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Continuous approximation formulas for location and hybrid routing/location problems

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Continuous approximation formulas for location and hybrid routing/location problems

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Content CONTINUOUS APPROXIMATION FORMULAS FOR LOCATION AND HYBRID ROUTING/LOCATION PROBLEMS by Bo John Jones A Dissertation Presented to the FACULTY OF THE USC GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulllment of the Requirements for the Degree DOCTOR OF PHILOSOPHY (INDUSTRIAL AND SYSTEMS ENGINEERING) August 2022 Copyright 2022 Bo John Jones Acknowledgements I would rst like to thank my advisor, Professor John Gunnar Carlsson, for his guidance and for provid- ing me with a model to aspire to as a researcher, teacher and mentor. Thank you also to my dissertation committee members Professor Maged M. Dessouky and Professor Vishal Gupta. In addition I owe thanks to qualifying committee members and wonderful teachers Professor Jong-Shi Pang and Professor Sheldon Mark Ross. I am very grateful to my fellow students for their friendship and for creating a work environ- ment that pushed me to accomplish what I have accomplished. In particular I would like to thank Ziyu He and my group-mates (in order of joining group) Siyuan Song, Ye Wang, Xiangfei Meng, Jiachuan Chen, Javad Azizi, Ying Peng, Shannon Sweitzer-Siojo, Haochen Jia and Julien Yu. Finally, I could not have done any of this without the love and support of my family. ii TableofContents Acknowledgements ii ListofTables v ListofFigures vi Abstract ix Chapter1: Introduction 1 1.1 Structure of this work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 The continuous approximation paradigm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Chapter2: Locationproblems 3 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2.2.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2.2.2 Lemmas for our analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2.3 Problem denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.3 Thek-medians problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.3.1 Bounds for thek-medians problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.3.2 Bounds for the balanced medians problem . . . . . . . . . . . . . . . . . . . . . . . 16 2.3.3 The non-uniformk-medians problem . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.4 Thek-dispersion problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.4.1 The uniform case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.4.2 The non-uniform case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 2.5 The generalized spanning tree problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 2.6 Computational experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 2.6.1 Medians problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 2.6.2 Thek-dispersion problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 2.6.3 The generalized spanning tree problem . . . . . . . . . . . . . . . . . . . . . . . . . 46 2.6.4 Proportionality analysis and conclusions . . . . . . . . . . . . . . . . . . . . . . . . 47 Chapter3: Thesidekickroutingproblem 50 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.1.1 Remark on notational conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 3.2 Related work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 3.2.1 Other autonomous vehicle delivery schemes and applications . . . . . . . . . . . . 55 iii 3.2.2 Exact solutions to sidekick problems . . . . . . . . . . . . . . . . . . . . . . . . . . 56 3.2.3 Heuristic solutions to sidekick problems . . . . . . . . . . . . . . . . . . . . . . . . 58 3.2.4 Theoretical results and continuous approximation models . . . . . . . . . . . . . . 59 3.3 Problem denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 3.4 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 3.4.1 Existing results from related work . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 3.4.2 Further notes on Theorem 33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 3.5 A continuous approximation analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 3.5.1 Naive asymptotic analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 3.5.2 A lower bound for (SK2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 3.5.3 An upper bound for (SK1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 3.5.4 Convergence analysis for (SK1) and (SK2) . . . . . . . . . . . . . . . . . . . . . . . 80 3.6 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 3.7 Computational results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 3.7.1 Uniformly distributed demand with Euclidean travel . . . . . . . . . . . . . . . . . 85 3.7.2 On a real road network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 3.8 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Chapter4: Conclusions 98 References 99 AppendixA 110 Proof of Lemma 38 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 iv ListofTables 2.1 Approximate constant by which to multiply the proportionality given in Section 2.6.4 to predict the objective value under the assumption of uniform point distribution. . . . . . . 48 3.1 A summary of work on sidekick problems. Papers are classied by their model assumptions as well as their contributions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 3.2 The ratio ((Time With Sidekicks)/(Time Without Sidekicks)) / p 0 =(maxf 0 ; 1 kg) on real Downtown Los Angeles road network withk = 3 and sidekick speed the given multiple of the driving speed. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 v ListofFigures 2.1 The curvec(1e 1=c ) = and its two approximations as! 0 and! 1. . . . . . . . 16 2.2 Partitioning the square into rectangles withd = 4. Note the “decient” rectangles to the right that contain fewer thand + 1 points. . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.3 The functiong(t) dened in (2.8). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.4 There are 14 non-empty squares in odd rows and columns, all of which are at least 1= p cn away from each other. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.5 Figure 2.5a is the curvec(1e 1=c )=4 = , and Figure 2.5b shows that the function 7! 0:3549 p (1)= is bounded above by one of 1= p c() and p (log)=. . . . . 34 2.6 The grid cells i and their interior components 0 i , all of which are a distance of at least apart from one another. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 2.7 Objective values for thek-medians problem plotted with the lower and upper bounds. . . 45 2.8 Objective values for the balanced medians problem plotted with the lower and upper bounds. 45 2.9 Objective values for thek-dispersion problem plotted with the lower and upper bounds. . 46 2.10 Objective values for the generalized minimum spanning tree problem plotted with the lower and upper bounds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 2.11 Objective value ratios. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 3.1 The gure on the left shows a travelling salesman tour of a set of client destinations and a central depot, that is, the shortest tour that visits a collection of points and starts and ends at the central depot. The gure on the right shows the solution to a “sidekick” problem in which the truck has a “helper” (such as a robot or a drone) that alternates between visiting the truck and visiting the customer locations. Note that the truck’s tour with helpers is about half the length of the tour with no helpers. . . . . . . . . . . . . . . . . . . . . . . . 53 3.2 Various hardware implementations of sidekick routing schemes. . . . . . . . . . . . . . . . 60 vi 3.3 Solutions to the sidekick problem forn = 30 customers,k = 3 sidekicks, and a sidekick speed, 1 , that is twice the speed, 0 , of the truck. The solid line is the truck tour; dotted lines are the sidekicks’ routes; the square represents the starting and nishing point of the truck. In the gure on the left all deliveries are made by the sidekicks (Problem SK1). In the gure on the right the truck is also allowed to make deliveries (Problem SK2). . . . . . 68 3.4 Constructing a TSP solution from a solution to problem (SK2). The horizontal line represents the tour of theu i in the (SK2) solution. We follow the tour, traveling fromu i top i and back for eachi2S. IfT n is the cost of the problem (SK2) solution then the total cost of the resulting TSP solution is less than or equal to ( 0 +k 1 )T n . . . . . . . . . . . 77 3.5 Constructing a balanced medians solution from a solution to problem (SK2). The horizontal line represents the tour of theu i in the (SK2) solution. Here we choosed = 4 and group every 5 points along the tour. We choose as the median for these 5 points the point which is closest to the tour. Using the paths pictured, it is clear that to connect all points to their medians we need travel at mostd times the length of the truck tour plus twice the total distance from the points to theiru i . IfT n is the cost of the problem (SK2) solution then the total cost of the resulting Bounded Medians solution is less than or equal to (d 0 +k 1 )T n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 3.6 The tour described in Lemma 37, assumingk = 3. . . . . . . . . . . . . . . . . . . . . . . . 79 3.7 The tour described in Lemma 38; it consists of the same kind of tour as in Figure 3.6, but “scaled” with respect to the probability distribution that is indicated by shading. . . . . . . 80 3.8 Value of the sidekick problem completion time times p 0 max( 0 ; 1 k)= p n plotted over a range of values of 1 = 0 andn for xedk. We expect this value to be constant for large n. We can see that our heuristic approach does not suciently capture the benet of introducing more and more sidekicks in the slow sidekick case, but otherwise these plots are near constant, particularly asn becomes large. . . . . . . . . . . . . . . . . . . . . . . . 88 3.9 Value of the sidekick problem completion time times p 0 max( 0 ; 1 k)= p n plotted over a range of values ofk andn for xed 1 = 0 . We expect this value to be constant for large n. As these values of 1 = 0 are small our heuristic approach is not capturing the benet due to introducing more and more sidekicks, and thus these plots are not constant though they fall within a range of about 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 3.10 Value of the sidekick problem completion time times p 0 max( 0 ; 1 k)= p n plotted over a range of values ofk andn for xed 1 = 0 . We expect this value to be constant for large n. We see the result is near constant asn becomes large. . . . . . . . . . . . . . . . . . . . 90 3.11 Value of the sidekick problem completion time,T , times p 0 max( 0 ; 1 k)= p n plotted over a range of values ofk and 1 = 0 for xedn. We expect this value to be constant for largen. When 1 = 0 is small our heuristic approach does not suciently capture the benet of introducing more and more sidekicks, but otherwise these plots are near constant. 91 vii 3.12 Value of the sidekick problem completion time times p 0 max( 0 ; 1 k) divided by the optimal time for a TSP with only the truck, plotted over a range of values ofk and 1 = 0 for xedn. We expect this value to be constant for largen. When 1 = 0 is small our heuristic approach does not suciently capture the benet of introducing more and more sidekicks, but otherwise these plots are near constant. . . . . . . . . . . . . . . . . . . . . . 92 3.13 Pseudocode for our solution on real road network data. . . . . . . . . . . . . . . . . . . . . 96 viii Abstract Continuous approximation is a powerful tool for predicting the objective value of optimization problems having a geographic component. Work in continuous approximation has largely focused on routing prob- lems. We will focus instead on location problems and a hybrid routing/location problem arising from a novel delivery system that utilizes a truck and unmanned aerial or ground vehicles in tandem. We provide new continuous approximation formulas and new bounds on constants appearing in existing formulas. Our results characterize the objective value’s dependence on problem parameters as the problems grow large. This allows practitioners to predict objective values and make decisions about those parameters. ix Chapter1 Introduction 1.1 Structureofthiswork One of the more novel recent innovations in the logistics world, both in theory and in practice, is the use of small autonomous vehicles to facilitate last-mile delivery. One particular scheme that has received considerable recent attention is the “sidekick” scheme, in which a large cargo truck acts as a mobile “base” that deploys smaller vehicles, such as drones or unmanned ground vehicles (UGVs). The problem of routing the truck and sidekicks in this system is really a hybrid of a routing problem and a location problem, as we must decide locations at which sidekicks are launched and retrieved along the route. For this reason the path we will take to insights about the sidekick routing scheme will require appealing to both a continuous approximation formula for the traveling salesman routing problem and to a continuous approximation formula for a degree-constrained variant of thek-medians location problem. The majority of research in the continuous approximation paradigm has emphasized routing prob- lems, such as the traveling salesman problem and the vehicle routing problem. Location problems, like thek-medians variant to be used in our sidekick analysis, are not as well-studied. In Chapter 2 we inves- tigate continuous approximation formulas for this problem as well as other problems involving location, specically thek-medians,k-dispersion, and generalized minimum spanning tree problems. We contribute bounds for constants that describe the growth rate of the cost of these problems as the number of demand 1 points becomes large. We conduct computational experiments that verify that they provide a good ap- proximation in practice. In Chapter 3, we use results from Chapter 2 to develop a continuous approximation model that esti- mates the improvements to eciency that the sidekick system provides, in the asymptotic limit as many demand points are drawn from a continuous probability distribution in the plane. We conduct computa- tional experiments to demonstrate the power of our model to predict service times. In Chapter 4 we provide concluding remarks. 1.2 Thecontinuousapproximationparadigm In the continuous approximation paradigm we approximate a discrete input like the locations of customers to be visited on a tour by a continuous demand distribution from which our customer points are then drawn. We can then arrive at results that amount to a strong law for your optimization problem. The most classical result for instance is that ifn points to be visited are drawn from a uniform distribution, then the length of an optimal traveling salesman tour divided by p n converges to a constant asn goes to innity [9]. We can thus predict that the optimal tour length will be this constant times p n when our number of customers is large and our demand is uniformly distributed. Such results also hold for nonuniform demand distributions. In this work we will arrive at similar results for an array of problems. We provide bounds on the relevant constants to allow for objective prediction. For each problem we will also characterize how the constants depend on the problem parameters. This allows us to assess the impact of changing those param- eters. For instance, in the case of the sidekick system we can predict the total service time and characterize its dependence on the composition of the eet, i.e. how many sidekicks are on the truck and how fast they and the truck can travel. 2 Chapter2 Locationproblems 2.1 Introduction The continuous approximation paradigm has been one of the most prevalent tools in logistics systems analysis since the 1950s [9, 34]. The vast majority of its applications have focused on vehicle routing applications, and have emphasized such problem aspects as trip length, the impact of capacities or time windows on quality of service [71], or the value of districting [40]. Such models are valuable because they enable one to make rapid managerial insights by making use of simple analytic functions as opposed to detailed models involving optimization problems that may be intractable [25]. The goal of this chapter is to introduce some bounds and continuous approximation formulas that arise for location problems, where the decision is “where to put things”, as opposed to “where should things move”. This work is certainly not the rst to do so [54, 53, 83], though here we will both improve on existing results and establish new formulas that have not previously been recognized, to the best of our knowledge. The remainder of this chapter is as follows: Section 3.4 establishes some preliminary lemmas that we refer to throughout, Sections 2.3, 2.4 and 2.5 establish our new formulas for thek-medians problem, thek- dispersion problem, and the generalized spanning tree, and Section 2.6 shows some computational results that verify the formulas and bounds derived herein. 