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Routing for Relief Efforts
Ann Melissa Campbell
Department of Management Sciences, University of Iowa
108 John Pappajohn Business Building, Iowa City, Iowa 52242-1000
ann-campbell@uiowa.edu
Dieter Vandenbussche
William Hermann
Mechanical and Industrial Engineering, University of Illinois
1206 West Green St., Urbana, IL 61801
dieterv@uiuc.edu
hermann2@uiuc.edu
Abstract
In the aftermath of a large disaster, the routing of vehicles carrying critical supplies can
greatly impact the arrival times to those in need. Since it is critical that the deliveries are
both fast and fair to those being served, it is not clear that the classic cost-minimizing routing
problems properly reflect the priorities relevant in disaster relief. In this paper, we take the
first steps in developing new methodologies for these problems. We focus specifically on two
alternative objective functions for the TSP and VRP: one that minimizes the maximum arrival
time (minmax) and one that minimizes the average arrival time (minavg). To demonstrate the
potential impact of using these new objective functions, we bound the worst case performance
of optimal TSP solutions with respect to these new variants and extend these bounds to include
multiple vehicles and vehicle capacity. Similarly, we examine the potential increase in routing
costs that result from using these alternate objectives. We present solution approaches for these
two variants of the TSP and VRP which are based on well known insertion and local search
techniques. These are used in a series of computational experiments to help identify the types
of instances where TSP and VRP solutions can be significantly different from optimal minmax
and minavg solutions.
1
1 Introduction
In recent years, several regions around the world have fallen victim to natural disasters of a mas-
sive scale. In 2004, an earthquake in the Indian Ocean spawned a series of tsunamis which caused
damage and loss of life as far away as 5000 miles from the epicenter of the earthquake [1]. In
2005, Hurricane Katrina affected a region of approximately 90,000 square miles which is about the
same size as Great Britain [2]. The enormous scale of these disasters has brought attention to the
need for methodology and technology for effectively managing relief supply chains. A recent report
published by the Fritz Institute, a nonprofit agency concerned with the logistics of relief efforts,
indicated that most aid organizations involved after the 2004 tsunami were significantly lacking in
logistics expertise and technology [26]. A European ambassador attending a UN-sponsored confer-
ence for donor nations said: “We don’t need a donors’ conference; we need a logistics conference”
[43]. While a great deal of research and technology is available for commercial supply chains, the
challenges associated with managing a humanitarian relief chain following a large-scale emergency
are often quite different than in commercial applications. Beamon [9] cites several examples of these
differences including the unpredictability of demand for humanitarian aid, where commercial supply
chains are designed around a known set of customers with relatively predictable demand patterns.
Another key difference is in the strategic goal of each supply chain. Where commercial supply
chains are focused on quality and profitability, humanitarian supply chains are usually focused on
minimizing loss of life and suffering. Also, when goods are distributed to the public, equity and
fairness become much more of a concern than in commercial applications.
With these ideas in mind, we became interested in exploring how such differences in strategic
goals could impact the routing of the vehicles delivering emergency aid or if it would impact the
routing at all. There are many existing tools for solving vehicle routing problems, and most focus
on minimizing the total distance travelled, which serves as a proxy for total cost. After a disaster,
the arrival time of relief supplies at the affected communities clearly impacts the survival rate of
the citizens and the amount of suffering. In the routes created by optimizing with respect to total
distance, some communities may be served significantly later than others in order to save on total
cost. Since it is critical that the deliveries are both fast and fair to those being served in a relief
context, such observations suggest that using service-based objective functions may better reflect
the different priorities and strategic goals found in delivering humanitarian aid. Specifically, we
examine two alternative objective functions for the classic traveling salesman problem (TSP) and
2
vehicle routing problem (VRP): one that minimizes the maximum arrival time (minmax routing)
and one that minimizes the average arrival time (minavg routing). These objective functions are
based on the literature on fairness, which has developed primarily outside of a routing context [33].
Even though these objective functions are by definition different than the traditional objectives,
it is not clear how much impact switching to one of these objectives will have on the solutions. Due
to the convenience of using existing cost-based routing tools, it is important to verify that such a
switch has the potential to make significant improvements in service to those affected by the disaster
in order to justify such a change. First, we will bound the worst case performance of optimal TSP
solutions for these new objectives and then extend these results to include multiple vehicles and
vehicle capacity. Next, we will present solution approaches for these two variants of the TSP and
VRP which are based on well known insertion [13] and local search techniques. These tools are used
in a series of computational experiments that help identify the types of instances where TSP and
VRP solutions can be significantly different from optimal minmax and minavg solutions. Both the
bounds and computational experiments will demonstrate that the optimal solutions for traditional
routing problems can be significantly different than those found with these alternative objective
functions, and that optimal solutions for the two alternatives can also differ. Importantly, both
in developing our theoretical bounds and in our computational experiments, we will also examine
the impact on total cost as a result of using these alternate objectives. We emphasize that the
tools developed in this paper are not intended to be sufficient for solving realistic routing problems
in a relief context, rather our intent is to demonstrate the potential impact of using alternative
objective measures when developing techniques to be used in practice. This is simply a first step in
developing better tools for the delivery of humanitarian aid. Our paper concludes with a discussion
of the some of the many issues that remain to be addressed.
2 Literature Review
This paper is part of a recent, emerging effort to apply operations research techniques to facilitate the
delivery of humanitarian aid. In addition to [9], examples include a study of the inventory systems
for disaster relief [10] and a worldwide facility location model to site warehouses in anticipation of
major emergencies [6]. ¨
Ozdamar et al. [36] describe a model that coordinates deliveries of supplies
between different supply depots in the context of a relief operation, and Barbarosoglu et al. [7] look at
how to effectively use helicopters in a relief operation. Long [30] discusses the strategic and tactical
3
issues that relief organizations face in preparing for and responding to disasters. Long [30] points
out some of the unique issues to relief supply chain management, including clearing supply routes
and the critical role of real-time non-computerized communications systems. Another key area of
concern in providing relief after a disaster is coordination of the organizations involved. Pettit and
Beresford [37], for example, model the relationships between participating bodies including military
and non-military organizations. Other efforts to examine the supply chains required by relief efforts
have been carried out by various nonprofit agencies, including the Fritz Institute [25].
There are papers that focus specifically on the dispatch of vehicles after a disaster. For example,
Shen et al. [44] analyze and develop solution methods for a stochastic vehicle routing problem
motivated by strategic planning for large-scale emergencies, where the total quantity of unmet
demand is minimized. The model is managed from a two-stage perspective, where pre-planned
routes are designed in the first stage, and adjustments to these routes are made in the second stage.
Desai et al. [16] propose a model that decides the number of emergency responders, such as police
or firefighters, to dispatch to each of several regions affected by a major disaster. The responders
sent to a particular region mitigate the risks resulting from the disaster, and the model’s objective
aims to distribute those risks equitably across regions.
In a relief context, fairness and equity are often important factors. Incorporating equity in
operations research models , though, is not new. Mandell [31], for example, provides an overview
of modeling equity in public systems, and Larson [29] discusses equity as a factor in the perception
of justice in queuing systems.
Equity has been a significant focus in the location of public facilities, as discussed in [33, 34].
The typical assumption in this literature is that each client is interested in minimizing his/her
own outcome fi(x) [34]. This function can measure a variety of outcomes in the location context,
but usually represents distance traveled or travel time to the nearest facility. For the weighted
location problem, both center and median objectives have been studied extensively, which minimize
the largest fivalue and the average fivalue, respectively [23, 20, 32, 40, 39]. Both of these
objectives capture the idea of minimizing individual outcomes while controlling inequity but remain
very simple to evaluate. They serve as the basis of our minmax and minsum routing objective
functions, which we will discuss in further detail in sections 5.1 and 5.2. In location studies,
inequality measures are often combined with a more traditional efficiency based objective function
in a bi-criteria approach, such as in [34]. This enables the study of location problems to incorporate
inequality measures developed originally for problems in the field of economics. These measures
4
were created to evaluate the fairness of a distribution of a particular commodity or resource. The
most popular of these measures is the Gini index [17]. For more details on inequality measurement
in economics, see the book by Sen [42].
In addition to location problems, equitable objectives are also considered in communication
networks, where users of the network are given allocations of bandwidth. In these applications,
equitable solutions are often characterized through a concept called proportional fairness, see e.g.
[27] for details. In the context of communication networks, Ogryczak et al. [35] discuss an axiomatic
approach to the question of fairness, presenting various objective functions that reflect the need for
both efficiency and equity.
Several of the objectives we will consider can also be found in the scheduling literature. For
instance, the analogue of minimizing the makespan in scheduling (see [38]) is minimizing the arrival
time at the last community to receive aid. Similarly, minimizing the sum of weighted completion
times is analogous to minimizing a weighted sum of arrival times.
There is little research on alternate objectives for vehicle routing, and we have not found any
work that compares using one of these alternate objectives to minimizing total travel time. In
[5], the authors consider the 2-TSP where two salesmen must together visit all of the nodes on
a tree. The objective is to minimize the length of the longest of the two tours. Even on a tree,
this problem is NP-hard, and an approximation algorithm is provided. Fran¸ca et al. [19] present a
heuristic for minimizing the length of the longest tour in an m-TSP. Similarly, the problem solved
in [3] is a particular instance of a VRP where the length of the longest of four routes is minimized.
