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A New Method for Determining Model Parameters of the k-Order Trend Curve
DOI: 10.12677/AAM.2019.88162, PDF, HTML, XML, 下载: 1,056  浏览: 1,452

Abstract: In the fields of economy, engineering, energy and environment, non-stationary series caused by deterministic factors can be fitted and predicted by k-order trend curve (or named polynomial function). Inspired by the method of solving the parameters of the modified exponential curve model, this paper proposes a method to determine the parameters of the k-order trend curve model. Compared with the least square method, the modelling process of this new method is simper and easier. Further, two examples in the textbook show that the proposed method has good accuracy and can be used in data modelling, fitting and prediction.

1. 引言

2. 修正的指数曲线

${Y}_{t}=a{b}^{t}+k$(1)

${S}_{1}=\underset{t=1}{\overset{m}{\sum }}{Y}_{t}$${S}_{2}=\underset{t=m+1}{\overset{2m}{\sum }}{Y}_{t}$${S}_{3}=\underset{t=2m+1}{\overset{3m}{\sum }}{Y}_{t}$(2)

$\left\{\begin{array}{l}{S}_{1}=mk+a+ab+a{b}^{2}+\cdots +a{b}^{m-1}\\ {S}_{2}=mk+a{b}^{m}+a{b}^{m+1}+\cdots +a{b}^{2m-1}\\ {S}_{3}=mk+a{b}^{2m}+a{b}^{2m+1}+\cdots +a{b}^{3m-1}\end{array}$(3)

$\left\{\begin{array}{l}b={\left(\frac{{S}_{3}-{S}_{2}}{{S}_{2}-{S}_{1}}\right)}^{\frac{1}{m}}\\ a=\left({S}_{2}-{S}_{1}\right)\frac{b-1}{{\left({b}^{m}-1\right)}^{2}}\\ k=\frac{1}{m}\left({S}_{1}-\frac{a\left({b}^{m}-1\right)}{b-1}\right)\end{array}$(4)

3. K阶趋势曲线

${Y}_{t}={a}_{k}{t}^{k}+{a}_{k-1}{t}^{k-1}+\cdots +{a}_{1}t+{a}_{0}$(5)

${S}_{i}=\underset{t=\left(i-1\right)m+1}{\overset{im}{\sum }}{Y}_{t},\text{\hspace{0.17em}}i=1,2,\cdots ,k+1$(6)

$\left\{\begin{array}{l}{S}_{1}={a}_{k}\underset{t=1}{\overset{m}{\sum }}{t}^{k}+{a}_{k-1}\underset{t=1}{\overset{m}{\sum }}{t}^{k-1}+\cdots +{a}_{1}\underset{t=1}{\overset{m}{\sum }}t+{a}_{0}m\\ {S}_{2}={a}_{k}\underset{t=m+1}{\overset{2m}{\sum }}{t}^{k}+{a}_{k-1}\underset{t=m+1}{\overset{2m}{\sum }}{t}^{k-1}+\cdots +{a}_{1}\underset{t=m+1}{\overset{2m}{\sum }}t+{a}_{0}m\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }⋮\\ {S}_{k+1}={a}_{k}\underset{t=km+1}{\overset{\left(k+1\right)m}{\sum }}{t}^{k}+{a}_{k-1}\underset{t=km+1}{\overset{\left(k+1\right)m}{\sum }}{t}^{k-1}+\cdots +{a}_{1}\underset{t=km+1}{\overset{\left(k+1\right)m}{\sum }}t+{a}_{0}m\end{array}$(7)

