﻿ 分数阶累加时滞GM(1,<i>N</i>,<i>τ</i>)模型及其应用
 文章快速检索 高级检索

1. 武汉理工大学 理学院, 武汉 430070;
2. 武汉理工大学 可靠性研究中心, 武汉 430070

Fractional order accumulation time-lag GM(1,N,τ) model and its application
MAO Shu-hua1,2, GAO Ming-yun1, XIAO Xin-ping1
1. College of Science, Wuhan University of Technology, Wuhan 430070, China;
2. Reliability Engineering Institute, Wuhan University of Technology, Wuhan 430070, China
Abstract:Based on the time lag effect of system, this paper constructs the time-lag GM(1,N,τ) model, and provides its least squares parameter estimation formula and analytical solution. Then introducing fractional order accumulation generation operator, GM(1,N,τ) is transformed into fractional order accumulation time-lag GM(1,N,τ), offering adjacent integer points weighted method when time-lag value is not integer, determining the time-lag value via particle swarm optimization. Finally, we test the original GM(1,N,τ) and fractional order accumulation time-lag GM(1,N,τ) by using the real data of R&D investment and GDP in Wuhan from 1995 to 2008. The example indicates that the accuracy of fractional order accumulation time-lag GM(1,N,τ) is more satisfactory.
Key words: grey model     GM(1,N,τ) model     fractional order accumulation     time-lag     particle swarm optimization

0 引言

GM$(1,N)$模型,可以看成是GM(1,1)模型的推广形式, 文献[3]假定因子项为常数或者变化很小, 进而退化为一个特殊的GM(1,1)模型, 再运用GM(1,1)模型求解公式进行拟合和预测. 文献[4]基于多因子灰色模型的精确级差分析,建立了MGM$(1,N)$模型; 文献[5]在此基础上,将误差融入级差格式, 基于理想状态时的相对误差提出了MGMp$(1,N)$模型; 文献[6] 结合自记忆方程,建立了GM(1,2)自记忆模型; 文献[7]在白化GM$(1,N)$方程中考虑因素间线性关系派生出了GM$(1,N,x_{i}^{(0)})$和GM$(1,N,x_{i}^{(1)})$模型.

20世纪70年代以来, 研究人员发现分数阶微积分可以作为一种很好的描述与刻画手段, 应用于分形几何、幂律现象与记忆过程等相关现象或过程. 文献[11] 将分数阶累加生成应用到灰色模型中, 建立了基于分数阶累加生成技术的灰色系统模型.

1 时滞GM$(1,N,\tau)$模型

GM$(1,N,\tau)$是多变量时滞灰色模型的符号. GM$(1,N,\tau)$ 中包括一个行为变量,记为$y$; $N-1$个因子变量,记为${{x}_{i}}$, $i=1,2,\cdots ,N-1$; 以及因子变量与行为变量的时间差, 即时滞值$\tau$.

 $${{y}^{(0)}}(k)+a{{z}_{y}}^{(1)}(k)=\sum\limits_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(1)}\left( k-\tau \right)}$$ (1)

 $\left\{ \begin{array}{l} \displaystyle\frac{{\partial s}}{{\partial a}} = 2\sum\limits_{k = \tau + 2}^n {\left( {{y^{(0)}}(k) - \sum\limits_{i = 1}^{N - 1} {{b_i} x_i^{(1)}(k - \tau )} + az_y^{(1)}(k)} \right)z_y^{(1)}(k) = 0}, \\ \displaystyle\frac{{\partial s}}{{\partial {b_1}}} = - 2\sum\limits_{k = \tau + 2}^n {\left( {{y^{(0)}}(k) - \sum\limits_{i = 1}^{N - 1} {{b_i}x_i^{(1)} (k - \tau )} + az_y^{(1)}(k)} \right)x_1^{(1)}(k - \tau ) = 0}, \\ \displaystyle\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \vdots \\ \displaystyle\frac{{\partial s}}{{\partial {b_{N - 1}}}} = - 2\sum\limits_{k = \tau + 2}^n {\left( {{y^{(0)}}(k) - \sum\limits_{i = 1}^{N - 1} {{b_i}x_i^{(1)}(k - \tau )} + az_y^{(1)}(k)} \right)x_{N - 1}^{(1)}(k - \tau ) = 0}. \end{array} \right.$

