《1 引言》

1 引言

《2 GM (1, 1) 模型背景值的优化》

2 GM (1, 1) 模型背景值的优化

X (1) 的紧邻均值生成序列Z (1) ={z (1) (1) , z (1) (2) , …, z (1) (n) }, 其中

$\begin{array}{l}z{}^{\left(1\right)}\left(k\right)=\left[x{}^{\left(1\right)}\left(k\right)+x{}^{\left(1\right)}\left(k-1\right)\right]/2,\\ k=2,3,\cdots ,n。\end{array}$

$\frac{\text{d}x{}^{\left(1\right)}}{\text{d}t}+ax{}^{\left(1\right)}=b\text{ }\text{ }\text{ }\left(1\right)$

$x{}^{\left(0\right)}\left(k\right)+az{}^{\left(1\right)}\left(k\right)=b\text{ }\text{ }\text{ }\left(2\right)$

$\stackrel{^}{x}{}^{\left(1\right)}\left(t\right)=\left(x{}^{\left(0\right)}\left(1\right)-\frac{b}{a}\right)\text{e}{}^{-a\left(t-1\right)}+\frac{b}{a}\phantom{\rule{0.25em}{0ex}}$

t=k (k=1, 2, …, n) 处的值来逼近或描述x (1) (k) 。

$\stackrel{^}{\mathbit{a}}=\left(\begin{array}{l}a\\ b\end{array}\right)=\left(\mathbit{B}{}^{\text{Τ}}\mathbit{B}\right){}^{-1}\mathbit{B}{}^{\text{Τ}}\mathbit{Y},\text{ }\text{ }\text{ }\left(3\right)$

$\mathbit{Y}=\left(\begin{array}{l}x{}^{\left(0\right)}\left(2\right)\\ x{}^{\left(0\right)}\left(3\right)\\ ⋮\\ x{}^{\left(0\right)}\left(n\right)\end{array}\right),\text{ }\mathbit{B}=\left(\begin{array}{cc}-z{}^{\left(1\right)}\left(2\right)& 1\\ -z{}^{\left(1\right)}\left(3\right)& 1\\ ⋮& ⋮\\ -z{}^{\left(1\right)}\left(n\right)& 1\end{array}\right)。$

${\int }_{k-1}^{k}\frac{\text{d}x{}^{\left(1\right)}}{\text{d}t}\text{d}t+a\underset{k-1}{\overset{k}{\int }}x{}^{\left(1\right)}\text{d}t=b,$

$x{}^{\left(1\right)}\left(k\right)-x{}^{\left(1\right)}\left(k-1\right)+a{\int }_{k-1}^{k}x{}^{\left(1\right)}\text{d}t=b,$

$x{}^{\left(0\right)}\left(k\right)+a{\int }_{k-1}^{k}x{}^{\left(1\right)}\text{d}t=b。\text{ }\text{ }\text{ }\left(4\right)$

《图1》

$z{}^{\left(1\right)}\left(k\right)=\left[x{}^{\left(1\right)}\left(k\right)+x{}^{\left(1\right)}\left(k-1\right)\right]/2$

$\phantom{\rule{0.25em}{0ex}}z{}^{\left(1\right)}\left(k\right)={\int }_{k-1}^{k}x{}^{\left(1\right)}\left(t\right)\text{d}t。\text{ }\text{ }\text{ }\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\left(5\right)$

x (1) (t) =BeAt, 其中A, B为待定常数, 且满足

$\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}x{}^{\left(1\right)}\left(k\right)=B\text{e}{}^{Ak},\left(k=1,2,\cdots ,n\right)。$

x (1) (t) =BeAt代入式 (5) 得

$\begin{array}{l}z{}^{\left(1\right)}\left(k\right)=\underset{k-1}{\overset{k}{\int }}B\text{e}{}^{At}\text{d}t=\frac{1}{A}\left[B\text{e}{}^{Ak}-B\text{e}{}^{A\left(k-1\right)}\right]=\\ \text{ }\text{ }\text{ }\frac{1}{A}\left[x{}^{\left(1\right)}\left(k\right)-x{}^{\left(1\right)}\left(k-1\right)\right]。\text{ }\text{ }\text{ }\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\left(6\right)\end{array}$

