Tolerable Arrival Rate for an M/M/1 Landing Queue: 23-Ind-A1 Operations Research, May 2016, Q6

Question 6: Tolerable Arrival Rate for an M/M/1 Landing Queue (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. Single landing runway; service (landing) time exponential with mean $1/\mu=1.5$ min $\Rightarrow \mu=2/3$ aircraft/min; arrivals occur "at random" ⇒ Poisson arrivals, so the landing runway is an $M/M/1$ queue; required $W_q\le 3$ min.

Find. The maximum arrival rate $\lambda$ (aircraft/min, and per hour) that keeps the average time waiting in the sky (queueing delay before landing) at or below 3 minutes.

Approach. Use the standard $M/M/1$ waiting-time formula $W_q=\rho/(\mu-\lambda)$, set it equal to the 3-minute limit, and solve the resulting equation for $\lambda$.

Write the M/M/1 mean-wait-in-queue formula. With utilization $\rho=\lambda/\mu$, $$W_q=\dfrac{L_q}{\lambda}=\dfrac{\rho^2/(1-\rho)}{\lambda}=\dfrac{\lambda}{\mu(\mu-\lambda)}.$$
Impose the 3-minute limit and solve for $\lambda$. Setting $W_q=3$ and clearing denominators: $$\dfrac{\lambda}{\mu(\mu-\lambda)}\le 3\ \Longrightarrow\ \lambda\le 3\mu(\mu-\lambda) \ \Longrightarrow\ \lambda(1+3\mu)\le 3\mu^2 \ \Longrightarrow\ \lambda\le \dfrac{3\mu^2}{1+3\mu}.$$ Substituting $\mu=2/3$: $$\lambda \le \dfrac{3(2/3)^2}{1+3(2/3)} = \dfrac{4/3}{3} = \boxed{\dfrac{4}{9}\approx 0.444\ \text{aircraft/min}}.$$
Convert to a practical rate and check. $4/9$ aircraft/min $\times\,60 = \boxed{26.7\ \text{aircraft/hr}}$. Checking: at $\lambda=4/9$, $W_q=(4/9)\big/\big[(2/3)(2/3-4/9)\big]=(4/9)/(2/3\cdot2/9)=(4/9)/(4/27)=3$ min exactly, confirming the bound is tight.

Final results — Question 6
Quantity	Value
Landing service rate $\mu$	0.667 aircraft/min
Maximum tolerable $\lambda$	4/9 ≈ 0.444 aircraft/min
Same, per hour	≈ 26.7 aircraft/hr
Resulting utilization $\rho=\lambda/\mu$	2/3 ≈ 0.667

Topic: M/M/1 queueing model — mean waiting time in queue.

Key relations: $L_q=\rho^2/(1-\rho)$, $W_q=L_q/\lambda=\rho/(\mu-\lambda)$, $W=W_q+1/\mu$ (Little's Law family), valid for $\lambda<\mu$.

Why this works: Exponential inter-arrival and service times make the number of aircraft either landed or waiting a birth–death Markov chain, whose steady-state probabilities collapse to the simple closed-form $L_q,W_q$ expressions above; because $W_q$ grows without bound as $\lambda\to\mu$, "tolerable" arrival rates are always found by capping $W_q$ well before saturation, exactly as this question does.

Common pitfall: Solving for $\lambda$ using $W$ (total time including the landing itself) instead of $W_q$ (time waiting in the sky before landing begins) — the question explicitly asks for the wait in the sky, i.e. the queueing delay only, not the 1.5-minute landing time itself.

Source: Hillier & Lieberman, Introduction to Operations Research (11th ed.), ch. 17 (queueing theory), the M/M/1 model.

Question 6 of 8: Tolerable Arrival Rate for an M/M/1 Landing Queue

Question 6: Tolerable Arrival Rate for an M/M/1 Landing Queue (20 marks)