The emphasis of [3] is on developing specialized cutting planes and a distributed search algorithm.
Equity has also appeared in the routing literature as a constraint. In [15], the authors propose a
model for an overnight delivery business that includes a constraint to limit the permissible deviation
from the average delivery time.
There are some related routing problems, including routing for hazardous materials and school
bus routing, that incorporate equity. In designing routes for the delivery of hazardous materials (see
e.g. [22]), the focus is on balancing risk to the regions visited on the routes. This focus makes the
resulting problems very different than the ones addressed here. There is a dual focus on efficiency
and equity in the case of school bus routing. It is important that school bus routing be efficient in
terms of total mileage, but it is also important that there are not students who spend an inordinate
amount of time on a bus. In [11], the authors optimize both efficiency and equity through a multi-
criteria objective function.
5
3 Assumptions and Definitions
In this paper, we make the following assumptions. We will be developing tours for customers labeled
1 to n. Since we may think of the customers as being located on a network, we will also refer to
them as nodes. Because the tours are designed to deliver a particular commodity to the customers,
we assume all tours will start from a designated depot, labeled 0. Thus, the arrival time at a
customer will be based on travel time from this depot, and the tour will be directed. For simplicity,
we will assume all tours start at time 0, and customers do not have time windows limiting delivery
times. We assume the travel times are nonnegative and satisfy the triangle inequality. We do
not explicitly model service times, but these may easily be included in the travel times. When
we consider capacity constraints, we will assume that each customer has unit demand, so that the
vehicle capacity indicates how many customers can be visited on one route. For convenience, rather
than minimizing the average arrival time, we will use the equivalent objective of minimizing the
sum of arrival times. Henceforth, we will refer to the two objectives of interest as minmax and
minsum.
The minmax routing problem minimizes the latest time any customer receives service, where
the traditional TSP or VRP focuses on the travel time or travel distance for the complete roundtrip
from the depot. Similarly, an optimal minsum route minimizes the sum of the arrival times for all
of the customers, which, as in the case of minmax, does not include the return trip from the last
customer back to the depot. In the delivery of aid supplies, it may be more important to quickly
deliver supplies to those in need than it is to get the truck back to the depot quickly.
The advantage of the minsum objective over minmax is that the arrival times at all customers
are reflected in the objective function. The minmax problem does not reflect, for example, the
second latest arrival time.
In this paper, we will address the performance of an optimal solution for each of the objectives
with respect to one or more of the other objectives. To do this, we require some notation. We use
c(T SP ) to denote the length of an optimal TSP tour. Similarly, we use c(V RP )k
Qto denote the
optimal value of the usual VRP problem with kvehicles and capacity Qon each vehicle. If either
kor Qare omitted, we assume their default values to be 1 and ∞, respectively.
la(M M )k
Qdenotes the latest arrival time in an optimal minmax routing when kvehicles with
capacity Qare available. Given a solution that minimizes the usual VRP objective, i.e. a solution
that achieves c(V RP )k
Q, we denote the latest arrival time as la(V RP )k
Q. Note this is a slight
6
0
C3
11
2
4
4 4
C1
C2
(a) TSP is different from min-
max and minsum
0
5
3
5
6
4
1
C3
C1
C2
(b) minmax is different from
minsum
Figure 1: Different objectives yield different solutions
abuse of notation, as alternate optima to the VRP objective may have different latest arrival times.
Unless stated otherwise, la(V RP )k
Qwill refer to any of the optimal solutions of the VRP. We define
la(T S P ) analogously.
We refer to sa(MS)k
Qas the minimum sum of arrival times when using at most kvehicles,
each with capacity Q. We can also consider quantities such as sa(V RP )k
Qand sa(T SP ). Other
quantities of interest will be c(MM)k
Q, which represents the value of an optimal minmax solution
with respect to the total duration objective. Similarly, we denote c(M S)k
Qas the total travel time
for an optimal minsum solution.
To illustrate our notation and to better illustrate the difference between minmax or minsum
routing and the traditional TSP, as well as between minmax and minsum, consider Figures 1(a)
and 1(b). Figure 1(a) is a graph where an optimal TSP tour is 0 −C1−C2−C3−0. The length
of the tour is 10, the latest arrival time is 9, and the sum of arrival times is 15, i.e. c(T SP ) = 10,
la(T S P ) = 9, and sa(T S P ) = 15. Optimal minmax and minsum routes traverse 0 −C1−C3−C2
instead, with c(MM ) = c(M S ) = 11, la(MM) = 7, and sa(MS) = 11.
Optimal minmax and minsum routes can also be different. In Figure 1(b), we show a graph
where the optimal minmax route is 0−C3−C2−C1(la(M M ) = 9, sa(MM ) = 20), but the optimal
minsum route is 0 −C1−C2−C3(la(M S) = 10, sa(M S) = 19). These examples illustrate that the
sets of optimal solutions for the different objectives need not be equal, although they may intersect.
7
4 Integer Programming Formulations
Like TSP and VRP, the minmax and minsum routing problems can be formulated as mixed integer
programming problems. We follow the notation and formulation introduced by Bard et al. [8].
Denote the set of customers with N:= {1, . . . , n}, the depot with 0, and let N0=N∪ {0}. Define
tij as the travel time between nodes iand jin N0. The variables xij are 0-1 variables that indicate if
a vehicle travels from node ito jand aidenotes the arrival time at customer i. The uncapacitated
VRP can be formulated as
min X
i,j∈N0
tij xij (1)
subject to X
j∈N0
xij = 1 ∀i∈N(2)
X
j∈N0
xij −X
j∈N0
xji = 0 ∀i∈N0(3)
tij +ai≤aj+T(1 −xij )∀i, j ∈N(4)
ai≥t0ix0i∀i∈N(5)
xij ∈ {0,1} ∀i, j ∈N0,(6)
where T > 0 is sufficiently large. The constraints (2) ensure that each customer is visited by a
vehicle. Equations (3) are standard flow balance constraints and ensure that all routes return to
the depot. Inequalities (4) and (5) make sure the variables arepresent the appropriate arrival times.
Furthermore, inequalities (4) ensure there are no subtours that do not pass through node 0. By
adding appropriate variables and constraints, we can easily enforce capacities on the vehicles.
In order to model the minsum objective, we replace the objective (1) with Pi∈Nai. To solve the
minmax problem, we add an auxiliary variable ¯ato represent the latest arrival time at a customer.
For the minmax problem, we will minimize ¯aand add the constraints ai≤¯a∀i∈N.
We found that these models are very difficult to solve with standard IP solvers. While more
advanced methods using cutting planes such as those presented in [8] would certainly improve the
performance, the goal of this paper is to obtain a better understanding of the impact of choosing one
of these alternative objective functions. Future work may consider developing more sophisticated
exact solution approaches for these alternate objectives.
8
0
M
M
M
C1
C2
C3
1
M
1
M
2
M
Figure 2: Bound la(T S P )≤2la(MM ) is tight
5 Bounds
5.1 Minmax
In this section, we examine the relationship between the total duration objective and the minmax
objective. We begin with the uncapacitated, one vehicle case.
5.1.1 Latest arrival: TSP versus minmax
Proposition 1. la(T S P )≤2la(MM).
Proof. Since the maximum arrival time in an optimal minmax tour represents the length of a
Hamiltonian path, which is a special case of a spanning tree, we know that la(M M )≥M S T ,
where MST is the total length of a minimum spanning tree. Under the triangle inequality, it is
well known that c(T SP )≤2M ST . Therefore, since la(T S P )≤c(T SP ), using an optimal TSP
tour to solve a minmax variant will yield a solution no more than 2 times the optimal solution to
the minmax problem.
We can get arbitrarily close to this bound using the example pictured in Figure 2. It is easy
to see that the optimal TSP tour in Figure 2 is 0 −C1−C2−C3−0. Note that regardless of the
orientation of the tour, the last customer will have an arrival time of 1
M+ 2Mwhereas the sequence
0−C1−C3−C2yields la(M M ) = 3
M+M. Taking the limit as M→ ∞, we get
lim
M→∞
1
M+ 2M
3
M+M= 2,
which shows that one can come arbitrarily close to this ratio of 2 by increasing M.
9
Ck
Ci
tij
tik
t0i
0
t0j
Cj
t0k
Figure 3: Optimal TSP tour in proof of Proposition 2
When put into the context of vehicle routing for disaster relief, Proposition 1 is significant,
indicating that an optimal TSP tour could double the time needed to reach the last group compared
to an optimal minmax solution.
5.1.2 Tour length: TSP vs. minmax
Next, we examine the quality of minmax tours with respect to the traditional TSP objective. This
will help us determine how much we can lose in overall efficiency, i.e. total travel time, by using
minmax as the objective. We show that the worst case ratio is 3
2, indicating that an optimal minmax
tour increases the total roundtrip traveled by at most 50 %.
Proposition 2. c(MM )≤3
2c(T SP ).