 $\sum\limits_{k = \tau + 2}^n {\left( {{y^{(0)}}(k) - \sum\limits_{i = 1}^{N - 1} {{b_i}x_i^{(1)}(k - \tau )} + az_y^{(1)}(k)} \right)} z_y^{(1)}(k) = 0,\\ \displaystyle\sum\limits_{k = \tau + 2}^n {\left( {{y^{(0)}}(k) - \sum\limits_{i = 1}^{N - 1} {{b_i}x_i^{(1)}(k - \tau )} + az_y^{(1)}(k)} \right)} x_1^{(1)}(k - \tau ) = 0,\\ \begin{array}{*{20}{c}} {}&{}&{}&{\begin{array}{*{20}{c}} {}&{}&{}& \qquad \vdots \end{array}} \end{array}\\ \displaystyle\sum\limits_{k = \tau + 2}^n {\left( {{y^{(0)}}(k) - \sum\limits_{i = 1}^{N - 1} {{b_i}x_i^{(1)}(k - \tau )} + az_y^{(1)}(k)} \right)} x_{N - 1}^{(1)}(k - \tau ) = 0.$

 $$\hat{{P}}={{\left( {{B}^{\text{T}}}\mathbf{B} \right)}^{-1}}{{B}^{\text{T}}}{Y}.$$

 $$\frac{{d}{{y}^{(1)}}}{{ d}t}+a{{y}^{(1)}}\left(t\right)=\sum\limits_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(1)}\left( t-\tau \right)}$$ (3)

 $${{y}^{(1)}}\left( t+1 \right)=\left( {{y}^{(1)}}(\tau +1)-\sum\limits_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(1)}(t+1)} \right){{\rm e}^{-a(t-\tau )}}+\frac{1}{a}\sum\limits_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(1)}(t+1)},\quad t\ge \tau +1$$ (4)

 \begin{eqnarray*} & \displaystyle\frac{{\rm d}{y^{(1)}}}{{\rm d}t}+a{y^{(1)}}\left(t\right)=\sum\limits_{i=1}^{N-1}{{b_{i}}x_{i}^{(1)}\left(t-\tau \right)},\\ & \displaystyle\frac{{\rm d}{{y}^{(1)}}}{a{{y}^{(1)}}+\sum_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(1)}\left( t-\tau \right)}}={\rm d}t,\quad \text{即} \int{\frac{{\rm d}{{y}^{(1)}}}{a{{y}^{(1)}}+\sum_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(1)}\left( t-\tau \right)}}}=\int{{\rm d}t}. \end{eqnarray*}

 $$\displaystyle a{{y}^{(1)}}+\frac{1}{a}\sum\limits_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(1)}\left( t-\tau \right)}={{C}_{1}}\exp \left( at \right),$$

 $\displaystyle{{\hat{y}}^{(1)}}\left( t+1 \right)=\left( {{y}^{(1)}}(\tau +1)-\sum\limits_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(1)}(t+1)} \right){{\rm e}^{-a(t-\tau )}}+\frac{1}{a}\sum\limits_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(1)}(t+1)}.$

 $${{\hat{y}}^{\left( 0 \right)}}\left( t+1 \right)={{\hat{y}}^{\left( 1 \right)}}\left( t+1 \right)-{{\hat{y}}^{\left( 1 \right)}}\left( t \right),\quad t\ge \tau +1$$ (5)

2 模型的误差计算

 $$\text{MAPE}=\frac{1}{n}\sum\limits_{k=1}^{n}{( {| {{y}^{\left( 0 \right)}}\left( k \right)-{{{\hat{y}}}^{\left( 0 \right)}}\left( k \right)|}/{{{y}^{\left( 0 \right)}}\left( k \right)} )\times 100\%}$$ (6)

