Solution
Exam #2
TTE 4004
Summer 2009
3:30 PM – 5:30PM, 7/15/09
1. At a signalized intersection, assume vehicles on a single-lane approach arrive at a uniform
rate of 720 vehicles per hour and the saturation flow rate (uniform) for the intersection is
1800 vehicles per hour. The cycle length is 120 sec with 60 sec (t = 0 – 60) of effective red
and 60 sec (t = 60 – 120) of effective green. Calculate
(1) Delay for a vehicle arriving at the intersection at t= 30 sec
(3 points)
(2) Average delay per vehicle?
(2 points)
Solution:
(1) Queue at t=30 sec
queue  t 
720
* 30  6
3600
When effective green starts, the time to clear 6 vehicles is
tg 
Therefore the delay d = (60 - 30) + 12 = 42 seconds
queue
6

 12
1800
u
3600
(2) Assume at time t, the queue is cleared
t   (t  60) 
720
1800
*t 
* (t  60)  0.3t  30  t  100
3600
3600
The total number of accumulated vehicles at t=100 sec
N  t 
720
*100  20
3600
60 * 20
 600 sec
2
The total delay:
d total 
Average delay:
d average 
d total d total
600


 25 sec/ veh
720
N cycle
C
*120
3600
2. What is the four-step transportation forecasting model? Please put the steps in a sequential
order.
(5 points)
Solution:
Trip Generation  Trip Distribution  Mode Choice  Trip Assignment
1
Solution
Exam #2
TTE 4004
Summer 2009
3:30 PM – 5:30PM, 7/15/09
3.For mode choice, a calibrated study resulted in the following utility function:
Uk = ak – 0.03X1 – 0.04X2 – 0.015X3 – 0.004x4
Where
X1 = access plus egress time, in min
X2 = waiting time, in min
X3 = line-haul time, in min
X4 = out-of-pocket cost, in cents
The trip-distribution forecast for a particular interchange was a target-year volume of Qij =
8000 person-trips per day. During the target trip-makers on this particular interchange will
have a choice between the private automobile (A) and a local bus system (B). The targetyear service attributes of the two competing modes have been estimated to be:
Attribute
Automobile
Local bus
X1
5
10
X2
0
10
X3
20
40
X4
350
150
aAuto = 0 and abus = – 0.20, Apply the Multinomial Logit (MNL) model to estimate
(1) The target-year market share of the two modes
(8 points)
(2) The resulting fare-box revenue of the bus system
(2 points)
Solution:
(1) The utility equation yields
U  A  0.03 * 5  0.04 * 0  0.015 * 20  0.004 * 350  1.85
U B  0.20  0.03*10  0.04 *10  0.015 * 40  0.004 *150  2.1
According to the Multinomial Logit model equation,
eU ( A)
e 1.85
0.157237


 0.5622  56.22%
U ( A)
U ( B)
1.85
2.1
e
e
e
e
0.157237  0.122456
eU ( A)
e 2.1
0.122456
p A  U ( A) U ( B )  1.55

 0.4378  43.78%
2.1
e
e
e
e
0.157237  0.122456
p A 
Therefore the market share of each mode is
QIJ  A  0.5622 * 8000  4,498trips / day
QIJ B   0.4378 * 8000  3,502trips / day
(2) The resulting fare-box revenue estimate of the bus system is
(3,502trips / day )($1.5 / trip )  $5,253 / day
2
Solution
Exam #2
3:30 PM – 5:30PM, 7/15/09
TTE 4004
Summer 2009
4. The probability that a vehicle will turn right at an intersection is known to be 0.25.
Assuming independence, calculate the probabilities of the following events:
(1) The ninth vehicle is not turning right.
(2 points)
(2) Exactly four out of ten vehicles will turn right.
(2 points)
(3) At least three out of ten vehicles will turn right.
(2 points)
(4) The first right-turning vehicle will be the fifth vehicle.
(2 points)
(5) The eighth vehicle will be the third to turn right.
(2 points)
Solution:
(1) The movement of each vehicle is a Bernoulli trail with p=0.25 and q=0.75
The probability that the ninth vehicle, and any other vehicle, will not turn right is q =
0.75
(2) According to the binomial distribution, the probability that four out of ten vehicles will
turn right is
px  
n!
10!
p x q nx 
0.254 0.756  0.146
x!(n  x)!
4!*6!
(3) The complement of event A: at least three out of ten is event B: Zero, or one or two out
of ten and
PA  PX  2  1  PB
For n=10
PA  1   p(0)  p(1)  p(2)  1  (0.0563  0.1877  0.2816)  0.4744
(4) This question may be answered by using either the geometric distribution or the negative
binomial (Pascal) distribution with k=1. Thus
p5  0.754 0.251  0.0791
(5) The negative binomial (Pascal) distribution provides the answer to the question of the
probability that the kth (i.e., third) left turner is the eighth vehicle, x, is the sequence
px  
( x  1)!
p k q x k
(k  1)!( x  k )!
For x=8 and k=3
p8 
(8  1)!
0.2530.755  0.0779
(3  1)!(8  3)!
3
Exam #2
Solution
3:30 PM – 5:30PM, 7/15/09
TTE 4004
Summer 2009
5.
(5 points)
6. A. Using an annual interest rate i = 6%, find the present worth of cash flows
(10 points)
Solution:
(a) P = 200(PWF, i, 10) + 100(PWF, i, 6) ( PWF ' , i, 14) =1689.511
(b) P = 200(PWF,i,10)+ 100(CAF,i,6)( PWF ' ,i,20)= 1689.511
4
Exam #2
Solution
3:30 PM – 5:30PM, 7/15/09
TTE 4004
Summer 2009
B.
(10 points)
Solution:
The 2nd, 5th and 6th equations are correct.
5
Solution
Exam #2
3:30 PM – 5:30PM, 7/15/09
7. List the standard procedure to identify and eliminate hazards.
TTE 4004
Summer 2009
(10 points)
Solutions:
8.
A. In one county, a 2-mile roadway segment has total 35 crashes in the past 3 years. The
average ADT volume for the segment at three year time period is 5,000 vehicles per
day. Based on the following formula, what is the crash rate per million involved
vehicles per mile?
(5 points)
R
1,000,000  A
365  T  V  L
R = crash rate per million involved vehicles per mile
A = total number of crashes on the roadway section
T = the time frame of the analysis (years);
V= the average ADT volume of the segment at three-year time period; and
L = the length of the selected roadway segment (miles).
Solution:
R
1,000,000 * A
1,000,000 * 35

