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FTA represents Federal Transportation Administration, a federal government agency within the US Department of Transportation. True or False? | False (Federal Transit Administration) | 1.1 | T/F | 0 | 0 | ||
In recent years, approximately 20\% of GDP is accounted for by expenses related to the transportation sector, and transportation covered nearly 10\% of household consumption expenditures. True or False? | nFalse (9.0\% of GDP in 2022; 16.8\% household consumption in 2022) | 1.2 | T/F | 0 | 0 | ||
In Year 2000, the U.S. transportation industry shares more than 50\% of the nation’s energy consumption. True or False? | False (27.4\%.) | 1.3 | T/F | 0 | 0 | ||
In year 2001, the U.S. transportation industry consumes more energy than the U.S. production industry. True or False? | False (Transportation: 27.3\%; Industrial: 33.7\% ) | 1.4 | T/F | 0 | 0 | ||
FHWA, FAA, AASHTO, IDOT, and AAR all stand for government transportation agencies. True or False? | False. The Association of American Railroads (AAR) is a (private) trade association. | 1.5 | T/F | 0 | 0 | ||
In Year 2003, only about four thousand fatal transportation accidents in total occurred in the United States. True or False? | False (42,884 fatal motor vehicle fatalities) | 1.6 | T/F | 0 | 0 | ||
Before the Interstate highway system was constructed, it took Lt. Col. Eisenhower’s convoy more than 60 days to travel from Washington DC to S.F in 1919. True or False. | True (62 days) | 1.7 | T/F | 0 | 0 | ||
The transportation industry is highly deregulated, and government influence has become negligible. True or False? Explain your answer. | False. The transportation industry is governed by many federal and local authorities. In aviation, for example, the FAA tightly controls and monitors the US airspace. Other governmental agencies which oversee the industry through laws and policies include the FMCSA, NHTSA, EPA, STB, FMC, and the FTA. State and local DOTs also regulate and/or operate the industry. | 1.8 | T/F | 0 | 0 | ||
Someone who never travels can benefit from the improvement of a transportation system. True or False? Explain your answer. | True. Improvement of transportation systems (e.g. by making them more efficient, resilient, and/or more sustainable) can benefit non-users by reducing cost/price of products and services. There are also indirect benefits to non-users from energy/environmental enhancements. | 1.9 | T/F | 0 | 0 | ||
Public carriers are owned and operated by public agencies to provide transportation service to the society. True or False? Explain your answer. | False. Public carriers provide service to the general public. They can be owned by public agencies or private companies. | 1.10 | T/F | 0 | 0 | ||
Public carriers (i.e., those serving the public), by definition, try to provide quality transportation service and maximize social welfare. True or False? Explain your answer. | False (public carriers include firms that are for-profit and are not necessarily concerned with maximizing social welfare) | 1.11 | T/F | 0 | 0 | ||
Private carriers are generally owned and operated to serve own needs. True or False? Explain your answer. | True. Private carriers are generally for-profit entities that are not obliged to serve the entire public. This is in contrast with public carriers who are obliged to serve the public. | 1.12 | T/F | 0 | 0 | ||
Transportation systems generally include two components: moving parts (e.g., vehicles) and fixed parts (e.g., physical guideway and abstract network). True or False? Explain your answer. | False. They should also include operating rules and control mechanisms (e.g. right of way). | 1.13 | T/F | 0 | 0 | ||
Railroad provides an energy-efficient and cheap mode for freight transportation. True or False? | True. Transporting freight by rail is often cheaper and much more energy efficient than by truck or air. | 1.14 | T/F | 0 | 0 | ||
List five critical issues in transportation engineering. | Below are some issues:
\begin{enumerate}
\item Improving efficiency (reducing congestion)
\item Improving safety
\item Responding to climate change
\item Promoting equity and inclusion
\item Building a strong economy
\item Advancing public health
\item Enhancing resiliency
\item Workforce development
\end{enumerate}
\noindent Source: Critical Issues in Transportation for 2024 and Beyond (2024) and earlier versions. | 1.15 | Q&A | 0 | 0 | ||
List five transportation modes in transportation engineering. | \begin{enumerate}
\item [a.] Railway transport
\item [b.] Roadway transport
\item [c.] Air transport
\item [d.] Pipeline transport
\item [e.] Maritime/waterway transport
\end{enumerate} | 1.16 | Q&A | 0 | 0 | ||
Briefly describe the components that constitute a typical transportation system, and give one example for each. | \begin{itemize}
\item Moving parts. The parts of a transportation system that move people/goods, e.g. cars, trains, aircraft;
\item Physical infrastructure. The parts of a transportation system that act as guideways for the moving parts, e.g. roadway, railway, waterway;
\item Control infrastructure. The parts of a transportation system that are put in place to manage or control the movement of the moving parts for reasons of safety, efficiency, e.g. right of way priorities.
\end{itemize} | 1.17 | Q&A | 0 | 0 | ||
Briefly describe how government is involved in the transportation industry (e.g., private and public carriers). | Below are some bullet points:
\begin{itemize}
\item The government enacts laws and policies that control the physical usage of transportation infrastructure by the public, and public and private carriers
\item The government sets goals for the transportation industry (e.g. improve public health, mitigate environmental damage)
\item The government invests in planning, construction, and operations of transportation infrastructure
\item The government invests in transportation research and development
\end{itemize} | 1.18 | Q&A | 0 | 0 | ||
Briefly describe the possible positive and negative effects of two emerging technologies in the transportation industry. | Below are just two examples.
\begin{itemize}
\item [1.] \textbf{Autonomous and connected vehicles} As autonomous and connected vehicle technology improves, the number of vehicle crashes will likely be reduced, and traffic flow efficiency can improve through the synchronization of connected autonomous vehicles. There may be further impacts on land use, urban form, and life style. However, autonomous vehicles might increase the risk of cyber-security, and disrupt the job security of drivers (e.g. taxi/truck/bus drivers).
\item[2.] \textbf{Electric vehicles} Electric vehicles have the potential to significantly reduce carbon emissions from the transportation sector, and increase sustainability, but the batteries are resource-intensive and their extra weight can increase the amount of non-tailpipe emissions (e.g. brake dust). EV also impose higher charging time and driver anxiety.
\end{itemize} | 1.19 | Q&A | 0 | 0 | ||
Briefly summarize how improved transportation systems can benefit the society as a whole. | Society benefits from improved transportation systems because investing in well-planned and well-designed transportation systems improves the quality of life for its users by many metrics. From an economic standpoint, improving transportation systems can increase the time/space utility of goods, giving rise to specialization and economies of scale. In terms of health, improving transportation systems (e.g. by building more bike lanes and public transit) can decrease negative externalities such as air/noise pollution. | 1.20 | Q&A | 0 | 0 | ||
Describe the roles of the transportation “non-users” (i.e., government, professionals, and general public) in the U.S. transportation industry (regarding public and private carriers). | \textbf{Government}
on public carriers:
\begin{itemize}
\item [a.] Own/operate vehicles and infrastructure
\item[b.] Subsidize transit systems
\item[c.] Regulate, e.g. controlling air pollution
\item[d.] Expedite/facilitate transportation research through grants
\end{itemize}
On private carriers:
\begin{itemize}
\item [a.] Regulate private carriers; e.g., via licensing, axle load limits, safety regulations, and taxes.
\end{itemize}
\textbf{Professionals}
The primary goal of professionals in the transportation industry is to improve operations and efficiency through research, planning, policy-making, and educating the public.
\textbf{Public} As "non-users", the public is still affected by the transportation industry through externalities such as noise/air pollution and taxation. The public also indirectly influences transportation policies. | 1.21 | Q&A | 0 | 0 | ||
Describe both positive and negative roles of transportation in the economy. | \textbf{Positive roles:}
\begin{itemize}
\item [a.] May improve time/space utility of goods. Note that time utility is the increase of the value of a good from reducing its transportation time (i.e. fast delivery of newspaper, no store sells one week old newspapers). The space utility is related to market expansion, urban sprawl etc.
\item [b.] Lead to changes in urban form, land use, e.g., by allowing people to work at jobs farther away.
\end{itemize}
\textbf{Negative roles:}
\begin{itemize}
\item [a.] Consumes user time, money, and other public resources
\item [b.] The physical infrastructure and vehicles can be harmful to safety (via crashes) and the environment (via emissions).
