Optimal deployment of multistatic radar for belt barrier coverage

Abstract

How to optimally construct belt barrier coverage to blanket the field of interest (FoI) is a critical and essential issue in wireless sensor networks. To solve this issue, an optimization method is proposed in this paper for deploying multistatic radar. The proposed method utilizes multiple unequal-width barriers consisting of different deployment sequences. First, the structure of the sequence consists of different deployment patterns, and it is challenging to determine it directly. We prove the mutual relations among different patterns and propose an optimization principle to simplify the sequence structure. In addition, one barrier coverage is usually inadequate to cover the whole FoI, thus multiple unequal-width barrier coverages are adopted. Then, we model the issue as a linear programming problem. To solve this model, integer linear programming (ILP) and exhaustive methods are jointly exploited to determine the sequence structure on one deployment line by minimizing the deployment cost. Subsequently, the ILP is adopted to explore the number of varied barriers and the corresponding minimum cost. Meanwhile, the optimal deployment parameters, such as the number of deployment lines, receivers, and transmitters, can be determined. Some numerical simulations are conducted to demonstrate that, compared with the existing method, the proposed method requires a lower deployment cost and fewer transmitters.

Appendices

Appendix 1

Proof of Lemma 1

Let $f(n) = \sqrt {n(\beta^{2} - n)}$, and it can be seen that $f(x) = \sqrt {x(\beta^{2} - x)}$ has $f^{\prime}(x) > 0,$ $f^{\prime\prime}(x) < 0$ when $x \in [0,{{\beta^{2} } \mathord{\left/ {\vphantom {{\beta^{2} } 2}} \right. \kern-\nulldelimiterspace} 2}]$. Thus, $f(n)$ is a single increasing concave function, and $x_{n + 1} - x_{n} = h\{ f(n + 2) + f(n) - 2f(n + 1)\} < 0$.

The proof is completed.

Proof of Lemma 2

(1) Without loss of generality, let $n = v - i,m = v + i$. First of all, it is assumed that $n,m,i$ are even. According to formula (5) and (6), we have

$$\begin{aligned} 2\sigma_{F} (h,v) &= \sum\nolimits_{k = 1}^{v/2} {{\text{8}d}_{k} } + 4x_{\frac{v}{2}} \\ &= \sum\nolimits_{k = 1}^{{\frac{v + i}{2}}} {{\text{4}d}_{k} } - \sum\nolimits_{{k = \frac{v}{2} + 1}}^{{\frac{v + i}{2}}} {{\text{4}d}_{k} } \\&+ \sum\nolimits_{k = 1}^{{\frac{v - i}{2}}} {{\text{4}d}_{k} } + \sum\nolimits_{{k = \frac{v - i}{2} + 1}}^{\frac{v}{2}} {{\text{4}d}_{k} } + 4x_{\frac{v}{2}} \\& = \sigma_{F} (h,m) + \sigma_{F} (h,n) - \sum\nolimits_{{k = \frac{v}{2} + 1}}^{{\frac{v + i}{2}}} {{\text{4}d}_{k} } + \sum\nolimits_{{k = \frac{v - i}{2} + 1}}^{{\frac{v}{2} + 1}} {{\text{4}d}_{k} } - 2x_{{\frac{v + i}{2}}} - 2x_{{\frac{v - i}{2}}} \hfill \\ &= \sigma_{F} (h,m) + \sigma_{F} (h,n) - \sum\nolimits_{{k = \frac{v}{2} + 1}}^{{\frac{v + i}{2}}} {{\text{4}d}_{k} } + \sum\nolimits_{{k = \frac{v - i}{2} + 1}}^{\frac{v}{2}} {{\text{4}d}_{k} } + 4x_{\frac{v}{2}} - 2x_{{\frac{v + i}{2}}} - 2x_{{\frac{v - i}{2}}} \hfill \\ &= \sigma_{F} (h,m) + \sigma_{F} (h,n) + 4\sum\nolimits_{k = 1}^{{\frac{i}{2} - 1}} {(d_{{k + \frac{v - i}{2} + 1}} - d_{{k + \frac{v}{2} + 1}} } ) + 2(x_{{\frac{v - i}{2}}} - x_{{\frac{v + i}{2}}} ) \hfill \\ & = \sigma_{F} (h,m) + \sigma_{F} (h,n) + 4\sum\nolimits_{k = 1}^{{\frac{i}{2} - 1}} {(x_{{k + \frac{v - i}{2}}} - x_{{k + \frac{v}{2}}} } ) + 2(x_{{\frac{v - i}{2}}} - x_{{\frac{v + i}{2}}} ) \hfill \\ \end{aligned}$$