Proof. Consider an optimal TSP tour as displayed in Figure 3. We denote the first and last cus-
tomers on this tour as Ckand Cj. Denote the last customer served on the optimal minmax route
as Ci, where c(MM ) = la(MM) + t0i. The minmax route is the shortest Hamiltonian path starting
from the depot. Since c(T S P )−t0jis the length of a Hamiltonian path starting from the depot, then
la(M M )≤c(T S P )−t0j. Similarly, la(MM )≤c(T S P )−t0k. Now proceed by contradiction and
assume c(MM )>3
2c(T SP ). This implies la(MM ) + t0i>3
2c(T SP ). Subtracting [la(MM ) + t0j]
from both sides and using the triangle inequality, we find:
tij ≥t0i−t0j>1
2c(T SP ) + c(T S P )−[la(MM ) + t0j]≥1
2c(T SP ).
Hence, tij >1
2c(T SP ) and a similar relation establishes tik >1
2c(T SP ). Adding these two inequal-
ities gives tij +tik > c(T SP ), which is a contradiction, since tij and tik are lower bounds on the
lengths of two disjoint subpaths of the tour. This establishes the result.
10
C1
C2N+1
C2N+1
C2
C2N
C2N
C2
minmax
TSP
0
0
C1
Figure 4: Bound of minmax versus TSP is tight
The optimal TSP and minmax routes given in Figure 4 demonstrate that this bound is tight. In
this example, all n= 2N+ 1 customers lie on a grid in the Euclidean plane where each vertical and
horizontal edge has length 1. The dashed segments in each route represent a portion of the route
that is repeated as many times as Nrequires. It is a simple matter to check that c(T SP ) = 2N+2√2
and c(MM ) = 2N+1+√N2+ 4. This gives the following ratio as we increase N:
lim
N→∞
2N+1+√N2+ 4
2N+ 2√2=3
2.
5.1.3 Multiple Vehicles Latest Arrival: VRP versus minmax
It is easy to see that if there are no capacity constraints, then the usual VRP objective has no
incentive to use more than one vehicle. On the other hand, the latest arrival time can be significantly
improved with additional vehicles since customers can be served simultaneously. The following
proposition bounds how much the minmax objective can improve as a function of the number of
vehicles, k.
Proposition 3. la(M M )k≥max{1
2k−1la(M M ),maxi∈Nt0i}.
Proof. Consider an optimal set of tours that achieves la(MM)k. No matter how many vehicles
there are, the optimal latest arrival time must be larger than the longest direct trip from the depot,
i.e. la(M M )k≥t0i∀i∈N. Note that the minmax objective has no incentive to use fewer than
ktours, so we can assume that 1, . . . , k are the last nodes on each of the ktours. Recall that ai
is the arrival time at node iand assume that ak=la(MM )k, that is k∈argmaxi{ai}. We can
construct a single tour whose minmax value is less than or equal to (2k−1)ak, which proves the
desired result. We traverse each tour in sequence: starting at the depot, traveling to its endpoint,
and return to the depot to start the next tour, leaving the tour with endpoint aklast. By the
11
2
M
M
M
M
2
2
2
M
22
C1
C4C3
C2
C8
C5
C6
C7
C7C8C1C2
C3
C4
C5
C6
M
√2
√2
2
√2√2
Figure 5: VRP versus minmax with 2 vehicles
triangle inequality and since aj≤ak∀j6=k, we have that the length of this single tour is less than
or equal to Pk−1
j=1 2aj+ak≤(2(k−1) + 1)ak= (2k−1)la(MM )k.
It is not difficult to see that this bound is tight. Consider an instance with kcustomers where
each customer is distance Mfrom the depot, and the distance between any pair of customers is
2M. Clearly, using kvehicles, the maximum arrival time is M. If we have only one vehicle, the
maximum arrival time is 2M(k−1) + M= (2k−1)la(M M )k.
Proposition 4. la(V RP )k=la(T S P )≤2k la(MM)k.
Proof. Suppose the optimal minmax solution consists of kroutes and Ncustomers with k≤N.
Using the same notation as given in Proposition 3, let ak=la(MM )kand aibe the arrival time at
the last node on tour ifor i∈ {1, . . . k}. Since akis the maximum arrival time, then ai≤akfor
all i. Summing over the arrival times of the last customers on each tour gives: Pk
i=1 ai≤kak=
k la(M M )k. Multiplying Pk
i=1 aiby 2 gives the total length of a tour that visits all customers
and returns to the depot ktimes by retracing each route. Clearly such a tour must have total
length bounded below by the optimal TSP solution. Hence, c(V RP )k≤2Pk
i=1 ai≤2k la(M M )k.
When the vehicles are uncapacitated, the total cost objective provides no incentive to use more
than one vehicle; hence we may assume that c(T S P ) = c(V RP )k. Since la(V RP )k≤c(V RP )kand
la(T S P )≤c(T SP ), the result follows.
The two graphs in Figure 5 demonstrate that this bound is tight for two uncapacitated vehicles.
The upper graph shows the two routes of the optimal minmax tour (0 −C1−C4−C3−C2) and
(0 −C8−C5−C6−C7). Here la(M M)2=M+ 4 + √2. The lower graph gives the optimal VRP
solution (0−C1−C2−C3−C4−C5−C6−C7−C8−0) with la(V RP )2=la(V RP ) = 4M+6 + √2.
12
As the distances Mgrow to ∞, we have
la(V RP )2
la(M M )2=4M+ 6 + √2
M+4+√2= 4 = 2k.
Propositions 3 and 4 demonstrate that the impact of optimizing with respect to alternative
objectives increases with the number of vehicles. If additional vehicles are available, the costs asso-
ciated with them may be justified by the huge potential reductions in arrival time. It is important
to carefully evaluate these tradeoffs when determining the appropriate fleet size. Next, we evaluate
the potential changes in route duration (cost) that may be required to create these reductions in
arrival time.
5.1.4 Multiple Vehicles Tour Length: VRP versus minmax
Proposition 5. c(MM )k≤k c(V RP )k=k c(T S P ).
Proof. We will take an optimal TSP tour and use it to construct two paths from the depot. Consider
the midpoint of an optimal TSP. If this halfway point occurs along an edge, then delete this edge from
the tour. We are left with two connected paths from the depot. If the midpoint occurs at a node,
then arbitrarily assign the first half of the TSP tour to be the first path with the second half becoming
the second path from the depot. Dividing the TSP tour in this fashion corresponds to creating two
paths starting from the depot where the length of each path is less than or equal to c(T SP )
2. For
k≥2, la(M M )kis a decomposition of nodes into kpaths which start from the depot such that the
longest path is minimized. Hence, la(MM )k≤c(T S P )
2for k≥2. Following the notation as given
in Proposition 4, akis the maximal arrival time of the optimal minmax solution with kroutes. By
use of the triangle inequality and the division of the TSP tour given above, we find the relation:
c(MM )k=Pk
i=1(ai+t0i)≤Pk
i=1 2ai≤2Pk
i=1 ak≤2k la(M M )k≤k c(T S P ) = k c(V R P )k.
It is easy to see this bound is tight. Consider an instance with kcustomers, all a large distance
Mfrom the depot and the distance between any two customers is some small ² > 0. The VRP
solution will use one vehicle, hence we have that c(V RP )k= 2M+ (k−1)², while the minmax
solution will send one vehicle to each customer, which results in c(M M )k= 2k M. Hence as M
grows large, c(MM)kapproaches k c(V RP )k.
Thus, using kvehicles may reduce arrival times by a factor of 2kbut increase total duration by
a factor of k.
13
5.1.5 Capacity
Next we will consider what happens when we explicitly consider capacity limitations for the vehicles.
Recall that we have assumed each customer has uniform demand, so capacity here will reflect the
maximum number of customers that can be served by each vehicle.
Since the proof of Proposition 3 does not change with the introduction of capacity, we have
Proposition 6. la(M M )k
Q≥max{1
2k−1la(M M ),maxi∈Nt0i}.
This bound is tight with groups of Qcustomers co-located at a distance Mfrom the depot and
2Mfrom each other.
However, this does not mean that capacity has no impact on minmax solutions. When there are
kvehicles, solutions achieving la(MM)kand la(M M)k
Qwill use all kvehicles, but the customers
may be allocated differently due to the capacity limitations. The following proposition provides a
bound on the difference between la(MM )kand la(MM)k
Q:
Proposition 7. la(M M )k
Q≤(2k−1)la(M M )k.
Proof. The value for la(MM )k
Qmust be less than la(M M ) since there are ktours, each visiting a
subset of the customers in the one route solution. With triangle inequality, each tour will have a
lower latest arrival time than the one complete tour. We know from Proposition 3 than la(M M )≤
(2k−1)la(M M )kwhich gives us our desired result.
Consider the following example with k= 2, Q= 2, and n= 4. Three of the four customers
are co-located at distance Mfrom the depot, and the fourth customer is Mfrom the depot and
2Mfrom the other three. With two uncapacitated trucks, we can visit the three with one truck
and the fourth with the other truck, yielding a latest arrival time of la(M M )2=M. Whereas with
capacitated trucks, we can only visit two of the three co-located points with a single truck. To reach
the other two, it will take la(M M )2
2=M+ 2M= 3M= 3la(M M )2= (2k−1)la(MM)k.