\end{itemize} | 1.22 | Q&A | 0 | 0 | ||
Suppose that a metropolitan transit authority needs to provide transit service in a huge linear city of length $L$. The city is homogeneous everywhere. The daily cost of providing transit service in this area is $c_1 x^{-1/2} + c_2 x^2$ (\$/day per station), where $x \ll L$ is the length of the catchment area served by one station along the line and constants $c_1, c_2 > 0$. Determine $n^*$, the optimal number of transit stations, that minimizes total cost incurred to the authority per day. Express it in terms of $L$, $c_1$, $c_2$. | Since there are $n = L/x$ bus stops in the transit system, each with cost $c_1x^{-\frac{1}{2}} + c_2x^2$, we must minimize the function
\[
\min \frac{L}{x}\left( c_1 x^{-\frac{1}{2}} + c_2 x^2 \right)
\]
\[
= \min L\left( c_1 x^{-\frac{3}{2}} + c_2 x \right)
\]
Minimizing the function by taking the derivative, setting it to 0, and solving for $x$, we have
\[
x^* = \left( \frac{9c_1^2}{4c_2^2} \right)^{\frac{1}{5}}
\]
\[
n^* = L \left( \frac{4c_2^2}{9c_1^2}\right)^\frac{1}{5}
\] | 2.1 | Q&A | 0 | 0 | ||
Upon successfully running the “widget” plant at Pittsburgh, you are promoted to manage a newly established overseas manufacturing plant at Arica, the most northern city in Chile. Chile can be considered a linear country that is 2500 miles in length. A competitor has already built a similar plant at Punta Arenas, the most southern city of Chile. Both you and the competitor produce these widgets at a rate of $C = 8.00$ dollars per unit.
Population of potential customers is uniformly distributed with a density of $\delta = 5$ people/mi. The prevailing market price of these widgets is $P = 10.00$ dollars per unit everywhere, and every customer will buy $D = 2$ widgets per person per year. A public carrier is hired by both plants to deliver the widgets to the customers at a cost rate of $f = 10^{-3}$ dollars per widget-mile. To most effectively penetrate into the market, you may choose to offer variable discount prices to certain customers. Your competitor may use the same strategy.
\begin{itemize}
\item [a.] While maximizing your own profit, what are your plant’s total revenue, total cost, and total profit?
\item [b.] Due to recent technological innovations (confidential from your competitor), you are able to produce these widgets at a rate of $C = 7.00$ dollars per unit next year. Nothing else will change. What will be your total revenue, cost, and profit?
\end{itemize} | \begin{itemize}
\item [a.] First, assume there is no competitor. The revenue, \(R\), will be
\[
R = L\delta PD
\]
where \(L\) is the distance to the end point of the serviced length (without the competitor).
The total cost will be the sum of the transportation cost, \(T\), and manufacturing cost, \(Z\):
\[
Z = L\delta CD
\]
\[
T = f \delta D \int_0^L x \, dx
\]
\[
T = f\delta D \frac{L^2}{2}
\]
The profit for a given \(L\) can then be found by the equation:
\[
\text{Profit} = R - (Z + T)= L\delta PD - L\delta CD - f\delta D \frac{L^2}{2}
\]
We can maximize the profit by optimizing the above formula over $L$. Specifically, we can take the derivative, setting it to 0, and solving for the optimal \(L^*\):
\[
L^* = \frac{P - C}{f}
\]
which yields a service length for both us and our competitor of 2000 miles. Alternatively, the value of $L^*$ can directly be found by equating the marginal revenue to the marginal cost per widget; i.e.,
$$ P = C+fL.$$
When the competitor is present, by symmetry, the competition will reduce your market to be between 0 and 1250 miles (past \(x = 1250\) our competitor can undercut us). From \(x = 0\) to \(x = 500\) (i.e., the area where our competitor does not serve --- farther than 2000 miles from its plant) we can charge the full price, but for \(x = 500\) to \(x = 1250\), we must offer a discount equal to the profit our competitor earns (in other words, our price must be equal to the competitor's cost). So, our profit will be the sum of profits from the ranges 0-500 and 500-1250:
\[
\text{Profit} = P_{0-500} + P_{500-1250}
\]
\[
\text{Profit}_{0-500} = 500P\delta D - 500C\delta D - f \delta D \frac{500^2}{2}
\]
\[
\text{Profit}_{500-1250} = 750P \delta D - 750C \delta D - f\delta D \int_{500}^{1250} x \, dx - \left( 750P\delta D - 750C\delta D - f\delta D \int_{1250}^{2000} x \, dx \right)
\]
Simplifying,
\[
\text{Profit}_{500-1250} = -f\delta D \int_{500}^{1250} x \, dx + f\delta D \int_{1250}^{2000} x \, dx
\]
We get \(\text{Profit}_{0-500} = \$8,750.0\), \(\text{Profit}_{500-1250} = \$5,625.0\) for a total profit of \$14,375.0. Our cost, \(Z\), is given by:
\[
Z = 1250C \delta D + f\delta D \frac{1250^2}{2}
\]
which yields \$107,812.5. Our revenue, \(R\), is the sum of the revenue from the two ranges:
\[
R = 500P\delta D + 750P\delta D - \left( 750P\delta D - 750C\delta D - f\delta D \int_{1250}^{2000} x \, dx \right)
\]
which yields a revenue of \$122,187.5.
\item [b.] First we must find the point at which both of us would pay the same cost so we set the equations for marginal costs equal to one another. Solving $7 + f x = 8 + f (2500-x)$, we have the market boundary would be at $x=1750$ miles, and the total revenue would be 167,187.5. Total cost would be 137,812.5. The total profit would be 29,375.
\end{itemize} | 2.2 | Q&A | 0 | 0 | ||
Upon successfully running the “widget” plant at Pittsburgh, you are promoted to manage a newly established overseas manufacturing plant at Arica, the most northern city in Chile. Chile can be considered a linear country that is 2500 miles in length. Your plant produces the widgets at $C = 8.00$ per unit. A public carrier is hired to deliver the widgets to the customers at a cost rate of $f = 10^{-3}$ dollars per widget-mile. Population of potential customers is uniformly distributed with a density of $\delta = 5$ people/mi. Every customer will need to buy $D = 2$ widgets per person per year if the price is no more than $10$ per widget.
\begin{itemize}
\item [a.] What are your plant’s total revenue, total cost, and total profit?
\item [b.] Due to local government’s policy, for any widget sold to the southern part of the country (beyond 1500 miles of length), you are now charged with a local tax of \$0.50. Nothing else changes. What will be your total revenue, cost, and profit?
\end{itemize} | \begin{itemize}
\item [a.] First, assume there is no competitor. The revenue, \(R\), will be
\[
R = L\delta PD
\]
where \(L\) is the distance to the end point of the serviced length. The total cost will be the sum of the transportation cost, \(T\), and manufacturing cost, \(Z\):
\[
Z = L\delta CD
\]
\[
T = f \delta D \int_0^L x \, dx
\]
\[
T = fD \delta \frac{L^2}{2}
\]
The profit for a given \(L\) can then be found by the equation:
\[
\text{Profit} = R - Z - T
\]
\[
\text{Profit} = L\delta PD - L\delta CD - f\delta D \frac{L^2}{2}
\]
We can maximize the profit by taking the derivative, setting it to 0, and solving for \(L\):
\[
L^* = \frac{P - C}{f}
\]
which yields a service length of 2000 miles.
The revenue is then \$200,000.00, the total cost is \$180,000.00, and the profit is \$20,000.00.
\item [b.] This tax will apply to 1000 miles of our current service. For a given \(L\), our new revenue is still
\[
R = L\delta PD
\]
But our cost is now the sum of the manufacturing cost, \(Z\), the tax, \(F\), and the transportation cost, \(T\):
\[
Z = L\delta CD
\]
\[
T = f\delta D \int_0^L x \, dx
\]
\[
T = f\delta D \frac{L^2}{2}
\]
\[
F = 0.5 (L - 1000)\delta D
\]
Again, taking the derivative of the profit equation, setting it to 0, and solving for \(L\) gives the equation for \(L^*\):
\[
L^* = \frac{P - C - 0.5}{f}
\]
which gives a service length of 1500 miles. Therefore, our revenue is
\[
R = 1500\delta PD = \$150,000.0
\]
Our cost is
\[
Z + T + F = 1500\delta CD + f\delta D \frac{1500^2}{2} + 0.5 (1500 - 1000)\delta D = \$133,750.0
\]
So our profit is then \$16,250.0.
\end{itemize} | 2.3 | Q&A | 0 | 0 | ||
Upon successfully running the “widget” plant at Pittsburgh, you are promoted to manage a newly established overseas manufacturing plant at Arica, the most northern city in Chile. Chile can be considered a linear country that is 2500 miles in length. You possess advanced production technology to produce the widgets at $C_1 = 6.00$ per unit. A public carrier is hired to deliver the widgets to the customers at a cost rate of $f = 10^{-3}$ dollars per widget-mile.
