Small- and medium-sized enterprise post-disaster reconstruction management patterns and application

Abstract

Earthquakes have become more frequent in the past few decades causing significant impact to regional economic development. Small- and medium-sized enterprises (SMEs), as a key component of modern business, often suffer more severely because of their vulnerability. This article proposes a common SME post-earthquake reconstruction management pattern. This pattern focuses on SME post-disaster reconstruction work, adopts corporate social responsibility as a management philosophy and applies a principal component analysis to analyse the SME operational situation in the pre-earthquake phase. Then, a SME hierarchical classification based on the cluster analysis method is presented according to extent of the earthquake damage, and an optimal time control for SME reconstruction is proposed. The SME post-disaster reconstruction implementation pattern includes self-help, hybrid and external help approaches. After theoretical research, the SME management reconstruction pattern is applied to businesses in Wenchuan. Through investigation, 182 SMEs were chosen from Chengdu, Ya’an, Guangyuan, Aba, Mianyang and Deyang as samples and were classified into 4 classes: minor damage, heavily damage, severely damage and completely destroyed. Applying the optimal time control model, the shortest time was calculated to be 36 months. For each class of damaged SMEs, the proposed three approaches for the SME reconstruction pattern were implemented separately.

Appendices

Appendix 1: Solution steps

The optimal time control model is:

$$ \left\{\begin{array}{l} \min z=\sum\limits_{k=1}^{n}[\frac{x_k(0)(1+u_k(0))}{1+v(0)}-\alpha_k]^2\\ \sum\limits_{k=1}^{n}x_k(0)u_k(0)=v(0)\\ u_k^{\min}(0)\leqslant u_j(0)\leqslant u_k^{\max}(0),\quad k=1, 2, \ldots, n \end{array}\right. $$

(19)

This model is a nonlinear programming with linear constraints, and it can be solved by different methods; in this paper, we apply a simple method to solve the model.

Step 1. Obtain the initial feasible solution, set

$$ \tilde{u}^1(0)=\left[ {\begin{array}{c} v(0)\\ \vdots\\ v(0) \end{array}} \right] $$

where $\sum\nolimits_{k=1}^{n}x_k(0)u_k(0)=v(0)$ and $u_k^{\min}(0)\leqslant u_j(0)\leqslant u_k^{\max}(0), k=1, 2, \ldots, n$, so the $\tilde{u}^1(0)$ is feasible solution.

Step 2. Find a feasible direction and step length

$$ d^1=\left[ {\begin{array}{c} d_1^1\\ \vdots\\ d_n^1 \end{array}} \right] $$

d ¹ _k is valued from:

$$ \left\{\begin{array}{l} d_k^1=0, \quad k\in \{r+1,\ldots, q\}\\ d_k^1=(1+v(0))E_k, \quad k\notin \{r+1, \ldots, q\} \end{array}\right. $$

(20)

where

$$ P_k=\frac{\alpha_k}{x_k(0)}-\frac{\sum\limits_{k\notin\{r+1,\ldots, q\}} \alpha_k}{\sum\limits_{k\notin\{r+1,\ldots, q\}} k(0)} $$

(21)

Set $u^2(0)=\tilde{u}^1(0)+ad^1$, so ∑ ⁿ _k=1 x _k (0)u ² _k (0) = v(0). If $\tilde{u}^1(0)$ is not the optimum solution, there must be a sufficiently small integer to make the following formula exist:

$$ \left\{\begin{array}{l} u_k^{\min}(0)\leqslant u_j^2(0)\leqslant u_k^{\max}(0)\\ \sum\limits_{k=1}^n\left[\frac{x_k(0)(1+u_k^2(0))}{1+v(0)} -\alpha_k\right]^2\\ < \sum\limits_{k=1}^n\left[\frac{x_k(0)(1+\tilde{u}_k^1(0))}{1+v(0)}-\alpha_k\right]^2 \end{array}\right. $$

(22)

so d ¹ is the feasible direction.