3 Literaturereview Prior non-probabilistic work characterizing the growth of location problem objectives includes Francis et al.’s [37] demonstration of partial convexity in k of the k-medians, k-center and more general location objectives. The seminal papers by Beardwood, Halton, and Hammersley [9] and Few [34] on the Euclidean TSP are the rst to our knowledge that employed continuous approximation to study the asymptotic behavior of a geometric combinatorial optimization problem. The books of Yukich [137] and Steele [112] present a general framework of convergence of subaddititive and superadditive Euclidean functionals which sub- sumes the analysis for the TSP. In addition, [137] collects convergence results for the power-weighted minimum spanning tree (MST) problem (that is, [136] extends the results of [111] and [4]), Steiner tree problem,k-nearest neighbors problem (due to McGivney [84]), minimal 2-matching problem, many trav- eling salesman problem, directed TSP (due to Talagrand [115]), minimal triangulation problem, and me- dians problems. Steele [111] collects convergence results for the Steiner tree problem, minimal matching problem and MST problem. Steele [112] also presents a method due to Avram and Bertsimas [8] for nd- ing central limit theorems for the lengths of thek-nearest neighbor graph, Delaunay triangulation and Voronoi diagram of random points. Penrose and Yukich [93] use the objective method (see [5]) to provide a general framework for establishing weak laws of large numbers for geometric problems that are not amenable to the subadditivity arguments found in these books. They apply it to MST,k-nearest neighbors, Delaunay and Voronoi, sphere of inuence and proximity graphs as well as a packing model. Jiménez and Yukich give a convergence result for objectives the sum of edge lengths weighted by some function in- cluding weightedk-nearest neighbor, Voronoi, and Delaunay graphs. Ajtai et al. [3] show the asymptotic behavior of the bipartite minimal matching problem and Talagrand [116] provides a general theory of bi- partite matching convergence that includes analysis of the problem of minimizing the maximum intrapair distance. Srivastav and Werth [109] show convergence for the degree constrained MST problem. Lee [73] 4 gives a convergence rate for Yukich’s power weighted MST result. Ganesan [41] shows the asymptotic behavior of the MST on a random geometric graph, having edges where the distance between points is below some threshold. Beyond Euclidean MST, Hutson and Shier [55] give bounds on the distribution and expectation of the minimum spanning tree objective given random edge weights. De Graaf et al. [47] show the convergence of a power assignment problem heuristic solution that uses the MST. Werth [133] studies the asymptotics of the multi-depot vehicle routing problem. Hwang et al. [56] show convergence of the shortest path between two points. Asymptotic results like these are useful for predicting how one’s solution to a given problem will scale as the number of points increases. However, many of the preceding papers do not concern themselves with the actualvalues of the constants to which their scaled objectives are converging. To make prediction possible, our work establishes bounds on these constants for a number of location problems. Bounds on the constant for the TSP can be found in the original paper of Beardwood et al. [9] and are improved in [113] and [43]. For an alternative statistical method of TSP objective prediction that depends on actual deterministic locations of points, see Golden [44] and [21]. For an approach to TSP objective estimation that uses regression and neural network models see Kwon et al. [70]. An analytical expression for, and bounds on, the MST constant can be found in Avram and Bertsimas [8] by way of an equivalence of models demonstrated by Jaillet [57]. In [35], Fisher and Hochbaum give a characterization, with constant bounds, of the objective’s growth for thek-medians problem withk that does not grow withn, whereas we will be concerned with the asymptotic behavior ofk-medians wherek scales linearly withn. Carlsson and Song [20] study the asymptotic behavior, with bounds on constants, of the “horsey routing problem”, and Carlsson et al. [18] study the generalized traveling salesman problem. Werth [133] provides bounds on the constant for the multi-depot vehicle routing problem. Beyond prediction, continuous approximation results are useful in the design of asymptotically optimal algorithms. They can be used to prove that a particular solution method yields a near-optimal solution 5 with high probability as the number of points becomes large. The rst to demonstrate this was the work of Karp [64, 63], which uses the original Beardwood, Halton and Hammersley result to analyze algorithms for the TSP. This type of analysis has been extended to location problems as well. For an asymptotically optimalk-medians heuristic see Fisher and Hochbaum [35]. For a survey of work on three heuristics for k-medians andk-centers see Hochbaum [53]. For both the discrete and continuous versions ofk-medians and k-centers see Piersma [95]. For the k-location problem see Foster and Vohra [36]. For the online traveling repairman problem and some online machine scheduling problems see Jaillet and Wagner [58]. For budgeted network design problems see Wong [134]. For the capacitated tree problem see Papadimitriou [92]. In the other direction, Jain [59] uses probabilistic models to demonstrate that linear programming relaxation bounds for the Steiner tree problem on networks are not tight asymptotically. 2.2 Preliminaries 2.2.1 Notation • We will sayf(x) =O(g(x)) asx!1 if there existsx 0 2R and positiveM2R such that jf(x)j<Mg(x) 8x>x 0 : • We will sayf(x) =o(g(x)) asx!1 if f(x) g(x) ! 0: • Forz with positive real part, the Gamma function is dened as (z) = R 1 0 x z1 e x dx. For positive integern, (n) = (n 1)!: 6 2.2.2 Lemmasforouranalysis We begin with some lemmas that we will refer to frequently throughout this chapter. Lemma 1. Letf : R! R be a real-valued function and letB D (r) R D be a ball of radiusr centered about the origin. We have Z B D (r) f(kxk)dx = Z r 0 S D1 (t)f(t)dt; whereS D1 (t) is the surface area of a (D 1)-sphere of radiust, which is given by S D1 (t) = 2 D=2 (D=2) t D1 : Proof. This follows from the denition of a Lebesgue integral, which involves integrating with respect to the level sets of a function [118]. Lemma2. The volume of aD-dimensional ball of radiusr is D=2 r D =(D=2 + 1): Lemma3. Let`> 0 and letDR 2n denote the set of alln-tuples (u 1 ;:::;u n ) of points inR 2 such that P n i=1 ku i k`: The volume ofD, Vol(D), satises Vol(D) = (2) n (2n + 1) ` 2n : (2.1) Proof. This is just the integral Z B 2 (`) Z B 2 (`kunk) Z B 2 (` P n i=3 ku i k) Z B 2 (` P n i=2 ku i k) 1du 1 du 2 du n1 du n ; which we can compute by induction. Forn = 1, Z B 2 (`) 1du 1 =` 2 = (2) 1 (2(1) + 1) ` 2(1) : 7 Suppose the relation holds for all` 0 for the set of all (n 1)-tuples such that P n1 i=1 ku i k` 0 . Then ifD is the set of alln-tuples (u 1 ;:::;u n ) of points inR 2 such that P n i=1 ku i k` , we have Vol(D) = Z B 2 (`) Z B 2 (`kunk) Z B 2 (` P n i=3 ku i k) Z B 2 (` P n i=2 ku i k) 1du 1 du 2 du n1 du n = Z B 2 (`) (2) n1 (2(n 1) + 1) (`jju n jj) 2(n1) du n (induction hypothesis) = Z ` 0 2t (2) n1 (2(n 1) + 1) (`t) 2(n1) dt (Lemma 1) = (2) n (2(n 1) + 1) Z ` 0 t(`t) 2(n1) dt = (2) n (2n 1) ` 2n 2n(2n 1) = (2) n (2n + 1) ` 2n : Lemma4 (Stirling’s formula). The gamma function (x) satises log (x + 1) =x logxx + 1 2 logx + 1 2 log 2 + 1 2 log +O(1=x) asx!1. Proof. This is Stirling’s formula. 2.2.3 Problemdenitions The problems of interest throughout this chapter are dened below: Denition 5 (Medians Problem). Given a collection of pointsx 1 ;:::;x n inR 2 and a positive integerk, the k-medians problem is given by KMed(x 1 ;:::;x n ;k) := min Sf1;:::;ng:jSjk n X i=1 min j2S kx i x j k ; 8 that is, the problem of selecting a subsetS f1;:::;ng of “median” points such thatjSj k, that minimizes the sum of the distances from all points to their nearest median. Denition6 (Balanced Medians Problem). ThebalancedmediansproblemBMed(x 1 ;:::;x n ;d) is a further- constrained variation of the k-medians problem that imposes an additional constraint that each center x j :j2S can be assigned to at mostd 2 points in addition to itself, where the number of mediansk is given byk =dn=(d + 1)e. That is, BMed(x 1 ;:::;x n ;d) = minimize S:jSjk; :f1;:::;ng7!S n X i=1 jjx i x (i) jj; (2.2) where k = l n d + 1 m ; (i) is the index of the median assigned to pointi, and for allj2S,(j) =j and(i) =j for up tod of thei6=j. Denition 7 (Dispersion Problem). Given a collection of pointsx 1 ;:::;x n inR 2 and a positive integer k, thek-dispersion problem is given by KDisp(x 1 ;:::;x n ;k) := max Sfx 1 ;:::;xng:jSjk min x i 6=x j 2S kx i x j k ; that is, the problem of selecting a subsetSfx 1 ;:::;x n g of dispersed points such that the minimum pairwise distance between selected points is maximized. We note, breaking from our Euclidean setting, that Shier [107] demonstrated the duality of ak-centers type objective and ak-dispersion type objective when the underlying network is a tree. 9 Denition 8. Given a collection of sets,X 1 ;:::;X n , each consisting of points in R 2 , the generalized minimum spanning tree problem is given by GMST(X 1 ;:::;X n ) = min x 1 2X 1 ;:::;xn2Xn min T (x 1 ;:::;xn) X e2T kek; where the second minimum is taken over all connected graphsT onx 1 ;:::;x n andkek is the cost (Eu- clidean distance) of the edgee ofT ; that is, the problem of selecting one point from each set and a spanning tree on these points such that the total cost of the spanning tree is minimized. The following sections present new approximation formulas for thek-medians problem, the balanced medians problem, thek-dispersion problem, and the generalized minimum spanning tree problem. 2.3 Thek-mediansproblem For thek-medians problem, Hochbaum and Steele [54] show that whenk grows linearly inn, we have the following: Theorem9 (Asymptotic convergence of thek-medians problem). IfX 1 ;X 2 ;:::isasequenceofpointsi.i.d. uniform on [0; 1] 2 then for any2 (0; 1), with probability one the costKMed(X 1 ;:::;X n ;bnc) satises lim n!1 KMed(X 1 ;:::;X n ;bnc) p n = KMed (); where KMed () depends only on. In introducing the constant KMed , the authors of [54] note that “it is usually impossible” to compute such quantities exactly. Note that for the case wherek does not grow withn, asn goes to innity the problem of interest essentially becomes the continuous Fermat-Weber problem, for which algorithmic techniques [19, 30] and asymptotic analyses [51] are known. 10 For the balanced medians problem, [83] shows a more general result allowing for nonuniform demand: Theorem10 (Asymptotic convergence of the balanced medians problem). IfX 1 ;X 2 ;::: is a sequence of randompointsi.i.d. accordingtoanabsolutelycontinuousprobabilitydensityfunctionf denedonacompact planar regionR andd 2 is xed, then with probability one, the costBMed(X 1 ;:::;X n ;d) satises lim n!1 BMed(X 1 ;:::;X n ;d) p n = BMed (d) ZZ R p f(x) dx where BMed (d) depends only ond. Our rst new results are to provide bounds for KMed and BMed . 2.3.1 Boundsforthek-mediansproblem Theorem11. The function KMed () satises 0:1780 (1) 3=2 p KMed () 0:75 (1) 3=2 p : Proof. The high level view of the lower bound is that we will use the union bound to bound the probability that some selection of medians and assignment of points to medians has cost less than what will be our lower bound. The total number of possible selections of medians is n k , and once those are xed, the number of possible assignments of those medians to other points isk nk , and therefore P(KMed(X 1 ;:::;X n ;k)<`) =P(some selection of, and assignment to, medians is of cost<`) sum over all selections and assignmentsP(cost of choice <`) = (# of ways to select)(# of ways to assign)P(cost of arbitrary choice <`) = n k k nk P(cost of arbitrary selection and assignment<`): (2.3) 11 To obtain an upper bound on the right-hand side of the above, we x our median indices,S, and our assignment map, which we will call, so thatX (i) is the median assigned to pointX i . Because we can reorder and adjust andS accordingly, we can assume without loss of generality that theX i are ordered such that the medians we have selected are the lastk points,X nk+1 ;:::;X n . We then see that the cost of a particular selection and assignment is given by P nk i=1 kX i X (i) k. Next, we suppose that the medians X nk+1 ;:::;X n take on xed valuesp 1 ;:::;p k , and that the assignment map is also xed, and consider the conditional probability P nk X i=1 kX i X (i) k<` X nk+1 =p 1 ;:::;X n =p k ! =P nk X i=1 kX i p (i) k<` ! ; where we have abused notation by dening p (i) to be the median assigned to point X i . DeneE R 2(nk) to be the set of all pointsx 1 ;:::;x nk whose total distances to their associated landmark points does not exceed`: E := ( x 1 ;:::;x nk 2R 2 : nk X i=1 kx i p (i) k<` ) : Then, since all pointsX i are drawn uniformly from the unit square, we see that the following holds for all p 1 ;:::;p k and all assignment maps: P nk X i=1 kX i p (i) k<` ! =P((X 1 ;:::;X nk )2E) = Vol(E\ [0; 1] 2 ) Vol(E): 12 To compute this volume, we make the volume-preserving transformationu i :=x i p (i) (that is, translate each median point to the origin and move its assigned set commensurately) and consider E 0 := ( u 1 ;:::;u nk 2R 2 : nk X i=1 ku i k<` ) ; because the aforementioned map is volume-preserving, we have Vol(E) = Vol(E 0 ), and note thatE 0 only depends on` (and in particular, not on any landmark points). By Lemma 3 we have Vol(E 0 ) = (2) nk (2(nk) + 1) ` 2(nk) ; and hence for all selections ofS and, we have P nk X i=1 kX i X (i) k<` ! Vol(E 0 ) = (2) nk (2(nk) + 1) ` 2(nk) P(cost of arbitrary selection and assignment<`) =) (2) nk (2(nk) + 1) ` 2(nk) (2.4) We now combine the preceding analysis into our original expression (2.3) to nd that P(KMed(X 1 ;:::;X n ;k)<`) n k k nk P(cost of arbitrary selection and assignment<`) (2.5) n k k nk (2) nk (2(nk) + 1) ` 2(nk) = (n + 1) (k + 1)(nk + 1) k nk (2) nk (2(nk) + 1) ` 2(nk) : 13 From this point on, we merely perform algebraic manipulations. Substituting` =c p n; taking logarithms, and applying Lemma 4, we have logP(KMed(X 1 ;:::;X n ;k)<c p n) log (n + 1) (k + 1)(nk + 1) k nk (2) nk (2(nk) + 1) (c p n) 2(nk) = [(log 2 log 2 log + 3 log(1) 2 logc 2) log 2 + log + log 3 log(1) + 2 logc + 2]n logn +O(1): It is easy to see that our probability of interest approaches zero if the above logarithmic expression ap- proaches1, which will happen if the coecient ofn in the above is negative. Isolating the coecient ofn, we have (log 2 log 2 log + 3 log(1) 2 logc 2) log 2 + log + log 3 log(1) + 2 logc + 2< 0 m p 2(1) 3=2 e p (21)=(22) >c: In addition, for2 (0; 1); we have (21)=(22) p e. To show this we rst note the equivalency, (21)=(22) p e () 2 1 2 2 log() 1 2 (1 + log()) () 1 log() 1 () log() 1 : 14 Then indeed, letting h() = log() 1 for> 0; we have h 0 () = 1 1 2 = 1 1 8 > > > > < > > > > : > 0 when> 1 < 0 when< 1: Thush() is minimized at = 1 with valueh(1) = 0 and log() ( 1)= for2 (0; 1). We then have the desired lower bound on KMed () because p 2= p e 3 > 0:1780. For the upper bound, we divide the unit square intocn square cells, wherec is to be determined later. At a high level, we will make our squares just large enough so as to haven nonempty squares. We can then assign each point to a median in its square. The distance to the median is then bounded in terms of the size of a square. The number of points in an arbitrary cell follows a binomial distribution withn trials and success probability 1=(cn), which approaches a Poisson distribution with mean 1=c asn!1. Our strategy is: for each non-empty cell, select a median point within that cell uniformly at random from one of the points therein. This assignment is feasible provided that the number of non-empty cells does not exceedn. If we letN Pois(1=c) denote the number of points within an arbitrarily selected cell, the law of large numbers says that the fraction of allcn cells that are non-empty approaches 1 Pr(N = 0) = 1e 1=c as n ! 