3 分数阶累加GM$(1,N,\tau)$模型 3.1 分数阶累加生成矩阵

 $$\displaystyle{{{A}}^{r}}=\left[\begin{matrix} 1 & 0 & 0 & \cdots & 0 \\ r & 1 & 0 & \cdots & 0 \\ \frac{r(r+1)}{2!} & r & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{r(r+1)\cdots (r+n-2)}{(n-1)!} & \frac{r(r+1)\cdots (r+n-3)}{(n-2)!} & \frac{r(r+1)\cdots (r+n-4)}{(n-3)!} & \cdots & 1 \\ \end{matrix} \right].$$

{\HT 注} 当$r$为正整数时,${{{A}}^{r}}$等同于矩阵运算, 是矩阵${A}$的$r$次方; 当$r$不是正整数时, ${{{A}}^{r}}$只是一个记号. 例如,当$n=4$时, 阶数为$\frac{1}{2},\frac{5}{4}$ 时,累加生成矩阵如下
 $${{{A}}^{\frac{1}{2}}}=\left[\begin{matrix} 1 & 0 & 0 & 0 \\ 0.5 & 1 & 0 & 0\\ 0.375 & 0.5 & 1 & 0 \\ 0.3125 & 0.375 & 0.5 & 1 \end{matrix} \right],\qquad {{{A}}^{\frac{5}{4}}}=\left[\begin{matrix} 1 & 0 & 0 & 0\\ 1.25 & 1 & 0 &0 \\ 1.4063 & 1.25 & 1 & 0\\ 1.5234 & 1.4063 & 1.25 &1 \end{matrix} \right].$$

 $${{{A}}^{-\frac{1}{2}}}=\left[ \begin{matrix} 1 & 0 & 0 & 0 \\ -0.5 & 1 & 0 & 0\\ -0.125 & -0.5 & 1 & 0 \\ -0.0625 & -0.125 & -0.5 & 1 \end{matrix} \right],\qquad {{{A}}^{-\frac{5}{4}}}=\left[ \begin{matrix} 1 & 0 & 0 & 0\\ -1.25 & 1 & 0 &0 \\ 0.1563 & -1.25 & 1 & 0\\ 0.0391 & 0.1563 & -1.25 &1 \end{matrix} \right].$$

3.2 分数阶累加GM$(1,N,\tau)$模型

 $$\frac{{\rm d}{{y}^{(r)}}}{{\rm d}t}+a{{y}^{(r)}}= \sum\limits_{i=1}^{N-1}{{{b}_{i}}}x_{i}^{(r)}\left( t-\tau \right)$$ (7)

 $${{y}^{\left( r-1 \right)}}\left( k \right)+a{{z}_{y}}^{\left( r \right)}\left( k \right)=\sum\limits_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(r)}\left( k-\tau \right)},\quad k=\tau +2,\tau +3,\cdots ,n$$ (8)

 $$\hat{{P}}={{( {{B}^{*}}^{\rm T}{{B}^{*}} )}^{\mathbf{-1}}}{{B}^{*}}^{\rm T}{{{Y}}^{*}}$$ (9)

$\text{其中}\,{{B}^{*}}=\left[\begin{matrix} {{z}_{y}}^{(r)}\left( \tau +2 \right) & x_{1}^{(r)}\left( 2 \right) & \cdots & x_{N-1}^{(r)}\left( 2 \right) \\ {{z}_{y}}^{(r)}\left( \tau +3 \right) & x_{1}^{(r)}\left( 3 \right) & \cdots & x_{N-1}^{(r)}\left( 3 \right) \\ \vdots & \vdots & \ddots & \vdots \\ {{z}_{y}}^{(r)}\left( n \right) & x_{1}^{(r)}\left( n-\tau \right) & \ldots & x_{N-1}^{(r)}\left( n-\tau \right) \\ \end{matrix} \right]$,${{{Y}}^{*}}=\left[\begin{matrix} {{y}^{(r)}}(\tau +2)-{{y}^{(r)}}(\tau +1) \\ {{y}^{(r)}}(\tau +3)-{{y}^{(r)}}(\tau +2) \\ \vdots \\ {{y}^{(r)}}(n)-{{y}^{(r)}}(n-1) \\ \end{matrix} \right].$

