16-Civ-A6 Highway Design, Construction, and Maintenance · May 2015

Question 2 of 7: Deterministic (D/D/1) Queueing at an Incident

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examination — 98-Civ-A6 Transportation Planning & Engineering, May 2015. Closed book (one two-sided aid sheet), 3 hours. Seven questions of equal value (20 marks); any five constitute a complete paper. All seven are solved here as a study resource.

Reference texts (subject):

Papacostas, C.S. & Prevedouros, P.D., Transportation Engineering and Planning, 3rd ed. (Prentice Hall).
Roess, R.P., Prassas, E.S. & McShane, W.R., Traffic Engineering, 4th ed. (Pearson).
Mannering, F.L. & Washburn, S.S., Principles of Highway Engineering and Traffic Analysis, 6th ed. (Wiley).
Ortúzar, J. de D. & Willumsen, L.G., Modelling Transport, 4th ed. (Wiley).
Sheffi, Y., Urban Transportation Networks (Prentice Hall).

Question 2: Deterministic (D/D/1) Queueing at an Incident (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A one-directional freeway with a constant demand and three successive service regimes triggered by an incident, analysed as a deterministic D/D/1 queue with the clock starting at the moment of closure (t = 0).

Given data
Quantity	Value
Arrival (demand) rate, λ	12 veh/min (constant)
Full capacity (both lanes)	18 veh/min
Service rate, 0 ≤ t < 10 min (fully closed)	0 veh/min
Service rate, 10 ≤ t < 35 min (one lane)	6 veh/min
Service rate, t ≥ 35 min (both lanes)	18 veh/min

Find. (a) the time the queue clears; (b) the maximum queue length; (c) the total and average delay.

Cumulative arrival A(t) and departure D(t) curves. The horizontal gap is delay per vehicle; the vertical gap is the queue length; the area between the curves is total delay.

Approach. Build cumulative arrival and departure functions, find where they re-intersect (queue clears), read the maximum vertical gap (max queue), and integrate the area between them (total delay).

Write the cumulative curves. Arrivals accumulate at the constant demand rate: $$A(t)=12\,t \quad\text{(veh)}.$$ Departures accumulate at whatever service rate is in force: $$D(t)=\begin{cases}0, & 0\le t\lt 10\\ 6\,(t-10), & 10\le t\lt 35\\ 150+18\,(t-35), & t\ge 35.\end{cases}$$ At the switch points $D(10)=0$ and $D(35)=6(25)=150$ veh.
Locate where the queue clears. During recovery the queue empties because departures (18) exceed arrivals (12). Set $A(t)=D(t)$ for $t\ge 35$: $$12\,t = 150 + 18\,(t-35)\;\Rightarrow\; 12t = 18t-480\;\Rightarrow\; 6t = 480.$$ $$\boxed{t_{\text{clear}} = 80\ \text{min after closure began.}}$$ Check: $A(80)=960$ and $D(80)=150+18(45)=960$ veh — the curves meet.
Find the maximum queue. The queue $Q(t)=A(t)-D(t)$ grows whenever arrivals exceed the current service rate, i.e. for all $t\lt 35$ (service is 0 then 6, both below 12), and shrinks after $t=35$. The maximum therefore occurs at $t=35$ min: $$Q_{\max}=A(35)-D(35)=12(35)-150 = 420-150.$$ $$\boxed{Q_{\max}=270\ \text{vehicles at }t=35\ \text{min.}}$$
Total delay = area between the curves. The queue profile is piecewise linear with vertices $(0,0),(10,120),(35,270),(80,0)$, where $Q(10)=12(10)-0=120$ veh. Summing three trapezoids (veh·min): $$D_{\text{tot}}=\tfrac12(0+120)(10)+\tfrac12(120+270)(25)+\tfrac12(270+0)(45)$$ $$=600+4875+6075.$$ $$\boxed{D_{\text{tot}} = 11\,550\ \text{veh}\cdot\text{min.}}$$
Average delay per vehicle. Every vehicle arriving before the queue clears is delayed; that is $N=\lambda\,t_{\text{clear}}=12(80)=960$ vehicles. Hence $$\bar d=\frac{D_{\text{tot}}}{N}=\frac{11\,550}{960}.$$ $$\boxed{\bar d \approx 12.0\ \text{min/veh.}}$$

Question 2 — Final Results
Quantity	Value
Time queue clears	80 min after closure began
Maximum queue length	270 veh (at t = 35 min)
Total vehicle delay	11 550 veh·min
Vehicles delayed	960
Average delay per vehicle	12.03 min ≈ 12 min

Topic: Deterministic (D/D/1) queueing analysis of a freeway incident with cumulative diagrams.

Key relations: $Q(t)=A(t)-D(t)$; queue clears when $A=D$; total delay $=\int[A(t)-D(t)]\,dt$ (area between curves); average delay $=$ total delay $\div$ number of vehicles delayed.

Why this works: When arrival and service rates are treated as deterministic and uniform within each interval, cumulative counts are straight-line segments, so the queue is exactly the vertical distance between them and the delay is exactly the enclosed area. The queue keeps growing as long as demand exceeds the momentary service rate; the maximum occurs at the instant service first overtakes demand — here at the end of the reduced-capacity period, not at the moment the road first reopens.

Common pitfall: Placing the maximum queue at t = 10 (reopening) instead of t = 35; and dividing total delay by the wrong vehicle count (use only the vehicles that actually queued, N = 960).

Source: Mannering & Washburn, Principles of Highway Engineering and Traffic Analysis, Ch. 5 (queueing); Roess, Prassas & McShane, Traffic Engineering, Ch. 8.

← Question 1 Next: Question 3 →