$A=\text{l}\text{n}x{}^{\left(1\right)}\left(k\right)-\text{l}\text{n}x{}^{\left(1\right)}\left(k-1\right)。\text{ }\text{ }\text{ }\left(7\right)$

$\begin{array}{l}z{}^{\left(1\right)}\left(k\right)=\frac{x{}^{\left(1\right)}\left(k\right)-x{}^{\left(1\right)}\left(k-1\right)}{\mathrm{ln}\phantom{\rule{0.25em}{0ex}}x{}^{\left(1\right)}\left(k\right)-\mathrm{ln}\phantom{\rule{0.25em}{0ex}}x{}^{\left(1\right)}\left(k-1\right)},\\ \text{ }k=2,3,\cdots ,n。\text{ }\text{ }\text{ }\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\left(8\right)\end{array}$

$\begin{array}{l}z{}^{\left(1\right)}\left(k\right)=\frac{x{}^{\left(1\right)}\left(k\right)-x{}^{\left(1\right)}\left(k-1\right)}{\mathrm{ln}\phantom{\rule{0.25em}{0ex}}x{}^{\left(1\right)}\left(k\right)-\mathrm{ln}\phantom{\rule{0.25em}{0ex}}x{}^{\left(1\right)}\left(k-1\right)}‚\\ \text{ }k=2,3,\cdots ,n‚\end{array}$

$\mathbit{Y}=\left(\begin{array}{l}\mathbit{x}{}^{\left(0\right)}\left(2\right)\\ \mathbit{x}{}^{\left(0\right)}\left(3\right)\\ ⋮\\ \mathbit{x}{}^{\left(0\right)}\left(\mathbit{n}\right)\end{array}\right),\text{ }\mathbit{B}=\left(\begin{array}{cc}-z{}^{\left(1\right)}\left(2\right)& 1\\ -z{}^{\left(1\right)}\left(3\right)& 1\\ ⋮& ⋮\\ -z{}^{\left(1\right)}\left(n\right)& 1\end{array}\right)‚$

1) 灰色微分方程x (0) (k) +az (1) (k) =b的最小二乘估计参数满足

$\stackrel{^}{\mathbit{a}}=\left(\begin{array}{l}a\\ b\end{array}\right)=\left(\mathbf{B}{}^{\text{Τ}}\mathbit{B}\right){}^{-1}\mathbit{B}{}^{\text{Τ}}\mathbit{Y};$

2) 灰色微分方程x (0) (k) +az (1) (k) =b的白化方程$\frac{\text{d}x{}^{\left(1\right)}}{\text{d}t}+ax{}^{\left(1\right)}=b$的时间响应式为

$\stackrel{^}{x}\left(t\right)=\left(x{}^{\left(0\right)}\left(1\right)-\frac{b}{a}\right)\text{e}{}^{-a\left(t-1\right)}+\frac{b}{a};$

3) 灰色微分方程 x (0) (k) +az (1) (k) =b时间响应式为

$\begin{array}{l}\stackrel{^}{x}{}^{\left(1\right)}\left(k+1\right)=\left(x{}^{\left(0\right)}\left(1\right)-\frac{b}{a}\right)\text{e}{}^{-ak}+\frac{b}{a}‚\\ \text{ }k=1,2,\cdots ,n；\end{array}$

4) 还原值为

$\begin{array}{l}\stackrel{^}{x}{}^{\left(0\right)}\left(k+1\right)=\stackrel{^}{x}{}^{\left(1\right)}\left(k+1\right)-\stackrel{^}{x}{}^{\left(1\right)}\left(k\right)‚\\ \text{ }k=1,2,\cdots ,n。\end{array}$

$\begin{array}{l}z{}^{\left(1\right)}\left(k\right)=\underset{x{}^{\left(1\right)}\left(k\right)\to x{}^{\left(1\right)}\left(k-1\right)}{\mathrm{lim}}\left[\left(x{}^{\left(1\right)}\left(k\right)-x{}^{\left(1\right)}\left(k-1\right)\right)/\\ \left(\mathrm{ln}\phantom{\rule{0.25em}{0ex}}x{}^{\left(1\right)}\left(k\right)-\mathrm{ln}\phantom{\rule{0.25em}{0ex}}x{}^{\left(1\right)}\left(k-1\right)\right)\right]=\\ \underset{x{}^{\left(1\right)}\left(k\right)\to x{}^{\left(1\right)}\left(k-1\right)}{\mathrm{lim}}\left(x{}^{\left(1\right)}\left(k\right)=x{}^{\left(1\right)}\left(k-1\right)。\end{array}$

