Revision as of 03:30, 9 June 2024 by Bot (Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> Consider a queueing process such that in each minute either 1 or 0 customers arrive with probabilities <math>p</math> or <math>q = 1 - p</math>, respectively. (The number <math>p</math> is called the ''arrival rate''.) When a customer starts s...")
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Jun 09'24

Exercise

[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

Consider a queueing process such that in each minute either 1 or 0 customers arrive with probabilities [math]p[/math] or

[math]q = 1 - p[/math], respectively. (The number [math]p[/math] is called the arrival rate.) When a customer starts service she finishes in the next minute with probability [math]r[/math]. The number [math]r[/math] is called the service rate.) Thus when a customer begins being served she will finish being served in [math]j[/math] minutes with probability [math](1 - r)^{j -1}r[/math], for [math]j = 1[/math], 2, 3, \ldots.

  • Find the generating function [math]f(z)[/math] for the number of customers who arrive in one minute and the generating function [math]g(z)[/math] for the length of time that a person spends in service once she begins service.