Question 2 of 7: Deterministic Queueing at a Signalised Approach

Question 2: Deterministic Queueing at a Signalised Approach (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.

Find. The instant $t^{*}$ at which the cumulative departure curve first catches the cumulative arrival curve, the queueing diagram over $0 \le t \le t^{*}$, and the total vehicle-seconds of delay accumulated over that interval.

Approach. This is a deterministic D/D/1 queue: integrate the arrival rate to obtain the cumulative arrival curve, build the cumulative departure curve as a stepped function that is flat through every red and rises at the saturation flow through every green, find the first intersection of the two curves, and take the total delay as the area enclosed between them.

1. Build the cumulative arrival function. The number of vehicles that have arrived by time $t$ is the integral of the arrival rate: $$A(t) = \int_0^{t}\lambda(\tau)\,d\tau = \int_0^{t}\left(0.9 - 0.005\,\tau\right)d\tau = 0.9\,t - 0.0025\,t^{2}$$ This is a downward-opening parabola. Its slope, the instantaneous arrival rate, starts at 0.9 veh/s and falls linearly, reaching zero at $t = 0.9/0.005 = 180$ s — a detail that turns out to matter in Step 5.
2. Build the cumulative departure function. No vehicle may move during red, and during green the stop line discharges at the saturation flow $\mu = 3600/3600 = 1.0$ veh/s. With the red interval starting at $t = 0$, the green intervals for this approach are $33\!-\!60$ s, $93\!-\!120$ s, $153\!-\!180$ s, and so on. Hence $D(t)$ is flat through each red and has unit slope through each green: $$D(t)=\begin{cases}0, & 0 \le t \le 33\\ t-33, & 33 \le t \le 60\\ 27, & 60 \le t \le 93\\ 27+(t-93), & 93 \le t \le 120\\ 54, & 120 \le t \le 153\\ 54+(t-153), & 153 \le t \le 180\end{cases}$$ The queue at any instant is the vertical separation $Q(t) = A(t) - D(t)$.
3. Test whether the queue clears in the first cycle. Setting $A(t) = D(t)$ over the first green gives $0.9t - 0.0025t^{2} = t - 33$, that is $t^{2} + 40t - 13200 = 0$, whose positive root is $t = 96.6$ s. That root lies outside the first green interval, which ends at 60 s, so it is not admissible. Checking the state at the end of the first green confirms this directly: $$A(60) = 0.9(60) - 0.0025(60)^{2} = 54.0\ \text{veh},\qquad D(60) = 27\ \text{veh}$$ $$Q(60) = 54.0 - 27 = 18.0\ \text{veh}$$ Eighteen vehicles are left standing when the light returns to red, so the queue survives cycle 1. This step is where most candidates lose marks: writing one equation and solving it without checking that the root falls inside the green interval it was derived for gives 96.6 s, which is wrong by 83 s.
4. Test the second cycle. Over the second green, $D(t) = 27 + (t-93) = t - 66$, and $A(t) = D(t)$ gives $t^{2} + 40t - 26400 = 0$ with positive root $t = 143.7$ s — again outside its own green, which ends at 120 s. At the end of green 2, $$A(120) = 0.9(120) - 0.0025(120)^{2} = 72.0\ \text{veh},\qquad D(120) = 54\ \text{veh},\qquad Q(120) = 18.0\ \text{veh}$$ The residual queue is again 18 vehicles. It is unchanged from cycle 1 because the arrival rate is falling: over cycle 2 the approach received exactly as many vehicles as the 27 s of green could discharge.
