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Date: Jul 28, 2006 By Christopher L. Tinker, David Carmichael, James Kirkland, Gregory L. Tinker. Sample Chapter is provided courtesy of Prentice Hall.
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This chapter covers basic concepts of SCSI over Fibre Channel Protocol (FCP) using raw/block device files and volume managers. In addition, it covers block size, multipath I/O drivers, and striping with a volume manager, and concludes with a discussion of filesystem performance and CPU loading. Examples are included for each topic throughout the chapter.
As a general discussion, performance is much too broad for a single book, let alone a single chapter. However, in this chapter we narrow the focus of performance to a single subject: I/O on a SCSI bus within a storage area network (SAN). SANs are growing in popularity because they assist with storage consolidation and simplification. The main discussion point within the computing industry with regards to storage consolidation is, as it has always been, performance.In this chapter, we cover basic concepts of SCSI over Fibre Channel Protocol (FCP) using raw/block device files and volume managers. In addition, we cover block size, multipath I/O drivers, and striping with a volume manager, and we conclude our discussion with filesystem performance and CPU loading. We include examples of each topic throughout the chapter.
A majority of the time, performance issues are related to I/O. However, assuming that a given performance problem is I/O-based is grossly oversimplifying the problem. With any filesystem I/O, there are middle-layer tasks that require resources which may be the source of an I/O contention, such as the volume manager, the volume manager's striping, the filesystem, a multipath I/O driver, or something similar. When troubleshooting a performance problem, always try to simplify the problem by removing as many middle layers as possible. For example, if a particular filesystem is slow, focus your attention first on the disk block or character device performance before considering the volume manager and filesystem performance.
Dissecting a volume with respect to physical device (aka LUN) or lvol into its simplest form is absolutely required when preparing to run any performance test or find a performance concern. In this section, we test the raw speed of a storage device by bypassing the filesystem and volume management layers. We bypass as many layers as possible by using a raw device, better known as a character device. A character device must be bound to a block device through the raw command. To describe "raw" with more detail would include the physical access to a block device bypassing the kernel's block buffer cache. Our first test performs a simple sequential read of a Logical Unit Number (LUN), which resides on a set of spindles, through a single path after we bind the block device to the character. We create a (LUN) character device because we want to test the speed of the disk, not the buffer cache.
Note
Today's large arrays define a data storage device in many ways. However, the best description is Logical Device (LDEV). When an LDEV is presented to a host, the device changes names and is referred to as a Logical Unit Number (LUN).
The components used throughout this chapter for examples and scenarios include:
The tools for examining the hardware layout and adding and removing LUNs are discussed in Chapter 5, "Adding New Storage via SAN with Reference to PCMCIA and USB." Performance tools were fully discussed in Chapter 3, "Performance Tools," and are used in examples but not explained in detail in this chapter. As stated previously, this chapter's focus is strictly on performance through a system's I/O SCSI bus connected to SAN. Let's look at how to find and bind a block device to a character device using the raw command.
The LUN, hereafter called disk, used throughout this example is /dev/sdj, also referred to as /dev/scsi/sdh6-0c0i0l2. Determine the capacity of the disk through the fdisk command:
atlorca2:~ # fdisk -l Disk /dev/sdj: 250.2 GB, 250219069440 bytes 255 heads, 63 sectors/track, 30420 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Device Boot Start End Blocks Id System /dev/sdj1 1 30421 244354559+ ee EFI GPT
Use lshw (an open source tool explained in more detail in Chapter 5) to show the device detail:
atlorca2:~ # lshw
~~Focus only on single disk test run~~~~
*-disk:2
description: SCSI Disk
product: OPEN-V*4
vendor: HP
physical id: 0.0.2
bus info: [email protected]:0.2
logical name: /dev/sdj
version: 2111
size: 233GB
capacity: 233GB
capabilities: 5400rpm ### <- just the drivers attempt to
guess the speed via the standard scsi lun interface...