Thus,

$$\sigma_{F} (h,m) + \sigma_{F} (h,n) = 2\sigma_{F} (h,v) - 4\sum\nolimits_{k = 1}^{{\frac{i}{2} - 1}} {(x_{{k + \frac{v - i}{2}}} - x_{{k + \frac{v}{2}}} } ) - 2(x_{{\frac{v - i}{2}}} - x_{{\frac{v + i}{2}}} )$$

(29)

Based on formula (29) and the descending property of the sequence $\{ x_{n} {\text{\} }}$, it can be seen that $\sigma_{F} (h,m) + \sigma_{F} (h,n) \le 2\sigma_{F} (h,v)$. Furthermore, if and only if i = 0, i.e., $n = m$, the equation holds.

Second, it is assumed that $n,m,i$ are odd. Similarly, we can derive that

$$\sigma_{F} (h,m) + \sigma_{F} (h,n) = 2\sigma_{F} (h,v) - 4\sum\nolimits_{k = 0}^{{\frac{i - 3}{2}}} {\left( {x_{{k + \frac{v - i + 1}{2}}} - x_{{k + \frac{v}{2} + 1}} } \right)} .$$

(30)

Based on formula (30) and the descending property of the sequence $\{ x_{n} {\text{\} }}$, it can be seen that $\sigma_{F} (h,m) + \sigma_{F} (h,n) \le 2\sigma_{F} (h,v)$. Furthermore, if and only if i = 1, i.e., $n = v - 1$,$m = v + 1$, the equation holds.

(2) Let $n = v - i,m = v + i$. First of all, it is assumed that $n,m$ are even and i is odd. According to formula (5) and (6),

$$\begin{gathered} 2\sigma_{F} (h,v) = \sum\nolimits_{k = 1}^{(v + 1)/2} {{\text{8}d}_{k} } = \sum\nolimits_{k = 1}^{{\frac{v + i}{2}}} {{\text{4}d}_{k} } - \sum\nolimits_{{k = \frac{v + 1}{2} + 1}}^{{\frac{v + i}{2}}} {{\text{4}d}_{k} } + \sum\nolimits_{k = 1}^{{\frac{v - i}{2}}} {{\text{4}d}_{k} } + \sum\nolimits_{{k = \frac{v - i}{2} + 1}}^{{\frac{v + 1}{2}}} {{\text{4}d}_{k} } \hfill \\ = \sigma_{F} (h,m) + \sigma_{F} (h,n) - \sum\nolimits_{{k = \frac{v + 1}{2} + 1}}^{{\frac{v + i}{2}}} {{\text{4}d}_{k} } + \sum\nolimits_{{k = \frac{v - i}{2} + 1}}^{{\frac{v + 1}{2}}} {{\text{4}d}_{k} } - 2x_{{\frac{v + i}{2}}} - 2x_{{\frac{v - i}{2}}} \hfill \\ = \sigma_{F} (h,m) + \sigma_{F} (h,n) + 4\sum\nolimits_{k = 0}^{{\frac{i - 3}{2}}} {{\text{(}d}_{{k + \frac{v - i}{2} + 2}} } - {d}_{{k + \frac{v + 1}{2} + 1}} ) + 2(x_{{\frac{v - i}{2}}} - x_{{\frac{v + i}{2}}} ) \hfill \\ = \sigma_{F} (h,m) + \sigma_{F} (h,n) + 4\sum\nolimits_{k = 0}^{{\frac{i - 3}{2}}} {{(}x_{{k + \frac{v - i}{2} + 1}} } - x_{{k + \frac{v + 1}{2}}} ) + 2(x_{{\frac{v - i}{2}}} - x_{{\frac{v + i}{2}}} ) \hfill \\ \end{gathered}$$

Thus,

$$\sigma_{F} (h,m) + \sigma_{F} (h,n) = 2\sigma_{F} (h,v) - 4\sum\nolimits_{k = 0}^{{\frac{i - 3}{2}}} {{(}x_{{k + \frac{v - i}{2} + 1}} } - x_{{k + \frac{v + 1}{2}}} ) - 2(x_{{\frac{v - i}{2}}} - x_{{\frac{v + i}{2}}} )$$

