Use iostat to check linux hard disk I/O performance

Source: Internet
Author: User
Recently, the company has installed several DELLPE2650 and 2850 Servers. the RHLE5.132 system is installed in a unified manner, while the server's SCSI hard disks are all RAID 1. The boss of the company asked for a unified detection report on hard disk I/O. many tools were found in Linux, and iostat was the most practical tool. this...
Recently, the company has installed several DELL PE2650 and 2850 Servers. the RHLE5.132 system is installed in a unified manner, while the server's SCSI hard disks are all RAID 1. The boss of the company asked for a unified detection report on hard disk I/O. he found many tools in Linux and found that iostat is the most practical. He needs to install sysstat first, that is, yum-y install sysstat; setting up yum servers in the company is not the focus of this article. I will not describe them in detail here. # Iostat-x 1 10 Linux 2.6.18-92. el5xen 03/01/2010 avg-cpu: % user % nice % system % iowait % steal % idle 1.10 0.00 4.82 39.54 Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s avgrq-sz avgqu-sz await svctm % utilsda 0.00 3.50 0.40 2.50 5.60 48.00 18.48 0.00 0.97 0.97 0.28sdb 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00sdc 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00sdd 0.00 0.00 0. 00 0.00 0.00 0.00 0.00 0.00 0.00 0.00sde 0.00 0.00 0.10 0.30 0.20 2.40 2.40 9.60 0.00 0.08sdf 1.60 1.60 17.40 0.50 102.00 0.20 12095.20 5.60 118.40 0.70 21.36sdg 6.81 2.09 232.40 1.90 379.70 0.50 76451.20 19.20 201.13 4.94 13.78 2.45 93.16 rrqm/s: the number of merge read operations per second. That is, delta (rmerge)/swrqm/s: Number of merge write operations per second. Delta (wmerge)/sr/s: the number of read I/O devices per second. That is, delta (rio)/sw/s: the number of write I/O devices completed per second. That is, delta (wio)/srsec/s: Number of read sectors per second. That is, delta (rsect)/swsec/s: number of write sectors per second. That is, delta (wsect)/srkB/s: the number of Read K bytes per second. It is half of rsect/s because the size of each slice is 512 bytes. (To be calculated) wkB/s: number of write K bytes per second. Half of wsect/s. Avgrq-sz: average data size (slice) of each device I/O operation ). Delta (rsect + wsect)/delta (rio + wio) avgqu-sz: average I/O queue length. That is, delta (aveq)/s/1000 (because aveq is measured in milliseconds ). Await: average wait time (in milliseconds) for each device I/O operation ). That is, delta (ruse + wuse)/delta (rio + wio) svctm: average service time per device I/O operation (MS ). That is, delta (use)/delta (rio + wio) % util: the percentage of time in one second for I/O operations, or the number of I/O queues in one second is not empty. That is, delta (use)/s/1000 (because the unit of use is millisecond) if % util is close to 100%, it indicates that too many I/O requests are generated and the I/O system is fully loaded, this disk may have a bottleneck. the idle is under a high I/O stress of less than 70%, and the reading speed is usually wait. You can also view the parameters B (number of processes waiting for resources) and wa in combination with vmstat (percentage of CPU time occupied by I/O wait, higher than 30% when I/O pressure is high) in addition, you can also refer to general: svctm <await (because the wait time of the simultaneously waiting request is calculated repeatedly). The size of svctm is generally related to disk performance: the CPU/memory load will also affect it, and too many requests will indirectly lead to the increase of svctm. Await: the size of await generally depends on the service time (svctm), the length of the I/O queue, and the mode in which I/O requests are sent. If svctm is close to await, there is almost no waiting time for I/O. if await is much larger than svctm, it indicates that the I/O queue is too long, slow application response time if the response time exceeds the user's allowable range, you can consider replacing a faster disk, adjusting the kernel elevator algorithm, optimizing the application, or upgrading the CPU. The queue length (avgqu-sz) can also be used as an indicator to measure the system I/O load. However, because avgqu-sz is based on the average per unit time, therefore, it cannot reflect the instantaneous I/O flood. A good example from others (I/O system vs supermarket queue): for example, how can we decide which payment platform to pay when we queue for checkout in the supermarket? First, let's look at the number of people in the queue. is the total number of five people faster than 20? In addition to the number of people, we often look at the number of items purchased by the previous person. if there is a big mom who has purchased food for a week, we can consider changing the team. There is also the cashier's speed. if you have a newbie who doesn't even know about the money, you will have to wait. In addition, the timing is also very important. it may be 5 minutes ago, but the money was crowded, and now people go to the building, but it is refreshing to pay the money. of course, the premise is that what has been done in the past five minutes is more meaningful than queuing (but I have not found anything boring than queuing ). The I/O system has many similarities with supermarket queues: r/s + w/s is similar to the average queue length of the total number of payers (avgqu-sz) similar to the average number of queues per unit time average service time (svctm) similar to the cashier's receipt speed average wait time (await) similar to the average wait time of each person, the average I/O data (avgrq-sz) is similar to the average I/O operation rate (% util) of the items bought by each person) it is similar to the time ratio when someone queues before the cashier. We can analyze the I/O request mode and the I/O speed and response time based on the data. The following is an analysis of the output of this parameter written by someone else # iostat-x 1avg-cpu: % user % nice % sys % idle16.24 0.00 4.31 Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm % util/dev/cciss/c0d00.00 44.90 1.02 27.55 8.16 579.59 4.08 289.80 20.57 22.35 78.21 5.00 14.29/dev/cciss/c0d0p10.00 44.90 1.02 27.55 8.16 579.59 4.08 289.80 20.57 22.35 78.21/dev/cciss/c0d0p000000 5.00 14.29 0.00 0.00 0.00 0.00 0. 00 0.00 0.00 0.00 0.00 0.00 0.00 The iostat output above indicates that the second has 28.57 device I/O operations: total IO/s = r/s (read) + w/s (write) = 1.02 + 27.55 = 28.57 (times/second) where write operations occupy the main body (w: r = 27 ). On average, only 5 ms is required for each device I/O operation, but 78 ms is required for each I/O request. why? Because too many I/O requests are sent (about 29 requests per second), assuming these requests are sent at the same time, the average wait time can be calculated as follows: average wait time = single I/O service time * (1 + 2 +... + total number of requests-1)/total number of requests applied to the above example: average wait time = 5 ms * (1 + 2 +... + 28)/29 = 70 ms, which is very close to the average waiting time of 78 ms given by iostat. This in turn indicates that I/O is initiated at the same time. There are many I/O requests per second (about 29), but the average queue is not long (only about 2). This indicates that the arrival of these 29 requests is uneven, i/O is idle most of the time. In one second, 14.29% of the time I/O queues have Requests. that is to say, the I/O system has nothing to do in 85.71% of the time, all 29 I/O requests are processed within 142 milliseconds. Delta (ruse + wuse)/delta (io) = await = 78.21 => delta (ruse + wuse)/s = 78.21 * delta (io) /s = 78.21*28.57 = 2232.8, indicating that I/O requests per second need to wait for a total of 2232.8 ms. Therefore, the average queue length should be 2232.8 ms/1000 ms = 2.23, while the average queue length (avgqu-sz) provided by iostat is 22.35. why? Because there is a bug in iostat, the avgqu-sz value should be 2.23 instead of 22.35. ※Note: When I use iostat to detect servers, I usually use the iostat-d command, I usually focus on tps, blk_read/s, and blk_wrth/s. I usually compare and test three servers of different models in the same environment. in this way, the performance is different, all of a sudden. Author: FutureInHands
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