SNR(A) ≥ γ, the target is in the coverage area of a bistatic radar; otherwise, the target is out of the coverage area. Therefore, the coverage area of the bistatic radar is defined by the formula d T (A)d R (A) ≤ K b /γ. For ease of presentation, we have l max = 4 K b /γ. It is noteworthy that a target in the coverage area can be detected with an acceptable probability [34].
According to [16], the shape of the coverage area is determined by the relation between d 1 and l max , as shown in Figure 1: i. d 1 > l max , the coverage is two unconnected areas, as plotted in Figure 1a; ii. d 1 = l max , the coverage is enclosed by a lemniscate of Bernoulli, as plotted in Figure 1b; iii. √ 2l max /2 ≤ d 1 < l max , the coverage is enclosed by a waist-liked area, as plotted in Figure 1c; iv. 0 < d 1 < √ 2l max /2, the coverage is enclosed by an elliptical curve, as plotted in Figure 1d.   , the coverage is enclosed by a waist-liked area, as plotted in Figure   1c; iv. 1 m a x 0 2 2 d l < < , the coverage is enclosed by an elliptical curve, as plotted in Figure  1d.
It is also assumed that transmitters work at orthogonal frequencies. By choosing a working frequency, a receiver can pair with a certain transmitter and avoid being interfered with by other transmitters. Thus, the coverage area of the multistatic radar F m is overlapped with that of m bistatic radars. If a target is covered by multiple bistatic radars, the maximum SNR at point A, i.e., max (A) SNR , is exploited to measure the coverage of the multistatic radar, which is formulated as: It is also assumed that transmitters work at orthogonal frequencies. By choosing a working frequency, a receiver can pair with a certain transmitter and avoid being interfered with by other transmitters. Thus, the coverage area of the multistatic radar F m is overlapped with that of m bistatic radars. If a target is covered by multiple bistatic radars, the maximum SNR at point A, i.e., SNR max (A), is exploited to measure the coverage of the multistatic radar, which is formulated as: where l(A) = d T j (A)d R i (A), T j denotes the jth transmitter. Based on this, for a target at arbitrary point A, it can be covered by the multistatic radar if SNR max (A) ≥ γ, and we have l(A) ≤ l 2 max