Take this at face value.
configuration: ansiversion=2
Linux does not allow raw access to a storage device by default. To remedy this problem, bind the device's block device file to a /dev/raw/rawX character device file to enable I/O to bypass the host buffer cache and achieve a true measurement of device speed through the host's PCI bus (or other bus architecture). This binding can be done by using the raw command. Look at the block device for /dev/sdj:
atlorca2:~ # ls -al /dev/sdj brw-rw---- 1 root disk 8, 144 Jun 30 2004 /dev/sdj
Take note of the permissions set on the device file brw-rw----. The b means block device, which has a major number 8, which refers to the particular driver in control. The minor number 144 represents the device's location in the scan plus the partition number. Refer to man pages on sd for more info. Continuing with our example, the next step is to bind the /dev/sdj to a raw character device, as depicted in the following:
atlorca2:~ # raw /dev/raw/raw8 /dev/sdj /dev/raw/raw8: bound to major 8, minor 144
Now, issue one of the following commands to view the binding parameters:
atlorca2:~ # raw -qa /dev/raw/raw8: bound to major 8, minor 144
or
atlorca2:~ # raw -q /dev/raw/raw8 /dev/raw/raw8: bound to major 8, minor 144
Now that we have bound a block device to a character device, we can measure a read by bypassing the block device, which in turn bypasses the host buffer cache. Recall that our primary objective is to measure performance from the storage device, not from our host cache.
Disclaimer
Throughout this chapter, we use a sequential read test provided by the dd command. By no means are we implying that this is the best performance benchmark tool. Every array has certain strengths with regards to I/O size and patterns of sequential versus random, so for benchmarks, we like IOzone (refer to http://www.iozone.org) for a nice benchmark tool. One must be aware that some arrays suffer during heavy, large sequential reads, whereas other arrays thrive in these situations, and the same is true for particular HBAs. In addition, those arrays that suffer on sequential read/writes usually excel at random read/writes, whereas the reverse can be said about the arrays that perform well under sequential read/write operations. In addition, hdparm impacts performance with respect to direct memory access (DMA) and read-ahead along with other parameters. In this chapter, we use the default settings of hdparm. So to reiterate, this chapter uses a simple sequential read/write test to help isolate a performance problem by comparison, and many examples are illustrated.
Our next step requires that we measure a sequential read and calculate time required for the predetermined data allotment. Throughout this chapter, our goal is to determine what factors dictate proper performance of a given device. We focus on average service time, reads per second, writes per second, read sectors per second, average request size, average queue size, and average wait time to evaluate performance. In addition, we discuss the I/Os per second with regard to payload "block" size. For now, we start with a simple sequential read to get our baseline.
Though a filesystem may reside on the device in question, as shown previously in the fdisk-l output, the filesystem cannot be mounted for the test run. If mounted, raw access is denied. Proceed with the following action as illustrated in the next section.
The dd command provides a simple way to measure sequential I/O performance. The following shows a sequential read of 1GB (1024MB). There are 1024 1MB (1024KB) reads:
atlorca2:~ # time -p dd if=/dev/raw/raw8 of=/dev/null bs=1024k count=1024 1024+0 records in 1024+0 records out real 6.77 user 0.00 sys 0.04
The megabytes per second can be calculated as follows:
1GB/6.77 sec = 151.25MBps
For those who are unfamiliar with high-speed enterprise servers and disk storage arrays, 151MBps may seem extremely fast. However, higher speeds can be achieved with proper striping and tuning of the filesystem across multiple spindles. Though we discuss some of those tuning options later, we first need to reduce the previous test to its simplest form. Let us begin with calculating MBps, proceeding with the blocking factors on each I/O frame and discussing service time for each round trip for a given I/O.
In the previous example, we saw 1024 I/Os, where each I/O is defined to have a boundary set to a block size of 1MB, thanks to the bs option on the dd command. Calculating MBps simply takes an arithmetic quotient of 1024MB/6.77 seconds, providing a speedy 151MB/sec. In our testing, cache on the array is clear, providing a nice 151MBps, which is not bad for a single LUN/LDEV on a single path. However, determining whether the bus was saturated and whether the service time for each I/O was within specifications are valid concerns. Each question requires more scrutiny.
Note
Different arrays require special tools to confirm that cache within the array is flushed so that a true spindle read is measured. For example, HP's largest storage arrays can have well over 100GB of cache on the controller in which a read/write may be responding, thereby appearing to provide higher average reads/writes than the spindle can truly provide. Minimum cache space should be configured when running performance measurements with respect to design layout.
