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# 数学代写|随机过程Stochastic Porcesses代考|AMATH562 Queues with a single server

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## 数学代写|随机过程Stochastic Porcesses代考|The model M/M/1

We first consider a queueing system with a single server, in which the customers arrive according to a Poisson process with rate $\lambda$, and the service times are independent exponential random variables, with mean equal to $1 / \mu$. We suppose that the system capacity is infinite, as well as the population from which the customers come. Finally, the queue discipline is that of first-come, first-served. We can therefore denote this model simply by $M / M / 1$.

The stochastic process ${X(t), t \geq 0}$ is an irreducible birth and death process. We will calculate the limiting probabilities $\pi_n$ as we did in Chapter 3 . The balance equations of the system (see p. 140) are the following:
\begin{aligned} &\text { state } j \text { departure rate from } j=\text { arrival rate to } j \ &0 \quad \lambda \pi_0=\mu \pi_1 \ &n(\geq 1) \quad(\lambda+\mu) \pi_n=\lambda \pi_{n-1}+\mu \pi_{n+1} \ & \end{aligned}
We have (see p. 141)
$$\Pi_0:=1 \quad \text { and } \quad \Pi_n=\underbrace{\frac{\lambda \lambda \cdots \lambda}{\mu \mu \cdots \mu}}{n \times}=\left(\frac{\lambda}{\mu}\right)^n \text { for } n=1,2, \ldots$$ If $\lambda<\mu$, the process ${X(t), t \geq 0}$ is positive recurrent, and Theorem $3.3 .4$ then enables us to write that $$\pi_n=\frac{(\lambda / \mu)^n}{\sum{k=0}^{\infty}(\lambda / \mu)^k}=\frac{(\lambda / \mu)^n}{[1-(\lambda / \mu)]^{-1}} \quad \text { for } n=0,1, \ldots$$

## 数学代写|随机过程Stochastic Porcesses代考|The model M/M/l/c

Although the $M / M / 1$ queue is very useful to model various phenomena, it is more realistic to suppose that the system capacity is an integer $c<\infty$. For $j=0,1, \ldots, c-1$, the balance equations of the system remain the same as when $c=\infty$. However, when the system is in state $c$, it can only leave it because of the departure of the customer being served. In addition, this state can only be entered from $c-1$, with the arrival of a new customer. We thus have
\begin{aligned} &\text { state } j \quad \text { departure rate from } j=\text { arrival rate to } j \ &\begin{aligned} 0 & \lambda \pi_0 &=\mu \pi_1 \ 1 \leq k \leq c-1 &(\lambda+\mu) \pi_k &=\lambda \pi_{k-1}+\mu \pi_{k+1} \ c & \mu \pi_c &=\lambda \pi_{c-1} \end{aligned} \ & \end{aligned}
The process ${X(t), t \geq 0}$ remains a birth and death process. Moreover, given that the number of states is finite, the limiting probabilities exist regardless of the values the (positive) parameters $\lambda$ and $\mu$ take.

As in the case when the system capacity is infinite, we find (see p. 319) that
$$\Pi_k=\left(\frac{\lambda}{\mu}\right)^k \quad \text { for } k=0,1, \ldots, c$$
It follows, if $\rho:=\lambda / \mu \neq 1$, that
$$\sum_{k=0}^c \Pi_k=\sum_{k=0}^c\left(\frac{\lambda}{\mu}\right)^k=\sum_{k=0}^c \rho^k=\frac{1-\rho^{c+1}}{1-\rho}$$

## 数学代写|随机过程STOCHASTIC PORCESSES代考|THE MODEL M/M/1

$\$ \\begin{aligned} } \& Itext { state } . j \backslash text { 出发率 from } j=\lfloor Itext { arrival rate to } . j \backslash \& 0 \backslash quad \backslash lambda \backslash pi_0 0=\backslash \mathrm{mu} \backslash \mathrm{pi} 1 \backslash \& n \geq 1 \backslash 四 \lambda+\mu \backslash pi_n n=\mid l a m b d a \backslash pi_{n-1}+\mu \backslash pi_{n+1} \backslash \& lendi对齐} Wehave(seep. 141) \mathrm{n}=1,2, 《ldots If \ \lambda<\mu \,theprocess $\$ X(t), t \geq 0$\$sispositiverecurrent, andTheorem $\$ 3.3 .4$\$thenenablesustowritethat
|pi_n=|frac $\left{\lambda / \mu^{\wedge} \mathrm{n}\right} \backslash \backslash$ sum{k=0}^{\infty} $\left.\lambda / \mu^{\wedge} \mathrm{k}\right}=\left{\right.$ frac $\left{\lambda / \mu^{\wedge} \mathrm{n}\right}$
$$1-(\lambda / \mu)$$
${{-1}} \backslash$ quad $\backslash$ text ${$ for $} n=0,1, \backslash$ ldots
$\$ \$$## 数学代写|随机过程STOCHASTIC PORCESSES代考|THE MODEL M/M/L/C 虽然 M / M / 1 队列对各种现象的建模非常有用，假设系统容量是一个整数更符合实际 c<\infty. 为了 j=0,1, \ldots, c-1, 系统的平衡方程保持不变 c=\infty. 但是，当 系统处于状态 c ，它只能因为被服务的顾客离开而离开它。另外，这个状态只能从 c-1, 随着新客户的到来。因此，我们有 \ \$$
\begin{aligned }
\&
$$0 \lambda \pi_0 \quad=\mu \pi_1 1 \leq k \leq c-1(\lambda+\mu) \pi_k \quad=\lambda \pi_{k-1}+\mu \pi_{k+1} c \mu \pi_c \quad=\lambda \pi_{e-1}$$
\&
lend{aligned $}$

$$\Pi_k=\left(\frac{\lambda}{\mu}\right)^k \quad \text { for } k=0,1, \ldots, c$$

$$\sum_{k=0}^c \Pi_k=\sum_{k=0}^c\left(\frac{\lambda}{\mu}\right)^k=\sum_{k=0}^c \rho^k=\frac{1-\rho^{c+1}}{1-\rho}$$

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