We note that Proposition 7 does not explicitly involve the value for Q. We expect the bound to
be tighter, though, when Qis close to its minimum value of dn
ke. This is something we will evaluate
in our computational experiments.
14
5.2 Minsum
5.2.1 Sum Arrival: TSP vs. minsum
In this section, we examine the performance of TSP solutions with respect to the minsum objective.
Recall that a minsum objective reflects the sum of arrival times, so service times to all customers
are considered. We will show that TSP gives a factor napproximation for the minsum objective.
Proposition 8. sa(T SP )≤n sa(M S).
Proof. Suppose that 0 −1−2−. . . −n−0 is an optimal TSP tour. Since the orientation in which
we traverse the tour affects the latest arrival time, the best minsum objective that can be obtained
with this tour is
sa(T SP ) = min {nt01 + (n−1)t12 +. . . +tn−1,n, nt0,n + (n−1)tn−1,n +. . . +t12 }
≤n
2(t01 +t12 +...+tn−1,n +tn0)
=n
2c(T SP )≤nM ST.
Clearly, MST is less than or equal to the optimal minsum objective. Thus, we can conclude that
using an optimal TSP tour to solve a minsum variant will yield a solution no more than ntimes
the optimal value of the minsum problem.
In Figure 6, we present an example of a graph that achieves a ratio of n
2. It is easy to check
that the optimal TSP tour is 0 −CN+1 −C1−...−CN−CN+2 −0 and the optimal minsum tour
is 0 −C1−. . . −CN−CN+2 −CN+1 . Hence, the performance ratio with respect to the minsum
objective is
(N+ 2)M+ (N+ 1)M+N² +. . . + 2²+M
(N+ 2)²+ (N+ 1)²+. . . + 3²+ 2M+ 2M
Taking the limit as M→ ∞, we get N+2
2=n
2.
Proposition 8 indicates that the sum of the arrival times in an optimal TSP solution can be
significantly worse than in an optimal minsum route. This again confirms that a TSP tour may not
be a very equitable solution and will likely yield much higher average service times to those in need.
We can also present an alternative bound to Proposition 8 that depends on the spread of the
edge lengths, tij .
Proposition 9. sa(T SP )≤maxi,j tij
mini,j tij sa(MS).
15
CN+1
M
M
²
2M
MM
CN+2
Clique of Nnodes (all edges length ²)
0
Figure 6: Factor n/2 ratio for TSP versus minsum
Proof. Suppose 0 −1−2−. . . −nis an optimal minsum tour. We have
sa(MS) =
n−1
X
i=0
(n−i)ti,i+1 ≥Ãn−1
X
i=0
(n−i)!min
i,j tij .
Similarly, we can show that sa(T SP )≤³Pn−1
i=0 (n−i)´maxi,j tij . The result follows from combin-
ing the two inequalities.
Note that an analogous result can be shown for the minmax objective. It is particularly relevant
here because it gives us a condition when sa(T SP ) could be much closer to sa(MS) than a factor
of n. Specifically, if the difference between the longest and shortest edge is small, the two objectives
can be close in value.
5.2.2 Tour Length: TSP vs minsum
We next show how much the tour length using the minsum solution may differ from the tour length
for the TSP solution. Again, we are interested in determining how much total efficiency, in terms
of total travel time, is lost by considering an alternative objective such as minsum. In particular,
we show that the ratio c(M S)
c(T SP )cannot be bounded by a constant ratio. To do this, we use the graph
pictured in Figure 7. In this graph, we find the depot and Knodes located in the plane. Each node
Ciwill refer to a clique of size Ni, where the edge lengths within the clique are assumed to be 0. All
other distances will be determined by assuming the cliques lie in the Euclidean plane. We assume
that Kis even, so that clique C1is on the right side. For M > 2 and integer and Ni=Mi−1, we
will show that the optimal minsum route will traverse the solid line depicted in Figure 7. Observe
that the cliques on the solid line are labeled in descending order, so that the first clique, CK, will
contain the largest number of nodes. Given these assumptions, we have the following lemma, which
we prove in the appendix:
16
0
1
C1
C3
CK−1
√M2−1
CK
C2
M
M
M
M
Figure 7: Illustration for tour length of minsum versus TSP
Lemma 1. The optimal minsum tour in Figure 7 is 0−CK−CK−1−. . . −C3−C2−C1, where
nodes within a clique are visited consecutively in any order.
Assuming this is the case, we have that c(MS) = M K +K. Furthermore, it is easy to see
that the optimal TSP tour is 0 −CK−1−CK−3−. . . −C1−C2−C4−. . . −Ckand has length
c(T SP ) = 2 K
2+M+ 2K−2
2+M= 2M+ 2K−2. This gives the ratio c(MS )
c(T SP )=M K+K
2M+2K−2which
goes to K
2as M→ ∞. By choosing a large enough number of cliques and a large enough integer
M, we can make the ratio c(MS)
c(T SP )arbitrarily large.
This indicates that if cost is a concern or if we want to be able to use a vehicles to make multiple
trips in a day, we will need to be very careful in choosing an appropriate objective function. We
may need to consider a combination of objectives to balance the needs for equity and efficiency.
This will be discussed further in section 8.
5.2.3 Multiple Vehicles
Just as with minmax, we can expect to reduce the minsum objective by increasing the number of
available vehicles. The following proposition shows how much better the minsum objective can be
as a result of being able to simultaneously serve customers.
17
Proposition 10. sa(MS)≤(2n(k−1) + 1)sa(MS)k.
Proof. Consider an optimal set of tours that achieves sa(MS)k=Pn
i=1 ai. Note that like the
minmax objective, there is no incentive to use fewer than ktours. Next let lm= maxj{lj}where lj
represents the arrival time at the end of the jth of the ktours. Note also that lm= maxi∈N{ai}.
We now construct a single tour from these ktours. We visit each tour in sequence: starting at
the depot, traveling to its endpoint, and returning to the depot to start the next tour, until we
reach the last customer on the last tour. The arrival time to the customers on the original first
(chosen arbitrarily) tour will be unchanged. The arrival time to the customers on the kth tour will
be increased by at most 2l1+ 2l2+· ·· + 2lk−1.
Since lm≥lj∀j∈1, . . . , k, the new arrival time at customer iis clearly less than or equal to
2(k−1)lm+ai. Hence, we have that
sa(MS)≤X
i∈N
(2(k−1)lm+ai) = 2n(k−1)lm+sa(MS)k.
Since lm< sa(MSk), we have the desired result.
Combining Propositions 8 and 10, we get
Proposition 11. sa(T SP )≤(n2(k−1)2 + n)sa(M S)k.
As discussed in section 5.1.3, it is important to evaluate the tradeoffs between the costs associated
with additional vehicles and the potential improvements in service.
5.2.4 Capacity
The impact of having additional vehicles is affected by the capacity of the vehicle. Unlike with
minmax, the following bounds explicitly includes Q.
Proposition 12. sa(MS)≤(k(k−1)Q+ 1)sa(MS)k
Q.
Proof. We proceed as in the proof of Proposition 10 and concatenate all the routes in the solution
that achieves sa(MS)k
Q. Since there are at most Qcustomers on the second route, the sum of arrival
times for those customers increases by at most 2l1Q. The sum of arrival times for the original third
18
tour increases by a total of 2(l1+l2)Q, etc. Thus
sa(MS)≤sa(MS)k
Q+ 2l1Q+. . . + 2(l1+l2+. . . lk−1)Q
≤sa(MS)k
Q+ 2lmQ+. . . + 2(k−1)lmQ
=sa(MS)k
Q+ (k−1)(k)lmQ
≤sa(MS)k
Q+ (k−1)(k)Qsa(MS)k
Q,
which is our desired result.
6 Heuristic Algorithms
In this section, we will discuss how to modify the popular insertion algorithm to handle these new
objective functions as well as discuss the use of improvement heuristics. It is nontrivial to modify
construction and improvement heuristics to achieve good solutions for these problems while keeping
them computationally efficient and simple to implement. Both of these factors are important for
these tools to be successfully used in practice.
Insertion heuristics have proven to be popular methods for solving a variety of vehicle routing and
scheduling problems. They were first introduced and analyzed, as many other popular optimization
techniques, for the TSP [41]. Insertion heuristics construct a feasible solution, i.e. a set of feasible
routes, by repeatedly and greedily inserting an unrouted customer into a partially constructed
feasible solution. Different variants of the insertion heuristic arise as a result of choices in how
routes are initially created and the criteria for selecting unrouted customers to insert and where
to insert them in the partial solution. In some variants, the heuristic starts with a series of null
routes, while others use seeds to initiate routes. Typically, the customer and insertion point are
selected that yield the least increase in the current objective function, but this procedure can be
modified to include randomization in the selection process [18]. For a review of insertion heuristics,
see Campbell and Savelsbergh [13].
