UNSW Sydney
SEMESTER 2 2017 EXAMINATIONS
CVEN9405: Urban Transport Planning Practice
1. TIME ALLOWED – 2 hours
2. READING TIME – 10 minutes
3. THIS EXAMINATION PAPER HAS 12 PAGES
4. TOTAL NUMBER OF QUESTIONS – 6 questions in two sections
5. ANSWER THE QUESTIONS OF EACH SECTION IN A SEPARATE ANSWER BOOKLET
6. TOTAL MARKS AVAILABLE – 100
7. MARKS AVAILABLE FOR EACH QUESTION ARE SHOWN IN THE EXAMINATION PAPER
7. ALL ANSWERS MUST BE WRITTEN IN INK. EXCEPT WHERE THEY ARE EXPRESSLY REQUIRED, PENCILS MAY BE
USED ONLY FOR DRAWING, SKETCHING OR GRAPHICAL WORK
8. THIS PAPER SHOULD BE RETURNED
Section 1
Question 1:
[10 Points]
Determine which one of the following statements is true and which one is false.
a) When converting number of trips from person-trip to vehicle-trips, the
vehicle occupancy rate should be rounded to the nearest integer as
decimals are not meaningful.
(2 points)
b) When determining Traffic analysis zone (TAZ) boundaries in urban
transport planning studies, TAZs should have identical size.
(2 points)
c) Accessibility is determined based on transport infrastructure.
(2 points)
Solution
Statement
𝑎)
𝑏)
𝑐)
True/False
False
False
True
Question 3:
[20 Points]
The following data is available from a travel survey in a city with 6 zones.
Zone
Trip generation (Y)
Population (𝑿𝟏 )
1
2
3
4
5
6
Mean
209
199
210
190
202
156
194.3
108000
94000
115000
101000
94000
78000
98333.3
Number of employees in industrial jobs
who have access to car (𝑿𝟐 )
20758
20038
20898
17192
20580
15239
19117.5
Assume the following three linear regression models are candidate trip generation models for this
region.
Model ID
Model specification
(i)
(ii)
(iii)
𝑌 = 𝛽0 + 𝛽1 𝑋1
𝑌 = 𝛼0 + 𝛼2 𝑋2
𝑌 = 𝜆0 + 𝜆1 𝑋1 + 𝜆2 𝑋2
The results of regression analysis for each model are provided in the next page. Based on the results,
answer the following questions.
a) Compare the goodness-of-fits in models. Explain which model better fits
the data and specify the measurement that you use for the comparison.
(4 points)
b) Considering a confidence level of 95 per cent, discuss the statistical
significance of coefficients in model (iii). Specify the values that you use
for this purpose.
(4 points)
c) Which model would you select as the best model for trip generation
prediction purposes? Explain your criteria and justify your answer.
(6 points)
d) Calculate trip generation for a zone with the mean values of 𝑋1 and 𝑋2
using each of the three models? Explain your answer and show all the
work.
(6 points)
Model (i)
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations
0.861
0.742
0.677
11.445
6
Intercept
Population (X1)
CoefficientsStandard Error
t Stat
61.85209 39.35941 1.571
0.001347 0.000397 3.390
P-value
0.191
0.028
Model (ii)
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations
0.949
0.900
0.875
7.131
6
CoefficientsStandard Error
t Stat
38.83076 26.11147 1.487
Intercept
Number of employees in industrial
jobs who have access to car (X2) 0.008134 0.001357 5.993
P-value
0.211
0.004
Model (iii)
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations
Intercept
Population (X1)
Numebr of employees in industiral
jobs who have access to car (X2)
0.983
0.966
0.943
4.812
6
CoefficientsStandard Error
t Stat
25.58579 18.46185 1.386
0.000579 0.000241 2.405
P-value
0.260
0.095
0.005848 0.00132
0.021
4.430
Solution
a) Since the models have different number of coefficients, a suitable
measurement to compare their goodness-of-fit is adjusted R square.