$\left(\begin{array}{ccccccc}\underset{t=1}{\overset{m}{\sum }}{t}^{k}& \underset{t=1}{\overset{m}{\sum }}{t}^{k-1}& \underset{t=1}{\overset{m}{\sum }}{t}^{k-2}& \cdots & \underset{t=1}{\overset{m}{\sum }}{t}^{2}& \underset{t=1}{\overset{m}{\sum }}t& m\\ \underset{t=m+1}{\overset{2m}{\sum }}{t}^{k}& \underset{t=m+1}{\overset{2m}{\sum }}{t}^{k-1}& \underset{t=m+1}{\overset{2m}{\sum }}{t}^{k-2}& \cdots & \underset{t=m+1}{\overset{2m}{\sum }}{t}^{2}& \underset{t=m+1}{\overset{2m}{\sum }}t& m\\ \underset{t=2m+1}{\overset{3m}{\sum }}{t}^{k}& \underset{t=2m+1}{\overset{3m}{\sum }}{t}^{k-1}& \underset{t=2m+1}{\overset{3m}{\sum }}{t}^{k-2}& \cdots & \underset{t=2m+1}{\overset{3m}{\sum }}{t}^{2}& \underset{t=2m+1}{\overset{3m}{\sum }}t& m\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮\\ \underset{t=\left(k-2\right)m+1}{\overset{\left(k-1\right)m}{\sum }}{t}^{k}& \underset{t=\left(k-2\right)m+1}{\overset{\left(k-1\right)m}{\sum }}{t}^{k-1}& \underset{t=\left(k-2\right)m+1}{\overset{\left(k-1\right)m}{\sum }}{t}^{k-2}& \cdots & \underset{t=\left(k-2\right)m+1}{\overset{\left(k-1\right)m}{\sum }}{t}^{2}& \underset{t=\left(k-2\right)m+1}{\overset{\left(k-1\right)m}{\sum }}t& m\\ \underset{t=\left(k-1\right)m+1}{\overset{km}{\sum }}{t}^{k}& \underset{t=\left(k-1\right)m+1}{\overset{km}{\sum }}{t}^{k-1}& \underset{t=\left(k-1\right)m+1}{\overset{km}{\sum }}{t}^{k-2}& \cdots & \underset{t=\left(k-1\right)m+1}{\overset{km}{\sum }}{t}^{2}& \underset{t=\left(k-1\right)m+1}{\overset{km}{\sum }}t& m\\ \underset{t=km+1}{\overset{\left(k+1\right)m}{\sum }}{t}^{k}& \underset{t=km+1}{\overset{\left(k+1\right)m}{\sum }}{t}^{k-1}& \underset{t=km+1}{\overset{\left(k+1\right)m}{\sum }}{t}^{k-2}& \cdots & \underset{t=km+1}{\overset{\left(k+1\right)m}{\sum }}{t}^{2}& \underset{t=km+1}{\overset{\left(k+1\right)m}{\sum }}t& m\end{array}\right)\left(\begin{array}{c}{a}_{k}\\ {a}_{k-1}\\ {a}_{k-2}\\ ⋮\\ {a}_{2}\\ {a}_{1}\\ {a}_{0}\end{array}\right)=\left(\begin{array}{c}{S}_{1}\\ {S}_{2}\\ {S}_{3}\\ ⋮\\ {S}_{k-1}\\ {S}_{k}\\ {S}_{k+1}\end{array}\right)$ (8)

$•$$k=1$ 时，求解线性趋势曲线 ${Y}_{t}={a}_{1}t+{a}_{0}$ 参数的表达式。首先，有

$\left\{\begin{array}{l}{S}_{1}={a}_{1}\underset{t=1}{\overset{m}{\sum }}t+{a}_{0}m,\\ {S}_{2}={a}_{1}\underset{t=m+1}{\overset{2m}{\sum }}t+{a}_{0}m.\end{array}$ (9)

$\left\{\begin{array}{l}{a}_{1}=\frac{{S}_{2}-{S}_{1}}{{m}^{2}},\\ {a}_{0}=\frac{{S}_{1}}{m}-\frac{\left(m+1\right)\left({S}_{2}-{S}_{1}\right)}{2{m}^{2}}.\end{array}$ (10)

