Using a Projected Subgradient Method to Solve a Constrained Optimization Problem for Separating an Arbitrary Set of Points into Uniform Segments

Transcription

1 Using a Projected Subgradient Method to Solve a Constrained Optimization Problem or Separating an Arbitrary Set o Points into Uniorm Segments Michael Johnson May 31, Background Inormation The Airborne Topographic Mapper [Kra09] is a circular scanning LIDAR that measures surace height along a track as the aircrat lies overhead. We desire to run analysis on how the topography is changing along the track, using bins o constant along-track length to do so. As the data is stored in long vectors o latitude, longitude, and height sorted only in time, the circular pattern in which the data is recorded does not allow simple sorting methods to be used to bin the data. We propose to use optimization methods to locate the points on the boundary o the track that yield the bins o constant length. Figure 1 illustrates a sample track o data being considered, which was taken over sea ice above Northern Greenland in Please reer to the project proposal and progress report or more background inormation on the problem and data. 2 Model Figure 2 illustrates a simpliied instance o this problem which assumes the track boundaries, B 1 and B 2, are straight, parallel lines. A track bin is deined as the area enclosed by a starting line segment connecting B 1 and B 2 and an ending line segment connecting B 1 and B 2. The segment length is deined as the distance between the center o the starting line segment, x s, and the center o the ending line segment, x. The points x 1 and x 2 are known, and we wish to ind the points x 3, x 4, and x (which are all related) that enclose the track bin. The circle o radius α deines the possible points or the center o the next line segment that is a distance o α away rom our starting point. In this problem instance, the obvious choice o x 3, x 4, and x are shown, as they are the points that create the most uniorm segment o length α. The length d o our optimal ending line segment has the minimum 1

2 83.7 Sample Track o ATM Measurements Latitude Longitude Figure 1: Sample track o ATM measurements length o any line segment ormed by the intersection o the boundaries with a line tangent to our circle. Thereore, solving this problem involves minimizing the length o this line segment. Figure 2: Simpliied Instance This simpliied instance will be used to solve this problem or the actual data. The nonconvex nature o the boundaries will be addressed by using an aine approximation which is updated as the solution progresses. A mathematical model o this problem is: minimize x 3 x 4 2 subject to x 3 = A 1 1 b 1 x 4 = A 1 2 b 2 x x s 2 = α The equations or x 3 and x 4 are ormed as the solution to the problem o inding the 2 (1)

3 intersection between the boundaries and the line tangent to the circle at point x. For simplicities sake, we will consider the point x s to be the origin o our rame so can be ignored in uture equations. Likewise, we will use x = [x y] T. In solving or the intersection o the two lines, we get: [ ] [ ] x y x T A 1 =, b m B1 1 1 = x (2) b B1 In equation (2), m B1 reers to the slope o our aine approximation to the track boundaries, and b B1 reers to the intercept. Because A and b are dependent on the point x, there is not a simple closed orm solution to this problem. The problem can be progressed by inputting the deinitions or x 3 and x 4 into the objective unction and splitting the vector into its constituents to give us F (x, y). The gradient can then be ormed as F (x, y) = [δf/δx δf/δy] T. Although in general F (x, y) is a non-convex unction, it can be shown to be convex when x is constrained to lie on the circle. The derivation or F (x, y) using (1) and (2) is as ollows. Using the deinitions or x 3 and x 4, the objective unction becomes: F (x, y) = x 3 x 4 2 = A 1 1 b 1 A 1 2 b 2 2 (3) Using the deinitions o A and b and x = [x y] T gives us an equation o the orm: Where: F (x, y) = 1 (x, y) + g(x, y) (4) h(x, y) h(x, y) = (x + m 1 y)(x + m 2 y) (x, y) = (x 2 + y 2 b 1 y)(x + m 2 y) (x 2 + y 2 b 2 y)(x + m 1 y) g(x, y) = (m 1 x 2 + b 1 x + m 1 y 2 )(x + m 2 y) (m 2 x 2 + b 2 x + m 2 y 2 )(x + m 1 y) Using these equations, we similarly get expressions or δf/δx and δf/δy. Dropping the (x, y) notation or simplicity, we have: δf/δx = (1/h 2 )(h[2(δ/δx) + 2g(δg/δx)] (δh/δx)[ 2 + g 2 ] δf/δy = (1/h 2 )(h[2(δ/δy) + 2g(δg/δy)] (δh/δy)[ 2 + g 2 ] Using the equations in (5), we get: δh/δx = 2x + (m 1 + m 2 )y δh/δy = 2m 1 m 2 y + (m 1 + m 2 )x δ/δx = 2x(x + m 2 y) + (x 2 + y 2 b 1 y) 2x(x + m 1 y) + (x 2 + y 2 b 2 y) δ/δy = (2y b 1 )(x + m 2 y) + m 2 (x 2 + y 2 b 1 y) (2y b 2 )(x + m 1 y) + m 1 (x 2 + y 2 b 2 y) δg/δx = (2m 1 x + b 1 )(x + m 2 y) + (m 1 x 2 + b 1 x + m 1 y 2 )... (2m 2 x + b 2 )(x + m 1 y) + (m 2 x 2 + b 2 x + m 2 y 2 ) δg/δy = 2m 1 y(x + m 2 y) + m 2 (m 1 x 2 + b 1 x + m 1 y 2 )... 2m 2 y(x + m 1 y) + m 1 (m 2 x 2 + b 2 x + m 2 y 2 ) (7) 3 (5) (6)

