Continuous location model of a rectangular barrier facility

Abstract

This paper develops a bi-objective model for determining the location, size, and shape of a finite-size facility. The objectives are to minimize both the closest and barrier distances. The closest distance represents the accessibility of customers, whereas the barrier distance represents the interference to travelers. The distributions of the closest and barrier distances are derived for a rectangular facility in a rectangular city where the distance is measured as the rectilinear distance. The analytical expressions for the distributions demonstrate how the location, size, and shape of the facility affect the closest and barrier distances. A numerical example shows that there exists a trade-off between the closest and barrier distances.

Appendices

Appendix 1: Distribution of the closest distance

The distribution of the closest distance is given by

$$\begin{aligned} f(r)=\frac{1}{S}\sum _{i=1}^8S_if_i(r), \end{aligned}$$

(20)

where

$$\begin{aligned} S&=a_1a_2-b_1b_2,\\ f_1(r)&= {\left\{ \begin{array}{ll} \frac{r}{S_1},&{}\quad 0<r\le \alpha _1,\\ \frac{\alpha _1}{S_1},&{}\quad \alpha _1<r\le \beta _1,\\ \frac{\gamma _1-r}{S_1},&{}\quad \beta _1<r\le \gamma _1, \end{array}\right. }\\ S_1&=\left( x_b-\frac{b_1}{2}\right) \left( y_b-\frac{b_2}{2}\right) ,\quad \alpha _1=\min \left\{ x_b-\frac{b_1}{2}, y_b-\frac{b_2}{2}\right\} ,\\ \beta _1&=\max \left\{ x_b-\frac{b_1}{2}, y_b-\frac{b_2}{2}\right\} ,\quad \gamma _1=x_b+y_b-\frac{b_1}{2}-\frac{b_2}{2},\\ f_2(r)&=\frac{b_2}{S_2},\quad 0<r\le x_b-\frac{b_1}{2},\\ S_2&=b_2\left( x_b-\frac{b_1}{2}\right) ,\\ f_3(r)&= {\left\{ \begin{array}{ll} \frac{r}{S_3},&{}\quad 0<r\le \alpha _3,\\ \frac{\alpha _3}{S_3},&{}\quad \alpha _3<r\le \beta _3,\\ \frac{\gamma _3-r}{S_3},&{}\quad \beta _3<r\le \gamma _3, \end{array}\right. }\\ S_3&=\left( x_b-\frac{b_1}{2}\right) \left( a_2-\frac{b_2}{2}-y_b\right) ,\quad \alpha _3=\min \left\{ x_b-\frac{b_1}{2}, a_2-\frac{b_2}{2}-y_b\right\} ,\\ \end{aligned}$$

$$\begin{aligned} \beta _3&=\max \left\{ x_b-\frac{b_1}{2}, a_2-\frac{b_2}{2}-y_b\right\} ,\quad \gamma _3=a_2-\frac{b_1}{2}-\frac{b_2}{2}+x_b-y_b,\\ f_4(r)&=\frac{b_1}{S_4},\quad 0<r\le y_b-\frac{b_2}{2},\\ S_4&=b_1\left( y_b-\frac{b_2}{2}\right) ,\\ f_5(r)&=\frac{b_1}{S_5},\quad 0<r\le a_2-\frac{b_2}{2}-y_b,\\ S_5&=b_1\left( a_2-\frac{b_2}{2}-y_b\right) ,\\ f_6(r)&= {\left\{ \begin{array}{ll} \frac{r}{S_6},&{}\quad 0<r\le \alpha _6,\\ \frac{\alpha _6}{S_6},&{}\quad \alpha _6<r\le \beta _6,\\ \frac{\gamma _6-r}{S_6},&{}\quad \beta _6<r\le \gamma _6, \end{array}\right. }\\ S_6&=\left( a_1-\frac{b_1}{2}-x_b\right) \left( y_b-\frac{b_2}{2}\right) ,\quad \alpha _6=\min \left\{ a_1-\frac{b_1}{2}-x_b, y_b-\frac{b_2}{2}\right\} ,\\ \beta _6&=\max \left\{ a_1-\frac{b_1}{2}-x_b, y_b-\frac{b_2}{2}\right\} ,\quad \gamma _6=a_1-\frac{b_1}{2}-\frac{b_2}{2}-x_b+y_b,\\ \end{aligned}$$

$$\begin{aligned} f_7(r)&=\frac{b_2}{S_7},\quad 0<r\le a_1-\frac{b_1}{2}-x_b,\\ S_7&=b_2\left( a_1-\frac{b_1}{2}-x_b\right) ,\\ f_8(r)&= {\left\{ \begin{array}{ll} \frac{r}{S_8},&{}\quad 0<r\le \alpha _8,\\ \frac{\alpha _8}{S_8},&{}\quad \alpha _8<r\le \beta _8,\\ \frac{\gamma _8-r}{S_8},&{}\quad \beta _8<r\le \gamma _8, \end{array}\right. }\\ S_8&=\left( a_1-\frac{b_1}{2}-x_b\right) \left( a_2-\frac{b_2}{2}-y_b\right) ,\\ \alpha _8&=\min \left\{ a_1-\frac{b_1}{2}-x_b, a_2-\frac{b_2}{2}-y_b\right\} ,\\ \beta _8&=\max \left\{ a_1-\frac{b_1}{2}-x_b, a_2-\frac{b_2}{2}-y_b\right\} ,\quad \gamma _8\!=\!a_1\!+\!a_2-\frac{b_1}{2}-\frac{b_2}{2}-x_b-y_b. \end{aligned}$$

Appendix 2: Total closest distance

The total closest distance is given by

$$\begin{aligned} T_a=\sum _{i=1}^8T_i, \end{aligned}$$

(21)

where

$$\begin{aligned} T_1&=\frac{1}{16}(2x_b-b_1)(2y_b-b_2)(2x_b+2y_b-b_1-b_2),\\ T_2&=\frac{b_2}{8}(2x_b-b_1)^2,\\ T_3&=\frac{1}{16}(2x_b-b_1)(2a_2-b_1-b_2+2x_b-2y_b)(2a_2-b_2-2y_b),\\ T_4&=\frac{b_1}{8}(2y_b-b_2)^2,\\ T_5&=\frac{b_1}{8}(2a_2-b_2-2y_b)^2,\\ T_6&=\frac{1}{16}(2a_1-b_1-2x_b)(2y_b-b_2)(2a_1-b_1-b_2-2x_b+2y_b),\\ T_7&=\frac{b_2}{8}(2a_1-b_1-2x_b)^2,\\ T_8&=\frac{1}{16}(2a_1-b_1-2x_b)(2a_2-b_2-2y_b)(2a_1+2a_2-b_1-b_2-2x_b-2y_b). \end{aligned}$$