Set

$$ \begin{aligned} a_k^1&=\min \left\{\frac{u_k^{\max}(0)-v(0)}{d_k^1}, \frac{(\alpha_k-1)(1+v(0))}{d_k^1}\right\},\\ k&=1, 2, \ldots, r\\ a_k^1&=\min \left\{\frac{u_k^{\min}(0)-v(0)}{d_k^1}, \frac{(\alpha_k-1)(1+v(0))}{d_k^1}\right\},\\ k&=q+1, \ldots, n \end{aligned} $$

(23)

Assume $a^1=\min \{a_k^1, k=1, 2, \ldots, r, q+1, \ldots, n\}$, set

$$ \tilde{u}^2(0)=\tilde{u}^1(0)+a^1d^1 $$

(24)

If there is

$$ \left[ {\begin{array}{c} \tilde{u}_1^2(0)\\ \vdots\\ \tilde{u}_n^2(0) \end{array}} \right]=\left[ {\begin{array}{c} \alpha_1\\ \vdots\\ \alpha_n \end{array}} \right] $$

so $u^*(0)=\tilde{u}^2(0)$, the calculation is over and N = 1; if Eq. 21 is not exist, the calculation go to the next step.

In the Step 3, the feasible direction of S ¹ _j values 0, other values are similar to or the same as Step 2, then take iteration till the calculation is finished. Using u*(0) value, the J ₁*(x(N − 1)) can be calculated, similarly, we can get u*(1), calculating J ₂*(x(N − 2)), so continue the $u^*(0), u^*(1), \ldots, u^*(N-1)$ and J ₀*(x(N)) will be calculated.

Appendix 2: Proof the convergence of the solution

According to the definition of the optimal time control model Eqs. 19 and 24, we transform $\tilde{u}^2(0)=\tilde{u}^1(0)+a^1d^1$ to x _t+1 = x _t  + a _t d _t , the proof is equivalent to proving: for any initial x ₁, the solution will generate an infinite sequence {x}, and each limit point is the stable point of the optimal time control model Eq. 19.

Hypothesis: a subsequence {x _t } of {x} converges to a point x*, and x* is NOT the stable point of Eq. 19.

As d _t is bounded and the index set are finite, so we assume {d _t } converges to a point d** and $J_N\equiv J^{\prime}$,

Set Z*, d* is the optimal solution of L(x*; J ₀), if Z* < 0, there is $\delta\in (0, \min{Y_t|t \notin J_0(x^*)}$ makes Z* =  − 2δ. If $J^{\prime}=J_0(x^*)$, so when Set Z*, d* is the optimal solution of L(x*; J ₀), if Z* < 0, there is $\delta\in (0, \min{Y_t|t \notin J_0(x^*)}$ makes Z* =  −2δ. If $J^{\prime}=J_0(x^*)$, so when t is sufficiently large, there is Z _t  <  −δ. If $J^{\prime}=J_0(x^*)$, there is Z _t  <  −δ. If $J^{\prime}\neq J_0(x^*)$, there is $J\in J^\prime J_0(x^*)$. So when t is sufficiently large, there is Y ^k _t  > δ. Thus

$$ Z_k< -Y_t^k< -\delta $$

So there is natural number N, when $t\geqslant N_1$, there is Z _k  <  −δ. As $J^{\prime}\supset J_0(x^*)$ and $x_k\rightarrow x^*$, so we can infer $\{Y_k|j\notin J^\prime, t=1, 2, \ldots\}$ has a positive lower bound. As {d _k } has a boundary and $ad_k\leqslant 0, j\in J^\prime$, so there is $\sigma\in [0, 1]$. For any $a\in [0, \sigma]$, there is

$$ x_k+ad_k\in R $$

here R is a nonempty set.

According to Z _t  <  −δ, we can get f*(x*; d**), so there is $a^*\in [0, \sigma]$ to set:

$$ f(x^*+a^*d^{**})< f(x^*)+\frac{1}{2}a^*f^*(x^*;d^{**}) $$

and then there is natural number N, let $t\geqslant N$,

$$ f(x_k^t+a^*d_k^t)< f(x_k^t)+\frac{1}{2}a^*f^*(x^*; d_k^t). $$

so when $t\geqslant N$, there is:

$$ f(x_k^t+1)\leqslant f(x_k^t)+\frac{1}{2}a^*f^*(x_k^t; d_k^t)<f(x_i^t)-\frac{1}{2}a^*\delta. $$

(25)

As the continuity of f(x), so there is

$$ \lim\limits_{t\rightarrow \sigma}f(x_k^t)=f(x^*). $$

As the monotonicity of f(x), so there is

$$ \lim\limits_{k\rightarrow \sigma}f(x_k)=f(x^*). $$

which is contradict to Eq. 25. So we prove that the conclusion of the solution is convergent.

Appendix 3: Integrative score and order of enterprises

See Table 8.

Table 8 Integrative score and order of enterprises