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# 数学代写|随机过程Stochastic Porcesses代考|AMATH562 The busy period; zero-avoiding transitions

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## 数学代写|随机过程代写Stochastic Porcesses代考|The busy period; zero-avoiding transitions

Suppose that initially the system contains $i(\geq 1)$ customer, and let $T_i$ be the next subsequent epoch of time at which the server is free; $T_i$ is called the busy period initiated by $i$ customers.
Then $T_i=\inf {t \mid Q(t)=0, Q(0)=i}$.
$(10.2 .31)$
Suppose that $T_i$ has d.f. $G_i(t)=P\left(T_i \leq t\right)(0 \leq t<\infty)$. Define the zero-avoiding transition probabilities of the Markov process $$Q(t) \text { as }{ }^0 p_{i j}(t)=P\left(Q(t)=j ; T_i>t \mid Q(0)=i\right)(i, j \geq 1) \text {. }$$
It is the probability that there will be $j$ customers busy throughout the interval $[0, t]$.

By (10.2.20), (10.2.31) can be written as $T_i=\inf {t \mid i+X(t) \leq 0}$ and hence
\begin{aligned} { }^0 p_{i j}(t) & =P[i+X(\tau)>0 \quad(0 \leq \tau \leq t) ; i+X(t)=j] \ & =P{i+\sigma(t)>0, i+X(t)=j], \end{aligned}
By $(10.2 .20)$ and (10.2.32), we get
\begin{aligned} P(Q(t) & 0 ; v+X(t)=\sum_{i+1}^{\infty}{ }^{\infty} p_{k j}(t) \quad(10.2 .33)\right. \end{aligned}
which is a relation connecting the zero-avoiding transition probability with the ordinary ones. In particular (putting $j=1$ in (10.2.33)), we get
$$p_{i 0}(t)=\sum_{i+1}^{\infty} 0 p_{k \mathrm{l}}(t)$$

## 数学代写|随机过程代写Stochastic Porcesses代考|The Model M /M /c

Here the input process and the service time distribution are the same as $M / M / 1$ model but there are $c$ servers where $1 \leq c<\infty$ customers are served on a ‘first come first served’ basis (FCFS). Here again the number of customers $Q(t)$ present in the system at time $t$ is a birth and death process, the transition densities are computed as follows.

Here for each of the c channels use i.i.d exponential service time distribution with mean rate $\mu$. If $n(\leq c)$ channels, are busy, the number of services completed in the whole system is a Poisson process with mean $n \mu$ and thus the time between two successive service completions is exponential with mean $1 /(n \mu)$ whereas if $n$ (>c) channels are busy, the time between two successive service completions is exponential with mean $1 / c \mu$.
\begin{aligned} P_{n n+1}(d t) & =P(Q(t+d t)=n+1 \mid Q(t)=n)=\lambda d t+o(d t) \ p_{n n-1}(d t) & =\mu_n d t+o(d t) \ p_{n n}(d t) & =1-\left(\lambda+\mu_n\right) d t+o(d t) \ p_{n m}(d t) & =P(Q(t+d t)=m \mid Q(t)=n)=o(d t) \text { if } m \neq n-1, n, n+1, \end{aligned}

## 数学代写|随机过程代写Stochastic Porcesses代考|The busy period; zeroavoiding transitions

$(10.2 .31)$

\$\$
Ibegin ${$ aligned
{}$\wedge 0 p_{-}{i \mathrm{ij}}(\mathrm{t}) \&=P[i+x(\backslash$ tau $)>0 \backslash$ quad $(0 \backslash$ leq $\$ tau $\backslash$ leq $t) ; i+x(t)=j] \backslash$
$\&=P{i+\mid \operatorname{sigma}(t)>0, i+X(t)=j]$,
lend{aligned
By $\$(10.2 .20) \$\operatorname{sand}(10.2 .32)$, weget

whichisarelationconnectingthezero – avoidingtransitionprobabilitywiththeordinaryones. Inparticular (pu
$p_{-}{\mathrm{i} 0}(\mathrm{t})=\backslash$ sum ${i+1} \wedge{\backslash$ infty $} 0 p_{-}{\mathrm{k} \backslash \operatorname{mathrm}{\mathrm{I}}}(\mathrm{t})$
\$\$

## 数学代写|随机过程代写Stochastic Porcesses代考|The Model M/M/c

$$P_{n n+1}(d t)=P(Q(t+d t)=n+1 \mid Q(t)=n)=\lambda d t+o(d t) p_{n n-1}(d t) \quad=\mu_n d t+o(d t) p_{n n}(d t)=1-\left(\lambda+\mu_n\right) d t+o(d t) p_{n m}(d t)$$

## MATLAB代写

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