Semi-autonomous Control of Multiple
Heterogeneous Vehicles for Intersection Collision
/4NmS
LITITUTE,
SO
T^C
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M
Avoidance
by
AUG15 2014
Heejin Ahn
LIBRARIES
Submitted to the Department of Mechanical Engineering
in partial fulfillment of the requirements for the degree of
Master of Science in Mechanical Engineering
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
June 2014
@ Massachusetts Institute of Technology 2014. All rights reserved.
Signature redacted
Author.......
...................
Department of Mechanical Engineering
.
May 27, 2014
Signature redacted
Certified by..................
Domtilla Del Vecchio
Associate Professor
Thesis Supervisor
Signature redacted'
Accepted by.........................................
David E. Hardt
Chairman, Department Committee on Graduate Theses
2
Semi-autonomous Control of Multiple Heterogeneous
Vehicles for Intersection Collision Avoidance
by
Heejin Ahn
Submitted to the Department of Mechanical Engineering
on May 27, 2014, in partial fulfillment of the
requirements for the degree of
Master of Science in Mechanical Engineering
Abstract
This paper describes the design of a supervisory controller (supervisor) that manages
multiple heterogeneous vehicles, i.e., multiple controlled and uncontrolled vehicles, to
avoid intersection collisions. Two main problems are addressed: verification of the
safety of all vehicles at an intersection, and management of the inputs of controlled
vehicles. For the verification problem, we employ an inserted idle-time scheduling
approach, where the "inserted idle-time" is a time interval when the intersection is
deliberately held idle for uncontrolled vehicles to safely cross the intersection. For the
management problem, we design a supervisor that is least restrictive in the sense that
it overrides controlled vehicles only when a safety violation becomes imminent. We
analyze computational complexity and propose an efficient version of the supervisor
with a quantified approximation bound. To mitigate the abrupt changes of control
inputs and to reduce the number of unnecessary interventions, we additionally design two optimization problems and provide the supervisor with a more conservative
bound.
Thesis Supervisor: Domitilla Del Vecchio
Title: Associate Professor
3
4
Acknowledgments
First and foremost, I would like to express a sincere appreciation to my thesis supervisor, Prof. Domitilla Del Vecchio. I am particularly thankful for her thoroughness
in discussing research directions and editting my naive papers. I always appreciate
her interest in my research problems and her effort to provide anything that can help
me tackle the problems. Her uncountable advice and encouragement have been a
constant driving force behind my research work.
I benefited from interaction with the past and current members of Prof. Del
Vecchio's group for their valuable comments and friendship. In particular, I wish to
acknowledge the help provided by Dr. Alessandro Colombo, who is now an assistant professor of Politecnico di Milano. His suggestions have inspired me to analyze
problems correctly and thoroughly. I would like to thank Dr. Daniel Hoehener for
his help on formulating optimization problems. Without his experience, I could not
complete the optimal approaches in this paper. I wish to thank Andrea for helping
me run experiments. Because of my initial ignorance of Linux and HTTP, it might
be tough for him to teach everything to me.
I am grateful to the members of the MIT Korean Graduate Student Association of
Mechanical Engineering (KGSAME) for making my life at MIT much more pleasant
and exciting. I would also like to express my gratitude to my friends who take classes
and study together, to my friends from First Korean Church in Cambridge for praying
for me, and to my friends in Korea for always being emotionally close to me. My
special thanks go to Minkyun, who encourages me and makes me always able to count
on him. Thank you for being here with me.
I owe everything to my family, particularly to my parents, Kookyoung Ahn and
Kyoungsoo Hong, for having given me unconditional love and support. I am also
thankful to my sister and brother, Heeyoon and Junhong, and to my grandmother
for their constant care. Without them, this thesis could not be possible.
5
6
Contents
1
Introduction
15
2 Basic Concepts
19
2.1
Mathematical Background ......
2.1.1
Scheduling .......
2.1.2
Equivalence .......
........................
19
.............................
19
............................
21
2.2
Intersection Model
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
2.3
Dynamic Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
3 An Exact Approach
3.1
3.2
3.3
25
Problem Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
3.1.1
Verification Problem . . . . . . . . . . . . . . . . . . . . . . .
25
3.1.2
Supervisory Problem . . . . . . . . . . . . . . . . . . . . . . .
29
Problem Solutions ............................
.
3.2.1
Solution of the Verification Problem . . . . . . . . . . . . . . .
30
3.2.2
Solution of the Supervisory Problem
32
Simulation Results .......
. . . . . . . . . . . . . .
............................
4 An Efficient Approach
4.1
4.2
30
Relaxed Verification Problem
36
39
. . . . . . . . . . . . . . . . . . . . . .
39
4.1.1
Relaxed Scheduling Problem . . . . . . . . . . . . . . . . . . .
39
4.1.2
Relaxed Verification Problem
42
Efficient Supervisors .......
. . . . . . . . . . . . . . . . . .
...........................
7
44
5.3
Motivations ...............
. . . . . . . . . . . . . . . . .
47
5.1.1
Abrupt Changes of Inputs ...
. . . . . . . . . . . . . . . . .
48
5.1.2
Unnecessary Intervention ....
..... .. ... .. .. ...
49
..... .. .. ... .. .. .
51
Input Optimization Problem ......
Discrete Scheduling Problem .
... .... .. .. .. ... .
52
5.2.2
Input Optimization Problem .
.... ... .. .. .. ... .
54
5.2.3
Simulation Results . . . . . .
... .... . ... . .... .
56
Schedule Optimization Problem . . .
.. .... .. ... .. .. ..
57
5.3.1
Totally Unimodular Matrix [20] . . . . . . . . . . . . . . . . .
58
5.3.2
Schedule Optimization Problem . . . . . . . . . . . . . . . . .
59
5.3.3
Simulation Results . . . . . . ..
61
.
.
5.2.1
.
5.2
44
47
Optimal Approaches
5.1
6
. . . . . . . . . . . . . . . . . . . . . . . . . . .
.
5
Simulation Results
.
4.3
. ..... .. .. .. .. ...
67
Conclusions and Future Works
69
A Proof for the discrete system
A.1 Existence of input ...................................
69
A.2 Relation between Discrete Scheduling Problem and Verification Problem 72
B Total modularity of the Schedule Optimization Problem
8
75
List of Figures
2-1
An example of 3 vehicles approaching an intersection. The traffic intersection (left) is modeled as one point at which unidirectional paths
intersect (right). Here, ai and Bi represent the location of the intersection on the longitudinal path i.
3-1
. . . . . . . . . . . . . . . . . . . . .
21
Positions of 5 controlled (black solid line) and 2 uncontrolled (red dotted line) vehicles controlled by the supervisor s in Section 3.2 (Algorithm 3.2). The blue boxes at the bottom represent the time steps
when the supervisor overrides the controlled vehicles. The shaded region represents the location of the intersection. . . . . . . . . . . . . .
3-2
37
Positions of 8 controlled (black solid line) and 2 uncontrolled (red dotted line) vehicles controlled by the supervisor s in Section 3.2 (Algorithm 3.2). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4-1
38
Positions of 8 controlled (black solid line) and 2 uncontrolled (red dotted line) vehicles controlled by spp,,,. in Section 4.2. The blue boxes at
the bottom represent the time steps when the supervisor overrides the
controlled vehicles. The darker shaded region represents the location
of the intersection while the lighter shaded region represents that of
the inflated intersection.
. . . . . . . . . . . . . . . . . . . . . . . . .
9
45
5-1
Positions of 4 vehicles (black solid line). The blue boxes at the bottom
represent the time steps when the supervisor overrides the controlled
vehicles. The darker shaded region represents the location of the intersection while the lighter shaded region represents that of the inflated
intersection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5-2
48
Control input (black solid line) for the last car with its desired input
(blue dotted line). The two horizontal dotted lines represent the upper
and lower bounds for the input. . . . . . . . . . . . . . . . . . . . . .
5-3
Positions of 4 vehicles where the last vehicle is far from the intersection
compared to the others.
5-4
49
. . . . . . . . . . . . . . . . . . . . . . . . .
50
Control input (black solid line) for the last car in Figure 5-3(a) with
its desired input (blue dotted line). The two horizontal dotted lines
represent the upper and lower bounds for the input. . . . . . . . . . .
51
5-5
Average cost on 60 samples of four vehicles . . . . . . . . . . . . . . .
57
5-6
Implementation of the two optimization problems (Problem 5.3 and
Problem 5.4) at stage k. First, the Discrete Scheduling Problem is
solved via Algorithm 4.4, and if there exists a feasible schedule, the
sequence of the schedule lets the Schedule Optimization Problem compute the optimal schedule T*. If there does not exist a feasible schedule, Algorithm 4.4 returns no and an empty schedule. Depending on
this answer, the supervisor either runs the Input Optimization Problem to store the safe input for the next step Uk+1,,afe or overrides the
system using the safe input Uk,safe.
10
. . . . . . . . . . . . . . . . . . .
62
5-7
The same initial condition and desired input are considered as in Figure 5-3. In Figure 5-7(a), the lighter shaded rectangle represents the
inflated intersection due to the discretization and approximation for
the efficient solution. The blue boxes represent the time steps when
the supervisor intervenes. Although the supervisor intervenes, the last
vehicle is not overridden, that is, the control input (black line in Figure 5-7(b)) remains equivalent to the desired input (blue dotted line in
Figure 5-7(b)).
5-8
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63
The same initial condition but a varying desired input are considered.
The control input (black line in Figure 5-8(b)) maintains the similar
level to the desired input (blue dotted line in Figure 5-8(b)). . . . . .
5-9
64
Average cost on 60 samples of four vehicles. Random schedule (blue
bars) represents the results from solving Problem 5.3 while Optimal
schedule (red bars) represents the results from solving Problem 5.3
and Problem 5.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
65
12
List of Algorithms
3.1
Verification of Problem 3.2 . . . . . . . . . . . . . . . . . . . . . . . .
31
3.2
Solution of Problem 3.3. . . . . . . . . . . . . . . . . . . . . . . . . .
34
4.3
Solution of DEC(1|r,p = 1ILmaxO) . . . . . . . . . . . . . . . . . . .
40
4.4
Verification of Problem 4.1 . . . . . . . . . . . . . . . . . . . . . . . .
42
13
14
Chapter 1
Introduction
In the United States, the most critical and frequent vehicle collisions occur at or
near intersections according to the National Motor Vehicle Crash Causation Survey
(NMVCCS) [2]. In response to this, there have been a number of attempts within
academia, industry, and government to reduce the number of accidents at intersections. One of the most conspicuous examples is Google's self-driving car, which has
surpassed 700,000 autonomous driving miles without collisions, including at or near
intersections [19]. However, the fully automated driving is still controversial [24];
one of the reasons is that people prefer having control of their vehicles. The alternative promising approach has been semi-autonomous driving, which guarantees the
freedom of the drivers and takes over the controls only when safety constraints are
violated [4,23].
In order to inform, warn, and eventually override human drivers to prevent crashes,
the auto industry and the government have been developing vehicle to vehicle (V2V)
and vehicle to infrastructure (V21) communication [1]. The potential for using multivehicle communication has stimulated research on networked vehicle systems to maintain safety since the 90's with the California PATH project [3], focusing mostly
on highways.
More recently, works employing job scheduling and optimal control
[9,10,16,17] for intersection collision avoidance have appeared. Works that mostly
focused on warning for collision avoidance at intersections have also appeared for
general traffic scenarios [8,22].
Driver behavior estimation and classification near
15
intersections has also been of interest [5,13].
