CO - Robert I. Carr, RICarr.com

CO3 TRAFFIC DEMAND, DELAY,
AND USER COST MODEL1
By Robert I. Carr,2 Fellow, ASCE
ABSTRACT: The Construction Congestion Cost System (CO3) is an integrated
set of tools to estimate the impact of traffic maintenance contract provisions on
congestion, road user cost, and construction cost. We use CO3 to produce realistic
budgets and select practical contracting methods that provide an acceptable balance between construction cost and congestion. This paper describes the underlying CO3 model of traffic demand, delay, and user cost, particularly the methods by
which CO3 calculates traffic backup and delay, diverted and cancelled trips, and
road user cost. It provides examples that illustrate the CO3 model and methods.
CO3 models the common characteristic of construction work zones by which road
capacity can vary from hour to hour as lanes close and open and work conditions
change. Traffic demand also varies with time of day, and drivers may divert to an
alternate route or cancel trips because of delays caused by construction. CO3 estimates traffic delay due to traffic congestion as a function of demand and capacity
and it estimates traffic cancellations and diversions as functions of traffic delay.
1 INTRODUCTION
The Construction Congestion Cost System (CO3) is an integrated set of tools to estimate the
impact of traffic maintenance contract provisions on congestion, road user cost, and construction
cost (Carr 1997). With it we estimate and calculate traffic input variables, traffic delay and user
cost impact, construction cost impact, integrated user cost and construction cost impact, period
costs for contract provisions, and traffic related contract payments. We use CO3 to produce realistic budgets and select practical contracting methods that provide an acceptable balance between
construction cost and congestion. We compare user cost and construction cost of alternative
contract provisions to help us select the best project alternative. Therefore, CO3 helps us select
among alternative methods of maintaining traffic during construction, and it helps us select contract period costs for contract provisions that provide incentives for reducing congestion impacts
during construction. This paper describes only the underlying CO3 model of traffic demand, delay,
and user cost, particularly the methods by which it calculates traffic backup and delay, diverted
and cancelled trips, and user cost.
CO3 models the common characteristic of construction work zones by which road capacity can
vary from hour to hour as lanes close and open and work conditions change. Traffic demand also
varies with time of day, and traffic backs up when demand exceeds capacity. Some motorists divert to an alternate route or cancel trips because of delays caused by construction. Road user cost
results from traffic delays, from increased travel time and distance of alternate routes, and from
trip cancellations caused by congestion. CO3 calculates backups, delays, diverted traffic, and can-
1
2
© Robert I. Carr, 1998, Department of Civil and Environmental Engineering, Ann Arbor, MI.
Professor, Department of Civil and Environmental Engineering, University of Michigan, 2340 G. G. Brown
Building, Ann Arbor, MI 48109-2125, (734) 764-9420, RICarr@umich.edu.
1
celled traffic to measure congestion. It calculates user cost as a measure of congestion that is integrated with construction cost in selecting among alternative contract provisions.
CO3 models the circular relationship between work zone demand and delay, in which diversions
and cancellations are functions of delay, work zone demand is a function of diversions and cancellations, and delay is a function of work zone demand. It models traffic as an integrated stream
of two classes of vehicles, trucks and cars, that share backups but differ in diversion and cancellation sensitivity. It models and estimates traffic delay due to traffic congestion as a function of demand and capacity and it estimates traffic cancellations and diversions as functions of traffic delay.
It calculates traffic backup that is due to actual demand on the work zone exceeding work zone
capacity.
The CO3 system is implemented in a Microsoft Excel spreadsheet. The spreadsheet and CO3
User Manual can be downloaded from the worldwide web at http://grader.engin.umich.edu/. CO3
was developed with financial support from the Michigan Department of Transportation to allow it
to reduce the impact of highway construction on Michigan travelers.
2 BACKGROUND
An analytical procedure to estimate traffic delay and congestion and assess the tradeoffs in
cost-effectiveness of alternative measures was described in (“Alternative” 1978). This was implemented in a microcomputer spreadsheet that is still widely used (Morales 1986). The underlying
model includes only delay and user cost of vehicles in the queue formed by demand exceeding capacity. The QUEWZ model provides a recent model of traffic flow through lane closures in freeway work zones and queue lengths and road user costs that result (Krammes, Ullman, Memmott,
and Dudek 1993). QUEWZ was designed for Texas urban freeways with continuous frontage
roads, and it diverts all traffic that would cause a queue greater than a specified maximum length
or motorist delay greater than a specified maximum delay. Microscopic traffic simulation models
can also be applied to freeway incident management (Cragg and Demetsky 1995). However, this
is generally impractical for broad application to work zone analysis for incentive contracts, particularly for project scope and budget decisions.
The FHWA has provided for incentive/disincentive and related clauses for highway construction based on road user costs since 1984 (“Incentive/disincentive” 1989), and practicalities of
such clauses were described by Riley (1984), and Newman and Heijl (1984). Construction work
zones can be considered a special case of freeway incident management in which interruptions of
normal traffic is planned and controlled, and much can be learned from freeway incident management literature (Reiss 1991). Traffic/congestion management recommendations are summarized in
(“Traffic” 1992), and alternative contract management techniques are described in (“Contract”
1997). Incentive-based highway contracting methods of various forms have been widely implemented with significant success (Jaraiedi, Plummer, and Aber 1995) (Herbsman, Chen, and Epstein 1995) (Herbsman 1995). The FHWA 1998 National Strategic Plan includes initiatives to
stabilize congestion and reduce delays caused by construction (“Federal” 1997).
3 DEMAND, ROUTES, AND DELAY
3.1 DESIGN DEMAND
Design demand (Vph) = number of vehicles that travel through the site under normal conditions when it is not a construction site. Design demand is generally based on historical demand =
sample(s) of number of vehicles per hour that traveled through the site under normal conditions at
2
the most recent sampling. Design demand can be estimated directly, but it normally will be directly calculated from historical demand, with adjustment for years of growth between the year
in which historical demand is determined and the year in which the work will be performed, based
on estimated annual traffic growth using input shown in Fig. 1. CO3 calculates design demand
using the following equation:
a
f
a
design demand = historical demand * 1+ annual traffic growth
f byears of growthg . . . . . . . . . . . (1)
For example, using values in Fig. 1, if a period's historical demand = 3,124 Vph, the period's
design demand is
design demand = 3,124 * 1 + 0.03 2 = 3,314 Vph
a
fa
f
These values of historical demand and design demand are shown for 8:00 am to 9:00 am (8A) in
Table 1, which also includes values for 9:00 am to 2:00 pm. [Note: For brevity, the example in
this paper is for six periods, as shown in Table 1, rather than the 24 periods of CO3 software.]
period length (min)
annual traffic growth (%)
years of growth
VEHICLE INPUT
cars
design demand (%)
90.0%
60
3.00%
2
trucks
10.0%
Fig. 1 - Input of Traffic and User Cost Constants3
Table 1. Period Capacity and Demand
period
(hr)
(1)
8A
9A
10 A
11 A
12 P
1P
Total
capacity
(V/hr)
(2)
3,400
1,400
1,400
1,400
1,400
3,400
12,400
historical
demand
(V/hr)
(3)
3,124
2,436
2,051
1,436
1,513
2,099
12,659
total
period
decrease
(V/hr)
(5)
0
571
810
431
282
0
2,095
design
demand
(V/hr)
(4)
3,314
2,584
2,176
1,523
1,605
2,227
13,430
car
actual
demand
(V/hr)
(6)
2,983
1,773
1,177
955
1,171
2,004
10,064
truck
actual
demand
(V/hr)
(7)
331
240
188
137
152
223
1,272
actual
demand
(V/hr)
(8)
3,314
2,013
1,366
1,092
1,323
2,227
11,335
3.2 WORK ZONE AND DIVERSION ROUTES
Fig. 2 shows travel routes that are important to calculations of extra distance and time that
may be required during construction, relative to normal travel when there is no work zone. We
differentiate between a method route and a normal route for work zone travel and for diversion
travel. Vehicles travel a method route while a construction method is being applied. Diversion
method distance = the route vehicles follow between leaving and returning to the normal route if
they divert around the work zone. Work zone method distance = zone method distance = distance
vehicles follow from one side of the work zone to the other if they do not divert. The normal
route is the comparable route vehicles travel when there is no construction. Therefore, diversion
normal distance = distance on the normal route between the points at which diverting vehicles
leave and return to the normal route. Work zone normal distance = zone normal distance = distance on the normal route between one side of the work zone to the other when there is no work
3
Fig. 1, Fig. 5, and Fig. 8 show the format of CO3 software input.
3
zone. Fig. 2 (a) shows the distances when there is no detour, in which case work zone method
distance = work zone normal distance. Fig. 2 (b) shows the distances where there is a detour and
no vehicles travel the normal route through the work zone, in which case generally the work zone
method distance > work zone normal distance.
(a) Without Detour
(b) With Detour
Fig. 2 - Work Zone and Diversion Distances
3.3 DELAY
A work zone can cause three types of delay:
(1) speed delay = difference in time to travel the method travel distance through the work
zone (or around it if a detour is required) during construction and the normal time
when there is no work zone..
(2) backup delay = time vehicles wait to enter the work zone when work and traffic
maintenance conditions reduce capacity below design demand, and
(3) diversion delay = difference in time to travel the diversion travel distance around the
work zone during construction and the normal time when there is no work zone.