6.1 Minmax
As indicated above, the basic version of the insertion algorithm repeatedly inserts the customer
creating the least expensive increase in total distance. For the minmax objective, it would appear
that the best insertion customer and insertion point would be the combination that leads to the
least increase in the latest arrival time. This is true for one vehicle, but not when multiple vehicles
19
are available. In this case, there may be several tours where some of the customers can be inserted
without creating an increase in the latest arrival time over all tours. Not all insertions that do not
increase the latest arrival time at some iteration will result in the same minmax objective at the
end of the algorithm. Some insertion points may lead to larger or smaller increases in the arrival
times to the last customer on a particular tour. We take this into consideration in our modification
of the insertion algorithm. Assume there exists at least one customer who can be inserted on some
tour without causing an increase in the maximum arrival time. We then choose the customer and
insertion point creating the least increase in the arrival time at the last customer on any of the
tours that do not determine the latest arrival time. Otherwise, the customer and insertion point
yielding the least increase in the latest arrival over all tours is selected. Algorithm 1 details this
approach.
To give the algorithms the best chance to “succeed” in terms of finding good solutions to these
problems, we will experiment both with initiating the algorithm with knull (empty) routes and k
routes started from seed customers. This choice should not impact the complexity of the algorithm
(it depends on the complexity of the seed selection procedure), but will likely impact the final
solution. If we begin with null routes, the LAT ES Trvalue will initially be 0 for all r. When using
seeds, this value will be the arrival time at the seed customer for each route.
In Algorithm 1, for each potential insertion point for each uninserted customer, G1reflects any
change in the latest arrival time over all routes as a result of the insertion. G2reflects the change
in the latest arrival time for the route under consideration. At each iteration, all of the uninserted
customers (N) are considered. If G?
2<∞at the end of an iteration, then this indicates there are
customers that can be inserted without increasing the objective function. The cheapest of these to
insert is customer j?
2in position i?
2on route r?
2. Similarly, if all customers create an increase in the
latest arrival time, the best customer is stored in j?
1, and the best insertion point is i?
1on route r?
1.
Algorithm 1 runs in O(n3), like the traditional insertion algorithm. For each of the ninsertions,
n2customer/insertion point combinations are considered. The updating procedure is in constant
time and thus does not impact the complexity.
6.2 Minsum
In adapting the basic insertion algorithm for minsum, we start by examining how a single insertion
impacts the objective function. An insertion that makes a vehicle arrive xminutes later to the
following customer also makes the arrival xminutes later to each subsequent customer on the tour.
20
Algorithm 1 Insertion Algorithm for Minmax Objective
1: N= set of unassigned nodes, R= set of routes;
2: LAT ES Tr= initial latest arrival time for route r∈R.
3: ROU T EM AX = maxr∈RLAT E STr;
4: while N6=∅do
5: LET G?
1=∞,G?
2=∞;
6: for j∈Ndo
7: for r∈Rdo
8: for (i−1, i)∈rdo
9: LAT ES T =LAT ESTr+ti−1j+tj i −ti−1i
10: G1= max(ROU T EM AX, LAT ES T )−ROU T EM AX
11: G2=LAT ES T −LAT E STr
12: if G1< G?
1then
13: update(G?
1,j?
1,i?
1, r?
1);
14: end if
15: if (G1= 0)and(G2< G?
2)then
16: update(G?
2,j?
2,i?
2, r?
2);
17: end if
18: end for
19: end for
20: end for
21: if (G?
2<∞)then
22: insert(j?
2,i?
2, r?
2);
23: LAT ES Tr?
2=LAT ES Tr?
2+G?
2;
24: N=N−j?
2;
25: else
26: insert(j?
1,i?
1, r?
1);
27: LAT ES Tr?
1=LAT ES Tr?
1+G?
1
28: ROU T EM AX =R OU T EM AX +G?
1;
29: N=N−j?
1;
30: end if
31: end while
This is key in the development of Algorithm 2.
In Algorithm 2, the Avariables represent arrival times and will have initial values other than 0
if we start from seeded routes. Gwill represent the change in the sum of arrival times as a result of
each proposed insertion, where |r|represents the number of customers on route r. The difference
in sums will include the arrival time for the added customer plus the lateness incurred at all of the
subsequent customers. At the end of each iteration, customer j?will be inserted in position i?on
route r?.
For each of the ninsertions, we again evaluate n2possible insertion combinations. After each
insertion is selected, O(n) operations are required to do the necessary updating. This makes the
algorithm O(n×(n2+n)) = O(n3), so there is no increase in complexity.
21
Algorithm 2 Insertion Algorithm for Minsum Objective
1: N= set of unassigned nodes, R= set of routes;
2: A(i, r) = initial arrival time at ith customer on route r;
3: while N6=∅do
4: LET G?=∞
5: for j∈Ndo
6: for r∈Rdo
7: for (i−1, i)∈rdo
8: ARR =A(i−1, r) + ti−1j
9: CH AN GE =ti−1j+tj i −ti−1i
10: G=ARR +CHAN GE ×(|r| −i+ 1)
11: if G < G?then
12: update(G?, ARR?, C HAN GE?,j?, i?, r?);
13: end if
14: end for
15: end for
16: end for
17: insert(j?, i?, r?); A(i?, r?) = ARR?;
18: for k∈r?after i?do
19: A(k, r?) = A(k−1, r?) + CH AN GE?;
20: end for
21: N=N−j?
22: end while
6.3 Seeding the Routes
The choice of seeds can have a large impact on the final routes. With the alternate objective func-
tions, we found the following algorithm successful in determining seeds for the routes. In Algorithm
3, we iteratively determine the seeds. In each iteration, any unrouted customer is considered a
potential seed. For each potential seed, we evaluate the cost to insert that customer on any of the
already seeded routes. We retain the cheapest insertion point for each potential seed with i?
2and
r?
2, and its cost is G?
2. The customer with the highest G?
2value is selected as the new seed since this
is the customer who can be served least efficiently by the existing routes.
6.4 Improvement heuristics
Insertion heuristics are often followed by iterative improvement heuristics. Various papers have
been written on how to efficiently implement iterative improvement heuristics in the presence of
complicating constraints. For a survey of these techniques, see Kindervater and Savelsbergh [28].
Before we describe the improvement heuristics used here, we will describe an interesting ob-
servation that illustrates the difference in the objective functions, even from a problem-solving
perspective. In the case of Euclidean problems, where the distances represent Euclidean distances
between points in the plane, exchange heuristics help eliminate edges in the tours that cross since
22
Algorithm 3 Partitioning Insertion Algorithm
1: N= set of unassigned nodes; AR = set of assigned routes; R=set of routes;
2: while |AR|<|R|do
3: LET G?=∞
4: for j∈Ndo
5: LET G?
2=∞
6: for r∈AR do
7: for (i−1, i)∈rdo
8: G= insertion cost for minmax or minsum objective;
9: if G < G?
2then
10: update(G?
2, i?
2, r?
2);
11: end if
12: end for
13: end for
14: if G?
2> G?then
15: update(G?, j?, i?,r?);
16: end if;
17: end for
18: N=N−j?;
19: end while
0
3√7
√2
C2
C1
3√3
√2
C3
C5
C4
Figure 8: Uncrossing may not improve the solution for minsum
this has been shown to be suboptimal for the standard VRP objective. For the minsum objective,
however, it is not always better to uncross two edges. Consider the graph given in Figure 8. All
unmarked edges have length one. One can verify that this problem can be embedded in the plane,
and distances are Euclidean. The sequence 0 −C1−C4−C3−C2−C5yields a smaller minsum
objective than uncrossing this sequence to get 0 −C1−C2−C3−C4−C5. This is because in the
second sequence, the order of the edges with length 3√3 and 3√7 is exchanged, putting the longer
edge earlier, which more than offsets the cost of taking the longer diagonal edges. Note also that an
optimal minmax tour may cross itself with the last edge of the tour since its length is not included
in the objective (see, for instance, the optimal minmax tour in Figure 1(a)).
23
C6
C2
C3
C4
C5
C1
Figure 9: 2-Edge exchange example.
In our computational experiments, we used two forms of local improvement: 2-edge exchange
and 2 node-relocation. The first of these seeks to find an improved solution by taking any two
edges within a single tour and replacing them. The 2 node-relocation removes any two nodes
from any tours and checks if inserting them elsewhere would create an improvement. These are
described in greater detail in [28]. We selected these two improvement schemes since they involve
different search spaces, and thus should work well together. In our experiments, we run the 2
node-relocation improvement heuristic first until no improving relocations can be found, then run
the 2-edge exchange improvement heuristic. These two heuristics are alternated until there are no
improving changes of either type.
Implementing the improvement heuristics is straightforward for the VRP objective, and only mi-
nor modifications are necessary when dealing with the minmax objective. For the minsum objective,
it is necessary to maintain extra information in order to efficiently test the possible improvements
to the tours. Consider, for instance, applying a 2-edge exchange in the example pictured in Figure
9, where edges (C5, C6) and (C1, C2) are replaced by (C1, C5) and (C2, C6). This exchange reverses
the order of customers C2, . . . , C5on the tour. In the case of the VRP objective or the minmax
objective, this has no impact since the change in objective is determined only by the exchanged
edges. To recalculate the minsum objective, however, a myopic approach may iterate through the
nodes whose order was reversed, increasing the overall complexity of the heuristic. By storing and
maintaining the arrival time at each node in the current solution, we are able to calculate the
change in the minsum value more efficiently.
