Question 5 of 6: Single-channel drive-up window with one and two servers

Question 5: Single-channel drive-up window with one and two servers (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 mean waiting time and the server idle fraction under each staffing arrangement, and a reasoned recommendation on the second hire.

Approach. Both cases are M/M/1 — the second person does not add a channel, it raises $\mu$ — so the whole question follows from the single-server formulae once $\rho = \lambda/\mu$ is evaluated for each staffing level.

1. Part (a) — the single-server queue. A mean service time of 2 min gives $\mu = 60/2 = 30$ veh/h, so $$\rho = \frac{\lambda}{\mu} = \frac{24}{30} = 0.80,\qquad W_q = \frac{\lambda}{\mu\,(\mu - \lambda)} = \frac{24}{30\,(30 - 24)}\ \text{h} = \frac{24}{180}\ \text{h}.$$ Converting to minutes, $$\boxed{\;W_q = 0.1333\ \text{h} = 8.0\ \text{min of queueing}\;}$$ with the corresponding mean queue $L_q = \rho^2/(1-\rho) = 3.2$ vehicles and a total time in the system of $W = W_q + 1/\mu = 10.0$ min.
2. Part (b) — idle fraction with one server. The probability that the system is empty is the complement of the utilisation: $$P_0 = 1 - \rho = 1 - 0.80 = 0.20 \;\Rightarrow\; \boxed{\;\text{the server is unoccupied }20\ \%\text{ of the time}\;}$$ Averaged over an hour that is 12 minutes with no car at the window — and yet the average customer still queues for eight minutes, which is the counter-intuitive signature of a system running at $\rho = 0.8$: idleness and long waits coexist because the arrivals are random, not evenly spaced.
3. Part (c) — the two-person queue. A service time of 1 min 15 s is 1.25 min, so $\mu = 60/1.25 = 48$ veh/h and $\rho = 24/48 = 0.50$. The same M/M/1 expression now gives $$W_q = \frac{24}{48\,(48 - 24)}\ \text{h} = \frac{24}{1152}\ \text{h} = 0.02083\ \text{h},$$ $$\boxed{\;W_q = 1.25\ \text{min of queueing}\;}$$ with $L_q = 0.5$ vehicles and a total time in the system of $W = 2.5$ min.
4. Part (d) — idle fraction with two people. Because the window is still a single channel, the system is empty whenever no customer is present, so $$P_0 = 1 - \rho = 1 - 0.50 \;\Rightarrow\; \boxed{\;\text{no one is occupied }50\ \%\text{ of the time}\;}$$ Note the honest reading of this figure: the pair is idle for half of every hour, but they are idle together, so the labour cost of that idleness has doubled while the number of customers served is unchanged.
5. Part (e) — make the recommendation. The case for the hire is the queueing improvement, which is large: $$\text{waiting time } 8.0 \rightarrow 1.25\ \text{min}\ (-84\ \%),\qquad \text{queue } 3.2 \rightarrow 0.5\ \text{vehicles},$$ saving $24(8.0 - 1.25)/60 = 2.7$ customer-hours of waiting in every hour of operation. An eight-minute wait at a drive-up window is beyond what customers tolerate; it drives balking (cars that see the queue and pass by) and it produces a 3.2-car standing queue that will often spill back onto the street, which is a traffic-engineering problem as much as a commercial one. The case against is that the second person is paid for a full shift to remove waiting that only exists at the peak, and is idle half the time. On balance, yes — hire the second person, but staff the position only through the peak periods. The decision is really a threshold problem: the second server is worth its wage while the value of the 2.7 customer-hours saved per hour, plus the revenue recovered from customers who would otherwise balk, exceeds the hourly wage. Because both staffing levels are stable ($\rho < 1$ either way), this is a service-quality and spill-back decision rather than a capacity one, and it should be revisited if arrivals rise above about 40 veh/h, at which point the single window becomes unstable even with two people.

Final Results