Vehicles that travel through the work zone experience
work zone delay = speed delay + backup delay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
whereas vehicles that divert around the work zone experience diversion delay.
3.4 ACTUAL DEMAND
Demand through the work zone (or detour) decreases when drivers divert to other routes or
cancel trips to avoid the work zone conditions and possible delays associated with it. Actual demand in a period consists of vehicles that arrive at the work zone during the period, which is de4
sign demand minus vehicles that cancel trips or divert around the work zone. At any capacity, the
decrease in traffic from diverted and canceled vehicles is modeled in four components for both
cars and trucks:
• decrease (with no delay) = diverted (with no delay) + canceled (with no delay) traffic
that occurs even in periods in which there is no significant delay at the work zone.
This is expressed in percent.
• decrease (with delay) = diverted (with delay) + canceled (with delay) traffic that occurs due to delay at the work zone. This is expressed in percent decrease per minute
of delay.
Therefore, demand = actual demand is a function of work zone delay:
decrease % = decrease (with no delay) (%)
. . . . . . . . . . . . . (3)
+ decrease (with delay) (% / min) * work zone delay (min)
af
where decrease (%) = per cent of design demand that diverts or cancels its trips. This is demonstrated in Fig. 3 (a) for diverted cars. Total decreases are the sum of diverted and cancelled cars
and trucks,
Demand decrease = decrease in design demand is a function of decrease (%) and design demand,
decrease Vph = decrease % * design demand Vph
a f
af
a f
R| decrease (with no delay) (%) U|
= S + decrease (with delay) (% / min) V * design demand aVph f
|T * work zone delay (min) |W
and demand = actual demand is calculated by
actual demand (Vph) = design demand (Vph) − decrease (Vph)
l
= design demand (Vph) * 1 − decrease (%)
q
. . . . . . . . . . . . . . (4)
. . . . . . . . . . . . . . . . . . . . . . . . (5)
4 SPEED DELAY
4.1 GENERAL
Vehicle speed through a construction zone may be lower than speeds outside the zone, due to
pavement conditions, traffic conditions, routing through a detour, or other work zone conditions.
If a detour is used, it adds distance and time. The increase in time required to traverse the work
zone is speed delay, and this section describes input variables and calculation of speed delay.
4.2 SPEED DELAY = FUNCTION OF DEMAND AND CAPACITY
Speed delay is the difference in time to travel the method travel distance through the work
zone (or around it if a detour is required) during construction and the time it would take if there
were no construction. Threshold capacity for speed delay is the capacity at and below which
speed delay occurs. Generally, threshold capacity for speed delay for a day is the capacity through
the work zone that day during construction. If there are different capacities at different times
during a day of construction, the threshold capacity for speed delay is generally the largest of
them. Speed delay is calculated for all periods with capacity ≤ (threshold capacity for speed delay), which allows speed delay of a vehicle to vary as work zone conditions and traffic conditions
5
diverted cars ( %)
change. Work zone conditions change from time to time as lanes or ramps are closed and opened
or as traffic is routed over temporary paving or detours. Therefore, capacity changes occur when
work zone conditions change. Traffic conditions change as work zone conditions change and as
demand changes.
threshold diverted cars (with delay)
diverted cars = 0 for capacity > threshold
threshold diverted cars
(without delay)
0
average delay (minutes)
diverted cars ( %)
(a) W ith threshold capacity for decreases
h
(wit
y)
dela
for
C1
la
h de
(wit
y)
rs
ars
d ca
ed c
t
r
e
iv
C
ge d
for 2
ran
lay)
e
d
ith
lay)
th de
rs (w
i
a
w
c
(
cars
rted
dive
erted
v
i
d
hold
thres
rte
dive
diverted cars (without delay), C 1
range diverted cars (without delay)
diverted cars (without delay), C 2
threshold diverted cars
(without delay)
diverted cars = 0 for capacity > threshold
average delay (minutes)
File Fig_5_3.CV5
(b) With threshold and rang e capacities for decreases
Fig. 3 Percent Decrease in Car Demand for Threshold and Range Values
Speed delay is modeled as a function of period capacity (C) and period zone demand (D),
where period zone demand = vehicles that enter the work zone during a period (as opposed to
period actual demand = vehicles that arrive at the work zone during a period). That is, each period with capacity ≤ (threshold capacity for speed delay) has a speed delay function. For each period the value of its speed delay is a function of the ratio D/C = (demand that enters the zone
during the period) / (period capacity), and D/C can have any value between 0 and 1. Fig. 4 shows
this relationship between speed delay and zone demand. Speed delay never exceeds its value at
D/C = 1, because traffic flow through the zone cannot exceed capacity, by definition, and all demand greater than capacity backs up and waits to enter the zone.
6
speed d elay ( min)
threshold
(when
threshold
(when
speed delay
D= C)
speed delay
D~0)
for all C ≤ threshold capacity
speed delay = 0 for capacity > threshold capacity
actual
dema nd
(Vph)
threshold capacity for speed delay (Vph)
speed d elay ( min)
(a) With threshold capacity for speed delay
extrapolation for C 1
for range capacity
range speed delay (D=C)
interpolation for C 2
threshold speed delay (D=C )
range speed delay (D~0)
threshold speed delay (D~0)
for threshold capacity
speed delay = 0 for capacity > threshold capacity
actual
demand
(Vph)
C1
C2
range capacity for speed delay (Vph)
threshold capacity for speed delay (Vph)
File FIG_4B.CV5
(b) With thres hold and range capacities for speed delay
Fig. 4 Speed Delay as Function of Capacity and Demand
The speed delay function for each period with capacity ≤ (threshold capacity for speed delay)
is therefore defined by three variables, which are shown in Speed Delay Input in Fig. 5 (a) for
threshold values:
• capacity for speed delay = capacity for the period.
• speed (when D ~ 0) = speed when demand is very low.
• speed (when D = C) = speed when the number of vehicles that enter the zone during
the period equals its capacity.
Fig. 5 (a) shows these values where (threshold capacity for speed delay) = 1400 Vph. For
cases where different periods during a day have different capacities and the different capacities
7
allow different speeds that differ from normal travel speed, another set of values, called range inputs, is needed in addition to threshold inputs. Fig. 5 (b) shows values of the three inputs for both
threshold values and range values. When there are range inputs, the computer calculates speed
delay by interpolating between the threshold inputs for the threshold capacity and the range inputs
for the range capacity.
4.3 THRESHOLD CAPACITY FOR SPEED DELAY
4.3.1 Normal Travel
Travel time is a function of travel distance and speed,
distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(6)
time =
speed
For the highway under study, we estimate normal traffic speed = 70 mph when there is no construction work zone. The project considered will require work zone distance = 2 mi, for which
normal travel time is
work zone normal distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
normal time =
(7)
work zone normal speed
=
2 mi
= 0.02857 hr = 1.71 min.
70 mph
4.3.2 Low Demand (D≈
≈ 0)
We estimate that closing one lane of the example two-lane highway during construction will
reduce capacity to 1400 Vph over the two miles of lane closure. At capacity = 1400 Vph, we estimate average vehicle speed at low traffic will be reduced from normal speed = 70 mph to
[method speed (when D ≈ 0)] = 50 mph due to construction conditions.
Travel time during construction at low traffic is calculated from:
work zone method distance
. . . . . . . . . . . . . . . . . . . . . . . (8)
method time (when D ≈ 0) =
work zone method speed (when D ≈ 0)
=
2 mi
= 0.04 hr = 2.40 min.
50 mph
and
delay = method time − normal time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (9)
a
f a
f
From this, speed delay at low traffic is calculated below and shown as work zone travel in Fig.
5.
