Group | Participating Group Members |
MONASH UNIVERSITY
DEPARTMENT OF CIVIL ENGINEERING
CIV2282: Transport and Traffic Engineering
Semester 2/2020, Week 9
Practical Class 9 – Travel Demand and Forecasting
Exercise 1 – Trip Generation (1)
Problem 8.3 from Mannering and Washburn (2013)
A large residential area has 1,100 households with an average household income of $40,000, an average household size of 4.8, and, on average, 1.5 working members. Using the model described below, predict the change in the number of peak-hour social / recreational trips if employment in the area increases by 25% and household income by 10%.
Number of peak-hour vehicle-based social/recreational trips per household
= 0.04 + 0.018 × (household size)
+ 0.009 × (annual household income in thousands of dollars)
+ 0.16 × (number of nonworking household members)
Before | After | |
Households | 1100 | 1100 |
HH Income | $40K | $40K +10% = $44K |
HH Size | 4.8 | 4.8 |
Workers | 1.5 | 1.5 + 25% = 1.875 |
Trips / HH | 0.04 + 0.018 x 4.8 + 0.009 x 40 + 0.16 x (4.8-1.5) = 1.0144 | 0.04 + 0.018 x 4.8 + 0.009 x 44 + 0.16 x (4.8-1.875) = 0.9904 |
Total Trips | 1.0144 x 1100 = 1116 | 0.9904 x 1100 = 1089 |
The model predicts a decrease of 27 peak-hour social / recreational trips
If the household described above has an average size of 6 and 1 working member, an income of $50k, how many peak-hour social/recreational trips are predicted?
The household in Example 8.1 had six members, an income of $50K, and 6-1 = 5 non-working members
The number of trips would be = 0.04 + 0.018 × (6) + 0.009 × (50) + 0.160 × (5) = 1.398
Exercise 2 – Trip Generation (2)
A neighbourhood has 205 retail employees and 700 households that can be categorized into four types, with each type having characteristics as follows:
Type | Household size | Annual Income | Number of non workers in the peak hour | Workers departing |
1 | 2 | $40,000 | 1 | 1 |
2 | 3 | $50,000 | 2 | 1 |
3 | 3 | $55,000 | 1 | 2 |
4 | 4 | $40,000 | 3 | 1 |
There are 100 type 1, 200 type 2, 350 type 3, and 50 type 4 households. Assuming that shopping, social/recreational, and work vehicle-based trips all peak at the same time.
Determine the total number of peak-hour trips (work, shopping, social/recreational) using the generation models described below (hint: work out the total trips per household per type first):
- Social/rec trips = 0.04 + 0.018 × HHsize + 0.009 × Income + 0.160 × Nonworking (from Exercise 1)
- Shopping trips = 0.12 + 0.09 × HHsize + 0.011 × Income – 0.150 × Neighb.Employ.
Type | HH size | Annual Income | Non workers in the peak hour | Workers departing | Shopping Trips / household | Social & Rec Trips / household | Work Trips/ household | Total Trips / household |
1 | 2 | $40K | 1 | 1 | 0.4325 | 0.596 | 1 | 2.0285 |
2 | 3 | $50K | 2 | 1 | 0.6325 | 0.864 | 1 | 2.4965 |
3 | 3 | $55K | 1 | 2 | 0.6875 | 0.749 | 2 | 3.4365 |
4 | 4 | $40K | 3 | 1 | 0.6125 | 0.952 | 1 | 2.5645 |
There are 100 type 1, 200 type 2, 350 type 3, and 50 type 4 households.
Total Trips = 100 × 2.0285 + 200 × 2.4965 + 350 × 3.4365 + 50 × 2.5645 = 2033
Exercise 3– Logit Model of Shopping Destination Choice (1)
Problem 8.8 from Mannering and Washburn (2013)
A survey found that 4000 car trips are generated in a large residential area from noon to 1:00 am on Saturdays for shopping purposes. Four major shopping centres have the following characteristics:
Shopping Centre | Distance from Residential Area (km) | Commercial Floor Space (thousands of m2) |
1 | 4 | 20 |
2 | 9 | 15 |
3 | 8 | 30 |
4 | 14 | 60 |
A Logit model is estimated with coefficients of -0.283 for distance (in kilometres) and +0.172 for commercial space (in thousands of m2).
How many shopping trips will be made to each of the four shopping centres?