Continuing with the dd command, we repeat the test but focus only on the data yielded by the sar command to depict service time and other traits, as per the following.
atlorca2:~ # sar -d 1 100 Linux 2.6.5-7.97-default (atlorca2) 05/09/05 14:19:23 DEV tps rd_sec/s wr_sec/s 14:19:48 dev8-144 0.00 0.00 0.00 14:19:49 dev8-144 0.00 0.00 0.00 14:19:50 dev8-144 178.00 182272.00 0.00 14:19:51 dev8-144 303.00 311296.00 0.00 14:19:52 dev8-144 300.00 307200.00 0.00 14:19:53 dev8-144 303.00 309248.00 0.00 14:19:54 dev8-144 301.00 309248.00 0.00 14:19:55 dev8-144 303.00 311296.00 0.00 14:19:56 dev8-144 302.00 309248.00 0.00
This sar output shows that the total number of transfers per second (TPS) holds around 300. rd_sec/s measures the number of read sectors per second, and each sector is 512 bytes. Divide the rd_sec/s by the tps, and you have the number of sectors in each transfer. In this case, the average is 1024 sectors at 512 bytes each. This puts the average SCSI block size at 512KB. This is a very important discovery; because the dd command requests a block size of 1MB, the SCSI driver blocks the request into 512 byte blocks, so two physical I/Os complete for every logical I/O requested. Different operating systems have this value hard coded at the SCSI driver at different block sizes, so be aware of this issue when troubleshooting.
As always, more than one way exists to capture I/O stats. In this case, iostat may suit your needs. This example uses iostat rather than sar to evaluate the dd run.
atlorca2:~ # iostat
Linux 2.6.5-7.97-default (atlorca2) 05/09/05
avg-cpu: %user %nice %sys %iowait %idle
0.00 0.01 0.02 0.08 99.89
Device: tps Blk_read/s Blk_wrtn/s Blk_read Blk_wrtn
sdj 0.06 57.76 0.00 15222123 72
sdj 0.00 0.00 0.00 0 0
sdj 0.00 0.00 0.00 0 0
sdj 98.00 102400.00 0.00 102400 0
sdj 298.00 305152.00 0.00 305152 0
sdj 303.00 309248.00 0.00 309248 0
sdj 303.00 311296.00 0.00 311296 0
sdj 301.00 307200.00 0.00 307200 0
sdj 302.00 309248.00 0.00 309248 0
sdj 302.00 309248.00 0.00 309248 0
sdj 141.00 143360.00 0.00 143360 0
Calculating MBps from iostat can be achieved by calculating KB from blocks read per second (Blk_read/s) and multiplying them by the transactions per second (TPS). In the previous example, 311296 Blk_read/s / (303 tps) = 1027.3 blocks x 512 bytes/block = 526018 bytes / 1024 bytes/KB = 513KB avg.
Before we explain the importance of the blocking size based on a given driver, let us demonstrate the same test results with a different block size. Again, we move 1GB of data through a raw character device using a much smaller block size. It is very important to understand that the exact same 1GB of data is being read by dd and written to /dev/null.
The I/O block size can impact performance. By reducing the dd read block size from 1024k to 2k, the FCP payload of 2k and the SCSI disk (sd) driver can deliver about 1/16 of the performance. Additionally, the I/O rate increases dramatically as the block size of each request drops to that of the FCP limit. In the first example, the sd driver was blocking on 512k, which put the I/O rate around 300 per second. In the world of speed, 300 I/O per second is rather dismal; however, we must keep that number in perspective because we were moving large data files at over 100 MBps. Though the I/O rate was low, the MBps was enormous.
Most applications use an 8K block size. In the following demonstration, we use a 2K block size to illustrate the impact of I/O payload (I/O size).
atlorca2:~ # time -p dd if=/dev/raw/raw8 of=/dev/null bs=2k count=524288 524288+0 records in 524288+0 records out real 95.98 user 0.29 sys 8.78
You can easily see that by simply changing the block size of a data stream from 1024k to 2k, the time it takes to move large amounts of data changes drastically. The time to transfer 1GB of data has increased 13 times from less than 7 seconds to almost 96 seconds, which should highlight the importance of block size to any bean counter.