speed delay (when D ≈ 0) = method time (when D ≈ 0) − normal time . . . . . . . . . . . . . . . . . . . (10)
= 2.40 − 1.71 = 0.69 min
8
VEHICLE INPUT
cars
trucks
user cost per hour ($/V hr)
$12.00
$30.00
user cost per mile, ($/V mi)
$0.30
$1.00
user cost per cancellation, ($/V)
$4.00
$10.00
distance
speed
work zone method travel
2.0
see delay
work zone normal travel
2.0
diversion method travel
10
DISTANCE AND SPEED INPUT
diversion normal travel
SPEED DELAY INPUT
capacity for speed delay (V/hr)
VEHICLE INPUT
user cost per mile, ($/V mi)
$0.30
$1.00
$4.00
$10.00
user cost per cancellation, ($/V)
DISTANCE AND SPEED INPUT
speed
70
work zone normal travel
2.0
70
45
diversion method travel
10
45
diversion normal travel
4.0
70
threshold
range
1400
1000
speed (when D~0) (mph)
50
45
speed (when D=C) (mph)
40
SPEED DELAY INPUT
1400
40
$30.00
see delay
70
speed (when D=C) (mph)
$12.00
2.0
range
WORK ZONE TRAVEL
threshold
method travel time (when D~0) (min)
2.40
normal travel time (min)
1.71
speed delay (when D~0) (min)
0.69
method travel time (when D=C) (min)
3.00
speed delay (when D=C) (min)
1.29
WORK ZONE SPEED DELAY USER COST
threshold
car speed delay user cost (when D~0)
$0.14
truck speed delay user cost (when D~0)
$0.34
car speed delay user cost (when D=C)
$0.26
truck speed delay user cost (when D=C)
$0.64
DIVERSION TRAVEL
method travel time (min)
13.33
normal travel time (min)
3.43
diversion delay (min)
9.90
extra diversion travel distance (mi)
6.0
DIVERSION USER COST
cars
diversion delay user cost
$1.98
diversion distance user cost
$1.80
diversion user cost
$3.78
backup delay balance (min)
17.62
user cost per hour ($/V hr)
work zone method travel
4.0
50
trucks
distance
threshold
speed (when D~0) (mph)
cars
capacity for speed delay (V/hr)
range
WORK ZONE TRAVEL
threshold
method travel time (when D~0) (min)
2.40
normal travel time (min)
1.71
speed delay (when D~0) (min)
0.69
method travel time (when D=C) (min)
3.00
speed delay (when D=C) (min)
1.29
WORK ZONE SPEED DELAY USER COST
threshold
car speed delay user cost (when D~0)
$0.14
truck speed delay user cost (when D~0)
$0.34
car speed delay user cost (when D=C)
$0.26
truck speed delay user cost (when D=C)
$0.64
DIVERSION TRAVEL
method travel time (min)
13.33
normal travel time (min)
3.43
diversion delay (min)
9.90
extra diversion travel distance (mi)
6.0
DIVERSION USER COST
cars
diversion delay user cost
$1.98
diversion distance user cost
$1.80
diversion user cost
$3.78
backup delay balance (min)
17.62
range
trucks
$4.95
$6.00
$10.95
20.62
(a) With threshold capacity
20
range
2.67
1.71
0.95
6.00
4.29
range
$0.19
$0.48
$0.86
$2.14
trucks
$4.95
$6.00
$10.95
20.62
(b) With threshold and range capacity
Fig. 5 - Speed Delay, Diversion, and User Cost
4.3.3 High Demand (D=C)
Travel time during construction at high traffic is calculated below and shown as work zone
travel in Fig. 5.
work zone method distance
. . . . . . . . . . . . . . . . . . . . . (11)
method time (when D = C ) =
work zone method speed (when D = C )
=
2 mi
= 0.05 hr = 3.00 min.
40 mph
speed delay (when D = C ) = method time (when D = C ) − normal time . . . . . . . . . . . . . . . . . . (12)
= 3.00 − 1.71 = 1.29 min
Col. 4 of Table 3 shows average speed delay = 1.29 min for 9:00 am to 11:00 am, where capacity = 1,400 Vph < actual demand = 2,013 Vph, and for 10:00 am to 12:00 pm, where there is
backup for the entire period and traffic flow through the zone = capacity = 1,400 Vph.
9
4.3.4 Intermediate Demand (0 < D < C)
The general equation for speed delay as a function of zone demand = D and capacity = C is
speed delay = speed delay (when D ≈ 0)
+
|RS speed delay (when D = C ) |UV * F D I
|T− speed delay (when D ≈ 0) |W H C K
ES
= speed delay (when D = C )
for D < C . . . . . . . . . . . . . . . . . . . (13)
for D ≥ C
where ES = speed delay exponent (default ES = 2). Fig. 4 illustrates Eq. (13).
Speed delay at an intermediate demand between zero and capacity, such as at D = 900 VPH at
which D/C = 900/1400 = 0.643, is calculated using Eq. (13),
900 I 2
F
f H 1400 K
a
= 0.69 + a0.69f * a0.643f2 = 0.69 + 0.29 = 0.98 min
speed delay (when D = 900) = 0.69 + 1.29 − 0.60 *
4.4 RANGE CAPACITY FOR SPEED DELAY
As described in Sec. 4.2 above, we sometimes need range inputs to describe vehicle speeds at
one or more capacities below the threshold capacity for speed delay. For example, outside of
working hours, capacity and speeds of a signed work zone are often lower than normal, and capacity and speeds during work periods are even lower. Or capacity and speed may be higher while
a lane is being closed than when work is proceeding close to traffic. An example of such range
input is shown in Fig. 5 (b). Fig. 4 (b) shows range speed delay and its relationship to threshold
speed delay. Range travel time and speed delay follow Eq. (6) to Eq. (12), which produces the
range travel times and speed delays shown for work zone travel in Fig. 5 (b). Range speed delay
for intermediate demand (0 < D < C) follows Eq. (13). For example, this produces [range speed
delay (when demand = 900 Vph)] = 3.66 min.
4.5 CAPACITY ≠ THRESHOLD CAPACITY OR RANGE CAPACITY
When there is range input as well as threshold input, CO3 calculates values of any variable xC =
value of x at capacity = C < threshold capacity by the following interpolation:
a
f a
f a
OP . . . (14)
athreshold capacityf − C
f LMathreshold
capacity f − a range capacityf Q
N
xC = threshold x + range x − threshold x *
where (threshold x) = value of x at threshold capacity and (range x) = value of x at range capacity.
The dashed lines in Fig. 4 (b) show interpolated values for capacity C2, where (threshold capacity) > C2 > (range capacity) and extrapolated values for capacity C1< (range capacity). Thus
for our example we can model any capacities below 1400 Vph. For example, we can calculate
travel time and speed delay for capacity = 900 Vph using Eq. (14),
1400 − 900
method time (when D ≈ 0) = 2.40 + 2.67 − 2.40 *
1400 − 1000
= 2.40 + 0.27 * 1.25 = 2.40 + 0.33 = 2.73 min
LM
N
10
OP
Q
1400 − 900 O
a f a f LMN1400
− 1000 PQ
method time (when D = C ) = 3.00 + 6.00 − 3.00 *
= 3.00 + 3.00 * 1.25 = 3.00 + 3.75 = 6.75 min
These are 2.0 mi travel times at [average speed (when D ≈ 0)] = (2.0 mi) / [(2.73 min)*(60
min/hr)] = 38.1 mph and [average speed (when D=C)] = 17.8 mph. Similarly, Eq. (14) calculates
speed delay for capacity = 900 Vph,
1400 − 900
= 1.02 min
speed delay (when D ≈ 0) = 0.69 + 0.95 − 0.69 *
1400 − 1000
LM
OP
N
Q
1400 − 900 O
speed delay (when D = C ) = 1.29 + 4.29 − 1.29 * LM
N1400 − 1000 PQ = 5.04 min
which agree with Eq. (12), from which [speed delay (when D ≈ 0)] = 2.73 – 1.71 = 1.02 min, and
[speed delay (when D = C)] = 6.75 – 1.71 = 5.04 min.
4.6 USER COST OF SPEED DELAY
User cost = monetary equivalent of any extra travel time and distance that vehicles incur due to
construction. We calculate speed delay user cost using Eq. (15), for which Fig. 5 shows an example of user cost per hour ($/V-hr) and user cost per mile ($/V-mi).
a f
a
user cost = delay hr * user cost per hour $ / hr
m
a f
f
a f r * user cost per mile a$ / mif
+ method distance mi − normal distance mi
. . . . (15)
For example,
a
f
car speed delay user cost when D = C =
a
a
1.29 min
* $12.00 / hr
60 min / hr
fa
f
f
+ 2.0 mi - 2.0 mi * $0.30 / mi = $0.26
Similarly, truck speed delay user cost (when D = C) = (1.29 min) * ($30.00 / hr) / (60 min/hr) =
$0.645. Speed delay user cost ($/V) for other conditions for cars and trucks is shown in Fig. 5.
5 BACKUP DELAY
5.1 GENERAL
A backup is a queue of vehicles that is waiting to enter a work zone, and backup delay is the
time a vehicle waits in the queue to enter the zone. CO3 assumes arrivals and departures are uniformly distributed over each period. Backup is the excess of demand over capacity, and all vehicles ahead of a particular vehicle must pass through the zone before it can. Therefore, a vehicle
that arrives when there is a backup must wait to enter the zone until the backup ahead of it has
entered the zone. If there are BN vehicles backed-up when vehicle N arrives, and the capacity
(V/period) of the zone is capacityN following the arrival of N, vehicle N is delayed by
backup delay B while it waits to enter the zone, where
a
f
backup delay amin f B
N
N
=
a
f
BN (V) 60 min/ hr
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (16)
capacity N (Vph )
11
These relationships are shown in Fig. 6, in which Bsop i = backup at start-of-period i, and
Beop i = backup at end-of-period i. Of course, the start-of-period backup of one period = end-ofperiod backup of the previous period, or
Bsop i +1 = Beop i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (17)
Vehicles arriving at the start-of-period 1 are behind backup Bsop 1. The last vehicle in that
backup enters the zone at time a, after which time the first vehicles arriving in period 1 will start
entering the zone. The number of period 1 vehicles that enter the zone during period 1 = e – d . If
period 1 demand = demand1,1, then end-of-period 1 backup = Beop 1,1 = start-of-period 2
backup = Bsop 2,1 , and the last arriving period 1 vehicle will be backed-up until time b, at which
time all of demand1,1 will have entered the zone. Thus, all period 1 vehicles will have entered the
zone before end-of-period 2, and end-of-period 2 backup = Beop 1,1 = 0. The total vehicleperiods period 1 vehicles will have spent in backups is the cross-hatched area
= (total
backup delay)1,1.