1, and therefore, our strategy is a valid one provided that cn(1 e 1=c ) n (note that then terms on both sides of the inequality cancel each other out). We denec() to be the solution to the equationc(1e 1=c ) =; this does not have an algebraic closed form, but it is routine to verify that c() as! 0 and thatc() 1 2(1) as! 1; see Figure 2.1. There are (1)n non-median points, and the expected cost due to each one isq= p c()n, whereq = p 2=15 + 1 3 log(1 + p 2) + 2=15 0:5214 15 0 0:2 0:4 0:6 0:8 1 0 2 4 c c(1e 1=c ) = c = c = 1 2(1) Figure 2.1: The curvec(1e 1=c ) = and its two approximations as! 0 and! 1. is the average distance between two points in a unit square [16]. The total cost of this assignment policy satises Cost p n ! (1)nq= p c()n p n = q(1) p c() with probability one asn!1. By considering the inverses of the monotonically increasing functions c() and (1=2)=(1) we nd c() 1 2 1 : Thus q(1) p c() 0:75 (1) 3=2 p ; which completes the proof. 2.3.2 Boundsforthebalancedmediansproblem We obtain bounds for the balanced medians problem using a similar argument as in the preceding: Theorem12. The function BMed (d) satises 0:2061 p d BMed (d) 2 3 p d: 16 The high level path to a lower bound will once again be to employ the union bound to bound the probability that some selection of medians and assignment of points to medians has cost less than what will be our lower bound. In order to do this it will be necessary to introduce a lemma concerning the number of ways we can assign points to medians given the balanced medians degree constraint. To obtain an upper bound we will divide our region into strips, further dividing those strips left to right into rectangles containingd + 1 points. It will be necessary to bound the cost of the balanced medians problem on the leftover rightmost rectangles that do not necessarily containd + 1 points. Having done this, we will then bound the cost of ourd + 1 point rectangles in terms of the size of the rectangle. Before we prove this result, we require a simple combinatorial lemma and a crude upper bound for the balanced medians problem. Lemma 13. Letp andq be positive integers. The number of ways to partitionf1;:::;pqg into an ordered collection ofp unordered sets of size exactlyq is (pq)!=(q!) p . Proof. As a point of clarication, when we write “an ordered collection of unordered sets”, we mean that the partitionf1; 2; 3g;f4; 5; 6g is distinct fromf4; 5; 6g;f1; 2; 3g, but equivalent tof3; 2; 1g;f6; 5; 4g. This is a textbook-level counting exercise: write out any permutation off1;:::;pqg, and let the rst block of q numbers be the rst set, let the second block ofq numbers be the second set, and so forth. Each one of these sets can be permutedq! dierent ways, and there arep sets, from which the result follows. Lemma14. For any set of pointsx 1 ;:::;x m in the unit square and anyd 2, we have BMed(x 1 ;:::;x m ;d) 3d p m. Proof. The length of the travelling salesman tour throughm points in the unit square cannot exceed p 2m+ 1:75 < 3 p m, as proven in [34]. Assume without loss of generality that the pointsx i are ordered with respect to their TSP tour. We can construct a balanced medians solution by using the induced clustering from the TSP tour, i.e. the rst cluster isfx 1 ;:::;x d+1 g, thenfx d+2 ;:::;x 2(d+1) g, and so forth; we will 17 designate the last point in each cluster as the median (the points of the formx q(d+1) for integersq). It is easy to see that the total cost of this assignment is at mostd times the length of the TSP tour, which itself cannot exceed 3 p m, which proves our desired result. We will now prove Theorem 12: Proof of Theorem 12. As BMed (d) is independent of the demand distribution, we can arrive at a lower bound by rst assuming we are in the case that theX i are i.i.d. Unif([0; 1] 2 ). We obtain the lower bound of BMed (d) in the same way as in the proof of Theorem 11, with only one minor change: whereas the union bound in Theorem 11 asserts that there arek nk possible assignment patterns between thek medians and thenk non-median points, the number of assignment patterns is less in the balanced medians problem due to the degree constraint. Since the convergence result in Theorem 10 holds with probability one, we are welcome to assume thatn is divisible byd + 1 so as to save wear and tear on oors and ceilings. This means that each of thek = n=(d + 1) medians is assigned to exactlyd points in addition to itself, and therefore Lemma 13 (withp =k =n=(d + 1) andq =d) tells us that the number of possible assignments is exactly(dk)!=(d!) k . Therefore, applying inequality (2.4), P(BMed(X 1 ;:::;X n ;d)<`) n k (dk)! (d!) k P(cost of arbitrary selection and assignment<`) n k (dk)! (d!) k (2) nk (2(nk) + 1) ` 2(nk) and we repeat the same procedure as in the proof of Theorem 11: we substitutek =n=(d+1) and` =c p n, take logarithms, and require the coecient ofn to be negative, so as to conclude that the probability of interest, P(BMed(X 1 ;:::;X n ;d)<c p n); 18 c= p n Figure 2.2: Partitioning the square into rectangles withd = 4. Note the “decient” rectangles to the right that contain fewer thand + 1 points. approaches zero provided that c< p 2d(d!) 1=(2d) p e(d + 1) (2d+1)=(2d) : We are done by noting that p 2d(d!) 1=(2d) p e(d+1) (2d+1)=(2d) > 0:2061 p d for all integersd 2, which completes the proof of the lower bound. To obtain the upper bound, divide the unit square into horizontal strips of heighth = c= p n, with c = p d + 1; the very bottom strip’s height may be shorter. There are m = d p n=ce p n=c + 1 such strips. Further subdivide each strip into rectangles so that each rectangle – except possibly, say, the rightmost in each strip – containsd+1 points. There are at mostm of these “decient” rectangles (one per row), and therefore at mostdm points that lie in decient rectangles; see Figure 2.2. We will isolate these dm points by using Lemma 14, which says that if we solve the balanced medians problem restricted to them, then the total cost due to them does not exceed 3d p dm 3d 3=2 p p n=c + 1. The purpose of this is merely to establish that the cost due to “decient” rectangles (provided thatc andd are xed) iso( p n), so we are free to ignore them for the rest of this proof. We now compute the costs due to each non-decient rectangle, of which there are at mostdn=(d + 1)en=(d+1)+1. The expected Euclidean distance between any two points in rectanglei cannot exceed (W i +h)=3 (in fact, this is exactly equal to the expected` 1 distance between them [39]), whereW i is the 19 (random) width of the rectangle. Therefore, sinceh =c= p n = p d + 1= p n, the total cost of assignment is at most d 3 X i (W i +h) = d 3 ( X i W i | {z } =m p n=c+1 + X i h) d 3 p n=c + 1 +h n d + 1 + 1 = d 3 2 p n p d + 1 + p d + 1 p n + 1 2d 3 p d + 1 p n 2 3 p dn asn!1, which completes the proof. 2.3.3 Thenon-uniformk-mediansproblem Our nal result in this section describes the non-uniform scaling behavior of thek-medians problem. At a high level what we will do is approximate our nonuniform demand distribution function by a step function that is constant on patches. We will take as a lower bound a sum overk-medians solutions on the patches. However, due to the possibility of interaction between patches in ak-medians solution, a sum of in-patch k-medians solutions over patches is not truly a lower bound. To overcome this, we introduce what is called thek-medians boundary functional. In this new problem points can be assigned to medians or to the boundary, which essentially serves as another median, accounting for the reduction in cost that could come from assigning through the boundary to a median in another patch. The sum over patches of the solution to this boundary functional problem on the binomially distributed number of points that end up in the patch serves as a lower bound for our nonuniformk-medians problem. We must only decide the minimum-cost allocation of number of medians in each patch that will yieldn total medians across the patches. For the upper bound, because of the possibility of interaction between patches, the cost of 20 the nonuniformk-medians problem is less than or equal to the sum over patches of in-patchk-medians solution for the binomially distributed number of points that end up in the patch. We again need only use the upper bound that has the minimum-cost allocation of number of medians in each patch that will yield n total medians across the patches. In order to describe the constants for KMed(X 1 ;:::;X n ;bnc) for the nonuniform case, we rst establish some technical observations that are needed in order to make a lower bound: Lemma15. LetKMed 0 (X 1 ;:::;X n ;bnc)denotearelaxationofthek-mediansproblemintheunitsquare, withtheadditionalfeaturethattheboundaryofthesquareitselfisregardedasamedian(inotherwords,each non-median point is either connected to a median or to the boundary of the square). LetB n denote the set of points that are assigned to the boundary in an optimal solution. Asn!1, we havejB n j=n! 0 with probability one. Proof. Functions likeKMed 0 are called “boundary functionals” in [137] and “rooted duals” in [112], and are standard lower bounding machinery in analysis of this kind. Suppose for a contradiction thatjB n j=n6! 0, i.e. that lim sup n!1 jB n j=n = t > 0. The distance between each pointX i and the boundary, which we will callD i , follows a distributiong(s) given byg(s) = 4 8s for 0 d 1=2. Further note that the t-quantiles of this distribution are equal to (1 p 1t)=2, i.e. that Z (1 p 1t)=2 0 4 8sds =t: The law of large numbers says that if we select thetn nearest points to the boundary, then the sum of total distances,Dist, to the boundary satises Dist n ! Z (1 p 1t)=2 0 s(4 8s)ds = (1t) 3=2 3 + t 2 1 3 > 0: 21 By our assumption, there exist innitely many indices n such thatjB n j=n t. However, Theorem 9 already proved that lim n!1 KMed(X 1 ;:::;X n ;bnc) p n ! KMed () asn!1, and our assumption would imply that lim sup n!1 KMed 0 (X 1 ;:::;X n ;bnc) p n = lim sup n!1 p n KMed 0 (X 1 ;:::;X n ;bnc) n = lim sup n!1 " (1t) 3=2 3 + t 2 1 3 # p n =1; a contradiction. Lemma16. LetKMed 00 (X 1 ;:::;X n ;k;m)withmndenotearelaxationofthek-mediansprobleminthe unitsquaredenedasfollows: asinthetraditionalk-mediansproblem,weselectasetofmediansofsizek out of the pointsX i . However, we are now permitted to discard a set of points of sizem, which do not contribute to any cost, so that we only incur a cost of assigning thek medians to the remainingnmk points. Then ifm(n) is a function such thatm(n) =o(n), we have lim inf n!1 KMed 00 (X 1 ;:::;X n ;bnc;m(n)) p n 0:1780 (1) 3=2 p (2.6) with probability one asn!1. Proof. The right hand side of the inequality is the same as in Theorem 11. Recall that in our proof of Theorem 11, we applied the union bound and established in inequality (2.5) that P(KMed(X 1 ;:::;X n ;k)<`) n k k nk (2) nk (2(nk) + 1) ` 2(nk) ; 22 where we will use k =bnc for notational convenience. If we apply the union bound to our present problem, we (i) select a set of of medians of sizek, (ii) identify a subset of size nk m < n m to discard, (iii) assign thosenmk remaining points to medians, and (iv) compute the sum of the distances from the nmk remaining points to their assigned medians. We therefore have P KMed 00 (X 1 ;:::;X n ;k;m(n))` < n k |{z} (i) n m |{z} (ii) k nmk | {z } (iii) (2) nmk (2(nmk) + 1) ` 2(nmk) | {z } (iv) = n m k m (2) m ` 2m (2(nk) + 1) (2(nmk) + 1) | {z } () n k k nk (2) nk (2(nk) + 1) ` 2(nk) | {z } right hand side of (2.5) : Since we have expressed our probability of interest in terms of the right hand side of (2.5), whose behavior we have already studied in the limit asn!1, it will suce to verify that the other term () makes a contribution that is small; specically we merely need to show that log() iso(n) asn!1. This is mere algebra once again. The rst term of log() is log n m , which satises log n m(n) =o(n) for the following reason: if we dene =m=n, then we have! 0, and hence Lemma 4 says that log n m = log n n = ( log(nn) log logn log (nn) + logn)n +o(n) = ( log(1) log log(1))n +o(n) 23 and we see that the three terms that comprise the coecient of n all approach zero as ! 0, which completes the proof. This takes care of the n m term, which frees us to address others. Substituting ` =c p n,k =bnc, and noting that (2(nk)+1) (2(nmk)+1) [2(nk)] 2m , we have log() = log n m | {z } o(n) m logkm log(2) 2m log` + log (2(nk) + 1) (2(nmk) + 1) | {z } [2(nk)] 2m m logkm log(2) 2m log` + 2m log 2 + 2m log(nk) +o(n) =m logm lognm log(2) 2m logcm logn + 2m log 2 + 2m log(1) + 2m logn +o(n) = (2 log 2 + 2 log(1) log log(2) 2 logc)m +o(n) =o(n) as desired. The fact that our asymptotic bound (2.6) holds with probability one is due to the Borel-Cantelli lemma, because we have established that the probability of our event occurring is bounded above bya n for somea< 1. Corollary17. The functionKMed 0 (X 1 ;:::;X n ;bnc) from Lemma 15 shares the same lower bound as in Theorem 11. that is, lim inf n!1 KMed 0 (X 1 ;:::;X n ;bnc) p n 0:1780 (1) 3=2 p Proof. This follows from the two preceding lemmas. If we denem =jB n j as in Lemma 15, then we note thatKMed 00 (X 1 ;:::;X n ;bnc;m) forms a lower bound ofKMed 0 (X 1 ;:::;X n ;bnc) in which all travel to the boundary incurs no cost. 24 Lemma18 (A lower bound for binomially distributedn). Let 0 0 to be the solution to 1 m 2 m 2 X i=1 f i 1 +f i = and set s i = 1=(1+f i ). We further choose> 0 such that< (1 p s i )=m. For each component i off, let 0 i i be an “interior component” consisting of all points in i that are at a distance greater than=2 40 Figure 2.6: The grid cells i and their interior components 0 i , all of which are a distance of at least apart from one another. from the boundary of i , so that 0 i has area (1=m) 2 , as in Figure 2.6. Within each interior component 0 i , we selectd s i f i n=m 2 e points as landmarks, which we are eventually able to do (with probability one for largen) thanks to our assumption that < (1 p s i )=m. Since any two dispersed points that lie in dierent interior components 0 i are at least a distance apart from one another, the objective valueCost of such an assignment is at least Cost min ; min i KDisp(fX 1 ;:::;X n g\ 0 i ;d s i f i n=m 2 e) and therefore, Cost p n min 8 < : p n |{z} !1 ; p n min i KDisp(fX 1 ;:::;X n g\ 0 i ;d s i f i n=m 2 e) 9 = ; ; and Lemma 23 tells us that for eachi, we have lim inf n!1 p nKDisp(fX 1 ;:::;X n g\ 0 i ;d s i f i n=m 2 eg (1=m) 1=m 0:3549 p f i r 1s i s i = (1m)0:3549 p with probability one, which completes the proof since we can make arbitrarily small. 41 2.5 Thegeneralizedspanningtreeproblem For the minimum spanning tree problem, Yukich [136] showed the following. Theorem24 (Asymptotic convergence of the minimum spanning tree problem). LetX 1 ;:::;X n beinde- pendently and identically distributed with values in [0; 1] 2 with densityf which is bounded away from zero and has support on [0; 1] 2 , then letting MST(x 1 ;:::;x n ) be the length of the minimum spanning tree on x 1 ;:::;x n , lim n!1 MST(X 1 ;:::;X n ) p n = MST Z [0;1] 2 p f(x)dx; almost surely for some constant MST . Avram and Bertsimas [8] provided an analytical expression for (and bounds on) such a constant when points are sampled i.i.d. on a torus. Jaillet [57] proved the equivalence of the torus and unit square con- stants. From this we have 0:600822 MST 1= p 2. Here, we give an analogous characterization of the scaling behavior of the generalized minimum span- ning tree problem in the uniform distribution case. Theorem 25. LetX 1 ;:::;X n each consist ofk points independently and identically distributed uniformly in [0; 1] 2 . Then r 2 e 2 lim inf n!1 GMST(X 1 ;:::;X n ) p n=k lim sup n!1 GMST(X 1 ;:::;X n ) p n=k p 2: with probability one, asn!1. 42 Proof. To obtain the lower bound we proceed as we did with thek-medians problem; we employ the union bound: P(GMST(X 1 ;:::;X n )<`) =P(some selection of points and tree on those points has cost <`) sum over all selections of points and trees on those pointsP(cost of choice <`) = (# of ways to select points)(# of ways to select tree)P(cost of arbitrary choice <`) =k n n n2 P(cost of arbitrary choice <`): where in the last step, we have used Cayley’s tree theorem [2] that the number of spanning trees onn points isn n2 . To obtain an upper bound on the probability the cost of an arbitrary choice is less than`, we x our selection of pointsX 1 ;:::;X n and our tree on those points. Root the tree atX n and letp(i) denote the index of the parent ofX i in the tree; the root is its own parent. Then consider E := ( x 1 ;:::;x n 2R 2 :x n 2 [0; 1] 2 ; n X i=1 kx i x p(i) k<` ) : Then clearly P(cost of arbitrary choice <`) =P((X 1 ;:::;X n )2E) = Vol(E\ [0; 1] 2n ) Vol(E): To compute the volume ofE we make the volume preserving transformation u i := x i x p(i) for i2 f1;:::;n 1g,u n :=x n and consider E 0 = ( u 1 ;:::;u n1 2R 2 : n1 X i=1 u i <` ) : 43 Then Vol(E) = Vol(fu n 2 [0; 1] 2 g) Vol(E 0 ) = 1 Vol(E 0 ). By lemma 3, Vol(E 0 ) = (2) n1 (2(n 1) + 1) ` 2(n1) : Combining the above and taking logarithms, and substituting` =c p n, P(GMST(X 1 ;:::;X n )<c p n)k n n n2 (2) n1 (2(n 1) + 1) (c p n) 2(n1) logP(GMST(X 1 ;:::;X n )<c p n) (2 logc + logk + 2 + + log log 2)n 2 log 2 3 2 log 2 logc 7 2 logn +O(1=n) where we have employed Lemma 4. The coecient ofn is negative if c< r 2 e 2 k ; which completes our lower bound; the fact that it holds with probability one is due to the Borel-Cantelli lemma, because we have established that the probability of our event occurring is bounded above bya n for somea< 1. To obtain the upper bound we compare to a xed selection of points from the sets. From each point setX i select the pointX i = (x 1 ;:::;x d ) that has the smallest rst coordinatex 1 . Then the chosen points X i have probability density f((x 1 ;:::;x d )) =k(1x 1 ) k1 : Theorem 24 tells us that the MST of such points satises MST(X 1 ;:::;X n ) p n ! MST Z [0;1] 2 p f(x)dx = MST 2 p k k + 1 MST 2 p k r 2 k ; 44 0 100 200 300 0 10 20 30 n Objective value Obj. val LB UB (a) = 0:1 0 100 200 300 0 5 10 15 n (b) = 0:3 0 100 200 300 0 2 4 6 n (c) = 0:5 0 100 200 300 0 1 2 n (d) = 0:7 0 100 200 300 0 0:2 0:4 n (e) = 0:9 Figure 2.7: Objective values for thek-medians problem plotted with the lower and upper bounds. 