This paper exploits the V21 communication so that a supervisor embedded in a
roadside infrastructure is able to measure the position and speed of all approaching
vehicles and give commands to some of them, called controlled vehicles. The controlled vehicles are equipped with driver assist systems that enable the supervisor to
communicate with the vehicles and override the human drivers. In this paper, the
design of a centralized supervisor is described to prevent intersection collisions for a
heterogeneous system, that is, in the presence of multiple controlled and uncontrolled
vehicles. We thus extend the work of [10], which applies only to cases when all vehicles
are controlled. To consider uncontrolled vehicles, a different problem formulation of
safety verification and a different state prediction scheme are required. In particular,
safety verification must consider all possible future states of uncontrolled vehicles to
estimate when they will cross an intersection since the inputs for the uncontrolled
vehicles are neither measured nor controlled.
An inserted idle-time (IIT) scheduling approach [15] is employed to detect upcoming collisions in the presence of uncontrolled vehicles. Depending on the outcome
of safety verification, the supervisor intervenes if necessary.
In general, computa-
tional complexity is a major concern in centralized control systems [6], and the IIT
scheduling used in this paper is known to be NP-hard [14], that is, the IIT scheduling
sometimes cannot be solved in short time. We address this problem by proposing an
efficient scheduling approach that solves a relaxed problem. We quantify the extent
of conservatism of the solution by determining an equivalent "Bad" set, which the
trajectories of the system are forced to avoid. Based on this efficient solution, we
further formulate optimization problems to generate a smoother input profile. To
implement the optimization problems, we consider the corresponding discrete system
dynamics and provide a more conservative Bad set.
This paper is organized as follows. In Chapter 2, we review the mathematical
framework of scheduling problems and define the system. Chapter 3 presents two main
problems, Verification and Supervisory, and provides the algorithms that exactly solve
these problems. To address the computational complexity of these exact solutions,
16
approximate problems and efficient solutions are introduced in Chapter 4. To generate
smoother control inputs, optimal approaches are presented in Chapter 5. The exact,
efficient, and optimal approaches are followed by their simulation results. Conclusions
and future works are also provided in Chapter 6.
17
18
Chapter 2
Basic Concepts
In this chapter, we introduce several basic concepts that are used through this paper.
In section 2.1, mathematical background of scheduling and equivalent relation of
problems are introduced. In section 2.2, we define an intersection with assumption
that all cars are moving on pre-determined longitudinal paths that are converging to
one point. In section 2.3, a dynamical system is modeled, and several notations and
properties of system variables are given.
2.1
2.1.1
Mathematical Background
Scheduling
Scheduling problems are described by jobs, machines, and processing characteristics [11, p. 4]. Jobs represent tasks which have to be executed; machines represent
scarce resources such as facilities where jobs are performed; and processing characteristics used in this paper are release times ri, deadlines di, and process times
pi. A schedule is a vector of starting times t, for all jobs such that all constraints
in a scheduling problem are satisfied. While most scheduling literature focuses on
non-delay schedules, we consider a more general class of schedules which is called
an inserted idle-time (IIT) schedule. This schedule is created by considering an IIT,
which is an open time interval when the machine is deliberately held idle while at
19
least one job is waiting to be processed [151. If we say (fy,p,) is an IIT, a necessary
condition for an IIT schedule t, to be feasible is for all i,
(tilts
+pi)
n (i-,)
=0.
In this paper, we consider the scheduling problem DEC(1IrsILm.,O), which aims
to find a schedule that makes the maximum lateness Lm. := maxi tj +pi - d smaller
than or equal to 0. This is stated in a formal manner as follows.
,
Definition 2.1 (DEC(1|riL,.,0)). Given a set of n jobs that have release times r
deadlines di, and process times p over one machine and given a set of m inserted
idle-times (f,,y), determine whether there exists a schedule T = t1...,t.) such
that for all i E {1, ...
, n},
for all i ) j E{1, ... , n}
and for all -7E{1,..
tj
,
ts :5 tj => t +p P
., m}
ti.
-Y =>ti
12PY.
Notice that if such a schedule exists, (ti, ti + pi) n (f,
pY)
=
0 for all i and -y
because for all t* E (ti,tj + pi), the last two constraints imply t* < , or t* > p-.
The problem DEC(lri,p, = 11L,.,0) has the same constraints as in Definition 2.1
except with the addition of the constraint pi = 1. It has been shown that a problem
with arbitrary process times such as DEC(1IriILm.,0) is NP-hard, while a problem
with unit process times such as DEC(lri,p = 11Lm.,O) is solved in polynomial
time [21]. The problem DEC(lri,p = 1L,.,O) is used to address the complexity
issue in Chapter 4.
20
a2
P2
a3
Figure 2-1: An example of 3 vehicles approaching an intersection. The traffic intersection (left) is modeled as one point at which unidirectional paths intersect (right).
Here, a and 3i represent the location of the intersection on the longitudinal path i.
2.1.2
Equivalence
To solve a problem relying on the solution to a different problem, concepts of "reducibility" and "equivalence" [18] are introduced. In general, an instance of a problem
is the information required to compute a solution to the problem, while satisfying all
given constraints [12, p. 5]. A problem P is reducible to a problem P2 if for an instance
12
of P2 that can be constructed in polynomial-bounded time for any instance
I1 of P1 , P2 accepts
12
if and only if P accepts 1. We write P1 oc P2 if P is reducible
to P2 . If P2 oc P and P1 cx P2 , P and P2 are equivalent.
2.2
Intersection Model
In this paper, we model the intersection as n vehicles moving along unidirectional
paths intersecting at one point (Figure 2-1). We assume that a subset of these vehicles
can be controlled while the remaining vehicles cannot be controlled. We identify
vehicles with a natural number from 1 to n. To distinguish between controlled and
uncontrolled vehicles, we define a controlled set C and an uncontrolled set C, which
21
contain n, and ne elements, respectively, as follows:
C := {i E {1, ... , n} : vehicle i is controlled}
C := {-y E {1,
..
, n} : vehicle 7 is uncontrolled}.
In this paper, i usually denotes a controlled vehicle while 7 denotes an uncontrolled
vehicle. For notational brevity, let C = {1, ...
, n,} and C = {n, + 1, ... , n} through
this paper.
2.3
Dynamic Model
Let xi represent the dynamic state of vehicle i with yi E R the position on its path.
For i E C, let ui represent the input to vehicle i and for -y E C, let d, represent
a disturbance, that is, an unknown input to vehicle -y. Each controlled vehicle is
modeled by
i = fi(xi, u2 ),
yj = hi(xi),
(2.1)
y- = hy (x).
(2.2)
while each uncontrolled vehicle is modeled by
S= f (xy , d-y),I
,
where xi E Xi g Rr, ui E U c R-, y E Y g R, xy E X c R', d, E Dy c R 8
and yy E Y, C R. The functional spaces of the piecewise continuous signals ui(t)
and dy (t) for t E [0, oo) are U2 and Thy, respectively. The symbols x, u, d, y denote
aggregate vectors for these states, inputs, disturbances, and outputs. For instance
x = (x 1,
.
., xn). The spaces of these vectors are denoted by X, Y, U, D,U, V. Notice
that X C (R')" and Y C Rn while U C (R ' )fc and D C (RB)ne. Moreover, the
output function hi(xi) and h.(xy) are continuous and the derivative of the output is
bounded, that is, ij E [yi,m, Yi,M] with
y4,m
> 0. We assume that sets U and D, are
intervals, that is, U = [ui,m, ui,M] and D, = [dy,m, df,m].
22
The parallel composition of (2.1) and (2.2) for all n vehicles describes the system
dynamics:
x
= f(x, u, d),
y = h(x).
(2.3)
We assume that this system has a unique solution that continuously depends on the
input and the disturbance. The state and output of vehicle i at time t starting from
xi(to) with input u, are denoted by xi(t, ui, xi(to)) and y1 (t, ui, xi(to)), respectively.
The corresponding aggregate state is denoted by x(t, u, d, x(to)) and the output by
y(t, u, d, x(to)). For notational brevity, we omit an initial condition when to = 0
and also omit the other arguments when they are not important. From now on, we
consider to = 0.
For any two states xi, xl' E Xi
; R'r, we say that xi < x' if xij < x'g for all
j E {1,...,r} where xij and x'. are the j-th elements of xi and x ', respectively.
Similarly, for any two time signals ui, u' E 14, we say that ui 5 u' if u (t)
u(t)
for all t > 0. We assume that the trajectories y,(t, uj, xi(0)) and yy,(t, d.,, xz(0)) are
order preserving functions of their arguments. That is, if u i<i'
dci.,
d, for d., d, E ',,, then y;(t,ui, xi(0))
for uj, u' E U1 and
yi(t,u', xi(0)) and y.,(t, dy, xy(0))
y7 (t, d,,x,(O)) for all t > 0. Also, if x2 (O) < x'(O), then y,(t, ui, xi(0)) 5 y,(t, ui, x'(O))
and if x.,(0) 5 x ,(0), then y.,(t, d., xy(O)) 5 y,(t, dy, x,(0)).
In the same sense,
the states x,(t, ui, xi(0)) and xy(t, d, x.y(0)) are order preserving functions of their
arguments.
23
24
Chapter 3
An Exact Approach
In this paper, two main problems are addressed: the Verification Problem and the Supervisory Problem. The Verification Problem determines whether the current drivers'
inputs will lead to a collision. If this is the case, the Supervisory Problem overrides
all controlled vehicles with safe inputs. In Section 3.1, these problems are formulated
in a formal manner. An exact solution for these problems are given in Section 3.2.
Finally, simulation results validate these solutions in Section 3.3 and show computational defects in these solutions.
3.1
Problem Statement
In this section, we formulate the two main problems, the Verification Problem and the
Supervisory Problem. Also, the Inserted Idle Time Scheduling Problem is introduced,
which is proved to be equivalent to the Verification Problem.
3.1.1
Verification Problem
As seen in Figure 2-1, an interval (ai, /i) is assigned to vehicle i for all i E C U C.
Let a := {ai, ... , ao} and / := {#1,..., #,}. If two or more vehicles, at least one of
which is a controlled vehicle, are inside the intersection at the same time, we consider
that a collision occurs. Accordingly, we define the bad set B C Y as the set of output
25
configurations corresponding to a collision as follows:
B := {y E Y : yi E (ai, Pi) and yj E (aj, #j)
(3.1)
(3.2)
for some i : j such that i E C and j E C UC}.
This implies that we focus on preventing collisions in which one or more controlled
vehicles axe involved. These include collisions between controlled and uncontrolled
vehicles and collisions among controlled vehicles.
Problem 3.1 (Verification Problem). Given an initial condition x(O), determine if
there exists an input signal u E U which guarantees that y(t, u, d)
B for all d E V
for all t > 0.
Input u E U ensures safety if and only if y(t, u, d)
g
B for all d E V for all
t > 0. An instance of Problem 3.1 is described by the initial condition x(O) and the
parameters E) = {f, h, X, Y, U, D, U, E), a,#B}. Thus, if there exists u which ensures
safety for a given instance {x(0),
e}, then we say {x(0), 6} E
Problem 3.1.
To solve Problem 3.1, we adapt the IIT scheduling problem DEC(1rsIL,., 0) in
Definition 2.1 to our problem. In particular, the intersection plays the role of the
machine, and the nr vehicles are the jobs. Furthermore, we define release times R,,
deadlines Di, process times P(T), and IITs (I.,, I 7) as follows.
Definition 3.1. For all i E C where y2 (0) < ai, let
R, :=min{t > 0 : y,(t, ui) = ai},
Di :=max{t > 0 : yi(t, ui) = ai}.
Given a non-negative real number Ti, let
#
Pi(T) := min{t > 0 : y,(t, ui) =
with constrainty (Ti,ui) = ai}.