We can use sar to determine the average I/O size (payload).
atlorca2:~ # sar -d 1 100| grep dev8-144 14:46:50 dev8-144 5458.00 21832.00 0.00 14:46:51 dev8-144 5478.00 21912.00 0.00 14:46:52 dev8-144 5446.00 21784.00 0.00 14:46:53 dev8-144 5445.00 21780.00 0.00 14:46:54 dev8-144 5464.00 21856.00 0.00 14:46:55 dev8-144 5475.00 21900.00 0.00 14:46:56 dev8-144 5481.00 21924.00 0.00 14:46:57 dev8-144 5467.00 21868.00 0.00
From the sar output, we can determine that 21868 rd_sec/s transpires, while we incur a tps of 5467. The quotient of 21868/5467 provides four sectors in a transaction, which equates to 2048 bytes, or 2K. This calculation shows that we are moving much smaller chunks of data but at an extremely high I/O rate of 5500 I/O per second. Circumstances do exist where I/O rates are the sole concern, as with the access rates of a company's Web site. However, changing perspective from something as simple as a Web transaction to backing up the entire corporate database puts sharp focus on the fact that block size matters. Remember, backup utilities use large block I/O, usually 64k.
With the understanding that small block I/O impedes large data movements, note that filesystem fragmentation and sparse file fragmentation can cause an application's request to be broken into very small I/O. In other words, even though a dd if=/file system/file_name of=/tmp/out_file bs=128k is requesting a read with 128k block I/O, sparse file or filesystem fragmentation can force the read to be broken into much smaller block sizes. So, as we continue to dive into performance troubleshooting throughout this chapter, always stay focused on the type of measurement needed: I/O, payload, or block size. In addition to considering I/O, payload, and block size, time is an important factor.
Continuing with our example, we must focus on I/O round-trip time and bus saturation using the same performance test as earlier. In the next few examples, we use iostat to illustrate average wait time, service time, and percent of utilization of our test device.
The following iostat display is from the previous sequential read test but with block size set to 4096k, or 4MBs, illustrating time usage and device saturation.
atlorca2:~ # dd if=/dev/raw/raw8 of=/dev/null bs=4096k & atlorca2:~ # iostat -t -d -x 1 100 Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdj 0.00 0.00 308.00 0.00 319488.00 0.00 159744.00 0.00 1037.30 4.48 14.49 3.24 99.80 sdj 0.00 0.00 311.00 0.00 319488.00 0.00 159744.00 0.00 1027.29 4.53 14.66 3.21 99.70
In this iostat output, we see that the device utilization is pegged at 100%. When device utilization reaches 100%, device saturation has been achieved. This value indicates not only saturation but also the percentage of CPU time for which an I/O request was issued. In addition to eating up CPU cycles with pending I/O waits, notice that the round-trip time (service time) required for each I/O request increased.
Service time, the time required for a request to be completed on any given device, holds around 3.2ms. Before we go into detail about all the items that must be completed within that 3.2ms, which are discussed later in this chapter, we need to recap the initial test parameters. Recall that the previous iostat data was collected while using the dd command with a block size of 4096k. Running the same test with block size set to 1024k yields identical block counts in iostat, as you can see in this example:
atlorca2:~ # dd if=/dev/raw/raw8 of=/dev/null bs=1024k & atlorca2:~ # iostat -t -d -x 1 100 Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdj 0.00 0.00 303.00 0.00 309248.00 0.00 154624.00 0.00 1020.62 1.51 4.96 3.29 99.80 sdj 0.00 0.00 303.00 0.00 311296.00 0.00 155648.00 0.00 1027.38 1.51 4.98 3.29 99.70 sdj 0.00 0.00 304.00 0.00 311296.00 0.00 155648.00 0.00 1024.00 1.50 4.95 3.26 99.00 sdj 0.00 0.00 303.00 0.00 309248.00 0.00 154624.00 0.00 1020.62 1.50 4.93 3.28 99.40 sdj 0.00 0.00 304.00 0.00 311296.00 0.00 155648.00 0.00 1024.00 1.50 4.93 3.28 99.60
As we have illustrated earlier in this chapter, block size greatly impacts an application's overall performance. However, there are limits that must be understood concerning who has control over the I/O boundary. Every application has the capability to set its own I/O block request size, but the key is to understand the limits and locations. In Linux, excluding applications and filesystems, the sd driver blocks all I/O on the largest block depending on medium (such as SCSI LVD or FCP). An I/O operation on the SCSI bus with any typical SCSI RAID controller (not passing any other port drivers, such as Qlogic, or Emulex FCP HBA) holds around 128KB. However, in our case, through FCP, the largest block size is set to 512KB, as shown in the previous example when doing a raw sequential read access through dd. However, it goes without saying that other factors have influence, as shown later in this chapter when additional middle layer drivers are installed for I/O manipulation.