However, if period 1 demand = demand1,2 , then period 1 vehicles will remain into period 3,
the latest arriving period 1 vehicles will be backed-up from end-of-period 1 until c, and the total
vehicle-periods of backup delay accumulated by period 1 vehicles will be the area
= (total
backup delay)1,2.
Thus, the total vehicle-periods of delay due to backups experienced by vehicles that arrive in a
period is a function of the backup at the start of the period, the period demand, and the period
capacity. It is also a function of the following periods’ capacities, if the initial start-of-period
backup and period demand exceed the period capacity.
5.2 BACKUP AS FUNCTION OF DEMAND AND CAPACITY
The relationship among these variables, as demand and capacity vary from period to period, is
shown in Fig. 7. The slopes of the dashed line along the top of the crosshatched areas represent
the period demands, and the slopes of the solid line along the bottom represent the period capacities. Each point along the demand line represents arrival of a vehicle at the zone on the vertical
axis and the time the vehicle arrives on the horizontal axis. A horizontal line drawn from that
point intersects the capacity line at the time the vehicle enters the zone. The horizontal distance
between the point of arrival on the demand line and the point of entering the zone on the capacity
line represents the backup delay for the vehicle. For example, the vehicle that arrives at 9 (point
A) will enter the zone as shown by point B, and it will be backed up for time AB.
A period’s demand does not start to enter the work zone (or other primary path) until all preceding backed-up vehicles have entered the zone. Thus, the first vehicles arriving 7-8 can enter
the zone immediately. However, no vehicles arriving 9-10 can enter the zone until well into time
11-12.
Similarly, the last vehicle to arrive in a period is backed-up until the earlier arriving vehicles,
including those backed-up from prior periods, have entered the zone. Thus, vehicles that arrive
just before 8 will be backed-up until almost 9, because demand7-8 significantly exceeds capacity79. Vehicles arriving just before 10 will be backed-up until past 12, because the backup at 9 plus
demand9-10 exceeds capacity9-12.
12
Area = (backup delay) 1,2
back up
B eop 1,1 = B sop 2,1
e
d
capacity 1 = capacity 2 = capacity 3
sop = start-of-period
eop = end-of-period
a
0
B eop 2,2
demand 1,1
B eop 1,2 = B sop 2,2
demand 1,2
back up
B eop 0 = B sop 1
Cumulative Demand and Capacity
(vehicles)
Area = (backup delay) 1,1
b
1
c
2
Fig_2b.CV5
Periods of Day
3
Fig. 6 Comparison of Backup Delay for Two Demands
3
The CO model determines the vehicle-periods of delay for the each period’s demand by
tracking the period demand through successive periods until all of its vehicles have entered the
zone. CO3 first calculates the period’s vehicles’ delays for the period in which they arrive. If vehicles remain at the end of the period, those vehicles’ delays are calculated for the succeeding period. If vehicles remain at the end of the succeeding period, their delays are calculated for the next
succeeding period. This continues until none of the initial period’s vehicles remain. The sum of the
delays for those periods is the total backup delay of the vehicles that arrived in the initial period. If
the periods are one hour in length, then total backup delay is measured in vehicle-hours (V-hr).
5.3 INITIAL PERIOD AND FOLLOWING PERIOD CONDITIONS
The initial and following period backup conditions for vehicles arriving during a period are indicated by the following terms:
I = Initial, indicating the initial period = the period in which the demand occurs.
F = Following, indicating a period following the period in which the demand occurs.
B indicates a start-of-period backup < (D = initial period demand).
D indicates a following period start-of-period backup ≥ (D = initial period demand).
R indicates 0 < end-of-period backup < (D = initial period demand).
O indicates there is no backup, i.e., backup = 0.
These terms are combined to describe the different possible conditions for which to calculate
period backup and delay. The start-of-period and end-of-period condition classes are described in
Table 2.
13
Cumulative demand and capacity (vehicles)
IBO
FDO
IBD
FR
O
FDD
IBD
demand
IBD
A
B
FRO
FRR
FRR
IBR
capacity
FRO
(backup delay)9 = AB
IOR
7
FRO
FDR
FDD
8
9
11
10
1
12
File FIG 3.CV5
Time of Day (hours)
Fig. 7. Backup, from Cumulative Demand and Capacity
Period 1, which is the initial period considered in Fig. 6, is IBR for both demand1,1 and demand1,2, because it has start-of-period backup = d > 0 and end-of-period backup > 0. Period 2,
which follows period 1, is FRO for demand1,1 , because period 2 capacity exceeds period 1 endof-period backup, and no period 1 demand is backed up at end-of-period 2. However, if period 1
demand = demand1,2 , then period 2 is FRR and period 3 is FRO, because the last part of period
1 demand is still backed up at end-of-period 2 but all of it has entered at end-of-period 3.
Each period in Fig. 7 is labeled to show its condition. Fig. 7 shows each of the nine possible
period conditions, with several possible condition sequences. For example, period 9-10 demand
proceeds through sequence IBD, FDD, FDR, and FRO in periods 9-10, 10-11, 11-12, 12-1 respectively.
For each period i, the sum of initial period demand and start-of-period backup is compared to
initial period capacity to determine the initial period class. The initial period delay and end-ofperiod backup are calculated using formulas appropriate to that class. If the initial period end-ofperiod backup is greater than 0 (i.e., if Beop I = Beop i > 0), then end-of-period backup is compared to the following period capacity to determine the following period class. The following period delay and end-of-period backup are calculated using formulas appropriate to that class. This
continues from following period to following period until all the initial period demand has been
accommodated (i.e., until Beop F = 0). At this point backup delay i = total backup delay for ve-
a
14
f
a
f
hicles that arrive during period i and average backup delay i = average backup delay per vehicle
that arrives during period i can be calculated.
backup delay i =
∑ backup delay i, j . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (18)
a
f
a
f
i, j
j = i,i +1,i + 2⋅⋅⋅
backup delayfi
abackup delayfi . . . . . . . . . . . . . . . . . . . . . . . . . . . (19)
=
aaverage backup delayfi = aaactual
demand fi
Di
where abackup delayfi, j = the delay during period j, for vehicles that arrived in period i.
Table 2. Period Backup Condition Relationships
Start of Period
End of Period
Class
Symbol
Condition
Symbol
Condition
(1)
(2)
(3)
(4)
(5)
(a) Start of period and end of period backup for I = initial period
IOR
R
DI > Beop·I > 0
O
Bsop·I = 0
IOO
O
Beop·I = 0
IBD
IBR
B
Bsop·I > 0
IBO
D
Beop·I ≥ DI
R
DI > Beop·I > 0
O
Beop·I = 0
(b) Start of period and end of period backup for F = following period
Beop·F ≥ DI
FDD
D
DI > Beop·F > 0
FDR
D
R
Bsop·F ≥ DI
Beop·F = 0
FDO
O
DI > Beop·F > 0
FRR
R
R
DI > Bsop·F ≥ 0
Beop·F = 0
FRO
O
5.4 PERIOD BACKUP FORMULAS
For each initial period, Beop·I = (end-of-period backup)I is calculated from
Beop⋅I = Bsop⋅I + D − C
forBsop⋅I + D > C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (20)
Beop⋅I = 0
for Bsop⋅I + D ≤ C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (21)
Classes IOR, IBD, and IBR follow Eq. (20), and class IBO follows Eq. (21) and has no endof-period backup. Similarly, we calculate Beop·F of each following period, using
Beop⋅F = Bsop⋅F − C
for Bsop⋅F > C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (22)
Beop⋅F = 0
for Bsop⋅F ≤ C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (23)
Classes FDD, FDR, and FRR follow Eq. (22), and classes FDO and FRO follow Eq. (23) and
have no end-of-period backup. Appendix 2 shows the period demand, capacity, start-of-period
15
backup, and end-of-period backup for each initial period and following backup condition described in Table 2.
5.5 BACKUP DELAY EXAMPLE
Table 1 shows an example of capacity and historical demand inputs into CO3 and calculated
values for design demand from 8:00 am (period 8A) through 2:00 pm (period 1P) using Eq. (1).
Normal capacity is 3,400 Vph, and work zone capacity is reduced to 1,400 Vph due to lane closure from 9A to 12P. Table 1 also shows calculated actual demand, and Table 3 shows backup
and delay times. CO3 calculation of backup demand is shown in this section, and calculation of
actual demand is described in Sec. 6.
CO3 calculates each period’s backup delay from its capacity and actual demand and from capacity in the following period(s) if vehicles are backed up at end-of-period. Drawings and equations in Appendix 2 shows relationships of actual demand, capacity, backup, backup delay, and
speed delay for the period classes described in Table 2.
Though Table 1 does not show values before 8A, (capacity before 9A) = 3,400 Vph, which
exceeds both threshold capacity for speed delay and design demand. Therefore, before 9A there is
no speed delay or backup delay, actual demand = design demand, and all periods before 9A are
class IOO.
