Question 2 of 7: Single-Channel Queue at a Bank Teller
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format.98-Civ-B10 Traffic Engineering, National Examinations, December 2015. Three-hour duration, OPEN BOOK, any non-communicating calculator permitted. Seven questions, all of equal value (20 marks each), with the mark split printed in the grading scheme; the paper requires a total of five solutions. All seven are worked below, because this document is a study resource rather than an exam script.
Reference texts. Garber, N. J. and Hoel, L. A., Traffic and Highway Engineering — Ch. 3 (driver characteristics and the PIEV process), Ch. 4 (traffic-engineering studies: spot-speed studies, time-mean and space-mean speed), Ch. 6 (fundamental principles of traffic flow, deterministic and stochastic queueing) and Ch. 8 (intersection control: saturation flow, change and clearance intervals, Webster’s green split and delay). Transportation Research Board, Highway Capacity Manual — signalised-intersection capacity, degree of saturation, control delay and level of service. AASHTO, A Policy on Geometric Design of Highways and Streets, and Transportation Association of Canada, Geometric Design Guide for Canadian Roads — stopping and passing sight distance, crest vertical-curve design and K-values. FHWA, Manual on Uniform Traffic Control Devices, and Transportation Association of Canada, Manual of Uniform Traffic Control Devices for Canada — traffic-signal warrants and no-passing-zone marking.
Assumptions declared under the paper’s NOTE 1 and NOTE 2 (“any data required, but not given, can be assumed”). These are used throughout and are not repeated in every question: bus passenger-car equivalent \(E_B = 2.0\); pedestrian walking speed \(S_p = 1.2\ \text{m/s}\); driver perception–reaction time 2.5 s and deceleration \(a = 3.4\ \text{m/s}^2\) for stopping sight distance; AASHTO/TAC metric sight-distance eye and object heights; first-in–first-out discipline in every queueing calculation.
Question 2: Single-Channel Queue at a Bank Teller (20 marks — 4 each)
Given. A single service channel (one teller window), Poisson arrivals and negative-exponential service times — the M/M/1 model.
Quantity
Symbol
Value
Mean arrival rate
\(\lambda\)
15 customers/h
Mean service rate
\(\mu\)
17 customers/h
Number of service channels
\(N\)
1
Queue discipline
—
first-in, first-out, unlimited waiting room
Find. The idle probability \(P_0\), the mean queue length \(L_q\), the mean number in the system \(L\), the mean waiting time \(W_q\) and mean time in the system \(W\), and the probability that more than four customers are present.
Approach. Confirm the system is stable (\(\rho < 1\)), then evaluate the standard M/M/1 steady-state results in the order \(\rho \to P_0 \to L_q \to L \to W_q, W \to P(n > N)\), cross-checking each against Little’s law.
Establish the traffic intensity and confirm stability. The utilisation factor is the ratio of demand to service capability,
\[ \rho = \frac{\lambda}{\mu} = \frac{15}{17} = 0.8824 \]
Since \(\rho < 1\) the teller can keep up on average and a steady state exists. The teller is busy 88.24 % of the time; the queue is nevertheless long, because \(\rho\) is close to unity and M/M/1 measures blow up as \(\rho \to 1\).
Part (a) — probability that the teller is idle. For a single channel the probability of zero customers in the system is the complement of the utilisation,
\[ P_0 = 1 - \rho = 1 - 0.8824 = \boxed{0.1176} \]
so the teller is free from processing a customer about 11.8 % of the time, roughly 7 minutes in every hour.
Part (b) — average number waiting to be processed. The mean queue length excludes the customer in service:
\[ L_q = \frac{\lambda^{2}}{\mu\,(\mu - \lambda)} = \frac{15^{2}}{17\,(17-15)} = \frac{225}{34} = 6.62\ \text{customers} \]
Equivalently \(L_q = \rho^2/(1-\rho) = 0.7785/0.1176 = 6.62\), which confirms the arithmetic.
Part (c) — average number in the bank. The mean number in the system adds the customer being served:
\[ L = \frac{\lambda}{\mu - \lambda} = \frac{15}{17-15} = 7.5\ \text{customers} \]
The consistency check is \(L = L_q + \rho = 6.618 + 0.882 = 7.500\) — the expected number in service is exactly the utilisation, because there is only one server and it is occupied a fraction \(\rho\) of the time.
Part (d) — waiting time and total time in the bank. Little’s law converts numbers into times by dividing by the arrival rate. Waiting in the queue,
\[ W_q = \frac{L_q}{\lambda} = \frac{6.618}{15} = 0.4412\ \text{h} = 26.5\ \text{min} \]
and total time in the bank,
\[ W = \frac{L}{\lambda} = \frac{7.5}{15} = 0.500\ \text{h} = 30.0\ \text{min} \]
The difference \(W - W_q = 3.5\) min is exactly the mean service time \(1/\mu = 1/17\) h \(= 3.53\) min, as it must be.
Part (e) — probability of more than four people present. For M/M/1 the number in the system is geometrically distributed, \(P(n) = \rho^{n}(1-\rho)\), so the tail probability collapses to a single power:
\[ P(n > N) = \rho^{\,N+1} \quad\Rightarrow\quad P(n > 4) = \rho^{5} = 0.8824^{5} = \boxed{0.535} \]
There is therefore a 53.5 % chance that more than four customers are in the bank at a randomly chosen instant — an unmistakable signal that one teller is inadequate for this demand.
Check — reading of “in line”. The result 0.535 counts everyone present, including the customer at the window, which is the convention behind the standard tail formula \(P(n>N)=\rho^{N+1}\) and the reading adopted here. If “in line” is instead read strictly as waiting, then more than four waiting means at least six in the system, giving \(P(n \ge 6) = \rho^{6} = 0.472\). Both readings are defensible; the answer is boxed on the system convention and the alternative is stated, as the paper’s NOTE 1 invites.