According to the results, model (iii) has the highest adjusted R square
(0.943). After that comes model (ii) with an adjuster R square of 0.875
and finally model (i) with 0.677.
b) This model has two explanatory variables of 𝑋1 and 𝑋2 . According to the
p-value, the coefficient of 𝑋2 is not statistically significant with 95
percent confidence (because 0.095 is greater than 0.05), while the
coefficient of 𝑋1 is statistically significant.
c) Model selection criteria should include: goodness-of-fit, statistical
significance of coefficients, and number of explanatory variables.
• Model (iii) has a non-significant coefficient (and probably
multicollinearity) so it should not be selected
• The decision between model (i) and (ii) boils down to the tradeoff between achieving higher goodness-of-fit and the difficulty of
predicting future values for explanatory variables.
• Students can argue either way and as long as they have
acknowledged the trade-off they get the full mark in this section
d) Regression line goes through the mean values. So the answer will be
194.3 for all the models. Students get full mark if they explain this fact or
if they do the correct calculation for all the models.
Section 2
Question 4: Trip Distribution
[20 marks]
a) (10 marks)
i)
ii)
In one sentence describe the concept of trip distribution in the context of the 4-step
transport model? (3 marks)
Trip distribution matrices can be presented in two formats, what are these formats? What is
the difference between the two methods? (4 marks)
Answer True or False for the following (3 marks)
iii)
iv)
v)
Growth models (e.g., Fratar Model) used for trip distribution are able to make provision for
zones which are at present undeveloped. (T/F)
Gravity models used for trip distribution are able to make provision for zones which are at
present undeveloped. (T/F)
Trip distribution can be completed after mode choice in the 4-step transport model. (T/F)
Solution
i)
ii)
iii)
iv)
v)
Trip distribution is the second step of the 4 step transport model and involves determining
the origins and destinations of the generated trips from the first step. (3 marks)
Origin-Destination Matrix, Production-Attraction Matrix (2 marks). OD-Matrices account for
trip direction while production-attraction matrices do not account for direction (are based
on land use alone) (2 marks)
F (1 mark)
T (1 mark)
T (1 mark)
b) (10 marks)
Approximately 80,000 people work in a city centre. In general, these workers live in four
neighbouring suburbs within 20 miles of the city centre. The number of households and
travel time from each travel zone to the city centre is shown below within part i) of the
question. The following function defines the gravity model used to distribute trips within the
region for a particular time of day, neglecting intra-zonal trips.
𝐴𝑗
2
𝑡𝑖𝑗
𝑇𝑖𝑗 = 𝑇𝑖
(
∑𝑛𝑖=1
𝐴𝑗
2
𝑡𝑖𝑗
)
𝑇𝑖𝑗 = number of trips generated in zone 𝑖 that terminates in attraction zone 𝑗
𝑇𝑖 = number of trips originating in zone 𝑖
𝐴𝑗 = number of trips attracted to zone 𝑗
𝑛 = number of zones
𝑡𝑖𝑗 = travel time between zone 𝑖 and zone 𝑗 (assume that 𝑡𝑖𝑗 = 𝑡𝑗𝑖 )
i)
Complete the following table and determine the total values for each column in
the final row. State any assumptions. (5 marks)
Suburb
Total
Attractions
to each
suburb (𝐴𝑗 )
Travel time
(𝑡𝑖𝑗 )
1
2
3
4
25,000
15,000
30,000
37,500
30
15
9
18
𝐴𝑗
2
𝑡𝑖𝑗
𝐴𝑗
2
