Big Triangle Small Triangle Method for the Weber Problem on the Sphere

Abstract

We propose the Big Triangle Small Triangle (BTST) Method for solving the Weber problem on the sphere (WPS). It can also be applied to other single facility location problems on the sphere. The WPS is a variation of the Weber problem which is a classic and well-studied location problem. We assume that the demand points are distributed on the surface of a sphere, and our problem is to find the location of a facility so as to minimize the sum of the distances from demand points to the facility. The distance is measured by the great circle distance. The objective function of the WPS is not convex, and the Weiszfeld-type algorithm is not guaranteed to find the facility’s optimal location. The BTST type algorithm divides the surface of the sphere into spherical triangles and applies a branch-and-bound method. We show that it finds the optimal solution within a short computational time.

Appendix

4.1.1 Partial Derivatives of G ₁

The calculation of the partial derivatives of G ₁(t ₁, t ₂, t ₃) is as follows:

$$\displaystyle \begin{aligned} \begin{array}{rcl} g_{1i}(t_{1}, t_{2}, t_{3}) &\displaystyle \equiv &\displaystyle \cos (t_{1}\phi^{1}+t_{2}\phi^{2}+t_{3}\phi^{3})\cos\phi_{i} \times \cos (t_{1}\theta^{1}+t_{2}\theta^{2}+t_{3}\theta^{3}- \theta_{i}) \\ &\displaystyle + &\displaystyle \sin (t_{1}\phi^{1}+t_{2}\phi^{2}+t_{3}\phi^{3}) \sin\phi_{i}. \end{array} \end{aligned} $$

Then

$$\displaystyle \begin{aligned}G_{1}(t_{1}, t_{2}, t_{3}) = \sum_{i \in \cap_{k=1}^{3} I^{k}} w_{i} g_{1i}(t_{1}, t_{2}, t_{3}). \end{aligned}$$

The partial derivatives of G ₁ are

$$\displaystyle \begin{aligned}{\partial G_{1} \over {\partial t_{k}}} = - \sum_{i \in \cap_{k=1}^{3} I^{k}} w_{i} (1- g_{1i}(t_{1}, t_{2}, t_{3})^{2})^{1 \over 2} {\partial \over {\partial t_{k}}} g_{1i}(t_{1}, t_{2}, t_{3}), k=1,2,3. \end{aligned}$$

The partial derivatives of g _1k are calculated as follows:

$$\displaystyle \begin{aligned} {\partial \over {\partial t_{k}}} g_{1i}(t_{1}, t_{2}, t_{3}) & = \cos\phi_{i} \{-\phi^{k} \sin{}(t_{1} \phi^{1} +t_{2} \phi^{2} +t_{3} \phi^{3} ) \cdot \cos{}(t_{1} \theta^{1} +t_{2} \theta^{2} +t_{3} \theta^{3}-\theta_{i} ) \\ & - \theta^{k} \cos{}(t_{1} \phi^{1} +t_{2} \phi^{2} +t_{3} \phi^{3} ) \cdot \sin (t_{1} \theta^{1} +t_{2} \theta^{2} +t_{3} \theta^{3}-\theta_{i} )\} \\ & + \sin\phi_{i} \{\phi^{k} \cos{}(t_{1} \phi^{1} +t_{2} \phi^{2} +t_{3} \phi^{3} )\}. \end{aligned} $$

4.1.2 Shortest Distance to the Triangle T

For the lower bound of F ₃(s), we need to calculate the shortest distance from a demand point to the triangle T. The shortest distance from a demand point is attained on the edges or the vertices of T. Therefore, we need to obtain the shortest distance from the demand point P(a _i) to an edge T ¹ T ² of the triangle T. We calculate the distance from the demand point to the three edges and set the minimum of them as the shortest distance from the demand point to the triangle T.

For the calculation of the distance from the demand point P(a _i) to an edge T ¹ T ², we need to consider two cases. Consider the spherical angles $\alpha = \angle PT^{1}T^{2}$ and $\beta = \angle PT^{2}T^{1}$ (see Fig. 4.3).

Fig. 4.3

The shortest distance from a demand point to the edge T ¹ T ²

The first case is that both α and β are less than π∕2. In this case, the shortest distance from P to T ¹ T ² is attained at a point T′ on T ¹ T ² see Fig. 4.3. The distance is calculated as

$$\displaystyle \begin{aligned} d(T', a_{i})=\arccos(\overrightarrow{OT'}\cdot\overrightarrow{OP}), \end{aligned}$$

where

$$\displaystyle \begin{aligned} \overrightarrow{OT'} = {{\overrightarrow{OP}-k(\overrightarrow{OT^{1}}\times \overrightarrow{OT^{2}})} \over {|| \overrightarrow{OP}-k(\overrightarrow{OT^{1}}\times \overrightarrow{OT^{2}})||}}, \end{aligned}$$

$$\displaystyle \begin{aligned} k= {{(\overrightarrow{OT^{1}}\times\overrightarrow{OT^{2}})\cdot\overrightarrow{OP}} \over {||\overrightarrow{OT^{1}}\times\overrightarrow{OT^{2}}||}}. \end{aligned}$$

The second case is that either α or β is greater than or equal to $\frac \pi 2$. If $\alpha \ge \frac \pi 2$, the shortest distance is attained at T ¹. If $\beta \ge \frac \pi 2$, the shortest distance is attained at T ².