Location of a line in the three-dimensional space Daniel Scholz - PowerPoint PPT Presentation

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results Location of a line in the three-dimensional space Daniel Scholz Institute for Numerical and Applied Mathematics, University of G¨ ottingen May 27, 2010 Doc-Course: Constructive Approximation, Optimization and Mathematical Modeling, Sevilla, Spain Supervised by Rafael Blanquero, Emilio Carrizosa, and Anita Sch¨ obel Daniel Scholz University of G¨ ottingen May 27, 2010 1 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results Outline 1. Introduction 3. Geometric branch-and-bound 1.1 The Weber problem 3.1 The algorithm 1.2 Line location in 2D 3.2 Initial box 1.3 Line location in 3D 3.3 Lower bounds 2. Problem formulation 4. Numerical results 2.1 Closed formula 4.1 Example instance 2.2 Properties 4.2 Random input data 2.3 Parametrization 4.3 Discussion Daniel Scholz University of G¨ ottingen May 27, 2010 2 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :The Weber problem :Line location in two dimensions :Line location in three dimensions Facility locations problems Figure: Some given demand points on the plane. Weber problem : Minimize the sum of the distances between a new facility and the demand points. Daniel Scholz University of G¨ ottingen May 27, 2010 3 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :The Weber problem :Line location in two dimensions :Line location in three dimensions Facility locations problems Figure: Some given demand points on the plane. Weber problem : Minimize the sum of the distances between a new facility and the demand points. Daniel Scholz University of G¨ ottingen May 27, 2010 3 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :The Weber problem :Line location in two dimensions :Line location in three dimensions Line location in the two-dimensional space Figure: Some given demand points on the plane. Median line problem : All optimal solutions pass through two demand points, see Korneenko and Martini (1993) or Sch¨ obel (1999). Daniel Scholz University of G¨ ottingen May 27, 2010 4 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :The Weber problem :Line location in two dimensions :Line location in three dimensions Line location in the two-dimensional space Figure: Some given demand points on the plane. Median line problem : All optimal solutions pass through two demand points, see Korneenko and Martini (1993) or Sch¨ obel (1999). Daniel Scholz University of G¨ ottingen May 27, 2010 4 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :The Weber problem :Line location in two dimensions :Line location in three dimensions Line location in the three-dimensional space 2.0 1.5 1.0 0.5 0.0 2.0 1.5 1.0 0.5 0.0 0.0 0.5 1.0 1.5 2.0 Figure: Some given demand points in the three-dimensional space. Objective : Find a line which minimizes the sum of the distances between such line and the given demand points. Daniel Scholz University of G¨ ottingen May 27, 2010 5 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :The Weber problem :Line location in two dimensions :Line location in three dimensions Line location in the three-dimensional space 2.0 1.5 1.0 0.5 0.0 2.0 1.5 1.0 0.5 0.0 0.0 0.5 1.0 1.5 2.0 Figure: Some given demand points in the three-dimensional space. Objective : Find a line which minimizes the sum of the distances between such line and the given demand points. Daniel Scholz University of G¨ ottingen May 27, 2010 5 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :The Weber problem :Line location in two dimensions :Line location in three dimensions Known results for the three-dimensional median line problem Some special cases are discussed in Brimberg, Juel, Sch¨ obel (2002) and in Brimberg, Juel, Sch¨ obel (2003): 1. Locating a vertical line can be reduced to a Weber problem in the two-dimensional space. 