*3.2. Analysis of the Applicability of the Network Model and Virus Model of a Prefabricated Building Supply Chain*


#### *3.3. The Existence of a Risk Balance Point*

In the infectious disease model, there are two modes: disease-free equilibrium and endemic equilibrium. The disease-free equilibrium is the extinction of the disease, and the endemic equilibrium means that the disease will develop in the population for a long time, forming a normalization [37]. Because this paper studies and analyses risk transmission based on the infectious disease model, it redefines the disease-free equilibrium point as the risk extinction point and the endemic equilibrium point as the long-term risk existence point. The risks in the supply chain can only be controlled but do not completely disappear. Therefore, this paper only explores the spread of the endemic equilibrium point, that is, the long-term existence point of risks, defined according to the supply chain:

In Model (1), there is always a disease-free equilibrium *P*<sup>0</sup> *μ <sup>Z</sup>* , 0, 0, 0, 0 . According to the method of the regeneration matrix in the literature [38], the basic regeneration number is obtained:

$$R\_0 = \frac{\varkappa\_2 Q b \beta}{\mu M\_1 M\_2} + \frac{\varkappa\_1 Z b (1 - \beta)}{\mu M\_0 M\_2} + \frac{Hb \beta M\_2 d\_1 + Hb (1 - \beta) M\_1 d\_2}{M\_0 M\_1 M\_2 (H + \beta)}$$

among *M*<sup>0</sup> = *b* + *μ*, *M*<sup>1</sup> = *μ* + *d*<sup>1</sup> + *c*1, *M*<sup>2</sup> = *μ* + *d*<sup>2</sup> + *c*2. There are the following conclusions about the existence of the equilibrium point of the model:

**Theorem 1.** *When R*<sup>0</sup> > 1*, Model (1) has a unique endemic equilibrium P*∗(*S*∗, *E*∗, *A*∗, *I*∗, *R*∗)

It is proven that if the right side of Model (1) is equal to 0 and an endemic equilibrium *P*∗ exists, the algebraic equations are satisfied:

$$\begin{cases} Q - a\_1 \mathbb{S}^\* A^\* - a\_2 \mathbb{S}^\* I^\* - \mu \mathbb{S}^\* = 0 \\ a\_1 \mathbb{S}^\* A^\* + a\_2 \mathbb{S}^\* I^\* - \mu E^\* - b \beta E^\* - b(1 - \beta) E^\* + HR^\* = 0 \\ b \beta E^\* - (c\_1 + \mu) A^\* - d\_1 A^\* = 0 \\ b(1 - \beta) E^\* - (c\_2 + \mu) I^\* - d\_2 I^\* = 0 \\ d\_1 A^\* + d\_2 I^\* - \mu R^\* - HR^\* = 0 \end{cases} \tag{2}$$

The following can be obtained from the third and fourth formulas of the equations:

$$\begin{aligned} b\beta E^\* &= (c\_1 + \mu)A^\* + d\_1 A^\* = M\_1 A^\* \\ b(1 - \beta)E^\* &= (c\_2 + \mu)I^\* + d\_2 I^\* = M\_2 A^\* \end{aligned}$$

Therefore,

$$A^\* = \frac{b\beta}{M\_1} E^\*, I^\* = \frac{b(1-\beta)}{M\_2} E^\*. \tag{3}$$

Substituting (3) into (2) yields

$$\frac{d\_1 b \beta}{M\_1} E^\* + \frac{d\_2 b (1 - \beta)}{M\_2} E^\* = (H + \mu) R^\* \lambda$$

Therefore,

$$R^\* = \frac{1}{H + \mu} (\frac{d\_1 b \beta}{M\_1} + \frac{d\_2 b (1 - \beta)}{M\_2}).\tag{4}$$

Combining Equation (4) and bringing the second equation in equation group (2) into the first equation, we obtain *<sup>Q</sup>* <sup>−</sup> *<sup>M</sup>*0*E*<sup>∗</sup> <sup>+</sup> *<sup>H</sup> <sup>H</sup>*+*<sup>μ</sup>* ( *<sup>d</sup>*1*b<sup>β</sup> <sup>M</sup>*<sup>1</sup> <sup>+</sup> *<sup>d</sup>*2*b*(1−*β*) *<sup>M</sup>*<sup>2</sup> )*E*<sup>∗</sup> = *<sup>μ</sup>S*<sup>∗</sup>

So

$$S^\* = \frac{Q}{\mu} - \frac{(H+\mu)M\_0M\_1M\_2 - d\_1b\beta HM\_2 - d\_2b(1-\beta)HM\_1}{\mu(H+\mu)M\_1M\_2}E^\* \tag{5}$$

Substituting (3), (4), and (5) into the second equation set of equation set (2) gives

$$
\mu\_1 S^\* A^\* + \partial\_2 S^\* I^\* - \mu E^\* - b\beta E^\* - b(1 - \beta)E^\* + HR^\* = 0
$$

$$\begin{cases} \frac{a\_1 b (1 - \beta)}{M\_2} + \frac{a\_2 b \beta}{M\_1} ) E^\* \times \left\{ \frac{Q}{\mu} - \frac{(H + \mu) M\_0 M\_1 M\_2 - d\_1 b \beta H M\_2 - d\_2 b (1 - \beta) H M\_1}{\mu (H + \mu) M\_1 M\_2} E^\* \right\} \\\ -M\_0 E^\* + \frac{H}{H + \mu} (\frac{d\_1 b \beta}{M\_1} + \frac{d\_2 b (1 - \beta)}{M\_2}) E^\* = 0. \end{cases}$$

Therefore,

$$E^\* = \frac{\mu M\_0 M\_1^2 M\_2^2 (H + \mu)}{\left\{ (H + \mu)M\_0 M\_1 M\_2 - \left[ H b\_1 \beta M\_2 d\_1 + H b\_2 (1 - \beta) M\_2 d\_2 \right] \right\} \left[ a\_1 b\_1 M\_1 (1 - \beta) + a\_2 b\_2 M\_2 \beta \right]} \left( R\_0 - 1 \right)$$

Therefore, this paper needs to prove

{(*H* + *μ*)*M*0*M*1*M*<sup>2</sup> − [*Hb*1*βM*2*d*<sup>1</sup> + *Hb*2(1 − *β*)*M*2*d*2]} > 0 Just do it.

The specific steps are as follows:

$$\begin{cases} (H+\mu)M\_0M\_1M\_2 - [Hb\_1\beta M\_2d\_1 + Hb\_2(1-\beta)M\_2d\_2] \big) & \\ > Hb\_1M\_2 - [Hb\_1\beta M\_2d\_1 + Hb\_2(1-\beta)M\_2d\_2] \big) & \\ = Hb[(\beta M\_1M\_2 - \beta d\_1M\_1M\_2) + (1-\beta)M\_1M\_2 - (1-\beta)d\_2M\_1] \big) & \\ > 0 \end{cases}$$

It can be proven that there is an endemic equilibrium point, so this situation will form a long-term risk in enterprises, that is, the risk that a supply chain will exist for a long time.
