LSM-Tree（43）

3.4 LSM-trees: Component Sizes（10）

Minimizing Total Cost

From Theorem 3.1, it can be seen that if we allow S0 to vary while R and SK remain constant and express the total I/O rate H as a function of S0, then since r increases with decreasing S0 by equation (3.5), and H is proportional to r by equation (3.6), clearly H increases with de- creasing S0. We can now minimize the total cost of the LSM-tree as in the two component case by trading off expensive memory for inexpensive disk. If we calculate the disk media needed to store the LSM-tree and the total I/O rate H that keeps these disk arms fully utilized, this be- comes a starting point in our calculation to determine the size for S0 that minimizes cost. From this point as we further decrease the size of C0 the cost of disk media goes up in inverse pro- portion, since we have entered the region where disk arm cost is the limiting factor. Example 3.3, below, is a numerically based illustration of this process for a two and three component LSM-tree. Prior to this example, we offer an analytic derivation for the two component case.
从定理3.1,可以看出,如果我们允许S0改变而R和SK保持不变和表达总I / O率H S0的函数,然后从R由方程(3.5),增加与减少S0和H正比于R方程(3.6),显然与de - H增加压痕S0。现在，我们可以将lsm树的总成本最小化，就像在两个组件的情况下一样，用昂贵的内存换取廉价的磁盘。如果我们计算存储lsm树所需的磁盘介质和使这些磁盘臂充分利用的总I/O速率H，那么这将成为我们确定使成本最小化的S0大小的计算起点。从这一点开始，随着我们进一步减小C0的大小，磁盘介质的成本以反比的比例上升，因为我们已经进入了磁盘臂成本是限制因素的区域。下面的例子3.3是一个基于数字的lsm树处理过程的说明。在这个例子之前，我们提供了一个解析推导的双分量情况。（有道翻译）

todo：自己翻译