From 0dab92f9b14e75f68dda8c556398ea2d55e27494 Mon Sep 17 00:00:00 2001
From: Douglas Rumbaugh <dbr4@psu.edu>
Date: Wed, 4 Jun 2025 16:01:20 -0400
Subject: updates

---
 chapters/design-space.tex | 40 +++++++++++++++++-----------------------
 1 file changed, 17 insertions(+), 23 deletions(-)

(limited to 'chapters/design-space.tex')

diff --git a/chapters/design-space.tex b/chapters/design-space.tex
index 952be42..7cfe6a3 100644
--- a/chapters/design-space.tex
+++ b/chapters/design-space.tex
@@ -192,14 +192,13 @@ analysis. The worst-case cost of a reconstruction is $B(n)$, and there
 are $\log_s(n)$ total levels, so the total reconstruction costs associated
 with a record can be upper-bounded by, $B(n) \cdot
 \frac{W(\log_s(n))}{n}$, and then this cost amortized over the $n$
-insertions necessary to get the record into the last level, resulting
+insertions necessary to get the record into the last level. We'lll also
+condense the multiplicative constants and drop the additive ones to more
+clearly represent the relationship we're looking to show. This results
 in an amortized insertion cost of,
 \begin{equation*}
-\frac{B(n)}{n} \cdot  \frac{1}{2}(s-1) \cdot ( (s-1)\log_s n + s)
+\frac{B(n)}{n} \cdot s \log_s n
 \end{equation*}
-Note that, in the case of $s=2$, this expression reduces to the same amortized
-insertion cost as was derived using Binomial Theorem in the original BSM
-paper~\cite{saxe79}.
 \end{proof}
 
 \begin{theorem}
@@ -361,9 +360,10 @@ and sum this over all of the levels.
 \end{equation*}
 To calculate the amortized insertion cost, we multiply this write amplification
 number of the cost of rebuilding the structures, and divide by the total number
-of records,
+of records. We'll condense the constant into a single $s$, as this best
+expresses the nature of the relationship we're looking for,
 \begin{equation*}
-I_A(n) \in \Theta\left(\frac{B(n)}{n}\cdot \frac{1}{2} (s+1) \log_s n\right)
+I_A(n) \in \Theta\left(\frac{B(n)}{n}\cdot s \log_s n\right)
 \end{equation*}
 \end{proof}
 
@@ -503,17 +503,10 @@ I(n) \in \Theta\left(B(n)\right)
 \end{theorem}
 \begin{proof}
 The worst-case reconstruction in tiering involves performing a
-reconstruction on each level. Of these, the largest level will
-contain $\Theta(n)$ records, and thus dominates the cost of the
-reconstruction. More formally, the total cost of this reconstruction
-will be,
+reconstruction on each level.  More formally, the total cost of this
+reconstruction will be,
 \begin{equation*}
-I(n) = \sum_{i=0}{\log_s n} B(s^i) = B(1) + B(s) + B(s^2) + \ldots B(s^{\log_s n})
-\end{equation*}
-Of these, the final term $B(s^{\log_s n}) = B(n)$ dominates the others,
-resulting in an asymptotic worst-case cost of,
-\begin{equation*}
-I(n) \in \Theta\left(B(n)\right)
+I(n) \in \Theta\left(\sum_{i=0}^{\log_s n} B(s^i)\right)
 \end{equation*}
 \end{proof}
 
@@ -600,8 +593,8 @@ reconstructions, one per level.
 & \textbf{Gen. BSM} & \textbf{Leveling} & \textbf{Tiering} \\ \hline
 $\mathscr{Q}(n)$ &$O\left(\log_s n \cdot \mathscr{Q}_S(n)\right)$ & $O\left(\log_s n \cdot \mathscr{Q}_S(n)\right)$ & $O\left(s \log_s n  \cdot \mathscr{Q}_S(n)\right)$\\ \hline
 $\mathscr{Q}_B(n)$ & $\Theta(\mathscr{Q}_S(n))$ & $O(\log_s n \cdot \mathscr{Q}_S(n))$ & $O(\log_s n \cdot \mathscr{Q}_S(n))$ \\ \hline
-$I(n)$ & $\Theta(B(n))$ & $\Theta\left(B\left(\frac{s-1}{s} \cdot n\right)\right)$ & $\Theta(B(n))$\\ \hline
-$I_A(n)$ & $\Theta\left(\frac{B(n)}{n} \frac{1}{2}(s-1)\cdot((s-1)\log_s n +s)\right)$ & $\Theta\left(\frac{B(n)}{n} \frac{1}{2}(s-1)\log_s n\right)$& $\Theta\left(\frac{B(n)}{n} \log_s n\right)$ \\ \hline
+$I(n)$ & $\Theta(B(n))$ & $\Theta\left(B\left(\frac{s-1}{s} \cdot n\right)\right)$ & $ \Theta\left(\sum_{i=0}^{\log_s n} B(s^i)\right)$ \\ \hline
+$I_A(n)$ & $\Theta\left(\frac{B(n)}{n} s\log_s n)\right)$ & $\Theta\left(\frac{B(n)}{n} s\log_s n\right)$& $\Theta\left(\frac{B(n)}{n} \log_s n\right)$ \\ \hline
 \end{tabular}
 
 \caption{Comparison of cost functions for various layout policies for DSPs}
@@ -699,7 +692,7 @@ due to cache effects most likely, but less so than in the MDSP case.
 \begin{figure}
 \centering
 \subfloat[ISAM Tree]{\includegraphics[width=.5\textwidth]{img/design-space/isam-tput.pdf} \label{fig:design-isam-tput}} 
-\subfloat[VPTree]{\includegraphics[width=.5\textwidth]{img/design-space/vptree-insert-dist.pdf} \label{fig:design-vptree-tput}} \\
+\subfloat[VPTree]{\includegraphics[width=.5\textwidth]{img/design-space/vptree-tput.pdf} \label{fig:design-vptree-tput}} \\
 \caption{Insertion Throughput for Layout Policies}
 \label{fig:design-ins-tput}
 \end{figure}
@@ -772,9 +765,10 @@ method shows similar trends to leveling.
 
 In general, the Bentley-Saxe method appears to follow a very similar
 trend to that of leveling, albeit with even more dramatic performance
-degradation as the scale factor is increased. Generally it seems to be
-a strictly worse alternative to leveling in all but its best-case query
-cost, and we will omit it from our tests moving forward as a result.
+degradation as the scale factor is increased and slightly better query
+performance across the board. Generally it seems to be a strictly worse
+alternative to leveling in all but its best-case query cost, and we will
+omit it from our tests moving forward.
 
 \subsection{Query Size Effects}
 
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