Population of potential customers is uniformly distributed with a density of $\delta = 5$ people/mi. Every customer will need to buy $D = 2$ widgets per person per year no matter how expensive the widgets are. However, the customers will choose to purchase widgets at the lowest available price. A competitor has already built a plant at Punta Arenas, the southern city of Chile. The competitor produces the widgets at a rate of $C_2 = 8.00$ dollars per unit, and its established rail transportation system allows negligible variable transportation cost.
\begin{itemize}
\item [a.] While maximizing your own profit, what are your plant’s total revenue, total cost, and total profit?
\item [b.] The northern part of the country (1000 miles) is mountainous. Due to fuel price increase, the public carrier now charges $1.5f$ per widget-mile for shipping in the mountainous area. Nothing else will change. What will be your total revenue, cost, and profit?
\end{itemize} | \begin{itemize}
\item [a.] Let \(P_1\) be the price we sell at, and \(P_2\) be the price our competitor sells at. Were there no competitor, our profit per unit distance at any given distance \(x\) away from Arica would be given by
\[
\text{Profit}_{\text{US}} = P_1 \delta D - C_1 \delta D - f \delta D x
\]
The competitor’s unit profit is independent of the distance, and is given by
\[
\text{Profit}_{\text{COMPETITOR}} = P_2 \delta D - C_2 \delta D
\]
In order for us to maximize our profit, we must set our price at every unit length such that our competitor cannot reduce their price below ours without losing money. Intuitively, we should be able to profit by setting our price below our competitor's lowest possible price up until some point where due to the increasing transportation cost, we can no longer reduce our price. At this point, both us and our competitor will have an equal price since neither can reduce the price without losing money. It will also be at this point where neither can profit, since both of us will have reduced our price such that the profit is 0. We can rewrite the two equations above in terms of the profit instead of price.
\[
P_1 = \frac{\text{Profit}_{\text{US}}}{\delta D} + C_1 + fx
\]
\[
P_2 = \frac{\text{Profit}_{\text{COMPETITOR}}}{\delta D} + C_2
\]
Setting them equal, and setting both profits to 0, we can solve for \(x\):
\[
x = \frac{C_2 - C_1}{f}
\]
Solving for \(x\) yields 2000 miles. Thus, our price (and our competitor's) at 2000 miles from Arica must be set to \$8. Our competitor will set their price to \$8 from \(x = 0\) to \(x = 2000\), so our price will be set to \$8 from \(x = 0\) to \(x = 2000\). Our revenue is then
\[
R = 2000 \cdot \delta \cdot P_1 D = \$160{,}000
\]
Our total cost, \(Z\), is
\[
Z = 2000 \cdot \delta \cdot C_1 D + \delta \cdot f D \int_0^{2000} x \, dx= \$140{,}000
\]
So our profit is \$20{,}000.
\item [b.] Because the transportation cost for the first 1000 miles increases, our extent of service will decrease. For widgets transported more than 1000 miles, the profit per unit distance is
\[
\text{Profit}_1 = P_1 \delta D - C_1 \delta D - D \delta f \left(1.5 \times 1000 + x - 1000\right)
\]
Our competitor’s profit is still
\[
\text{Profit}_2 = P_2 \delta D - C_2 \delta D
\]
Again, rewriting both equations above in terms of the profit instead of price:
\[
P_1 = \frac{\text{Profit}_{\text{US}}}{\delta D} + C_1 + f(x + 500)
\]
\[
P_2 = \frac{\text{Profit}_{\text{COMPETITOR}}}{\delta D} + C_2
\]
Setting the above equal to each other and setting the profits to 0 gives:
\[
x = \frac{C_2 - C_1}{f} - 500
\]
Solving for \(x\) yields 1500 miles. Thus, our price (and our competitor's) at 1500 miles from Arica must be set to \$8. Our competitor will set their price to \$8 from \(x = 0\) to \(x = 1500\), so our price will be set to \$8 from \(x = 0\) to \(x = 1500\). Our revenue is then:
\[
R = 1500\delta P_1 D = \$120{,}000
\]
Our total cost, \(Z\), is
\[
Z = 1500\delta C_1 D + 1.5 \delta f D \int_0^{1000} x \, dx + 1.5 f \cdot 1000 \cdot 500 \cdot \delta \cdot D + \delta \cdot f D \int_{1000}^{1500} x \, dx
\]
\[
Z = \$111{,}250
\]
So our profit is \$8{,}750.
\end{itemize} | 2.4 | Q&A | 0 | 0 | ||
Suppose that you have one widget plant at the center of a sufficiently large country of size $S$ mi$^2$, where population is uniformly distributed with a density of $\delta$ people/mi$^2$. The distances are measured along straight lines (Euclidean metric). The production cost of widgets is $C$ dollars per unit, and the selling price is $P$ per unit, and knows that it can sell $D$ widgets per person per year everywhere in the country. Transportation costs include a fixed portion of $F$ per unit and a variable cost of $f$ dollars per widget-mile. Suppose $P > C + F$.
\begin{enumerate}
\item[a.] Determine the shape and size of your market (the area your plant will serve).
\item[b.] What will be the profit of this one widget plant when $f = 0$?
\end{enumerate} | \begin{itemize}
\item [a.] The shape of the market will be circular since it is a Euclidean metric.
The boundary of the market will be at the point where the revenue from selling one widget is equal to the cost of producing and transporting it, i.e. $P = C + F + fr$ where $r$ is the radius. Therefore, $r = (P – C – F)/f$.
\item[b.] If there is no variable transportation cost, then it will be profitable to serve the entire country under a fixed cost. So the total annual profit is $M= (P-C-F) \delta S D$.
\end{itemize} | 2.5 | Q&A | 0 | 0 | ||
Decision sight distance is introduced to engineering practice because stopping sight distance may preclude drivers from performing evasive maneuvers. True or False? Explain your answer. | True. There are situations where drivers have to use a longer reaction/decision time (than what is normally used in computing stopping sight distance) to determine (and conduct) complex evasive maneuvers. | 3.1 | T/F | 0 | 0 | ||
Generally, stopping sight distance (the distance necessary for the vehicle to come to a full stop) is longer than the decision sight distance. True or False? Explain your answer. | False. Stopping sight distance is the minimum distance required to perceive-interpret-react and brake to a full stop. Decision sight distance is longer because it is the distance the driver needs to perceive and react to more complex situations. In these cases, the driver must digest navigational or hazard information and select an appropriate response (which may not be just braking), the distance traveled is larger than the stopping sight distance. | 3.2 | T/F | 0 | 0 | ||
Briefly state the definition/components of perception-reaction time, their normal design values. | Reaction to a stimulus is usually divided into three stages
\begin{itemize}
\item \textit{Perception}: \, negligible if a person is warned; about 1 sec if not warned, or 0 sec if the driver is fully alerted.
\item \textit{Interpretation}: \, depends on complexity; e.g. ~ 1 sec for 3 words
\item \textit{Decision/Reaction}: \, ~ 0.6 – 1.5 sec (use high end for design to be on the safe side).
