Question 2 of 7: Deterministic queueing at a loading dock

Question 2: Deterministic queueing at a loading dock (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 deterministic (D/D/1) queue at a single dock, measuring time $t$ in minutes after 8:00 am.

Find. The time the queue clears, the maximum queue length, the longest individual waiting time, and the total and average delay from 8:00 am until the queue clears.

Approach. Build the cumulative arrival and departure functions, then read the queue length as the vertical gap and the delay as the area between the curves; the queue clears where cumulative arrivals equal cumulative departures.

1. Write the cumulative arrival curve. Arrivals accumulate at 8/min then 3/min:
   $$A(t)=\begin{cases}8t, & 0\le t\le 30\\ 240+3\,(t-30), & t\gt 30\end{cases}$$
   so $A(30)=240$ trucks have arrived by 8:30.
2. Write the cumulative departure curve. The dock is closed until $t=15$, then serves 6/min:
   $$D(t)=\begin{cases}0, & 0\le t\le 15\\ 6\,(t-15), & t\gt 15\end{cases}$$
3. Find when the queue clears. After 8:30 arrivals (3/min) fall below service (6/min), so the queue drains. Setting $A(t)=D(t)$ on that branch:
   $$240+3(t-30)=6(t-15)\;\Rightarrow\;150+3t=6t-90\;\Rightarrow\;t=80\ \text{min}.$$
   Check: $A(80)=150+240=390$ and $D(80)=6(65)=390$. The queue clears
   $$\boxed{t=80\ \text{min after 8:00 am} = 9\!:\!20\ \text{am}.}$$
4. Maximum queue length (b). The queue grows while arrivals exceed departures. For $15\le t\le 30$ the net rate is $8-6=+2$/min; for $t\gt 30$ it is $3-6=-3$/min, so the queue peaks at the break point $t=30$:
   $$Q_{\max}=A(30)-D(30)=240-6(15)=240-90=\boxed{150\ \text{trucks}.}$$
5. Longest waiting time (c). With first‑in‑first‑out service the wait is the horizontal gap. The critical truck is the last to arrive during the build‑up, the 240th (arriving at $t=30$). It departs when $D=240$: $6(t_d-15)=240\Rightarrow t_d=55$. Hence
   $$w_{\max}=t_d-t_a=55-30=\boxed{25\ \text{min}.}$$
6. Total delay (d‑1). Total delay is the area between the two curves. Using the queue function $Q(t)=A(t)-D(t)$ over its three linear segments:
   $$W=\int_0^{15}\!8t\,dt+\int_{15}^{30}\!(2t+90)\,dt+\int_{30}^{80}\!(240-3t)\,dt=900+2025+3750.$$
   $$\boxed{W=6675\ \text{truck-minutes}.}$$
7. Average delay (d‑2). A total of $A(80)=390$ trucks pass through, so
   $$\bar w=\frac{W}{N}=\frac{6675}{390}=\boxed{17.12\ \text{min per truck}.}$$