Conceptions of CBC
As for a closed area with irregular boundaries, it can be represented by a minimum circumscribed circle, and a circular FoI is built around the circle to protect the closed area from infiltrating. As shown in Figure 2a, the gray circle area represents the circular FoI with a width of H, and the center of the FoI is O. c i and c o respectively denote the inner and outer circle boundary of the FoI. Besides, R min represents the radius c i .
In this paper, the whole FoI is divided into multiple sub-circles with equal width, and each sub-circle is covered by a sub-CBC. By constructing a sub-CBC with a width of 2 h, multiple multistatic radars are deployed along the ring in the middle of a sub-circle. This ring is called the deployment curve, and it is denoted as l k , k = 1, . . . , q. The radius of l k , k = 1, . . . , q is denoted as r k . Here, 2h ≥ 2η > 0, where 2η is the minimum width threshold of a sub-circle, and 2 h is the minimum width of a sub-CBC.
For instance, an FoI is divided into three sub-circles in Figure 2b. Without loss of generality, all the sub-CBCs are numbered from inside to outside. Correspondingly, l k , k = 1, 2, 3 represents the deployment curve of each sub-CBC. Moreover, the nodes are deployed in counterclockwise order. The transmitter is represented by a square, and the receiver is represented by a hexagram. k k threshold of a sub-circle, and 2 h is the minimum width of a sub-CBC. For instance, an FoI is divided into three sub-circles in Figure 2b. Without loss of generality, all the sub-CBCs are numbered from inside to outside. Correspondingly, It is desirable to study the deployment optimization for multistatic radars on a deployment curve. For ease of presentation, two definitions are given: , , , , . Without ambiguity, it can be denoted as n P for short. Note that n receivers are deployed between two transmitters, i.e., T i and +1 T i . The node coverage angle It is desirable to study the deployment optimization for multistatic radars on a deployment curve. For ease of presentation, two definitions are given: Definition 1. Let P n i denote a deployment pattern comprised of n receivers and two transmitters, i.e., P n i = (T i , R 1 , . . . , R n , T i+1 ). Without ambiguity, it can be denoted as P n for short. Note that n receivers are deployed between two transmitters, i.e., T i and T i+1 . The node coverage angle (NCA) in P n i is denoted as 2θ j , j = 1, . . . , n + 1, where According to [33], NCA can be calculated as: where, r denotes the radius of the deployment curve l, and R denotes the radius of the outer boundary of the sub-circle. Besides, R j is a valid receiver if θ j > 0, j = 1, 2, . . . , n + 1.
Furthermore, according to Formula (3), by using algebraic operation, we have Let ω P (n, h, r) denote the central angle formed by P n , and it is referred to as pattern coverage angle (PCA). Usually, a PCA is inadequate to cover the whole FoI. Thus, we have can be calculated as follows: Sensors 2021, 21, 6573 7 of 21 Definition 2. Let S denote a deployment sequence formed by k varied types of P n j , j = 1, 2, . . . , k. Let s j , j = 1, 2, . . . , k denote the number of types. Specifically, we have S = s 1 , P n 1 1 , . . . , s k , P To build a CBC, ∑ k j=1 s j ω P n j (n j , h, r) ≥ 360 • should hold.
As shown in Figure 3, in the case where r = 3.833 km, l max = 2 km, R= 4.667 km, there is a sub-circle covered by the deployment sequence of P 3 1 , P 3 2 , P 2 3 , and all patterns are deployed on the deployment curve l. According to (5), the total angle formed by these three patterns equals to 367.2 • , which is a bit larger than 360 • . In theoretical, the transmitter T 4 should be placed where the blue square is plotted. However, in practice, P 3 1 , P 2 3 share the transmitter T 1 . , , P P P , and all patterns are deployed on the deployment curve l. According to (5), the total angle formed by these three patterns equals to 367.2°, which is a bit larger than 360°. In theoretical, the transmitter T4 should be placed where the blue square is plotted. However, in practice, 3 2 1 3 , P P share the transmitter T1.

Description of Optimization Deployment Problem
The problem can be described as follows. Let D denote the FoI, and FoI is equally divided into q sub-circles with a width of 2 h. The CBC is exploited to cover each sub-circle. Let M k and N k , (k = 1, . . . , q) denote the number of transmitters and receivers on each deployment curve, respectively. Besides, it is assumed that all the transmitters are identical, so are the receivers. In addition, the unit costs of the transmitters and receivers are denoted as c T and c R , respectively. Let α = c T /c R denote the transmitter-receiver cost ratio, and it is reasonable to set α > 1.
Under the minimum deployment cost, c tot , the optimization problem model can be formulated as follows: In this paper, k rounds k to the nearest integer towards minus infinity; k rounds k to the nearest integer towards positive infinity; Φ denotes an empty set; N + and N * denote the positive integer set and natural integer set, respectively. For point A and A , |AA | denotes the length of the line AA .

The Upper Threshold of Width for a Sub-Circular Barrier
As mentioned above, 2µ is the lower threshold width of the CBC. In this section, a novel formula is introduced to determine the upper threshold width of the CBC. Like the conditions in [33], the waist-liked area in the bistatic radar is exploited first. Then, Lemma 1 is adopted to determine h sup .

Lemma 1.
As for the width-equipartition strategy, h sup denotes the upper threshold of h, and it can be calculated as follows: Proof of Lemma 1. The proof of Lemma 1 is given in the Appendix A.

The Maximum Number of Receivers and the Property of NCA
In this section, Lemma 2 is introduced to improve the formula for determining the maximum number of receivers in a P n . Then, Lemma 3 is exploited to uncover the property of NCA.