7 Computational Results
We have implemented the above algorithms and will next explore the impact of different data
characteristics on the resulting solutions using these algorithms.
24
7.1 Data
Our computational experiments used data representing several different geographical distributions.
The first dataset, hereafter referred to as Augerat-A, is available through http://branchandcut.
org/VRP/data/ and was introduced in [4]. Augerat-A consists of 27 different instances with 31-
79 customers. In each instance, customers are distributed rather uniformly, but the depot is not
necessarily in the center. For example, in certain instances, all of the customers may be to the lower
left of the depot on a graph. The second dataset, hereafter referred to as Augerat-B, is also available
through http://branchandcut.org/VRP/data/ and was introduced in [4]. Augerat-B consists of
23 datasets with between 30 and 77 customers each. In Augerat-B, unlike A, customers are grouped
in clusters. The use of the two datasets should help us understand the impact of the clustering
on the relative performance of the different objectives. The third dataset, hereafter referred to
as Golden is available through http://neo.lcc.uma.es/radi-aeb/WebVRP/ and included in [21].
These instances all are based on special structures. The special structures include concentric circles,
diamonds, squares, and stars. We have selected 11 of the smaller instances to test here. We
further eliminated customers from these selected instances so that the structure of the instance was
maintained but the number of customers is less than 100.
7.2 Heuristics Quality
In an effort to verify that our heuristics provide reasonably good solutions, we generated small test
instances and compared the results from our heuristics with the solutions obtained from the MIP
formulations presented in section 4. A total of 16 instances were derived from the Augerat-A and
-B collection, by removing customers such that the instances were of size ranging between 15 and 20
customers, but such that the overall distribution pattern of the customers was maintained. For each
instance, the MIP solver [24] was given 11,000 CPU seconds, and the best solution found during the
branch-and-bound process is compared to the best solution found using the heuristics discussed in
section 6. For each instance, we solved two routing problems: the first allows exactly one vehicle,
the second problem allows two. To keep the MIP model simple, the vehicles were uncapacitated.
We decided to use a time limit of 11,000 CPU seconds for a number of reasons. One was
because of the number of tests we were doing, but second, and more importantly, was the fact that
experiments with longer time limits either created no change in the solution, or at best, matched
the solution found by our heuristics. We did experiments with longer time limits for six different
instances including two where the MIP objective was worse than the heuristics after 11,000 CPU
25
seconds, two where the MIP performed better than the heuristics after 11,000 CPU seconds, and
two where the MIP and the heuristics reached similar objective values after 11,000 CPU seconds.
For each of these six instances, we let the solver run for 10 hours. Running them longer is infeasible
because after 10 hours the branch-and-bound tree requires more than 1 Gb of memory. For the
two instances where the MIP objective was worse after 3 hours (11,000 CPU seconds), the solutions
after 10 hours of CPU time were the same solutions as created by our heuristics. For the other
four instances, none of the objective values improved from the solution found after 3 hours, and
optimality gaps barely changed as well. Thus, even these relatively small problems are extremely
difficult to solve, but 11,000 CPU seconds gives us a good picture of the solution quality that is
possible with an MIP approach.
The results for the full set of instances can be found in figure 10, where we plot for each
instance the value found by the heuristic against the best MIP solution found. One can see that the
heuristic solutions are comparable to, and often better than, the MIP solutions, both for minsum
and minmax and with 1 or 2 vehicles. The MIP obviously can eventually find a better solution than
the heuristics, but often not before the computer runs out of memory, as indicated by our tests.
Here, the heuristics required only fractions of a second, while the MIP solver was terminated at
the time limit on each instance. The MIP solver did not terminate prematurely on any of the 32
problems.
7.3 Results
In the experiments, we solve each instance using two different methods and select the best solu-
tion for our tables. For the minmax and minsum objectives, initial solutions are constructed using
the basic insertion algorithm without seeded routes as well as the insertion algorithm seeded using
the partitioning algorithm described earlier. For the traditional TSP and VRP objectives, initial
solutions were constructed using the basic insertion algorithm without seeded routes and the well
known Clarke-Wright algorithm [14]. For fairness, we wanted to create the best solutions possible
for each objective, and Clarke-Wright proved to be more successful than partitioning in our exper-
iments with the traditional TSP and VRP objectives. After each initial solution was constructed,
improvement heuristics were applied as described earlier. The best solution after improvement was
the solution used in the tables.
In reporting the results, we have used two methods to compare arrival times in different solutions.
The first method evaluates the objective value of a routing solution with respect to the other
26
200
250
300
350
400
200 250 300 350 400
MIP
Heuristic
(a) M M 1: minmax, 1 vehicle
120
140
160
180
200
120 140 160 180 200
MIP
Heuristic
(b) M M 2: minmax, 2 vehicles
2000
2500
3000
3500
4000
2000 2500 3000 3500 4000
MIP
Heuristic
(c) M S1: minsum, 1 vehicle
1000
1200
1400
1600
1800
2000
1000 1200 1400 1600 1800 2000
MIP
Heuristic
(d) M S2: minsum, 2 vehicles
Figure 10: Objective values for heuristic solutions versus best MIP solutions
27
Table 1: Ratios for k= 1 route comparing objective functions and upper semideviations
la(T SP )
la(MM )
la(MS )
la(MM )
sa(T SP )
sa(MS )
sa(MM )
sa(MS )
c(MM )
c(T SP )
c(MS )
c(T SP )
us(T SP )
us(MM )
us(T SP )
us(MS )
us(MM )
us(MS )
Augerat-A MIN 0.948 0.970 0.974 0.973 1.004 1.027 0.926 0.902 0.850
MAX 1.051 1.257 1.229 1.270 1.246 1.431 1.114 1.138 1.234
AVG 1.013 1.114 1.089 1.124 1.075 1.184 1.016 1.016 1.003
Augerat-B MIN 0.912 0.922 0.852 0.938 0.995 0.963 0.822 0.730 0.728
MAX 1.146 1.275 1.424 1.321 1.233 1.384 1.391 1.565 1.315
AVG 1.020 1.092 1.061 1.074 1.047 1.114 1.041 1.068 0.987
Golden MIN 0.977 0.966 1.006 1.020 1.012 1.007 0.951 0.955 0.901
MAX 1.049 1.263 1.185 1.210 1.095 1.303 1.089 1.080 1.052
AVG 1.014 1.115 1.105 1.101 1.055 1.167 1.003 1.017 0.987
objective functions. For example, in the notation from section 5, la(T S P ) represents the latest
arrival time for the best TSP routing solution. Note this is a slight abuse of notation, as these
values now correspond to heuristic rather than optimal solutions. We then compute the ratio of the
appropriate routing solutions using the same objective measure. For example, la(T SP )
la(M M)is the ratio of
the latest arrival time for a TSP solution as compared to a minmax solution. The further this ratio
deviates from 1 is in an indicator of how much the objective function changes the solutions. We
average these ratios over all of the instances within a particular dataset. This weights the instances
within a dataset identically and allows easy comparison among different datasets. We also provide
the minimum and maximum ratios within a given dataset to capture the variance in these values.
The second method for comparing routing solutions is intended to measure the inequity in the
arrival times. These comparisons are based on the mean absolute upper semideviation (see e.g.
[33]), which is a measure of the deviation of arrival times that have a higher value than the average
arrival time. In particular, for a given solution using nnodes, denote the arrival times as aifor
i∈ {1, . . . , n}, and denote the mean arrival time as µ=1
nPn
i=1 ai. The mean absolute upper
semideviation is computed as: 1
nPai≥µ(ai−µ). Solutions with large upper semideviations indicate
that the arrival times induced by the solution may not be equitable. As an example of notation,
the mean absolute upper semideviation of a TSP route will be denoted as us(T S P ). Further, the
mean absolute upper semideviation of a VRP solution using kroutes and capacity Qis denoted
as us(V RP )k
Q. Similar to the first method of comparing routes, ratios of mean absolute upper
semideviations have been computed for all instances and objectives on a dataset. The minimum,
maximum, and average of these ratios are reported alongside the appropriate objective ratios in the
tables.
In Table 1, we compare the relative performance of the different objective functions when one
vehicle is available to serve the customers. For Augerat-B, we see that, on average, the latest arrival
28
time in a TSP solution is 2% higher than in a minmax solution, while the upper semideviation is 4.1%
higher. This comes at the cost of a 4.7 % increase in total route duration. There is approximately
a 6 % difference between the sum of arrival times in a TSP solution and in a minsum solution, with
a 6.8% increase in upper semideviation. These improvements in sum of arrival times and upper
semideviation come at cost of an 11% increase in total route duration.
We note that the minimum values for the minmax and minsum ratios are less than one, but this
is due to the fact the problems are solved using heuristics. For the Augerat-A and Golden datasets,
we find similar results with both yielding larger improvements in the minsum objective than with
minmax when compared to TSP solutions. Overall, we do not observe the significant differences
indicated by the worst-case results presented earlier.
