exercise:0201e77b01: 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> In the queueing problem of Exercise Exercise, let <math>S</math> be the total service time required by a customer and <math>T</math> the time between arrivals of the customers. <ul><li> Show that <math>P(S = j) = (1 - r)...") |
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\newcommand{\mathds}{\mathbb}</math></div> In the queueing problem of | \newcommand{\mathds}{\mathbb}</math></div> In the queueing problem of [[exercise:D094f2f7d6 |Exercise]], let <math>S</math> be the total service time required by a customer and <math>T</math> the time between arrivals of the customers. | ||
let <math>S</math> be the total service time required by a customer and <math>T</math> the time | <ul style="list-style-type:lower-alpha"><li> Show that <math>P(S = j) = (1 - r)^{j - 1}r</math> and <math>P(T = j) = (1 - p)^{j - | ||
between | |||
arrivals of the customers. | |||
<ul><li> Show that <math>P(S = j) = (1 - r)^{j - 1}r</math> and <math>P(T = j) = (1 - p)^{j - | |||
1}p</math>, for <math>j > 0</math>. | 1}p</math>, for <math>j > 0</math>. | ||
</li> | </li> | ||
Latest revision as of 22:01, 17 June 2024
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In the queueing problem of Exercise, let [math]S[/math] be the total service time required by a customer and [math]T[/math] the time between arrivals of the customers.
- Show that [math]P(S = j) = (1 - r)^{j - 1}r[/math] and [math]P(T = j) = (1 - p)^{j - 1}p[/math], for [math]j \gt 0[/math].
- Show that [math]E(S) = 1/r[/math] and [math]E(T) = 1/p[/math].
- Interpret the conditions [math]s \lt 1[/math], [math]s = 1[/math] and [math]s \gt 1[/math] in terms of these expected values.