*2.1. Grey Model GM (1,1)*

In recent years, the grey prediction model has been applied in many research fields thanks to its popularity and computational efficiency. GM (1,1) represents the time-series prediction model in the first order of placing a variable [28], and is the most popular forecast model used by scientists because a part of GM (1,1) can deliver a relatively high predictive rate while not requiring an entire set of historical data except for a small amount of input data (at least four). This is the reason why GM (1,1) is suitable for almost all fields and different areas.

In this study, as the period of the past data collection was only five years, the selection of this model to predict future results is perfectly appropriate.

The model structure of GM (1,1) is described as follows:

Denote the original form of GM (1,1) as in Equation (1)

$$\mathbf{x}^{(0)} = \begin{pmatrix} \mathbf{x}^{(0)}(1), \ \dots, \ \mathbf{x}^{(0)}(n) \end{pmatrix}, \ n \ge 4 \tag{1}$$

The one-time accumulated generating operation (1-AGO) of the original sequence *x*(0) is defined as:

$$\mathbf{x}^{(1)} = \begin{pmatrix} \mathbf{x}^{(1)}(1), \ \dots, \ \mathbf{x}^{(1)}(n) \end{pmatrix}, n \ge 4 \tag{2}$$

Consider Equation (3) as the original form of the GM (1,1) model, where the symbol GM (1,1) stands for the first order grey model in variables.

$$\mathbf{x}^{(0)}(k) + a\mathbf{x}^{(1)}(k) = b \tag{3}$$

Consider Equation (4) as the basis form of this model.

$$x^{(0)}(k) \, + \, az^{(1)}(k) = b, \, k = 1, \, 2, \, \dots \,, n \tag{4}$$

where *<sup>z</sup>*(1)( *<sup>k</sup>*) = 0.5*x*(1)(*k*) + 0.5*x*(1)(*<sup>k</sup>* <sup>−</sup> <sup>1</sup>), *<sup>k</sup>* = 1, 2, ... , *<sup>n</sup>*. a, b are the coefficients; in grey system theory terms, a is said to be a developing coefficient and b the grey input; and *x*(0)(*k*) is the grey derivative that maximizes the information density for a given series to be modeled.

According to the least square method, we have aˆ = *a b* = (*BTB*) −1 *BTYN*. Therefore,

$$\mathbf{B} = \begin{bmatrix} -z^1(2) & 1 \\ -z^1(3) & 1 \\ \vdots & \vdots \\ -z^1(n) & 1 \end{bmatrix} \\ \mathbf{Y}\_N = \begin{bmatrix} \mathbf{x}^{(0)}(2) \\ \mathbf{x}^{(0)}(3) \\ \vdots \\ \mathbf{x}^{(0)}(n) \end{bmatrix} \tag{5}$$

Here, **B** is called a data matrix.

By considering the following equation *dx*(1)/*dt* + *ax* = *b* as a shadow for *x*(0)(*k*) + *az*(1)(*k*) = *b*, then the response equations for GM (1,1) are as follows:

$$\hat{\mathbf{x}}^{(1)}(k+1) = \left( \left( \mathbf{x}^{(0)}(1) - \frac{b}{a} \middle| \mathbf{c}^{-ak} \right. \\ \left. + \frac{b}{a} \right\vert, k = 1, 2, 3, \dots, n. \tag{6}$$

$$\mathfrak{x}^{(0)}(k+1) = \mathfrak{x}^{(1)}(k+1) - \mathfrak{x}^{(1)}(k) \tag{7}$$
