5分钟规则 - 中盘的一个接入价格的I / O [英] five minutes rules - the price of one access of disc I/O
问题描述
这是很有意思的话题，他们用下面的公式来计算访问间隔时间：
This is very interesting topic, they use following formula to compute access interval time:
BreakEvenIntervalinSeconds =（PagesPerMBofRAM / AccessesPerSecondPerDisk）×（PricePerDiskDrive / PricePerMBofRAM）。
BreakEvenIntervalinSeconds = (PagesPerMBofRAM / AccessesPerSecondPerDisk) × (PricePerDiskDrive / PricePerMBofRAM).
这是使用RAM成本公式在缓冲池中持有页面和 A（分数）磁盘的成本执行I / O每次需要一个页面时，等同衍生这两项费用，并求解访问之间的间隔式。
It is derived using formulas for the cost of RAM to hold a page in the buffer pool and the cost of a (fractional) disk to perform I/O every time a page is needed, equating these two costs, and solving the equation for the interval between accesses.
所以盘I / O的每个接入成本就是 PricePerDiskDrive / AccessesPerSecondPerDisk ，我的问题是，为什么我的光盘每次访问O开销计算/这样吗？
so the cost of disc I/O per access is PricePerDiskDrive / AccessesPerSecondPerDisk, My question is why disc I/O cost per access is computed like this?
推荐答案
的基本假设是，限制盘的寿命是多少磁盘寻求有，而RAM具有用于它的大小的固定费用，和一个固定一辈子不管它的访问频率。这是合理的，因为希望将磁盘造成身体磨损，而当盘的话，你失去了整个磁盘。相比之下RAM没有物理移动部件，因此不会磨损搭配使用。
The underlying assumption is that the limit to the life of a disk is how many disk seeks there are, while RAM has a fixed cost for its size, and a fixed lifetime regardless of how often it is accessed. This is reasonable because seeking to disk causes physical wear and tear, and when the disk goes, you lose the whole disk. By contrast RAM has no physical moving parts, and so does not wear out with use.
使用这样的假设，保持在磁盘上数据的成本取决于访问的频率和盘的成本。在RAM中保存数据的成本取决于你多少内存使用。他们正在努力寻找的是盈亏平衡之间的地方是便宜保持在磁盘或RAM中的数据点。
With that assumption, the cost of keeping data on disk depends on the frequency of access and the cost of the disk. The cost of keeping data in RAM depends on how much RAM you're using. What they are trying to find is the break even point between where it is cheaper to keep data on disk or in RAM.
的然而的作为给出的公式是不完整的。虽然该方程确定了相关的因素外，还有相称缺少一个重要的常数。有多少可以访问的平均硬盘支持？多久RAM最后平均？那些进入保持在硬盘和RAM数据的成本，没有他们，你是比较苹果和桔子。
However the equation as given is incomplete. While that equation identifies relevant factors, there is an important constant of proportionality missing. How many accesses can the average hard drive sustain? How long does RAM last on average? Those enter into the costs for keeping data on hard drives and RAM, and without them you are comparing apples and oranges.
这是表明整个纸我IM pression的。它说了很多连篇累牍，关于一个重要话题，但分析是马虎。他们喷溅，并留下关键的东西出来，并没有做足够的帮助人们了解他们在想什么，当他们的分析是合适的，你在做什么。例如，如果你想保持低延时系统，你必须保持所有数据在RAM中。期。如果你正在处理大型数据集，并且不希望支付保持它的所有在RAM中，那么你将磁盘流数据/。如果你保持数据冗余的格式，例如RAID，你在做每读比他们接纳更多的寻求。
This is indicative of my impression of the whole paper. It says a lot at great length, about an important topic, but the analysis is sloppy. They are slopping and leave critical things out, and don't do enough to help people understand what they are thinking and when their analysis is appropriate what you are doing. For instance if you are trying to maintain a low latency system, you have to keep all of your data in RAM. Period. If you're processing large data sets and don't want to pay to keep it all in RAM, then you will be streaming data to/from disk. If you're keeping data in a redundant format, for instance RAID, you are doing more seeks per read than they admit.
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