0 50 100 150 200 250 0 10 20 n Objective value Obj. val LB UB (a)d = 5 0 50 100 150 200 250 0 10 20 30 n (b)d = 10 0 50 100 150 200 250 0 20 40 n (c)d = 25 0 50 100 150 200 250 0 20 40 60 80 n (d)d = 50 Figure 2.8: Objective values for the balanced medians problem plotted with the lower and upper bounds. with probability one, where we have used the known upper bound on MST . 2.6 Computationalexperiments 2.6.1 Mediansproblems We draw points i.i.d. uniformly in the unit square and solve the integer programming formulations of the k-medians and balanced medians problems using Gurobi. For a survey of other solution methods for the k-medians problem see Reese [103]. In Figures 2.7 and 2.8, we plot the objective values versusn for various parameter values andd, together with our upper and lower bounds. Results are averaged over 5 draws of points. 45 0 100 200 300 0 0:5 1 n Objective value Obj. val LB UB (a) = 0:1 0 100 200 300 0 0:2 0:4 0:6 n (b) = 0:3 0 100 200 300 0 0:2 0:4 n (c) = 0:5 0 100 200 300 0 0:1 0:2 n (d) = 0:7 0 100 200 300 0 5 10 2 0:1 n (e) = 0:9 Figure 2.9: Objective values for thek-dispersion problem plotted with the lower and upper bounds. 0 100 200 300 0 5 10 15 n Objective value Obj. val LB UB (a)k = 3 0 100 200 300 0 5 10 n (b)k = 5 0 100 200 300 0 5 10 n (c)k = 7 0 100 200 300 0 2 4 6 8 n (d)k = 10 Figure 2.10: Objective values for the generalized minimum spanning tree problem plotted with the lower and upper bounds. 2.6.2 Thek-dispersionproblem We draw points i.i.d. uniformly in the unit square and solve the integer programming formulation of thek- dispersion problem using Gurobi. In Figure 2.9 we plot the objective values versusn for various parameter values. We also plot our upper and lower bounds. Results are averaged over 5 draws of points. 2.6.3 Thegeneralizedspanningtreeproblem We draw points i.i.d. uniformly in the unit square and solve the generalized spanning tree problem using the prize collecting generalized spanning tree solver of Contreras-Bolton and Parada [23], letting each node simply carry the same prize. Other GMST solution methods can be found in [32, 31, 88, 46]. The prize collecting GMST was rst introduced, with heuristic and exact solution methods, by Golden et al. [45]. In Figure 2.10 we plot the objective values versusn for various parameter valuesk. We also plot our upper and lower bounds. Results are averaged over 5 draws of points. 46 2.6.4 Proportionalityanalysisandconclusions Our analysis suggests that we have the following proportionality for the four problems of interest: KMed(X 1 ;:::;X n ;bnc) (1) 3=2 p p n BMed(X 1 ;:::;X n ;d) p dn KDisp(X 1 ;:::;X n ;bnc) r 1 n GMST(X 1 ;:::;X n ) r n k : In Figure 2.11, we show the ratio between the right-hand side and the left-hand side of the four above expressions; ideally, each plot would be a constant, indicating a near-perfect prediction without any further dependency on the input parameters. It appears that such is not quite the case; for example, Figure 2.11a suggests that KMed(X 1 ;:::;X n ;b0:1nc) 0:4 (1) 3=2 p p n; but KMed(X 1 ;:::;X n ;b0:9nc) 0:5 (1) 3=2 p p n; because the bottom plot and the top plot are separated from one another by 0:1. This impact is the least signicant in the balanced medians problem (2.11b). Fortunately, none of these dierences is terribly drastic, suggesting that our bounds give rise to reliable predictors for the costs of the various problems discussed in this chapter. In Table 2.1, as a quick-reference means of predicting our objective value under the assumption of uniformly distributed points, we provide tables of the approximate constant by which to multiply the proportionalities above. These constants are obtained for a range of parameter values using the computed 47 constant 0.1 0.36572 0.3 0.38614 0.5 0.43222 0.7 0.48958 0.9 0.48499 (a)k-medians d constant 5 0.33528 10 0.34768 25 0.36249 50 0.38438 (b) Balanced medians constant 0.1 1.0623 0.3 0.91845 0.5 0.86512 0.7 0.86656 0.9 0.84978 (c)k-dispersion k constant 3 0.78739 5 0.90177 7 1.014 10 1.1511 (d) Generalized MST Table 2.1: Approximate constant by which to multiply the proportionality given in Section 2.6.4 to predict the objective value under the assumption of uniform point distribution. objective values for our largestn. For instance, if we have ak-medians problem with = 0:3, we would consult Table 2.1a and nd our objective approximation is KMed(x 1 ;:::;x n ;b0:3nc) 0:38614 (1 0:3) 3=2 p 0:3 p n: 48 0 100 200 300 0:2 0:4 0:6 0:8 n = 0:1 = 0:3 = 0:5 = 0:7 = 0:9 UB/LB (a)k-medians 0 50 100 150 200 250 0:2 0:4 0:6 n d = 5 d = 10 d = 25 d = 50 UB/LB (b) Balanced medians 0 100 200 300 0:5 1 1:5 2 n = 0:1 = 0:3 = 0:5 = 0:7 = 0:9 UB/LB (c)k-dispersion 0 100 200 300 0:5 1 1:5 n k = 3 k = 5 k = 7 k = 10 UB/LB (d) Generalized MST Figure 2.11: Objective value ratios. 49 Chapter3 Thesidekickroutingproblem 3.1 Introduction One logistical paradigm that has received considerable attention in recent years is the sidekick routing scheme. A sidekick routing scheme is a logistical framework in which a large vehicle, such as a truck or van, serves as a mobile base for a eet of small vehicles (the “sidekicks”), such as autonomous ground vehicles (AGVs) or unmanned aerial vehicles (UAVs). The sidekicks alternate between visiting the truck to pick up items and visiting the customers, and the overall objective is to determine a coordinated set of routes for all vehicles in order to optimize system eciency, such as minimizing the time to completion or the vehicle miles travelled (VMT). A sketch of such a system is shown in Figure 3.1. Schemes of this kind have been deployed by many public and private sector organizations very recently, as described below and pictured in Figure 3.2: • The California-based startup companies Kiwibot, Dispatch (acquired by Amazon [52]), and Starship all use ground vehicles to deliver food and groceries. Kiwibot’s delivery system is really a sidekick system that relies on tricycles to carry and launch the ground robots [129]. Amazon’s Scout ground delivery robot, launched in 2019, relies on vans at which its robots pick up their deliveries [72]. In 50 2016 Starship partnered with Mercedes-Benz to design a van that could deploy its ground robots [15]. • The Ohio-based company AMP Electric Vehicles, now Workhorse, has introduced a system that they call the “horsey” scheme, in which a drone ies back and forth with a delivery van. Though com- panies like Zipline and the California-based startup Matternet have had wide success with drone healthcare deliveries and companies like Alphabet’s Wing and Flytrex have commercial drone de- livery operations in limited regions, commercial aerial delivery still faces regulatory hurdles and as such there are no sidekick-system UAVs currently in permanent use making deliveries. How- ever, the horsey system was tested by UPS in 2017 [94]. The drones were launched from trucks to make deliveries. MatterNet has also partnered with Mercedes-Benz in designing a similar system for transporting blood samples; they call their system the “Vision Van” [85]. The Vision Van hardware was tested for commercial delivery in a three week trial in 2017, in Zurich. Drones made trips from retailers to stock trucks which then made deliveries [127]. • More recently, the Connecticut-based rm Target Arm has developed commercially available tech- nology for launching and retrieving rotary or xed-wing drones from a moving vehicle [120]. They are interested in both military and commercial applications and have showcased their technology for the Army in 2019 [24] and received multiple Air Force contracts in 2021 [121, 119]. In the commer- cial space, in 2019 they announced a partnership with Autonodyne and Valqari to develop package delivery with their system [7] and in 2020 announced a partnership with BIB Technologies for food delivery [11]. Although the hardware for these systems is fairly mature, the problem of determining ecient routes has not been considered until very recently. From the perspective of routing these systems pose an excep- tionally dicult challenge due to the need to synchronize multiple vehicles that can all be traveling at the 51 same time and at dierent speeds. We cannot consider vehicles’ routes separately as we must include the possibility of vehicles carrying other vehicles for periods of time and the need for intermittent meetings of vehicles at the same position at the same point in time. Thus we see that individual vehicles’ routes are highly interdependent, and any reasonable objective will be impacted by this interdependence, making the optimization very hard. Furthermore, the high-level attributes of these systems are not at all clear: how much more ecient can they be? When are they useful? What are the trade-os inherent in such a scheme? We employ a continuous approximation analysis as a means of answering these questions. This chapter is organized as follows. In Section 3.2 we provide an overview of related work. In Section 3.3 we formally dene the sidekick routing problem. In Section 3.4 we introduce preliminary results that will be of use in our analysis. In Section 3.5 we derive our main result concerning the asymptotic behavior of the sidekick routing problem. In Section 3.6 we summarize the operational implications of this result. In particular, we consider how much improvement in eciency can be gained by switching to the sidekick system and the tradeos that must be weighed in implementing the system. Finally, in Section 3.7 we consider the scaling behavior, and dependence on the conguration of the sidekicks, that our result tells us we should expect. We empirically demonstrate that actual tour times, obtained by heuristically solving the sidekick problem on simulated sets of customer points, corroborate our expectations. 3.1.1 Remarkonnotationalconventions When it is necessary to specify that a map is injective, we use,!. For optimization problems we use sans serif problem names to denote the optimal objective value of the problem, e.g. the length of the optimal TSP tour is denotedTSP. 52 (a) (b) Figure 3.1: The gure on the left shows a travelling salesman tour of a set of client destinations and a central depot, that is, the shortest tour that visits a collection of points and starts and ends at the central depot. The gure on the right shows the solution to a “sidekick” problem in which the truck has a “helper” (such as a robot or a drone) that alternates between visiting the truck and visiting the customer locations. Note that the truck’s tour with helpers is about half the length of the tour with no helpers. 3.2 Relatedwork In 2018 Otto et al. compiled a comprehensive review, [91], of work on optimization approaches to systems employing drones for a wide range of applications, to include package delivery and specically package delivery using drones as sidekicks for trucks. We cover much of the same territory to give a complete picture of the drone, and particularly sidekick, landscape and survey even more recent developments in the area. We are primarily interested in prior work in the area of continuous approximation and theoretical results bounding the objective or characterizing the improvement due to sidekick introduction. For work on solving sidekick problems we concern ourselves principally with these papers’ formulations of the problem. There are many variants, each diering in the assumptions that are made about the delivery system. Three critical questions, the answers to which change from model to model, that need to be posed are given below. • Does the truck also deliver packages or are packages only delivered by the sidekicks? • Can the truck carry multiple sidekicks capable of making simultaneous deliveries? 53 • Are the sidekick launch and pickup locations restricted to customer points, or otherwise to a discrete set of points that is specied a priori? Table 3.1 provides a summary of how previous formulations have answered these questions and the work that was done on the resulting problem. We can see that our formulation has the least restrictive answers to these questions and thus addresses the problem in the greatest generality. That is, in our model we have the following. • We allow for both the case that deliveries must be made by sidekicks and the case that the truck can also make deliveries. • There can be any number of sidekicks on the truck and they are free to be launched and picked up in any order. • The sidekick launch and pickup locations can be any point in the plane. Other factors that distinguish the models surveyed here are the treatment of a restricted drone range and the way that the objective, be it completion time or some measure of energy consumed, is determined. This work assumes unlimited drone range and that drones make a single delivery per trip from the truck. We take as our objective completion time. We assume for simplicity that the time spent actually dropping a package at a customer node as well as the time spent capturing a sidekick and preparing it for relaunch are negligible. Both the truck and the sidekicks travel at xed speeds along Euclidean distances. The specication of their relative speeds does however allow one to build some knowledge of the underlying network into the objective. One additional assumption that adds to the robustness of our formulation is that the sidekicks are allowed to be slower than the truck. We are thus able to accurately model systems like the truck-AGV schemes discussed in the introduction, whereas some papers surveyed require that the sidekicks be faster. 54 3.2.1 Otherautonomousvehicledeliveryschemesandapplications A lot of work is being done in the area of optimal coordination of autonomous vehicles. Applications beyond just delivery include power network surveillance in response to extreme weather events [75], re detection and extinguishing [128], facilitating intervehicle communications after disasters [61], and performing earth observation [135]. Other work has focused on drone-specic operational aspects of making deliveries. These include routing that considers the possibility of drones failing along their routes [124], routing that takes into account uncertain air temperature’s impact on battery duration [66], and depot location and route planning that takes into account battery consumption rates dependent on payload [125]. Examples of routing for delivery using a eet of only autonomous vehicles can be found in [108] and [22]. Kim et al. [67] give a drone eet routing problem for medical deliveries with the additional facet of depot location planning. Heterogeneous vehicle delivery systems where the dierent types of vehicles work completely in par- allel, both making deliveries from a depot rather than being synchronized, are also studied. Murray and Chu [86] give formulation of one such problem, the “Parallel Drone Scheduling Traveling Salesman Prob- lem (PDTSP).” A truck makes a tour of some points while drones serve the others by single-customer trips from the depot. Murray and Chu provide a heuristic solution to this problem. Another solution approach can be found in [82]. Li et al. [74] use a continuous approximation model to derive the expected cost of a similar problem where trucks move rst from a distribution center to depots that each serve a region. A PDTSP-like system is then used within the regions. In [122] drones provide an alternative delivery method to a cross-docking system. In [126] a Markov Decision Model is developed for the problem of meeting dynamic customer demand with either trucks or drones. A problem that is similar to the sidekick problem in that it seeks to take advantage of the long range of a truck as well as the benets of autonomous vehicles appears often in the literature as the Two Echelon 55 Routing Problem. In it trucks start at a distribution center then take tours of, or direct routes to, secondary drone depots. Trucks bring packages or packages and drones. Drones then make deliveries from the secondary depots. In some cases trucks also visit customers on their tour. Variations of this system can be found in [106, 65, 89, 29]. There are a number of problems that involve tandem delivery that do not quite fall into our classica- tion of a sidekick routing problem due to some restrictive assumptions. These assumptions are the truck moves along a linear course [104], the truck’s route is predetermined [90], the visit sequence is predeter- mined [42], and sidekicks must be picked up where they are launched [79, 33]. A nondelivery application of a sidekick system is presented in [123]. A ground robot and a drone are used in tandem to monitor nitrogen levels in an agricultural plot. 3.2.2 Exactsolutionstosidekickproblems In 2015 Murray and Chu [86] were the rst to formally describe a sidekick system. They provide a mixed integer linear programming (MILP) formulation of the “Flying Sidekick Traveling Salesman Prob- lem (FSTSP).” In it a single truck carries a single drone. The truck can, and for some customers must, make deliveries. Sidekick launch and pickup locations are limited to customer points and the depot. A time endurance range is imposed on the sidekicks. The objective is to minimize the time to complete all deliveries. Another similar, widely cited, integer programming (IP) formulation, the “Traveling Salesman Problem with Drone (TSP-D)” is given by Agatz et al. [1]. Ha et al. [50] introduce a cost based objective to these single-drone models. Jeong et al. [60] present a mixed integer programming (MIP) formulation extending this problem by letting drone range depend on payload and introducing no y zones. Addi- tional modications to the MIP model can be found in [27] and [130]. Bouman et al. [12] give a dynamic programming solution to the problem. 