M/M/s Queueing System

Definition

The M/M/s Queueing System is a multi-server queue where:

• Customers arrive according to a Poisson process with rate $\lambda$
• Service times are i.i.d. $\text{Exp}(\mu )$
• There are $s$ servers, each working at rate $\mu$
• Queue discipline is FIFO

$X(t)$ is a birth and death process with:

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

${\mu }_n=\{\begin{array}{ll}n\mu ,& 1\le n\le s\\ s\mu ,& n>s\end{array}$

Interpretation

When $n\le s$ customers are in the system, $n$ servers are busy and each works at rate $\mu$, giving total departure rate $n\mu$. When $n>s$, all $s$ servers are busy, giving total departure rate $s\mu$. Customers beyond $s$ wait in queue.

Key Quantities

• Number of customers in service: $\min \{X(t),s\}$
• Number of customers in queue: $\max \{X(t)-s,0\}$

Traffic Intensity

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

Steady state exists only when $\rho <1$ (the arrival rate is less than the maximum service rate).

Steady-State Probabilities

The auxiliary quantity:

${\theta }_j=\{\begin{array}{ll}\frac{(\lambda /\mu {)}^j}{j!},& j\le s\\ \frac{(\lambda /\mu {)}^s}{s!}{\left(\frac{\lambda }{s\mu }\right)}^{j-s},& j>s\end{array}$

Then ${\pi }_0=1/{\sum }_{j=0}^{\infty }{\theta }_j$ and ${\pi }_k={\theta }_k{\pi }_0$.

Performance Metrics

• Mean queue length: $L_Q={\sum }_{k=s+1}^{\infty }(k-s){\pi }_k$
• Mean waiting time in queue: $W_Q=L_Q/\lambda$
• Mean time in system: $W=W_Q+1/\mu$
• Mean # in system: $L=\lambda W=L_Q+\lambda /\mu$

Related

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

Exercises

Back to Roadmap 📖 → 🃏 → ✏

M/M/2 — konstruksi Q. Dua server identik, $\lambda =4$/jam, $\mu =3$/jam per server. Tentukan matriks generator $Q$ untuk state $\{0,1,2,3\}$.

Jawaban: ${\lambda }_n=4$ untuk semua $n$. ${\mu }_1=3$, ${\mu }_2=6$, ${\mu }_3=6$. $Q=\left[\begin{array}{cccc}-4& 4& 0& 0\\ 3& -7& 4& 0\\ 0& 6& -10& 4\\ 0& 0& 6& -10\end{array}\right]$.

Cek kondisi steady-state. Kapan M/M/2 mencapai steady-state?

Jawaban: $\rho =\frac{\lambda }{s\mu }=\frac{4}{2\cdot 3}=\frac{2}{3}<1$ → steady-state exists.