In Col. 2 of Table 3 we see period 9A has no start-of-period backup, because 8A end-ofperiod backup = 0. We use Eq. (20) to calculate 9A end-of-period backup = Beop 9A = 0 + 2,013
– 1,400 = 613 V, as shown in Col. 2 in Table 3 for 9A. Period 9A is class IOB, because it is an
initial period, Bsop 9A = 0, and demand = D9A = 2,013 > Beop 9A = 613 > 0. From Eq. (38),
D − C 2,013 - 1,400
backup delay 9A,9A =
=
= 307 V - hr
2
2
By Eq. (23), For 9A demand, period 10A is class FRO, because it follows 9A, it has capacity = C
= 1,400 > Bsop 10A = Beop 9A = 613, and by Eq. (23) Beop 10A = 0 because Bsop 10A < capacity10A = 1,400. Therefore, from Eq. (53),
a
f
2
a613f2 = 134 V - hr
=
abackup delayf9A,10A = 2Bsop
* C 2 * 1, 400
From Eq. (18) and (19),
backup delay 9A = backup delay 9A,9A + backup delay 9A,10A = 307 + 134 = 441 V - hr
a
f a
f
a
f
delayf9A a441 V - hr f * a60 min / hr f
=
= 1314
. min
aaverage backup delayf9A = abackupD9A
2013 V / hr
as shown in Col. 3 of Table 3.
Similarly, we use Eq. (20) to calculate Beop 10A = 613 + 1,366 – 1,400 = 579 Vph, as shown
for 10A in Table 3. Therefore, for 10A demand, period 10A is class IBR, because it is an initial
period, Bsop 10A = Beop·9A = 613 > 0, and demand = D10A = 1,366 > Beop 10A = 579 > 0.
From Eq. (43), (backup delay)10A,10A = 0.5 * [1,366 – (1,400 – 613)2 / 1,400] = 462 V. For
10A demand, period 11A is class FRO, because it follows 10A, it has capacity = C10A = 1,400 >
Bsop 11A = Beop 10A = 579, and by Eq. (23) Beop 10A = 0 because Bsop 11A = 579 < capacity11A = 1,400. Therefore, from Eq. (53), (backup delay)10A,11A = (579)2 / (2 * 1,400) = 120
16
V-hr. From Eq. (18) and (19), (backup delay)10A = 462 + 120 = 582 V-hr, and (average backup
delay) 10A = (582 V-hr) * (60 min/hr) / (1,366 Vph) = 25.55 min. as shown in Col. 3 of Table 3.
5.6 USER COST OF BACKUP DELAY
In parallel with speed delay user cost, we calculate backup delay user cost using Eq. (15).
Method distance and normal distance have no meaning for backup delay. Therefore, backup delay
user cost = (backup delay) * (user cost per hour). From this, (car backup user cost)9A = (backup
delay)9A * (car user cost per hour) = (13.14 min) * ($12.00 / hr) / (60 min/hr) = $2.63. From
Col. 6 of Table 1 for (car actual demand)9A = 1,773 Vph, (total car backup user cost)9A =
(1,733) * 2.63 = $4,558. Similarly (truck backup user cost)9A = (backup delay)9A * (truck user
cost per hour) = (13.14 min) * ($30.00 / hr) / (60 min/hr) = $6.57. From Col. 7 of Table 1 for
(truck actual demand)9A = 240 Vph, (total truck backup user cost)9A = (240) * 6.57 = $1,577.
Total backup user cost is the sum of car and truck backup user cost = 4,558 + 1,577 = $6,135.
Table 3. Period Backup, Delay, and User Cost
period
(hr)
(1)
8A
9A
10 A
11 A
12 P
1P
Total
end of
period
backup
(V)
(2)
0
613
579
271
194
0
average
backup
delay
(min)
(3)
0.00
13.14
25.55
18.22
9.63
0.28
average
speed
delay
(min)
(4)
0.00
1.29
1.29
1.29
1.10
0.00
average
delay
(min)
(5)
0.00
14.43
26.83
19.50
10.73
0.28
total
period
delay
(V hr)
(6)
0
484
611
355
237
11
1,697
delay
cost
($)
(7)
$0
$6,846
$8,846
$5,065
$3,329
$145
$24,231
decrease
cost
($)
(8)
$0
$2,325
$3,316
$1,760
$1,146
$0
$8,547
user
cost
($)
(9)
$0
$9,172
$12,162
$6,825
$4,475
$145
$32,778
5.7 AVERAGE AND TOTAL DELAY AND USER COST
In accord with Eq. (2), average delay for vehicles that arrive at the work zone during a period
= average work zone delay = sum of speed delay and backup delay. Period delay = total period
delay is calculated,
period delay i = average delay i * actual demand i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (24)
a
f a
f a
f
and period delay cost is calculated from Eq. (15). From Table 3, (average delay)9A = 13.14 +
1.29 = 14.43 min = 0.2404 hr. Then (period delay)9A = (0.2404 hr) * 2,013 = 484 hr. Using car
and truck actual demand from Table 1, (delay cost)9A = (0.2404 hr) * [(1,773 cars) * ($12.00 /
car-hr) + (240 trucks) * ($30.00 / truck-hr)] = (0.2404) * [21,276 + 7,200] = (0.2404) * [28,476]
= $6,846, as shown in Col. 7 of Table 3.
6 DECREASES TO DESIGN DEMAND
6.1 DEMAND DECREASE = FUNCTION OF DELAY
CO3 provides for cars and trucks to divert to an alternate route or cancel their trips as vehicles
are delayed by backups and speed reductions at work zones. Eq. (3) describes the general equa-
17
tion for decrease (%) = % decrease in demand, which is demonstrated in Fig. 3 (a) for diverted
cars. Total decrease is the sum of diverted and cancelled cars and trucks,
U|
LM decrease a%f OP = R|S car decrease a%f awith no delayf
V
+ truck decrease a%f awith no delayf |W
Nawith no delayfQ |T
R|
U|
LR
cars awith no delay f + U| O
P
%
a| % carsf * MM|S diverted
a
f
V| P | . . . . . . . . . . . . (25)
cancelled
cars
with
no
delay
a
f
|
W Q |V
NT
|
=S
||+a% trucksf * LMR| diverted trucks awith no delayf +U|a%fOP||
MNST| cancelled trucks awith no delayf W|V PQ|W
|T
U|
LMdecrease a% / minfOP = R|S car decrease awith delayf a% / minf
V
+ truck decrease awith delay f a% / min f |W
N (with delay) Q |T
R|
LMR| diverted cars awith delayf +U|F % I OP U| . . . . . . . . .
(26)
||a% carsf * MNS|T cancelled cars awith delayf V|WH min K PQ ||
=S
||+a% trucksf * LMR| diverted trucks awith delayf +U|F % I OPV||
MN|TS cancelled trucks awith delayf V|WH min K PQ|W
|T
Magnitudes of the decrease (%) variables are functions of users’ knowledge of conditions (including diversion routes), users’ access to diversion routes, and delays encountered by users. Decrease (Vph) = decrease in design demand is a function of decrease (%) and design demand, as
shown in Eq. (4), from which demand = actual demand is calculated by Eq. (5).
6.2 THRESHOLD CAPACITY FOR DEMAND DECREASE
If (1) capacity is the same for all periods during construction or if (2) all capacities during construction have the same demand decrease value, then we input only threshold values for demand
decreases, as shown in Fig. 8 (a). This shows threshold capacity for decreases to design demand =
1,400 Vph. Fig. 3 (a) shows the relationship between diverted cars and delay for the threshold
capacity. To demonstrate the CO3 model, we first start with (car design demand)9A = (car design
demand % = 90%) * [(design demand)9A = 2,584] = 2,326 Vph. For capacity9A = threshold capacity for demand decrease = 1,400 Vph, we input diverted car values from Fig. 8 (a) and the
(average delay)9A = 14.43 min into Eq. (3),
a f
a
f
diverted cars % 9A = diverted cars with no delay = 3%
a
f
a
f
+ diverted cars with delay = 1% / min * delay 9A = 14.43 min
= 3% + 14.43% = 17.43%
Therefore, (diverted cars)9A = (17.43%) * (2,326) = 406 V. Similarly, (cancelled cars)9A =
[2.0% + (0.3%)(14.43 min) = 2.0 + 4.33 = 6.33%] * (2,326) = 147 V. From this, (car de-
18
crease)9A = 406 + 147 = 553 Vph, and (actual car demand)9A = 2,326 – 553 = 1,773 Vph, as
shown in Col 6, Table 1.
Similarly, using values from Fig. 8 (a) for total traffic, from Eq. (25) [decrease (%) (with no
delay)] = (90 %) * (3 % + 2 %) + (10%) * (0 + 0) = 4.5 % and from Eq. (26) [decrease (%/min)
(with delay) ] = (90 %) * (1 %/min + 0.3 %/min) + (10 %) * (0.5 %/min + 0) = 1.22 %/min.
From Eq. (3), total traffic [decrease (%)]9A = 4.5 % + (1.22 %/min) * (14.43 min) = 4.5 + 17.6 =
22.1 %. From Eq. (4), traffic (decrease)9A = 22.1 % * 2,584 V = 571 V, and (actual demand)9A
= 2,584 – 571 = 2,013 V as shown in Col. 8 of Table 1.
6.3 RANGE CAPACITY FOR DEMAND DECREASE
In parallel with Sec. 4.4 above, we sometimes need range inputs to describe vehicle decreases
at one or more capacities below the threshold capacity for demand decrease. An example of such
range input is shown in Fig. 8 (b). Fig. 3 (b) shows range diverted cars and its relationship to
threshold diverted cars. Range decreases and actual demand follow Eq. (3) and (5) in parallel with
threshold decrease and actual demand calculations.