4 The equations presented in (5), (6), and (7) allow us to orm the gradient o the objective unction. Figure 3: Unconstrained objective unction Figure 3 shows a plot o the unconstrained objective unction. The large increase in value is attributed to when the line tangent to the circle is parallel to the boundaries. The unction is convex on each halplane and using the gradient will lower the objective value and ensures that x stays on the proper side o x s ; we do not want the segment to go to a section o the track that is already segmented. 3 Algorithm It is proposed to solve this problem using a projected subgradient method [Boy11] because o the constraint that the solution lie on the circle centered around x s. Do while not converged: Generate linear approximation o boundaries Do while not converged: Calculate F (x (k) ), the gradient at F (x(k) Take x (k+1) = P(x (k) γ k F (x (k) )) ) where P is the projection onto the circle centered around x s, γ k is the step length (could be 1/k) Update linear approximation o boundaries based on solution 4

5 As The objective is dierentiable, the subgradient is the gradient o the unction which simpliies the procedure. The constraint that x lie on the circle centered at x s with radius α requires the projection, as the gradient o the unconstrained objective unction will not keep the value o x (k) constrained to the circle. Projection onto the circle can be done in the ollowing manner. Using x s = [0 0] T, we calculate θ = tan 1 (y (k) /x (k) ) to determine the angle that the current value o x (k) makes with x s. Projecting the point onto the circle is then: [ ] x (k) αcosθ = (8) αsinθ Equation (8) will give the point on the circle that is closest to the x calculated using the gradient o the objective. 4 Results and Conclusion The method and algorithm described in this paper was applied to the set o data shown in igure 1. This dataset was chosen as it has multiple traits that are to be tested. The track contains near-straight sections, curved sections, kinks, sections where the track width narrows, and sections o missing measurements. The latter traits introduce non-uniormity to the boundaries which will test the robustness o the model and algorithm Figure 4: The entire track segmented. 5

6 (a) Detail (b) Detail (c) Detail (d) Detail 4 Figure 5: Details o segmented track Figure 4 shows the entire track ater segmentation to show the method worked on the entire track. The details in igure 5 show close-ups o several segments o the track to illustrate the perormance o the algorithm. In the igures, the dashed lines connect x 3 and x 4 that are calculated or each segment, which are shown as stars along the track boundary. Figure 5a shows a relatively straight section o track ater segmentation. Figure 5b shows the portion o the track where the boundary narrows due to bad measurements. The algorithm is able to successully segment and pass through this region. Likewise, igure 5c shows the portion o the track that has a large gap in both boundaries, again due to bad measurements. Even though the boundaries are missing, the algorithm makes a visually good approximation to what the segments may be like. Figure 5d shows the area o the track with modest kinks. The algorithm segments this region without problem, and the pattern o the segments at the kinks visually makes: adjacent boundary points are closer together on the inside o the kink and arther apart on the outside o the kink. The results presented in this section suggest the model and algorithm described is adequate at solving this problem. Ater careul selection o problem parameters, the method is shown to be robust, eicient, and accurate. 6

7 Reerences [Boy11] S. Boyd. Subgradient Methods or Constrained Problems, EE364B Lecture Notes. Stanord Univeristy, [Kra09] William B Krabill. IceBridge ATM L1B Qit Elevation and Return Strength. Boulder, Colorado USA: National Snow and Ice Data Center. Digital Media,