(31)

According to formula (31), it can be seen that $\sigma_{F} (h,m) + \sigma_{F} (h,n) < 2\sigma_{F} (h,v).$

Second, it is assumed that $n,m$ are odd and $i$ is an even. It can be derived that

$$\begin{aligned} \sigma _{F} (h,m) + \sigma _{F} (h,n) = 2\sigma _{F} (h,v) \\ - 4\sum\nolimits_{{k = 0}}^{{\frac{{i - 2}}{2}}} {\left( {x_{{k + \frac{{v - i + 1}}{2}}} - x_{{k + \frac{{v + 1}}{2}}} } \right)} \\ \end{aligned}$$

(32)

From formula (32), it can be seen that $\sigma_{F} (h,m) + \sigma_{F} (h,n) \le 2\sigma_{F} (h,v)$. Furthermore, if and only if $i$ = 0, i.e., $n = m$, this equation holds.

(3) Without loss of generality, let $n = v + i$,$m = v + 1 - i$ and m is odd. Thus, it is assumed that $v,i$ are odd. Correspondingly, $n$ is even. Then, according to formula (5) and (6),

$$\begin{gathered} \sigma_{F} (h,v) + \sigma_{F} (h,v + 1) = \sum\nolimits_{k = 1}^{(v + 1)/2} {{\text{8}d}_{k} } + 2x_{{\frac{v + 1}{2}}} = \sum\nolimits_{k = 1}^{{\frac{v + i}{2}}} {{\text{4}d}_{k} } - \sum\nolimits_{{k = \frac{v + 1}{2} + 1}}^{{\frac{v + i}{2}}} {{\text{4}d}_{k} } + \sum\nolimits_{k = 1}^{{\frac{v + 2 - i}{2}}} {{\text{4}d}_{k} } + \hfill \\ \sum\nolimits_{{k = \frac{v + 2 - i}{2} + 1}}^{{\frac{v + 1}{2}}} {{\text{4}d}_{k} } + 2x_{{\frac{v + 1}{2}}} = \sigma_{F} (h,m) + \sigma_{F} (h,n) + 4\sum\nolimits_{k = 0}^{{\frac{i - 3}{2}}} {(x_{{k + \frac{v + 2 - i}{2}}} - x_{{k + \frac{v + 1}{2}}} } ) + 2(x_{{\frac{v + 1}{2}}} - x_{{\frac{v + i}{2}}} ) \hfill \\ \end{gathered}$$

Then,

$$\begin{aligned} \sigma _{F} (h,m) + \sigma _{F} (h,n) & = \sigma _{F} (h,v) + \sigma _{F} (h,v + 1) \\ - 4\sum\nolimits_{{k = 0}}^{{\frac{{i - 3}}{2}}} {\left( {x_{{k + \frac{{v + 2 - i}}{2}}} - x_{{k + \frac{{v + 1}}{2}}} } \right)} \\ - 2(x_{{\frac{{v + 1}}{2}}} - x_{{\frac{{v + i}}{2}}}) \\ \end{aligned}$$

(33)

According to formula (33), it is clear that $\sigma_{F} (h,m) + \sigma_{F} (h,n) \le \sigma_{F} (h,v) + \sigma_{F} (h,v + 1)$. Furthermore, if and only if $i$ = 1, i.e., $n = v + 1,m = v$, the equation holds.

Second, it is assumed that $v,i$ are even. Similarly, we have

$$\begin{aligned} \sigma_{F} (h,v) + \sigma_{F} (h,v + 1) = \sum\nolimits_{k = 1}^{(v + 2)/2} {{\text{8}d}_{k} } - 2x_{\frac{v}{2}} & = \sum\nolimits_{k = 1}^{{\frac{v + i}{2}}} {{\text{4}d}_{k} } - \sum\nolimits_{{k = \frac{v}{2} + 1}}^{{\frac{v + i}{2}}} {{\text{4}d}_{k} } + \sum\nolimits_{k = 1}^{{\frac{v + 2 - i}{2}}} {{\text{4}d}_{k} } \hfill \\ & \quad\,\,+ \sum\nolimits_{{k = \frac{v + 2 - i}{2} + 1}}^{{\frac{v + 2}{2}}} {{\text{4}d}_{k} } + 2x_{\frac{v}{2}} = \sigma_{F} (h,m) + \sigma_{F} (h,n) \hfill\\ & \quad\,\, + 4\sum\nolimits_{k = 0}^{{\frac{i - 2}{2}}} {(x_{{k + \frac{v + 2 - i}{2}}} - x_{{k + \frac{v}{2}}} } ) + 2(x_{\frac{v}{2}} - x_{{\frac{v + i}{2}}} ) \hfill \\ \end{aligned}$$