26
For all - E C where y,(0) < a,, let
-
R:= min {t > 0 : y(t, dy) = ay},
d-ED-y
P7 := max{t > 0 : y(t, dy) = 8}.
For i E C, if yt(O)
/3l, then set R, = 0, Di = 0, and P(T) = 0. If y,(0)
then set R, = 0, Di = 0, and P(T) = miniEui{t : yi(t, uj) =
#l}.If
yi(T, uj) = ai is not satisfied, set P(T) = oo. For ^ E J, if y7 (0)
and PR = 0. If yy(0)
aj,
the constraint
8,, set R, = 0
ay, set R, = 0 and P, = maxEy.,{t : yy(t, dy) = # }.
These variables axe well-defined since it is assumed that the system (2.3) has a
unique solution and
y;
> 0. Notice that R, Di, R, P
for all i E C and -yE C are
fixed once an initial condition is provided. We can view (Ry, Py) for -y E j as an IT
because the intersection needs to be kept empty as long as an uncontrolled vehicle
can occupy it. We introduce the following scheduling problem.
Problem 3.2 (IT Scheduling Problem). Given an initial condition x(O), determine
whether there exists a schedule T = (T1,..., Tn) where C = {1,... ,nc}, such thatfor
all i E C,
R45 T !- Ds,7(3.3)
for alli
j EC,
T < T => P;(TO)
(3.4)
T,
for all i E C and y E C,
T
T
R
, P(T) <
,=> T
,
(3.5)
P,.
(3.6)
This problem is also described by an instance I = {x(0), 0}. We say {x(0), e} E
Problem 3.2 if there exists a schedule T that satisfies all constraints in Problem 3.2.
A feasible schedule for Problem 3.2 ensures (T, P(T)) n(T,, PY)
=
0 by (3.5) and
(3.6). The IT scheduling determines if a sequence of controlled vehicles exists such
27
that controlled vehicles cross an intersection before or after IIT. We now prove that
Problem 3.2 is equivalent to Problem 3.1 and then propose a solution to Problem 3.2
in Section 3.2.
Theorem 3.1. Problem 3.1 and Problem 3.2 are equivalent.
Proof. We prove that the Verification Problem (Problem 3.1) is reducible to the IIT
Scheduling Problem (Problem 3.2) and vice versa. Notice that Problem 3.1 and 3.2
have the same instance I E {x(0), E}. Thus, equivalence between the two problems
follows by showing that for any given instance I = {x(0), 9},
I E Problem 3.1 -> I E Problem 3.2.
(=>) Given an initial condition x(0), there is an input fi which ensures safety, that is,
y(t,f, d)
B for all d and for allt > 0. For y(0) < a for i E C, define T as the time
t when y,(t, fi) = aj, and for y,(0) < 8i, define P,(T) as the time when y,(t, fi) = 63.
If yj (0)
: ai , set T = 0. Set
Pi(T)
By definitions of R, and Di, R,
=0 if yj (0) ! #8.
T
Di (Condition (3.3)). Suppose T :5 T for
some i $ j E C. Since fij and fii prevent an intersection collision between vehicles i
and j, when y,(t, s,) = a3 , we have that y,(t, fi)
/3. Since
i(Ti) 5 T. By definition of P(T), it follows that P(T)
y,
> 0 by assumption,
i (T) 5 T (Condition
(3.4)). Also, with the assumption that the input fi ensures safety for all d, if T 5 Ry
for i E C, - E C, when maxs y7 (t, dy) = ay, it must be true that y,(t, fii)
/3. Since
y,(t) is increasing with time, {t: max 4 yy(t, dy) = ay} becomes min,{t : y7 (t, dy) =
a7 } which is equal to R, by definition. Combining this with the definition of P(T),
we have that P(T) 5 F(T)
Ry (Condition (3.5)). If T
kBy, when y (t,fii) = as,
the vehicle y must stay away from the intersection, i.e., mind y,,(t, dy)
because of the increasing property of the output with respect to t, T
8,. Again,
P, (Condition
(3.6)).
(.=) Given an initial condition x(O), there exists a schedule T E Rnc that satisfies
all conditions of Problem 3.2. We start assuming that y,(O) < ai for all i E C U C,
which does not break generality since vehicles after the intersection axe not of interest.
28
According to Lemma 5.1 in [10], if T E [Ri, Dj] for yi < as, there exists an input ui E
Uj such that y (T, ui) = aj. Then, we can construct ui E Uj such that y (Ti, u) = ai
for all i E C.
Now, we show that these inputs ui for i E C ensure safety, if T satisfy conditions
When T < Tj for i # j E C, condition (3.4) indicates P(T)
(3.3)-(3.6).
which implies that when y (t, Uj) = fl, we have y,(t, u,)
a so that uj and u,
prevent collisions between controlled vehicles. From condition (3.5), when T
for i E C,7-y E j, P(T)
have yi(t, u)
3l, that is, y,(t, dy)
ay and yi(t, u)
aj, it is true for all d, that yy(t, d)
/,l.
#li. If
T > R,,, condition
yy(t, dy)
f,. Thus, when
Hence, ui for i E C prevent
collisions between controlled and uncontrolled vehicles for all d.
3.1.2
RY
R.. This implies that when maxdy,(t, d) = ay, we
(3.6) indicates that when yi(t, u) = aj, we have min
y (t, uj)
T,
0
Supervisory Problem
We now design a supervisor operating in discrete time. At each time step kr, a
current state x(k-r) E X and a desired input ak E U for the controlled vehicles are
given, where the desired input represents the input applied by a driver. Then, the
supervisor returns an input based on the partial state prediction for the next step.
We define the j-th entry of this partial state prediction as follows:
(xj(r,ak3 ,x 3 (kr))
j EC
xj (kr)
jEC
[xk+1(ak)]3 -=
(3.7)
where akj E U for j E C is the j-th entry of the vector ak, and it it known since
it is for a controlled vehicle. Notice that the current disturbance d E V does not
contribute to the partial state prediction since uncontrolled vehicles are not equipped
with driver assist systems that can monitor the input of the drivers. This is why we
called xk+1(ak) the partial state prediction.
Because a supervisor interacts with human operators, it has to be least restrictive
in the sense that it intervenes only if a collision is guaranteed to occur in the future
29
for some d E V. To describe future trajectories, define a continuous input profile
u'|(t) defined for t E [kr, oo) and let uk(t) be uj|(t) restricted to t E [kr, (k + 1)r].
During this time period, let
Uk(t) =
ak, where ak is the desired input at t = kr and is
considered constant on [kr, (k + 1)r]. Similarly, let u',,e(t) be a safe input defined
for t E [kr, oo) and let
Uk,,afe(t)
be u00,fe(t) restricted to t E [kr, (k + 1)r]. We now
formalize the Supervisory Problem.
Problem 3.3 (Supervisory Problem). Design a supervisor s(x(k-r), ak) at time step
kr such that
B
if 3uj|*: y(t, uj|, d)
for all d E D for t > kr
uk,safe
otherwise,
=
and such that it is non-blocking: if s(x(kr), ak)
#
0, then for any ak+1, s(x((k
+
s(x(k'r), ak)
IUk
1)-r), ak+1) = 0.
3.2
Problem Solutions
By virtue of Theorem 3.1, we can solve the Verification Problem by solving the IIT
Scheduling Problem. In this section, we thus focus on providing an exact solution to
the IT Scheduling Problem, which we then use to solve the Supervisory Problem.
3.2.1
Solution of the Verification Problem
Assume, without loss of generality, that yi > a for m controlled vehicles. Let P
be the set of all permutations of controlled vehicles such that yi < ai. Then, each
element of P is an (ne - m)-tuple denoted by ir. Let fr be an n--tuple composed of
all -f E d in an increasing order of R.. The notation -r represents the i-th element of
7r, and fri represents the i-th element of ir. Parameter d represents a time delay, as
explained in Section 3.2.2.
Algorithm 3.1 with 6 = 0 solves Problem 3.2 (IIT Scheduling Problem), and,
hence, Problem 3.1 (Verification Problem) by Theorem 3.1. Also, notice that the
30
Algorithm 3.1 Verification of Problem 3.2
procedure EXACTSOLUTION(X(0), 6)
if y(O) E B then return {0, no}
for all i E C and -y E C do
given x,(0) calculate R, Di
given x,(0) calculate I?, P,
Ih +- max(Ih - 6,0)
P1 +- max(P1 - J, 0)
for i E C such that yi(O) ;> ai do
T +- 0
calculate P(T) and P. +- max, P(T)
for all 7r E P do
for j +- 1 to n, - m do
Tr1 +-max(Rri,
P
Tr+-
max(RK,
) forj
=
1
-,,_(T,,_,))
given T,, calculate P3
for r +- 1 to nE do
if T 1 . > R,. then
Tr, +-max(T,,, Ifr)
given T, calculate P,
else if Pj, > R*, then
Tr +-P,
given T, calculatePr
if T < Di for all i E C then
return {T, yes}
31
running time of Algorithm 3.1 increases in a factorial manner with (nc - m) since the
algorithm checks all permutations in P to return no. This is consistent with the fact
that Problem 3.2 is known to be NP hard [141.
Example 3.1. Suppose we have C = {1, 3,4} and C = {2, 5} that are modeled for
simplicity as f(xi, u) = uj, h(xi) = xi for i E C, and f (zy, d-,) = d., h(x1,) = Xy for
- E C. A given initialcondition is x(0) = (44,26,20,5,2) with J = 0 with parameters
um = 3 ,U
15, dm = 6, dM = 12 and (ci,#6l) = (50,53). The release times become
=
(R1 , R3 , R 4 ) = (, 2,3), the deadlines (D1 , D 3 , D4 ) = (2,10,15), and the idle-times
Consider r = (1,3, 4) and i = (2,5). Forj = 1 andr, = 1, we have T1 = R,=
and P1 = T1 +
-.
that T1 = 2 and P1 =I
.
Since T < R2 < R5 and P1 < 12< &1, we determine
='
For j = 2 and then r2 = 3, we have T =max(R3,P1) = 2
However,
and Pa= 2.
,
(?2, 7A) = (2,4), and (P2,iP
5)= (4-,8}).
for r
= 1, we have Tr1 = 2 and T3
12.
Thus, T3 =
max(Ta,P2 ) = 41. This leads to P = 4-. For r = 2, we have T2 = 5. Because
T3 <11 5 and P3
>
R5 ,
we have T 3 = P5 = 8. and P3 = 8-L. Finally, for j = 3, we
have r3 = 4 so that T4 = max(R4 , P) = 8-L and P4 = 8 . Even though for r = 1
we have T4
R 1, notice that T4 does not change because max(T4 , T1) = T4 . This is
the same for r = 2.
Since Ti ; Di for alli E {1,3,4}, we have found a set of schedule T = (2,81,8-6 ).
If there was no feasible schedule for this -x,Algorithm 3.1 would try the other permutations such as -r = (1, 4, 3).
3.2.2
Solution of the Supervisory Problem
A supervisor is designed to override controlled vehicles with a safe input Uk,,afe if a
desired input ak leads to an intersection collision at some future time. To achieve
this goal, we define an input operator for y;(O) < a for i E C as follows:
a- (xi(0), T) := arg inf {t > 0 : y (t, u,) = f8 for i E C with constraint y (T, uj) = ai}.
uEU,
(3.8)
32
This operator returns an input ui E U2 for i E C such that y (t, uj) reaches a at TI
and 3 at P(T). Set ai (xi(0), T) = uj,m if y (0)
y1 (0) >
#j.
aj, and set a1 (x1 (0), T) = 0 if
If the constraint y2 (T, ui) = ac cannot be satisfied, set a,(x1 (0), T) = 0.