 $${{\hat{y}}^{(r)}}\left( t+1 \right)=\left( {{y}^{(r)}}(\tau +1)-\sum\limits_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(r)}(t+1)} \right){{\rm e}^{-a(t-\tau )}}+\frac{1}{a}\sum\limits_{i=1}^{N-1}{{{b}_{i}}x_{i}^{(r)}(t+1)}$$ (10)

 $${{\hat{y}}^{(0)}}\text{=}{{{A}}^{-r}}{{\hat{y}}^{(r)}}$$ (11)

3.3 非整数时滞值下模型完善

 $$x\left( p \right)=\left( q+1-p \right)x\left( q \right)+(p-q)x\left( q+1 \right)$$ (12)

 $$\displaystyle x(1.8)=(2-1.8)x(1)+(1.8-1)x(2)=0.2x(1)+0.8x(2).$$

 $$\begin{array}{r@{~}l} {{\hat{y}}^{(r)}}\left( t+1 \right)=& \left( {{y}^{(r)}}(\tau +1)-\sum\limits_{i=1}^{N-1}{{{b}_{i}}\left( {{k}_{1}}x_{i}^{(r)}\left( t-\left[ \tau \right] \right)+{{k}_{2}}x_{i}^{(r)}\left( t-\left[\tau \right]+1 \right) \right)} \right){{\rm e}^{-a(t-\tau )}}+ \\ & \displaystyle{+\frac{1}{a}\sum\limits_{i=1}^{N-1}{{{b}_{i}}\left( {{k}_{1}}x_{i}^{(r)}\left( t-\left[\tau \right] \right)+{{k}_{2}}x_{i}^{(r)}\left( t-\left[\tau \right]+1 \right) \right)}} \end{array}$$ (13)

3.4 模型阶数的确定

 $$\begin{array}{r@{~}l} & \min \ \operatorname{MAPE}(r) \\ & {\rm s.t.}\ \quad r>0 \\ \end{array}$$

4 实例分析

4.1 时滞值的估计

4.2 $r$阶GM$(1,N,\tau)$结果比较

 \begin{eqnarray} {{\hat{y}}^{(r)}}(k\!+\!1)=(0.4308\!-\!0.2264( {{x}^{(r)}}(k\!-\!1 )+{{x}^{(r)}}(k\!-\!2 )))\\ {\operatorname{e}}^{-0.5357(k-2)}+\text{0.2264}( {{x}^{(r)}}(k\!-\!1 )+{{x}^{(r)}}(k\!-\!2)) \end{eqnarray} (14)
 \begin{eqnarray} {{\hat{y}}^{(0)}}={{{A}}^{-0.039659}}{{\hat{y}}^{(r)}} \end{eqnarray} (15)