Then,

$$\sigma_{F} (h,m) + \sigma_{F} (h,n) = \sigma_{F} (h,v) + \sigma_{F} (h,v + 1) - 4\sum\nolimits_{k = 0}^{{\frac{i - 2}{2}}} {(x_{{k + \frac{v + 2 - i}{2}}} - x_{{k + \frac{v}{2}}} } ) - 2(x_{\frac{v}{2}} - x_{{\frac{v + i}{2}}} )$$

(34)

According to formula (34), it is clear that $\sigma_{F} (h,m) + \sigma_{F} (h,n) \le \sigma_{F} (h,v) + \sigma_{F} (h,v + 1)$. Furthermore, if and only if $i = 0$, i.e., $n = v,m = v + 1$, the equation holds.

The proof is completed.

Proof of Lemma 3

Without loss of generality, let $m_{1} \le m_{2}$. Then, it needs to prove that the following equation holds if $m_{1} + m_{2} = 2l.$

$$\sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) = \sigma_{F} (h,m_{1} + m_{2} ) - 2\sum\nolimits_{{k^{\prime} = 0}}^{{\frac{{m_{2} - m_{1} - 2}}{2}}} {(x_{{m_{1} + 1 + k^{\prime}}} } - x_{{l + k^{\prime}}} ) - (x_{{m_{1} }} - x_{{m_{2} }} {) }$$

(35)

Similarly, if $m_{1} + m_{2} = 2l + 1$, we have

$$\sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) = \sigma_{F} (h,m_{1} + m_{2} ) - 2\sum\nolimits_{k = 0}^{{\frac{{m_{2} - m_{1} - 3}}{2}}} {(x_{{m_{1} + 1 + k}} } - x_{l + 1 + k} ) - (x_{{m_{1} }} - x_{{m_{2} }} {)}$$

(36)

Actually, according to formula (5) and (6), it can be seen that

$$\begin{aligned} \sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) & = \sum\nolimits_{k = 1}^{{m_{1} }} {2{d}_{k} } + x_{{m_{1} }} + \sum\nolimits_{k = 1}^{{m_{1} }} {2{d}_{k} } + \sum\nolimits_{{k = m_{1} + 1}}^{{m_{2} }} {2{d}_{k} } + x_{{m_{2} }} \hfill \\ & = \sum\nolimits_{k = 1}^{{m_{1} }} {{4}d}_{k} + 4d_{{m_{1} + 1}} + \sum\nolimits_{{k = m_{1} + 2}}^{l} {{\text{4}d}_{k} } - \sum\nolimits_{{k = m_{1} + 2}}^{l} {{\text{2}d}_{k} } + \sum\nolimits_{k - l + 1}^{{m_{2} }} {{\text{2}d}_{k} } + x_{{m_{2} }} - x_{{m_{1} }} \hfill \\ & = \sum\nolimits_{k = 1}^{l} {{\text{4}d}_{k} } + 2x_{l} + \sum\nolimits_{k = l + 1}^{{m_{2} }} {2{\text{}d}_{k} } - \sum\nolimits_{{k = m_{1} + 2}}^{l + 1} {{\text{2}d}_{k} } + x_{{m_{2} }} - x_{{m_{1} }} = \sigma_{F} (h,2l) + \hfill \\ & \quad \sum\nolimits_{k = l + 1}^{m_2} {2{\text{}d}_{k} } - \sum\nolimits_{k = m_1+2}^{l + 1} {{\text{2}d}_{k} } + x_{m_2} - x_{m_1} = \sigma_{F} (h,2l) - (x_{{m_{1} }} - x_{{m_{2} }} ) - \hfill \\ & \quad 2\sum\nolimits_{k = 0}^{{\frac{{m_{2} - m_{1} - 2}}{2}}} {{(}d_{{k + m_{1} + 2}} } - d_{k + l + 1} { )}{\text{.}} \hfill \\ \end{aligned}$$