\end{itemize} | 3.3 | Q&A | 0 | 0 | ||
Suppose a student uses a falling ruler to measure reaction time. He noticed a significant difference across trials (i.e., sometimes the ruler fell to the ground, while some other times it did not). What can we infer from that observation? | The reaction time of the student has high variance, possibly because the subject of the test may or may not be fully focused. | 3.4 | Q&A | 0 | 0 | ||
Suppose a student uses a falling ruler to measure reaction time. Before the ruler is released, the 0-inch mark is at the same height of the student’s hand. Suppose a student is able to grab the falling ruler at the 10-inch mark. Ignore air resistance. What is the student’s reaction time? | First, convert 10 inches to feet $\rightarrow$ 0.83 ft. Using the kinematic equation:
\[
x = x_0 + v_0 t + \frac{1}{2}at^2,
\]
where \(x_0=0, v_0=0, a = 32.2 \, \text{ft/s}^2\). We see that the reaction time was 0.228 seconds. | 3.5 | Q&A | 0 | 0 | ||
Suppose a student uses a falling ruler to measure reaction time. Assume the length of the ruler is 10 inch. Before the ruler is released, the 0-inch mark is at the same height of the student’s hand. Suppose a student is not able to grab the falling ruler. Ignore air resistance. What is the minimum value of the student’s reaction time? | First, convert 10 inches to feet $\rightarrow$ 0.83 ft. Using the kinematic equation:
\[
x = x_0 + v_0 t + \frac{1}{2}at^2 > 0.83 \text{ ft},
\]
where \(x_0=0, v_0=0, a = 32.2 \, \text{ft/s}^2\). We see that the reaction time $t >0.228$ seconds. | 3.6 | Q&A | 0 | 0 | ||
Due to grade resistance (from gravity), a vehicle will have a higher fuel consumption rate (e.g., per mile) when its tires are flat. True or False? Explain your answer. | False (A vehicle will have a higher fuel consumption rate with flat tires, but due to increased rolling resistance; grade resistance will remain the same) | 4.1 | T/F | 0 | 0 | ||
The rolling resistance force (on moving vehicles) is related to the slope/grade of the road. True or False? Explain your answer. | False (rolling resistance is primarily due to uneven deformation of the tire and ground surface while the tire rolls on the ground) | 4.2 | T/F | 0 | 0 | ||
In steady-state (stationary) traffic, density = flow $\times$ speed. True or False? Explain your answer. | False. flow = density $\times$ (space-mean speed) | 4.3 | T/F | 0 | 0 | ||
At a signalized intersection, everything else being the same, the presence of a downhill grade will increase the likelihood for a driver to be in a dilemma zone. True or False? Explain your answer. | True (Downhill grade increases minimum stopping distances, which tends to widen the dilemma zone.) | 4.4 | T/F | 0 | 0 | ||
If the speed of a vehicle keeps increasing, the aerodynamic resistance will eventually exceed the rolling resistance, because the aerodynamic resistance depends on the velocity of the vehicle while the rolling resistance does not. True or False? Explain your answer. | False (rolling resistance increases with speed; e.g., when speed is zero, there is no rolling resistance.) | 4.5 | T/F | 0 | 0 | ||
The magnitude of aerodynamic resistance force (on moving vehicles) is directly proportional to the vehicle weight, air density, and vehicle speed. True or False? Explain your answer. | False (not directly related to vehicle weight, and proportional to the square of vehicle speed; also dependent on the frontal cross-sectional area size) | 4.6 | T/F | 0 | 0 | ||
Aerodynamic resistance is typically the most important factor affecting train resistance at train speeds below 40 mph. True or False? Explain your answer. | False: The aerodynamic resistance of a train is normally not the most prominent at speeds below 40 mph. Aerodynamic resistance increases with the square of speed, so it becomes the most important factor at higher speeds (e.g., > 65 mph). | 4.7 | T/F | 0 | 0 | ||
Describe basic resistance forces of vehicles, and for each resistance force, list one or more factors that influence its value. | \begin{itemize}
\item[a.] Rolling resistance
\begin{itemize}
\item[i.] Weight or mass
\item[ii.] Road surface rigidity
\item[iii.] Tire rigidity/inflation
\item[iv.] Speed
\end{itemize}
\item[b.] Aerodynamic resistance (drag)
\begin{itemize}
\item[i.] Speed
\item[ii.] Air density
\item[iii.] Frontal cross-sectional area size
\item[iv.] Shape/size of vehicle (due to turbulence)
\end{itemize}
\item[c.] Grade resistance (gravity)
\begin{itemize}
\item[i.] Weight or mass
\item[ii.] Grade of road
\end{itemize}
\end{itemize} | 4.8 | Q&A | 0 | 0 | ||
At time \( t_0 \), a vehicle is with velocity \( v_0 \) at location \( x_0 \). Derive the location profile of a vehicle, \( x(t) \), for \( t \ge t_0 \), when acceleration is constant; i.e., \( a(x, v, t) \equiv a \). Show intermediate steps. | The location profile will depend on the initial velocity, initial position, and acceleration constant. To understand how the position will change due to acceleration, it is helpful to plot velocity versus time which we can do because the acceleration is constant.
\begin{figure}[h!]
\centering
\includegraphics[width=0.5\linewidth]{img/P4_9.png}
\end{figure}
The equation for this line is simply \( v\left(t\right)=v_0+a\left(t-t_0\right) \). Because position is the integral of velocity, we can obtain the location profile by integrating this equation with respect to time
\[
x\left(t\right)=x_0+v_0(t-t_0)+a\left(t-t_0\right)^2/2
\] | 4.9 | Q&A | 0 | 1 | img/P4_9.png |
|
A motorcycle’s skid mark at an accident site is 70 feet long. Collision (onto a parked work-zone construction vehicle) occurred at 20 mile/hour. The road is level. The friction coefficient is 0.35. Help the police to determine the motorcycle’s speed when its brakes were first applied. Ignore air resistance. | First, convert 20 mph to $29.33$ ft/s. We can use the kinematic equation below to find the initial speed:
\[\
u^2 = u_0^2 + 2a\Delta x.
\]
Noting that the deceleration due to braking is $fg$ , we have
\[
u_0 = \sqrt{u^2 + 2fg \Delta x},
\]
where \(g = 32.2 \, \text{ft/s}^2\), \(f = 0.35\), \(u = 29.33 \, \text{ft/s}\), and \(x = 70 \, \text{ft}\). Solving for \(u_0\), we have that the initial speed is 49.38 ft/s or 33.66 mph. | 4.10 | Q&A | 0 | 0 | ||
A motorcycle’s skid mark at an accident site is 80 feet long. Impact occurred at 20 mile/hour. The road is level. The friction coefficient is 0.3. Determine the motorcycle speed when its brakes were first applied. Ignore air resistance. | First, convert 20 mph to $ 29.33$ ft/s. We can use the kinematic equation below to find the initial speed:
\[\
u^2 = u_0^2 + 2a\Delta x.
\]
Noting that the deceleration due to braking is $fg$ , we have
\[
u_0 = \sqrt{u^2 + 2fg \Delta x},
\]
where \(g = 32.2 \, \text{ft/s}^2\), \(f = 0.3\), \(u = 29.33 \, \text{ft/s}\), and \(x = 80 \, \text{ft}\). Solving for \(u_0\), we have that the initial speed is 49.05 ft/s or 33.44 mph. | 4.11 | Q&A | 0 | 0 | ||
In horizontal curve design, for a given design speed, the larger the road radius, the larger the rate of super-elevation. True or False? Explain your answer. | False. A larger road radius will decrease the centrifugal force experienced by a vehicle as compared to a smaller radius with the same design speed, thus reducing the required super-elevation. | 5.1 | T/F | 0 | 0 | ||
While designing an equal-tangent vertical curve, given initial and final roadway grades and PVI, the parameters in the parabolic function can normally be determined once we find the curve length. True or False? Explain your answer. | True. The equation for an equal-tangent vertical curve is
\[
y = \frac{g_2-g_1}{L}x^2 + g_1x + e_{PVC},
\]
where $g_1$ and $g_2$ are the grades expressed as a decimal,
$L$ is the curve length, and $e_{PVC}$ is the elevation of the PVC. The value of $e_{PVC}$ can be easily found from the elevation of the PVI, as PVC is $L/2$ upstream of the PVI (at $g_1$ grade). | 5.2 | T/F | 0 | 0 | ||
In horizontal curve design, for a given design speed, the larger the road radius, the smaller the rate of super-elevation. True or False? Explain your answer. | True. A larger road radius will decrease the centrifugal force experienced by a vehicle as compared to a smaller radius with the same design speed, thus reducing the required super-elevation. | 5.3 | T/F | 0 | 0 | ||
You are given a railroad curve and a highway curve. You find that the “deflection of angle” of the railroad curve (chord definition) and that of the highway curve (arc definition) have the same numerical value. Hence, the highway curve has a smaller radius. True or False? Explain your answer. | True. The deflection angle is defined as the angle subtended by a chord (for railroads), or by a curve (for highways) 100 ft in length. If the deflection angles are the same, then the highway curve would have a smaller radius because two points connected by a 100 ft curve on a circular arc would necessarily be closer together than two points connected by a 100 ft chord. | 5.4 | T/F | 0 | 0 | ||
Parabolic functions are often used in horizontal curve design to yield a constant curvature (and hence a constant centrifugal force). True or False? Explain your answer. | False. Simple circular arcs are often used in horizontal curve design, which are distinct from parabolic curves which are often used for vertical curves. | 5.5 | T/F | 0 | 0 | ||
Minimum lengths of crest vertical curves as determined by sight distance requirements generally are satisfactory from the standpoint of safety, comfort, and appearance. True or False? Explain your answer. | True. Generally on crest curves the sight distance is long enough to ensure drainage, driver comfort, and aesthetic needs. | 5.6 | T/F | 0 | 0 | ||
Railroad trains stopping distance generally exceeds sight distance. True or False? Explain your answer. | True. While the stopping distance varies with speed and weight, it is often a mile or more for railroad trains. Often the train operator's sight distance is less than a mile. | 5.7 | T/F | 0 | 0 | ||
Cyclists and pedestrians share a 6 ft wide path (level, with a zero grade) between the roadway and the inside (convex) wall of a long curved tunnel. The radius of the inside tunnel wall is 200 ft. Local cyclists ride in the middle of the path. Pedestrians (who also walk in the middle of the path, and will stop as soon as a cyclist comes into view) will be injured if hit by a cyclist with a speed greater than or equal to 5 mph.