 3.196
365 * T * V * L 365 * 3 * 5000 * 2
6
Solution
Exam #2
TTE 4004
Summer 2009
3:30 PM – 5:30PM, 7/15/09
B. Assume the crash rates in this county follow a normal distribution. If the average crash
rate per million involved vehicles per miles is 2.6 and the standard deviation s of crash
rates for the county is 0.4, what is the threshold value to determine the highest 5% of
crash rates in the county?
(5 points)
Solution:
For 95% z = 1.645
z
xi  x
s
xi  1.645s  x  xi  1.645 * 0.4  2.6  xi  3.258
The threshold value is 3.32
C. Based on the threshold value you compute from Question 7B, is the crash rate of the
roadway segment described in 7A within the top 5% in the County? Why? (5 points)
Solution:
No. From question 7A, crash rate is 3.196.
Since 3.196<3.258, it is not in the top5% in the county.
9. For evaluation of treatments at crash locations, please describe the meaning of Cross
Sectional Comparison study.
(5 points)
Solution:
Two sets of sites should be identified with similar control factors except for
the improvement countermeasures.
10. Please draw the national ITS architecture including subsystems and the communication
between the subsystems.
(10 Points)
Solution:
Wireline Communications
7
Solution
Exam #2
3:30 PM – 5:30PM, 7/15/09
11. What do the following ITS teams stand for?
TTE 4004
Summer 2009
(5 points)
a. ATMS
b. ATIS
c. APTS
d. ETTM
e. AHS
Solution:
a. ATMS – Advanced Traffic Management Systems
b. ATIS – Advanced Traveler Information Systems
c. APTS – Advanced Public Transportation Systems
d. ETTM – Electronic Toll & Traffic Management
e. AHS – Automatic Highways Systems
12. ITS is made up by 17 technology based systems. These systems can be divided into
Intelligent Infrastructure Systems and Intelligent Vehicle Systems. Which one does
Commercial Vehicle Operations (CVO) belong?
(5 points)
Solution:
CVO is belong to the Intelligent Infrastructure Systems.
8
Solution
Exam #2
TTE 4004
Summer 2009
3:30 PM – 5:30PM, 7/15/09
n
N
 xi

 
2
  2 
X
i 1
x
n
N
( xi   ) 2
N

i 1
s2 
N
( xi   ) 2
N

i 1
n

i 1
( xi  X ) 2
n 1
n

i 1
( xi  X ) 2
s  s2 
N
i
i 1
n 1
N
COV p ( X , Y ) 
 ( x   )( y  
i
i 1
x
i
y
)
N
n
COVs ( X , Y ) 
CV 
SD
X
d average 
d total d total

N cycle C

 ( x  X )( y  Y )
i 1
i
i
n 1
COV p ( X , Y )
 XY
r
COVs ( X , Y )
s X sY
PA  B   PA  PB   PA  B 
P A  B 
PB 
eU ( A )
p  A  U ( A ) U ( B )
e
e
P A | B  
eU ( B )
p B   U ( A ) U ( B )
e
e
PA  B   PA PB 
9
(Independent Events)
Solution
Exam #2
TTE 4004
Summer 2009
3:30 PM – 5:30PM, 7/15/09
px  
n!
p x q n x
x!n  x !
E X   np and V X   npq
px   q x 1 p
EX   p 1 and V X   qp 2
E X   1 p  0q  p and V X   pq  p1  p 

t x t
px  
e
p  xn   p  xm  
1
, n  m, for all n  
N
N 1
N 2 1
EX  
and V X  
2
12
x!
E  X   V  X   t
 1

f x    b  a
 0
for a  x  b
f t   e  t
otherwise
1
xa
F x   
dx 
ba
ba
a
t
x
E x  
f x  
r
for a  x  b
b  a 
ab
and V x  
2
12
1
 1 x   2
exp  (
) 
 2
 2 

F t    e t dt  1  e t

E t  
2


x  1! p k q xk
k  1!x  k !
EX   kp1 and V X   kqp2
px  
PF   q  1  PS   1  p
1

and V t  
z
for    x  
z
( t ) n   t
pn (t ) 
e
n!
10
1
2
x

xi  x
s
p0 (t )  e  t
Exam #2
Solution
3:30 PM – 5:30PM, 7/15/09
11
TTE 4004
Summer 2009