《3 数据模拟精度的比较》

3 数据模拟精度的比较

1) 原始序列

2) 以上述原始序列分别建立原GM (1, 1) 模型及新GM (1, 1) 模型, 并求出相应的时间响应式。

GM (1.1) 模型时间响应式:

$\begin{array}{l}\begin{array}{l}\stackrel{^}{x}{}_{1}^{\left(1\right)}\left(k+1\right)=10.509647\text{e}{}^{0.09907k}-9.509647‚\\ \stackrel{^}{x}{}_{2}^{\left(1\right)}\left(k+1\right)=5.516431\text{e}{}^{0.1993401k}-4.516431‚\\ \stackrel{^}{x}{}_{3}^{\left(1\right)}\left(k+1\right)=3.85832\text{e}{}^{0.297769k}-2.858321‚\\ \stackrel{^}{x}{}_{4}^{\left(1\right)}\left(k+1\right)=3.033199\text{e}{}^{0.394752k}-2.033199‚\\ \stackrel{^}{x}{}_{5}^{\left(1\right)}\left(k+1\right)=2.541474\text{e}{}^{0.4898382k}-1.541474‚\\ \stackrel{^}{x}{}_{6}^{\left(1\right)}\left(k+1\right)=2.216359\text{e}{}^{0.582626k}-1.216359‚\\ \stackrel{^}{x}{}_{7}^{\left(1\right)}\left(k+1\right)=1.8159718\text{e}{}^{0.7598991k}-0.8159718‚\end{array}\\ \begin{array}{l}\stackrel{^}{x}{}_{8}^{\left(1\right)}\left(k+1\right)=1.581973\text{e}{}^{0.9242348k}-0.5819733‚\\ \stackrel{^}{x}{}_{9}^{\left(1\right)}\left(k+1\right)=1.287182\text{e}{}^{1.270296k}-0.2871283‚\\ \stackrel{^}{x}{}_{10}^{\left(1\right)}\left(k+1\right)=1.198197\text{e}{}^{1.432596k}-0.1981966。\end{array}\end{array}$

Table 1 The original datum

《表1》

 -a i x(0)i${}_{i}^{\left(0\right)}$ (1) x(0)i${}_{i}^{\left(0\right)}$ (2) x(0)i${}_{i}^{\left(0\right)}$ (3) x(0)i${}_{i}^{\left(0\right)}$ (4) x(0)i${}_{i}^{\left(0\right)}$ (5) x(0)i${}_{i}^{\left(0\right)}$ (6) 0.1 1 1.0 1.105 2 1.221 4 1.349 9 1.491 8 1.648 7 0.2 2 1.0 1.221 4 1.491 8 1.822 1 2.225 5 2.718 3 0.3 3 1.0 1.349 9 1.822 1 2.459 6 3.320 1 4.481 7 0.4 4 1.0 1.491 8 2.225 5 3.320 1 4.953 0 7.389 0 0.5 5 1.0 1.648 7 2.718 3 4.481 7 7.389 0 12.182 5 0.6 6 1.0 1.822 1 3.320 1 6.049 6 11.023 2 20.085 5 0.8 7 1.0 2.225 5 4.953 0 11.023 2 24.532 5 54.598 2 1.0 8 1.0 2.718 3 7.389 0 20.085 5 54.598 2 148.413 2 1.5 9 1.0 4.481 7 20.085 5 90.017 1 403.428 8 1 808.042 4 1.8 10 1.0 6.049 6 36.598 2 221.406 4 1 339.430 8 8 103.083 9