Also in parallel with speed delay, when there is range input for demand decrease, as well as
threshold input, CO3 calculates values of any demand decrease value at capacity = C < threshold
capacity using Eq. (14). The dashed lines in Fig. 3 (b) show interpolated values for capacity C2,
where (threshold capacity) > C2 > (range capacity) and extrapolated values for capacity C1<
(range capacity). Thus for our example we can model any capacities below 1400 Vph. For example, using Eq. (14) we can calculate [diverted cars (with no delay) for capacity = 900 Vph] = 3%
+ (6% – 3%) * [(1,400 – 900) / (1,400 – 1,000)] = 3% + (3%) * (1.25) = 6.75%.
DECREASE TO DEMAND INPUT
threshold
DECREASE TO DEMAND INPUT
threshold
range
capacity for decreases to design
demand (V/hr)
1400
range
capacity for decreases to design
demand (V/hr)
1400
1000
canceled cars (with no delay) (%)
2.0%
canceled cars (with no delay) (%)
2.0%
4.0%
0.3%
0.3%
3.0%
6.0%
canceled trucks (with no delay) (%)
canceled cars (with delay) (%/min)
canceled trucks (with no delay) (%)
0.3%
canceled cars (with delay) (%/min)
canceled trucks (with delay) (%/min)
diverted cars (with no delay) (%)
canceled trucks (with delay) (%/min)
3.0%
diverted cars (with no delay) (%)
diverted trucks (with no delay) (%)
diverted trucks (with no delay) (%)
diverted cars (with delay) (%/min)
1.0%
diverted cars (with delay) (%/min)
1.0%
1.0%
diverted trucks (with delay) (%/min)
0.5%
diverted trucks (with delay) (%/min)
0.5%
0.5%
(a) With threshold capacity
(b) With threshold and range capacity
Fig. 8 Input for Decrease to Demand
7 DELAY, DEMAND DECREASE, AND ACTUAL DEMAND
7.1 GENERAL
Speed delay and backup delay are functions of actual demand, which is a function of traffic decrease, and traffic decrease is a function of speed delay and backup delay. This circular dependency requires an iterative solution for work zone delay and actual demand for many scenarios.
19
CO3 iterative solutions calculate and test successive trial values of actual demand and work zone
delay until they converge to an acceptable solution.
However, speed delay occurs only when capacity ≤ (threshold capacity for speed delay). Traffic decreases = diversions + cancellations occur only when capacity ≤ (threshold capacity for demand decrease). Otherwise, (actual demand) = (design demand). Backup delay occurs only when
(actual demand) > capacity. Therefore, we can have speed delay without backup delay or traffic
decreases. We can have backup and backup delay without speed delay or traffic decreases. We
can have traffic decreases due to [decreases (with no delay)] when there is no speed delay or
backup delay. Because of constraints and relationships such as these, we can directly calculate
work zone delay, decreases, and actual design for several scenarios without an iterative solution.
For example, Table 1 shows capacity8A = 3,400 Vph, which is larger than (1) (design demand)8A, (2) (threshold capacity for speed delay), and (3) (threshold capacity for demand decrease). Considering that (start-of-period backup)8A = 0, then from the above we know that in
period 8A there will be no backup, no speed delay, and no demand decrease.
7.2 COMMON CASE
The most common case of backup, speed delay, and demand decrease is (1) where a period’s
[(demand) + (start-of-period backup)] > capacity, and (2) where capacity is constant during the
time all the period’s vehicles enter the zone. We call periods that exhibit these two conditions the
Common Case. The hatched area in Fig. 9 and both hatched areas in Fig. 6 satisfy the Common
Case. These conditions are also met for demands of periods 7, 10, and 11 in Fig. 7. Periods 8 and
9 are not Common Case, because each demands enters the work zone when there is more than
one capacity. For example, demand8 enters the work zone with capacities from 8 to 12, and capacity8 ≠ capacity9 = capacity10 ≠ capacity11. Period 12 is not Common Case, because demand12 + (start-of-period backup)12 < capacity12.
Similarly, in our example, Table 1 shows demand9A = 2,013 Vph and capacity9A = 1,400
Vph. From this we determine that demand9A + (start-of-period backup) 9A = 2,013 + 0 = 2,013
> capacity9A = 1,400 Vph. This satisfies Common Case condition 1. We calculate (end-of-period
backup)9A = 2,013 – 1,400 = 613 V, as is also shown in Col. 2 of Table 3. We also see capacity10A = 1,400 Vph > (end-of-period backup)9A = 613, so the 613 V of demand9A remaining at
10A enters the work zone in period 10A. Therefore, all of demand9A enters the work zone under
the same capacity9A,10A = 1,400 Vph, which satisfies Common Case condition 2. Therefore, period 9A is Common Case, as are 10A and 11A.
However, 12P is not Common Case. From Eq. (20) we calculate Beop12P = 271 + 1,323 –
1,400 = 194 V. Thus the first 1,323 – 194 = 1,129 V of demand12P enters the zone at capacity12P = 1,400 Vph and the last 194 V of demand12P enters the zone at capacity1P = 3,400 Vph
≠ 1,400 Vph. Therefore, Common Case condition 2 is not satisfied, and period 12P is not Common Case.
20
capacity
deman d
backup delay
capacity
backup
B eop 1
slope = capacity
time = 1
time = 1
Fig. 9 Constant Capacity for Period’s Demand
Each of the hatched areas of Fig. 6, Fig. 7, and Fig. 9 shows total backup delay. For the Common Case periods identified above, the hatched area is a rhomboid, whose area is
demand * demand + 2 * backup − capacity
. . . . . . . . . . . . . . . . . . . . (27)
total backup delay hr =
2 * capacity
a f a
fa
a
f
f
from which we can calculate average backup delay per vehicle,
total backup delay * 60 min / hr
backup delay (min / V) =
demand
. . . . . . . . . . . . . . . . . . . . . . . . . . . (28)
30 * demand + 2 * backup − capacity
=
capacity
a
fa
a
a
f
f
f
We calculate demand = (actual demand) by combining Eq. (28) with Eq. (2), (3), and (5) to
produce
LM R % decrease awithout delayf
U|OP
|
MM | F % decrease awith delayf
PP
|
I
|
JJ |VP
demand = adesign demand f * M1 − S G Laspeed delayf
O
G
MM ||+ G * M demand + 2 * backup − capacity PJ ||PP . . . . . . . . (29)
PPJJ |P
MN |T GGH MMN+ 30
capacity
QK WQ
which resolves to a direct solution for demand,
LMmadesign demandf * acapacityfr
OP
MM R|1 − % decrease awithout delayf − % decrease awith delayf U|PP
MM* S*Laspeed delayf + 60 * F backup I − 30O
V|PP
|
G
J
MN
PQ
H
capacity K
M
WPQ . . . . . . . . . . . . . . (30)
T
N
demand =
capacity + 30 * adesign demand f * % decrease awith delayf
21
We can calculate demand9A by inserting [speed delay (when D = C)] = 1.29 min calculated in
Sec. 4.3.3, [decrease (with no delay)] = 4.5 % and [decrease (with delay)] = 1.22 %/min calculated in Sec. 6.2, (design demand)9A = 2,584 Vph and capacity9A-1P = 1,400 Vph from Table 1,
and (start-of-period backup)9A = (end-of-period backup)8A = 0 from Table 3 into Eq. (30),
LMla2,584f * a1,400fq
OP
U|P
MM R|1 − 0.045 − 0.0122
MM* S|*LMa1.29f + 60 * FG 0 IJ − 30OPV|PPP
H 1,400 K QWQ = 2,013 Vph
TN
ademandf9A = N 1,400
+ 30 * a2,584f * a0.0122f
as shown in Col. 8 of Table 1. From Eq. (28), (backup delay)9A = 30 * (2,013 + 2 * 0 – 1,400) /
(1,400) = 13.14 min, as shown in Col 2 of Table 3. By Eq. (2), (work zone delay)9A = 1.29 +
13.14 = 14.43 min, as shown in Col 5 of Table 3.
Similarly, for period 10A, which has (start-of-period backup)10A = (end-of-period backup)9A
= 613 V from Table 3,
LMla2,176f * a1,400fq
OP
U|P
MM R|1 − 0.045 − 0.0122
MM* S|*LMa1.29f + 60 * FG 613 IJ − 30OPV|PPP
H 1,400 K QWQ = 1,366 Vph
TN
ademandf10A = N 1,400
+ 30 * a2,176f * a0.0122f
as shown in Col. 8 of Table 1. (backup delay)10A = 30 * (1,366 + 2 * 613 – 1,400) / (1,400) =
25.55 min, as shown in Col 3 of Table 3. By Eq. (2), (work zone delay)10A = 1.29 + 25.54 =
26.83 min, as shown in Col 5 of Table 3.
7.3 CONVERGING ON ACTUAL DEMAND AND DELAY
As stated in Sec. 7.1 above, CO3 solves iteratively for actual demand and work zone delay
when a direct solution is not possible. Fig. 11 shows the iterative steps used to calculate demand
and vehicle delay for a period. In summary, a trial demand = demandtrial is calculated as an
estimate of actual demand. A trial delay = delaytrial is calculated from demandtrial, a new demand
= demandnew is calculated from delaytrial, and a new delay = delaynew is calculated from
demandnew. Demandnew is then compared to demandtrial, and delaynew is compared to
delaytrial. If either new value differs significantly from its respective trial value, another
demandtrial is calculated, and the steps are repeated until new values are not significantly different
from the trial values on which they are based. This section describes the method of calculating
trial demands such that the overall iterative process converges efficiently on accurate values of
actual demand and vehicle delay.