Let a(x(O), T) be the parallel composition of (3.8) for all i E C. Then, a(x(0), T)
indicates the safe input uf'f,e introduced in Section 3.1.
At each time step, Algorithm 3.1 determines whether a predicted state from a
desired input leads to an intersection collision. To account for the fact that in the
partial state prediction xk+1 (ak) we do not predict the states for the uncontrolled
vehicles since their inputs are not known, a time delay r is considered for vehicle -Y
with -y E C. In particular, let (R',
P4) denote an IIT for a given initial condition
xz(kr). That is,
Pk:= min {t : y,,(t, d7 , x-(kr)) = ay}
47EV7 y
and
P
:= max{t: y7 (t, dy, x-,(kr))
=
d-fEV-y181
}.
Also, let (R+, PI+') denote the predicted idle-time defined as follows:
min{t : y,(t, d, x,(r, dy,m, x,(kr))) =
Iy}
and
PY+'
:= max{t : yy(t, d, x,(,dy,m, ,(kr)))
=
From these definitions, it can be verified that
(
+)
P;+1
-r).
-
(3.9)
We propose Algorithm 3.2 to solve the Supervisory Problem (Problem 3.3). Notice
that Algorithm 3.1 takes J = 0 when x(0) is a measured state and J =
ir when
x(0)
is a partial state prediction. The next lemmas and theorem prove that Algorithm 3.2
solves Problem 3.3.
Lemma 3.1. If EXACTSOLUTION(x(k-r), 0) = {T, yes}, then u := u(x(kr), T)
33
#0.
Algorithm 3.2 Solution of Problem 3.3
procedure s(x(kr),ak)
Uk(t) +- ak Vt E [kr, (k + 1)r]
{T, answer} +- ExACTSOLUTION(xk+1 (Uk), r)
if answer = yes and y(t, uk, d) 0 B
for all d E V for all t E [kr, (k + 1)r] then
(Xk+1(Uk), T)
,,l,safe +U
C+,,sa0erestricted to[kr, (k + 1)r]
Uk+1,aafe
return uk
else
{T, answer}
+- EXACTSOLUTION(Xk+1(Uk,safe), r)
U01,safe +-
0(Xk+1(Uk,,afe), T)
Uk+1,afe
Cu+0,saferestricted
return Uk,safe
to[kr, (k + 1)7]
Moreover, EXAcTSOLUTION(Xk+1(uk), r) = {T, yes} where uk restricts u to [kr, (k+
1)7].
Proof. Given an initial condition x(kr), if a schedule T exists, then {x(kr), e} E
Problem 3.1 by Theorem 3.1, indicating that there exists an input u E U such that
y(t, u, d, x(kr)) 0 B for all d for t E [kr, oo). Thus, a(x(kr), T) = 0. Let fi be u
restricted to [(k + 1)-r, oo). By assumption, the schedule T is feasible, which implies
that it satisfies the conditions that Ti E [Ri, Dj] for all i E C, that (T, P2(T)) does
not intersect each other, and that (T, P(T)) n (RJ, P4) = 0 for all 7 E C. Define
T' as Tj
=
{t : yi(t,iii, [xk+1(uk)] 1 ) = ai} for all i E C where fi, is the entry of 6i
corresponding to vehicle i. Then Tj' = T -,r and P (Ti') = P(T) - -rsince uk equals
u on [kr, (k + 1)r]. Consequently, intervals (T, P(Ti')) do not overlap for all i E C
and (T,P (T1 )) n (Rk+, P+') = 0 for all y E j due to (3.9). Also Tj' E [R, Dj]
by construction. Therefore, EXACTSOLUTION(Xk+l(Uk), -r)has a feasible schedule T'
0
and thus returns yes.
Lemma 3.2. Given a desired input ak, if ExACTSOLUTION(xk+1(ak), -r) = {T, yes},
then o(xk+1(ak), T) # 0.
Proof. Define xk+1(ak) such that its j-th element [kk+1(ak)] is either x (-r, akj, x 3 (kr))
for j E C where akj is the j-th element of ak, or xj(r, dj, xj(kr)) for j E j for any
34
disturbance dj E [dj,m, dj,M] on [kr, (k + 1)r]. Let (R+1, P4+') denote the IIT corresponding to [kk+,(ak)]y. Because the state is an order preserving function of the
disturbance, we have
x-y (r, dy,m, xy(kr)) 5 [kk+,(ak)] 5 x y(7, dy,M, xY(kr)).
Also, since the output is an order preserving function with respect to the initial
condition, we have
y (t, dy, [kk+,(ak)]y) 5 yy(t, d,x, x7 (r, d7 ,m, zy(kr))),
and
y-y (t, dy, x-, (7, df,m , x-y (kr))) :5 y-y (t, d-y, [:kk+1(ak)].Y)These two inequalities and the definitions of 5f+1 and P4+l further imply that since
y
> 0 by assumption, R+1 "+1 and pk+1
that (R+,P4 +1 ) ; (pk+1,
Jk+1,
respectively. Thus, we have
P+ 1 ).
If EXAcTSOLUTION(xk+1(ak),r) returns yes, a schedule T satisfies that Ti E
[Ri, DI], that (Ti, P(T)) does not intersect each other for all i E C, and that for all
y E C, (T,P(T)) n (R+1, pk+1)
=
0. The last condition implies that (T, P(T)) n
(R +1 Pk+1) = 0. Therefore, this schedule T makes EXACTSOLUTION(kk+1(ak),0)
return yes. From Lemma 3.1, this implies -(^k+1(ak), T) = 0. Notice that [ik+,(ak)],
is equal to [xk+l(ak)]i for all i E C. Thus, o-([xk+,(ak)]i,T) = og([kk+,(ak)]i, T).
Since o(xk+1(ak), T) is composed of ai([xk+l(ak)] , T) for all i E C, it follows that
o(xk+,(ak), T) = 0.
0
Theorem 3.2. The supervisor s(x(kr), ak) defined in Algorithm 3.2 solves Problem
3.3.
Proof. If EXACTSOLUTION(xk+1(ak), r) returns yes, by Lemma 3.2 there exists an
input signal -(xk+1(ak), T)
$
0 that ensures safety. Thus, Algorithm 3.2 returns uk
which is identical to ak. If ExAcTSOLUTION(xk+1(ak), r) returns no, Algorithm 3.2
35
returns Uk,safe.
We next show that Algorithm 3.2 is non-blocking. We use mathematical induction to prove this. Suppose at t = k-r, s(x(kr), ak)
=
Uk,,,t and
Uk,out
# 0. Then we
must prove that s(x((k + 1)r), ak+1) =#4 0 for any ak+1. Notice from Algorithm 3.2
that this is possible when
Uk+1,safe
As seen in Algorithm 3.2,
Uk,mt
ACTSOLUTION(xk+1 (Uk), r) =
empty.
exists, that is, u+1afe =
is either uk or
Uk,,afe.
O(Xk+1(Uk,out),
T)
#
0.
In the former case, Ex-
{T, yes}, and by Lemma 3.2,
-(xk+1(uk), T) is non-
In the latter case, because we assume that upafe exists, ExACTSOLU-
TION(Xk+i(Uk,aafe),
-)
returns yes by Lemma 3.1. In turn, Lemma 3.2 implies that
U+00 1,safe exists. Therefore, the supervisor is non-blocking.
3.3
Simulation Results
Consider the design of the supervisor s, which is described in Section 3.2, for multiple
vehicles converging to an intersection in the presence of uncontrolled vehicles. The
dynamic state of each vehicle is xi = (pi, vi) where pi is the position and vi is the
speed. With an input u, and a disturbance dy for i E C and -y E C, the longitudinal
dynamics of each vehicle are modeled as follows:
u= vi E (Vi,m, VM)
A.
0
otherwise,
-,
=
0
Vi E(V-.y,m,7V"yM)
otherwise.
We implement the algorithms using MATLAB with a time step r = 0.1s. Each vehicle has the same parameters: (a1 , fi) = (50m, 53m) and [vi,m, Vi,M] = [3m/s, 15m/s].
Also, the input ui E [-2m/s2 , 2m/s 2 ] for i E C and the disturbance dy E [-lm/s 2 , 1m/s 2]
for -f E j. We consider the desired input as lm/s2 for all controlled vehicles and construct the disturbance using a random number generator in MATLAB.
In Figure 3-1 and Figure 3-2, Algorithm 3.2 is simulated, based on EXACTSOLU36
44"
.
30
140
20
0
1
2
3
4
5
6
7
time (S)
Figure 3-1: Positions of 5 controlled (black solid line) and 2 uncontrolled (red dotted
line) vehicles controlled by the supervisor s in Section 3.2 (Algorithm 3.2). The
blue boxes at the bottom represent the time steps when the supervisor overrides the
controlled vehicles. The shaded region represents the location of the intersection.
TION
(Algorithm 3.1). Notice that the supervisor s guarantees that only one vehicle
crosses the shaded region at a time, that is, the trajectories avoid the bad set. The
overall simulation time when using the exact supervisor s in Figure 3-2 is 8000 seconds (7.3 seconds per iteration on average). This computation time is greater than
the time step
T
=
U.s so that this algorithm is unable to be implemented in a real
time.
37
50
40
30
w20
-10
0
1
2
3
4
5
6
time (s)
7
a
9
1
Figure 3-2: Positions of 8 controlled (black solid line) and 2 uncontrolled (red dotted
line) vehicles controlled by the supervisor s in Section 3.2 (Algorithm 3.2).
38
Chapter 4
An Efficient Approach
As seen from Chapter 3, Problem 3.1 is NP-hard. To address this computational
complexity problem, we introduce a relaxed version of Problem 3.2, called the Relaxed Scheduling Problem, which can be solved in polynomial time. Based on this
scheduling problem, we construct a modified version of Problem 3.1, called the Relaxed Verification Problem in Section 4.1. The corresponding supervisor, called the
Efficient Supervisor, is designed in Section 4.2. Finally, in Section 4.3, simulation
results verify that computation complexity problem is resolved by this efficient approach.
4.1
Relaxed Verification Problem
To formulate the Relaxed Verification Problem, we first introduce the Relaxed Scheduling Problem, whose solution is solved in polynomial time as can be found from the
scheduling literature. Then, we modify Problem 3.1 by inflating the bad set in light
of the relaxation for the Relaxed Scheduling Problem.
4.1.1
Relaxed Scheduling Problem
Garey et al. [14] proposed an efficient algorithm to solve DEC(1|r,pi = 1JL,O..,) in
0(n2 ) where n is the number of jobs. To design the algorithm, they introduce the
39
concept of a forbidden region F, a time interval when no task is allowed to start to
have a feasible schedule. Let F, C R be the k-th interval in F. Algorithm 4.3 is
Algorithm 4.3 Solution of DEC(ljr,,p = 1JL,.,0)
procedure POLYNOMIAL(F, r, d)
Sort r, d into an increasing order of r
for i +- n to 1 do
for all j such that d4 di do
if c3 = 0 then c +- dj - 1
else c3 +- c, - 1
if cj E F for some k then c+- inf(Fk)
if i = 1 or r.- 1 < ri then
c +- min{c,}
if c < ri then return {0, no}
else if r, _< c < r + 1 then
F +- F U (c - 1, ri)
for all it +- 1 to n do
if {i : r5 < s and t not defined}= 0 then
s +- min r,
while s E F for some k do s +-sup(F)
for all k such that rk :; s and tk = 0 do
d = mink (dk)
t-
s and set s +- s + 1
Sort T = {t 1 ,..., t,} in the original order
return {T, yes}
adapted from Algorithm A in [14]. The difference between these two algorithms is
that Algorithm A in [14] initially declares F an empty set, while here we allow to
introduce F arbitrarily.