Thus, $\begin{aligned} & \sigma _{P} (h,m_{1} ) + \sigma _{P} (h,m_{2} ) = \sigma _{F} (h,m_{1} + m_{2} ) \hfill \\ & - 2\sum\nolimits_{{k^{\prime } = 0}}^{{\left( {m_{2} - m_{1} - 2} \right)/2}} {\left( {x_{{m_{1} + 1 + k^{\prime } }} - x_{{l + k^{\prime } }} } \right)-\left( {x_{{m_{1} }} - x_{{m_{2} }} } \right)} \hfill \\ \end{aligned}$ holds.

Similarly, formula (36) can be proven.

Finally, since $n + m_{1} + m_{2} = 2v$, for ease of presentation, let $m = m_{1} + m_{2}$. Without loss of generality, let $m \ge n$,$m_{2} \ge m_{1}$. Firstly, there are two cases where $v$ is even:

• ① if $n,m_{1} + m_{2}$ are odd, according to formula (30) and (36), we have

$$\begin{gathered} \sigma_{F} (h,n) + \sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) = 2\sigma_{F} (h,v) - 4\sum\nolimits_{k = 0}^{{\frac{m - n - 3}{2}}} {(x_{{k + \frac{n + 1}{2}}} - x_{{k + \frac{v}{2} + 1}} } ) - \hfill \\ 2\sum\nolimits_{k = 0}^{{\frac{{m_{2} - m_{1} - 3}}{2}}} {(x_{{m_{1} + 1 + k}} } - x_{{(m_{1} + m_{2} + 1)/2 + k}} ) - (x_{{m_{1} }} - x_{{m_{2} }} {)} \hfill \\ \end{gathered}$$

Furthermore, we have $\sigma_{F} (h,n) + \sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) < 2\sigma_{F} (h,v)$.

• ② if $n,m_{1} + m_{2}$ are even, according to formula (29) and (35), we have

$$\begin{aligned} & \sigma_{F} (h,n) + \sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) \hfill\\ &= 2\sigma_{F} (h,v) - 4\sum\nolimits_{k = 1}^{{\frac{m - n}{2} - 1}} {(x_{{k + \frac{n}{2}}} - x_{{k + \frac{v}{2}}} } ) - 2(x_{\frac{n}{2}} - x_{\frac{m}{2}} ) \hfill \\ & - 2\sum\nolimits_{{k^{\prime} = 0}}^{{\frac{{m_{2} - m_{1} - 2}}{2}}} {(x_{{m_{1} + 1 + k^{\prime}}} } - x_{{l + k^{\prime}}} ) - (x_{{m_{1} }} - x_{{m_{2} }} {) } \hfill \\ \end{aligned}$$

(37)

Furthermore, if and only if $n = m$, $m_{2} = m_{1}$, the maximum value of $\sigma_{F} (h,n) + \sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} )$ can be obtained, and it is $2\sigma_{F} (h,v)$. In this case, we have $m_{2} = m_{1} = \frac{v}{2}$. Also, it can be written as $m_{2} = m_{1} = \left\lfloor \frac{n}{2} \right\rfloor .$

Also, there are two cases where $v$ is odd:

• ① It is assumed that both $n$ and $m_{1} + m_{2}$ are odd. Let $n = v - i$,$m_{1} + m_{2} = v + i$, and it can be seen that $i$ is even. According to formula (34) and (36), we have

$$\begin{gathered} \sigma_{F} (h,n) + \sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) = 2\sigma_{F} (h,v) - 4\sum\nolimits_{k = 0}^{{\frac{i - 2}{2}}} {(x_{{k + \frac{v - i + 1}{2}}} - x_{{k + \frac{v + 1}{2}}} } ) - \hfill \\ 2\sum\nolimits_{k = 0}^{{\frac{{m_{2} - m_{1} - 3}}{2}}} {(x_{{m_{1} + 1 + k}} } - x_{{(m_{1} + m_{2} + 1)/2 + k}} ) - (x_{{m_{1} }} - x_{{m_{2} }} {)}{\text{.}} \hfill \\ \end{gathered}$$

(38)