In Tables 2 and 3, we see how the results change as additional vehicles are considered, in
particular for the number of vehicles k∈ {1,5,10}. In these tables, all of the vehicles have the
minimum capacity such that there is a feasible solution, i.e. Q=§n
k¨. For all of the datasets, we
find that the la(V RP )k
Q/la(M M )k
Qratio is significantly larger with multiple vehicles, while the cost
increases are not as significant. For instance, for Augerat-A, having 5 vehicles provides a 36.5%
improvement in latest arrival time, with an average increase in total cost of 12%. The results for
minsum are similar. For Augerat-A, we obtained an average increase of 20% in sum of arrival times,
with a 13 % increase in total cost. Under tight capacity, the additional vehicles did not seem to
provide significant improvements in upper semideviation.
It is interesting to note that for Augerat-A and B, the increase to 5 vehicles creates a significant
difference in the average latest arrival ratio (e.g. 1% to 36.5% for Augerat-A), but there is not
much of a jump when 10 vehicles are considered (e.g. 36.5% to 39% for Augerat-A). The results
for the upper semideviation measure are similar. For the specially structured Golden dataset, the
impact of choosing a minmax objective appears to grow more steadily with the increasing number
of vehicles with increases from 1.5% with one vehicle to 36.4% with 5 vehicles to 68.3% with 10
vehicles. This behavior is similar with the minsum objective.
29
Table 2: Effect of multiple vehicles for minmax routing using tight capacity, Q=§n
k¨
la(V RP )k
Q
la(MM )k
Q
c(MM )k
Q
c(V RP )k
Q
us(V RP )k
Q
us(MM )k
Q
k= 1 k= 5 k= 10 k= 1 k= 5 k= 10 k= 1 k= 5 k= 10
Augerat-A MIN 0.948 1.050 1.170 1.004 1.044 1.067 0.926 0.942 0.993
MAX 1.051 1.618 1.627 1.246 1.220 1.370 1.114 1.376 1.305
AVG 1.013 1.365 1.390 1.075 1.122 1.212 1.016 1.163 1.150
Augerat-B MIN 0.912 0.890 1.066 0.995 1.013 0.994 0.822 0.622 0.771
MAX 1.146 2.129 2.047 1.233 1.340 1.403 1.391 1.880 1.629
AVG 1.020 1.390 1.390 1.047 1.129 1.180 1.041 1.180 1.126
Golden MIN 0.977 1.032 1.233 1.012 0.972 0.925 0.951 0.968 1.086
MAX 1.049 1.723 2.880 1.095 1.234 1.288 1.089 1.324 2.178
AVG 1.014 1.364 1.683 1.055 1.120 1.138 1.003 1.141 1.340
Table 3: Effect of multiple vehicles for minsum routing using tight capacity, Q=§n
k¨
sa(V RP )k
Q
sa(MS )k
Q
c(MS )k
Q
c(V RP )k
Q
us(V RP )k
Q
us(MS )k
Q
k= 1 k= 5 k= 10 k= 1 k= 5 k= 10 k= 1 k= 5 k= 10
Augerat-A MIN 0.974 1.015 1.137 1.027 1.008 1.061 0.902 0.949 0.999
MAX 1.229 1.351 1.287 1.431 1.242 1.420 1.138 1.430 1.382
AVG 1.089 1.202 1.203 1.184 1.132 1.190 1.016 1.184 1.188
Augerat-B MIN 0.852 1.003 1.026 0.963 0.974 1.029 0.730 0.777 0.891
MAX 1.424 1.304 1.188 1.384 1.229 1.269 1.565 2.038 1.440
AVG 1.061 1.136 1.099 1.114 1.107 1.136 1.068 1.256 1.156
Golden MIN 1.006 1.068 1.179 1.007 1.015 0.925 0.955 0.994 1.170
MAX 1.185 1.317 1.671 1.303 1.223 1.206 1.080 1.339 2.178
AVG 1.105 1.183 1.312 1.167 1.142 1.116 1.017 1.186 1.405
Table 4: Effect of capacity on minmax routing using 5 vehicles
la(V RP )5
Q
la(MM )5
Q
c(MM )5
Q
c(V RP )5
Q
us(V RP )5
Q
us(MM )5
Q
C= 1 C= 2 C= 3 C= 1 C= 2 C= 3 C= 1 C= 2 C= 3
Augerat-A MIN 1.050 2.175 3.914 1.044 1.329 1.355 0.942 2.081 4.080
MAX 1.618 3.769 4.953 1.220 1.816 2.083 1.376 3.507 5.361
AVG 1.365 2.796 4.347 1.122 1.593 1.717 1.163 2.709 4.714
Augerat-B MIN 0.890 1.686 2.067 1.013 1.395 1.509 0.622 1.392 2.363
MAX 2.129 3.313 4.475 1.340 2.245 2.602 1.880 3.849 8.658
AVG 1.390 2.453 3.630 1.129 1.708 2.072 1.180 2.637 4.462
Golden MIN 1.032 2.337 4.350 0.972 1.203 1.242 0.968 2.097 3.802
MAX 1.723 2.820 4.928 1.234 1.697 1.806 1.324 2.716 5.135
AVG 1.364 2.573 4.587 1.120 1.410 1.477 1.141 2.488 4.706
30
Next, Tables 4 and 5 address what happens if capacity is not as “tight” and 5 vehicles are
available. The columns identified with C= 1 represent tight capacity (Q=§n
k¨), C= 3 represents
uncapacitated (Q=n), and C= 2 represents the results when capacity is halfway between tight
and uncapacitated (Q=§1
2§n
k¨+1
2n¨). We see the anticipated substantial impact by considering
additional capacity. For example, with C= 2, using the traditional VRP objective can yield a latest
arrival time on average almost 2.8 times larger than when latest arrival time is explicitly considered
for Augerat-A. Similar improvements occur for the upper semideviation measure. It is interesting
to note that the total length ratios are not nearly as large, indicating that significant improvements
in latest arrival do not necessarily come at equally significant increases in total route duration. The
results are also similar with the Augerat-B and Golden datasets. For the minsum objective, we
also find significant improvements in the sum of arrival times, with smaller increases in total route
duration.
Tables 6 and 7 present the effect of increases in capacity when 10 vehicles are available. For
example, with the Golden dataset and 10 uncapacitated vehicles, explicitly minimizing latest arrival
times yields solutions that differ on average by a factor of 8. For the minsum objective, we find
very similar results as capacity loosens. We note that Augerat-B has the least impact with both
objectives when C= 3. This indicates that clustering may limit the contribution of loosening
capacity at some point. Just as with 5 vehicles, Tables 6 and 7 show that the magnitudes of the
improvements in latest arrival or sum of arrival times are significantly larger than the increases in
total route length when the alternate objectives are used.
31
Table 5: Effect of capacity on minsum routing using 5 vehicles
sa(V RP )5
Q
sa(MS )5
Q
c(MS )5
Q
c(V RP )5
Q
us(V RP )5
Q
us(MS )5
Q
C= 1 C= 2 C= 3 C= 1 C= 2 C= 3 C= 1 C= 2 C= 3
Augerat-A MIN 1.015 1.869 3.035 1.008 1.370 1.410 0.949 2.181 4.179
MAX 1.351 2.665 4.613 1.242 1.789 2.052 1.430 3.553 5.366
AVG 1.202 2.273 4.036 1.132 1.561 1.683 1.184 2.741 4.778
Augerat-B MIN 1.003 1.251 1.582 0.974 1.383 1.497 0.777 1.455 2.378
MAX 1.304 2.344 4.619 1.229 1.951 2.630 2.038 4.197 6.590
AVG 1.136 1.839 2.942 1.107 1.651 2.005 1.256 2.574 4.306
Golden MIN 1.068 2.083 3.994 1.015 1.197 1.208 0.994 2.218 4.022
MAX 1.317 2.860 5.729 1.223 1.678 1.786 1.339 3.080 5.729
AVG 1.183 2.499 4.846 1.142 1.393 1.460 1.186 2.597 4.905
Table 6: Effect of capacity on minmax routing using 10 vehicles
la(V RP )10
Q
la(MM )10
Q
c(MM )10
Q
c(V RP )10
Q
us(V RP )10
Q
us(MM )10
Q
C= 1 C= 2 C= 3 C= 1 C= 2 C= 3 C= 1 C= 2 C= 3
Augerat-A MIN 1.170 2.989 4.340 1.067 1.791 1.825 0.993 2.579 4.688
MAX 1.627 4.680 8.318 1.370 2.776 3.185 1.305 4.563 9.237
AVG 1.390 2.796 4.347 1.212 2.382 2.609 1.150 3.668 6.742
Augerat-B MIN 1.066 1.746 2.076 0.994 1.663 1.809 0.771 1.654 2.514
MAX 2.047 3.715 5.595 1.403 3.531 4.683 1.629 6.724 13.951
AVG 1.390 2.819 4.178 1.180 2.647 3.269 1.126 3.193 5.267
Golden MIN 1.233 3.148 5.946 0.925 1.497 1.572 1.086 3.592 6.850
MAX 2.880 6.846 11.181 1.288 2.688 2.888 2.178 7.136 12.379
AVG 1.683 4.770 8.163 1.138 1.929 2.036 1.340 4.903 9.067
Table 7: Effect of capacity on minsum routing using 10 vehicles
sa(V RP )10
Q
sa(MS )10
Q
c(MS )10
Q
c(V RP )10
Q
us(V RP )10
Q
us(MS )10
Q
C= 1 C= 2 C= 3 C= 1 C= 2 C= 3 C= 1 C= 2 C= 3
Augerat-A MIN 1.137 2.141 3.477 1.061 1.802 1.836 0.999 2.713 4.932
MAX 1.287 3.682 7.261 1.420 2.969 3.405 1.382 4.784 9.308
AVG 1.203 2.273 4.036 1.190 2.316 2.540 1.188 3.838 7.043
Augerat-B MIN 1.026 1.332 1.639 1.029 1.946 2.193 0.891 1.494 2.473
MAX 1.188 2.786 5.256 1.269 3.432 4.704 1.440 5.669 11.761
AVG 1.099 2.093 3.342 1.136 2.714 3.352 1.156 3.211 5.294
Golden MIN 1.179 2.668 5.075 0.925 1.489 1.554 1.170 3.749 7.149
MAX 1.671 5.766 11.044 1.206 2.662 2.860 2.178 7.136 12.379
AVG 1.312 4.169 7.902 1.116 1.889 1.994 1.405 5.189 9.598
32
0
0.5
1
1.5
2
2.5
3
3.5
4
0246810
c(MM )k
Q−c(V RP )k
Q
c(V RP )k
Q
la(V RP )k
Q−la(M M)k
Q
la(MM )k
Q
k= 5, C= 3
k= 10, C= 3
Figure 11: Improvements in latest arrival versus total duration
Figures 11 and 12 further illustrate the trade-off between improving latest arrival time or
sum of arrival times and the total route duration. In Figure 11, for example, we plot the point
µla(V RP )k
Q−la(M M)k
Q
la(M M)k
Q
,c(MM)k
Q−c(V RP )k
Q
c(V RP )k
Q¶for each instance. This plots the percentage increase in lat-
est arrival time by using the VRP objective against the percentage increase in total duration by
using the minmax objective. We do this for k= 5 and 10, uncapacitated vehicles. Since almost all
points lie below the 45 degree line and the majority lie below the line y=1
2x, we can remark that
the relative increase in latest arrival is consistently and often significantly larger than the relative
increase in total route duration. Similar conclusions can be drawn for minsum from Figure 12.