What is the fastest safe speed the cyclists should use that can avoid pedestrian injury? Take the coefficient of braking friction as 0.35 and cyclists’ reaction time as 1 sec. | The middle ordinate, \(M_{SSD}\), is the distance between the wall and the center of the path, which is 3 ft. We also know that the radius of the curve that the cyclist travels, $R$, is 203 ft. Using these values, we can calculate the central angle of the curve, $\Delta_{SSD}$, through the equation below:
\[
M_{SSD} = R \left(1 - \cos\left(\frac{\Delta_{SSD}}{2}\right)\right),
\]
and hence
\[
\Delta_{SSD} = 2 \arccos \left(1 - \frac{M_{SSD}}{R}\right) \cdot \frac{180}{\pi} = 19.7^{\circ}.
\]
The distance at which the cyclist first sees the pedestrian can be found using the equation below:
\[
x_{SSD} = \frac{\pi}{180} R \cdot \Delta_{SSD} = 69.86 \, \text{ft}.
\]
Since the cyclist slows down to 5 mph as opposed to coming to a complete stop, we use the following kinematic equation below for the braking distance term of the SSD:
\[
x_{SSD} = u_0 t + \frac{u^2 - u_0^2}{-2fg},
\]
where \(a\) is the braking deceleration (\(a = -fg\)). First, convert the final final impact speed of 5 mph, $u$, to 7.35 ft/s.
Solving for \(u_0\) yields a max speed of 30.64 ft/s or 20.83 mph. | 5.8 | Q&A | 0 | 0 | ||
You are determining the speed limit for a mountainous one-way highway segment along the bottom of a cliff. The road segment is level (i.e., zero grade) but contains a horizontal curve. The center line of the lane is 20 ft from the inside (convex) rock wall of the cliff. The radius of the inside wall is 600 ft.
What should the speed limit be (round down to the nearest multiple of 5 mph) to ensure enough stopping sight distance on the curve? Take the coefficient of braking friction as 0.35 and drivers’ reaction time as 1.5 sec. | The middle ordinate, \(M_{SSD}\), is the distance between the wall and the center of the highway, which is 20 ft. We also know that the radius of the curve that the vehicle travels, $R$, is 620 ft. Using these values, we can calculate the central angle of the curve, $\Delta_{SSD}$, through the equation below:
\[
M_{SSD} = R \left(1 - \cos\left(\frac{\Delta_{SSD}}{2}\right)\right),
\]
\[
\Delta_{SSD} = 2 \arccos \left(1 - \frac{M_{SSD}}{R}\right) \cdot \frac{180}{\pi} = 29.19^{\circ}.
\]
The stopping sight distance is calculated using the equation below:
\[
x_{SSD} = \frac{\pi}{180} R \cdot \Delta_{SSD} = 315.8 \, \text{ft}.
\]
We can use the stopping sight distance equation to find the speed limit by first solving for \(u\) (mph), the initial speed:
\[
x_{SSD} = 1.47ut + \frac{u^2}{30 \times 0.35},
\]
where $t$ (s) is the driver's reaction time
\[
u = 47.16 \, \text{mph}.
\]
Rounding down to the nearest multiple of 5 gives a speed limit of 45 mph. | 5.9 | Q&A | 0 | 0 | ||
You are designing an arterial two-way highway in the hills where the speed limit is 45 miles per hour. At one crest point, you need to design a vertical curve that connects a +5\% grade to a -3\% grade. Let’s assume that the drivers’ eyes are generally 3.5 feet from the ground, while the potential obstructing objects are 2 feet tall. The perception-to-reaction time is 2.5 sec. Can you determine the minimum curve length? |
The equation for stopping sight distance, in feet, is
\[
SSD = 1.47ut + \frac{u^2}{30\left(\frac{11.2}{32.2} \pm \frac{G}{100}\right)},
\]
where $u$ (mph) is the speed limit, and $t$ (s) is the perception-to-reaction time. We have $G_1 = 5 \%$ and $G_2 = -3 \%$. We use the worst-case SSD, which will occur on the steepest downgrade. In case of a two-way arterial highway, the steepest downgrade is $-5\%$. So the SSD is
\[
SSD = 1.47 \times 45 \times 2.5 + \frac{45^2}{30\left(\frac{11.2}{32.2} - \frac{5}{100}\right)} = 392 \, \text{ft}.
\]
The distance, $S$, at which the driver with eye height $H_1$ first sees the object with height $H_2$ will dictate the minimum curve length, $L_{min}$.
\[
L_{min} = \frac{AS^2}{200(\sqrt{H_1}+\sqrt{H_2})^2},
\]
where $A = |G_2 - G_1|$. Substituting in the heights and setting $S = SSD$ gives
\[
L_{min} = \frac{AS^2}{2158} = 569.7 \, \text{ft}.
\]
Because \(S < L_{min}\), we are done. The minimum curve length is 569.7 ft. | 5.10 | Q&A | 0 | 0 | ||
In one signal cycle at an intersection, the effective green time for one approach often equals its actual green phase time plus its yellow phase time. True or False. Explain your answer. | True. The effective green time is the amount of time vehicles actually can utilize to clear the intersection, which includes the yellow phase. | 6.1 | T/F | 0 | 0 | ||
Greenshields' model (1930s) conjectures that the average travel speed on a freeway remains relatively constant for all low density (uncongested) traffic. True or False? Explain your answer. | False (Greenshields conjectured (simply, but incorrectly) that the average speed linearly decreases with density) | 6.2 | T/F | 0 | 0 | ||
In all stationary traffic, time-mean speed equals space-mean speed. True or False? Explain your answer. | False (Note that stationary traffic does not require that all vehicles move at the same speed, hence time and space mean speeds do not need to be equal.) | 6.3 | T/F | 0 | 0 | ||
In all stationary traffic, time-mean speed is larger than or equal to space-mean speed. True or False? Explain your answer. | True. Time-mean speed is larger than (or equal to) space-mean speed in stationary traffic because a stationary observer (the observer who measures time-mean speed) will count more (or the same amount of) fast vehicles per unit time than an overhead observer (the observer who measures space-mean speed) will count per unit distance. | 6.4 | T/F | 0 | 0 | ||
In traffic, density = 1 / headway. True or False? Explain your answer. | False. The inverse of the (average) headway is flow. Density is equal to the inverse of the (average) spacing. | 6.5 | T/F | 0 | 0 | ||
In traffic flow dynamics, we use conservation of vehicles to prove the following fact: ``The line/interface in the time-space diagram separating two traffic states has the same slope as that of the line connecting the corresponding traffic state points on the flow-density fundamental diagram.” True or False? Explain your answer. | True. Let $\Delta x$ be a small portion of the road where all vehicles are initially in state A. Let $\Delta t$ be the amount of time it takes for all vehicles in $\Delta x$ to end up in state B. Note that in this time, all vehicles entering $\Delta x$ will enter in state A, but will leave $\Delta x$ in state B. The speed of the interface between the two states, $u_{AB}$, is then $\Delta x / \Delta t$. Conservation of vehicles requires that the number of vehicles in $\Delta x$ at time $t_0$ + the number of vehicles entering $\Delta x$ in $\Delta t$ time must equal the number of vehicles leaving $\Delta x$ in $\Delta t$ time + the number of vehicles in $\Delta x$ at time $t + \Delta t$. This situation would be reflected on a time-space diagram as a line/interface between the two traffic states.
\begin{align*}
\Delta x k_A + \Delta t q_A = \Delta x k_B + \Delta t q_B
\end{align*}
where $k_A$ and $k_B$ are the densities of traffic states $A$ and $B$ respectively, and $q_A$ and $q_B$ are the flows of traffic states $A$ and $B$ respectively. Rearranging terms we have
\begin{align*}
u_{AB} = \frac{\Delta x}{\Delta t} = \frac{q_A - q_B}{k_A - k_B}.
\end{align*}
It is easy to see that the above slope is the same slope as the slope of the line connecting two traffic states on the q-k fundamental diagram. | 6.6 | T/F | 0 | 0 | ||
In a cumulative diagram for a traffic network, the horizontal separation between the departure and the arrival curves gives the number of vehicles between the two observers. True or False? Explain your answer. | False -- cumulative diagram refers to the cumulation of vehicles over time (or sometimes, space). Hence, only the vertical separation should be about the number of vehicles. | 6.7 | T/F | 0 | 0 | ||
Railroad traffic control systems must use electric track circuits to monitor and control train position along a line. True or False? Explain your answer. | False. Block-based control does not need to know exact locations of the train along all points in the line. | 6.8 | T/F | 0 | 0 | ||
Write a formula for the average delay per car for an approach to a traffic signal in terms of arrival flow $\lambda$, saturation flow $\mu$, cycle length $C$, effective red time $R$, and effective green time $G$. State the condition under which it applies. |
\begin{figure}[h!]
