Question 4 of 6: Parking-Space Availability as a Multi-Channel Queue
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2014 — 98-Civ-B10 Traffic Engineering, 3-hour duration, OPEN BOOK (any non-communicating calculator permitted). Six questions are set; the paper requires a total of five solutions and states that all questions are of equal value (20 marks each, split as shown in the paper's own grading scheme). Because the paper is a study resource, all six questions are solved below. The paper's NOTE 1 invites a clear statement of any assumptions made, and NOTE 2 states that any data required but not given may be assumed — every assumed factor is therefore declared explicitly where it is first used.
Reference texts. Garber, N. J. and Hoel, L. A., Traffic and Highway Engineering, 5th ed. — Ch. 4 (traffic engineering studies: volume studies, peak-hour factor, spot-speed studies), Ch. 6 (fundamental principles of traffic flow: the two mean speeds, the Greenshields model, Poisson arrivals, negative-exponential headways, single- and multi-channel queueing) and Ch. 8 (intersection control: saturation flow, change and clearance intervals, cycle length). This is the principal reference for the subject. Transportation Research Board, Highway Capacity Manual — saturation-flow adjustment factors and the passenger-car-equivalent convention. Transportation Association of Canada, Geometric Design Guide for Canadian Roads and Manual of Uniform Traffic Control Devices for Canada (MUTCDC) — Canadian lane widths, crosswalk practice and clearance-interval policy. Institute of Transportation Engineers, Traffic Engineering Handbook — queueing applications to parking and drive-up facilities. Webster, F. V. and Cobbe, B. M., Traffic Signals, Road Research Laboratory Technical Paper No. 56 — the optimum-cycle relation quoted as a closing check in Question 5.
Question 4: Parking-Space Availability as a Multi-Channel Queue (20 marks)
Given. A parking facility with four spaces, exponential occupancy times of mean 6 minutes and Poisson (exponentially spaced) arrivals at 20 customers per hour in the design hour.
Parameter
Symbol
Value
Number of parking spaces (service channels)
N
4
Mean arrival rate
λ
20 veh/h
Mean occupancy (service) time
1/μ
6 min = 0.100 h
Service rate per space
μ
10 veh/h
Total service capacity
Nμ
40 veh/h
Find. The probability that an arriving customer finds all four spaces occupied, that is $P(n \ge 4)$ where n is the number of customers in the system.
State-transition diagram for the four-space facility. Arrivals move the system one state right at rate λ; departures move it left at rate nμ until all four spaces are busy, after which the departure rate saturates at 4μ. The shaded states are those in which no space is free.
Approach. Both the arrival and the service processes are exponential and the four spaces serve one queue, so this is a multi-channel M/M/N system with N = 4. Form the traffic intensity, confirm the system is stable, obtain the empty-system probability from the normalising sum, and then sum the state probabilities from n = 4 upwards.
Form the traffic intensity and test stability. The offered load in erlangs is the arrival rate divided by the service rate of one channel:
$$\rho = \frac{\lambda}{\mu} = \frac{20}{10} = 2.00, \qquad \frac{\rho}{N} = \frac{2.00}{4} = 0.500$$
Since $\rho/N = 0.5 < 1$, the four spaces can serve the demand on average and steady-state results are valid. Physically, the lot is busy half the time: two of its four spaces are occupied on average.
Obtain the probability that the lot is completely empty. For M/M/N the normalising condition $\sum_{n\ge 0} P_n = 1$ gives
$$P_0 = \left[\sum_{n=0}^{N-1}\frac{\rho^n}{n!} + \frac{\rho^N}{N!\left(1 - \rho/N\right)}\right]^{-1}$$
The first four terms are $\rho^0/0! = 1$, $\rho^1/1! = 2$, $\rho^2/2! = 2$ and $\rho^3/3! = 1.3333$, summing to 6.3333, and the tail term is $\rho^4/[4!(1-0.5)] = 16/(24 \times 0.5) = 1.3333$. Therefore
$$P_0 = \frac{1}{6.3333 + 1.3333} = \frac{1}{7.6667} = 0.1304$$
Sum the states in which every space is taken. The tail term already is that sum, scaled by $P_0$:
$$P(n \ge N) = \frac{\rho^N}{N!\left(1 - \rho/N\right)}\,P_0 = \frac{16}{24 \times 0.500} \times 0.1304 = 1.3333 \times 0.1304$$
$$P(n \ge 4) = \boxed{0.1739 \;\; (17.4\ \%)}$$
So on roughly one arrival in six during the busiest hour — about 3.5 of the 20 arrivals — the driver finds no space free.
Tabulate the individual state probabilities as a check. For $n \le N$ the state probabilities are $P_n = (\rho^n/n!)P_0$, giving $P_1 = 0.2609$, $P_2 = 0.2609$, $P_3 = 0.1739$ and $P_4 = 0.0870$. These five sum to 0.9130, so $P(n \ge 5) = 0.0870$ and
$$P(n \ge 4) = P_4 + P(n \ge 5) = 0.0870 + 0.0870 = 0.1739$$
which reproduces the boxed answer by an independent route. The most likely single state is one or two spaces occupied, each at about 26 %.
Quantify the consequence of the shortfall. The expected number waiting for a space and the expected wait follow from the same $P_0$:
$$L_q = \frac{\rho^N (\rho/N)}{N!\,(1-\rho/N)^2}\,P_0 = \frac{16 \times 0.500}{24 \times 0.250}\times 0.1304 = 0.174\ \text{veh}$$
$$W_q = \frac{L_q}{\lambda} = \frac{0.174}{20}\ \text{h} = 0.52\ \text{min} \approx 31\ \text{s}$$
The average customer therefore waits about half a minute for a space, and total time on site averages $W = W_q + 1/\mu = 0.52 + 6.00 = 6.52$ min. A 17 % chance of arriving to a full lot with only a 31-second average wait is a defensible level of service for a convenience store; the owner does not need a fifth space, but the aisle must be able to hold one waiting vehicle without blocking the street.
Check: two models are admissible here and the paper's NOTE 1 requires the choice to be stated. The answer above treats the lot as an M/M/4 system with an unlimited waiting area, so a customer who finds all four spaces full waits in the aisle; “will not find an open parking space” is then $P(n \ge 4) = 0.174$. If instead a customer who finds the lot full is assumed to drive away (a blocked-calls-cleared loss system, M/M/4/4), the Erlang-B formula $P_B = (\rho^N/N!)/\sum_{n=0}^{N}(\rho^n/n!) = 0.6667/7.0$ gives 0.0952, or 9.5 %. The M/M/N value is the one boxed because it is the multi-channel model developed in the standard traffic-engineering treatment of this problem class, and because a driver who cannot park immediately at a convenience store typically waits rather than leaves. The two figures bracket the answer, and either is creditable if the assumption is declared.
