57 lines
1.8 KiB
TeX
57 lines
1.8 KiB
TeX
% Preamble
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\documentclass[11pt,fleqn]{article}
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\usepackage{amsmath, amsthm, amssymb}
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\usepackage{fancyhdr}
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\oddsidemargin -0.25in
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\textwidth 6.75in
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\topmargin -0.5in
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\headheight 0.75in
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\headsep 0.25in
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\textheight 8.75in
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\pagestyle{fancy}
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\renewcommand{\headrulewidth}{0pt}
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\renewcommand{\footrulewidth}{0pt}
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\fancyhf{}
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\lhead{HW Chap. 4\\\ \\\ }
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\rhead{Josh Holtrop\\2008-12-03\\CS 677}
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\rfoot{\thepage}
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\begin{document}
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\noindent
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\begin{enumerate}
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\item[1.]{
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In store-and-forward communication, the communication cost is given by
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$$T_{\mathrm{comm}} = t_s + (t_h + mt_w)\ell = t_s + \ell t_h + \ell mt_w$$
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In cut-through routing, the communication cost is given by
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$$T_{\mathrm{comm}} = t_s + \ell t_h + mt_w$$
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Since the ``header'' is the only part of the communication that is
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encountering overhead for the $\ell$ links in the communication network,
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cut-through routing can save communication time on the order of
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$(\ell - 1) mt_w$.
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Obviously, this makes cut-through routing only advantageous on
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architectures with $\ell > 1$, meaning non-fully-connected networks.
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}
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\vskip 1em
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\item[2.]{
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We are to transpose an $n \times n$ matrix that is initially
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rowwise block-decomposed among $p$ processes.
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Assume that $p \leq n$ and if any process $i$ owns $k > 1$ rows of
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the matrix, then the $k$ rows that it owns are contiguous.
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Then, the procedure to transpose the matrix is as follows:
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For each process $i$, a gather operation is performed.
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In this gather operation, each process $j$ sends to $i$ the $k$
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values it owns in column $i$ (where $k$ is the number of rows
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assigned to process $j$).
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Thus, at the end of each gather operation process $i$ has received
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the entire contents of column $i$ of the matrix.
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This algorithm takes $p \log p$ steps to complete.
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}
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\end{enumerate}
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\end{document}
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