M/M/∞ Queueing System

Definition

The M/M/∞ Queueing System is a queue with infinite servers where:

• Customers arrive according to a Poisson process with rate $\lambda$
• Service times are i.i.d. $\text{Exp}(\mu )$
• There are infinitely many servers, so arriving customers are served immediately (no waiting)

This is a birth and death process with:

${\lambda }_n=\lambda ,\forall n$ ${\mu }_n=n\mu ,\forall n$

Interpretation

Since there are always enough servers, every arriving customer goes directly into service. The departure rate is proportional to the number currently being served because each server works independently at rate $\mu$.

Steady-State Probabilities

${\pi }_0=e^{-\lambda /\mu }$ ${\pi }_k=\frac{(\lambda /\mu {)}^k{e}^{-\lambda /\mu }}{k!},k\ge 0$

This is a Poisson distribution with mean $\lambda /\mu$.

Performance Metrics

Metric	Formula	Reasoning
$L$	$\lambda /\mu$	Mean of Poisson($\lambda /\mu$)
$W$	$1/\mu$	Average service time (no waiting)

These satisfy Little’s formula: $L=\lambda W$.

Related

• 1 Queueing System
• s Queueing System
• Poisson Process

Exercises

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M/M/∞ — steady-state. $\lambda =3$/jam, $\mu =2$/jam. Tentukan ${\pi }_n$.

Jawaban: ${\pi }_n=\frac{(3/2{)}^n{e}^{-3/2}}{n!}$. Poisson dengan mean $\lambda /\mu =1.5$. ${\pi }_0=e^{-1.5}\approx 0.223$, ${\pi }_1=1.5e^{-1.5}\approx 0.335$, ${\pi }_2=\frac{2.25}{2}{e}^{-1.5}\approx 0.251$, dst.

M/M/∞ vs M/M/1. Bandingkan $L$ untuk M/M/∞ dan M/M/1 dengan $\lambda =3$, $\mu =2$.

Jawaban: M/M/∞: $L=\lambda /\mu =1.5$ (tidak ada antrian — selalu langsung dilayani). M/M/1: $\rho =3/2>1$ → tidak mencapai steady-state! (sistem meledak). Perlu $\mu >\lambda$ untuk M/M/1.