M/M/1 Queue Example

Example

Consider an 1 queue with arrival rate $\lambda =4$ customers per hour and service rate $\mu =6$ customers per hour.

Step 1: Check Steady-State Condition

Traffic intensity: $\rho =\frac{\lambda }{\mu }=\frac{4}{6}=\frac{2}{3}<1$ ✓

Step 2: Steady-State Probabilities

$P_0=1-\rho =1-\frac{2}{3}=\frac{1}{3}$

$P_n=(1-\rho ){\rho }^n=\frac{1}{3}{\left(\frac{2}{3}\right)}^n,n\ge 1$

Step 3: Performance Metrics

Metric	Formula	Value
$L$ (avg. # in system)	$\frac{\rho }{1-\rho }$	$\frac{2/3}{1/3}=2$ customers
$W$ (avg. time in system)	$\frac{1}{\mu -\lambda }$	$\frac{1}{6-4}=0.5$ hours (30 min)
$L_Q$ (avg. # in queue)	$\frac{{\rho }^2}{1-\rho }$	$\frac{4/9}{1/3}=\frac{4}{3}\approx 1.33$ customers
$W_Q$ (avg. wait in queue)	$\frac{\rho }{\mu -\lambda }$	$\frac{2/3}{2}=\frac{1}{3}$ hour (20 min)

Step 4: Verify Little’s Formula

$L=\lambda W⟹2=4\times 0.5$ ✓ $L_Q=\lambda W_Q⟹\frac{4}{3}=4\times \frac{1}{3}$ ✓

Step 5: Busy Period

Mean busy period: $\mathbb{E}[B]=\frac{1}{\mu -\lambda }=\frac{1}{2}=0.5$ hours

Mean idle period: $\mathbb{E}[I]=\frac{1}{\lambda }=\frac{1}{4}=0.25$ hours

Proportion of empty time: $\frac{\mathbb{E}[I]}{\mathbb{E}[I]+\mathbb{E}[B]}=\frac{0.25}{0.75}=\frac{1}{3}=P_0$ ✓

NOTE

The server is busy $\frac{2}{3}$ of the time and idle $\frac{1}{3}$ of the time. On average, there are 2 customers in the system, with typical waiting time of 30 minutes total (20 minutes in queue, 10 minutes in service).

Related

• 1 Queueing System
• Queueing Formula L = λW