In looking at the full set of tables, it appears that geography of the datasets does have some im-
pact on our results. For both Augerat-A (randomly distributed) and Augerat-B (clustered), almost
all of the improvement in service from having additional vehicles is felt by expanding to 5 vehi-
cles, but steady improvements in service are possible with increasing capacity for both objectives.
Interestingly, increasing capacity seems to have as much or more impact on improving service for
Augerat-A as for Augerat-B, but it usually accompanies lower increases in cost for Augerat-A for
both minmax and minsum objectives. The Golden datasets (special structures) yield quite different
results than Augerat-A and B. For both objectives, the improvement in service found with increas-
ing capacity is much more dramatic for Golden than found with the other datasets, and the cost
increases are often less. Also, the Golden sets exhibit steady increases in service with the increase
in vehicles.
33
0
0.5
1
1.5
2
2.5
3
3.5
4
0246810
c(MS )k
Q−c(V RP )k
Q
c(V RP )k
Q
sa(V RP )k
Q−sa(MS )k
Q
sa(MS )k
Q
k= 5, C= 3
k= 10, C= 3
Figure 12: Improvements in sum of arrival times versus total duration
8 Conclusions and Next Steps
This paper provides strong evidence that much better service times to customers are possible than
those created by traditional routing problems and algorithms. In a situation where service time
equates to survival, such as after a disaster, it is important to have appropriate routing technology
that addresses these concerns. We realize that this effort represents only a first step in developing
new methodology for routing humanitarian aid. There are many other issues in a relief context
that need to be considered, including additional factors in the objective function and additional
constraints. In terms of objectives, it is clearly important to consider combinations of cost and
service, since both are indeed relevant in practice. Future work may consider an objective function
that combines a traditional cost-based objective with the service-based objectives considered here.
It is also important to consider the fact that the different customers in our datasets represent
different size groups, so customer weighting schemes may be worth examining.
Key constraints to be considered involve the reliability of arcs in the network after the disaster.
After a disaster, existing roads may suddenly become closed or the travel time may become drasti-
cally increased, so routes may need to be designed to incorporate detours or alternate paths. For
some roads, their availability or travel time in the network may be dynamically changing during
the days following the disaster, so it may be necessary to dynamically change the routes as well.
These arc reliability issues similarly impact the location decisions for the distribution centers of
relief supplies to ensure reliable and efficient routes can be created. We are currently studying how
34
arc reliability can be incorporated into these location decisions [12].
Another area of future research is to consider the complications that result from delivering
multiple commodities. Consider what happens if the vehicles delivering relief are aid-specific, such
a water truck and a food truck. If we know a tour that is efficient and equitable for one commodity,
the natural choice is to send both trucks on the same route. If they arrive at the same distribution
point at the same time, there may be a large delay before the second truck can be unloaded due to
limited staff or security concerns. Such delays are not good from an efficiency perspective. Also, the
distribution point that is last on the route would receive both types of supply last, which does not
seem equitable. Thus, we would like to consider the issue of multiple arrival times in conjunction
with multiple products/arrival times. There are many questions yet to be addressed in developing
appropriate routing tools for disaster relief.
Acknowledgments
This work was partially supported by the National Science Foundation through grant number
0237726(Campbell).
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APPENDIX
We now complete the proof of Lemma 1 from section 5.2.2. Following Figure 7, it is required to
show that the optimal minsum route will select cliques of nodes in descending order. Note that the
order of nodes within a clique is arbitrary as long as the nodes are visited consecutively.
Proof. Assume that a route is given R: 0 −Ci1−Ci2−. . . −CiP−CK−CiP+1 −. . . where CK
is the highest indexed clique in the graph and is not first in route R. We will show that route
R0: 0 −CK−Ci1−Ci2−...−CiP−CiP+1 −. . . obtained by placing CKfirst has a strictly better
minsum objective. Let tij represent the distance between Ciand Cj. The sum of arrivals for Rand
R0are:
sa(R) = (N)t0i1+ (N−Ni1)ti1i2+. . . + (N−
P
X
j=1
Nij)tiPK+ (N−NK−
P
X
j=1
Nij)tKiP+1 +. . .
sa(R0) = (N)t0K+(N−NK)tKi1+. . . +(N−NK−
P−1
X
j=1
Nij)tiP−1iP+(N−NK−
P
X
j=1
Nij)tiPiP+1 +. . .
where N=PK
j=1 Nj. Note that the final portion of these sums is the same for both routes. We
wish to show that the difference sa(R)−sa(R0) is strictly positive. This difference can be simplified
by adding and subtracting the quantity (NK)tiPKto obtain:
sa(R)−sa(R0) = sa(R)−(NK)tiPK+ (NK)tiPK−sa(R0)
= (N−NK−
P
X
j=1
Nij)(tiPK+tKiP+1 −tiPiP+1 ) + N(t0i1−t0K) + NK(tiPK+
P−1
X
j=1
tijij+1 )−(N−NK)ti1K
≥N(t0i1−t0K) + NK(tiPK+
P−1
X
j=1
tijij+1 )−(N−NK)ti1K
The inequality follows from dropping the first term, which is positive since N≥NK+PP
j=1 Nij
and tiPK+tKiP+1 ≥tiPiP+1 by the triangle inequality. This holds for any route where CKis not
first. If CKis last, then omitting terms involving CiP+1 shows this inequality to still hold. By using
39
the fact that t0K=Min the graph, we arrive at:
sa(R)−sa(R0)≥N(t0i1−M)+NK(tiPK+
P−1
X
j=i
tijij+1 )−(N−NK)ti1K≥Nt0i1−N M +(2NK−N)ti1K
(A-1)
This last inequality in (A-1) holds because tiPK+PP−1
j=1 tijij+1 ≥ti1Kdue to the triangle inequality.
Recall that the number of nodes in clique CKis given by NK=MK−1and the total number of
nodes in the graph is N=MK−1
M−1. Hence for M > 2 the last coefficient, (2NK−N), in (A-1) is
positive. The smallest possible value of Nt0i1−N M + (2NK−N)ti1Koccurs when Ci1=CK−1,
where t0i1= 2 and ti1K=M. These values show sa(R)−sa(R0)≥2.
Thus if CKis not first on a route, then the route can be improved by placing clique CKfirst.
This argument serves as the basis for an induction proof. We assume that a route is given where the
first jcliques are in descending order as R1: 0 −CK−CK−1−. . . −CK−j+1 −. . . , but clique CK−j
is not the successor of CK−j+1. Similar to before, let route R0
1have clique CK−jfollowing CK−j+1.
The difference sa(R1)−sa(R0
1) will eliminate the portions of the two routes that are the same, which
now include the descending cliques CK, . . . , CK−j+1. We can remove nodes CK, . . . , CK−j+2 from
the graph and let CK−j+1 represent the depot instead. By symmetry of the graph, the distances
involved will be the same and the remaining graph is a smaller instance of the original. This allows
the basis argument to be used to establish the result that route R0
1is an improvement. Hence, the
only route that cannot be improved for the minsum objective on this graph is the route that visits
all cliques in descending order.
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