Example 4.1. Consider 3 jobs on a machine with r = (7,7,8), d = (12,12,10),
and unit process time. An initially declaredforbidden region is F = {(81, 101)}. The
,
increasingorder of r is (1, 2, 3), so Algorithm 4.3 startsfrom i = 3. Ford1 , d2 , d3 > d3
since di - 1 = d 2 - 1
F, we have cl = 11, c 2 = 11, while c3 = inf(Fl) = 82
because d3 - 1 E F1 . Since r2 < r3 , we find c = 81 which is inside [r3 , r3 + 1),
and thus we update F = {(71,8),(8-,10)}.
c=
For i = 2, we have d1 ,d 2 > d2 and
c2 = 81, and neither i $ 1 nor r1 < r2 . Fori = 1, since ci = c2 = 7., we obtain
F = {(61,7), (7, 8), (81, 101)}. Then, the next step allocates a schedule to each job.
40
For it = 1, we have s = 7 and t1 = 7. For it = 2, we have that j = 3 since d3 is the
least due date and r3
s, we have t3 = 8. For it = 3, s E F3 so that s becomes 101
thus t2 = 10 1.
Algorithm 4.3 has a complexity that is polynomial with respect to the number of
jobs n. We next relax Problem 3.2 to DEC(1|ri,pi = 1L.,,0) in order to address
the complexity problem. To make the constraint p = 1, we define the maximum
process time and normalize with it the other problem variables in Definition 3.1. Let
0,.z := max
max
iEC zx(O)EXi:y(O)=ac
{t : yi(t, ui,m) = #l},
(4.1)
which determines the largest time required to cross the intersection among all controlled vehicles. Considering this as the fixed process time, we rewrite the relaxed
version of Problem 3.2 as follows.
Problem 4.1 (Relaxed Scheduling Problem). Given an initial condition x(0), determine whether there exists a schedule T = (T1 ,... ,T), where C = 1, ... ,ne}, such
that for all i E C,
Rt :5 T 5 Di
(4.2)
for alli A j E C and y E C if T > 0,
T
T= T + maz
T
T
R => T + O=
R- I(4.4)
T
(4.3)
(4.5)
=> T ! R
if yt(0) > a1 , i.e., T = 0, condition& (3.4) and (3.5) need to be satisfied instead of
conditions (4.3) and (4.4).
Notice that (4.4) and (4.5) in Problem 4.1 imply that T for any i E C cannot
start during (T - Oma, P) for all -y E C. Thus, we normalize these time intervals
with
m., and let them be the initial forbidden region F in Algorithm 4.3. Assume
that yi(0)
ai for i E {1, ... , m} and yi(0) < ai for i E {m+ 1, ...
41
ne}.
Algorithm 4.4 Verification of Problem 4.1
procedure APPROXSOLUTION(X(0), 6)
if y(O) E B then return {0, no}
for all i E C and y E d do
given x,(0) calculate R4, D, 0,,.
given xy(0) calculate Ry, Py
A 7 +- max(Iy - 6,0)
S-max(P - 6, 0)
for all -y E j such that P > 0 do
F +- F U (max(RI/G,,. - 1,0), P7/9max)
for i - 1 to m do
calculate P(T) and P,,, +- maxi P (T)
for i +- m+1 to n, do R, +- max(R, P,.)
r +-(R:4+1/Omax, ..
- -,
./Om=)
d+- (Dm+/Omax+1, ... , Dnc/9maz+1)
[tm+1,... , tn, answer] +- POLYNOMIAL(F, r, d)
for i +- m + 1 to n, do T +- mxi
return {T, answer}
Algorithm 4.4 is the solution for Problem 4.1 when 6= 0.
Relaxed Verification Problem
4.1.2
Notice that Algorithm 4.4 may return no even when the solution for Problem 3.2
exists. We quantify how conservative Algorithm 4.4 is, which will be considered to
formulate the Relaxed Verification Problem.
Lemma 4.1. ConsiderT and T for some i, j E C with y,(0) < a, and y,(O) < a and
a, and y (T,uj) = aj. If T > T and T - T < OM=
ui and u, such that y (T,u )
=
or if (T,T +8m=) n (R-,P
7 )
$0
for some 7 E j, then there exists t* E [T;, T + 0'.]
such that eithery (t*, u3 ) E (a3 , a, + 0raxc.j,M) or yy(t*, dy) E (ay,
) respectively.
Proof. The condition T - T < Omax implies a violation of (4.3). Let t* = T. Since
T > Tj, we have y,(t*, u,) > aj. Also, the assumptions that T < T + On= and
Y,,M
imply y,(t*, u,) < a3 + 9 ma4 j,M. Thus, t* satisfies y (t*, u,) E (aj, aj
+
j <
Ocyj,m). The next condition (T, T + 0,O,) n (Iy, I 7 )
42
$
0 implies that T does not
satisfy (4.4) and (4.5). That is, for an uncontrolled vehicle -y, there is t* such that
t* E (Ti, Ti + ,m=)
n (I,
P). Since
.
> 0, for t* > I?., and dy = d,m, we have
yY(t*,dy) > %y; for t* < P. and dy = dy,m, we have yy(t*,d1 ) < 8y. Therefore, for
some d., E D.,, we have that yy (t*, c4) E (ay 8,).0
Taking this lemma into account, we inflate the intersection witha := ai+,.
yi,M
for i E C and Ay :=.y for -y E C. Then, an inflated bad set is defined as follows:
B :={y E Y : yi E (ai,/ji) and y E (aj, 8j)
for some i 3 j such that i E C and j E C U}.
(4.6)
We then modify Problem 3.1 by substituting the inflated bad set B in place of
the bad set B.
Problem 4.2 (Relaxed Verification Problem). Given an initial condition x(O), determine if there exists an input signal u E U which guarantees that y(t, u, d)
B for
all d E V for all t > 0.
Theorem 4.1. Given x(0) and 3 = {0, r}, if APPROXSOLUTION(X(O), 3) returns
no, then there is no input u such that y(t, u, d)
B for all d for all t > 0.
Proof. In Algorithm 4.4, APPROXSOLUTION(X(0), J) returns no for two cases: y(O) E
B or POLYNOMIAL(F, r, d) returns no. In the first case, y(O) E B because B C B
so that at t = 0, y(t, u, d) E B for any u. In the second case, according to Algorithm
4.3, POLYNOMIAL(F, r, d) returns no when the conditions described in Lemma 4.1
hold. Thus, by Lemma 4.1, if APPROXSOLUTION(X(0), 6) returns no, then there exist
some d E E) and some t > 0 such that y(t, u, d) E B for any u. Therefore, there is
no input u such that y(t, u, d) V B for all d for all t > 0.
Theorem 4.1 indicates that if Algorithm 4.4 returns no, then there is no solution
to Problem 4.2, thereby letting the efficient supervisor introduced in the next section
intervene.
43
4.2
Efficient Supervisors
We now define an efficient supervisor s(x(kr), ak) by employing B in place of B in
Problem 3.3. Also, define the input operator for i
EC
as follows:
ai(xi(O), T) := arg inf {t > 0: y(t,u,) = [ with constraint y;(T,uj) = a;}. (4.7)
Ui EUi
Moreover, define another efficient supervisor sa,,,.x(x(kr), a,) as in Algorithm 2 in
which APPROXSOLUTION is used in place of EXACTSOLUTION and & is used in place
of a. Here, we prove that s is no more restrictive than sap,,.', that is, we prove that
if S^ is non-empty, sapp,.x is also non-empty.
#
Lemma 4.2. If APPROxSOLUTION(X(kr), 0) = {T, yes}, then u := &(x(kr),T)
0. Furthermore,for uk
defined as u on [k-r, (k+1)'rI, APPRoxSOLUTION(Xk+1(Uk),-)=
{T, yes}.
Proof. Same procedure as in proof of Lemma 3.1.
0
Lemma 4.3. If APPRoxSOLUTION(xk+1(ak), r) = {T, yes}, then (xk+1(ak), T) 3
0.
Proof. Same procedure as in proof of Lemma 3.2.
0
Theorem 4.2. The supervisors.,,.(x(kr), ak) is no more restrictivethan S(x(k-r), ak
and non-blocking.
Proof. If 8(x(k-r), ak) is non-empty, then APPROxSOLUTION(xk+1(ak), r) returns yes
by Theorem 4.1 and definition of S. Then Lemma 4.3 implies that &(xk+1(ak), T) is
non-empty, which in turn implies that sappro.,,(x(kr), ak) is non-empty by definition
of sa,,,.x. Therefore, sap,,(x(kr), ak) is no more restrictive than S(x(kr), ak). The
non-blockingness follows the same procedure as in proof of Theorem 3.2.
4.3
0
Simulation Results
The same parameters and dynamic model are considered as in Section 3.3. The initial
condition used in this section is equivalent to the initial condition used in Figure 3-2
44
in Section 3.3.
50
J
40
130
20
~4e
10
-10
-20
0
2
4
6
time (s)
8
i
10
12
Figure 4-1: Positions of 8 controlled (black solid line) and 2 uncontrolled (red dotted
line) vehicles controlled by sa,,ppo in Section 4.2. The blue boxes at the bottom represent the time steps when the supervisor overrides the controlled vehicles. The darker
shaded region represents the location of the intersection while the lighter shaded
region represents that of the inflated intersection.
In Figure 4-1, Algorithm 3.2 using APPROXSOLUTION (Algorithm 4.4) in place
of EXACTSOLUTION (Algorithm 3.1) is simulated to design sajprox for an example
with 8 controlled and 2 uncontrolled vehicles. Figure 4-1 shows that the trajectories
cross the darker shaded region without time overlapping, which implies the system's
output avoids the bad set. Meanwhile, saprox does not guarantee that only one vehicle
stays inside the lighter shaded region at a time, that is, the trajectories are inside the
inflated bad set. This, together with the result of Theorem 4.2, shows that sapprox is
strictly less restrictive than
s.
Recall that the simulation time per iteration for Figure 3-2 is 7.3 seconds on
average. However, the simulation in Figure 4-1 takes 0.001 seconds per iteration on
average and 0.07 seconds for the worst iteration. In light of the fact that 0.1 seconds
is a usual practical sampling time, this algorithm is fast enough to be implemented
in real-time applications.
45
46
Chapter 5
Optimal Approaches
This chapter designs two optimization problems on the top of the efficient solution,
which is introduced in Chapter 4. By virtue of solving the optimization problems,
new smoother inputs are achieved in the sense that the abrupt change of the inputs and unnecessary intervention are minimized. These problems must be solved in
polynomial time to maintain fast computation time for real-time applications.
We specify the motivations of these optimization problems in Section 5.1 and
formulate the problems called the Input Optimization Problem and the Schedule
Optimization Problem in Section 5.2 and Section 5.3, respectively. The Input Optimization Problem is designed to replace the input operator & defined in (4.7), and the
Schedule Optimization Problem is designed to optimize the solutions of Problem 4.1.
Since we only override controlled vehicles with designed inputs, uncontrolled vehicles
are not considered in this chapter.
5.1
Motivations
In this section, we present two issues of the control inputs that are generated by
the Efficient Supervisor in Chapter 4. The first issue is the abrupt changes of inputs,
which may cause inconvenience to drivers. The other issue is unnecessary intervention,
which can be seen when there is one vehicle that is not involved in any collisions.
47
5.1.1
Abrupt Changes of Inputs
Consider an example with 4 controlled vehicles. The intersection is located at [30m, 33m],
and speed bounds are [lm/s, 10m/s] with a sampling time 0.1sec. The other parameters are equivalent to the example in Section 3.3. As been noticed in Chapter 4, one
iteration takes at most 0.07sec using the efficient supervisor.
