Deterministic Queueing at a Toll Booth: 16-Civ-B7 Transportation Planning and Engineering, December 2019, Q2

Question 2: Deterministic Queueing at a Toll Booth (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 single toll booth served by a three-stage deterministic arrival profile, with time $t$ measured in minutes after 8:00 a.m.

Arrival and service rates at the toll booth
Interval	Clock time	Arrival rate $\lambda$ (veh/min)	Service rate $\mu$ (veh/min)
$0 \le t \le 10$	8:00 – 8:10	10	6
$10 < t \le 25$	8:10 – 8:25	8	6
$t > 25$	after 8:25	2	6

The queue discipline is first-in–first-out, the booth serves at its maximum rate whenever a queue is present, and there is no vehicle waiting at 8:00 a.m.

Find. The cumulative arrival and departure curves up to the instant the queue dissipates, the maximum queue length, the total vehicle delay, and the average delay per vehicle.

Figure 2.1 — Cumulative arrival and departure curves at the toll booth. The vertical gap is the queue length, the horizontal gap is an individual vehicle’s wait, and the shaded area is the total delay of 1637.5 vehicle-minutes.

Approach. Build the cumulative arrival function $A(t)$ piecewise and the cumulative departure function $D(t) = \mu t$, find the clearance time where $A(t) = D(t)$, read the maximum queue as the largest vertical gap between the two curves, and obtain the total delay as the area enclosed between them.

Write the cumulative arrival function. Integrating each constant arrival rate and carrying the running total forward, $$A(t)=\begin{cases}10t, & 0 \le t \le 10\\[2pt] 100+8(t-10), & 10 < t \le 25\\[2pt] 220+2(t-25), & t > 25\end{cases}$$ so $A(10)=100$ vehicles and $A(25)=220$ vehicles have arrived by 8:10 and 8:25 respectively.
Write the cumulative departure function. Because $\lambda = 10 > \mu = 6$ from the first instant, a queue exists continuously from $t = 0$ and the booth is never starved until the queue clears. Departures therefore accumulate at the full service rate, $$D(t)=\mu t = 6t \qquad \text{(veh, while a queue is present).}$$
Track the queue through each interval. The queue length is the vertical separation $Q(t) = A(t) - D(t)$, which grows or shrinks at the difference of the two rates:
- $0 \le t \le 10$: grows at $10-6 = +4$ veh/min, so $Q(10) = 100 - 60 = 40$ veh.
- $10 < t \le 25$: grows at $8-6 = +2$ veh/min, so $Q(25) = 220 - 150 = 70$ veh.
- $t > 25$: shrinks at $2-6 = -4$ veh/min.
The queue therefore peaks at the rate crossover $t = 25$ min, not at the end of the heaviest arrival period — the arrival rate stays above the service rate right through the second interval.
Locate the clearance time. Setting $A(t) = D(t)$ on the third branch, $$220+2(t_c-25)=6t_c \;\Longrightarrow\; 170 = 4t_c - 50 \;\Longrightarrow\; t_c = \frac{70}{4}+25$$ $$\boxed{t_c = 42.5 \text{ min after 8:00 a.m.} \;=\; 8\!:\!42\!:\!30 \text{ a.m.}}$$ The root lies on the branch it was derived from ($t_c > 25$), so it is valid. Check: $A(42.5) = 220 + 2(17.5) = 255$ veh and $D(42.5) = 6(42.5) = 255$ veh — the two curves close, as they must.
(b) Maximum queue length. From step 3 the largest vertical gap occurs at the rate crossover, $t = 25$ min: $$Q_{\max}=A(25)-D(25)=220-150$$ $$\boxed{Q_{\max}=70 \text{ vehicles, at 8:25 a.m.}}$$ The corresponding physical storage requirement, at a typical 7.5 m per queued vehicle, is about 525 m of approach lane — the number a designer actually needs.
(c1) Total vehicle delay. The total delay is the area between $A(t)$ and $D(t)$ from 0 to $t_c$. Splitting it at the two rate breaks gives a triangle, a trapezoid and a triangle: $$W=\underbrace{\tfrac{1}{2}(10)(40)}_{0-10\text{ min}}+\underbrace{\tfrac{1}{2}(40+70)(15)}_{10-25\text{ min}}+\underbrace{\tfrac{1}{2}(17.5)(70)}_{25-42.5\text{ min}}$$ $$W=200+825+612.5$$ $$\boxed{W=1637.5 \text{ vehicle-minutes} = 27.29 \text{ vehicle-hours}}$$
(c2) Average delay per vehicle. Every vehicle that arrives before clearance is delayed, so the divisor is the total arriving in $[0, t_c]$, namely $A(t_c) = 255$ vehicles — the vehicles arriving during the recovery period must be included, because they still queue behind the residual. $$\bar{w}=\frac{W}{A(t_c)}=\frac{1637.5}{255}$$ $$\boxed{\bar{w}=6.42 \text{ min/veh} = 385 \text{ s/veh}}$$ The worst individual delay is the horizontal gap at the crossover: the 220th vehicle arrives at $t=25$ min and is served when $D=220$, i.e. at $t = 220/6 = 36.67$ min, giving a maximum wait of $11.67$ min.

