Assessing indirect economic losses of landslides along highways

Abstract

Landslides along highways not only induce direct losses related to cost of replacement, repair or maintenance due to physical damage, but also induce indirect losses due to reduction in production in different sectors in an economic system. In this study, a method is suggested to assess the indirect losses of landslides along highways, where the impact of landslides on the transportation network is evaluated through the gravity model, and the output losses of different sectors are computed based on the dynamic input–output model. The suggested method is used to assess the indirect losses of landslides that occur along highways related to Yingxiu Town, Sichuan Province, China, which is the epicenter of the 2008 great Wenchuan earthquake. It is found that the impact of a landslide hazard on different sectors is not the same depending on their dependence relationship with the transportation sector. The indirect losses of a landslide increase as the repair time increases and decrease as the transportation network becomes more robust. Also, the location of a landslide will affect the caused indirect losses, and the indirect losses will increase as the transportation network is attacked by more landslides. The suggested method provides a useful instrument for estimating the indirect losses of landslides along highways, which were seldom addressed in previous studies.

Appendices

Appendix A

Equation (11) is a system of linear first-order partial differential equations. Given the initial condition which is denoted as x_l(0), it can be solved as follows (Edwards and Penney, 2000):

$${\mathbf{x}}_{l} (t) = e^{{ - {\mathbf{K}}({\mathbf{I}} - {\mathbf{A}})t}} {\mathbf{x}}_{l} (0) + \int\limits_{0}^{t} {{\mathbf{K}}e^{{ - {\mathbf{K}}({\mathbf{I}} - {\mathbf{A}})(t - z)}} {\mathbf{c}}_{l} (z)} dz$$

(21)

Based on the assumption that the final demand is stationary, i.e., c_l(t) = c_b. Substituting c_l(t) = c_b into second term of the right side of Eq. (21), the following relationship exists:

$$\int\limits_{0}^{t} {{\mathbf{K}}e^{{ - {\mathbf{K}}({\mathbf{I}} - {\mathbf{A}})(t - z)}} {\mathbf{c}}_{l} (z)} {\text{d}}z{ = }{\mathbf{Kc}}_{b} \int\limits_{0}^{t} {e^{{ - {\mathbf{K}}({\mathbf{I}} - {\mathbf{A}})(t - z)}} } {\text{d}}z = ({\mathbf{I}} - {\mathbf{A}})^{ - 1} {\mathbf{c}}_{b} - e^{{ - {\mathbf{K}}({\mathbf{I}} - {\mathbf{A}})t}} ({\mathbf{I}} - {\mathbf{A}})^{ - 1} {\mathbf{c}}_{b}$$

(22)

Substituting Eq. (22) into Eq. (21), x_l(t) can then be calculated as follows:

$${\mathbf{x}}_{l} (t) = ({\mathbf{I}} - {\mathbf{A}})^{ - 1} {\mathbf{c}}_{b} - e^{{ - {\mathbf{K}}({\mathbf{I}} - {\mathbf{A}})t}} [({\mathbf{I}} - {\mathbf{A}})^{ - 1} {\mathbf{c}}_{b} - {\mathbf{x}}_{l} (0)]$$

(23)

Substituting Eqs. (3) and  (9) into Eq. (23) yields Eq. (12).

Appendix B

Integrating Eq. (13) with time, the indirect losses in different sectors can be obtained as follows:

$${\mathbf{L}} = \int_{0}^{\infty } {e^{{ - {\mathbf{K}}({\mathbf{I}} - {\mathbf{A}})t}} {\mathbf{q}}_{l} (0)} dt = ({\mathbf{I}} - {\mathbf{A}})^{ - 1} {\mathbf{K}}^{ - 1} {\mathbf{q}}_{l} (0)$$

(24)

where L = {L₁, L₂, …, L_n}^T. Substituting K = diag{k₁, k₂, …, k_n} and q_l(0) = {x_b1w_l(0), 0, 0, 0, 0, …, 0} ^T into the above equation, L can be calculated as follows:

$${\mathbf{L}}{ = }({\mathbf{I}} - {\mathbf{A}})^{ - 1} {\mathbf{K}}^{ - 1} {\mathbf{q}}_{l} (0) = ({\mathbf{I}} - {\mathbf{A}})^{ - 1} \left[ {\begin{array}{*{20}c} {{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 {k_{1} }}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{${k_{1} }$}}} & 0 & 0 & {...} & 0 \\ 0 & {{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 {k_{2} }}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{${k_{2} }$}}} & 0 & {...} & 0 \\ 0 & 0 & {{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 {k_{3} }}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{${k_{3} }$}}} & {...} & {...} \\ {...} & {...} & {...} & {...} & 0 \\ 0 & 0 & {...} & 0 & {{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 {k_{n} }}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{${k_{n} }$}}} \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {x_{b1} w_{l} \left( 0 \right)} \\ 0 \\ 0 \\ {...} \\ 0 \\ \end{array} } \right] = ({\mathbf{I}} - {\mathbf{A}})^{ - 1} \left[ {\begin{array}{*{20}c} {\frac{{x_{b1} w_{l} \left( 0 \right)}}{{k_{1} }}} \\ 0 \\ 0 \\ {...} \\ 0 \\ \end{array} } \right]$$

(25)

Let b_i1 denote the element of (I—A)⁻¹, in the ith row and 1st column. The ith element of L can then be calculated as Eq. (14).