CE 361 Introduction to Transportation Engineering
Out: Fri. 5 September 2008
Homework 2 (HW 2) Solutions
Due: Mon. 15 September 2008
TRAFFIC FLOW DATA


1-3 CE361 students. If 2-3, signatures required.
ID each problem by its number and name. State your methodology.
1. (10 points) Mr. Posey’s letter. The
largest standard deviation in FTE Table 6.15 is S =
4.378. Use z = 1.96 unless otherwise told. In FTE Table 2.7, U = 1.04 for the
85th percentile speed.
FTE (2.4)
S 2 z 2  2  U 2  4.378 21.96 2  2  1.04 2 




=
N
2
2
2E
2(1)
= 113.45 = 114 vehicles.
2. Occupancy and density from loop detectors. Effective length of 7.0 feet, actuated for 4.8 percent
of one-hour. 632 vehicles detected, mean speed 63.2 mph, mean length 18.24 feet.
A. (10 points) Convert the apparent occupancy rate to the actual occupancy rate.
L V  EL 
18 .24  7.0* 3600 = 0.2732 second
(2.6) t (P)  distance 
seconds
rate
S * 5280 / 3600 
63 .2 * 5280
(2.7)
t ' ( P)  t ( P) 
EL * 3600
7.0 * 3600
seconds 0.2732 
 0.2732  0.0755
S * 5280
63 .2 * 5280
= 0.1977 second
(2.8) Actual Occupancy Oact = q * t’(P) = (632 * 0.1977)/3600 = 0.035 or 3.5%.
B. (10 points) What was the density (vpm) on the roadway during the occupancy measurements?
Use any one of three equations:
O
* 5280 ft / mi 0.035 * 5280 ft / mi

(2.9) D  act
= 9.99 veh/mi
LV
18 .24 ft
Calculate Oapp as in FTE Example 2.10A, then use (2.10)
D
O app * 5280 ft / mi
(2.12)
L V  EL
D

q 632 vph

S 63 .2 mph
((632 * 0.2732 ) / 3600 ) * 5280 ft / mi
(18 .24  7.0) ft
= 10.00 veh/mi
= 10.03 veh/mi
CE361 08HW2 Solutions
-2-
Fall 2008
3. Speed-Flow-Density relationships. Do FTE Exercise 2.40, using the data emailed to you.
A. (15 points) Speed-density plot.
Obsvn
Flow
(vph)
Speed
(kph)
Occup
(%)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1260
1620
1440
1620
1440
1800
1620
2160
1260
1800
1800
2160
1440
1800
2160
1800
1800
1800
1980
1260
1800
1440
1800
1620
1260
900
1260
1440
1800
1980
62
59
55
56
59
54
43
36
18
32
35
33
20
25
36
22
33
37
35
29
20
25
36
24
18
12
17
25
33
35
10
13
14
15
12
18
19
33
37
27
26
33
38
35
28
39
28
24
27
24
41
29
23
33
32
47
32
30
27
28
Density
(vpk)
20.3
27.5
26.2
28.9
24.4
33.3
37.7
60.0
70.0
56.3
51.4
65.5
72.0
72.0
60.0
81.8
54.5
48.6
56.6
43.4
90.0
57.6
50.0
67.5
70.0
75.0
74.1
57.6
54.5
56.6
 Using Equation 2.12, calculate D = q/S
values for the Density column. Figure 3A
is the plot of 30 data points of S vs. D.
 Fit a straight line through the data points.
Free-flow speed Sf = 73.362 km/hr.
Jam density Dj = 73.362/0.7161 = 102.45
veh/km.
 The fitted line is consistent with the
plotted data points.
CE361 08HW2 Solutions
B. (15 points) Flow-density
plot.
 See Figure 3B.
 Use Equation 2.19 to
convert D = 0, 5, 10, …,
95, 100, 102.45 veh/km
into corresponding
“model” values of q.
 Sample calculation for
D = 30 veh/km:

D 2 
q  Sf  D 
.

Dj 



30 2 
q  73 .362  30 
.

102 .45 


q = 1556 veh/hr
 Plot the model values
as a smooth parabolic
curve.
 The smooth curve is
not inconsistent with the
plotted points.
C. (15 points) Speed-flow plot.
 See Figure 3C.
 Use Equation 2.18 to
convert S = 0, 10, 20,
…, 70, 73.36 kph into
corresponding
“model” values of q.
 Sample calculation for
S = 30 km/hr:

S 2 
q  D j S 

Sf 



S 2 
q  102 .45  S 

73 .362 


q = 1817 veh/hr
 The smooth curve is
not inconsistent with
the plotted points.
-3-
Fall 2008
CE361 08HW2 Solutions
-4-
Fall 2008
4. The road to Shreveport. 343 miles = 552 km) from New Orleans to Shreveport. Normally 5 hours
and 21 minutes; before Hurricane Gustav, 25 hours. 2 lanes each direction on a 10-kilometer
segment. Metric distances.


D 
D 2 
S 2 

(2.13) S  127 .23 1 
; (2.18) q  82 .32 S 
 ; (2.19) q  127 .23  D 

 127 .23 
82 .32 
 82 .32 




A. (10 points) Normal mean speed of the trip to Shreveport? What is the corresponding traffic
density and flow rate on the 10-kilometer segment, according to the Greenshields equations?
552 km
552
= 103.18 km/hr.

5.35
21 

 5  hr
60 

FTE (2.13) 103 .18  127 .23 1 

D  103 .18
D
1 
;
82 .32  127 .23
82 .32
; D = 15.56 veh/km.

103 .18 2 
= 1606 veh/hr
q  82 .32 103 .18 

127 .23 


B. (10 points) Mean speed of the pre-Gustav trip to Shreveport? Corresponding traffic density and
flow rate on the 10-kilometer segment, according to the Greenshields equations?
552 km 552

= 22.1 km/hr.
25 hr
25
FTE (2.18)
FTE (2.13)
D  22 .1
D

22 .1  127 .23 1 
1 
;
82 .32
 82 .32  127 .23
; D = 68.0 veh/km.

22 .12 
= 1503 veh/hr
q  82 .32 22 .1 

127 .23 


C. (5 points) Do all the values you calculated make sense? (Are any suspicious?)
 103.18 km/hr = 64.11 mph, which is a little low for an Interstate
FTE (2.18)


segment
under normal conditions, but it is not unreasonable.
A density 68.0 veh/km. equates to a vehicle every 1000 m/68 = 14.7 meters
or 48.2 feet. If an average vehicle is about 15 feet long, the D=68 veh/km
value represents a situation in which each vehicle (on the average) has a little
more than two vehicle lengths behind it until the next vehicle. This is
possible in an evacuation situation.
The flow rates of 1606 veh/hr and 1503 veh/hr are reasonable.