56 Other extensions to the single-drone problem involve allowing for drone launch and pickup at a dis- crete set of points rather than, or in addition to, the customer points. Mathematical programming formula- tions of such problems can be found in [78] and [76]. Matthew et al. [81] present a solution to this problem without truck deliveries by way of a reduction to the generalized traveling salesman problem. Poikonen and Golden [97] go a step further in freeing launch and pickup locations. They formulate the “Mothership and Drone Routing Problem,” in which launches and pickups can occur anywhere in the plane and drones are allowed to make multiple deliveries per trip from the truck. The authors give a branch and bound method for solving this problem exactly. Murray and Raj [87] extend the FSTSP model of [86] to allow for multiple heterogeneous sidekicks. They further develop a more sophisticated treatment of endurance that takes into account payloads. A MILP model is presented. Kitjacharoenchai et al. [68] give a MIP formulation for a more limited multi- drone problem in which only one drone launch or pickup can occur at each customer point. Karak and Abdelghany [62] formulate a MIP that extends the multi-sidekick model to allow for a discrete set of non- customer launch and pickup points. Boysen et al. [14] provide an MIP formulation for a multi-sidekick problem in which robots are launched from a predetermined launch sites and return to a predetermined set of robot depots. The truck can rell to its capacity of robots at any robot depot. Another class of sidekick problems is sidekick vehicle routing problems (VRPs) involving eets of multiple, often-capacitated, trucks and sidekicks. Wang et al. [131] provided the rst formulation, the “Vehicle Routing Problem with Drones,” but their work is not concerned with exact solutions. Wang and Sheu [132] give a MIP formulation. MIP formulations of the problem with the addition of customer time windows can be found in [102] and [101]. All of these exact solutions are intractable on problems of practical size and are generally only able to solve problem instances up to size 10 customer points in a reasonable amount of time. Thus heuristic approaches are needed. 57 3.2.3 Heuristicsolutionstosidekickproblems Many of the papers given in the previous section also develop heuristic methods for the problems they pose. Murray and Chu [86] give a solution to the FSTSP that begins with a truck TSP tour and iteratively reassigns customers to drones. Likewise Agatz et al.’s [1] solution to the TSP-D begins with a TSP tour and then partitions the tour into truck customers and drone customers. Poikonen et al. [98] develop a heuristic method for solving the TSP-D that uses a branch and bound procedure to explore possible sequences of deliveries. Tang et al. [117] present a constraint programming formulation of the TSP-D. Other heuristic solutions to the versions of the problem with a single-drone and launch and pickup locations restricted to customer points can be found in [50, 48, 38, 77, 49, 100, 130, 60]. Heuristic solutions to the problem with the extension of launch and pickups allowed at a discrete, predetermined set of non-customer points are given in [78] and [76]. Marinelli et al. [80] provide a heuristic algorithm for the “en-route” truck-drone delivery system. In it the drone can be launched and picked up at any point along the lines that make up the truck’s tour of the customers that it visits. Poikonen and Golden [97] develop a heuristic for the Mothership and Drone Routing Problem which has totally free launch and pickup sites. Murray and Raj give a heuristic solution to their FSTSP with multiple sidekicks model. Other heuristic solutions to multi-sidekick models with launch and pickup restricted to customer points can be found in [13] and [68]. Karak and Abdelghany [62] and Boysen et al. [14] heuristically solve their problems with multiple drones and a xed set of non-customer launch and pickup sites. Poikonen [96] provides a heuristic solution to a new problem, the “Multi-Visit Drone Routing Problem.” A single truck can carry multiple drones, though their launch and pickup order is restricted to all must be launched then all must be picked up. Launch and pickup locations are a given discrete set. Most notably, a drone can deliver multiple packages per trip from the truck. 58 For heuristic algorithms to solve sidekick VRPs, Daknama and Kraus [26] assume launch and pickups at customers, Wang et al. [132] assume launch and pickups at discrete set in addition to customers, and Schermer et al. [105] assume launch and pickups at discrete locations along the direct paths the truck travels between truck-delivery points. 3.2.4 Theoreticalresultsandcontinuousapproximationmodels Wang et al. [131] consider the Vehicle Routing Problem with Drones in which multiple vehicles each carry multiple drones. They derive upper bounds on the improvement to be gained over the optimal TSP and VRP solutions without drones as well as the improvement to be gained by introducing faster drones. Poikonen et al. [99] extend the model of [131]. A battery life (time limit) is imposed on the drones; the possibility of using dierent distance metrics for the truck and drone and the possibility of using cost rather than time based objectives are considered; and there is an extension to the close-enough vehicle routing problem. Their results are bounds on improvement due to introduction of drones and due to dierent drone congurations. Agatz et al. [1] produce a result that is a generalization of the results of [131] when applied to the TSP-D. That is they give an upper bound on the improvement over just-truck routing allowing dierent distance metrics to be used for the truck and drone distances. The authors further give a lower bound to the TSP-D and an approximation algorithm using minimum spanning trees. Campbell et al. [17] study a continuous approximation model for a sidekick problem with a truck carrying multiple drones. Demand is modeled as a continuous spatial density. Customer points are visited in rectangular swaths. The authors provide the expected cost of delivery in terms of the customer density and the truck and drone per-unit-distance and dropo costs. Comparison is made to the expected cost without drones. Unlike in our model, drone launch and pickup locations are limited to customer points, 59 (a) BoxBot (b) Dispatch (c) Kiwi Campus (d) HorseFly (e) Starship (f) Swiss Post (g) Mercedes Vision Van Figure 3.2: Various hardware implementations of sidekick routing schemes. and the sequence of deliveries is xed to a truck delivery at which all drones are launched followed by another truck delivery at which all drones are picked up and relaunched. In [20] Carlsson and Song consider the sidekick problem as formulated in this chapter except restricted to only one sidekick and assuming that the sidekick is faster than the truck. Using a continuous approxi- mation model that assumes a smooth demand distribution they are able to derive the asymptotic behavior of the optimal tour as the number of customers goes to innity. This then yields a characterization of the improvement to be gained by introducing a sidekick and how this improvement depends on the relative speeds of the truck and sidekick. 60 Table 3.1: A summary of work on sidekick problems. Papers are classied by their model assumptions as well as their contributions. Publication TSP/VRP Truck Delivers Multiple Drones Launch/Pickup Sites Exact Solutions Heuristic Algorithms Continuous Approximation Theoretical Results [86], [50] TSP X - Customer Points MILP X - - [49], [100], [48], [38], [77], [98] TSP X - Customer Points - X - - [81] TSP - - Fixed Set GTSP Reduction - - - [78] TSP - - Fixed Set IP X - - [97] TSP - - Completely Free Branch & Bound X - - [12] TSP X - Customer Points DP - - - [96] TSP - X Fixed Set - X - - 61 Table 3.1: (continued) Publication TSP/VRP Truck Delivers Multiple Drones Launch/Pickup Sites Exact Solutions Heuristic Algorithms Continuous Approximation Theoretical Results [80] TSP X - Lines Between Truck Deliveries - X - - [1] TSP X - Customer Points IP X - Bounds on Improvement [13] TSP - X Customer Points MIP X - - [117] TSP X - Customer Points Constraint Programming X - - [27] TSP X - Customer Points MILP - - - [87] TSP X X Customer Points MILP X - - 62 Table 3.1: (continued) Publication TSP/VRP Truck Delivers Multiple Drones Launch/Pickup Sites Exact Solutions Heuristic Algorithms Continuous Approximation Theoretical Results [76] TSP - - Fixed Set MIP X - - [68] TSP X X Customer Points MIP X - - [62] TSP - X Fixed Set MIP X - - [130], [60] TSP X - Customer Points MIP X - - [131], [99] VRP X X Customer Points - - - Bounds on Improvement [26] VRP X X Customer Points - X - - 63 Table 3.1: (continued) Publication TSP/VRP Truck Delivers Multiple Drones Launch/Pickup Sites Exact Solutions Heuristic Algorithms Continuous Approximation Theoretical Results [102] VRP X X Customer Points MIP - - - [132] VRP X X Fixed Set MIP X - - [105] VRP X X Path Between Truck Deliveries - X - - [101] VRP X 1 Per Truck Customer Points MIQP X - - [17] TSP X X Customer Points - - X Expected Cost [20] TSP X - Completely Free - - X Asymptotics, Improvement Over TSP 64 Table 3.1: (continued) Publication TSP/VRP Truck Delivers Multiple Drones Launch/Pickup Sites Exact Solutions Heuristic Algorithms Continuous Approximation Theoretical Results This Work TSP X X Completely Free - - X Asymptotics, Improvement Over TSP 65 3.3 Problemdenitions We begin by formally dening the problem of sidekick routing with multiple sidekicks. We assume that a single, uncapacitated truck must provide service to a collection ofn customers in the plane, using the assistance ofk sidekicks having unit capacity, and that the goal is to minimize the time to completion. To simplify exposition, we will rst formulate our problem with an additional constraint that the truck itself is not permitted to visit any customers: Denition26. Letp 1 ;:::;p n be a collection of points in the plane. Letk denote the number of sidekicks. Let 0 denote the speed of the truck, and let 1 denote the speed of each sidekick ( 1 can be greater or less than 0 ). Let variablesx 1 ;:::;x n be the launch points for the sidekicks, and let variablesy 1 ;:::;y n be the pickup points for the sidekicks. That is, pointp i is visited by a sidekick that is launched at pointx i and is retrieved at pointy i . Let variablesz j ,j2f1;:::; 2ng, be the location of thejth sidekick launch or pickup event, and let variablest j ,j2f1;:::; 2ng, be the time of thejth sidekick launch or pickup event. Thez j ’s take the same values as thex i ’s and they i ’s; we introduce them only to make indexing easier in the formulation. We letz 0 be the initial position of the truck and letz 2n+1 be its nal position. We require that the truck’s tour be a loop, i.e.z 2n+1 =z 0 . We lett 0 , equaling zero, be the time at which the truck starts its loop and lett 2n+1 be the time at which the truck completes its loop. Let : f1;:::;ng ,! f1;:::; 2ng map customer index i to the place of that customer’s sidekick launch event in the ordering of all launch and pickup events. Similarly let :f1;:::;ng,!f1;:::; 2ng map customer indexi to the place of that customer’s sidekick pickup event in the ordering of all launch 66 and pickup events. LetF be the set of all pairs of mappings (;) that induce a valid sidekick tour. The conditions for inclusion inF are (i)<(i) 8i2f1;:::;ng (launches occur before corresponding pickups) jfi :(i)<jgjjfi :(i)<jgjk 8j (never more thank sidekicks in use) ; are injective (a launch and a pickup for each customer) (i)6=(i 0 ) 8i;i 0 2f1;:::;ng: (one event per place in the ordering) The last two conditions say that the maps and have to jointly form a bijection betweenf1;:::;ng andf1;:::; 2ng (to be precise, the two maps actually form a bijection between the multisetf1;:::;ng] f1;:::;ng andf1;:::; 2ng, where] denotes the multiset union [69]). The sidekick problem problem is then given by minimize x;y;z;t;; t 2n+1 s:t: (SK1) t j t j1 + 1 0 kz j z j1 k 8j2f1;:::; 2n + 1g (3.1) t (i) t (i) + 1 1 kx i p i k + 1 1 kp i y i k 8i2f1;:::;ng (3.2) z (i) =x i 8i2f1;:::;ng z (i) =y i 8i2f1;:::;ng t 0 = 0 z 2n+1 =z 0 (;)2F; 67 (a) (b) Figure 3.3: Solutions to the sidekick problem for n = 30 customers, k = 3 sidekicks, and a sidekick speed, 1 , that is twice the speed, 0 , of the truck. The solid line is the truck tour; dotted lines are the sidekicks’ routes; the square represents the starting and nishing point of the truck. In the gure on the left all deliveries are made by the sidekicks (Problem SK1). In the gure on the right the truck is also allowed to make deliveries (Problem SK2). where the objective value is the time at which the truck completes its loop, (3.1) captures the time needed for the truck to travel between launch and pickup points, and (3.2) captures the time needed for a sidekick to travel from its launch point, to a customer, and then to its pickup point. To extend (SK1) to the case where the truck is permitted to visit customers, some additional notation is required: Denition 27. We partition the set of customers into two setsSf1;:::;ng, representing those cus- tomers visited by a helper, and its complementT = S, representing those customers visited by the truck (these sets are optimization variables because we can choose which customers are visited by the truck). The number of events is now equal tom := 2jSj +jTj because a truck visiting a customer counts as only one event. This necessitates a third map :T ,!f1;:::;mg, in addition to the maps; :S ,!f1;:::;mg. LetF be the set of all (;;) that induce a valid sidekick tour. We have the same conditions as in the previous problem that ensure and do not pickup before launching or use more thank sidekicks. In addition, in this case we must require that each sidekick-visited customer has a launch and a pickup event 68 and each truck-visited customer has a truck visit event, with each of these events being mapped to a unique place in the ordering of events. That is, ; :S ,!f1;:::;mg :T ,!f1;:::;mg (S);(S);(T ) are pairwise disjoint. Put another way, the maps,, and have to jointly form a bijection betweenS[T andf1;:::;mg (to be precise, the three maps actually form a bijection between the multisetS]S[T andf1;:::;mg). The extension to (SK1) is then a natural one: minimize x;y;z;t;;; t m+1 s:t: (SK2) t j+1 t j + 1 0 kz j+1 z j k 8j2f1;:::;mg (3.3) t (i) t (i) + 1 1 (kx i p i k +kp i y i k) 8i2S (3.4) z (i) =x i 8i2S z (i) =y i 8i2S z (i) =p i 8i2T t 0 = 0 z m+1 =z 0 (;;)2F; whereS is dened as the domain of and andT is the domain of. 69 Figure 3.3 shows examples of solutions to the problems dened above for 30 customers with multiple sidekicks that are faster than the truck. 