16-Civ-A6 Highway Design, Construction, and Maintenance · December 2015
Question 2 of 7: Deterministic Queueing at a Freeway Bottleneck
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: 98-Civ-A6 Transportation Planning & Engineering, National Examination, December 2015. Seven questions, each of equal value (20 marks); any five constitute a complete examination. Closed book, one two-sided aid sheet permitted. Three hours. All seven questions are solved below as a study resource.
Reference texts. Mannering, Washburn & Kilareski, Principles of Highway Engineering and Traffic Analysis (Wiley); Papacostas & Prevedouros, Transportation Engineering and Planning (Prentice Hall); Roess, Prassas & McShane, Traffic Engineering (Pearson); Ortúzar & Willumsen, Modelling Transport (Wiley) for the demand-model chapters (trip generation, distribution, mode choice, assignment).
Question 2: Deterministic Queueing at a Freeway Bottleneck (20 marks)
Given. A deterministic (D/D/1) bottleneck with time-varying arrivals and a fixed service capacity.
Given data (time measured in minutes after 8:00 am)
Interval
Arrival rate λ
Service capacity μ
0–10 min (8:00–8:10)
100 veh/min
60 veh/min
10–25 min (8:10–8:25)
80 veh/min
60 veh/min
> 25 min (after 8:25)
25 veh/min
60 veh/min
Find. The cumulative arrival/departure diagram, the maximum queue, the total delay and the average delay per vehicle.
Cumulative arrival A(t) and departure D(t) curves. The queue is the horizontal-then-vertical gap between the two; it is widest at t = 25 min and closes where the curves meet at t = 45 min.
Approach. Build the cumulative arrival curve $A(t)$ piecewise and the cumulative departure curve $D(t)=\mu t$ (the server runs at capacity the whole time a queue exists). The queue is $Q(t)=A(t)-D(t)$; the maximum queue occurs at the last instant arrivals still exceed capacity, and total delay is the area between the curves.
Cumulative arrivals at the rate breaks. With $\lambda$ constant over each interval,
$$A(10)=100(10)=1000,\qquad A(25)=1000+80(15)=2200.$$
After 8:25 the arrival rate (25/min) drops below capacity (60/min), so no further vehicles join the queue — the queue can only shrink beyond $t=25$.
Queue at each break. The server discharges at $\mu=60$/min throughout, so $D(t)=60t$ while the queue is non-empty:
$$Q(10)=1000-60(10)=400,\qquad Q(25)=2200-60(25)=700.$$
The queue grows at $100-60=40$/min then at $80-60=20$/min, so it keeps building right up to $t=25$.
Maximum queue. Because arrivals fall below capacity exactly at $t=25$ min, the queue peaks there:
$$\boxed{Q_{\max}=700\ \text{vehicles at } t=25\ \text{min (8:25 am)}}$$
Dissipation time. Beyond $t=25$, $A(t)=2200+25(t-25)$ and $D(t)=60t$. Setting $A=D$ (queue $=0$):
$$2200+25(t-25)=60t\;\Rightarrow\;1575=35t\;\Rightarrow\;t=45\ \text{min}.$$
The queue clears at 8:45 am, by which time $D(45)=60(45)=2700$ vehicles have been served.
Total delay = area between the curves. Summing the three trapezoids of $Q(t)$:
$$W=\underbrace{\tfrac12(10)(400)}_{2000}+\underbrace{\tfrac12(400+700)(15)}_{8250}+\underbrace{\tfrac12(20)(700)}_{7000}$$
$$\boxed{W=17{,}250\ \text{veh-min of total delay}}$$
Average delay per vehicle. All $N=2700$ vehicles that pass through by 8:45 experience the queue:
$$\bar w=\frac{W}{N}=\frac{17{,}250}{2700}=\boxed{6.39\ \text{min/vehicle}}$$
