Read/write-optimized tree indexing for solid-state drives

Abstract

Flash-memory-based solid-state drives (SSDs) are used widely for secondary storage. To be effective for SSDs, traditional indices have to be redesigned to cope with the special properties of flash memory, such as asymmetric read/write latencies (fast reads and slow writes) and out-of-place updates. Previous flash-optimized indices focus mainly on reducing random writes to SSDs, which is typically accomplished at the expense of a substantial number of extra reads. However, modern SSDs show a narrowing gap between read and write speeds, and read operations on SSDs increasingly affect the overall performance of indices on SSDs. As a consequence, how to optimize SSD-aware indices by reducing both write and read costs is a pertinent and open challenge. We propose a new tree index for SSDs that is able to reduce both writes and extra reads. In particular, we use an update buffer and overflow pages to reduce random writes, and we further exploit Bloom filters to reduce the extra reads to the overflow nodes in the tree. With this mechanism, we construct a read/write-optimized index that is capable of offering better overall performance than previous flash-aware indices. In addition, we present an analysis of the proposed index and show that the read and write costs of the operations on the index can be balanced by only tuning the false-positive rate of the Bloom filters. Our experimental results suggest that our proposal is efficient and represents an improvement over existing methods.

Appendix A: transformation from Eqs. 5.8 to 5.9

$$\begin{aligned}&f^{x BF }=\left( {\begin{array}{c}s\\ x\end{array}}\right) \cdot (f^\mathrm{BF})^x \cdot (1-f^\mathrm{BF})^{s-x} \end{aligned}$$

(5.7)

$$\begin{aligned}&E_{ BF \text {-}cost}=\sum \limits _{x=0}^{s}f^{x BF } \cdot (x+2) \end{aligned}$$

(5.8)

$$\begin{aligned}&E_{ BF \text {-}cost}= f^\mathrm{BF} \cdot s +2 \end{aligned}$$

(5.9)

Proof.

$$\begin{aligned} \sum \limits _{x=0}^{s}f^{x BF } \cdot (x+2)=\sum \limits _{x=0}^{s}f^{x BF } \cdot x + \sum \limits _{x=0}^{s}f^{x BF } \cdot 2 \end{aligned}$$

Then, we can simplify $\sum _{x=0}^{s}f^{x BF } \cdot x$ as shown in the following steps.

$$\begin{aligned}&\sum \limits _{x=0}^{s}f^{x BF } \cdot x \\&\quad =\sum \limits _{x=1}^{s}\left( {\begin{array}{c}s\\ x\end{array}}\right) \cdot (f^\mathrm{BF})^x \cdot (1-f^\mathrm{BF})^{s-x} \cdot x \\&\quad =\sum \limits _{x=1}^{s}x \cdot \frac{s!}{x!(s-x)!}\cdot (f^\mathrm{BF})^x \cdot (1-f^\mathrm{BF})^{s-x} \\&\quad =s \cdot f^\mathrm{BF} \cdot \sum \limits _{x=1}^{s} \frac{(s-1)!}{(x-1)!(s-x)!} \\&\quad \quad \cdot (f^\mathrm{BF})^{x-1} \cdot (1-f^\mathrm{BF})^{[(s-1)-(x-1)]} \\&\quad =s \cdot f^\mathrm{BF} \cdot \sum \limits _{x=1}^{s} \left( {\begin{array}{c}s-1\\ x-1\end{array}}\right) \\&\quad \quad \cdot (f^\mathrm{BF})^{x-1} \cdot (1-f^\mathrm{BF})^{[(s-1)-(x-1)]} \\&\quad =s \cdot f^\mathrm{BF} \cdot \sum \limits _{x=0}^{s-1} \left( {\begin{array}{c}s-1\\ x\end{array}}\right) \\&\quad \quad \cdot (f^\mathrm{BF})^x \cdot (1-f^\mathrm{BF})^{[(s-1)-x]} \\&\quad =s \cdot f^\mathrm{BF} \cdot (f^\mathrm{BF}+1-f^\mathrm{BF})^{s-1} \\&\quad =s \cdot f^\mathrm{BF} \end{aligned}$$

Next, we can infer that $\sum _{x=0}^{s}f^{x BF } \cdot 2=2$ through the following transformation.

$$\begin{aligned}&\sum \limits _{x=0}^{s}f^{x BF } \cdot 2 \\&\quad =2 \cdot \sum \limits _{x=0}^{s}\left( {\begin{array}{c}s\\ x\end{array}}\right) \cdot (f^\mathrm{BF})^x \cdot (1-f^\mathrm{BF})^{s-x}\\&\quad =2 \cdot (f^\mathrm{BF}+1-f^\mathrm{BF})^{s} \\&\quad =2 \end{aligned}$$

This gives the following equation.

$$\begin{aligned} \sum \limits _{x=0}^{s}f^{x BF } \cdot (x+2)=s\cdot f^\mathrm{BF}+2 \end{aligned}$$

Thus, we have shown that Eq. 5.9 can be derived from Eq. 5.8.