Deterministic Queueing behind a Freeway Incident: 16-Civ-B10 Traffic Engineering, December 2015, Q3

Question 3: Deterministic Queueing behind a Freeway Incident (20 marks — 4 each)

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 two-lane eastbound carriageway, one lane blocked from 5:05 PM for 20 minutes. All rates are converted to vehicles per minute, which keeps the arithmetic clean because every rate divides exactly by 60.

Quantity	Symbol	Value
Arrival (demand) flow	\(\lambda\)	2100 veh/h = 35 veh/min
Capacity during the blockage	\(\mu_1\)	960 veh/h = 16 veh/min
Capacity after the tow	\(\mu_2\)	2400 veh/h = 40 veh/min
Duration of the blockage	\(T_1\)	20 min (5:05 PM to 5:25 PM)

Find. Whether a queue forms and its maximum length; the time it clears; the longest individual wait; the total and average delay; and the position and wait of a vehicle joining at 5:23 PM.

Cumulative arrival and departure curves for the Highway 401 incident. The vertical gap is the queue, the horizontal gap an individual vehicle’s delay, and the shaded area the total delay of 18 240 veh·min.

Approach. Draw the cumulative arrival and departure curves with time measured from 5:05 PM. The vertical gap between them is the queue, the horizontal gap is an individual vehicle’s delay, and the enclosed area is the total delay — so all five parts are read off one diagram.

Part (a) — test whether a queue forms and size it. Demand exceeds the restricted capacity, \(\lambda = 35 > \mu_1 = 16\) veh/min, so a queue must form and it grows at the difference of the two rates. Over the whole 20-minute blockage, \[ Q_{\max} = (\lambda - \mu_1)\,T_1 = (35 - 16)(20) = \boxed{380\ \text{vehicles}} \] The queue is longest at the instant the tow truck clears the lane, 5:25 PM, because from that moment capacity (40) exceeds demand (35) and the backlog begins to shrink. During the blockage \(D(T_1) = \mu_1 T_1 = 16 \times 20 = 320\) vehicles have been discharged while \(35 \times 20 = 700\) have arrived, and \(700 - 320 = 380\) confirms the queue length independently.
Part (b) — time at which the queue clears. After the tow the surplus capacity is \(\mu_2 - \lambda = 40 - 35 = 5\) veh/min, so the 380-vehicle backlog takes \[ t_d = \frac{Q_{\max}}{\mu_2 - \lambda} = \frac{380}{5} = 76\ \text{min} \] to dissipate. Measured from 5:05 PM the queue clears at \(T_2 = 20 + 76 = 96\) min, that is at 6:41 PM. The check is that the two cumulative curves meet there: arrivals \(35 \times 96 = 3360\) vehicles and departures \(320 + 40 \times 76 = 3360\) vehicles. Note that only 20 minutes of blockage produced 96 minutes of congestion — the recovery is governed by the small surplus \(\mu_2 - \lambda\), not by the size of the disruption.
Part (c) — longest individual wait. Under first-in–first-out the wait of a vehicle is the horizontal distance between the curves. That distance is greatest for the last vehicle released while the restriction is still in force — vehicle number \(D(T_1) = 320\), which departs exactly at \(T_1\). It arrived when \(35\,t = 320\), i.e. at \(t = 9.14\) min, so \[ w_{\max} = T_1 - \frac{\mu_1 T_1}{\lambda} = 20 - \frac{320}{35} = 20 - 9.14 = \boxed{10.86\ \text{min}} \] or 10 min 51 s, for the vehicle that joined at about 5:14 PM. Vehicles arriving later wait less, because they are discharged at the higher rate \(\mu_2\); a scan over every arrival instant confirms 10.86 min is the maximum.
Part (d) — total and average delay. Total delay is the area between the curves, which is a triangle on each limb. On the growth limb the queue rises linearly from 0 to 380 over 20 min, and on the recovery limb it falls linearly from 380 to 0 over 76 min: \[ D_{\text{tot}} = \tfrac{1}{2}(380)(20) + \tfrac{1}{2}(380)(76) = 3800 + 14\,440 = 18\,240\ \text{veh}\cdot\text{min} \] that is 304 vehicle-hours. Every vehicle arriving in the 96 minutes before the queue clears is delayed, so the number affected is \(N = \lambda T_2 = 35 \times 96 = 3360\) vehicles and \[ \bar{d} = \frac{D_{\text{tot}}}{N} = \frac{18\,240}{3360} = 5.43\ \text{min/veh} \ (=326\ \text{s/veh}) \] Note that 79 % of the delay (14 440 of 18 240 veh·min) accrues after the lane is physically clear.
Part (e) — the vehicle joining at 5:23 PM. At \(t = 18\) min the queue has grown for 18 min at 19 veh/min, so the number of vehicles ahead of the newcomer is \[ Q(18) = (\lambda - \mu_1)(18) = 19 \times 18 = \boxed{342\ \text{vehicles}} \] This vehicle is arrival number \(35 \times 18 = 630\), and it departs when the departure curve reaches 630: two more minutes of restricted discharge deliver \(16 \times 2 = 32\) vehicles, leaving \(342 - 32 = 310\) to be cleared at 40 veh/min, which takes 7.75 min. Its wait is therefore \[ w = 2.0 + 7.75 = \boxed{9.75\ \text{min}} \] or 9 min 45 s, departing at 5:32:45 PM. The same figure follows from the departure curve directly, \(t_{\text{dep}} = 20 + (630-320)/40 = 27.75\) min, giving \(27.75 - 18 = 9.75\) min.