Lemma 2.
Let 2 h denote the width of a sub-circular barrier, and r denote the radius of the deployment curve. Let n max denote the maximum number of receivers in P n . Then n max is calculated as follows: (1) If h sup ≥ h > r 2 + l 2 max − r and u is the largest natural number that satisfies the inequality: h sup and u is the largest natural number that satisfies the inequality: Proof of Lemma 2. The proof of Lemma 2 is given in the Appendix A.
By contrast, it is claimed in [33] ). However, this formula is ineffective in some conditions. To prove this, given r = 4 km, R = 4.3 km, l max = √ 3 km, based on [33], we have θ max = acos(−1.9044). According to Formula (5), the more receivers are placed in P n , the larger a central angle can be formed. However, Lemma 3 states that the efficiency of placing receivers decreases as the number of receivers increases. Lemma 3. For P n it is assumed that ∑ u k=1 θ k < ϕ, u = (n + 1)/2 . Then, the NCA sequence θ k , k = 1, 2, . . . , u is strictly monotonous decreasing. Where, Proof of Lemma 3. The proof of Lemma 3 is given in the Appendix A.
For P n , let λ j = θ j /c R denote the effectiveness-cost ratio (ECR) of R j , j = 1, 2, . . . , n. It reveals the coverage efficiency of a receiver. The larger λ j , the better coverage efficiency of a receiver. Besides, let λ = 1 n ∑ n j=1 λ j denote the average ECR of P n , and it reveals the coverage efficiency of P n .
Moreover, Lemma 3 states that λ j decreases as j increases. In other words, the farther a receiver is placed away from the transmitter, the lower ECR is obtained. Meanwhile, the larger n, the lower average ECR of P n .

The Conditions of the Optimal Deployment Patterns for a CBC
According to the equipartition strategy, it is critical to find the optimization conditions of deployment patterns on a sub-CBC. These conditions are essential to establishing the optimization model of the deployment problem.
Given h and r, for ease of presentation, ω P (n, h, r) can be abbreviated as ω P (n).

Lemma 4.
Assuming two varied patterns P n 1 and P n 2 in the CBC and max{n 1 , n 2 } ≤ n max , the following statements holds.

Proof of Lemma 4. The proof of Lemma 4 is given in the Appendix A.
Lemma 4 provides the necessary and sufficient conditions for an optimal CBC. Moreover, it reveals that the PCA of P 2n−1 and P 2n+1 can be equally replaced by two PCAs of P 2n . Thus, P 2n−1 , P 2n , and P 2n+1 can be replaced with three P 2n to simplify the structure of the deployment sequence and make it easier for engineering. Furthermore, due to ω P (2n) + ω P (2n + 2) < 2ω P (2n + 1), it is necessary to deploy three P 2n+1 instead of P 2n , P 2n+2 , and P 2n+1 in an optimal CBC. Supposing that n 1 = 2, n 2 = 4, r = 4 km, R = 4.8 km, l max = √ 3 km, according to [33], the coverage angle is ω P (2) + ω P (4) = 188.67 • . By contrast, based on Lemma 4, we have 2ω P (3) = 197.75 • . It means that the proposed conditions can contribute to a better optimization result.

Model and Solution for Deployment Optimization Problem
It is revealed by Lemma 4 that, for a deployment sequence, the maximum coverage angle can be achieved by using two varied patterns at most. Moreover, the difference in the number of receivers in patterns should not be larger than 1. This deployment sequence is referred to as the optimal one. Besides, it is assumed that the whole circular area is equally divided into q sub-circular barriers. Let S k opt = s k 1 , P n k , s k 2 , P n k +1 , 1 ≤ k ≤ q denote the optimal deployment sequence for the kth sub-circular barrier. Without loss of generality, k = 1 represents the innermost sub-circular barrier and k = q represents the outmost sub-circular barrier. Then, the deployment optimization model can be represented as follows: max , (k = 1, 2, . . . , q), H/q = 2h q , s k i ∈ N 0 , n k ∈ N * , r k = R min + (2k − 1)h q , q min ≤ q ≤ q max (8) However, the nonlinearity and multiple parameters make the solution to this model difficult. To overcome these obstacles, the solution to this model is decomposed to a two-stage optimization problem: where c k s denotes the minimum deployment cost for the kth sub-CBC. In this stage, it is assumed that the width of each sub-circle is given. Then, q min and q max can be calculated. For each q, (q min ≤ q ≤ q max ) and q ∈ N + , 2h q = H/q, the following sub-optimization problem is solved: For each sub-CBC, there is a model depicted by Formula (9). min In this stage, supposing that n is determined, c k,n s denotes the deployment cost of the kth sub-CBC in this condition. n is the number of receivers in a pattern. Besides, based on Lemma 2, the maximum number of receivers n k max for the patterns in the kth sub-CBC can be calculated. The following sub-optimization problem can be solved: Formula (10) can be solved by using the ILP. The specific process of solving the deployment optimization model is given: 1: Calculate h sup , q min and q max through Lemma 1. Initialize the deployment curve set Q = Φ and the total deployment cost set C tot = Φ 2: For q = q min : q max 3: Calculate h = H/2q.