\centering
\includegraphics[width=0.5\linewidth]{P6_9.png}
\caption{Cumulative arrival/departure curve}
\label{fig:P6_9}
\end{figure}
The average delay is found by dividing the total delay (the shaded area, in units of $\text{vehicles} \cdot \text{time}$) by the total number of vehicles served in a cycle (which should be $\lambda C$, not $n$). By noting that $C = R + G$, the average delay is then
\begin{align*}
\text{Avg delay} &= \frac{R^2}{2C(1 - \lambda / \mu)}.
\end{align*}
For this equation to apply, the total number of vehicles arriving at the intersection during the cycle must NOT exceed the number of vehicles that can potentially leave the intersection during the effective green time.
\begin{align*}
\text{Condition: } & \lambda C < \mu (C - R).
\end{align*} | 6.9 | Q&A | 0 | 1 | img/P6_9.png |
|
One approach to a traffic signal has constant arrival flow $\lambda$, saturation flow $\mu$, effective red time $R$, and effective green time $G$. Suppose $G$ is sufficiently large to clear the queue at the end of each cycle. In one cycle, what is the number of vehicles that are actually delayed by the intersection? |
\begin{figure}[h!]
\centering
\includegraphics[width=0.5\linewidth]{img/P6_10.png}
\caption{Cumulative arrival/departure curve}
\label{fig:P6_10}
\end{figure}
The cumulative arrival/departure curve in Figure \ref{fig:P6_10} demonstrates the delay vehicles experience at the traffic signal. The total vehicle delay experienced (vehicles $\times$ time) is the shaded area, and the total number of vehicles delayed is $n$. By geometry, we have
\[
n=\frac{R \lambda\mu}{\mu -\lambda}.
\] | 6.10 | Q&A | 0 | 1 | img/P9_10.png |
|
Given Greenshields’ speed-density equation: $v(k) = v_f(1 - k/k_j)$, where $v_f$ is the free flow speed and $k_j$ is the jam density.
\begin{enumerate}
\item[a.] Derive the relationship (equation) and sketch the $q – k$ fundamental diagram. Mark important quantities on the diagram (such as $v_f$ and $k_j$).
\item[b.] What may be the flaw of Greenshields’ theory that makes its fundamental diagram unrealistic?
\end{enumerate} | \begin{itemize}
\item [a.] We have \( q = k v = k\ v_f(1-\frac{k}{k_j}) \) . This is a parabolic function similar to the following:
\begin{figure}[h!]
\centering
\includegraphics[width=0.5\linewidth]{img/P6_11.png}
\caption{Flow-density fundamental diagram}
\label{fig:P6_11}
\end{figure}
\item[b.] Speed decreases with density k even when $k \rightarrow 0^+$. This is not realistic as drivers can maintain a near-constant free-flow speed for a range of low density values.
\end{itemize} | 6.11 | Q&A | 0 | 1 | img/P6_11.png |
|
Briefly explain why the slope of the line connecting two traffic condition points in the flow-density fundamental diagram equals the interface shockwave speed between the corresponding traffic conditions in the time-space diagram.\\ | \begin{figure}[h!]
\centering
\includegraphics[width=0.75\linewidth]{img/P6_12.png}
\caption{Flow-density fundamental diagram and time-space diagram}
\label{fig:P6_5}
\end{figure}
The figure on the left is a time-space diagram. The one on the right is a flow-density fundamental diagram. Each state of traffic (area in the time-space diagram) corresponds to a point in the fundamental diagram.
\begin{itemize}
\item The line in the time-space diagram separating two states (e.g., A and B) has a defined slope. Geometrical considerations would show that the slope of this line is
\[
u_{AB} = \frac{q_A - q_B}{k_A - k_B},
\]
where $q_A$ and $q_B$ are the flows of traffic conditions A and B respectively, and $k_A$ and $k_B$ are the densities of of traffic conditions A and B respectively.
\item Conveniently for us, this is also the slope of the line going through points A and B of the flow-density fundamental diagram, based on simple geometry.
\end{itemize} | 6.12 | Q&A | 0 | 1 | img/P6_12.png |
|
Consider a two-lane highway with stationary traffic. There are two types of vehicles, cars and trucks, each occupying one lane. Cars travel with constant speed $v_c$ and the time headway between any two consecutive cars is $h_c$. Trucks travel with constant speed $v_t$, and the headway between any two consecutive trucks is $h_t$.
\begin{enumerate}
\item[a.] What is the total density (regarding all vehicles) on this highway?
\item[b.] What is the space-mean speed?
\end{enumerate} | \begin{itemize}
\item [a.] The total density is the sum of the density of each lane
\[
k = k_{car} + k_{truck},
\]
where \(k_{car} = \frac{1}{v_c h_c}\); \(k_{truck} = \frac{1}{v_t h_t}\).
So
\[
k = \frac{1}{v_c h_c} + \frac{1}{v_t h_t}.
\]
\item [b.] In 1 space unit, we have \(\frac{1}{v_c h_c}\) many cars and \(\frac{1}{v_t h_t}\) many trucks. We can use these numbers as weights to compute the average speed of all vehicles. This gives the space-mean speed:
\[
\bar{v_s} = \frac{v_c \frac{1}{v_c h_c} + v_t \frac{1}{v_t h_t}}{\frac{1}{v_c h_c} + \frac{1}{v_t h_t}} = \frac{\frac{1}{h_c} + \frac{1}{h_t}}{\frac{1}{v_c h_c} + \frac{1}{v_t h_t}}.
\]
\end{itemize} | 6.13 | Q&A | 0 | 0 | ||
Consider a one-lane highway with stationary traffic. There are two types of vehicles, cars and trucks, traveling at the same speed $v$. Every car keeps a time headway $h_c$ from its preceding vehicle (the preceding vehicle could be a truck or a car). Due to safety concerns, every truck will keep a different time headway $h_t$ from the preceding vehicle.
Suppose that the proportion of trucks in the vehicle stream is $p$. Express the total vehicle flow and total vehicle density on this highway, in terms of the above parameters. | The average headway in this one lane, \(\bar{h}\), is \((1-p)h_c+ph_t\). Flow is \(1/\bar{h}\), and hence
\[
\text{Flow } q = \frac{1}{(1-p)h_c + ph_t}.
\]
Since all vehicles on the road have the same speed, the time mean speed and the space mean speed are the same. Density is simply
\[
\text{Density } k = \frac{q}{v}= \frac{1}{v((1-p)h_c + ph_t)}.
\] | 6.14 | Q&A | 0 | 0 | ||
Traffic on a two-lane, one-way tunnel obeys a triangular fundamental diagram with the following characteristics: capacity = 4000 veh/hr, free-flow speed = 100 km/hr, jam density = 280 veh/km. A vehicle runs out of gas at 10:00 pm, and parks on the shoulder 1 km into the tunnel, when traffic was freely flowing at 3600 veh/hr. Due to “rubbernecking”, capacity at that incident point reduces to $\mu = 3000$ veh/hr.
\begin{enumerate}
\item[a.] Draw the fundamental diagram for the tunnel and show all the relevant parameters.
\item[b.] Determine when the back of the queue will emerge at the mouth (entrance) of the tunnel. Show your work.
\item[c.] What is the headway downstream of the parked vehicle?
\end{enumerate} | \begin{figure}[h!]
\centering
\includegraphics[width=0.75\textwidth]{img/P6_15.png}
\caption{Fundamental diagram}\label{fig:P6_15}
\end{figure}
\begin{enumerate}
\item [a.] See Fig. \ref{fig:P6_15} (a) for the fundamental diagram.
\item [b.] We want to find the speed at which the interface between the two traffic states, U (uncongested) and Q (congested), \(u_{UQ}\), moves backwards. Let \(q_U\) be the uncongested flow, and \(q_Q\) be the congested flow, \(k_U\) be the uncongested density and \(\mu\) be the congested density. Then we have
\[
u_{UQ} = \frac{q_U - \mu}{k_U - k_Q}.
\]
And we find that \(u_{UQ} = 9.375 \, \text{km/h}\). Since the mouth of the tunnel is 1 km away, it will take 6.4 minutes for the back of the queue to reach the mouth, or at 10:06 pm.
\item [c.] Flow downstream is 3000 vehicles per hour. Therefore, headway = 1/3000 hr = 3600/3000 sec = 1.2 sec.