𝑡𝑖𝑗
𝐴
∑𝑛𝑖=1 2𝑗
𝑡𝑖𝑗
𝑇𝑖𝑗
Total
ii)
iii)
Solution
i)
Describe in one sentence what the above table calculates from the perspective of
trip distribution. (2 marks)
Does the total of the “Total Attractions to each suburb” column equal the total
value of the 𝑇𝑖𝑗 column? If not, which column had a greater total and why? (3
marks)
Assume the number of trips originating in the city during an afternoon peak
period: Ti = 80,000 (1 mark, 1 mark for each column correctly calculated, ½ marks for
partially correct columns)
Suburb
A
B
C
D
Total
Total
Attractions to
each suburb
(𝐴𝑗 )
Travel time
(𝑡𝑖𝑗 )
25,000
15,000
30,000
37,500
107,500
30
15
9
18
𝐴𝑗
2
𝑡𝑖𝑗
𝐴𝑗
2
𝑡𝑖𝑗
∑𝑛𝑖=1
∑
𝐴𝑗
2
𝑡 𝑡𝑖𝑗
=
27.78
66.67
370.37
115.74
580.56
0.048
0.115
0.638
0.199
1
∑=1
ii)
iii)
𝑇𝑖𝑗
𝐴𝑗
2
𝑡𝑖𝑗
3,828
9,187
51,036
15,949
80,000
∑ = 80,000
The above table calculates the number of trips originating from the city zone that are
destined to suburb zones A, B, C and D, reflecting afternoon commuter trips. (1 mark for
stating where trips originate, 1 mark for stating the destination)
No the column values are not equivalent. The “Total Attractions to each suburb” has a
value of 107,500 which is greater than the 𝑇𝑖𝑗 column which totals to 80,000. This is the
case as the 𝑇𝑖𝑗 column represents trips originating from only the city zone; there would
be trips between each suburb which would form the entire trip distribution of the
region. (1 mark for stating ‘No’, 1 mark for indicating which column had a greater value,
1 mark for reasoning)
Question 5: Mode Choice
[20 marks]
a) (10 marks)
i)
Using the graph below, determine the number of public transport trips per day in a
zone which has 5000 people living on 50 hectares. The private vehicle (autos)
ownership statistics reveal that 40% of the households do not own a private vehicle
and the remaining households own 1 vehicle per household. (5 marks)
ii)
There are numerous factors which influence mode choice, please list 5 of the factors
that you believe affect mode choice. (5 marks)
Solution
i)
ii)
•
Number of persons per hectare = 5000/50 =100 (1 mark)
According to the graph:
0 autos/HH – 510 trips/day/1000 population (1 mark)
1 auto/HH – 250 trips/day/1000 population (1 mark)
Total public transport trips = 0.4*510*5 +0.6*250*5 = 1020+750 = 1770 trips per day (2
marks)
Note that 510 and 250 are simply read from the graph, so it’s fine that someone reads, for
example, 500 and 200.
(1 mark for each correctly identified factor)
Characteristics of the trip maker:
o Car availability
o Possession of a driving licence
o Age/Household structure
o Income
o Residential density/Accessibility
Characteristics of the journey
o Trip purpose
o Time of the day
o Travelling alone or in groups
Characteristics of the transport facility
o Travel time
o Fares and monetary costs
o Parking costs and availability
o Reliability of the mode
o Comfort
o Safety
•
•
b)
(10 marks)
Tourists to a particular region take advantage of three exotic modes of transport at a resort;
helicopter, hovercraft and balloon. The utility function is given by
𝑈𝑚 = 𝑎𝑚 − (0.009 × 𝐶𝑜𝑠𝑡) − (0.057 × 𝐼𝑉𝑇𝑇) − (0.061 × 𝑂𝑉𝑇𝑇)
where 𝐶𝑜𝑠𝑡 is the price in dollars of the ticket, 𝐼𝑉𝑇𝑇 is the in vehicle travel time and 𝑂𝑉𝑇𝑇
is the out of vehicle travel time (time spent waiting). The variables for the three modes are
as follows, where 𝑚 provides a subscript identifier for each mode.