Simulations
In this section, simulations are performed to verify the effectiveness of the proposed method. In the proposed method, the range of q can be determined by Lemma 1. Subsequently, given q, each width of sub-CBC is equal to H/q. Then, according to Lemma 2, n k max can be calculated for the kth sub-CBC. Given n ≤ n k max , a sequence c k,1 s , c k,2 s , . . . , c k,n k max s can be obtained by using the Formula (10). Thus, c k s is found via the minimum cost principle. Furthermore, the total minimum deployment cost is computed by ∑ q k=1 c k s for q equipartition strategy.  It can be seen that the value of κ is always larger than or equal to zero when  Then, the impact of H on the total number of transmitters is evaluated. The results are plotted in Figures 6-8. Correspondingly, , 1,2 j j τ = denotes the optimal number of transmitters in formula (11). The change rate of the number of transmitters is illustrated in Figure 6. The change rates for 2,10 α = are almost larger or equal to zero. By contrast, the change rates are constant and equal to zero for =50,80 α . The change rate κ is exploited to measure the performance achieved by the proposed method and the method mentioned in [33]; κ is defined as follows: where τ 1 and τ 2 represents the optimal deployment parameter obtained by [33] and this paper, respectively. First, the influence of H on the minimum deployment cost is investigated. Correspondingly, τ j , j = 1, 2 denotes the minimum deployment cost. In this case, the optimal deployment cost change rate revealed by Formula (11) is illustrated in Figure 4.
It can be seen that the value of κ is always larger than or equal to zero when α = 2, 10, 50, 80, indicating that the proposed method requires less deployment cost. Specifically, when α = 2, 10, the value of κ is much larger. The larger κ is the better costsaving effect can be achieved. Besides, in most cases, the optimal results under α = 2, 10 are superior to those under α = 50, 80. Therefore, the proposed method performs better when α = 2, 10. Although the change rate decreases in the case of α = 50, 80, Figure 4 shows that the proposed method can still obtain less minimum deployment cost.
It can be seen from Figure 5 that the magnitude of reduced cost for α = 50, 80 has the same level as that of α = 2, 10. However, c * tot is dramatically large for α = 50, 80, making the change rate descend.  It can be seen that the value of κ is always larger than or equal to zero when  Then, the impact of H on the total number of transmitters is evaluated. The results are plotted in Figures 6-8. Correspondingly, , 1,2 j j τ = denotes the optimal number of transmitters in formula (11). The change rate of the number of transmitters is illustrated in Figure 6. The change rates for 2,10 α = are almost larger or equal to zero. By contrast, Then, the impact of H on the total number of transmitters is evaluated. The results are plotted in Figures 6-8. Correspondingly, τ j , j = 1, 2 denotes the optimal number of transmitters in Formula (11). The change rate of the number of transmitters is illustrated in Figure 6. The change rates for α = 2, 10 are almost larger or equal to zero. By contrast, the change rates are constant and equal to zero for α = 50, 80. Besides, the relative change of the number of transmitters is illustrated in Figure 7. It can be seen from Figure 7 that the relative changes of the numbers of transmitters are equal to zero for =50,80 α leading the change rates equal to zero. Since a transmitter is far more expensive than a receiver, both methods exploit as few transmitters as possible. Actually, the number of transmitters cannot be further reduced by the proposed method under these conditions. Additionally, it can be observed from Figure 8 that the number of transmitters increases dramatically when H becomes large, leading to a decrease in the change rate.
Moreover, a tendency that the number of transmitters decreases with the increase of α is shown in Figure 8. Because of the considerable cost of a transmitter, it is economical to deploy more receivers to pair with one transmitter for extending the PCA. Consequently, the number of transmitters decreases with the number of patterns. Besides, the relative change of the number of transmitters is illustrated in Figure 7. It can be seen from Figure 7 that the relative changes of the numbers of transmitters are equal to zero for =50,80 α leading the change rates equal to zero. Since a transmitter is far more expensive than a receiver, both methods exploit as few transmitters as possible. Actually, the number of transmitters cannot be further reduced by the proposed method under these conditions. Additionally, it can be observed from Figure 8 that the number of transmitters increases dramatically when H becomes large, leading to a decrease in the change rate.
Moreover, a tendency that the number of transmitters decreases with the increase of α is shown in Figure 8. Because of the considerable cost of a transmitter, it is economical to deploy more receivers to pair with one transmitter for extending the PCA. Consequently, the number of transmitters decreases with the number of patterns. Besides, the relative change of the number of transmitters is illustrated in Figure 7. It can be seen from Figure 7 that the relative changes of the numbers of transmitters are equal to zero for α = 50, 80 leading the change rates equal to zero. Since a transmitter is far more expensive than a receiver, both methods exploit as few transmitters as possible. Actually, the number of transmitters cannot be further reduced by the proposed method under these conditions.
As illustrated in Figures 6 and 7, though the proposed method exploits one more transmitter and obtains the change rate of −0.93% in the case of α = 10, H = 19km, the results indicate that the proposed method requires fewer transmitters in most case.
Additionally, it can be observed from Figure 8 that the number of transmitters increases dramatically when H becomes large, leading to a decrease in the change rate. Moreover, a tendency that the number of transmitters decreases with the increase of α is shown in Figure 8. Because of the considerable cost of a transmitter, it is economical to deploy more receivers to pair with one transmitter for extending the PCA. Consequently, the number of transmitters decreases with the number of patterns. Subsequently, the impact of H on the relative change number of receivers is discussed in Figure 9. In this case, 1 τ and 2 τ represent the optimal number of receivers, respectively. α . The high cost of a transmitter makes it challenging to improve the optimal result on transmitters, but the proposed method can still reduce the optimal result of receivers. Taking  , and the total cost of these sequences is 642. It is worth emphasizing that there is a large peak for In this condition, although the proposed method exploits one more transmitter, it requires 32 fewer receivers. Therefore, it still reduces the cost.