Demandtrial,1 for the first iteration j =1 is calculated based on Eq. (30). Demandtrial,2 for the
second iteration, j = 2, is the average of demandtrial,1 and demandnew,1, in accord with
22
demand trial, j +1 =
demand trial, j + demand new, j
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (31)
2
This averaging formula is generally not applied to succeeding iterations, because it does not
always converge. High values of [decrease (with delay)] can cause repeated flip-flopping between
two alternative sets of trial and new values. Fig. 10 demonstrates such a sensitive case. In Fig. 10
the vertical scale indicates the magnitude of demand, and the arrows indicate that demandtrial, j
produced demandnew,j, and demandtrial, j–1 produced demandnew, j–1. When demand and delay
are related, Fig. 10 illustrates that the lower the value of demandtrial, the higher the value of demandnew. Therefore, the two arrows cross. To provide dependable and efficient convergence,
CO3 sets demandtrial, j+1 = the value of demandtrial with which to start the next iteration, j+1, is
the value of demand at which the two arrows intersect,
demand trial, j * demand new, j -1 − demand trial, j -1 * demand new, j
. . . . . . . (32)
demand trial, j +1 =
demand trial, j + demand new, j -1 − demand trial, j -1 − demand new, j
new j -1
new j - new j-1
trial j - trial j-1
de mand
trial j
trial j +1
trial j -1
new j
Fig. 10 Calculation of New Trial Actual Demand [change i to j above]
For each iteration j, we consider a new value is significantly different from its corresponding
trial value if its absolute difference exceeds the allowable difference we select,
∆ x , j = xnew, j − xtrial, j −1 > ∆ x allowable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (33)
where x = “demand” or “delay”.
To test significance of differences of trial and new values of demand and delay, we input into
CO3 values for ∆demand·allowable = maximum allowable difference in demand values and
∆delay·allowable = maximum allowable difference in work zone delay values. For our example
problem, we set ∆demand·allowable = (0.1 %) * (maximum design demand) = 0.001 * [(design
demand)8A = 3,314 Vph] = 7 Vph and ∆delay·allowable = 0.1 min = 6 sec.
For iteration j = 1 for the 12P values of our example, CO3 uses Eq. (30) to calculate
demandtrial,1 = 1,315.75 Vph. From this CO3 calculates delaytrial,1 = 0.17643 hr = 10.586 min.
From this CO3 calculates demandnew,1 = 1,325.88 Vph, from which delaynew,1 = 0.17936 hr =
10.762 min. The absolute values of differences between trial and new values of demand and delay
are calculated using Eq. (33), ∆demand,1 = 1,315.75 – 1,325.88 = 10.13 Vph >
23
∆demand·allowable = 7 Vph and ∆delay,1 = 10.586 – 10.762 = 0.176 min > ∆delay·allowable
= 0.1 min. We must perform another iteration, because ∆demand,1 exceeds ∆demand·allowable
and also because ∆delay,1 exceeds ∆delay·allowable.
We use Eq. (31) for iteration j = 2 to calculate demandtrial,2 = (1,315.75 + 1,325.88) / 2 =
1,320.81 Vph. From this CO3 calculates delaytrial,2 = 0.17789 hr = 10.763 min, from which
demandnew,2 = 1,324.17 Vph, from which delaynew,2 = 0.17886 hr = 10.732 min. The absolute
values of differences between trial and new values of demand and delay are ∆demand,2 =
1,320.81 – 1,324.17 = 3.36 Vph < ∆demand·allowable = 7 Vph and ∆delay,2 = 10.763 –
10.732 = 0.031 min < ∆delay·allowable = 0.1 min. We need not perform another iteration,
because differences between trial and new values of both demand and delay are within their
allowables and are therefore not significant. If either ∆demand,2 or ∆delay,2 exceeded its
allowable, CO3 would have used Eq. (32) to calculate demandtrial,3,
1,320.81*1,325.88 − 1,315.75 * 1,324.17
= 1,323.32 Vph
1,320.81 + 1,325.88 − 1,315.75 − 1,324.17
For its output, CO3 uses the latest value calculated. Therefore, delay12P = delaynew,2 =
0.17886 hr = 10.73 min as shown in Col. 5 of Table 3. From this CO3 calculates demand12P =
1,323 Vph, as shown in Col. 8 of Table 1.
demand trial,3 =
7.4 OVERALL VIEW OF CONVERGENCE
As stated above, demand decrease, actual demand, and delay cannot always be calculated directly, because demand decrease is dependent on average delay per vehicle, and average delay per
vehicle is dependent on actual demand, which is dependent on demand decrease. CO3 uses the
following iterative method where necessary to calculate these variables for each period, which is
diagrammed in Fig. 11.
1. Estimate trial value of actual demand = demandtrial for period i, based on input data:
2. Calculate trial values of delay: (backup delay)trial and (speed delay)trial, based on
demandtrial for i:
3. Calculate a trial average delay per vehicle = delaytrial = (backup delay)trial and (speed
delay)trial.
4. Calculate new estimate of decrease in demand = decreasenew based on delaytrial.
5. Calculate new estimate of actual demand = demandnew based on decreasenew.
6. Calculate new estimates of delay: (backup delay)new and (speed delay)ndw, based on
demandnew.
7. Calculate new estimate of average delay per vehicle = delaynew = (backup delay)new
and (speed delay)new.
8. Determine whether the new estimates of demand or delay are significantly different
from the trial estimates. I.e., is demandnew significantly different from demandtrial or
is delaynew significantly different from delaytrial?
24
•
•
If either new estimate is significantly different, calculate a trial value of demand =
demandtrial based on the current and immediate-past values of demandtrial and
demandnew, and repeat steps 1-8 for the new demandtrial.
If neither new estimate is significantly different, then set the values of actual demand, average delay, and other demand and delay variables for period i to the
new estimated values, and perform steps 1-8 above for time period (i + 1) = period following i.
8 USER COST OF DEMAND DECREASES
8.1 GENERAL
Decreases in design demand consist of cancellations and diversions of cars and trucks. User
cost per vehicle of decrease is
af
a f . . . . . . . . . . . . . . . . . . . . . . . . . . . (34)
decrease user cost $ = user cost per decreased vehicle $ / V
af
* demand decrease V
8.2 USER COST OF CANCELLATIONS
User cost per cancellation is input, as shown in Fig. 5 (a). For example, in Fig. 5 (a) (user cost
per car cancellation) = $4.00 /V and (user cost per truck cancellation) = $10.00 / V. Inputting
canceled car values from Fig. 8 into Eq. (4) for period 9A, (canceled cars)9A = [(2 %) + (0.3
%/min) * (14.43 min)] * [(90 %) * (2,584)] = [6.33 %] * [2,326] = 147 V. From Eq. (34), (canceled car user cost)9A = ($4.00 / V) * (147 V) = $588, (canceled truck user cost)9A = 0, because
Fig. 8 shows that no trucks cancel due to delay.
8.3 USER COST OF DIVERSIONS
For calculating diversion user cost, diversion travel times are calculated by Eq. (6), into which
we input values from Fig. 5 (a),
diversion normal distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
diversion normal time =
(35)
diversion normal speed
=
4 mi
= 0.05714 hr = 3.43 min.
70 mph
diversion method time =
=
diversion method distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(36)
diversion method speed
10 mi
= 0.22222 hr = 13.33 min.
45 mph
25
F o r p e r iod i: Esti m a t e d e m a n d tr ial
vehicles
A r e a = (total ba ckup delay) trial
(speed delay ) trial
d e m and trial
capacity i+1
d e m and trial
actual
demand
B sop.i
i
i+1
i+2
c a p a c ity i
time (hr)
Calculate ( s p e e d d e l a y ) trial = [s p e e d d e l a y ( w h e n D ~ 0 ) ]
(Vph)
+ [spe e d d e l a y ( w h e n D = C ) − s p e e d d e l a y ( w h e n D ~ 0)]
Calculate ( b a c k u p d e l a y ) trial
*(d e m a n d tria l / c a p a city ) E m s
= [ (tot al bac k u p dela y) trial ] / (de m a n d trial )
C a l c u late delay trial = (back up delay) trial + (s peed dela y) trial
(car d ecrea se %) ne w
cars
(truck decre ase % ) new
trucks
delay trial
de lay ( m in.)
