exercise:25488e32c5: Difference between revisions
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(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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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''.) | <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 | When a customer starts service she finishes in the next minute | ||
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Thus when a customer begins being served she will finish | 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, | being served in <math>j</math> minutes with probability <math>(1 - r)^{j -1}r</math>, for <math>j = 1</math>, 2, | ||
3, | 3, .... | ||
<ul><li> Find the generating function <math>f(z)</math> for the number of customers who | <ul style="list-style-type:lower-alpha"><li> 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 | 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. | that a person spends in service once she begins service.</li> | ||
<li> | |||
Consider a ''customer branching process'' by considering the offspring of a customer to be the customers who arrive while she is being served. Using [[exercise:6eeb09869c |Exercise]], show that the generating function for our customer branching process is <math>h(z) = g(f(z))</math>. | |||
</li> | |||
<li> | |||
If we start the branching process with the arrival of the first | |||
customer, then the length of time until the branching process dies out will be | |||
the ''busy period'' for the server. Find a condition in terms of the | |||
arrival rate and service rate that will assure that the server will | |||
ultimately have a time when he is not busy. | |||
</li> | |||
</ul> | |||
Latest revision as of 23:54, 14 June 2024
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, ....
- 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.
- Consider a customer branching process by considering the offspring of a customer to be the customers who arrive while she is being served. Using Exercise, show that the generating function for our customer branching process is [math]h(z) = g(f(z))[/math].
- If we start the branching process with the arrival of the first customer, then the length of time until the branching process dies out will be the busy period for the server. Find a condition in terms of the arrival rate and service rate that will assure that the server will ultimately have a time when he is not busy.