BUS606: Operations and Supply Chain Management

The queueing system M/M/1 is the simplest non-trivial queue where the customers arrive according to a Poisson process with rate $\lambda$, that is, the inter-arrival times are independent, exponentially distributed random variables with parameter $\lambda$. The service times are assumed to be independent and exponentially distributed with parameter $\mu$. Furthermore, all the involved random variables are supposed to be independent of each other.

Let $\rho=\frac{\lambda}{\mu} < 1$, then $\mathrm{C}_{\mathrm{N}}=\left(\frac{\lambda}{\mu}\right)^{\mathrm{N}}=\rho^{\mathrm{N}}$ for $\mathrm{N}=1,2,3, \ldots$

Therefore, $P_{N}=C_{N} P_{0}$. Now, the normalizing condition is

$\begin{aligned} &\sum_{\mathrm{N}=0}^{\infty} \mathrm{P}_{\mathrm{N}}=1 \\ &\Rightarrow \quad\left(1+\sum_{\mathrm{N}=1}^{\infty} \mathrm{C}_{\mathrm{N}}\right) \mathrm{P}_{0}=1 \\ &\Rightarrow \quad \mathrm{P}_{0}=\frac{1}{\left(1+\sum_{\mathrm{N}=1}^{\infty} \mathrm{C}_{\mathrm{N}}\right)} \\ &\Rightarrow \quad \mathrm{P}_{0}=\frac{1}{\left(1+\sum_{\mathrm{N}=1}^{\infty} \rho^{\mathrm{N}}\right)} \\ &\Rightarrow \quad \mathrm{P}_{0}=\frac{1}{\left(\rho^{0}+\sum_{\mathrm{N}=1}^{\infty} \rho^{\mathrm{N}}\right)} \\ &\Rightarrow \quad \mathrm{P}_{0}=\frac{1}{\left(\rho^{0}+\sum_{\mathrm{N}=1}^{\infty} \rho^{\mathrm{N}}\right)} \\ &\Rightarrow \quad \mathrm{P}_{0}=\frac{1}{\sum_{\mathrm{N}=0}^{\infty} \rho^{\mathrm{N}}} \\ &\Rightarrow \quad \mathrm{P}_{0}=\left(\frac{1}{1-\rho}\right)^{-1} \\&\Rightarrow \quad \mathrm{P}_{0}=1-\rho \end{aligned}$

Thus, $P_{N}=(1-\rho) \rho^{N}, \quad$ for $N=0,1,2, \ldots$

Consequently, average number of customers in the system is
$\begin{aligned} &\mathrm{L}_{\mathrm{s}}=\sum_{\mathrm{N}=0}^{\infty} \mathrm{NP}^{N} \\ &\Rightarrow \quad \mathrm{L}_{\mathrm{s}}=\sum_{\mathrm{N}=0}^{\infty} \mathrm{N}(1-\rho) \rho^{N} \\ &\Rightarrow \quad \mathrm{L}_{\mathrm{s}}=\rho(1-\rho) \sum_{\mathrm{N}=0}^{\infty} \frac{\mathrm{d}}{\mathrm{d} \rho} \rho^{N} \\ &\Rightarrow \quad \mathrm{L}_{\mathrm{s}}=\rho(1-\rho) \frac{\mathrm{d}}{\mathrm{d} \rho}\left(\sum_{\mathrm{N}=0}^{\infty} \rho^{N}\right) \\ &\Rightarrow \quad \mathrm{L}_{\mathrm{s}}=\rho(1-\rho) \frac{\mathrm{d}}{\mathrm{d} \rho}\left(\frac{1}{1-\rho}\right) \\ &\Rightarrow \quad \mathrm{L}_{\mathrm{s}}=\rho(1-\rho) \frac{1}{(1-\rho)^{2}} \\ &\Rightarrow \quad \mathrm{L}_{\mathrm{s}}=\frac{\rho}{1-\rho}=\frac{\lambda}{\mu-\lambda} \end{aligned}$

Summarizing the results, we have following conclusions:

i. The probability of having zero customers in the system

$\mathrm{P}_{0}=1-\rho$

ii. The probability of having $N$ customers in the system

$P_{N}=\rho^{N} P_{0}$

iii. Average number of customers in system

$\\mathrm{L}_{\mathrm{s}}=\frac{\rho}{(1-\rho)}$

iv. Average number of customers in the queue

$L_{q}=\frac{\rho^{2}}{(1-\rho)}$

v. Average waiting time in the system

$\mathrm{W}_{\mathrm{s}}=\frac{\rho}{\lambda(1-\rho)}$

vi. Average waiting time in the queue

$\mathrm{W}_{\mathrm{q}}=\frac{\rho}{\mu(1-\rho)}$