3.4 Preliminaries Having dened two variants of sidekick routing, we now turn to some preliminary results that will be useful in our analysis of these problems. This section presents existing results from prior work as well as some additional analysis of our own. 3.4.1 Existingresultsfromrelatedwork The following classical theorem, originally stated in [9] and further developed in [110, 112], is one of the fundamental results of the continuous approximation paradigm; it relates the length of a TSP tour of a sequence of points to the distribution from which they were sampled: Theorem 28 (BHH Theorem). Suppose that X 1 ;X 2 ;::: is a sequence of random points i.i.d. according to an absolutely continuous probability density functionf dened on a compact planar regionR. Then with probabilityone,thelengthTSP(X 1 ;:::;X n )oftheoptimaltravellingsalesmantourthroughallX i ’ssatises lim n!1 TSP(X 1 ;:::;X n ) p n = TSP ZZ R p f(x) dx where TSP is a positive constant. Although the exact value of TSP is unknown, it has been shown that 0:6277 TSP 0:9204; see [6, 9, 43] The concept of asubadditiveEuclideanfunctional was introduced in [110], which provides a key insight that we will use in this chapter: 70 Denition 29. A functionL() from the set of nite subsets ofR 2 to the non-negative real numbers is said to be a monotone subadditive Euclidean functional onR 2 if it satises the following properties: 1. L(;) = 0. 2. Homogeneity:L(x 1 ;:::;x n ) =L(x 1 ;:::;x n ) for all real> 0. 3. Translation invariance:L(x 1 +x;:::;x n +x) =L(x 1 ;:::;x n ) for allx2R 2 . 4. Monotonicity:L(x[A)L(A) for anyx2R 2 and nite subsetAR 2 . 5. Geometric subadditivity: There exists a constantC > 0, such that for all positive integersm;n and fx 1 ;:::;x n g2 [0; 1] 2 , we have L(x 1 ;:::;x n ) m 2 X i=1 L(fx 1 ;:::;x n g\Q i ) +Cm wherefQ i g, 1im 2 is the partition of [0; 1] 2 into squares of edge length 1=m. Examples of subadditive Euclidean functionals include the TSP tour and the Steiner tree. The minimum spanning tree, the minimum matching, and the nearest neighbor graph are all “close” to being subadditive Euclidean functionals, but violate the monotonicity requirement (though it turns out that this can easily be overcome for all relevant applications). The monographs [112, 137] are devoted to more general settings for Theorem 28, with the most prominent generalization being the following: Theorem 30 (basic theorem of subadditive Euclidean functionals). SupposeL is a monotone subadditive EuclideanfunctionaldenedonR 2 . IftherandomvariablesfX i gareindependentwiththeuniformdistribu- tion on [0; 1] 2 , then with probability one, we have L(X 1 ;:::;X n ) p n ! L 71 asn!1, where L 0 is a constant. We conclude with some additional problem denitions and convergence results that will also prove key to our analysis: Denition31 (Medians Problem). Given a collection of pointsx 1 ;:::;x n inR 2 and a positive integerp, the thep-medians problem is given by PMed(x 1 ;:::;x n ;p) := min Sf1;:::;ng:jSjp n X i=1 min j2S kx i x j k ; that is, the problem of selecting a subsetSf1;:::;ng ofmedian points such thatjSjp, that minimizes the sum of the distances from all points to their nearest median. Denition 32 (Balanced Medians Problem). The balanced medians problem BMed(x 1 ;:::;x n ;d) is a further-constrained variation of thep-medians problem. We can equivalently expressp-medians as the problem of selecting a set of mediansSf1;:::;ng and an assignment of the pointsx i to medians such that the sum of the distances from the points to their assigned medians is minimized. With no constraint on our assignment selection we have that in thep-medians problem the optimal assignment for any median set is simply to assign a point to its nearest median. The balanced medians problem imposes an additional constraint on the assignment selection, namely medianx j 2S can have at mostd 2 non-median points assigned to it. It is further required that each median is assigned to itself. That is, BMed(x 1 ;:::;x n ;d) := min Sf1;:::;ng:jSj=p :f1;:::;ng7!S n X i=1 kx i x (i) k; (3.5) where p = l n d + 1 m ; 72 x (i) is the median assigned to pointx i , and for allj such thatj2S,x (j) = x j andx (i) = x j for at mostd of thei6=j. The following result is due to [83]: Theorem33 (Asymptotic convergence of the balanced medians problem). Thebalancedmediansproblem satises the same convergence as in Theorem 30; that is, ifX 1 ;X 2 ;::: is a sequence of random points i.i.d. according to an absolutely continuous probability density functionf dened on a compact planar regionR andd 2 is xed, then with probability one, the costBMed(X 1 ;:::;X n ;d) satises lim n!1 BMed(X 1 ;:::;X n ;d) p n = BMed (d) ZZ R p f(x) dx where BMed (d) depends only ond. 3.4.2 FurthernotesonTheorem33 This section describes a lower bound on the function BMed (d) from Theorem 33. Theorem34. The function BMed (d) satises BMed (d) p 2d(d!) 1=(2d) p e(d + 1) (2d+1)=(2d) : That is, with probability one, lim n!1 BMed(X 1 ;:::;X n ;d) p n p 2d(d!) 1=(2d) p e(d + 1) (2d+1)=(2d) ZZ R p f(x) dx: Proof. We obtain this from a lower bounding technique that we employ for multiple location problems in Chapter 2. For the balanced medians problem see Section 2.3.2, proof of Theorem 12. 73 3.5 Acontinuousapproximationanalysis This section describes a continuous approximation analysis of the sidekick routing problems (SK1) and (SK2). 3.5.1 Naiveasymptoticanalysis Relying solely on Theorem 30, we can obtain the following partial characterization of the asymptotic behavior of both problems (SK1) and (SK2). Claim 35. For xed values ofk, 0 , and 1 , letT (p 1 ;:::;p n ) denote the optimal objective value of problem (SK1). Then if the customer pointsp i consist of random samplesP i independently drawn from a uniform distribution on the unit square, then there exists a non-negative constantc SK1 =c SK1 (k; 0 ; 1 ) such that T (P 1 ;:::;P n ) p n !c SK1 with probability one asn!1. The same statement holds whenT () is the optimal objective value of problem (SK2), with a dierent constantc SK2 c SK1 . Proof. This follows immediately from Theorem 30 because it is entirely straightforward to verify thatT () is a monotone subadditive Euclidean functional as dened in Denition 29. Claim 35 describes the scaling behavior of our problem asn!1, namely that the cost scales pro- portionally to p n, but it tells us nothing about c SK1 (or c SK2 ). For example, it is obvious that both are decreasing with respect to the three xed parameters 0 , 1 , and k (since making things faster or in- creasing the number of helpers can only improve eciency), and routine scaling arguments establish that c SK1 (k; 0 ; 1 ) = 0 c SK1 (k; 1; 1 = 0 ) for allk; 0 ; 1 (and similarly forc SK2 ). We devote the remainder of this section to a more precise analysis ofc SK1 andc SK2 . 74 3.5.2 Alowerboundfor(SK2) Of course, problem (SK2) is itself a lower bound of (SK1) by construction, so it will suce to consider (SK2) only. We derive a lower bound for (SK2) in terms of the Traveling Salesman tour and solution to a balanced medians problem on thep i . Lemma36. LetT n denote the optimal objective value for Problem (SK2). We have 1. TSP(p 1 ;:::;p n ) ( 0 +k 1 )T n 2. BMed(p 1 ;:::;p n ;d) (d 0 +k 1 )T n for alld 2. Proof. For the rst claim, we can construct a TSP solution from the (SK2) solution as follows. Consider an optimal solution (x;y;z;t;;;;S;T ) to (SK2). For eachi2S, the set of customers visited by the sidekicks, let u i := argmin u2fx i ;y i g jjup i jj; that isu i is the closer to the customer of its sidekick launch and pickup points. For eachi2T let u i :=p i : We then construct a TSP tour of the points as follows. Let the tour follow the path of the truck, visiting the customers inT along the tour. Whenever we reach one of theu i fori2S, let the tour travel fromu i top i and back, then continue along the truck path. It is clear that the length added to our TSP tour coming from the truck’s route is less than or equal to 0 T n , the truck’s speed times to total time for our sidekick tour. 75 To bound the length from visiting points inS we letP j be the set of points visited by sidekickj. Then T n 1 1 X i:p i 2P j jjx i p i jj +jjp i y i jj 8j2f1;:::;kg: That is the total sidekick tour time exceeds the time any given sidekick travels. Dividing by the above by k and summing over allj yields T n 1 k 1 X i2S jjx i p i jj +jjp i y i jj 2 k 1 X i2S jju i p i jj: Twice the sum of thejju i p i jj is precisely what we add to our TSP tour to visitS. Thus the contribution of this part of our TSP tour is bounded byk 1 T n . Adding together our truck and sidekick pieces of the TSP tour and applying the triangle inequality gives the result. For the second bound we can construct a balanced median solution from the (SK2) solution as follows. Think of the truck as completing a tour on ouru i dened as above. Group everyd + 1 of the customer points associated with theu i along this tour and choose as their median the point which is closest to the tour. This construction is pictured in Figure 3.5. By the triangle inequality, the distance from a point to its assigned median is less than or equal to the distance of traveling from that point to its correspondingu i , then traveling along the truck tour to the median’s correspondingu i , then traveling out to the assigned median. The cost of this balanced medians solution, i.e. sum of these distances, is then less than or equal to the sum of the distances from the non-median points to their correspondingu i , plusd times the length of the tour of theu i , plus the sum, over all medians, ofd times the distance from the median’s corresponding 76 Figure 3.4: Constructing a TSP solution from a solution to problem (SK2). The horizontal line represents the tour of theu i in the (SK2) solution. We follow the tour, traveling fromu i top i and back for each i2S. IfT n is the cost of the problem (SK2) solution then the total cost of the resulting TSP solution is less than or equal to ( 0 +k 1 )T n . u to the median. By our selection of the medians it is clear that this last sum is less than or equal to the sum of all of the distances from non-median points to their correspondingu. Then, noting once again that 0 T n truck tour of theu i ; and k 1 2 T n X i2S jju i p i jj = n X i=1 jju i p i jj; (u i =p i fori2T ) the length of the balanced medians solution is less than or equal to k 1 2 +d 0 + k 1 2 T n : 77 Figure 3.5: Constructing a balanced medians solution from a solution to problem (SK2). The horizontal line represents the tour of theu i in the (SK2) solution. Here we choosed = 4 and group every 5 points along the tour. We choose as the median for these 5 points the point which is closest to the tour. Using the paths pictured, it is clear that to connect all points to their medians we need travel at mostd times the length of the truck tour plus twice the total distance from the points to theiru i . IfT n is the cost of the problem (SK2) solution then the total cost of the resulting Bounded Medians solution is less than or equal to (d 0 +k 1 )T n . 3.5.3 Anupperboundfor(SK1) To bound the objective value of (SK1), we describe a simple “zig-zagging” heuristic in the unit square: Lemma 37. For xed 0 , 1 , and k and pointsp 1 ;:::;p n lying in the unit square, there exists a routing strategy for problem (SK1) whose time to completionT (p 1 ;:::;p n ) satises T (p 1 ;:::;p n ) 2 p 3 p 0 1 k p n +C whereC is a constant that depends only on 0 , 1 , andk. Proof. Assume without loss of generality that 0 = 1 and divide the unit square into strips of height h = p 3 1 k=n (there may be one strip whose height is less than this due to rounding). There arem = l p n=(3 1 k) m p n=(3 1 k) + 1 such strips. Further subdivide each strip into rectangles so that each rectangle (except possibly the rightmost in each strip) contains k points. There are at most m +n=k 78 (a) Input h (b) Output Figure 3.6: The tour described in Lemma 37, assumingk = 3. rectangles in total. Finally, construct a tour for the truck and all helpers by traversing each rectangle three times, releasing the helpers on the rst traversal and retrieving the helpers on the third traversal, as illustrated in Figure 3.6. It is easy to see that for a rectangle having width w (and height h), it is possible to perform three horizontal traversals and release and retrieve the helpers in at most 3w +h= 1 time units. It is also easy to see that the only remaining time needed is for the truck to perform vertical moves to move from one strip to the next, which is a constant amount of 2 time units, plus whatever time is needed for the truck to return to its point of origin, which is also at most p 2 time units. Hence, if we letw i denote the width of rectanglei, then the total amount of time to complete this tour is at most (2 + p 2) + X i (3w i +h= 1 ) (2 + p 2) + 3 X i w i | {z } =m +(m +n=k)h= 1 (2 + p 2) + 3 r n 3 1 k + 1 + 1 1 r n 3 1 k + 1 +n=k r 3 1 k n = 2 p 3 p 1 k p n + s 3k 1 n + 1 1 + (5 + p 2) 79 (a) yo (b) yo Figure 3.7: The tour described in Lemma 38; it consists of the same kind of tour as in Figure 3.6, but “scaled” with respect to the probability distribution that is indicated by shading. as desired. Lemma 37 is deterministic, but also implies the following: Lemma 38. Let 0 , 1 , andk be xed and letP 1 ;:::;P n be independent samples from an absolutely con- tinuous probability densityf with compact supportR. The optimal time to completionT (P 1 ;:::;P n ) for problem (SK1) satises lim sup n!1 T (P 1 ;:::;P n ) p n 3:47 p 0 1 k ZZ R p f(x) dx with probability one. Proof. This is a routine scaling argument, together with the law of large numbers and the factT () is a subadditive Euclidean functional (see Claim 35); see Appendix A for details. 3.5.4 Convergenceanalysisfor(SK1)and(SK2) We have now collected enough supporting evidence for our main claim: 80 Theorem39. Let 0 , 1 , andk be xed. LetT n denote the optimal objective value to problem (SK1), where inputpointsP 1 ;:::;P n areindependentuniformsamplesintheunitsquare. Thenthereexistsaconstant SK1 satisfying 0:1368< SK1 < 3:47 such that T n p n ! SK1 p 0 maxf 0 ; 1 kg (3.6) with probability one asn!1. Moreover, the same statement holds for a dierent constant SK2 SK1 whenT n is the optimal objective value to problem (SK2), which also satises 0:1368< SK2 < 3:47. Finally, whenthepointsP 1 ;:::;P n areindependentsamplesfromanabsolutelycontinuousprobabilitydensityf with compact supportR, we have 0:1368c lim inf T n p n lim sup T n p n 3:47c withprobabilityoneasn!1,whereT n istheoptimalobjectivevaluetoeitherproblem(SK1)or(SK2),and c = RR R p f(x) dx p 0 maxf 0 ; 1 kg : Proof. To simplify notation, assume without loss of generality that 0 = 1 and sett = 1 k, and rewrite the desired result (3.6) equivalently as p maxf1;tg T n p n ! SK1 : 81 The existence of SK1 and SK2 was already established in Claim 35 (set SK1 = c SK1 p 0 maxf 0 ; 1 kg and so forth); the real work lies in computing the bounds on these constants. Since SK2 SK1 , it will suce to show that 0:1368< SK2 and that SK1 < 3:47. To show that 0:1368< SK2 , Lemma 36 says that T n TSP(P 1 ;:::;P n ) 1 +t (3.7) =) lim n!1 T n p n lim n!1 TSP(P 1 ;:::;P n ) (1 +t) p n = TSP 1 +t (3.8) =) p maxf1;tg lim n!1 T n p n p maxf1;tg TSP 1 +t 0:6277 p maxf1;tg 1 +t =) SK2 0:6277 p maxf1;tg 1 +t > 0:1368 whenevert< 19; (3.9) where in (3.8) we are justied in taking limits as we have seen such limits exists for problem (SK2) (Claim 35) and for the TSP (Theorem 28). In addition Lemma 36 tells us that, providedt 2, T n BMed(P 1 ;:::P n ;btc) btc +t (3.10) =) lim n!1 T n p n lim n!1 BMed(P 1 ;:::P n ;btc) (btc +t) p n = BMed (btc) btc +t (3.11) =) p maxf1;tg lim n!1 T n p n BMed (btc) p maxf1;tg btc +t p 2btc(btc!) 1=(2btc) p maxf1;tg p e(btc + 1) (2btc+1)=(2btc) (btc +t) (3.12) =) SK2 p 2btc(btc!) 1=(2btc) p maxf1;tg p e(btc + 1) (2btc+1)=(2btc) (btc +t) > 0:1368 whenevert 19; (3.13) where in (3.11) we are justied in taking limits as we have seen such limits exists for problem (SK2) (Claim 35) and for the balanced medians problem (Theorem 33), and in (3.12) we have applied Theorem 34. 82 The upper bound SK1 < 3:47 is very simple. From Lemma 37, we have T n 2 p 3 p t p n +C (3.14) =) p maxf1;tg lim n!1 T n p n p maxf1;tg lim n!1 2 p 3 p t + C p n ! =) SK1 2 p 3< 3:47 fort 1; and for t < 1, we simply eschew the helpers altogether and visit all of the P i ’s with the truck (to be precise, since the truck is not allowed to visit any points in (SK1), we bring the truck within arbitrarily small distance from eachP i and release and retrieve one of the helpers): lim n!1 T n p n TSP =) p maxf1;tg lim n!1 T n p n p maxf1;tg TSP = TSP =) SK1 TSP 0:9204 fort< 1 as desired. This completes the proof of the uniform case of Theorem 39. The non-uniform case of Theorem 39 follows the exact same logic; the only distinction is that we are no longer guaranteed thatT n = p n has a limit, so we merely replace all instances of “lim n!1 T n = p n” with either a “lim inf n!1 ” or a “lim sup n!1 ” depending on whether we are bounding from above or below. 83 For example, the lower bound (3.7) becomes T n TSP(P 1 ;:::;P n ) 1 +t =) lim inf n!1 T n p n lim n!1 TSP(P 1 ;:::;P n ) (1 +t) p n = TSP RR R p f(x) dx 1 +t 0:6277 RR R p f(x) dx 1 +t =) p maxf1;tg lim inf n!1 T n p n 0:6277 p maxf1;tg 1 +t ZZ R p f(x) dx> 0:1368 ZZ R p f(x) dx whenevert< 19 =) lim inf n!1 T n p n 0:1368c whenevert< 19: The same reasoning is applied for the balanced-medians-derived lower bound fort 19. The upper bound that lim supT n = p n 3:47c is also immediate; we already proved this fort 1 in Lemma 38, and whent< 1, we again eschew the helpers altogether and use the truck: lim sup n!1 T n p n TSP ZZ R p f(x) dx 0:9204 ZZ R p f(x) dx< 3:47c; which completes the proof. 