40
I
. .......
...
............
......
3530
ram
25
C
S
20
10-
1
2
3
4
t0 5
time (s)
6
7
8
9
Figure 5-1: Positions of 4 vehicles (black solid line). The blue boxes at the bottom represent the time steps when the supervisor overrides the controlled vehicles.
The darker shaded region represents the location of the intersection while the lighter
shaded region represents that of the inflated intersection.
Recall from Section 3.3 that the control input for this system is a longitudinal
acceleration of the vehicles. The accelerations that we apply to vehicles usually show
many oscillations as in Figure 5-2.
Because of this flutter, the actuator could be
damaged, and the car sways back and forth by inertia. Moreover, the drivers would
feel the discrepancy between the input that they apply to the vehicles and the input
that the supervisor imposes.
To address this issue and provide a more convenient driving environment to
drivers, we propose the Input Optimization Problem in Section 5.2, where the difference between a control effort and a desired input is minimized for every step.
48
control effort
.
--desired input
2 .................................................................
1
________---___
----
CL
-1-
-3
0
1
1
1
2
3
4
5
- ------------ --
------
-
-2 --------------- ..
1
1
6
7
1
8
1
9
time
Figure 5-2: Control input (black solid line) for the last car with its desired input (blue
dotted line). The two horizontal dotted lines represent the upper and lower bounds
for the input.
5.1.2
Unnecessary Intervention
Suppose a different initial condition of 4 vehicles is given such that one vehicle travels
far from an intersection compared to the other vehicles. All the parameters are the
same as in Section 5.1.1. The Efficient Supervisor in Chapter 4 is used to control the
vehicles.
The positions of the vehicles over time are shown in Figure 5-3(a). All of the
vehicles are overridden when the blue boxes are present. However, in Figure 5-3(b),
the last vehicle is not overridden and travels with its desired input. This still does
not cause any collision because the last vehicle is not involved in any collision.
49
-
40
20 --
-
S10
C
.
0
-
-10
-
-20
0
1
2
3
4
5
6
7
8
9
10
time (s)
(a) All four vehicles are controlled by the Efficient Supervisor in Chapter 4
40
-
- --
30 - - - 20-
10
E
C
U
-10-20-30
0
1
2
3
4
5
6
time (s)
7
8
9
10
(b) The last vehicle is not in control
Figure 5-3: Positions of 4 vehicles where the last vehicle is far from the intersection
compared to the others.
50
3
control effort
- - - desired input
0-
. . .. . . .. . .. . . .. ..-. .-
-3 0
0
2
-
- -- -.
- 2 -.. .. . . .-.. . . . ..-
4
6
8
10
time
Figure 5-4: Control input (black solid line) for the last car in Figure 5-3(a) with its
desired input (blue dotted line). The two horizontal dotted lines represent the upper
and lower bounds for the input.
Figure 5-4 shows the input signal for the last vehicle in Figure 5-3(a). Although
the intervention in the last vehicle is unnecessary as can be seen in Figure 5-3(b),
the Efficient Supervisor overrides not only the other vehicles but also the last vehicle,
which is not involved in collisions. In this individual perspective, the Efficient Supervisor is not least restrictive, that is, it violates that it intervenes only if necessary.
This issue mainly originates from the design of the decision version of scheduling problem as in Definition 2.1. Therefore, to address this problem, we propose
optimizing a schedule to be closest to a desired schedule which is defined in Section
5.3.
5.2
Input Optimization Problem
To take advantage of a dynamic model in the design of an optimization problem, we
consider a linear discrete system. Based on this model, we formulate the Discrete
51
Schedule Problem and the Input Optimization Problem.
5.2.1
Discrete Scheduling Problem
Recall the dynamic model of vehicle i:
(2.1)
yi = hi(xi).
i = fA(xi, ui),
For a non-negative integer k E Z>o where Z>o is a space of non-negative intgers and
for a sampling time At, let the dynamical state xi(kAt) be denoted by xi[k] and
the output yi(kAt) by yi[k]. The input ui(t) for t E [kAt, (k + 1)At) is considered
constant and denoted by u [k]. Through this chapter, we consider the corresponding
linear discrete time-invariant system to (2.1) denoted as follows:
yi[k] = Cixi[k],
xi[k + 1] = Aixi [k] + Biu [k],
(5.1)
where Ai is the system matrix, Bi is the input matrix, and Ci is the output matrix
for each vehicle i. The input signal for t E [0, (N + 1)At) is a vector denoted by
Ui := (ui[0],... , u[N]).
Similar to the continuous system,
9; [k]
E [yi,m,l,M] and
ui[k] E [ui,m, ui,M] for all k E Z>o. The output trajectory yi[k, ui, xi[ko]] and state
xi[k, ui, xi[ko]] denote the output and state reached at t = (k + ko)At with an input
signal ui starting from xi [ko], respectively.
We assume that both yi [k, ui, xi [ko])
and xi[k, ui, xi[ko]] are order preserving functions of their arguments. For notational
brevity, we omit an initial condition when ko = 0. The aggregate state, output, and
input are denoted by x[k], y[k], and u[k], respectively.
Based on the dynamic model (5.1), we define the variables of discrete scheduling.
To ensure the existence of these variables, a truncation error Aai is considered.
Definition 5.1. For all i where yi[0] < ai - Aai, let
Ri := min{k E Z
Ut
: ai - Aai
52
yi[k, ui]
ai},
D
:= max{k E Z>o : ai - Aai
y [k, u4]
ai}.
U,
Given a non-negative integer Ti E Z>o,
where
9m, := max min{k E Z>o : y,[k, uj
t
If ai - Aaj
y [k,u,]
y4[0] < ai, let R := 0, and
ai}. If ai
B3}.
U,
y[0] < Pi, let
i := max,,{k E Z>o : ai - Aai
, := 0 and Di := 0. For both cases, that is,
a - Aaj 5 y[0] < i, given a non-negative integer Ti E Z>o,
Pi (T) :=min{k E Z>o : y [k, u.] >
#3i with contrainty,[7j]
- ai}.
Ui
If the constrainty [T, u]
a, is not satisfied, set Pi(T) = oo. If y
[0]
f3i, then set
R = 0, Vi = 0, and A (M)= 0.
We now formulate the Discrete Scheduling Problem. This problem requires the
same constraints as Problem 5.1 except the idle-time constraints.
We consider n
vehicles approaching an intersection.
Problem 5.1 (Discrete Scheduling Problem). Given an initial condition x[0], determine whether there exists a schedule T = (T1 , ... T,) where 7; E Z>o for all i such
that for all i E {1...,n},
for all i 0 j,
Notice that a truncation error Aaj appears when R, and Vi are defined. This
implies that a feasible schedule
7; obtained
from Problem 5.1 satisfies
ai - Aai 5 y4[T]
53
ai.
Similarly, from the definition of P(TK), a truncation error A,3i can be introduced
where A/3, satisfies
#5 y[Pi(T),
4
/38+ A 0.
Notice that we consider Pi to take the definition of 19,
into account, where fl:
a, + 9 vna.xj,M defined from a result of Lemma 4.1.
Therefore, the inflated bad set from discretization is defined as follows:
$:= {y E Y :3t > 0 such that y,(t) E (a, - Aaj,,fl + 4#3l) and
(5.2)
y 3 (t) E (aj - Aaj, #l + A#3) for some i = j}.
where Aa,
0 and Ai > 0 for all i E {1,...,n}. Notice that B considers the
continuous dynamics while Problem 5.1 considers the discrete dynamics.
Similar to Problem 4.2, we then modify Problem 3.1 by substituting the inflated
bad set from discretization B in place of the bad set B and eliminating the disturbance
of uncontrolled vehicles.
Problem 5.2 (Relaxed Verification Problem for Discretization). Given an initial
condition x(0), determine if there exists an input signal u E U which guarantees that
y(t,u) V $ for all t > 0.
The relation between this verification problem (Problem 5.2) and the Discrete
Scheduling Problem (Problem 5.1) is that if Problem 5.1 returns no, that Problem
5.2 always returns no. The converse is not true. Moreover, if Problem 5.1 returns
yes, then there exists an input to avoid the original bad set B, that is, Problem 3.1
returns yes. These are proved in Appendix A.2. Thus, we design a discrete supervisor
such that it overrides the system when Problem 5.1 returns no.
5.2.2
Input Optimization Problem
In the design of a discrete supervisor, the most important part is to define an input
generator such as a in (3.8) and & in (4.7) because the safety is directly guaranteed
from an input operator. While o and & randomly select the input that satisfies their
54
definitions, in this section an input operator is defined to minimize the difference
between a control input and a desired input by a driver, as mentioned in Section
5.1.1.
To implement an optimization problem, we use model predictive control (MPC),
which is widely used for the on-line computations of optimization problems with
constraints. The MPC encompasses open-loop feedback control and certainty equivalence [7]. The open-loop feedback control is a classical suboptimal scheme that selects
the present control input at each period r as if future measurements will be disregarded while indeed it updates the measurements in future step. When a disturbance
is applied to a system, the computation is simplified via certainty equivalence, which
at each period assumes a fixed disturbance at a nominal value.
In the formulation of the Input Optimization Problem, a desired input plays a
role of a disturbance, which can be regarded constant as in the certainty equivalence
scheme. Let a desired input vector ak,i be a constant vector of a desired input ak,i
in a (Pi('7) - 1)-dimensional space for a given process time Pi(7). Also, we assume
the time step for the supervisor r is a multiple of the sampling time for the discrete
dynamic At, that is, -r = m * At where m E Z>o. The weighting matrix Q is a
symmetric positive semi-definite and diagonal matrix. Since the MPC selects the
present control input, we choose Q = diag([1,0,...,0]) to weight only the present
input.
Problem 5.3 (Input Optimization Problem). For each vehicle i at time step K and
for given a schedule T and a process time Pi(7), find u, such that
min(ui - aKi)Q(ui - aK,i)
u<
subject to
for all k E {0,
Pi(T) - 1},
Uim
<ui[k] 5 yi,M,
ui,m < ui [k] < ui,m,
55
and
cei - ack 5 yL[u, uI
ai,
Notice that due to the constraints, the solution of Problem 5.3 guarantees safety
if ['7, 1'( 7 )] does not overlap for all i. Indeed, Problem 5.1 finds such a schedule.
Provided that the weighing matrix
Q is positive semi-definite,
Problem 5.3 is a convex
quadratic optimization problem, which can be easily solved.
The existence of a feasible input for Problem 5.3 is presented in Appendix A.1.
5.2.3
Simulation Results
The input vector solving Problem 5.3 minimizes the cost at every step. However, this
input vector and the actual control input are different due to the change from future
measurements, the uncertainties from a disturbance, and the construction of the
supervisor. To visualize the improvement achieved by solving the Input Optimization
Problem, we generate 60 different initial conditions for 4 vehicles and compute the
square of the difference between the actual control input and the desired input for
each vehicle. For each sample of initial conditions, we find a stage 1' when vehicle i
exits an intersection and then compute the average cost,
'Pi-1
E (ui[k] - ak,i) 2
k=O
'P
(5.3)
for each vehicle. Thus, each sample contains 4 average costs for each vehicle. Finally,
we sort these 4 costs in an increasing order and call the one that has the minimum
cost by the best performance and the one that has the maximum cost by the worst
performance. The following figures show the average cost across 60 samples from the
best to worst performance.
In Figure 5-5, we observe that the worst performance deteriorates in the optimal
56
34
-
Random input
Optimal input
2.5-
2-
cc
1
0.5-
0
1
2
3
from best to worst performance
4
Figure 5-5: Average cost on 60 samples of four vehicles
control that is obtained from Problem 5.3 while the others improve in the optimal
control input.
