3.4 LSM-trees: Component Sizes（3）
Given an LSM-tree of K components with a fixed total size S and memory component size S0,
给一个K个组件的LSM-tree，一个固定总size为S，内存组件size为S0，
the tree is totally described by the variables ri, i = 1, . . . ,K, representing size ratios between adjacent pairs of components, ri = Si/Si-1,
树描述为ri， i = 1, . . . ,K，表示组件之间size比例，ri=Si/Si-1，
As detailed below, the total page I/O rate to perform all ongoing merge operations between component pairs (Ci-1, Ci) can be expressed as a function of R, the rate of insertions into C0, and the ratios ri.
如下描述，总的page I/O速率为所有进行中的merge操作表示为R，插入为C0，速率为ri。
We assume that blocks of the different components are striped across different disk arms in a mixed way to achieve a balance in utilization, so that minimizing H is the same as minimizing the total disk arm cost
我们假设不同组件的块以混合的方式跨不同的磁盘臂进行条带化，以实现利用率的平衡，因此最小化H与最小化总磁盘臂成本相同（有道翻译）
(at least in any range where disk arms rather than media capacity constitute the gating cost).
(至少在磁盘臂而不是媒体容量构成闸门成本的任何范围内)。（有道翻译）
It is a standard calculus minimization problem to find the values for ri which minimize the total I/O rate H for a given R.
对于给定的R，找出使总I/O率H最小化的ri值是一个标准的微积分最小化问题。（有道翻译）
It turns out that the assumption that the total size S is fixed leads to a rather difficult problem with a somewhat complex recurrence relation between the ri values.
结果表明，总大小S是固定的假设导致了一个相当困难的问题，ri值之间有某种复杂的递归关系。（有道翻译）
However, if we make the comparable assumption that the largest component size SK is fixed (along with the memory size S0), as we will show in Theorem 3.1, this minimization problem is solved when all of the values ri are equal to a single constant value r.
然而，如果我们做一个可比较的假设，即最大的组件大小SK是固定的(以及内存大小S0)，如定理3.1所示，当所有ri值等于一个常数r时，这个最小化问题就得到了解决。（有道翻译）
We show in Theorem 3.2 the slightly more precise solution relating the ri values where the total size S is held constant, and argue that the constant value r for ri gives similar results in all areas of real interest.
我们在定理3.2中给出了稍微精确一点的关于ri值的解，其中总大小S保持不变，并论证了ri的常数值r在所有实际感兴趣的领域中给出了相似的结果。（有道翻译）
Assuming such a constant value r for all ri factors, we have Si = ri.S0.
假设ri都是r，Si=ri.S0.
Thus the total size S is given by the sum of the individual component sizes, S = S0 + r.S0 + r2.S0 + . . . + rK.S0, and we can solve for r in terms of S and S0.
因此，总尺寸S由单个组件尺寸之和给出，S = S0 + r.S0 + r2。S0 +…+ rK。我们可以用S和S0来表示r。（有道翻译）
翻译体会：
句子是真的长，竟然还有数学计算过程。