Furthermore, if and only if $n = m_{1} + m_{2} = v$,$m_{1} = m_{2} - 1$, the maximum value of $\sigma_{F} (h,n) + \sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} )$ can be obtained, and it is $2\sigma_{F} (h,v) - (x_{{m_{1} }} - x_{{m_{2} }} )$. In this case, $m_{1} = \frac{v - 1}{2} = \left\lfloor \frac{n}{2} \right\rfloor ,m_{2}= \frac{v + 1}{2} = \left\lfloor {\frac{n + 1}{2}} \right\rfloor .$

• ② It is assumed that $n,m_{1} + m_{2}$ are even and $n \ne m_{1} + m_{2}$. Besides, let $n = v - i,$$m_{1} + m_{2} = v + i$,$m_{2} \ge m_{1}$. For ease of presentation, we have $m = m_{1} + m_{2}$. According to formula (31) and (35), the following expression can be obtained

$$\begin{aligned} & \sigma_{P} (h,m_{1} ) + \sigma_{F} (h,m_{2} ) + \sigma_{F} (h,n) \hfill\\ & = 2\sigma_{F} (h,v) - 4\sum\nolimits_{k = 0}^{{\frac{i - 3}{2}}} {{(}x_{{k + \frac{v - i}{2} + 1}} } + x_{{k + \frac{v + 1}{2}}} ) - \hfill \\ & 2(x_{{\frac{v - i}{2}}} - x_{{\frac{v + i}{2}}} ) - 2\sum\nolimits_{{k^{\prime} = 0}}^{{\frac{{m_{2} - m_{1} - 2}}{2}}} {(x_{{m_{1} + 1 + k^{\prime}}} } - x_{{l + k^{\prime}}} ) - (x_{{m_{1} }} - x_{{m_{2} }} {) } \hfill \\ \end{aligned}$$

(39)

Furthermore, if and only if i = 1 ($|m_{1} + m_{2} - n| = 2$) and $m_{1} = m_{2}$, the maximum value of $\sigma_{P} (h,m_{1} ) + \sigma_{F} (h,m_{2} ) + \sigma_{F} (h,n)$ can be obtained. Specifically, the maximum value is $2\sigma_{F} (h,v) - 2(x_{{{{\left( {v - 1} \right)} \mathord{\left/ {\vphantom {{\left( {v - 1} \right)} 2}} \right. \kern-\nulldelimiterspace} 2}}} - x_{{{{\left( {v + 1} \right)} \mathord{\left/ {\vphantom {{\left( {v + 1} \right)} 2}} \right. \kern-\nulldelimiterspace} 2}}} )$. In addition, it can be seen that $x_{(v - 1)/2} - x_{(v + 1/2)} < 2(x_{(v - 1)/2} - x_{(v + 1/2)} )$. Thus, $\sigma_{P} (h,m_{1} ) + \sigma_{F} (h,m_{2} ) + \sigma_{F} (h,n) < 2\sigma_{F} (h,v) - (x_{(v - 1)/2} - x_{(v + 1)/2} )$.

Thus, if v is odd, the maximum value of $\sigma_{P} (h,m_{1} ) + \sigma_{F} (h,m_{2} ) + \sigma_{F} (h,n)$ can be obtained if and only if $n = m$,$m_{1} = m_{2} - 1$; If v is even, the maximum value of $\sigma_{P} (h,m_{1} ) + \sigma_{F} (h,m_{2} ) + \sigma_{F} (h,n)$ can be obtained if and only if $n = m$, $m_{2} = m_{1} .$

(2) For $n + m_{1} + m_{2} = 2v + 1$, it is assumed that $n$ is odd. Thus $m_{1} + m_{2}$ is even. According to formula (35), we have

$$\begin{aligned} & \sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) + \sigma_{F} (h,n) \hfill \\ &= \sigma_{F} (h,n) + \sigma_{F} (h,m_{1} + m_{2} ) \hfill \\ & - 2\sum\nolimits_{{k^{\prime} = 0}}^{{\frac{{m_{2} - m_{1} - 2}}{2}}} {(x_{{m_{1} + 1 + k^{\prime}}} } - x_{{l + k^{\prime}}} ) - (x_{{m_{1} }} - x_{{m_{2} }} {) } \hfill \\ \end{aligned}$$

(40)