𝑚
mode
𝑎𝑚
Cost
IVTT
OVTT
1
2
3
Helicopter
Hovercraft
Balloon
0.0
0.4
1.1
85.40
88.16
70.59
5.35
15.7
38.0
33
34
42
Answer the following questions, considering a multinomial logit model used to describe
mode choice which has the following general functional form:
𝑃(𝐾) =
𝑒 𝑈𝑘
∑𝑛𝑥=1 𝑒 𝑈𝑥
Where: 𝑃(𝐾) = probability of option K and 𝑛 = number of modes
i)
ii)
iii)
What are the probabilities each mode will be chosen? (3 marks)
Due to poor use, the balloon will be eliminated during the next season. What are the
new probabilities for the two remaining modes? (2 marks)
Considering just the helicopter and the hovercraft again and using the original costs
from the table, what value of OVTT for the hovercraft would make the probability
equal? (4 marks)
Solution
i)
(1 mark per mode)
𝑈𝑚
-3.087
-3.362
-4.263
Helicopter
Hovercraft
Balloon
𝑒 𝑈𝑚
0.046
0.035
0.014
0.095
P
0.484
0.368
0.147
ii)
Remove balloon as an option and determine the proportions (1 mark for each proportion)
0.046
Helicopter: 𝑃(1) = 0.046+0.035 = 0.568
0.035
Hovercraft: 𝑃(2) = 0.046+0.035 = 0.432
iii)
Goal: 𝑈1 = 𝑈2
𝑈1 = 0 − (0.009 × 85.40) − (0.057 × 5.35) − (0.061 × 33)
𝑈1 = −3.0866
𝑈2 = 0.4 − (0.009 × 88.16) − (0.057 × 15.7) − (0.061 × 𝑂𝑉𝑇𝑇2 )
𝑈2 = −1.2883 − (0.061 × 𝑂𝑉𝑇𝑇2 )
Equate and solve: (let Δ𝑂𝑉𝑇𝑇 = 𝑂𝑉𝑇𝑇2 − 𝑂𝑉𝑇𝑇1) (2 marks)
0.061 × 𝑂𝑉𝑇𝑇2 = −1.2883 + 3.0866
𝑂𝑉𝑇𝑇2 = 29.4789 (2 marks)
→ Assuming that 𝑂𝑉𝑇𝑇1 = 33, the OVTT for the hovercraft would have to be 29.48 (1
mark)
Question 6: Traffic Assignment
[10 marks]
a) Briefly describe the process of traffic assignment in the context of the 4-step transport
model. Please present 3 objectives of the traffic assignment process and why they are
important in transport appraisal. (5 marks)
Solution
•
•
Selection of routes between origins and destination of a transport network (2 marks)
Objectives: (3 marks – 1 for each objective)
o Estimate travel costs for each link of a road network.
o To obtain aggregate network performance measures to inform planners and
engineers in managing road traffic and transit operations
o Gain an understanding of route utilisation and redundancies within a network.
o To obtain microscopic information (turning movements and link volumes) to design
the future of the network.
b) Imagine you are a traffic engineer and you are faced with the following road network
presented in Figure 1 below. Due to heavy congestion in the network, there is a proposal to
include a new 2-way link between node B and node C. Discuss the implications of building a
new links on a road network in terms of route choice as well as overall urban planning. What
are the advantages and disadvantages? Are there better alternatives to mitigate congestion?
(5 marks)
Figure 1: Road Network for part b)
Solution
•
•
The construction of a new link between B and C creates additional routes for travelers on
the network. (1 mark)
Advantages: (1 mark)
o Adding capacity provides more supply which can cater for the demand
•
•
o Road construction results in employment and potentially drives economic growth.
Disadvantages: (2 marks)
o Induced demand
o Mode shift towards private transport
o Unsustainable as the induced demand may exacerbate congestion.
Alternatives: (1 mark)
o Invest in public and active transport (create a rail network, add a bike path)