2. The case that all demand points are contained in a hyperplane yields basically a two-dimensional median line problem. However, there is no solution method for the general median line problem in the three-dimensional. Daniel Scholz University of G¨ ottingen May 27, 2010 6 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :Closed formula :Properties :Problem parametrization Closed formular for the three-dimensional median line problem (1) A line r in R 3 has the form r = r ( x , d ) = { x + td : t ∈ R } where d ∈ R 3 \ { 0 } is the direction of r and x ∈ R 3 . Notation For any a ∈ R 3 and x , d ∈ R 3 with d � = 0 denote by δ ( r ) = δ a ( x , d ) = min t ∈ R � x + td − a � 2 the Euclidean distance from a to the line r = r ( x , d ). Hence, the median line problem can be formulated as follows: n � min δ a k ( x , d ) . x , d ∈ R 3 k =1 d � =0 Daniel Scholz University of G¨ ottingen May 27, 2010 7 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :Closed formula :Properties :Problem parametrization Closed formular for the three-dimensional median line problem (2) Lemma 1 For any a ∈ R 3 and x , d ∈ R 3 with d � = 0 we find � 2 − ( d T ( a − x )) 2 � x − a � 2 δ a ( x , d ) = . d T d Hence, the median line problem becomes � n 2 − ( d T ( a k − x )) 2 � � x − a k � 2 min d T d x , d ∈ R 3 k =1 d � =0 where A = { a 1 , . . . , a n } ⊂ R 3 are the given demand points. Daniel Scholz University of G¨ ottingen May 27, 2010 8 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :Closed formula :Properties :Problem parametrization Properties (1) Obviously, the line r ( x , d ) is not uniquely defined by the pair ( x , d ). One finds r ( x , d ) = r ( x + ν d , d ) for any ν ∈ R . Hence, we assume that x is the intersection of r with the hyperplane H d = { y : d T y = 0 } . Corollary 2 For any a ∈ R 3 and x , d ∈ R 3 with d � = 0 and d T x = 0 we find � 2 − ( d T a ) 2 � x − a � 2 δ a ( x , d ) = d T d . Daniel Scholz University of G¨ ottingen May 27, 2010 9 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :Closed formula :Properties :Problem parametrization Properties (2) Moreover, we have r ( x , d ) = r ( x , τ d ) for any τ � = 0. Hence, it can be assumed that � d � = 1. Corollary 3 For any a ∈ R 3 and x , d ∈ R 3 with � d � = 1 and d T x = 0 we find � � x − a � 2 2 − ( d T a ) 2 . δ a ( x , d ) = Daniel Scholz University of G¨ ottingen May 27, 2010 10 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :Closed formula :Properties :Problem parametrization Properties (3) Lemma 4 The (three-dimensional) median line problem with fixed direction d ∈ R 3 \ { 0 } is equivalent to a (two-dimensional) Weber problem. Figure: Solution method for the median line problem with fixed direction d . Daniel Scholz University of G¨ ottingen May 27, 2010 11 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :Closed formula :Properties :Problem parametrization Properties (4) Corollary 5 It exists an optimal solution ( x ∗ , d ∗ ) ∈ R 6 to the median line problem such that the line r = r ( x ∗ , d ∗ ) intersects the convex hull of A . This is a direct consequence of Lemma 4 since it is well-known that there exists an optimal solution for the Weber problem which intersects the convex hull of the (projected) demand points, see Wesolowsky (1975). Daniel Scholz University of G¨ ottingen May 27, 2010 12 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :Closed formula :Properties :Problem parametrization Problem parametrization (1) If we consider only d = ( d 1 , d 2 , d 3 ) ∈ R 3 such that d 3 = 1 as well as x = ( x 1 , x 2 , x 3 ) ∈ R 3 such that d T x = 0, we obtain x 3 = − ( x 1 d 1 + x 2 d 2 ) . Hence, Corollary 2 yields n � � g k f 1 ( x 1 , x 2 , d 1 , d 2 ) = 1 ( x 1 , x 2 , d 1 , d 2 ) k =1 and with a k = ( α k , β k , γ k ) for k = 1 , . . . , n we have 1 ( x 1 , x 2 , d 1 , d 2 ) = ( x 1 − α k ) 2 +( x 2 − β k ) 2 +( x 1 d 1 + x 2 d 2 + γ k ) 2 − ( d 1 α k + d 2 β k + γ k ) 2 g k . d 2 1 + d 2 2 + 1 Daniel Scholz University of G¨ ottingen May 27, 2010 13 / 34

:Introduction :Problem formulation :Geometric branch-and-bound :Numerical results :Closed formula :Properties :Problem parametrization Problem parametrization (2) In the same way we can also fix d 1 = 1 and d 2 = 1 which yields n � � g k f 2 ( x 1 , x 2 , d 1 , d 2 ) = 2 ( x 1 , x 2 , d 1 , d 2 ) , k =1 n � � g k f 3 ( x 1 , x 2 , d 1 , d 2 ) = 3 ( x 1 , x 2 , d 1 , d 2 ) . k =1 To sum up, the six-dimensional median line problem is equivalent to the four-dimension problem min f i ( x 1 , x 2 , d 1 , d 2 ) . x 1 , x 2 , d 1 , d 2 ∈ R i ∈{ 1 , 2 , 3 } Daniel Scholz University of G¨ ottingen May 27, 2010 14 / 34