Part	Quantity	Result
(a)	Queue forms? Maximum queue length	Yes — 380 vehicles at 5:25 PM
(b)	Time the queue clears	6:41 PM (96 min after 5:05 PM)
(c)	Longest wait of any vehicle	10.86 min (10 min 51 s)
(d)	Total delay	18 240 veh·min = 304 veh·h
(d)	Average delay per vehicle (3360 vehicles)	5.43 min/veh (326 s/veh)
(e)	Vehicles ahead of the 5:23 PM entrant	342 vehicles
(e)	Wait for that driver	9.75 min (9 min 45 s)

Topic. Deterministic (D/D/1) queueing at a temporary freeway bottleneck, solved by cumulative arrival–departure curves.

Key relations. \(Q(t) = A(t) - D(t)\); \(Q_{\max} = (\lambda-\mu_1)T_1\); dissipation time \(= Q_{\max}/(\mu_2-\lambda)\); \(w_{\max} = T_1 - D(T_1)/\lambda\); total delay \(= \int [A(t)-D(t)]\,dt\); \(\bar{d} = D_{\text{tot}}/(\lambda T_2)\).

Why this works. When demand is deterministic and known, no probability is needed: the cumulative arrival curve is a straight line of slope \(\lambda\), the departure curve is a straight line of slope equal to whatever capacity is available at the time, and the entire problem is plane geometry. The vertical separation is the queue because it counts vehicles that have arrived but not departed; the horizontal separation is an individual delay because, under first-in–first-out, the \(n\)th vehicle to arrive is the \(n\)th to leave; and the enclosed area, having units of vehicles multiplied by time, is total delay. This is also why the incident’s cost is so disproportionate to its duration — the area of the recovery triangle scales with the reciprocal of the surplus capacity, which here is only 5 veh/min out of 40.

Common pitfall. Assuming the queue clears when the obstruction is removed, or that the longest wait belongs to the vehicle at the back of the longest queue. Both are wrong: the queue keeps growing until capacity exceeds demand, and the worst-off driver is the last one released under the reduced capacity, who arrived long before the queue reached its maximum. A third error is dividing the total delay by only the vehicles queued at the peak rather than by every vehicle that was delayed.

Source. Garber & Hoel, Traffic and Highway Engineering, Ch. 6, “Deterministic queueing and the analysis of freeway incidents”; HCM, freeway-facility oversaturated-flow procedure.

Question 3 of 7: Deterministic Queueing behind a Freeway Incident

Question 3: Deterministic Queueing behind a Freeway Incident (20 marks — 4 each)