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authorMatt Strapp <matt@mattstrapp.net>2021-09-20 22:03:15 -0500
committerMatt Strapp <matt@mattstrapp.net>2021-09-20 22:03:15 -0500
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parentadd stuff heeh (diff)
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\begin{document}
\maketitle
\section{Question 1}
- \subsection{Overlapping Intervals}
-
- \subsection{Non-Overlapping Intervals}
- \(t_{transfer} = t_s+t_w*d*m/k*k\) or \(\sum_{i=1}^{k} t_s+t+w*d*m/k\)
+ \subsection*{Overlapping Intervals}
+ The time for one message to be sent 1 hop is \(t_s+t_w*m/k\). After the first message is sent forward one hop, we account for the final message \((d-1)*t_w*m/k\), making the final expression
+ \[k(t_s+t_w+m/k)+((d-1)t_w*m/k)\]
+ \subsection*{Non-Overlapping Intervals}
+ \[t_{transfer} = t_s*k+t_w*d*m\]
+ \subsection*{}
+ For both cases, as \(k\) goes to \(m\), the time to transfer will increase greatly. If \(t_s\) is very large, the optimal value of \(k\) is 1. In other words, it is better to transfer the message all at once instead of in \(k\) parts. If \(t_s\) is 0 it has little to no effect on overall transmission time regardless of \(k\) being large or small.
\section{Question 2}
\subsection*{A}
Shared memory has all of the processors access one large pool of memory while distributed memory has each processor have a section of the memory.