5. Solve within the third cycle. Over the third green, $153 \le t \le 180$, the departure curve is $D(t) = 54 + (t - 153) = t - 99$. Equating: $$0.9\,t - 0.0025\,t^{2} = t - 99 \;\Longrightarrow\; 0.0025\,t^{2} + 0.1\,t - 99 = 0 \;\Longrightarrow\; t^{2} + 40\,t - 39600 = 0$$ $$t = \frac{-40 + \sqrt{40^{2} + 4(39600)}}{2} = \frac{-40 + \sqrt{160000}}{2} = \frac{-40 + 400}{2}$$ $$\boxed{t^{*} = 180\ \text{s after the start of red}}$$ The root does lie inside its green interval, $153 \le 180 \le 180$, so it is admissible. It falls exactly at the end of the third green, and exactly at the instant the arrival rate itself reaches zero — the problem is constructed so that the last arriving vehicle is also the last vehicle served. Verifying the balance: $$A(180) = 0.9(180) - 0.0025(180)^{2} = 162 - 81 = 81\ \text{veh} = D(180) = 54 + 27 = 81\ \text{veh}\;\checkmark$$ The queue therefore persists through three red intervals and clears 180 s — three full cycles — after the arrival process begins, having served 81 vehicles.
6. (b) Draw the queueing diagram. The arrival curve is the parabola of Step 1, concave downward and flattening toward $t = 180$ s. The departure curve is the staircase of Step 2: horizontal through each red, rising at $45^{\circ}$ on this scale (1 veh/s) through each green. The vertical gap between the curves is the queue length in vehicles, the horizontal gap is the delay to an individual vehicle, and the enclosed area is the total delay in vehicle-seconds.
   Figure 2.1 — Cumulative arrival curve $A(t)=0.9t-0.0025t^{2}$ and cumulative departure curve $D(t)$ for the signalised approach. Red intervals are shaded pink, green intervals green. The vertical gap is the queue, the horizontal gap the delay to an individual vehicle, and the enclosed area the total delay of 3766.5 veh·s. The curves meet at $t=180$ s, after three full cycles.
   The largest vertical gap occurs at the end of the second red, at $t = 93$ s: $$Q(93) = A(93) - D(93) = \left[0.9(93) - 0.0025(93)^{2}\right] - 27 = 62.08 - 27 = 35.1\ \text{veh}$$ The largest horizontal gap belongs to the vehicle that arrives at that same instant. It is the 62nd vehicle, it cannot be discharged within green 2 (which ends after only 27 more departures at cumulative count 54), and it is finally served at $t = 153 + (62.08 - 54) = 161.1$ s, having waited 68.1 s. Both figures are useful design output: the queue governs the storage bay length, the individual delay governs the level of service experienced by the worst-off driver.
7. (c) Total delay as the enclosed area. Total delay is $$D_{\text{tot}} = \int_0^{t^{*}}\left[A(t) - D(t)\right]dt = \int_0^{180}A(t)\,dt - \int_0^{180}D(t)\,dt$$ The arrival integral is available in closed form: $$\int_0^{180}\left(0.9t - 0.0025t^{2}\right)dt = \left[0.45\,t^{2} - \tfrac{0.0025}{3}t^{3}\right]_0^{180} = 14580 - 4860 = 9720\ \text{veh}\cdot\text{s}$$ The departure integral is taken segment by segment, each green contributing a triangle of area $g^{2}/2 = 27^{2}/2 = 364.5$ on top of the rectangle carried forward from the previous cycles: $$\int_0^{180}D\,dt = \underbrace{364.5}_{33-60} + \underbrace{27(33)}_{60-93} + \underbrace{27(27)+364.5}_{93-120} + \underbrace{54(33)}_{120-153} + \underbrace{54(27)+364.5}_{153-180}$$ $$= 364.5 + 891 + 1093.5 + 1782 + 1822.5 = 5953.5\ \text{veh}\cdot\text{s}$$
8. Combine and convert. Subtracting the two integrals, $$\boxed{D_{\text{tot}} = 9720 - 5953.5 = 3766.5\ \text{veh}\cdot\text{s} = 1.046\ \text{veh}\cdot\text{h}}$$ Dividing by the 81 vehicles served gives the average delay per vehicle, which is the quantity that maps onto a level-of-service grade: $$\bar{d} = \frac{3766.5}{81} = 46.5\ \text{s/veh}$$ Under the HCM 6th edition criteria for a signalised intersection, a control delay of 46.5 s/veh places this approach in level of service D (the D band runs from 35 to 55 s/veh). That is a defensible operating condition for a peak period in an urban area but leaves no margin, which is consistent with a queue that takes three cycles to clear.