EEL6507sp09L38

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EEL6507 Spring 2009, Lecture 38, Monday 2009/04/20 (Notes created by Xiaoyuan Li)

  • Conditional Distribution on Queue size, G/M/m

Conditional Distribution on Queue size

From last time,

\begin{align} P_{i,j}& =Prob[q_{n+1}'=j|q_n'=i] \\ & = Prob[i+1-j\ departs|q_n'=i] \\ & = \int_{0}^{\infty}\binom{i+1}{j}(1-e^{-\mu t})^{i+1-j}e^{-\mu tj}a(t)\,dt, for j\le i+1 \le m , \end{align}

Refer to the Regions of pij

  • All busy:
Prob[k\ customer\ served|time=t, m\ busy] = \frac{(m\mu t)^k}{k!}e^{-m\mu t}

When m\le j\le i+1

P_{i,j}=\int_{0}^{\infty}\frac{(m\mu t)^{i+1-j}}{(i+1-j)!}e^{-m\mu t}a(t)\,dt
\beta_n = \int_{0}^{\infty}\frac{(m\mu t)^n}{n!}e^{-m\mu t}a(t)\,dt
P_{i,j}=\beta_{i+1-j} ,m\le j\le i+1
  • Transition Period: All Busy \to Some Busy
P_{i,j}= \int_{0}^{\infty}\binom{m}{j}e^{-j\mu t} \times \left[\int_{0}^{\infty}\frac{(m\mu t)^{i-m}}{(i-m)!}(e^{-\mu y}-e^{-\mu t})^{m-j}m\mu\,dy\right]a(t)dt

We want to get

Prob[k\ customer\ served|time=t, m\ busy] = \frac{(m\mu t)^k}{k!}e^{-m\mu t}

When m\le j\le i+1

\gamma_k\ =\ Prob[\ arrival\ finds\ k\ in\ system]
\vec \gamma = \vec \gamma P \ gives\ steady\ state
\gamma_k = \sum_{i=0}^{\infty}\gamma_i\ P_{ik}

Given arrival finds m, there is queue

Conditional Queue Length (Arrival Queues)

\begin{align} \gamma_k &= \sum_{i=0}^{\infty}\gamma_i\ P_{ik}, if\ k\ge m \\ &= \sum_{i=0}^{k-2}\gamma_i\ P_{ik}+ \sum_{i=k-1}^{\infty}\gamma_i\ P_{ik} \ (P_{ik}=0, if \ k>i+1)\\ &=\sum_{i=k-1}^{\infty}\gamma_i\ P_{ik}\\ &=\sum_{i=k-1}^{\infty}\gamma_i\ \beta_{i+1-k} \end{align}

Refer to the [Chain Diagram ]

N_k\ (t)= number of arrivals in (0,t) that finds k in the system

\begin{align} \sigma_k &= E[number\ of\ visits\ to\ E_{k+1}\ between\ visits\ to\ E_{k}]\\ &= \beta_0\sum n\gamma^{n-1}(1-\gamma)\\ &=\frac{\beta_0}{1-\gamma}\\ &=\sigma_k \end{align}

Observation Prob\ [zero\ return]=1-\beta_0

\begin{align} \gamma&=Prob[leave\ E_{k+1}\ and\ return\ without\ passing\ E_j\ where\ j\le k]\\ &=Prob[leave\ E_{k+1}\ and\ return\ without\ passing\ E_k] \end{align}
\to independent\ of\ k\ because\ P_{ij}\ depends\ only\ on\ i-j, for\ i,j\ge m
Prob[n-1 \ visits\ to\ E_{k+1}\ without\ E_{k}, 1\ with\ E_k ]=\gamma^{n-1}(1-\gamma)\beta_0
\sigma_k=\lim_{t \to \infty}\frac{N_{k+1}(t)}{N_{k}(t)}=\frac{\gamma_{k+1}}{\gamma_{k}}
\gamma_{k+1}=\sigma_k\gamma_k=\sigma\gamma_{k}, k\ge m
\gamma_{m},\ \gamma_{m+1}=\sigma\ \gamma_m,\gamma_{m+2}=\sigma^2\ \gamma_m
\vec \gamma=[\gamma_0,\gamma_1,...\gamma_{m},\sigma \gamma_m,\sigma^2 \gamma_m,... ]
\begin{align} \sigma &= \sum_{i=k-1}^{\infty}\sigma^{i+1-k}\beta_{i+1-k}\\ &=\sum_{n=0}^{\infty}\sigma^{n}\beta_{n}\\ &=A^*(\sigma\ m\ \mu)=\sigma(solve\ for\ \sigma) \end{align}

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