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diff --git a/assignment7/assignment7.tex b/assignment7/assignment7.tex new file mode 100644 index 0000000..489ca60 --- /dev/null +++ b/assignment7/assignment7.tex @@ -0,0 +1,81 @@ +\documentclass[10pt,a4paper]{article} + +\usepackage[margin=2cm]{geometry} +\usepackage{pdfpages} + +% textcomp package is not available everywhere, and we only need the Copyright symbol +% taken from http://tex.stackexchange.com/a/1677/23992 +\DeclareTextCommandDefault{\textregistered}{\textcircled{\check@mathfonts\fontsize\sf@size\z@\math@fontsfalse\selectfont R}} + +\usepackage{fancyhdr} +\renewcommand{\headrulewidth}{0pt} +\renewcommand{\footrulewidth}{0pt} +\fancyhead{} +\fancyfoot[C]{Copyright {\textcopyright} 2015 Camil Staps} +\pagestyle{fancy} + +\usepackage{amsmath} +\usepackage{amsfonts} +\usepackage{mathtools} + +\usepackage{array} +\newcolumntype{C}[1]{>{\centering\let\newline\\\arraybackslash\hspace{0pt}}m{#1}} + +\usepackage{enumitem} +\setenumerate[1]{label=\alph*.} +\setenumerate[2]{label=(\roman*)} + +\parindent0pt + +\title{Operating Systems - assignment 7} +\author{Camil Staps\\\small{s4498062}} + +\begin{document} + +\maketitle +\thispagestyle{fancy} + +\section*{9.1} +\begin{enumerate} + \item See Table \ref{tab:91a}. + \begin{table}[h] + \centering + \footnotesize + \begin{tabular}{| r | *{21}{c|}} \hline + ms & 0 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 & 20 & 22 & 24 & 26 & 28 & 30 & 32 & 34 & 36 & 38 \\\hline + SPN & P1 & P1 & P2 & P2 & P2 & P2 & P2 & P2 & P4 & P4 & P4 & P3 & P3 & P3 & P3 & P5 & P5 & P5 & P5 & P5 \\\hline + SRT & P1 & P1 & P2 & P3 & P3 & P3 & P3 & P2 & P4 & P4 & P4 & P2 & P2 & P2 & P2 & P5 & P5 & P5 & P5 & P5 \\\hline + \end{tabular} + \caption{Applying the SPN and SRT algorithms to five processes} + \label{tab:91a} + \end{table} + + \item See Table \ref{tab:91b}. + \begin{table}[h] + \centering + \footnotesize + \begin{tabular}{| r | *{4}{>{$}C{10mm}<{$} |}} \hline + & \multicolumn{2}{c|}{SPN} & \multicolumn{2}{c|}{SRT} \\\hline + Process & T_r & T_r/T_s & T_r & T_r/T_s \\\hline + P1 & 4 & 1 & 4 & 1 \\ + P2 & 14 & 1.167 & 28 & 2.333 \\ + P3 & 24 & 3 & 8 & 1 \\ + P4 & 6 & 1 & 6 & 1 \\ + P5 & 20 & 2 & 20 & 2 \\\hline + Mean & 13.2 & 1.633 & 13.2 & 1.467 \\\hline + \end{tabular} + \caption{TAT and relative delay for the processes from Table \ref{tab:91a}} + \label{tab:91b} + \end{table} +\end{enumerate} + +\section*{9.6} +The diagram shows the situation where processes only move down to a lower priority queue when preempted. The first two time units allocated to A are actually two dispatches, each with time unit $2^0=1$. Then the process is preempted because of B and moves down the priority queues. + +However, if we move a process down the priority queues any time it finishes a time unit, A will continue running for three consecutive time units. First, one time unit in RQ0, and then two in RQ1. + +\section*{9.7} +A steep line typically corresponds to a low service time, while a more gentle line corresponds to a high service time. The response ratio of a process with a low service time increases faster than that of a process with a high service time. Therefore, we should schedule the processes with a high service time last. If we schedule them earlier, the response times of other jobs increase rapidly and over a long time (that is, the service time of the running job). If we first schedule processes with a short service time however, the other response ratios don't increase that rapidly (because they belong to processes with a higher service time), and not for such a long time. + +\end{document} + |