deletions | additions       diff --git a/Problem 2.tex b/Problem 2.tex  index 1b8181d..a647093 100644  --- a/Problem 2.tex  +++ b/Problem 2.tex 
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Given that we're given frequency in Ghz we'll change $CT$, which is clock cycles in seconds, to clock speed. This means our equation will change to $L = \frac{CPI}{f}$. Table one shows the results of calculting $L$ for each processor.  \begin{tabular}[t]{|l|c|c|}  Processor&Latency Calc&Latency\\ \hline % Please add the following required packages to your document preamble:  \usepackage{booktabs}  \begin{table}[h]  \begin{tabular}{@{}lll@{}}  \toprule  Processor & Latency Calc & Latency \\ \midrule  A & $\frac{1.2}{1*10^9}$ & $1.2*10^-9$\\ \hline $1.2*10^-9$ \\  B & $\frac{2}{2*10^9}$ & $1*10-9$\\ \hline  \hline\end{tabular} $1*10-9$ \\ \bottomrule  \end{tabular}  \caption{My caption}  \label{my-label}  \end{table}  To determine the improvement from one over the other, we leverage the Speed up equation $Speedup =\frac{Latency_{old}}{Latency_{new}}$