$\begin{array}{l}\stackrel{^}{x}{}_{1}^{\left(1\right)}\left(k+1\right)=10.634472\text{e}{}^{0.099318k}-9.634472‚\\ \stackrel{^}{x}{}_{2}^{\left(1\right)}\left(k+1\right)=5.594064\text{e}{}^{0.199117k}-4.594064‚\\ \stackrel{^}{x}{}_{3}^{\left(1\right)}\left(k+1\right)=3.922068\text{e}{}^{0.299149k}-2.922068‚\\ \stackrel{^}{x}{}_{4}^{\left(1\right)}\left(k+1\right)=3.091401\text{e}{}^{0.399299k}-2.091401‚\\ \stackrel{^}{x}{}_{5}^{\left(1\right)}\left(k+1\right)=2.59745\text{e}{}^{0.49947k}-1.59745‚\\ \stackrel{^}{x}{}_{6}^{\left(1\right)}\left(k+1\right)=2.271684\text{e}{}^{0.599626k}-1.271684‚\\ \stackrel{^}{x}{}_{7}^{\left(1\right)}\left(k+1\right)=1.872098\text{e}{}^{0.799838k}-0.872098‚\\ \stackrel{^}{x}{}_{8}^{\left(1\right)}\left(k+1\right)=1.639822\text{e}{}^{0.999937k}-0.639822‚\\ \stackrel{^}{x}{}_{9}^{\left(1\right)}\left(k+1\right)=1.348329\text{e}{}^{1.499996k}-0.348329‚\\ \stackrel{^}{x}{}_{10}^{\left(1\right)}\left(k+1\right)=1.25953\text{e}{}^{1.799999k}-0.25953。\end{array}$

3) 两类GM (1, 1) 模型的模拟精度 (平均相对误差) 和预测精度 (误差) 比较见表2和表3。

《4 结论》

4 结论

1) 应用实际曲线在区间上的面积作为背景值, 重构了背景值的计算公式, 并保持了原GM (1, 1) 模型建模简单、计算简便及易于应用的优点。

2) 经大量数据模拟可知, 优化GM (1, 1) 模型既适用于低增长指数 (即发展系数的绝对值较小) 序列建模, 也适用于高增长指数 (即发展系数的绝对值较大) 序列建模, 尤其是对高增长指数序列优化GM (1, 1) 模型, 可用于做中长期预测且精度较高, 具有重要的理论意义和较高的应用价值。

Table 2 Contrast of the optimum one to the GM (1, 1) about the simulation

《表2》

 -a 0.1 0.2 0.3 0.4 0.5 0.6 0.8 1.0 1.5 1.8 原GM误差 0.105 963 2 0.499 163 1.300 909 2.613 955 4.520 585 9 7.074 289 9 14.156 851 23.544 004 51.032 934 65.453 743 新GM误差 0.337 920 5 0.731 469 1.147 005 1.558 524 1.955 963 5 2.334 233 3 3.027 305 4 3.633 131 2 4.733 841 3 5.132 772 9

Table 3 contrast of the optimum one to the GM (1, 1) about the prediction

《表3》

 -a 0.1 0.2 0.3 0.4 0.5 0.6 0.8 1.0 1.5 1.8 原GM1步误差 0.128 9 0.696 0 1.960 4 4.137 8 7.397 0 11.820 2 24.009 3 39.436 9 76.667 0 89.937 2 新GM1步误差 0.133 320 1 0.464 9 0.889 0 1.345 1 1.794 0 2.219 464 4 2.977 2 3.613 5 4.721 3 5.132 5 原GM2步误差 0.136 7 0.761 5 2.179 1 4.639 6 8.333 2 13.339 0 26.996 3 43.855 9 81.455 6 93.031 2 新GM2步误差 0.065 052 5 0.376 2 0.803 2 1.274 1 1.740 0 2.181 241 5 2.960 6 3.607 0 4.717 1 5.132 4 原GM5步误差 0.160 1 0.957 8 2.832 2 6.129 2 11.085 5 17.740 2 35.271 1 55.270 8 90.690 3 97.685 4 新GM5步误差 0.139 471 3 0.110 7 0.546 2 1.061 3 1.578 4 2.066 658 4 2.910 5 3.587 4 4.704 5 5.13 20 原GM10步误差 0.199 1 1.284 1 3.911 0 8.560 3 15.490 3 24.584 4 47.031 2 69.375 5 97.047 8 99.631 3 新GM10步误差 0.280 8 0.330 3 0.119 3 0.707 7 1.309 6 1.875 972 1 2.827 2 3.554 8 4.683 6 5.131 5