5.3
Schedule Optimization Problem
Consider a vehicle i travelling far from an intersection while the others are not, as in
Section 5.1.2. Recall that the Relaxed Scheduling Problem (Problem 4.1) is a decision
problem, which returns a binary answer. Thus, Algorithm 4.4 searches the soonest
possible schedule, more precisely, the schedule that minimizes Lmax, and checks if
the schedule violates T < Di. Because of this soonest schedule scheme, vehicle i is
assigned the soonest possible schedule at each stage, thereby being forced to accelerate
when collisions among the others become urgent.
To address this issue of unnecessary intervention, we formulate another optimization problem. Since in this chapter, we consider the discrete dynamic model and thus
an integer schedule, optimizing a schedule becomes an integer linear optimization
57
problem. To solve this problem in polynomial time, we introduce a special class of
matrix in Section 5.3.1 and transform the integer linear optimization problem into the
equivalent linear optimization problem in Section 5.3.2. In Section 5.3.3, simulation
results are given.
5.3.1
Totally Unimodular Matrix [20]
Since a schedule obtained from the Discrete Schedule Problem is integer, the Schedule
Optimization Problem becomes an integer linear optimization problem, which can be
formulated as follows:
min{c'x : Ax < b, x integers}.
(5.4)
Because of the integer constraint, this problem is known to be NP-hard. However,
for a special class of A, c and b, this problem can be transformed to a linear optimization problem.
The same problem without the integer constraints, that is,
min{c'x: Ax < b}, is called the LP-relaxation of the problem (5.4).
Consider a feasible set {x : Ax < b}. If every element of a matrix A is integer,
A is called an integral matrix. If every element of a vector b is integer, b is called
an integral vector. A square integral matrix B is "unimodular" if detB = t1. An
integral matrix A is "totally unimodular" if every square non-singular submatrix of
A is unimodular.
Proposition 5.1 (p. 269). Let A be a totally unimodular matrix, and let b be an
integral vector. Then, the polyhedron {x: Ax < b} has only integral vertices.
Proposition 5.2 (p. 267). If A is a totally unimodular matrix, and b and c are
integral vectors, then the problem
min{c'x: Ax < b}
has only integral optimum solutions.
Proposition 5.2 implies that given a totally unimodular matrix A and integral
vectors b and c, the LP relaxation of (5.4) retains only integral solutions, which
58
satisfy the integer constraint in (5.4). That is, the solutions for the LP relaxation
become the solutions for the original problem. Since a linear optimization problem can
be solved in polynomial time, the corresponding integer linear optimization problem
can be solved in polynomial time.
5.3.2
Schedule Optimization Problem
As we define a desired input vector aKi for vehicle i at stage K as a constant vector
to simplify computation, we define a desired schedule Td, for y[0] < a, - Aa as
follows:
Td,
:= {k E Z>O: ai - Aai
yi[k,aKI < ai}.
(5.5)
The goal of this section is to minimize the difference between an actual schedule T
and a desired schedule Td, where Td is an aggregate vector of Td,. By solving such
an optimization problem, the schedule T7 for vehicle i is equivalent to T,i if vehicle i
does not have to change its input, thereby, minimizing the occurrence of unnecessary
intervention. For the Schedule Optimization Problem, we only consider the vehicles
that satisfy y [0] < ai - Aai.
Let 7r be a given sequence of n vehicles. This sequence can be obtained from
Algorithm 4.4. If Algorithm 4.4 returns yes, we take the feasible sequence; otherwise,
there is no feasible schedule so that we do not have to solve a further optimization
problem. The ith element of 7r is denoted by 7ri.
Problem 5.4 (Schedule Optimization Problem). Find a schedule T = (T1 ,..., Tn)
such that
n
subject to
R, Z T 5 VI Vi E {1, ..., n},
'P7Q(T ) :5T%,,
Vi < j,
T E Z>0,
Vi.
Notice that via the constraints, a feasible schedule must solve the Discrete Scheduling Problem (Problem 5.1). That is, Problem 5.4 chooses the best schedule in terms
59
of the objective function among all possible schedules that satisfy Problem 5.1. Since
all vehicles of interest satisfy y [0] < ai - Aai, we have Pi (T) =
'7 +
V,..
In general, the absolute objective function can not be solved in polynomial time
[looking for a reference]. To modify the objective function into a solvable one, define
non-negative integers T,+ and T~ as follows:
+ := max((
- Td,i), 0) and
Then they also satisfy Ti - T, = %+ - 7'
Let T+ := (77, ... ,7)
in Z2" is denoted by
Ti :=
'Ti~
:= max(O, -(7i - Td,)).
while minimizing I Ti -
and T- := (T,-, ...
,
T-).
'd,i= 7;+
+ 7~.
The aggregate column vector
(T+, T-)'. Now we reformulate Problem 5.4 into a linear
programming format.
Problem 5.5 (Reformulation of Problem 5.4). Find a vector T
that
= (7+, T-)' such
n
min Z(T++T-)
subject to
7i <(T7 -WT) + Td,i
Vii
'Ti+, T E Z>O,
Vi
Vi.
By the definition of the LP relaxation of an linear integer programming, given in
Section 5.3.1, the LP relaxation of Problem 5.5 is as follows.
Problem 5.6 (LP relaxation of Problem 5.5). Find a vector T* = (T+, T-)' such
that
min Z(T++'Ti~)
'r+7ii=1
60
subject to
7ZT7-) + Tdh< Di,
(T7
-T7,)+Td,r
+
max !5 (77
Vi
T7) +Td7r.,
Vi <j
Vi.
7j+ >01T->O0
Problem 5.6 can be rewritten as min{c'T
: AT
< b}, where the explicit
expressions of A, b , and c are given in Appendix B(under construction...). Given
the fact that the variables RiDi,di,, and id
are integers, it can be verified also in
Appendix B that A is a totally modular matrix, and b and c are integral vectors. By
Proposition 5.2, Problem 5.6 has only integral optimum solutions. These solutions
also satisfy the integer constraints in Problem 5.5. That is, the solutions for Problem
5.6 become the solutions for Problem 5.5.
Using these optimal solutions for Problem 5.6 denoted by 7"+* and 7-* for all i,
the solutions for Problem 5.4 can be obtained by T* = (
5.3.3
_ 7-*) + T,.
-+*
Simulation Results
In this section, we implement the optimal inputs derived from the Input Optimization
Problem and the Schedule Optimization Problem, as described in Figure 5-6.
61
Verification
Supervisor
If answer = yes,
Algorithm 4.4
Input
yes
n
--answer
7ak
Optimization
-0
ak
Problem
:_1 r
Sche dule
Store Uk+1,safe
If answer = no,
OptirnzaItion
Prob lerm
Uk,safe
Figure 5-6: Implementation of the two optimization problems (Problem 5.3 and Problem 5.4) at stage k. First, the Discrete Scheduling Problem is solved via Algorithm
4.4, and if there exists a feasible schedule, the sequence of the schedule lets the Schedule Optimization Problem compute the optimal schedule T*. If there does not exist
a feasible schedule, Algorithm 4.4 returns no and an empty schedule. Depending on
this answer, the supervisor either runs the Input Optimization Problem to store the
safe input for the next step Uk+1,safe or overrides the system using the safe input
Uk,safe-
In the first example, the same initial condition and desired input as in Figure
5-3 in Section 5.1.2 are considered. By solving the Input Optimization Problem and
the Schedule Optimization Problem at each period r, the unnecessary intervention
that used to be observed in Figure 5-4 does not appear as can be seen in Figure
5-7(b). However, although all the problems can be solved in polynomial time, this
simulation takes 0.13 sec per iteration on average and 2.46 sec on the worst case. A
larger sampling time for the supervisor r than the computation time can tackle this
problem but make the supervisor conservative.
62
20
110-310
.9 0 --10-20
0
1
2
3
4
5
6
7
8
9
10
time (s)
(a) Positions of the optimally controlled vehicles
3
control
- -desired
effort
input]
*1
-2
.........
.3
0
1
'
-
2
...........-........
---.............................
3
4
5
time
p
6
7
8
9
I
10
(b) Control input of the last car
Figure 5-7: The same initial condition and desired input are considered as in Figure
5-3. In Figure 5-7(a), the lighter shaded rectangle represents the inflated intersection
due to the discretization and approximation for the efficient solution. The blue boxes
represent the time steps when the supervisor intervenes. Although the supervisor
intervenes, the last vehicle is not overridden, that is, the control input (black line
in Figure 5-7(b)) remains equivalent to the desired input (blue dotted line in Figure
5-7(b)).
To confirm that this optimal scheme also works with a general desired input, a
varying desired input is considered in the next example.
Recall that the definition of a desired schedule Td,i in (5.5) is defined with a
constant desired input because the future desired input is unknown. As can be seen
in Figure 5-8(b), unnecessary intervention for the last vehicle, which is not involved
63
40
,in
20-
g
10
-
0
-20 -2-
0
2
4
6
10
8
lime (s)
(a) Position of the optimally controlled vehicles
3
control 9Wor
- -desired input
-2. .. .................................
o
2
4
6
....... I - ......
8
10
time
(b) The input of the last car
Figure 5-8: The same initial condition but a varying desired input are considered.
The control input (black line in Figure 5-8(b)) maintains the similar level to the
desired input (blue dotted line in Figure 5-8(b)).
in a collision, can be prevented even for an unpredicted varying desired input. Also
it is noteworthy that at 4sec, the supervisor intervenes, as known from the presence
of the blue box in Figure 5-8(a), and there is one step delay for the control input
to follow the change of the desired input. This is because the change of the desired
input is unknown at the time when the control input is generated.
To provide a global quantification, we generate 60 random initial conditions of
4 vehicles, where one vehicle starts between 50 m and 60 m before an intersection,
and the others start between 20 m and 30 m before the intersection. That is, one
64
vehicle travels farther from an intersection than the others. Then, using the metric
of the difference between a control input and a desired input in (5.3), we compare the
inputs obtained by solving the Input Optimization Problem and the inputs obtained
by solving both the Input Optimization Problem (Problem 5.3) and the Schedule
Optimization Problem (Problem 5.4).
1.5
Random schedule
Optimal schedule
1
0
a)
a)
0.5
0,
1
2
3
from best to worst performance
4
Figure 5-9: Average cost on 60 samples of four vehicles. Random schedule (blue
bars) represents the results from solving Problem 5.3 while Optimal schedule (red
bars) represents the results from solving Problem 5.3 and Problem 5.4.
In Figure 5-9, the results present that solving Problem 5.4 along with Problem
5.3 generates smoother control inputs for most of vehicles than solving only Problem
5.4.
65
66
Chapter 6
Conclusions and Future Works
In this paper, we have designed a supervisor to prevent collisions among multiple
heterogeneous vehicles. The design of the supervisor is based on two main problems:
verification of the system safety, and construction of a safe input for the controlled
vehicles. An inserted idle-time (IIT), scheduling is employed to determine safety in the
presence of uncontrolled vehicles. The supervisor is least restrictive by construction,
that is, it intervenes only if necessary.
Since the exact problem is NP-hard, we also introduced a relaxed problem and
used an efficient scheduling algorithm with polynomial complexity to solve it. A quantification of conservatism of the approximate solution was also provided. To mitigate
the abrupt changes of control inputs and to reduce the number of unnecessary interventions, additional optimization problems were formulated, which can be solved in
polynomial time. The simulation results showed that the computation time exceeded
0.1 sec. Thus, to apply the problems to real-time applications, we need to enlarge
the step size of the supervisor.