4.3 结论及分析

 [1] 邓聚龙.灰理论基础[M].武汉:华中科技大学出版社, 2008: 282-300.Deng Julong. Grey theory foundation[M]. Wuhan: Huazhong University of Science & Technology Press, 2008: 282-300. [2] 肖新平,宋中民,李峰.灰技术基础及其应用[M].北京:科学出版社, 2005: 117-123.Xiao Xinping, Song Zhongmin, Li Feng. Grey technology foundation and its application[M]. Beijing: Science Press, 2005: 117-123. [3] 刘思峰,党耀国,方志耕,等.灰色系统理论及其应用[M]. 5版.北京:科学出版社, 2010: 169-171.Liu Sifeng, Dang Yaoguo, Fang Zhigeng, et al. Grey system theory and its application[M]. 5th ed. Beijing: Science Press, 2010: 169-171. [4] 翟军,盛建明,冯英浚. MGM(1,N)灰色模型及应用[J].系统工程理论与实践, 1997, 17(5): 109-113.Zhai Jun, Sheng Jianming, Feng Yingjun. The grey model MGM(1,N) and its application[J]. Systems Engineering —— Theory & Practice, 1997, 17(5): 109-113. [5] 李小霞,同小军,陈绵云. 多因子灰色MGMp(1,N)优化模型[J]. 系统工程理论与实践, 2003, 23(4): 47-51.Li Xiaoxia, Tong Xiaojun, Chen Mianyun. MGMp(1,N) optimization model[J]. Systems Engineering —— Theory & Practice, 2003, 23(4): 47-51. [6] 张高锋,姜国辉,付永锋,等. 灰色 GM(1,N)自记忆模型研究[J]. 沈阳农业大学学报, 2009, 40(2): 210-214.Zhang Gaofeng, Jiang Guohui, Fu Yongfeng, et al. Establishment of the grey GM(1,N) self-memory model[J]. Journal of Shenyang Agricultural University, 2009, 40(2): 210-214. [7] 周伟,方志耕.非线性优化GM(1,N)模型及其应用研究[J].系统工程与电子技术, 2010, 32(2): 317-320.Zhou Wei, Fang Zhigeng. Nonlinear optimization method of gray GM(1,N) model and application[J]. Systems Engineering and Electronics, 2010, 32(2): 317-320. [8] 翟军,冯英俊,盛建明.带有时滞的GM(1,2)模型及应用[J].系统工程, 1996, 14(6): 66-68.Zhai Jun, Feng Yingjun, Sheng Jianming. A time lag existing GM(1,2) model and its application[J]. Systems Engineering, 1996, 14(6): 66-68. [9] 郝永红,黄登宇,高红波,等. 娘子关泉水流量的GM(1,2)时滞预测模型[J].中国岩溶, 2004(3): 43-47.Hao Yonghong, Huang Dengyu, Gao Hongbo, et al. GM(1,2) time lag model for discharge of Niangziguan Spring[J]. Carsologica Sinica, 2004(3): 43-47. [10] Hao Y H, Wang Y J, Zhao J J, et al. Grey system model with time lag and application to simulation of karst spring discharge[J]. Grey Systems: Theory and Application, 2011, 1(1): 47-56. [11] Wu L F, Liu S F, Yao L G, et al. Grey system model with the fractional order accumulation[J]. Communications in Nonlinear Science and Numerical Simulation, 2013, 18(7): 1775-1785. [12] 屈婉玲.组合数学[M].北京:北京大学出版社, 1989, 11: 48-51.Qu Wanling. Combinatorial mathematics[M]. Beijing: Beijing University Press, 1989, 11: 48-51. [13] 武汉市统计局.武汉市统计年鉴[R].北京:中国统计出版社, 2010: 3-20, 56-70.Wuhan Statistics Bureau. Wuhan statistical yearbook[R]. Beijing: China Statistics Press, 2010: 3-20, 56-70. [14] 武汉市科学技术局,武汉市统计局.武汉科技统计年鉴[R].武汉:武汉市统计局, 2008: 29-36.Wuhan Science and Technology Bureau, Wuhan Statistics Bureau. Wuhan statistical yearbook of science and technology[R]. Wuhan: Wuhan Statistics Bureau, 2008: 29-36. [15] 肖新平, 毛树华. 灰预测与决策方法[M].北京:科学出版社, 2013: 86-88.Xiao Xinping, Mao Shuhua. The method of grey prediction and decision[M]. Beijing: Science Press, 2013: 86-88.
0

#### 文章信息

MAO Shu-hua, GAO Ming-yun, XIAO Xin-ping

Fractional order accumulation time-lag GM(1,N,τ) model and its application

Systems Engineering - Theory & practice, 2015, 35(2): 430-436.