M/M/1 Queueing System

Definition

The M/M/1 Queueing System is a single-server queue where:

• Customers arrive according to a Poisson process with rate $\lambda$
• Service times are i.i.d. $\text{Exp}(\mu )$
• There is one server
• Queue discipline is FIFO

The number of customers in the system $X(t)$ is a birth and death process with:

${\lambda }_n=\lambda ,n\ge 0$ ${\mu }_n=\mu ,n\ge 1$

Steady-State Condition

The system reaches steady state only when $\lambda <\mu$ (or equivalently, $\rho =\lambda /\mu <1$).

Steady-State Probabilities

The long-run probability of $n$ customers in the system:

$P_n=\left(1-\frac{\lambda }{\mu }\right){\left(\frac{\lambda }{\mu }\right)}^n,n\ge 0$

This is a geometric distribution with parameter $\rho =\lambda /\mu$.

Performance Metrics

Metric	Formula
${\pi }_0$ (system empty)	$1-\rho$
$L$ (avg. # in system)	$\frac{\lambda }{\mu -\lambda }=\frac{\rho }{1-\rho }$
$W$ (avg. time in system)	$\frac{1}{\mu -\lambda }$
$L_Q$ (avg. # in queue)	$\frac{{\rho }^2}{1-\rho }$
$W_Q$ (avg. wait in queue)	$\frac{\rho }{\mu -\lambda }$

NOTE

The waiting time distribution in M/M/1 is $T\sim \text{Exp}(\mu -\lambda )$. This follows because the total time in system for a customer finding $n$ ahead is $\text{Gamma}(n+1,\mu )$, and unconditioning over the geometric distribution of $n$ yields $\text{Exp}(\mu -\lambda )$.

Busy Period

The mean busy period is $\mathbb{E}[B]=\frac{1}{\mu -\lambda }$. The mean idle period is $\mathbb{E}[I]=\frac{1}{\lambda }$.

The proportion of time the system is empty: ${\pi }_0=\frac{\mathbb{E}[I]}{\mathbb{E}[I]+\mathbb{E}[B]}=1-\rho$.

Traffic Intensity

$\rho =\frac{\lambda }{\mu }$

As $\rho \to 1$, $L\to \infty$ (the system “blows up”). A well-functioning system requires $\rho <1$.

Related

• Queueing System
• s Queueing System
• ∞ Queueing System
• Birth and Death Queueing Models

Exercises

Back to Roadmap 📖 → 🃏 → ✏

Kuis 2 2025 No. 10. M/M/1 dengan kapasitas 2 (finite capacity). $\lambda =2$/menit, $\mu =3$/menit. Hitung $\rho$, cek kondisi steady-state, dan tentukan ${\pi }_n$ untuk $n=0,1,2$.

Jawaban: $\rho =2/3<1$ ✓. Untuk finite capacity, steady-state selalu ada. ${\pi }_1=\rho {\pi }_0$, ${\pi }_2={\rho }^2{\pi }_0$ (dengan ${\lambda }_2=0$). ${\pi }_0=\frac{1}{1+\rho +{\rho }^2}=\frac{9}{19}$, ${\pi }_1=\frac{6}{19}$, ${\pi }_2=\frac{4}{19}$.

M/M/1 — metrik performa. Untuk M/M/1 dengan $\lambda =4$/jam, $\mu =6$/jam, hitung $L$ dan $W$.

Jawaban: $\rho =2/3$. $L=\frac{\rho }{1-\rho }=\frac{2/3}{1/3}=2$ pelanggan. $W=\frac{1}{\mu -\lambda }=\frac{1}{2}=0.5$ jam.