3.6 Remarks Informally, Theorem 39 says that the time to completion of a sidekick routing problem satises Time With Sidekicks/ p n p 0 maxf 0 ; 1 kg : We see immediately that for (k 1 )= 0 < 1, both the lower and upper bounds on the Sidekick problems that produce our result are constant multiples of the optimal TSP objective. We conclude that the asymptotic behavior of the sidekick problem looks more or less like that of the TSP in this case. This yields the man- agerial insight that there is no real benet to introducing sidekicks if we are not guaranteed (k 1 )= 0 1. 84 If we will have suciently fast or suciently numerous sidekicks to guarantee this then Theorem 39 tells us as that as we add more and more customer points we can essentially say that Time With Sidekicks/ p n p 0 1 k : On the other hand in the limit the tour with just the truck has time Time Without Sidekicks = TSP 0 / p n 0 : So there is certainly a boost in eciency to be had by introducing sidekicks. The amount of improvement due to using sidekicks is captured by Time With Sidekicks Time Without Sidekicks / s 0 1 k : We note that all of the above remarks hold in both the uniform and non-uniform cases because, as we have also seen in Theorem 39, the dierence between these two cases merely amounts to multiplication by a factor of RR R p f(x) dx. 3.7 Computationalresults 3.7.1 UniformlydistributeddemandwithEuclideantravel We run simulations with points drawn from a uniform distribution in the unit square to see how the sidekick problem tours compare to our asymptotic expectations. In our computations the number of points, number of sidekicks and the ratio of the truck and sidekick speeds vary. We assume that the truck speed is always 1, letting the sidekick speed capture the ratio of the speeds. We solve Problem (SK1), in which the truck does not make deliveries. Each result is an average over 5 draws of customer points. 85 In order to approximately solve Problem (SK1) we use a heuristic to obtain and specifying the ordering in which we launch and pickup the sidekicks for each customer, and then solve the resulting problem in the variablesx,y,z, andt. To obtain the ordering, we rst compute the optimal TSP tour of the customers and break this tour into consecutive chunks of sizek, the number of sidekicks. To determine the launch and pickup ordering on each chunk we begin by assuming as in [96] and [17] that in the case of fast sidekicks the improvement due to multiple sidekicks is captured well by a sequence in which all sidekicks are launched and then all sidekicks are picked up. Within each chunk the launches for all the customer points occur in the order specied by the TSP tour, then the pickups occur in the order specied by the TSP tour. However, we recognize that in the case of slow sidekicks, this strategy breaks down and we expect an optimal tour to often use fewer than the full number of sidekicks at a given time, allowing the truck to cover large distances with no sidekicks out. For this reason we also consider the ordering in which we follow the TSP tour, launching for a customer and then picking up for that customer one at a time. We take the minimum result of these two approaches to be our service time. It will be clear from our plots that neither of these strategies suciently captures the benets of having more and more sidekicks when the sidekicks are slower than the truck. In fact the one launch one pickup option reduces us to the one sidekick case. However, it is unreasonable to think that such a simplistic heuristic solution to this dicult problem would come close to optimality in all cases. LettingT denote the optimal service time for the Problem (SK1), Theorem 39 tells us that we should have TC p n p 0 max( 0 ; 1 k) ; for some constantC. So we plot F :=T p 0 max( 0 ; 1 k) p n ; 86 for ranges of parameter values, expecting this value to be constant. • In Figure 3.8 we x the number of sidekicks and plotF over a range of values of 1 = 0 andn. • In Figure 3.9 we x the ratio of the speeds and plotF for a range of values ofk andn. • In Figure 3.10 we once again x the ratio of the speeds and plotF for a range of values ofk andn, now for higher values of the ratio. • In Figure 3.11 we xn and plotF over a range of values ofk and 1 = 0 . In addition, since the TSP can be approximated by a constant times p n for largen we would expect that, lettingTSP denote the time for the truck to complete a TSP of the same points we should have TC 0 TSP p 0 max( 0 ; 1 k) ; for some constantC 0 . So we plot F 0 :=T p 0 max( 0 ; 1 k) TSP ; for ranges of parameter values, expecting this value to be constant. • In Figure 3.12 we xn and plotF 0 over a range of values ofk and 1 = 0 . 3.7.2 Onarealroadnetwork We obtained driving times between 1500 discrete points in Downtown Los Angeles. We samplen of them at random and solve the sidekick routing problem subject to these travel times. Travel times for the truck are dictated by the driving times and for the sidekicks we let their travel time be some constant factor times either the driving times. 87 0 10 0.2 8 0.4 250 6 200 0.6 150 0.8 4 100 2 50 0 0 0 10 0.2 0.4 8 250 0.6 0.8 6 200 1 150 4 1.2 100 2 50 0 0 0 10 0.2 0.4 8 0.6 250 0.8 6 200 1 1.2 150 4 1.4 100 2 50 0 0 0 10 0.5 8 250 1 6 200 1.5 150 4 100 2 50 0 0 Figure 3.8: Value of the sidekick problem completion time times p 0 max( 0 ; 1 k)= p n plotted over a range of values of 1 = 0 andn for xedk. We expect this value to be constant for largen. We can see that our heuristic approach does not suciently capture the benet of introducing more and more sidekicks in the slow sidekick case, but otherwise these plots are near constant, particularly asn becomes large. 88 0 10 0.2 0.4 8 250 0.6 6 200 0.8 1 150 4 100 2 50 0 0 0 10 0.5 8 250 1 6 200 1.5 150 4 100 2 50 0 0 0 10 0.5 8 250 1 6 200 1.5 150 4 100 2 50 0 0 0 10 0.5 8 250 1 6 200 1.5 150 4 100 2 50 0 0 Figure 3.9: Value of the sidekick problem completion time times p 0 max( 0 ; 1 k)= p n plotted over a range of values ofk andn for xed 1 = 0 . We expect this value to be constant for largen. As these values of 1 = 0 are small our heuristic approach is not capturing the benet due to introducing more and more sidekicks, and thus these plots are not constant though they fall within a range of about 1. 89 0 10 0.2 0.4 8 250 0.6 6 0.8 200 1 150 4 1.2 100 2 50 0 0 0 10 0.2 0.4 8 250 0.6 6 200 0.8 1 150 4 100 2 50 0 0 0 10 0.2 0.4 8 250 0.6 6 200 0.8 150 4 1 100 2 50 0 0 0 10 0.2 8 0.4 250 0.6 6 200 0.8 150 4 1 100 2 50 0 0 Figure 3.10: Value of the sidekick problem completion time times p 0 max( 0 ; 1 k)= p n plotted over a range of values ofk andn for xed 1 = 0 . We expect this value to be constant for largen. We see the result is near constant asn becomes large. 90 0 10 0.5 8 10 1 6 8 6 1.5 4 4 2 2 0 0 0 10 0.5 8 10 1 6 8 6 1.5 4 4 2 2 0 0 0 10 0.5 8 10 1 6 8 6 4 1.5 4 2 2 0 0 0 10 0.5 8 10 1 6 8 6 1.5 4 4 2 2 0 0 Figure 3.11: Value of the sidekick problem completion time, T , times p 0 max( 0 ; 1 k)= p n plotted over a range of values ofk and 1 = 0 for xedn. We expect this value to be constant for largen. When 1 = 0 is small our heuristic approach does not suciently capture the benet of introducing more and more sidekicks, but otherwise these plots are near constant. 91 0 10 0.5 8 10 1 6 8 1.5 6 2 4 4 2 2 0 0 0 10 0.5 8 10 1 6 8 1.5 6 4 2 4 2 2 0 0 0 10 0.5 8 10 1 6 8 1.5 6 4 2 4 2 2 0 0 0 10 0.5 8 10 1 6 8 1.5 6 2 4 4 2 2 0 0 Figure 3.12: Value of the sidekick problem completion time times p 0 max( 0 ; 1 k) divided by the optimal time for a TSP with only the truck, plotted over a range of values ofk and 1 = 0 for xedn. We expect this value to be constant for largen. When 1 = 0 is small our heuristic approach does not suciently capture the benet of introducing more and more sidekicks, but otherwise these plots are near constant. 92 Solving the sidekick problem exactly on a discrete set of points is intractable. So, the way we solve the problem is to solve a modied version of Problem SK1, letting our launch and pickups occur anywhere in the plane, then snap their locations to the closest of our discrete set of points. We incorporate the travel times of the underlying network using speeds. Since we are looking at a city road network, instead of travel along Euclidean distances, we let the travel in Problem SK1 be along the` 1 distance, and instead of having a xed speed for the truck and a xed speed for the sidekicks, we have a dierent speed associated with each leg of travel. That is, there is a speed for the leg of the truck’s tour that follows the second pickup that is dierent from the speed for the leg of the truck’s tour that follows the third launch. Likewise, the sidekicks travel at a dierent speed for each customer’s delivery and each customer’s trip back to the truck. We initialize these speeds to be the average travel speeds over all pairs of points for which we have driving times. We then update the speeds iteratively throughout the solution process. That is, letting (i)l 0 be the speed of the truck for the leg of travel after the ith customer’s sidekick launch, (i)r 0 be the speed of the truck for the leg after theith customer’s sidekick retrieval, (i)l 1 be the sidekick’s speed for its travel from its launch location to customeri, and (i)r 1 be the sidekick’s speed for its travel from customeri to its retrieval location, we are solving the continuous problem: 93 minimize x;y;z;t;; t 2n+1 s:t: (3.15) t 1 t 0 + 1 ( 1 (1))l 0 kz 1 z 0 k 1 t j+1 t j + 1 ( 1 (j))l 0 kz j+1 z j k 1 8j2flaunch event indicesg t j+1 t j + 1 ( 1 (j))r 0 kz j+1 z j k 1 8j2fpickup event indicesg t (i) t (i) + 1 (i)l 1 kx i p i k 1 + 1 (i)r 1 kp i y i k 1 8i2f1;:::;ng z (i) =x i 8i2f1;:::;ng z (i) =y i 8i2f1;:::;ng t 0 = 0 z 2n+1 =z 0 (;)2F: Note that 1 (j) is just the customer for which j is the index of the launch event and 1 (j) is the customer for whichj is the index of the pickup event. Our constraints then ensure the leg of truck travel after the customer’s launch or pickup has the given speed, even if the launch-pickup ordering changes. This resulting problem is equivalent to a linear program when the launch and pickup orderings are xed. However, unlike in our unit square experiments of the preceding section, we do not only x a launch-pickup ordering and solve this linear program. We use the blackbox optimizer LocalSolver [10] to do a local search over feasible launch-pickup orderings. We initialize the launch-pickup ordering to follow the TSP tour sequence, taking chunks of sizek and launching allk sidekicks and then picking up allk sidekicks. We run LocalSolver for a time (6 minutes per iteration) then look at the resulting tour. We nd 94 the closest actual points to our launch and pickup points and then use the true travel times for each leg of travel to update the travel speeds. The new speeds are given by the` 1 distance divided by the true travel time. Thus the speeds are reective of the travel times in the general area of the legs, which we are then re-optimizing. The ordering search then continues until a specied number of speed update iterations (30 forn = 25, 45 forn = 50, 60 forn = 100) are nished. We express this more compactly in pseudocode. LetU be the set of coordinates of all points between which we have driving time data. Foru 1 ;u 2 2U letd(u 1 ;u 2 ) be the driving time fromu 1 tou 2 . LetP be the set ofn customer locations that we sampled fromU. Set a factor by which to multiply truck speed to get sidekick speed. Set a number of speed update iterations. Set a launch-pickup ordering search time. Let x;y;z;t SK fp 1 ;:::p n g;;;f( (i)l 0 ; (i)r 0 ; (i)l 1 ; (i)r 1 ) :i2f1;:::;ngg be the solution to Problem (3.15) dened above onp 1 ;:::;p n with xed launch pickup ordering maps, and xed speeds. Our algorithm is given in Figure 3.13. We perform these experiments withk = 3 sidekicks and variousn and sidekick speeds. Results are averaged over 5 draws of points. In Table 3.2 we give the value of Time With Sidekicks Time Without Sidekicks , s 0 maxf 0 ; 1 kg ; which we expect to be constant. Within a range of about 0.2 this value is constant, suggesting our predic- tion is a good one for real road network data. 3.8 Conclusions We have studied the limiting behavior of sidekick-assisted routing problems in the Euclidean plane and found that the improvements introduced by adding sidekicks depend on p 0 and p 1 k. There remain 95 P =fp 1 ;:::;p n g n points drawn uniformly fromU average_driving_speed P u 1 ;u 2 2U ku 1 u 2 k 1 P u 1 ;u 2 2U d(u 1 ;u 2 ) (i)l 0 average_driving_speed 8i2f1;:::;ng (i)r 0 average_driving_speed 8i2f1;:::;ng (i)l 1 sk_speed_factor average_driving_speed 8i2f1;:::;ng (i)r 1 sk_speed_factor average_driving_speed 8i2f1;:::;ng tsp_time, tsp_tour Solve the TSP onP using driving times ; result of launching and retrieving chunks of sizek along order of the TSP tour for iteration in 1:num_speed_update_iterations do for time ordering_search_time do Have a black box solver perform: Local search over; by checking objective ofSK P;;;f( (i)l 0 ; (i)r 0 ; (i)l 1 ; (i)r 1 )g end ; result of black box solver’s search x;y;z;t SK P;;;f( (i)l 0 ; (i)r 0 ; (i)l 1 ; (i)r 1 )g x 0 i argmin u2U kux i k 1 8i2f1;:::;ng y 0 i argmin u2U kuy i k 1 8i2f1;:::;ng z 0 j argmin u2U kuz j k 1 8j2f0;:::; 2n + 1g (i)l 0 kz 0 (i)+1 z 0 (i) k 1 d(z 0 (i) ;z 0 (i)+1 ) 8i2f1;:::;ng (i)r 0 kz 0 (i)+1 z 0 (i) k 1 d(z 0 (i) ;z 0 (i)+1 ) 8i2f1;:::;ng (i)l 1 sk_speed_factor kp i x 0 i k 1 d(x 0 i ;p i ) 8i2f1;:::;ng (i)r 1 sk_speed_factor ky 0 i p i k 1 d(p i ;y 0 i ) 8i2f1;:::;ng end Compute time of sidekick tour induced byx 0 ;y 0 ;z 0 ;;: t 0 0 forj in 1:2n+1do truck_arrival_time t j1 +d(z 0 j1 ;z 0 j ) if9i such that(i) =j (jth event is a pickup)then i i such that(i) =j sidekick_arrival_time t (i ) + 1 sk_speed_factor d(x 0 i ;p i ) +d(p i ;y 0 i ) t j maxftruck_arrival_time, sidekick_arrival_timeg end else t j truck_arrival_time end end time_with_sidekicks t 2n+1 0 P n i=1 (i)l 0 + P n i=1 (i)r 0 =2n, 1 P n i=1 (i)l 1 + P n i=1 (i)r 1 =2n return time_with_sidekicks, tsp_time, 0 , 1 Figure 3.13: Pseudocode for our solution on real road network data. 96 n sidekick speed factor 1 2 3 25 1.7091 1.6310 1.6248 50 1.7575 1.7732 1.7311 100 1.8237 1.7519 1.8634 Table 3.2: The ratio ((Time With Sidekicks)/(Time Without Sidekicks)) / p 0 =(maxf 0 ; 1 kg) on real Downtown Los Angeles road network withk = 3 and sidekick speed the given multiple of the driving speed. many open questions: for example, what happens when sidekicks are able to visit more than one customer node before returning to the truck? What happens when the truck is itself capacitated and must make returns to the depot? What happens when sidekick battery life considerations come into play? We hope to resolve these questions in future work. 97 Chapter4 Conclusions We have developed new continuous approximation formulas and bounds on constants for existing formulas for a number of location problems and for the sidekick routing problem. Notably, we are the rst to describe the improvement to be gained by introducing sidekicks to a delivery system in the setting that allows multiple sidekicks that can be faster or slower than the truck and can be launched and picked up at any location. We are also the rst to describe the scaling behavior of thek-medians problem in the case of a general, nonuniform continuous demand distribution. For each of our problems we have characterized the objective growth’s dependence on the problem parameters; for a summary see Section 2.6.4 and Section 3.6. 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To be more spe- cic, we assume thatf(x) = P m i=1 f i i (x), where i (x) is an indicator function representing membership in a square grid celli. Let denote the area of each grid cell, so that RR R f(x) dx = P m i=1 f i = 1, and let N i denote the number of samples offP 1 ;:::;P n g that belong to celli (so that P m i=1 N i =n). It is clear that we can construct a feasible tour by applying Lemma 37 to each grid cell and then “stitch- ing” the tours within each grid cell together. Certainly, the amount of time needed to visit allN i points in grid celli is at most p 2 p 3 p 0 1 k p N i +C i ! for some constantC i , and the amount of additional time needed to “stitch” all of the tours together is a constantC 0 that does not depend onn. Summing all of these together and lettingC = P m i=0 C i , we have T (P 1 ;:::;P n ) p m X i=1 2 p 3 p 0 1 k p N i +C i ! +C 0 =) T (P 1 ;:::;P n ) p n p m X i=1 2 p 3 p 0 1 k r N i n ! + C p n 110 and sinceN i =n!f i with probability one, we see that lim sup n!1 T (P 1 ;:::;P n ) p n p m X i=1 2 p 3 p 0 1 k p f i = 2 p 3 p 0 1 k m X i=1 p f i = 2 p 3 p 0 1 k ZZ R p f(x) dx as desired. 111

