Give the steady state parameters of M/G/1 queue and derive M/M/1 from M/G/1.
1 Answer


Single-Server Queues WithPoisson Arrivals and Unlimited Capacity: M/G/1

  • Suppose that service times have mean 1/µ and variance σ ⋇ σ and that there is one server.

  • If ρ=λ/µ<1, thenthe M/G/I queue has a steady-state probability distribution with steady-state characteristics, as given inTable 6.3.

  • In general, there is no simple expression for the steady-state probabilities P0, P1, P2,…

Table: Steady state parameters of the M/G/I Queue

$\rho$ $\frac{\lambda}{\mu}$
$\mathbf{L}$ $\rho+\frac{\lambda^{2}\left(1 / \mu^{2}+\sigma^{2}\right)}{2(1-\rho)}=\rho+\frac{\rho^{2}\left(1+\sigma^{3} \mu^{2}\right)}{2(1-\rho)}$
$\omega$ $\frac{1}{\mu}+\frac{\lambda\left(1 / \mu^{2}+\alpha^{2}\right)}{2(1-\rho)}$
$\omega_0$ $\frac{\lambda\left(1 / \mu^{2}+\sigma^{2}\right)}{2(1-\rho)}$
$L_0$ $\frac{\lambda^{2}\left(1 / \mu^{2}+\sigma^{2}\right)}{2(1-\rho)}=\frac{e^{2}\left(1+\sigma^{2} \mu^{2}\right)}{2(1-\rho)}$
$P_0$ $1-\rho$
  • When λ < µ, the quantity ρ = λ/µ is the server utilization, or long proportion of time the server is busy.

  • As is seen in Table 6.3, I - P0 = ϱ can also be interpreted as the steady-state probability that the system contains one or more customers.

Notice also that L - LQ = ϱ is the time-average number of customers being served.

The M/M/1 queue

  • Suppose that service times in an M/G/I queue are exponentially distributed, with mean 1/µ; then the variance as given as σ = l/µ.

  • The mean and standard: deviation of the exponential distribution are equal, so the M/M/1 queue will often be a useful approximate model when service times have standard deviations approximately equal to their means.

  • The steady-state parameters, given in Table 6.4, may be computed by substituting σ = l/µ into the formulas in Table 6.3. Alternatively,.given in Table 6.4, and then w, wQ, and LQ may be computed .

  • The student can show that the two expressions for each parameter are equivalent by substituting ρ= λ/µ into the right-hand side of each equation in Table 6.4.

Table: Steady-state parameters of M/M/1 Queue

$\mathbf{L}$ $\frac{\lambda}{\mu-\lambda}=\frac{\rho}{1-\rho}$
$\omega$ $\frac{1}{\mu-\lambda}=\frac{1}{\mu(1-\rho)}$
$\omega_Q$ $\frac{\lambda}{\mu(\mu-\lambda)}=\frac{\rho}{\mu(1-\rho)}$
$L_Q$ $\frac{\lambda^{2}}{\mu(t-\lambda)}=\frac{\rho^{2}}{1-\rho}$
$P_m$ $\left(1-\frac{\lambda}{\mu}\right)\left(\frac{\lambda}{\mu}\right)^{\prime}=(1-\rho) \rho^{*}$


$\begin{aligned} L &=\sum_{n=0}^{\infty} n P_{n} \\ &= \sum_{i=0}^{\infty} n \rho^{n}(1-p) \\ &=p(1-p) \sum_{n=0}^{\infty} n \rho^{n-1} \\ &= \rho(1-\rho) \sum_{n=0}^{\infty}\left(\frac{\partial}{\partial p} \rho^{n}\right) \\ &=\rho(1-\rho) \frac{\partial}{\partial p} \left( \begin{array}{c}{\infty} \\ {\sum} \\ {n=0} \end{array} \rho^n \right) \\ &=\rho(1-\rho) \frac{\partial}{\partial p}\left(\frac{1}{(1-\rho)}\right) \\ &=\rho(1-\rho) \frac{1}{(1-\rho)^{2}} \\ &=\frac{\rho}{(1-\rho)} \end{aligned}$

Average customer time in the system $w=\frac{L}{\lambda}$

As $L=\frac{p}{(1-\rho)} \quad w=\frac{L}{\lambda}=\frac{p}{\lambda(1-\rho)}$

put $\quad \rho=\frac{\lambda}{\mu} \quad w=\frac{\frac{\lambda}{\mu}}{\lambda\left[1-\frac{\lambda}{\mu}\right]}$

Solving we get $\quad w=\frac{1}{\mu-\lambda}$

We see that as mean arrival rate λ approaches mean service rate, the waiting time become very large.

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