\end{enumerate} | 6.15 | Q&A | 0 | 1 | img/P6_15.png |
|
In practice, we often use gravity model to determine the number of trips that will be produced in (and attracted to) a certain geographical zone. True or False? Explain your answer. | False. The gravity model determines the number of trips going between zone pairs, but does not determine the total number of trip productions or attractions in each zone. The total productions and attractions for each zone must be known before using the gravity model. | 7.1 | T/F | 0 | 0 | ||
The well-known gravity model originates from an analogy to the gravity force between two planets; it utilizes advanced human behavioral theory with regard to trip destination decisions. True or False? Explain your answer. | False. The basic assumption of the gravity model is that the interaction between two locations decreases with their distance, but increases with the level of activity at each location. The gravity model is not based on human behavioral theory. | 7.2 | T/F | 0 | 0 | ||
In practice, we often use cross-classification models to determine the number of trips from an origin zone to a destination zone. True or False? Explain your answer. | False. The cross-classification model is mainly intended for trip productions, i.e. used in the first step of the demand analysis process. Determining the number of trips from an origin zone to a destination zone is the second step of the demand analysis process. | 7.3 | T/F | 0 | 0 | ||
The cross-classification model often assumes a linear additive structure to simplify complex trip decisions. True or False? Explain your answer. | False. The cross-classification model is data driven and can capture any nonlinear relationships. | 7.4 | T/F | 0 | 0 | ||
In the context of transportation engineering, list the four steps commonly used in demand analysis processes, and briefly describe the inputs/outputs of each step. | \begin{enumerate}
\item Trip generation
\begin{itemize}
\item Inputs: Land uses, demographics, socioeconomic factors, zone partition
\item Outputs: Trip production and attraction for each zone
\end{itemize}
\item Trip distribution
\begin{itemize}
\item Inputs: Trip production and attraction for each zone
\item Outputs: Trip number between each OD zone pair
\end{itemize}
\item Mode choice
\begin{itemize}
\item Inputs: Trip number between each OD zone pair
\item Outputs: Proportions of demand for each OD that use each available mode choices
\end{itemize}
\item Trip assignment/network equilibrium
\begin{itemize}
\item Inputs: Proportions of demand for each OD that use each available mode choices
\item Outputs: For each mode and OD pair, equilibrium flow on the available network routes/links
\end{itemize}
\end{enumerate} | 7.5 | Q&A | 0 | 0 | ||
Briefly state, in your own words, the strength and weakness of regression model and cross-classification model. | \textbf{Regression:} Simple, easy to implement. But the model specification may be unrealistic or lacks theoretical foundation, and parameters may turn out to have “wrong” signs;
\textbf{Cross-classification:} Highly flexible, data driven, and allow complex relationships; but the number of classification types can be large, and thus time-consuming to model. | 7.6 | Q&A | 0 | 0 | ||
The gravity model takes a set of given zonal productions and attractions as input. The direct output is a set of trip numbers for all origin/destination zone pairs. We will find that for each origin zone, the summation of the calculated trip numbers across all destination zones equals the given production number. Is this coincidental? Briefly explain. | It is not coincidental because for each attraction zone j, the model multiplies the production of zone i by the fraction of demand going to zone j. So summing over all j’s will sum to the production of i. | 7.7 | Q&A | 0 | 0 | ||
Trip productions and attractions are generally computed separately. At the final stage of trip generation for N zones, what is the one condition that the zonal production numbers ($P_k \,\ k=1,2,\hdots, N$) and zonal attraction numbers ($A_k \,\ k=1,2,\hdots, N$) should theoretically satisfy (but often do not)? What do we do if that condition is not satisfied? | The sum of all zonal production numbers should theoretically equal the sum of all zonal attraction numbers. This is from the conservation of trips with the study region. This condition is often violated because independent/different methods are often used to calculate attraction and production. To solve this, some simple scaling ``correction" of attraction numbers can be conducted. | 7.8 | Q&A | 0 | 0 | ||
In traditional microeconomics theory, a rational human is analytic, well-informed, and willing to sacrifice for others. True or False? Explain your answer. | False – a rational person is greedy/selfish/self-interest-driven, and hence not willing to sacrifice for others. | 8.1 | T/F | 0 | 0 | ||
In traditional economics theory, utility is a cardinal quantification of satisfaction/welfare/happiness. True or False? Explain your answer. | False – it is ordinal. | 8.2 | T/F | 0 | 0 | ||
In traditional economics theory, a rational person is greedy, analytic, and well-informed. True or False? Explain your answer. | True. A fundamental assumption in traditional economics theory is that people are rational, and that rational people are greedy, analytic, and well-informed. | 8.3 | T/F | 0 | 0 | ||
According to Wardrop’s Principle, when traffic equilibrium is reached, some routes between an origin/destination pair may have unequal travel costs. True or False? Explain your answer. | True. Those unused routes could have a larger cost. | 8.4 | T/F | 0 | 0 | ||
In the utility theory, what do we mean by “rational human”? Is it realistic? Explain briefly why. | In traditional economics theory, a rational person is greedy, analytic, and well-informed. This assumption is for simplicity of analysis, but not fully realistic because in reality people show very complex behaviors (e.g., affected by other factors such as their emotions) and lack access to information or analysis skills. | 8.5 | Q&A | 0 | 0 | ||
Given a small island $A$ with $P$ family units, each unit $j$ derives $a_j$ daily in economic benefit by sending a commuter to a city $B$ on the mainland. Island $A$ is connected to city $B$ via both a ferry (with round trip ferry fare $F$) and also a toll bridge (with round trip toll $\delta$ plus other out-of-pocket round trip driving cost $c$). The ferry trip lasts $t_F$ hours (for each way) and the car trip $t_c$ hours (for each way). This commuter values driving time at a rate of $\alpha_j$ \$/hr and riding time at $\beta_j$ \$/hr.
Every family is required to send a commuter to $B$ every day, using one and only one of the two modes.
\begin{enumerate}
\item[a.] Write the utility functions of a round trip for family unit $j$, for both modes.
\item[b.] Derive an expression for the demand function for ferry service as a function of $F$; assume that: (i) daily economic benefit $a_j$ and driving time value $\alpha_j$ are the same for every family/commuter ($a_j = a$ and $\alpha_j = \alpha$); and (ii) commuters value ferry riding time $\beta_j$ differently, and for any positive value $\gamma$, the commuters with $\beta_j > \gamma$ form a fraction $e^{-\gamma/4}$ (note that $0 < e^{-\gamma/4} < 1$) of the total population.
\item[c.] Estimate approximately the percentage change in demand when the fare is reduced 1\%, from \$2 to \$1.98. Try to do this as simple as possible.
\end{enumerate} |
\begin{enumerate}
\item [a.] \begin{align*}
U_{ferry} &= a_j – 2 t_f \beta_j – F, \,\,
U_{drive} = a_j – 2 t_c \alpha_j – \delta – c
\end{align*}
\item [b.]
\begin{align*}
\text{Pr} \left(U_{ferry}>U_{drive} \right) &= \text{Pr}\left( \beta_j < [2 t_c \alpha_j + \delta + c – F] / (2 t_f) \right) =1 – e^{– \frac{(2 t_c \alpha_j + \delta + c – F) }{8 t_f} }.
\end{align*}
Therefore the demand is given by $P-Pe^{– \frac{(2 t_c \alpha_j + \delta + c – F) }{8 t_f} }$.
\item[c.] Take the derivative of demand w.r.t. $F$, and let\(\gamma = \frac{2t_c \alpha_j + \delta + c - F}{2t_f}\), we have
\begin{align*}
\frac{dD}{dF} &= -\frac{P}{8t_f}e^{\frac{-\gamma}{4}}.
\end{align*}
\begin{align*}
\text{Change in demand} \approx \frac{dD}{dF} \cdot -0.02
\end{align*}
\begin{align*}
\text{\% change in demand} &= \frac{\frac{dD}{dF} \cdot -0.02}{P-Pe^{\frac{-\gamma}{4}}} \times 100
= \frac{0.02\frac{P}{8t_f}e^{\frac{-\gamma}{4}}}{P-Pe^{\frac{-\gamma}{4}}} \times 100
= \frac{\frac{1}{400t_f}e^{\frac{-\gamma}{4}}}{1-e^{\frac{-\gamma}{4}}} \times 100
\end{align*}.
\end{enumerate} | 8.6 | Q&A | 0 | 0 | ||
Briefly state Wardrop’s Principle. | This principle is the basis of user equilibrium: each user will take the route which minimizes their own time resulting in all USED routes eventually having the same travel time – unused routes may have a higher cost. | 9.1 | Q&A | 0 | 0 | ||
When Nash Equilibrium is reached in a transportation network, no traveler can reduce its travel cost by unilaterally changing its route choice. Use your own words, briefly explain why the Wardrop’s Principle describes Nash Equilibrium in transportation networks. | \begin{itemize}
\item All used routes (given an origin-destination pair) have equal cost $\rightarrow$ there is no benefit from changing to a used route.
