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| -rw-r--r-- | csci5451/ass1.tex | 11 |
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diff --git a/csci5451/ass1.tex b/csci5451/ass1.tex index 6dbdfc1..24f2e0c 100644 --- a/csci5451/ass1.tex +++ b/csci5451/ass1.tex @@ -7,10 +7,13 @@ \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. |