The Influence of min
R on the Performance Subsequently, the impact of H on the relative change number of receivers is discussed in Figure 9. In this case, τ 1 and τ 2 represent the optimal number of receivers, respectively. Subsequently, the impact of H on the relative change number of receivers is discussed in Figure 9. In this case, 1 τ and 2 τ represent the optimal number of receivers, respectively. .Therefore, the total cost of these sequences is 646. By contrast, the optimal deployment sequences obtained by the method in [33]  , and the total cost of these sequences is 642. It is worth emphasizing that there is a large peak for In this condition, although the proposed method exploits one more transmitter, it requires 32 fewer receivers. Therefore, it still reduces the cost.

The Influence of min
R on the Performance The relative change number of receivers fluctuates around zero in the condition of α = 2, 10, by contrast, the curves are constantly larger than zero in the condition of α = 50, 80. The high cost of a transmitter makes it challenging to improve the optimal result on transmitters, but the proposed method can still reduce the optimal result of receivers.
Taking α = 50, H = 5km as an example, the optimal deployment sequences obtained by the proposed method are S 1 opt = 1, P 2 , 2, P 3 , S 2 opt = 1, P 4 , 3, P 3 and S 3 opt = 4, P 4 , 1, P 5 . Therefore, the total cost of these sequences is 646. By contrast, the optimal deployment sequences obtained by the method in [33] are S 1 opt = 1, P 2 , 2, P 3 , S 2 opt = 1, P 3 , 3, P 4 and S 3 opt = 1, P 3 , 4, P 5 , and the total cost of these sequences is 642. It is worth emphasizing that there is a large peak for α = 10, H = 19km. In this condition, although the proposed method exploits one more transmitter, it requires 32 fewer receivers. Therefore, it still reduces the cost.

