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diff --git a/chapters/design-space.tex b/chapters/design-space.tex index f85883c..22773e5 100644 --- a/chapters/design-space.tex +++ b/chapters/design-space.tex @@ -127,72 +127,89 @@ be $N_B (s - 1) \cdot s^i$, where $N_B$ is the size of the buffer. The resulting structure will have at most $\log_s n$ shards. The resulting policy is described in Algorithm~\ref{alg:design-bsm}. -Unfortunately, the approach used by Bentley and Saxe to calculate the -amortized insertion cost of the BSM does not generalize to larger bases, -and so we will need to derive this result using a different approach. +Analyzing the amortized insertion performance of BSM is slightly +complicated by the fact that each record is \emph{not} written on +every level. For the purposes of our analysis, establishing a reasonable +upper bound on the amortized insertion cost is sufficient, however, so +we will settle for a looser bound to keep things simple. \begin{theorem} The amortized insertion cost for generalized BSM with a growth factor of -$s$ is $\Theta\left(\frac{B(n)}{n} \cdot s\log_s n)\right)$. +$s$ is $O\left(\frac{B(n)}{n} \cdot s\log_s n)\right)$. \end{theorem} \begin{proof} +In generalized BSM, each record will be written at most $s$ times +per level. We will use this result to provide an upper-bound on the +amortized insertion performance. The worst case cost associated with a +reconstruction in BSM is a full compaction of the structure, which will +require $B(n)$ time to complete. Further, there are $O(\log_s n)$ levels +in the decomposition. As a result, the amortized insertion cost can be +bounded above by, +\begin{equation} +I_A(n) \in O\left(\frac{B(n)}{n} \cdot s \log_s n\right) +\end{equation} +\end{proof} -In order to calculate the amortized insertion cost, we will first -determine the average number of times that a record is involved in a -reconstruction, and then amortize those reconstructions over the records -in the structure. -If we consider only the first level of the structure, it's clear that -the reconstruction count associated with each record in that structure -will follow the pattern, $1, 2, 3, 4, ..., s-1$ when the level is full. -Thus, the total number of reconstructions associated with records on level -$i=0$ is the sum of that sequence, or -\begin{equation*} -W(0) = \sum_{j=1}^{s-1} j = \frac{1}{2}\left(s^2 - s\right) -\end{equation*} -Considering the next level, $i=1$, each reconstruction involving this -level will copy down the entirety of the structure above it, adding -one more write per record, as well as one extra write for the new record. -More specifically, in the above example, the first "batch" of records in -level $i=1$ will have the following write counts: $1, 2, 3, 4, 5, ..., s$, -the second "batch" of records will increment all of the existing write -counts by one, and then introduce another copy of $1, 2, 3, 4, 5, ..., s$ -writes, and so on. +% \begin{proof} -Thus, each new "batch" written to level $i$ will introduce $W(i-1) + 1$ -writes from the previous level into level $i$, as well as rewriting all -of the records currently on level $i$. +% In order to calculate the amortized insertion cost, we will first +% determine the average number of times that a record is involved in a +% reconstruction, and then amortize those reconstructions over the records +% in the structure. -The net result of this is that the number of writes on level $i$ is given -by the following recurrence relation (combined with the $W(0)$ base case), +% If we consider only the first level of the structure, it's clear that +% the reconstruction count associated with each record in that structure +% will follow the pattern, $1, 2, 3, 4, ..., s-1$ when the level is full. +% Thus, the total number of reconstructions associated with records on level +% $i=0$ is the sum of that sequence, or +% \begin{equation*} +% W(0) = \sum_{j=1}^{s-1} j = \frac{1}{2}\left(s^2 - s\right) +% \end{equation*} -\begin{equation*} -W(i) = sW(i-1) + \frac{1}{2}\left(s-1\right)^2 \cdot s^i -\end{equation*} +% Considering the next level, $i=1$, each reconstruction involving this +% level will copy down the entirety of the structure above it, adding +% one more write per record, as well as one extra write for the new record. +% More specifically, in the above example, the first ``batch'' of records in +% level $i=1$ will have the following write counts: $1, 2, 3, 4, 5, ..., s$, +% the second ``batch'' of records will increment all of the existing write +% counts by one, and then introduce another copy of $1, 2, 3, 4, 5, ..., s$ +% writes, and so on. -which can be solved to give the following closed-form expression, -\begin{equation*} -W(i) = s^i \cdot \left(\frac{1}{2} (s-1) \cdot (s(i+1) - i)\right) -\end{equation*} -which provides the total number of reconstructions that records in -level $i$ of the structure have participated in. As each record -is involved in a different number of reconstructions, we'll consider the -average number by dividing $W(i)$ by the number of records in level $i$. - -From here, the proof proceeds in the standard way for this sort of -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. We'll 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 s \log_s n -\end{equation*} -\end{proof} +% Thus, each new ``batch'' written to level $i$ will introduce $W(i-1) + 1$ +% writes from the previous level into level $i$, as well as rewriting all +% of the records currently on level $i$. + +% The net result of this is that the number of writes on level $i$ is given +% by the following recurrence relation (combined with the $W(0)$ base case), + +% \begin{equation*} +% W(i) = sW(i-1) + \frac{1}{2}\left(s-1\right)^2 \cdot s^i +% \end{equation*} + +% which can be solved to give the following closed-form expression, +% \begin{equation*} +% W(i) = s^i \cdot \left(\frac{1}{2} (s-1) \cdot (s(i+1) - i)\right) +% \end{equation*} +% which provides the total number of reconstructions that records in +% level $i$ of the structure have participated in. As each record +% is involved in a different number of reconstructions, we'll consider the +% average number by dividing $W(i)$ by the number of records in level $i$. + +% From here, the proof proceeds in the standard way for this sort of +% 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. We'll 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 s \log_s n +% \end{equation*} +% \end{proof} \begin{theorem} The worst-case insertion cost for generalized BSM with a scale factor @@ -586,7 +603,7 @@ reconstructions, one per level. \hline & \textbf{Gen. BSM} & \textbf{Leveling} & \textbf{Tiering} \\ \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 +$I_A(n)$ & $O\left(\frac{B(n)}{n} s\log_s n)\right)$ & $\Theta\left(\frac{B(n)}{n} s\log_s n\right)$& $\Theta\left(\frac{B(n)}{n} \log_s n\right)$ \\ \hline $\mathscr{Q}(n)$ &$O\left(\log_s n \cdot \mathscr{Q}_S(n)\right)$ & $O\left(\log_s n \cdot \mathscr{Q}_S(n)\right)$ & $O\left(s \log_s n \cdot \mathscr{Q}_S(n)\right)$\\ \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 \end{tabular} |