Question 4 of 7: Single-Server Queueing at a Campus Cashier
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations — May 2019, 16-Civ-B10 Traffic Engineering, 3-hour duration, OPEN BOOK (any non-communicating calculator permitted). Seven questions over four pages; a total of five solutions is required and all questions are of equal value (20 marks each). Grading scheme printed on page 1: Q1 (a)–(e) 4 marks each; Q2 20 marks; Q3 20 marks; Q4 (a)–(e) 4 marks each; Q5 (a) 6 marks, (b) and (c) 7 marks each; Q6 (a)–(e) 4 marks each; Q7 (a)–(h) 2.5 marks each. All seven questions are solved here, because the set is a study resource rather than a timed attempt. The paper's own NOTE 1 invites a clear statement of assumptions and NOTE 2 permits any datum not given to be assumed — both are used below and every such assumption is flagged.
Reference texts. Garber & Hoel, Traffic and Highway Engineering, 5th ed. (the EGBC-recommended reference for this exam code) — Ch. 6 (traffic studies and the moving-vehicle method), Ch. 8 (queueing and D/D/1 signal delay), Ch. 8/9 (signal timing and the Webster method), Ch. 3 (sight distance and vertical curves). Transportation Research Board, Highway Capacity Manual — signalised-intersection methodology and the pedestrian minimum-green relation. Transportation Association of Canada, Geometric Design Guide for Canadian Roads — Canadian practice for sight distance and vertical alignment. AASHTO, A Policy on Geometric Design of Highways and Streets — the 2001 stopping-sight-distance table reproduced on page 3 of this paper.
Question 4: Single-Server Queueing at a Campus Cashier (5 × 4 = 20 marks)
Given. A single cashier (one server), arrivals \(\lambda = 5\) customers per 10 min \(= 0.5\) customers/min, mean service rate \(\mu = 6\) customers per 10 min \(= 0.6\) customers/min.
Find. The idle probability, the mean queue and system contents, the mean waiting and system times, and the probability that the queue trigger for a second cashier is exceeded.
The cashier modelled as an M/M/1 system: Poisson arrivals at 0.5 customers/min, negative-exponential service at 0.6 customers/min, one server, FIFO discipline and an unlimited waiting area.
Approach. Treat the counter as an M/M/1 system — Poisson arrivals, negative-exponential service times, one server, first-in-first-out discipline and no limit on the waiting area — confirm it is stable, then read the standard results off the traffic-intensity \(\rho\).
Establish the traffic intensity and confirm stability. Both rates must be in the same time units:
$$\rho=\frac{\lambda}{\mu}=\frac{0.5}{0.6}=\frac{5}{6}=0.8333$$
Since \(\rho < 1\) the system is stable and steady-state formulas apply. The cashier is busy 83.3% of the time — heavily loaded, which is why the queue results below are large.
Part (a) — Probability that the cashier is free. The cashier is free when the system is empty, and for M/M/1 the probability of \(n\) customers in the system is \(P_n = \rho^{n}(1-\rho)\), so
$$P_0=1-\rho=1-0.8333=\boxed{0.1667}$$
The cashier is idle about 1 minute in every 6, or 16.7% of the time.
Part (b) — Average number waiting to be processed. "Waiting to be processed" excludes the customer currently at the till, so this is the mean queue length \(L_q\):
$$L_q=\frac{\rho^{2}}{1-\rho}=\frac{\lambda^{2}}{\mu(\mu-\lambda)}=\frac{(0.8333)^{2}}{0.1667}=\boxed{4.17\ \text{customers}}$$
Part (c) — Average number of customers in line. Taken as the mean number in the system — the queue plus the one being served — which is the quantity complementary to part (b):
$$L=\frac{\rho}{1-\rho}=\frac{\lambda}{\mu-\lambda}=\frac{0.8333}{0.1667}=\boxed{5.00\ \text{customers}}$$
The cross-check is exact and worth stating: \(L = L_q + \rho = 4.1667 + 0.8333 = 5.00\), i.e. the extra customer over part (b) is the fraction of the time someone is actually at the till.
Part (d) — Average waiting time and average time being processed. Little's law converts the counts to times, \(W = L/\lambda\):
$$W_q=\frac{L_q}{\lambda}=\frac{4.1667}{0.5}=\boxed{8.33\ \text{min}}\qquad\text{(waiting in line, before reaching the till)}$$
$$\frac{1}{\mu}=\frac{1}{0.6}=\boxed{1.67\ \text{min}}\qquad\text{(mean time actually being served)}$$
$$W=W_q+\frac{1}{\mu}=8.33+1.67=\boxed{10.00\ \text{min}}\qquad\text{(total time in the system)}$$
The check \(W = L/\lambda = 5.00/0.5 = 10.0\) min closes exactly.
Part (e) — Probability that a second cashier is opened. The trigger is a line longer than five customers. For M/M/1 the probability that the system contains more than \(N\) customers is the geometric tail
$$P(n>N)=\rho^{\,N+1}$$
$$P(n>5)=\left(\tfrac{5}{6}\right)^{6}=\boxed{0.335}$$
So a second cashier is warranted about 33.5% of the time — roughly one-third of the peak period, which is a strong operational argument for scheduling a second server rather than opening one reactively.
Check — two readings declared under NOTE 1. (1) Parts (b) and (c) overlap in wording; they are answered as the two standard complementary results, \(L_q\) (waiting, excluding the customer in service) for (b) and \(L\) (in the system, including the one being served) for (c). Read strictly, "in line" could repeat (b), in which case both answers are 4.17 customers — both values are reported above so either marking scheme is satisfied. (2) In part (e), "the line of customers is longer than five" is taken as more than five in the system, giving \(\rho^{6} = 0.335\). If "line" is read as the waiting line only, the trigger is six in the system and the probability is \(\rho^{7} = 0.279\). The system reading is the textbook convention for \(P(n > N)\) and is boxed; the alternative differs by 5.6 percentage points and does not change the operational conclusion.