Calculate decrease n e w = ( c a r d e c r e a s e % ) n e w )*(des i g n d e m a n d ) c a r
+ (truck de c r e a s e % ) n e w )*(d e s i g n d e m a n d ) truc k
Calculate demandnew = (design demand)*(1 − decreasenew )
vehicles
A r e a = (total backup d elay) n e w
(speed delay ) ne w
(s pe ed d elay) trial
de m a n d ne w
de m and trial
capacity i+1
dem and trial
actual
dem a nd n e w
B sop.i
i
i+1
i+2
c a p a c ity i
time (hr)
Calculate ( s p e e d d e l a y ) n e w = [ s p e e d d e l a y ( w h e n D ~ 0 ) ]
demand
(Vph)
+ [sp e e d d e l a y ( w h e n D = C ) − s p e e d d e l a y ( w h e n D ~ 0)]
Calculate ( b a c k u p d e l a y ) n e w
*(d e m a n d n e w / ca paci ty ) E m s
= [ (tot al ba c k u p dela y) n e w ] / ( d e m a n d n e w )
C a l c ula te d elay n e w = (bac kup delay) n e w + (spe ed dela y) new
RICarr FLOW3.CVS
Test f o r C h a n g e : If [abs(del a y n e w − delay tria l ) ≥ e del o r a b s ( d e m a n d n e w − d e m a n d tria l ) ≥ e D ]
t h e n esti m a t e n e w d e m a n d trial b a s e d o n old d e m a n d trial a n d o l d d e m a n d n e w , a n d R e c y c l e f o r i,
else s e t ( a c t u a l d e m a n d ) = d e m a n d n e w , de l a y = d e lay n e w , D o c a l culatio n s f o r p e r i o d i+ 1
Fig. 11 Steps in Calculating Actual Demand and Delay for Period i.
26
By Eq. (9), (diversion delay) = 13.33 – 3.43 = 9.90 min, as shown in Fig. 5 (a). User cost per
diversion is calculated using Eq. (15). For example,
9.90 min
car diversion user cost =
* $12.00 / hr + 10 mi - 4 mi * $0.30 / mi
60 min / hr
= 0.165 * (12.00) + 6 * (0.30) = $1.98 + $1.80 = $3.78 / V
a
f a
fa
f
as shown in Fig. 5 (a). Similarly, truck diversion cost uses the same (diversion delay) = 9.90 min
and distances, but different costs per hour and per mile to produce (truck diversion user cost) =
0.165 * (30.00) + 6 * (1.00) = 4.95 + 6.00 = $10.95 / V.
Inputting diverted car values from Fig. 8 into Eq. (4) for period 9A, (diverted cars)9A = [(3
%) + (1 %/min) * (14.43 min)] * [(90 %) * (2,584)] = [17.43 %] * [2,326] = 405 V. From Eq.
(34), (diverted car user cost)9A = ($3.78 / V) * (405 V) = $1,531. Also, (diverted trucks)9A =
[(0 %) + (0.5 %/min) * (14.43 min)] * [(10 %) * (2,584)] = [7.22 %] * [258] = 19 V, and (diverted truck user cost)9A = ($10.95 / V) * (19 V) = $208. Adding these costs produces
decrease cost = canceled costs + diversion cost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (37)
and (decrease cost)9A = 588 + 0 + 1,531 + 208 = $2,327 ≅ $2,325 as shown in Col. 8 of Table 3,
within round-off errors.
9 SUMMARY AND CONCLUSIONS
Design demand is the normal demand when there is no work zone. Capacity of a work zone is
less than normal, and maximum speed in a work zone is less than normal. Demand is closer to
work zone capacity than normal, because capacity is less than normal, which reduces work zone
speed further. Reduced work zone speed produces speed delay. Backup occurs in the amount by
which work zone demand exceeds capacity. Backup delay occurs as vehicles in the backup queue
wait to enter the work zone. Work zone delay is the sum of speed delay and backup delay.
When work zone delay causes traffic to divert or cancel trips, actual demand through the work
zone becomes less than design demand. Traffic that diverts has its own delay due to the additional
distance and lower speeds of alternate routes. Work zone delay, diverted and cancelled traffic,
and work zone demand share a circular dependency.
As CO3 users, we estimate traffic distances and speeds under normal and work zone conditions
and along alternate routes. We also estimate the dependency between work zone delay and the
fractions of car and truck traffic that divert and cancel trips. From this input, CO3 calculates delays, diversions, cancellations, and actual demand for each period. CO3 is unique in resolving the
interdependency between work zone delay and work zone demand.
Based on input of car and truck user cost per hour and mile of travel and per cancellation, CO3
calculates road user cost for work zone demand, diversions, and cancellations. For each period
CO3 provides detailed output of car and truck diversions, cancellations, and work zone demand;
backup, backup delay, and speed delay; and user costs of delay and traffic decreases. Output provides a detailed view of construction impact for each period and each day. We can input alternative lane closure patterns and calculate and compare the impact of alternative traffic maintenance
plans. By integrating traffic impacts and user costs with construction costs of alternative plans
(not shown here), we can select a preferred method of performing construction and maintaining
traffic that is an acceptable balance of public cost and benefit.
27
APPENDIX 1
REFERENCES
“Alternative surveillance concepts and methods for freeway incident management.” (1978).
FHWA-RD-77-58/63, Federal Highway Administration, U.S. Dept. of Transportation, Washington, D.C., March.
Carr, R. I. (1997). “Construction congestion cost – CO3 user manual.” Department Of Civil
and Environmental Engineering, University of Michigan, Ann Arbor, 252 pp.
“Contract management techniques for improving construction quality.” (1997). FHWA-RD97-067, U.S. Dept of Transportation, Washington, D.C.
Cragg, C. A., and Demetsky, M. J. (1995). “Simulation analysis of route diversion strategies
for freeway incident management.” VTRC 95-R11, Virginia Transportation Research Council,
Charlottesville, Virginia, February.
“Federal Highway Administration 1998 national strategic plan.” (1997). Federal Highway Administration, U.S. Dept of Transportation, Washington, D.C.
Herbsman, Z. J., Chen, W. T., and Epstein, W. C. (1995). “Time is money: innovative contracting methods in highway construction.” Journal of Construction Engineering and Management, ASCE 121(3), 273–281.
Herbsman, Z. J. (1995). “A+B bidding method—hidden success story for highway construction.” Journal of Construction Engineering and Management, ASCE 121(4), 430–437.
“Incentive/disincentive (I/D) for early completion.” (1989). FHWA Technical Advisory
T5080.10, Federal Highway Administration, U.S. Dept. of Transportation, Washington, D.C.,
February 8.
Jaraiedi, M., Plummer, R. W., and Aber, M. S. (1995). “Incentive/disincentive guidelines for
highway construction contracts.” Journal of Construction Engineering and Management, ASCE,
121(1), 112-120.
Krammes, R. A., Ullman, G. L., Memmott, J. L., and Dudek, C. L. (1993). “User’s manual for
QUEWZ-92.” Research Report (and Software) No. FHWA/TX-92/1108-7, Texas Transportation
Institute, College Station, Texas, January.
Newman, R. B. and Hejl, F. D. (1984). “Contracting practices and payment procedures.”
Transportation Research Record, 986, 50-59.
Reiss, R. A. and Dunn, W. M., Jr. (1991). “Freeway incident management handbook.” FHWASA-91-056, Federal Highway Administration, U.S. Dept. of Transportation, Washington, D.C.
Riley, O. (1984). “An overview of incentives and disincentives.” (1984). Transportation Research Record, 986, 57-64.
“Traffic/congestion management during highway construction.” (1992). Report of the Secretary of Transportation to the United States Congress, U.S. Dept. of Transportation, Washington,
D.C., September.
28
APPENDIX 2
A.1
PERIOD DELAY EQUATIONS
Class IOO: C > D; Bsop = 0
IOO
C
D
time = 1
A.2
Class IOR: D > C; Bsop = 0
IOR
B eop
D
C
time = 1
abackup delayfi = D2- C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (38)
aspeed delayfi = C * speed delay a D = Cf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (39)
A.3
Class IBD: Bsop + D – C; ≥ D
IBD
D
B eop
B sop
C
time = 1
abackup delayfi = D2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (40)
29
A.4
Class IBO: Bsop + D – C < 0
IBO
D
DB
C
B sop
time = 1
DB =
D * Bsop
C−D
2
D * Bsop
DB Bsop
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (41)
total backup delay =
*
=
a f
total speed delay = DB * speed delay a D = C f
+ b D − DB g * speed delay a0 < D < C f
2
A.5
2C C − D
C
. . . . . . . . . . . . . . . . . . . . . . . . . . . (42)
Class IBR: 0 < (Bsop + D – C) < D
IBR
B eop
D
C
B sop
time = 1
2O
LM d
i
PP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (43)
a
f M
MN
PQ
aspeed delayfi = dC - Bsop i * speed delay a D = Cf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (44)
C − Bsop
1
backup delay i = D −
2
C
30
A.6
Class FDD: (Bsop – C) ≥ D
FDD
B eop
D
B sop
C
time = 1
abackup delayfi = D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (45)
A.7
Class FDR: 0 < (Bsop – C) < D ≥ Bsop
FDR
B eop
D
B sop
Dz
C
time = 1
Dz = D + C − Bsop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (46)
abackup delayfi = D − 2D*zC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (47)
aspeed delayfi = Dz * speed delay a D = Cf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (48)
2
31
A.8
Class FDO: D ≤ Bsop ≤ C
FDO
D
C
B sop
time = 1
abackup delayfi = CD * FH Bsop − D2 IK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (49)
aspeed delayfi = D * speed delay a D = Cf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (50)
A.9
Class FRR: C < Bsop < D
FRR
B eop
B sop
C
time = 1
abackup delayfi = Bsop − C2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (51)
aspeed delayfi = C * speed delay a D = Cf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (52)
A.10
Class FRO: C < Bsop < D
FRO
B sop
C
time = 1
2
Bsop
abackup delayfi = 2 * C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (53)
aspeed delayfi = Bsop * speed delay a D = Cf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (54)
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