According to formula (40) and Lemma 2, if and only if $|n - m_{1} + m_{2} | = 1$,$m_{1} = m_{2}$, the maximum value of $\sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) + \sigma_{F} (h,n)$ can be obtained, and it is $\sigma_{F} (h,v) + \sigma_{F} (h,v + 1)$. In this case, if $v$ is odd, then $v$ = $n$,$m_{1} + m_{2} = v + 1$,$m_{1} = m_{2} = \frac{n + 1}{2} = \left\lfloor {\frac{n + 1}{2}} \right\rfloor$. Otherwise, if $v$ is even, then $n = v + 1$,$m_{1} + m_{2} = v$, $m_{1} = m_{2} = \frac{n}{2} = \left\lfloor \frac{n}{2} \right\rfloor$.

Second, it is assumed that $n$ is even. Thus, $m_{1} + m_{2}$ is odd. According to formula (36), we have

$$\begin{gathered} \sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) + \sigma_{F} (h,n) = \sigma_{F} (h,m_{1} + m_{2} ) + \hfill \\ \sigma_{F} (h,n) - 2\sum\nolimits_{k = 0}^{{\frac{{m_{2} - m_{1} - 3}}{2}}} {(x_{{m_{1} + 1 + k}} } - x_{l + 1 + k} ) - (x_{{m_{1} }} - x_{{m_{2} }} {)} \hfill \\ \end{gathered}$$

(41)

Since $m_{1} + m_{2}$ is odd, according to formula (41) and Lemma 2, we have $\sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) + \sigma_{F} (h,n) < \sigma_{F} (h,v) + \sigma_{F} (h,v + 1)$. It indicates the maximum value of $\sigma_{P} (h,m_{1} ) + \sigma_{P} (h,m_{2} ) + \sigma_{F} (h,n)$ cannot be obtained in this case.

The Proof is completed.

Proof of Lemma 4

According to formula (5) and (6), we have

$$\sigma_{P} (h,n + 1) = \sigma_{P} (h,n) + x_{n + 1} + x_{n}$$

(42)

$$\begin{gathered} \sigma_{F} (h,n + 1) - \sigma_{F} (h,n)) = \left\{ \begin{gathered} 2x_{{\frac{n + 1}{2}}} ,n{\text{ is an odd}} \hfill \\ 2x_{\frac{n}{2}} ,n{\text{ is an even}} \hfill \\ \end{gathered} \right.. \hfill \\ \hfill \\ \end{gathered}$$

(43)

(1) From formula (42) and (43), we have $\sigma_{F} (h,n + 1) + \sigma_{F} (h,n) = 2\sigma_{F} (h,n + 1) - 2x_{\frac{n}{2}}$ and $\sigma_{P} (h,m + 1) + \sigma_{P} (h,m) = 2\sigma_{P} (h,m) + x_{m + 1} + x_{m}$. Thus,

$$\begin{gathered} \sigma _{F} (h,n) + \sigma _{F} (h,n + 1) + \sigma _{P} (h,m + 1) + \sigma _{P} (h,m) \hfill \\ = 2\left( {\sigma _{F} (h,n + 1} \right) + \sigma _{P} (h,m)) - \left( {2x_{{n/2}} - x_{{m + 1}} - x_{m} } \right) \hfill \\ \end{gathered}$$

(44)

Besides, we have $m = \frac{n}{2}$. Furthermore, according to Lemma 1 and formula (44), it can be seen that formula (9) holds.

Similarly, we have

$$\sigma_{F} (h,n) + \sigma_{P} (h,m + 1) + \sigma_{P} (h,m) = \sigma_{F} (h,n + 1) + 2\sigma_{P} (h,m)) - (2x_{n/2} - x_{m + 1} - x_{m} ).$$

Also, it can be seen that formula (10) holds.

(2) Similarly, we have

$$\begin{aligned} & \sigma_{F} (h,n) + \sigma_{F} (h,n + 1) + \sigma_{P} (h,m + 1) + \sigma_{P} (h,m) \hfill \\ & = 2(\sigma_{F} (h,n) + \sigma_{P} (h,m + 1)) - (x_{m + 1} + x_{m} - 2x_{(n + 1)/2} ) \hfill \\ \end{aligned}$$

(45)

According to Lemma 1 and formula (45), it can be seen that formula (11) holds.

The proof is completed.