Abstract (if available)

Abstract Continuous approximation is a powerful tool for predicting the objective value of optimization problems having a geographic component. Work in continuous approximation has largely focused on routing problems. We will focus instead on location problems and a hybrid routing/location problem arising from a novel delivery system that utilizes a truck and unmanned aerial or ground vehicles in tandem. We provide new continuous approximation formulas and new bounds on constants appearing in existing formulas. Our results characterize the objective value's dependence on problem parameters as the problems grow large. This allows practitioners to predict objective values and make decisions about those parameters.

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Creator Jones, Bo John (author)

Core Title Continuous approximation formulas for location and hybrid routing/location problems

School Viterbi School of Engineering

Degree Doctor of Philosophy

Degree Program Industrial and Systems Engineering

Degree Conferral Date 2022-05

Publication Date 04/29/2022

Defense Date 04/11/2022

Publisher University of Southern California (original), University of Southern California. Libraries (digital)

Tag continuous approximation,drones,location,OAI-PMH Harvest,Robots,routing

Format application/pdf (imt)

Language English

Contributor Electronically uploaded by the author (provenance)

Advisor Carlsson, John Gunnar (committee chair), Dessouky, Maged (committee member), Gupta, Vishal (committee member)

Creator Email bojohn.jones@gmail.com,bojones@usc.edu

Permanent Link (DOI) https://doi.org/10.25549/usctheses-oUC111159689

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