The Influence of R min on the Performance
The influence of R min on the proposed method is studied in this section. Let l max = 2 km, µ= 0.1 km, H = 19 km. R min varies from 1 km to 20 km with a step size of 1 km. Besides, M and N denote the total number of transmitters and receivers, respectively (see Figure 10). The influence of min R on the proposed method is studied in this section. Let , 19km H = . min R varies from 1 km to 20 km with a step size of 1 km. Besides, M and N denote the total number of transmitters and receivers, respectively (see Figure 10). In Figure 11, the relation between min R and the ratio of N to M is illustrated. It can be seen that the ratio in the condition of 50,100 α = is almost twice that of 2,10 α = . When α becomes larger, the pattern is composed of more receivers. Also, the more receivers contained in a pattern, the narrower barrier coverage is built [32]. Thus, in the case of 50,100 α = , the whole area may need to be divided into more sub-circles, resulting in the fluctuation of q*. Additionally, as depicted in Figure 11, it is worth emphasizing that the number of transmitters is almost equal to that of receivers when 2 α = . In other words, the 1 P is mainly used in the deployment. This result is caused by two reasons. One reason is that For α = 2, 10, the value of q * varies at first and tends to be stable as the R min increases; for α = 50, 100, q * fluctuates as the R min increases. It reveals that for a small α and a large R min , q * does not change with the increase of R min .
In Figure 11, the relation between R min and the ratio of N to M is illustrated. It can be seen that the ratio in the condition of α = 50, 100 is almost twice that of α = 2, 10. When α becomes larger, the pattern is composed of more receivers. Also, the more receivers contained in a pattern, the narrower barrier coverage is built [32]. Thus, in the case of α = 50, 100, the whole area may need to be divided into more sub-circles, resulting in the fluctuation of q*.
The influence of min R on the proposed method is studied in this section. Le max 2 km, =0.1 km l μ = , 19km H = . min R varies from 1 km to 20 km with a step size of 1 km. Besides, M and N denote the total number of transmitters and receivers, respectively (see Figure 10). In Figure 11, the relation between min R and the ratio of N to M is illustrated. It can be seen that the ratio in the condition of 50,100 α = is almost twice that of 2,10 α = When α becomes larger, the pattern is composed of more receivers. Also, the more re ceivers contained in a pattern, the narrower barrier coverage is built [32]. Thus, in the case of 50,100 α = , the whole area may need to be divided into more sub-circles, result ing in the fluctuation of q*. Additionally, as depicted in Figure 11, it is worth emphasizing that the number o transmitters is almost equal to that of receivers when 2 α = . In other words, the 1 P i mainly used in the deployment. This result is caused by two reasons. One reason is tha Additionally, as depicted in Figure 11, it is worth emphasizing that the number of transmitters is almost equal to that of receivers when α = 2. In other words, the P 1 is mainly used in the deployment. This result is caused by two reasons. One reason is that the first receiver obtains the maximum ECR, according to lemma 3. The other reason is that the difference in the cost between a transmitter and a receiver is little. Therefore, it is economical and efficient to adopt as many p 1 as possible.
Next, the relationship between R min and M is investigated, and the result is shown in Figure 12. It can be seen that the number of transmitters increases approximately linearly with R min for α = 2, 10, 50, 100. According to Formula (5), the increase of R min leads to an increase in the radius of the deployment curve and a decrease in the PCA. This change means that more transmitters are required as R min increases. Moreover, given R min , the result in Figure 12 reveals that the number of transmitters decreases as α increases. Thus, it is worth deploying more receivers instead of a transmitter because the receivers are highly cheap when α = 50, 100. the first receiver obtains the maximum ECR, according to lemma 3. The other reason is that the difference in the cost between a transmitter and a receiver is little. Therefore, it is economical and efficient to adopt as many p 1 as possible. Next, the relationship between min R and M is investigated, and the result is shown in Figure 12. It can be seen that the number of transmitters increases approximately linearly with min R for 2,10,50,100 α = . According to formula (5), the increase of min R leads to an increase in the radius of the deployment curve and a decrease in the PCA. This change means that more transmitters are required as min R increases. Moreover, given min R , the result in Figure 12 reveals that the number of transmitters decreases as α increases. Thus, it is worth deploying more receivers instead of a transmitter because the receivers are highly cheap when =50,100 α . Furthermore, the relationship between min R and N is studied, and the result is illustrated in Figure 13. The number of receivers increases with min R for 2,10,50,100 It should be noted that there are three peaks for 100, 2,5,10 H α = = . It is because more sub-circles are divided in these cases, as plotted in Figure 10. Moreover, it is noteworthy that N increases significantly when increases. It can be explained by the slope of the transmitter numbers that decrease in this condition, as depicted in Figure 12. The decreased slope indicates that the magnitude of the increased number of transmitters decreases. To enlarge the PCA, more receivers are exploited in the patterns. Furthermore, the relationship between R min and N is studied, and the result is illustrated in Figure 13. The number of receivers increases with R min for α = 2, 10, 50, 100. It should be noted that there are three peaks for α = 100, H = 2, 5, 10. It is because more sub-circles are divided in these cases, as plotted in Figure 10. Moreover, it is noteworthy that N increases significantly when α = 10, 7 ≤ H ≤ 12 and then tends to be stable as the R min increases. It can be explained by the slope of the transmitter numbers that decrease in this condition, as depicted in Figure 12. The decreased slope indicates that the magnitude of the increased number of transmitters decreases. To enlarge the PCA, more receivers are exploited in the patterns.