Shopping Centre | Utility, Ui | Exp(Ui) | Proportion of Trips | Total Trips |
#1 | -0.283 × 4 + 0.172 × 20 = 2.308 | 10.0543 | 10.0543 / 606.28 = 1.7% | 1.7% × 4000 = 66 |
#2 | -0.283 × 9 + 0.172 × 15 = 0.033 | 1.0335 | 0.2% | 7 |
#3 | -0.283 × 8 + 0.172 × 30 = 2.896 | 18.1016 | 3.0% | 119 |
#4 | -0.283 × 14 + 0.172 × 60 = 6.358 | 577.091 | 95.2% | 3807 |
Total = 606.28 | 100% | Total = 4000 |
Shopping Centre #1 = 66 trips, SC #2 = 7 trips, SC #3 = 119 trips, SC #4 = 3807 trips
Exercise 4 – Logit Model of Shopping Destination Choice (2)
Problem 8.9 from Mannering and Washburn (2013)
Consider the shopping trip situation described in Exercise 3. Suppose that shopping centre 3 goes out of business and shopping centre 2 is expanded to 50,000 m2 of commercial space. What would be the new distribution of the 4000 Saturday afternoon shopping trips?
Shopping Centre | Utility, Ui | Exp(Ui) | Proportion of Trips | Total Trips |
#1 | -0.283 × 4 + 0.172 × 20 = 2.308 | 10.0543 | 10.0543 / 1012.533 = 1.0% | 1.0% × 4000 = 40 |
#2 | -0.283 × 9 + 0.172 × 50 = 6.053 | 425.3873 | 42.0% | 1680 |
#3 | Out of business | |||
#4 | -0.283 × 14 + 0.172 × 60 = 6.358 | 577.091 | 57.0% | 2280 |
Total = 1012.533 | 100% | Total = 4000 |
Shopping Centre #1 = 40 trips, SC #2 = 1680 trips, SC #3 = 0 trips, SC #4 = 2280 trips
Exercise 5 – Logit Model of Shopping Destination Choice (3)
Problem 8.10 from Mannering and Washburn (2013)
If shopping centre 3 is closed (see Exercise 3), how much commercial floor space is needed in shopping centres 1 and 2 to ensure that each of them has the same probability of being selected as shopping centre 4?
Shopping Centre | Utility, Ui | Exp(Ui) | Proportion of Trips | Total Trips |
#1 | -0.283 × 4 + 0.172 × A1 = 6.358 | 577.091 | 577.091 / 1731 = 1/3 | 1/3 × 4000 = 1333 |
#2 | -0.283 × 9 + 0.172 × A2 = 6.358 | 577.091 | 1/3 | 1333 |
#3 | Out of business | |||
#4 | -0.283 × 14 + 0.172 × 60 = 6.358 | 577.091 | 1/3 | 1333 |
Total = 1731.273 | 1 | Total = 4000 |
Solve utility equations for A1 = 43.546 and A2 = 51.773
For shopping centres #1 and #2 to have the same probability of being selected as centre #4, the model predicts that shopping centre #1 requires 43,546 m2 of commercial floor space and shopping centre #2 requires 51,773 m2 of commercial floor space.
Exercise 6 – Logit Model of Mode Choice
Problem 8.38 from Mannering and Washburn (2013)
A work-mode-choice model is developed from data acquired in the field in order to determine the probabilities of individual travellers selecting various modes. The mode choices include car drive alone (DL), car shared-ride (SR), and bus (B).
The utility functions are estimated as:
UDL = 2.6 ‑ 0.3 × (costDL) ‑ 0.02 × (travel timeDL)
USR = 0.7 ‑ 0.3 × (costSR) ‑ 0.04 × (travel timeSR)
UB = ‑0.3 × (costB) – 0.01 × (travel timeB)
where cost is in dollars and travel time is in minutes.
The cost of driving a car is $5.50 with a travel time of 21 minutes, while the bus fare is $1.25 with a travel time of 27 minutes. How many people will use the shared-ride mode from a community or 4500 workers, assuming the shared-ride option always consists of three individuals sharing costs equally?
Mode | Cost | Travel Time (minutes) | Utility, Ui | Exp(Ui) | Proportion of Trips | Total Trips |
Drive alone | $5.50 | 21 | 2.6 – 0.3 × 5.5 – 0.02 × 21 = 0.53 | 1.699 | 1.699 / 2.725 = 62.3% | 62.3% × 4500 = 2805 |
Shared ride (x3) | $5.50 / 3 = $1.83 | 21 | 0.7 – 0.3 × 1.83 – 0.04 × 21 = -0.69 | 0.502 | 18.4% | 828 |
Bus | $1.25 | 27 | -0.3 × 1.25 – 0.01 × 27 = -0.645 | 0.525 | 19.3% | 866 |
Total = 2.725 | 100% | Total = 4500 |
The model predicts that a total of 828 people will use the shared-ride mode.
- 866 people
- 2805 people
- 828 people
- 314 people
Exercise 7 – Discuss and briefly explain the role of demand forecasting
See slides 6-7 of Travel Demand and Forecasting (1), also see Monday workshop slides
Exercise 8 – Discuss and briefly explain four-step analysis and the connections between each step
See slide 8-11 of Travel Demand and Forecasting (2) , also see Monday workshop slides