While the extension that we have presented in this paper to consider uncontrolled
vehicles and to generate smoother control inputs makes the scheduling approach more
practical, a number of real-world issues still need to be tackled. These include considering also rear-end collisions, the fact that real intersections include multiple conflict
points, that a vehicle's path may be not known a priori, and that vehicle positions
and speeds may be subject to measurement uncertainty. The issue of the multiple
67
conflict points, which relaxes the assumption of the intersection model in Section 2.2,
is currently being investigated.
68
Appendix A
Proof for the discrete system
A.1
Existence of input
Proposition A.1. If there exists a schedule T; E [.,'
an input vector u
:= (Us([0],.. .,uj[
Vi] C Z>o, then there exists
- 1]) such that u,[k] E [ui,,,uim] for all
k = {0, ...,% - 1} and ai - Aaj < yi[Ti, ufi] < aj.
Proof. Consider an input u[0], ... ,u[N - 1], where N E Z>o. From the discrete
system dynamics (5.1), the future state xi[N] for vehicle i at stage N can be written
for a given initial condition xi [0] E Rn as follows:
x4[1] = Axi[0] + Bjui[0],
x [2] = Axi[1] + Bju[1] = A?xi [0] + ABju [0] + B u [1],
x,[N] = Afx [O] + A- 1 Bju,[0] +... + Bju[N - 1],
(u[0]
= Aixj[0I + [IA -'B...,Bj]
u[N - 1]
Afvxj[0] + Qvu,
where we define a matrix '$
as [Av- 1 Bj, A69
2
B4, ... ,Bi] and a vector uf as (u [0], ... ,[N-
1])'. Thus, the output at stage N becomes
yj [N] = CiAlxi [0] + C
u.
Recall that for an input vector uf, the output y [N, uj, xi[0]] is an order preserving
N2 E Z>o, we have
function with respect to the arguments. That is, for N
yi[N1 ,u' , xj[0]]
yi[N2 ,U2, X,[O]], where U2 contains the same elements of U1
from 1st to Nith elements and is augmented by (N2 -N1) random inputs. By choosing
U1 = 0 and u
2
= 0, we have
(A.1)
CjAfvixj[0] : CiAjxj[O].
Furthermore, by choosing x1 [O] = 0, we have C4',lufi < Cjx$J2u$2. By letting
Cjxv = (Co,..., CN-1), we have
k=N 2 -1
k=N1-1
Zk=O
which becomes 0 <
kN-
E
CkUikI
ckUi~k],
k=O
ckui[k] for any ui[k]. This implies that ck for all k is
non-negative.
By the definitions of 7Z and Vi in Definition 5.1, there must exist input vectors
yj[R,u,] < a1
ai - Aaj < CjA xj[0] + CjxFsu
<a
,
ai - Ace
,
respectively. For N =
ys[RI i,us] 5 ai and a, - Aaj 5 y[Vi, ,u'] 5 a,
,
u~' and upi that satisfy ai - Aaj
a - Aaj - CiA 1xi[0] 5 CPiV-u% 5 a - C A
xj[0].
Notice that Ci'44' is a vector so that let Ci{z = (co,.. ., C,.i-1).
Let the 1-norm |ICjxP
ck
for all k E {0,..
.,
|i|1 represent Icol + ... + |cg,_ 1 = co +
...
+ ci-1 because
- 1} are non-negative, and let urs = (u[0], .. u[,
70
- 1]) in
an abuse of notation. Then,
=
cou[0+...+c
(1C IO
=|jC ~it
:=
llCily'&
[7- 1]
u[O]+ ...
+
i'
1Ci QIi
-
l
j
11)
iUR.-
The last equation comes from the mean-value theorem. Since all elements of u~'
are inside [u,m, ui,M], we have uR E [Ui,m, Ui,MI C R.
Thus, for 'R and Vi,
a, - Aa- - C A 'xi[O]
<u <
a - Aaj - C A "xj[O]
-<D
<5
ai - C Ai x[0]
ai - CA 'x[O]
IICp" i1
(A.2)
(A.3)
II ci'i I I<
Let u-r; E R be a corresponding input to ui such that a - Aaj
< yj[7', uT ] <
aj,
and then,
ai - Aa1 - CiA xi [O]
<-
||ci q; Ill
<r a, -
C A xi [0]
< <i iny-
(A.4)
For the existence of such uT, we must show that u-r; is inside [ui,m, ui,M]. In other
words, the proof completes if we can find any uw; that satisfy U-; E [ui,m, u,,M] and
(A.4).
We prove this by contradiction. Suppose any u-r; satisfying (A.4) is not feasible,
that is, wr; < uim or u; > ui,m.
If u-; < ui,m, then from (A.4),
a,
-
Ci[A]
z
<[0]
< Ui'M-
IICit
%,1
Since Di
l,
we have ICi,'l||
IICj1FT'II because each element of Ca'IT
71
is
non-negative and CAi"xi[O] > C.Af x[O] by (A.1). Using (A.3), we have
D.
a, - CA'x[O] a - CA x[O]
This contradicts the fact that
up,
E [ui,m, ui,M]. The same procedure for the case
of w7; > ui,M leads to ul, > ui,M, which contradicts the fact that up, E [ui,m, ui,M]-
y (T, u0)
Aa
ui,M]
such that ai
-
Therefore, there must exist uT corresponding to uWi E [ui,m,
ai..5
By the definition of ?(7) in Definition 5.1, P (T) := min,{k E Z>o : y,[k, u,] >
#
6
with contraint y,[R]
fli and yi [Ti, uf I
A.2
ai}, we can obtain the input ur' such that y[-P(T), u']>
ai.
Relation between Discrete Scheduling Problem and Verification Problem
In Chapter 5, the Discrete Scheduling Problem (Problem 5.1) and the corresponding Relaxed Verification Problem (Problem 5.2) are formulated. In this section, we
also consider the Verification Problem (Problem 3.1) to see what the answers of the
Discrete Scheduling Problem imply.
Proposition A.2. If there is no schedule for Problem 5.1, neither is an input u(t)
for Problem 5.2.
Proof. The condition implies that there exist T E [RT,I)i] and Tj E [Ri, 2)] for
i 7 j such that satisfy T7
Pi(T)
5 T and P (T) > Tj, which violates the constraint
7. From Proposition A.1, there are input vectors uj and u, that satisfy
a, - Aaj 5 yi[T7, iii] 5 a, and a3 - Aaj 5 y [T, uj] 5 aj, respectively. Since T, T7
and ?,(T7) are integers, we have
T <7 < Tj+ 1
72
Pi(T).
Now consider the continuous outputs and prove that they are inside the inflated
bad set from discretization $ in (5.2). The corresponding continuous input ui(t) to
u- is a piecewise constant with ui(t) = ui[k] for t E [kAt, (k + 1)At). Similary, the
continuous input u,(t) corresponds to u. At t* = (T7 + !)At,
a3 - Aaj +
fjy,mAt
Since (T7 + -)At < t* < (Pi(7') ai - Aaj +
1
< y,(t*, Uj(t))
a3 + gyy'At.
})At,
imAt < Ys(t*, u(t))
1
3 + A3 - 3i9,mAt.
These inequalities hold because the output function yi(t) is non-decreasing in time.
9j4,mAt.
Then, y,(t*,uj(t)) E (aj
-
Assume that for all i, A43 are greater than
Aaj, 38+ A 43) and y (t*, ui) E (ai - Aai, /i + A#i3). That is, if there is no feasible
schedule for Problem 5.1, there always exists t* such that y(t*) E B. This completes
0
the proof.
Proposition A.3. If a feasible schedule T for Problem 5.1 exists, then an input u(t)
for Problem 3.1 always exists.
Proof. For any i
Pi(T)
#
j, if we have
R
5 'T for 1 E [R, 7i] and T7 E [R1, Vj], then
7Tj. From Proposition A.1, there are input signals u,(t) and u,(t) as defined
in the proof of Proposition A.2.
At t* = Pi(7)At, vehicle i exits the intersection by definition of Pi(7), i.e.,
y (t*, u2(t))
> 3.
At the same time, because t* = Pi(T)
enter the intersection, i.e., y,(t*, u,(t)) 5 y(TjAt, u)
and u,(t) have the vehicles avoid tha bad set B in (3.1).
73
T, vehicle j still does not
a3 . This implies that u1(t)
0
74
Appendix B
Total modularity of the Schedule
Optimization Problem
Recall the LP relaxation of the Schedule Optimization Problem:
Problem 5.6 [LP relaxation of Problem 5.5] Find a vector T
that
= (T+, T-)' such
n
min
subject to
Vi
R<'+
( , >-) Td, !5
Vi <j,
Ti-+>O0 T->O0
We rearrange Problem 5.6 to write it as min{c'TI
Vi.
: AT
5 b}. Since the
objective function is the sum of T+ and Ti- for all i, we have c = (1,1,
75
... ,1).
The
constraints can be written as follows:
(B.1)
T~ < -Rs +,Td, Vi,
-;++
T+ - T-! E9 -
(T+ -7T-)
-Ti+:!
Ti,
(B.2)
Vi,
(B.3)
-ima - Td,7, +T,,, Vi < j,
0,
-:-5
(B.4)
0, Vi.
Constraint (B.3) contains (n - 1) inequalities with {(7r,, 7r2), (72 , 7 3 ), .... , (Irn-.1, rn)}To generalize Problem 5.6 by considering preceding vehicles that are y [0]
ai - Aaj,
we define Po as the maximum process time 'Pi(7) among the vehicles satisfying y[0]
- T;)
-P +
Td,, 1
.
ai - Aaj. Thus, we consider an additional inequality -(T+
For the aggregate vectors T+ and T- and for the column vector T* := (T+, T-)',
bi for i = {1, 2,3, 4}, respec-
these constraints (B.1)-(B.4) can be notated by A.T
tively. Notice that
A1 = [-InynInxn],
A2
A 3 = [L, -L],
A 4 = I2nx2n,
=
[Inxn, -Inxn],
where
L:=
-1
0
0
0
1
-1
0
0
0
1
-1
0
0
---
(B.5)
0
Let bi, denote the jth element of a vector bi for i E {1, 2, 3,4}. Then, for j E
n},
bij = -Rr
+ Td,rj,
b2j =
76
hr, -
,
{1, ...
and
b3,1 = -PO +
Td,,i,,
b3, = -
-
Td,7 ..1_
+ Td,,, for j E {2, ... , n},
b 4 , = 0, for j E {1, ... , 2n}.
Notice that for all i E {1, 2, 3, 4}, A. are integral matrices with elements of
1,0,or -1, and bi are integral vectors.
To prove that the aggregate matrix A =
[Ai; A 2 ; A 3 ; A 4 ] is totally modular, we employ the following proposition from [20].
Proposition B.1. Let A be a totally unimodularmatrix. Then the following matrices
are all totally unimodular:
-A,
AT, [A,I], [A,-A].
By virtue of Proposition B.1, we can confirm that if L is totally unimodular,
then A is totally unimodular. This is because a totally unimodular matrix L leads
sequentially [I; L], [-I; I; L], and then [-I, I; I, -I; L, -L] to be totally unimodular.
Therefore, A becomes a totally unimodular matrix.
Lemma B.1. The square matrix L is a totally unimodularmatrix.
Proof. From (B.5), notice that L has -1 on the diagonal and 1 on the subdiagonal.
The other elements are 0. Thus, for any submatrix of L is always a triangular matrix.
Since all elements in L are either 1, -1, or 0, any subdeterminant of L is 1, -1, or
0.
0
Therefore, given the optimization problem min{c'T* : AT
; b}, which is equiv-
alent to Problem 5.6, we confirm that the matrix A is totally unimodular, and the
vectors b and c are integral.
77
78
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