Conclusions
A deployment optimization method is proposed in this paper to construct circular barrier coverage with an equipartition strategy. The concepts of multistatic radar coverage and CBC are introduced. Based on the equipartition strategy, a circular FoI is equally divided into q sub-circles. Each sub-circle is blanketed by a sub-CBC composed of sever-

Conclusions
A deployment optimization method is proposed in this paper to construct circular barrier coverage with an equipartition strategy. The concepts of multistatic radar coverage and CBC are introduced. Based on the equipartition strategy, a circular FoI is equally divided into q sub-circles. Each sub-circle is blanketed by a sub-CBC composed of several multistatic radar deployment patterns. Moreover, the optimization conditions for determining the optimal deployment patterns in a sub-CBC are studied. The conditions reveal that at most two varied types of patterns are needed by an optimal deployment sequence. Besides, a model based on minimum deployment cost is proposed to optimize the deployment on the whole circular FoI. The model is divided into two sub-models, and a two-stage optimization algorithm is proposed to solve the model. In the inner model, it is assumed that the width of a sub-circle is given. Based on the optimization conditions, ILP and EM are jointly exploited to obtain the types and numbers of deployment patterns. Besides, a novel formula is derived to improve the result of the maximum number of receivers in a pattern and reduce the search scope of EM. In the outer model, the improved formula is proposed to determine the number range of sub-circles and the width of a sub-circle. Then, EM is adopted to determine the minimum deployment cost and the optimal deployment patterns in each sub-CBC. Finally, simulations are conducted to validate the effectiveness of the proposed method. It is worth mentioning that the equipartition strategy results in a locally optimal solution. To improve the optimization results, the non-equipartition strategy should be considered in the future. Moreover, there are many overlapping areas in CBC, making it challenging to improve the coverage effectiveness. Therefore, it is beneficial to explore new coverage models, such as the regular polygonal model.
It is easy to know that h is a monotonically decreasing function concerning d. Let h max denotes maximum width. Additionally, waist-liked area means d 1 ≥ l max / √ 2. According to Formula (A1), we obtain h max = l max / √ 2 + r 2 − 0.5l 2 max − r. Furthermore, h max increases as r increases, and the innermost CBC obtains the smallest radius of R min + h. Therefore, in width-equipartition strategy, substituting r with R min + h, then h sup is calculated as follows: It is easy to know that h is a monotonically decreasing function concerning d. Let max h denotes maximum width. Additionally, waist-liked area means 1 m a x / 2 d l ≥ . According to formula (A1), we obtain  Figure A1. Schematic diagram of the upper limit of the sub-circular barrier width.
The proof of Lemma 1 is completed.
By contrast, in [33], it claims the width range of CBC is ( max 0, 2l   . However, max 2l is the maximum value of the AA′ . When max 2 2 h l = it means points C and B are coincident, which is impossible in this scenario.

Proof of Lemma 2.
(1) In the beginning, The proof of Lemma 1 is completed.
By contrast, in [33], it claims the width range of CBC is 0, √ 2l max . However, √ 2l max is the maximum value of the |AA |. When 2h = √ 2l max it means points C and B are coincident, which is impossible in this scenario.
The proof of Lemma 2 is completed. With the algebraic operation, we have