At 385 s of delay per vehicle the booth is far beyond any acceptable service standard — the HCM level-of-service F threshold for control delay is 80 s/veh — which is the engineering conclusion the arithmetic supports: a single booth is grossly under-capacity for this demand profile. Two booths ($\mu = 12$ veh/min) would clear the peak arrival rate outright and eliminate the queue entirely.

Question 2 — final results
Quantity	Symbol	Value
Queue clearance time	$t_c$	42.5 min after 8:00 a.m. (8:42:30 a.m.)
Total vehicles served to clearance	$A(t_c)$	255 veh
Maximum queue length	$Q_{\max}$	70 veh (at 8:25 a.m.)
Total vehicle delay	$W$	1637.5 veh-min (27.29 veh-h)
Average delay per vehicle	$\bar{w}$	6.42 min/veh (385 s/veh)
Maximum individual delay	$w_{\max}$	11.67 min

Topic. Deterministic (D/D/1) queueing analysis using cumulative arrival–departure curves.

Key relations. $Q(t)=A(t)-D(t)$; the queue clears where $A(t)=D(t)$; total delay $W=\int_0^{t_c}[A(t)-D(t)]\,dt$ is the area between the curves; average delay $\bar{w}=W/A(t_c)$; the horizontal gap at any cumulative count is that individual vehicle's wait.

Why this works. When arrivals and service are treated as deterministic, the whole problem becomes plane geometry on the cumulative plot. The vertical distance between the curves counts the vehicles physically present, the horizontal distance is the time an individual vehicle spends waiting, and the enclosed area — being vehicles multiplied by time — is total delay. Because the rates are piecewise constant, both curves are polylines and the area decomposes into triangles and trapezoids that need no calculus. The deterministic model is exact whenever demand genuinely exceeds capacity for a sustained period, which is the case here; stochastic queueing formulas are needed only when the degree of saturation approaches but does not exceed unity.

Common pitfall. Two errors dominate. First, placing the maximum queue at the end of the highest arrival rate (8:10, 40 veh) instead of at the rate crossover where arrivals finally drop below service (8:25, 70 veh) — the queue keeps growing as long as $\lambda > \mu$. Second, dividing the total delay only by the vehicles that arrived during the over-saturated period (220) instead of by all vehicles delayed up to clearance (255), which overstates the average by about 16 per cent. A free check on the whole solution is that $\mu t_c$ must equal total arrivals at $t_c$, because the two curves must meet.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §7.3 (deterministic queueing, D/D/1); Papacostas & Prevedouros, Transportation Engineering and Planning, 3rd ed., §5.6; TRB, Highway Capacity Manual, 6th ed., Ch. 19 (control-delay LOS thresholds).

Interval	Clock time	Arrival rate \(\lambda\) (veh/min)	Service rate \(\mu\) (veh/min)
\(0 \le t \le 10\)	8:00 – 8:10	10	6
\(10 < t \le 25\)	8:10 – 8:25	8	6
\(t > 25\)	after 8:25	2	6

Quantity	Symbol	Value
Queue clearance time	\(t_c\)	42.5 min after 8:00 a.m. (8:42:30 a.m.)
Total vehicles served to clearance	\(A(t_c)\)	255 veh
Maximum queue length	\(Q_{\max}\)	70 veh (at 8:25 a.m.)
Total vehicle delay	\(W\)	1637.5 veh-min (27.29 veh-h)
Average delay per vehicle	\(\bar{w}\)	6.42 min/veh (385 s/veh)
Maximum individual delay	\(w_{\max}\)	11.67 min

Question 2 of 7: Deterministic Queueing at a Toll Booth

Question 2: Deterministic Queueing at a Toll Booth (20 marks)