Queueing Theory — Finite-Capacity Port (M/M/1/K): 23-Ind-A1 Operations Research, December 2013, Q10

Question 10: Queueing Theory — Finite-Capacity Port (M/M/1/K) (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. Poisson arrivals $\lambda=7$/week; exponential service $\mu=8$/week; 1 berth in service + 3 mooring spaces, so system capacity $K=4$ (arrivals finding the system full balk to the Port of Poop). This is a birth–death $M/M/1/K$ queue.

Finite-capacity birth–death chain: every state has the same $\lambda$ forward and $\mu$ backward rate since there is exactly one server.

Find. $L_q$, the expected number of ships waiting for service (excluding the one being served).

Approach. Use the standard finite-capacity $M/M/1/K$ state probabilities $P_n=P_0\rho^n$ (truncated at $K=4$), compute $L$ (expected number in system) and $1-P_0$ (probability the server is busy), then $L_q=L-(1-P_0)$.

Traffic intensity and state probabilities. $\rho=\lambda/\mu=7/8=0.875$. For $M/M/1/K$ with $K=4$: $$P_0=\frac{1-\rho}{1-\rho^{K+1}}=\frac{1-0.875}{1-0.875^5}=\frac{0.125}{1-0.5129}=\frac{0.125}{0.4871}=0.2566,$$ and $P_n=P_0\rho^n$ for $n=0,\dots,4$: $P_0{=}0.2566,\,P_1{=}0.2245,\,P_2{=}0.1965,\,P_3{=}0.1719,\,P_4{=}0.1504$ (sum $=1.0000$ ✓).
Expected number in system. $$L=\frac{\rho}{1-\rho}-\frac{(K+1)\rho^{K+1}}{1-\rho^{K+1}}=\frac{0.875}{0.125}-\frac{5(0.5129)}{0.4871}=7-5.265=\boxed{1.735\text{ ships (in system)}}$$ (cross-checked against $L=\sum_{n=0}^{4}nP_n=1.735$ — matches).
Probability the server is busy. $1-P_0=1-0.2566=0.7434$ (expected number in service, since there is exactly one server).
Expected number waiting. $$\boxed{L_q=L-(1-P_0)=1.735-0.743=0.992\text{ ships waiting.}}$$

Quantity	Result
$\rho=\lambda/\mu$	$0.875$
$P_0$ (port idle)	$0.2566$
$L$ (expected number in system)	$1.735$ ships
$L_q$ (expected number waiting)	$\approx0.99$ ships

Concept & Theory

Topic. Finite-capacity single-server queue ($M/M/1/K$) with balking beyond capacity.

Key relations. Truncated geometric state distribution $P_n=P_0\rho^n$, $n=0,\dots,K$; $L=\rho/(1-\rho)-(K+1)\rho^{K+1}/(1-\rho^{K+1})$; $L_q=L-(1-P_0)$ (since exactly one unit is in service whenever the system is non-empty).

Why this works. Because arrivals that find all 4 slots occupied balk to the Port of Poop, the usual (infinite-capacity) $M/M/1$ formulas $L=\rho/(1-\rho)$ do not apply directly — the state space must be truncated at $K$ and $P_0$ renormalized over the finite set of states.

Common pitfall. Using the infinite-queue formula $L_q=\rho^2/(1-\rho)$ despite the explicit 3-mooring capacity limit — that formula assumes $\rho<1$ guarantees a proper (unbounded) steady state, which is irrelevant here since the finite $K$ makes a proper distribution exist for any $\rho$, even $\rho\ge1$.

Source. Hillier & Lieberman, Introduction to Operations Research, ch. 17 (Queueing Theory) — finite-capacity birth–death queueing models.

Question 10 of 10: Queueing Theory — Finite-Capacity Port (M/M/1/K)

Question 10: Queueing Theory — Finite-Capacity Port (M/M/1/K) (20 marks)