(未完成)I/O--翻译:ANALYZING I/O PERFORMANCE IN LINUX

来源: http://www.cmdln.org/2010/04/22/analyzing-io-performance-in-linux/

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Great so now that I know what IOPS are how do I cal­cu­late them? IOPS are a func­tion of rota­tional speed (aka spin­dle speed), latency and seek time. The equa­tion is pretty sim­ple, 1/(seek + latency) = IOPS. Scott Lowe has a good exam­ple on his techreplublic.com blog.

“Sam­ple drive:

• Model: West­ern Dig­i­tal Veloci­Rap­tor 2.5″ SATA hard drive
• Rota­tional speed: 10,000 RPM
• Aver­age latency: 3 ms (0.003 seconds)
• Aver­age seek time: 4.2 ®/4.7 (w) = 4.45 ms (0.0045 seconds)

Cal­cu­lated IOPS for this disk: 1/(0.003 + 0.0045) = about 133 IOPS

[pope译] 很好那IOPS 如何计算？IOPS是 rotational speed (转速 :主轴旋转速度)，latency(延时)和 seek time (查找时间) 。这个公式很简单 1/(seek + latency) = IOPS. Scott Lowe 在他的techreplublic.com blog. 上有一个很不错的例子:
" 如下硬盘:

• 型号: West­ern Dig­i­tal Veloci­Rap­tor 2.5″ SATA hard drive
• Rota­tional speed(转速):10,000 RPM
• Aver­age latency(平均延时):3 ms (0.003 seconds)
• Aver­age seek time(平均查询时间):4.2 ®/4.7 (w) = 4.45 ms (0.0045 seconds)
• 计算此硬盘的IOPS:1/(0.003 + 0.0045) = about 133 IOPS

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It’s great to know how to cal­cu­late a disks IOPS but for the most part you can get by with com­monly accepted aver­ages. Of course sources vary but from what I have seen.

Rota­tional Speed (rpm)	IOPS
5400	50–80
7200	75–100
10k	125–150
15k	175–210

“

Should I use SATA, SAS or FC? That’s a loaded ques­tion. As with most things the answer is “depends”. I don’t want to get into the SATA vs SAS debate you can do your own research and make your own deci­sions based on your needs, but I will point out a few things.

• SATA only gets up to 10k (at the time of this writing)
• SATA is only 1/2 duplex (From Tomak in comments)
• Dif­fer­ences in reli­a­bil­ity (MTBF, BER) inter­est­ing arti­cle on Real Life Raid Reliability
• See dif­fer­ences in Native Com­mand Queu­ing (NCQ) and Com­mand Tag Queu­ing (CTQ)

These fac­tors are  key con­sid­er­a­tions when choos­ing what kind of dri­ves to use.

[pope译]这样很好计算磁盘IOPS ,大部分你可以得到的数值是可接受的平均值。但根据我看到的源的不同而改变。

Rota­tional Speed (rpm)	IOPS
5400	50–80
7200	75–100
10k	125–150
15k	175–210

我可以使用SATA,SAS 或者FC 吗?（【popexizhi】sata，sas，fc 参见:http://blog.csdn.net/tianlesoftware/article/details/6009110）这个是加载问题.多数答案是"依赖"。我不想在 SATA 或者SAS 中讨论你的使用和满足你自己需求的决策，但是我给出一些内容:

• SATA 只能用到 10k （只个是写速度）
• SATA 只是半双工(1/2 duplex) （参见Tomak 的意见）
• 不同之处是真实存在的(MTBF,BER) 感兴趣的可以参考 真实情况中Raid的可靠性 [popexizhi:很可惜，测试了一下，这个地址的文章现在找不到]
• 查看 Native Com­mand Queu­ing (NCQ) and Com­mand Tag Queu­ing (CTQ) 的不同

选择何种类型的设备成为注意考虑的因素。

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What RAID level should I use? You know what IOPS are, how to cal­cu­late them and deter­mined what kind of dri­ves to use, the next log­i­cal ques­tion is com­monly RAID 5 vs RAID 10. There is dif­fer­ence in reli­a­bil­ity, espe­cially as the num­ber of dri­ves in your raid-set increases but that is out­side the scope of this post.
[pope译] 那我该用什么级别的RAID呢？你知道什么是IOPS，如何计算它们 以及 决定使用什么类型的设备，下一个分析的是RAID5 或 RAID 10 的选择。这真的很不一样，特别是当你的raid-设置数量的增长超出这个对应位置要求的范围时。[popexizhi:好吧！ 自己翻译的也感觉有点儿晕，再想想 increases  [?]]

Raid Level	Write Oper­a­tions	Read Oper­a­tions	Notes
0	1	1	Write/Read: high through­put, low CPU uti­liza­tion, no redundancy [pope译]Write/Read:高速输出，低水平CPU占用，没有冗余
1	2	1	Write: only as fast as sin­gle driveRead: Two read schemes avail­able. Read data from both dri­ves, or data from the drive that returns it first. One is higher through­put the other is faster seek times. [pope译]Write:与单个设备一样快 driveRead:两种读取方法，从两个设备同时读取，或读取返回数据最快的那个。其中一个的定位输出时间比另一个快。
5	4	1	Write: Read-Modify-Write requires two reads and two writes per write request. Lower through­put higher CPU if the HBA doesn’t have a ded­i­cated IOprocessor. Read: High through­put low CPU uti­liza­tion nor­mally, in a failed state per­for­mance falls dra­mat­i­cally due to par­ity cal­cu­la­tion and any rebuild oper­a­tions that are going on. [pope译]Write: Read-Modify-Write  每次写请求使用两个读取和写入请求。如果HBA 没有dedicated IO处理，就会有较低的输出 较高的CPU使用。[?]这里HBA的dedicated IOprocessor 如何翻译有问题，再想想吧！Read: 正常是低CPU使用，高输出。但在错误的预估状态时， 性能急剧下降同时引起大量计算和重新进行下面的操作。[popexizhi:这里的a failed state 是磁盘缓存对预读内容缓存的估算错误吗？]
6	5	1	Write: Read-Modify-Write requires three reads and three writes per write request. Do not use a soft­ware imple­men­ta­tion if it is avail­able Read: High through­put low CPU uti­liza­tion nor­mally, in a failed state per­for­mance falls dra­mat­i­cally due to par­ity cal­cu­la­tion and any rebuild oper­a­tions that are going on. [pope译]Write: Read-Modify-Write  每个写操作请求使用三个读取和三个写入操作。如果可行不使用软件实现。 Read:正常时低CPU使用，高输出。但在错误的预估状态时， 性能急剧下降同时引起大量计算和重新进行下面的操作。

As you can see in the table above, writes are where you take your per­for­mance hit. Now that the penalty or RAID fac­tor is known for dif­fer­ent raid lev­els we can get a good esti­mate of the the­o­ret­i­cal max­i­mum IOPS for aRAID set (exclud­ing caching of course). To do this you take the prod­uct of the num­ber of disks and IOPS per disk divided by the sum of the %read work­load and the prod­uct of the raid fac­tor (see write oper­a­tions col­umn) and %write workload.

[pope译] 正如你在上面表中看到的，写操作决定着你的操作性能。现在通过RAID设置，我们可以很好的预估理论上每个RAID设置的最大的IOPS （当然不包括缓存）[?]Now that the penalty or RAID fac­tor is known for dif­fer­ent raid lev­els we can get a good esti­mate of the the­o­ret­i­cal max­i­mum IOPS for aRAID set (exclud­ing caching of course). 这句翻译的有问题，再想想！