CIVL2700 · Transport Systems
Deterministic (D/D/1) Queueing & Departure-Time Choice
Week 5 of CIVL2700 Transport Systems at the University of Sydney puts two ideas on the same diagram: the deterministic (D/D/1) bottleneck queue and departure-time choice. Plotting cumulative arrivals A(t) against cumulative departures D(t) lets you read the queue length, when it clears, each vehicle's delay and the total delay straight off the picture; the Vickrey bottleneck then explains why commuters spread the peak, trading queueing time against arriving early or late. It is core content for the supervised In-class Test 2 and for the comprehensive final exam.
What this chapter covers
- 01The D/D/1 model: Deterministic arrivals, Deterministic service, 1 server, FIFO
- 02Cumulative curves A(t) and D(t) drawn on one time-vs-vehicles chart
- 03Queue length Q(t) = A(t) − D(t) as the vertical gap; it grows while λ > μ
- 04Maximum queue at the kink where the arrival rate drops to capacity (λ = μ)
- 05Clearance time from A(t) = D(t); maximum individual delay = the horizontal gap (FIFO)
- 06Total delay = area between the curves = ½·base·height for a triangular queue
- 07Average delay per vehicle (÷N) versus average queue length (÷time) — two denominators
- 08Unit discipline: convert every rate to one basis (veh/min) before drawing or integrating
- 09Vickrey departure-time utility U(t) = −c(t) − αE(t) − βL(t)
- 10Departure-time equilibrium: every used departure time carries the same generalised cost
D/D/1 ramp-meter queue — queue, clearance and total delay
- +1Common units and rates. Everything is already in veh/min: λ₁ = 8 (t ≤ 10), λ₂ = 3 (t > 10), μ = 5. Since λ₁ = 8 > μ = 5, a queue starts building at t = 0.
- +1Cumulative curves. A(t) = 8t for t ≤ 10, so A(10) = 80; then A(t) = 80 + 3(t − 10). The server stays busy, so D(t) = 5t.
- +2Maximum queue. The queue peaks at the kink t = 10, where λ drops from 8 to 3: Q_max = A(10) − D(10) = 80 − 5·10 = 80 − 50 = 30 veh.
- +2Clearance time. Set A(t) = D(t): 80 + 3(t − 10) = 5t ⇒ 50 = 2t ⇒ t = 25 min. Check: 5·25 = 125 veh served equals 80 + 3·15 total arrivals.
- +2Maximum individual delay (FIFO). The worst-off vehicle is the 80th, arriving at the kink t = 10; it is served when D(t) = 80 ⇒ 5t = 80 ⇒ t = 16, so its delay = 16 − 10 = 6 min.
- +1Total delay = area between the curves. The queue-vs-time profile is a triangle: base = 25 min, height = 30 veh, so total delay = ½·25·30 = 375 veh·min.
- +1Average delay. Total delay ÷ vehicles processed = 375 / 125 = 3 min/veh (and average queue = 375 / 25 = 15 veh).
Key terms
- D/D/1 queue
- A queue with Deterministic arrivals, Deterministic service and 1 server, worked first-in-first-out. Because nothing is random, the queue length, clearance time and delays are exact numbers read from cumulative curves rather than probabilities.
- Cumulative arrival curve A(t)
- The running total of vehicles that have arrived by time t. Its slope is the arrival rate λ; a step change in λ shows up as a kink in the curve.
- Departure / service curve D(t)
- The running total of vehicles served (released) by time t. While a queue exists the server is busy, so D(t) rises at the constant capacity rate μ and its slope equals μ.
- Queue length Q(t)
- The number waiting at time t, equal to the vertical gap A(t) − D(t) (never negative). It grows while λ > μ and shrinks while λ < μ, peaking where the arrival rate falls back to capacity.
- Total delay
- The area enclosed between the arrival and departure curves, ∫[A(t) − D(t)] dt, measured in veh·min (or veh·h). Dividing by the number of vehicles gives average delay per vehicle; dividing by the time period gives the average queue length.
- Capacity (service rate) μ
- The maximum rate at which the single server can discharge vehicles, μ = 1 / (service time per vehicle). A queue forms only when demand λ temporarily exceeds μ.
- Vickrey bottleneck model
- A departure-time model in which travellers passing a fixed-capacity bottleneck choose when to leave, trading queueing (travel) time against schedule delay from arriving early or late relative to a preferred time t*.
- Schedule delay (early / late penalty)
- The cost of not arriving exactly at the preferred time t*: α per minute early and β per minute late, with β usually larger than α. In the utility U(t) = −c(t) − αE(t) − βL(t) only one of the early (E) or late (L) terms is non-zero for any arrival.
Deterministic (D/D/1) Queueing & Departure-Time Choice FAQ
Where on the cumulative-curve diagram is the maximum queue, and why not at the end?
The queue is the vertical gap A(t) − D(t), and it is largest at the kink where the arrival rate drops back to capacity (λ = μ). Before that instant the gap is still widening because arrivals outrun the server; after it the server is catching up, so the gap shrinks. At clearance the two curves meet and the gap is zero, and at t = 0 the queue has not yet formed — so evaluate the maximum at the kink, not at either end.
What is the difference between the queue length and a vehicle's delay on the same diagram?
They are two different gaps between the same two curves. The queue length at a moment is the vertical gap (how many vehicles are waiting), while an individual vehicle's delay is the horizontal gap (how long that vehicle waits from arriving to being served). Under FIFO the largest horizontal gap belongs to the vehicle arriving at the kink, and the area enclosed between the curves is the total delay of everyone combined.
Can AI help me with deterministic queueing and departure-time choice in CIVL2700?
Yes — Sia can explain it step by step. Paste a D/D/1 or Vickrey question and Sia will walk through converting the rates to one unit, writing A(t) and D(t), locating the maximum queue at the kink, solving A(t) = D(t) for clearance, and taking the area for total delay, so you learn the method for the exam. Sia explains each step and checks your own working; it never hands over answers to assessed work, completes your assignment for you, or guarantees a grade, and you should confirm results against your Canvas materials.
Exam move
Practise the cumulative-curve routine until it is automatic: convert every rate to one time basis (veh/min), write A(t) and D(t) as cumulative functions, then sketch them. Read the maximum queue as the vertical gap at the kink where λ drops to μ, get the clearance time from A(t) = D(t), take the maximum individual delay as the horizontal gap for the kink vehicle, and get the total delay as the area (½·base·height for a single triangular queue). Keep the two averages straight — average delay divides the area by the number of vehicles, average queue divides the same area by the time period. For departure-time choice, minimise the generalised cost C = c + αE + βL and remember the equilibrium idea that every used departure time ends up costing the same. Because these are method-marked, show your cumulative functions and your working so an arithmetic slip costs one mark, not the question. The final exam runs 2.5 hours and is a hurdle (you must score at least 40% on it to pass the unit); budget time in proportion to marks (about 1.5 minutes per mark on a 100-mark paper) and confirm the exact date, room and open- or closed-book status on Canvas and the University exam timetable.
Working through Deterministic (D/D/1) Queueing & Departure-Time Choice in CIVL2700? Sia is AskSia’s AI Engineering tutor — ask any CIVL2700 Deterministic (D/D/1) Queueing & Departure-Time Choice question and get a clear, step-by-step explanation grounded in how CIVL2700 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