\item All unused routes (given an origin-destination pair) have higher or equal cost $\rightarrow$ there is no benefit from changing to an unused route.
\end{itemize}
As such, every user is stuck with the current route choice being the best option. | 9.2 | Q&A | 0 | 0 | ||
Travel from $O$ to $D$ during the middle of the day can be done by highway (car, drive alone) or transit line. The transit fare and the out-of-pocket driving costs (parking, gas…) are $F$ (\$) and $P$ (\$), respectively. The travel times by transit and car are $T$ (hr) and $C$ (hr), respectively. The latter increases with flow, $x$ (cars/hr), as a known function, $C(x)$. The mid-day travel demand (desired flow) from $O$ to $D$ is steady at $q$ (people/hr). We assume that all people value time equally at a rate of $v$ (\$/hr), and they are rational.
\begin{enumerate}
\item[a.] Define a network equilibrium problem that will identify the equilibrium modal split: (draw the network and label its links and nodes; identify the O/D flow, link flows, and link cost function for each traveler (in terms of \$).
\item[b.] Write Wardrop’s equilibrium conditions in this context.
\item[c.] Suppose 50\% of the travelers stay at home if the generalized cost of their trip is greater than $M$ (\$). Define a revised network (links, nodes, link costs and origin/destination table) that will jointly identify the equilibrium trip choices.
\end{enumerate} |
\begin{enumerate}
\item [a.]
\begin{tikzpicture}
% Nodes
\node[circle, draw, minimum size=1.2cm] (O) at (0,0) {O};
\node[circle, draw, minimum size=1.2cm] (D) at (10,0) {D};
% Edges
\draw[-, thick, bend left] (O) to node[midway, above] {Transit Cost: $C_T = Tv + F$} (D);
\draw[-, thick, bend right] (O) to node[midway, below] {Highway Cost: $C_c= C(x)v + P$} (D);
\end{tikzpicture}
\item[b.]
\begin{align*}
C(x)v + P &= Tv + F \quad \text{if} \quad 0 < x < q
C(0)v + P &> Tv + F \quad \text{if} \quad x = 0
C(q)v + P &< Tv + F \quad \text{if} \quad x = q
\end{align*}
\item[c.]
\begin{figure}
\centering
\includegraphics[width=0.5\linewidth]{img/P9_3.png}
\caption{Updated network}
\label{fig:P9_13}
\end{figure}
\end{enumerate} | 9.3 | Q&A | 0 | 1 | img/P9_3.png |
|
In designing a simple bus corridor using the standards approach, the optimal stop spacing is the same regardless of whether the door-to-door travel time is minimized for the average or worst-case passenger. True or False? | False.
As an example, assume $v_w$ is the walking speed, $s$ is the stop spacing, $l$ is an average travel distance, and $t_d$ is the average vehicle dwell/lost time at each stop.
The worst-case passenger travel time is \( \frac{s}{v_w} + \frac{lt_d}{s}\), and the optimal spacing is
\( s^\ast = \left( l t_dv_w \right)^\frac{1}{2} \),
The average-case passenger travel time is \( \frac{s}{2 v_w} + \frac{lt_d}{s} \), and the optimal spacing \( s^\ast = \left(2 l t_d v_w \right)^\frac{1}{2}
\).
So the optimal spacing varies by a factor of \(\sqrt{2}\). | 10.1 | T/F | 0 | 0 | ||
While plotting Vickery diagram for individual transportation shuttle system with $0<e<1\le L<\infty$, it is possible that arrival curve $V(t)$, departure curve $D(t)$, and wish curve $W(t)$ all overlap. True or False? | True – when demand is sufficiently low. | 10.2 | T/F | 0 | 0 | ||
Suppose there is a two-point shuttle bus system serving passengers with a non-constant arrival rate. All the buses have large enough capacity. To minimize the average waiting time of passengers, the headway between consecutive dispatched should always be equal. True or False? | False. Normally the optimal headway is inversely proportional to the square root of the demand arrival rate. | 10.3 | T/F | 0 | 0 | ||
Consider a city served by a fleet of identical taxis, and all customer trip origins/destinations are uniformly distributed across the city. If the operator decides to add more vehicles into the fleet, the average in-vehicle travel time per passenger is always to decrease. True or False? | False. The average in-vehicle travel time will either remain the same or increase (due to more vehicles/traffic) – as the average in-vehicle travel distance remains the same. | 10.4 | T/F | 0 | 0 | ||
For a ride-hailing company serving a service region, additional investment that increases the number of operating taxis will always reduce the average passenger waiting time. True or False? | False. Despite a large number of vehicles, the system can be stuck in an inefficient equilibrium known as the ``wild goose chase" in which the average passenger waiting time increases with the number of fleet (as a larger number of vehicles are in the deadheading state). | 10.5 | T/F | 0 | 0 | ||
When we use the standards approach to design a simple bus corridor, the optimal stop spacing would be doubled if the door-to-door travel time in the design standards is based on an average passenger instead of the worst-case passenger (while everything else remain the same). True or False? | False. The optimal stop spacing would increase by a factor of $\sqrt{2}$. | 10.6 | T/F | 0 | 0 | ||
You are going to design bus networks for two cities of similar size and shape, under the same investment budget. The ideal type of network (Loop, Hub-and-spoke or Grid) for both cities must be similar. True or False? | False. There are other factors which might affect the ideal network such as demand density, population's value of time. | 10.7 | T/F | 0 | 0 | ||
Suppose there is a shuttle bus system serving passengers with a constant arrival rate. All the buses have large enough capacity. If we want to minimize the average waiting time of the passengers, the headway between consecutive dispatches should always be equal. True or False? | True. Total waiting time increases with the square of the headway, so the total waiting time for two unequal headways will be larger than the total waiting time if the two headways instead both take the average value. | 10.8 | T/F | 0 | 0 | ||
In a dial-a-ride system, if each vehicle alternates between pickups and deliveries while have a target occupancy level of 7 passengers, then on average each passenger experiences 7 pickup-and-delivery cycles until his/her own drop-off. True or False? | True. Expectation of a geometric distribution, with the probability of success (drop-off) being 1/7. | 10.9 | T/F | 0 | 0 | ||
A smooth differentiable univariate function $f(x)$ may not reach a local optimum at point $x_0$ even if the first order condition holds at $x_0$, i.e. $f' (x_0) =0$. True or False? | True. The point may be a saddle point. | 10.10 | T/F | 0 | 0 | ||
Sufficiently large buses (infinite capacity) are used to provide shuttle service, and we always optimize dispatch frequency based on the demand. As the temporal demand distribution becomes more uneven (i.e., heterogeneous over time, but no change in total number of passengers), the total (agency + user) cost increases. True or False? | False. With a fixed total demand, time-dependent demand variations actually reduces the total cost, because such demand is easier to aggregate into shared vehicles. | 10.11 | T/F | 0 | 0 | ||
The agency is trying to decide whether to design a hub-and-spoke network or a grid network to serve a city with uniformly distributed passenger O/D demand. As agency investment (total service route length per unit area) approaches infinity, these two types of networks eventually yield the same expected door-to-door passenger travel time. True or False? | False. A grid network will yield a lower asymptotic door-to-door passenger travel time. | 10.12 | T/F | 0 | 0 | ||
If the daily cost for a transportation system to serve a steady demand of rate $\lambda$ [passengers/day] is $ Z= C\lambda^{1.05}$ [\$/day], where $C$ is a constant, then we say this system has ``economies of scale". True or False? | False. The exponent of $\lambda$ would need to be less than 1, so that increase in $\lambda$ would lead to decrease in cost per unit demand, $Z/\lambda$. | 10.13 | T/F | 0 | 0 | ||
For all types of transit networks, the average passenger travel time decreases as the total route length per unit area increases. True or False? | False. For a loop type of network, the longer the route, possibly there could be more passenger in-vehicle riding time due to circuitous routes. | 10.14 | T/F | 0 | 0 |
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in Dataset Viewer.
TransportBench is collection of 141 undergraduate transportation system engineering problems that span a broad spectrum of topics including transportation economics, driver characteristics, vehicle motion, road geometry design, traffic flow/control, transportation planning, utility/modal split, transportation networks, and public transit systems. TransportBench consists of both the true or false problems and the general Q&A problems. The purpose of thise dataset is to benchmark the capabilities of both commercial and open source large language models.
More details about this dataset and performance comparison of the current sate of the art language models can be found at: https://agi4engineering.github.io/TransportBench/
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