$•$$k=2$ 时，求解线性趋势曲线 ${Y}_{t}={a}_{2}{t}^{2}+{a}_{1}t+{a}_{0}$ 参数的表达式。首先，有

$\left\{\begin{array}{l}{S}_{1}={a}_{2}\underset{t=1}{\overset{m}{\sum }}{t}^{2}+{a}_{1}\underset{t=1}{\overset{m}{\sum }}t+{a}_{0}m,\\ {S}_{2}={a}_{2}\underset{t=m+1}{\overset{2m}{\sum }}{t}^{2}+{a}_{1}\underset{t=m+1}{\overset{2m}{\sum }}t+{a}_{0}m,\\ {S}_{3}={a}_{2}\underset{t=2m+1}{\overset{3m}{\sum }}{t}^{2}+{a}_{1}\underset{t=2m+1}{\overset{3m}{\sum }}t+{a}_{0}m.\end{array}$ (11)

${a}_{2}=\frac{\left(6{S}_{1}-6{S}_{3}\right)m\left(3m+1\right)-\left(6{S}_{1}-6{S}_{2}\right)m\left(5m+1\right)-6m\left(m+1\right)\left({S}_{2}-{S}_{3}\right)}{2{m}^{3}\left(-12{m}^{2}+m+1\right)}$

$\begin{array}{l}{a}_{1}=\frac{2m\left({S}_{1}-{S}_{2}\right)\left(38{m}^{2}+15m+1\right)-2m\left({S}_{1}-{S}_{3}\right)\left(14{m}^{2}+9m+1\right)}{2{m}^{3}\left(-12{m}^{2}+m+1\right)}\\ \text{ }\text{ }+\frac{4m\left({S}_{2}-{S}_{3}\right)\left(m+1\right)\left(m+\frac{1}{2}\right)}{2{m}^{3}\left(-12{m}^{2}+m+1\right)}\end{array}$

$\begin{array}{l}{a}_{0}=\frac{-m\left[{S}_{1}\left(3m+1\right)-{S}_{2}\left(m+1\right)\right]\left(38{m}^{2}+15m+1\right)}{2{m}^{4}\left(-12{m}^{2}+m+1\right)}\\ \text{ }\text{ }+\frac{m\left[{S}_{1}\left(5m+1\right)-{S}_{3}\left(m+1\right)\right]\left(14{m}^{2}+9m+1\right)}{2{m}^{4}\left(-12{m}^{2}+m+1\right)}\\ \text{ }\text{ }-\frac{2{m}^{2}\left[{S}_{2}\left(5m+1\right)-{S}_{3}\left(3m+1\right)\right]\left(m+1\right)\left(m+\frac{1}{2}\right)}{2{m}^{4}\left(-12{m}^{2}+m+1\right)}\end{array}$

4. 数值验证

$\text{APE}=|\frac{{Y}_{t}-{\stackrel{^}{Y}}_{t}}{{Y}_{t}}|×100%,t=1,2,\cdots ,n$(12)

$\text{MAPE}=\frac{1}{v-l+1}\underset{t=l}{\overset{v}{\sum }}|\frac{{Y}_{t}-{\stackrel{^}{Y}}_{t}}{{Y}_{t}}|×100%,v\le n$(13)

${Y}_{t}^{book}=52305.1982\text{\hspace{0.17em}}t+112731.7927$(14)

${Y}_{t}=53933.9440t+104911.7880$(15)

Table 1. The application of linear trend curve in China’s GDP

Figure 1. The linear trends of the China’s GDP

${Y}_{t}^{book}=-0.0078{t}^{4}+0.3268{t}^{3}-4.2017{t}^{2}+19.1011t-4.0453$(16)

${Y}_{t}=-0.0084{t}^{4}+0.3557{t}^{3}-4.6523{t}^{2}+21.6510t-7.6325$(17)

Table 2. Application of the fourth order trend curve in the output of metal cutting machine

Figure 2. The fourth order trend curve in the output of metal cutting machine

5. 结论

NOTES

*通讯作者。

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