To determine the maximum blocking factor, or max I/O size, of a request at the SD/FCP layer through a raw sequential read access, we must focus on the following items captured by the previous dd request in iostat examples.
The following example explains how to calculate block size.
Device: tps Blk_read/s Blk_wrtn/s Blk_read Blk_wrtn sdj 303.00 311296.00 0.00 311296 0
As the output shows, the number of blocks read per second is 311296.00, and the number of transactions per second is 303.00.
(sectors read/sec)/(read/sec) =~ 1024 sectors
Note
=~ means approximation.
Recall that a sector has 512 bytes.
(1024 sectors) x (512 bytes/sector) = 524288 bytes
Now convert the value to KB.
(524288 bytes) / (1024 bytes/KB) = 512KB
Another way to calculate the block size of an I/O request is to simply look at the avgrq-sz data from iostat. This field depicts the average number of sectors requested in a given I/O request, which in turn only needs to be multiplied by 512 bytes to yield the block I/O request size in bytes.
Now that we have demonstrated how to calculate the in-route block size on any given I/O request, we need to return to our previous discussion about round-trip time and follow up with queue length.
Service time only includes the amount of time required for a device to complete the request given to it. It is important to keep an eye on svctm so that any latency with respect to the end device can be noted quickly and separated from the average wait time. The average wait time (await) is not only the amount of time required to service the I/O at the device but also the amount of wait time spent in the dispatch queue and the roundtrip time. It is important to keep track of both times because the difference between the two can help identify problems with the local host.
To wrap things up with I/O time and queues, we need to touch on queue length.. If you are familiar with C programming, you may find it useful to look at how these values are calculated. The following depicts the calculation for average queue length and wait time found in iostat source code.
nr_ios = sdev.rd_ios + sdev.wr_ios;
tput = ((double) nr_ios) * HZ / itv;
util = ((double) sdev.tot_ticks) / itv * HZ;
svctm = tput ? util / tput : 0.0;
/*
* kernel gives ticks already in milliseconds for all platforms
* => no need for further scaling.
*/
await = nr_ios ?
(sdev.rd_ticks + sdev.wr_ticks) / nr_ios : 0.0;
arqsz = nr_ios ?
(sdev.rd_sectors + sdev.wr_sectors) / nr_ios : 0.0;
printf("%-10s", st_hdr_iodev_i->name);
if (strlen(st_hdr_iodev_i->name) > 10)
printf("\n ");
/* rrq/s wrq/s r/s w/s rsec wsec rkB wkB rqsz qusz await svctm %util */
printf(" %6.2f %6.2f %5.2f %5.2f %7.2f %7.2f %8.2f %8.2f %8.2f %8.2f %7.2f %6.2f %6.2f\n",
((double) sdev.rd_merges) / itv * HZ,
((double) sdev.wr_merges) / itv * HZ,
((double) sdev.rd_ios) / itv * HZ,
((double) sdev.wr_ios) / itv * HZ,
((double) sdev.rd_sectors) / itv * HZ,
((double) sdev.wr_sectors) / itv * HZ,
((double) sdev.rd_sectors) / itv * HZ / 2,
((double) sdev.wr_sectors) / itv * HZ / 2,
arqsz,
((double) sdev.rq_ticks) / itv * HZ / 1000.0,
await,
/* The ticks output is biased to output 1000 ticks per second */
svctm,
/* Again: ticks in milliseconds */
util / 10.0);
Though it is nice to understand the calculations behind every value provided in performance tools, the most important thing to recall is that a large number of outstanding I/O requests on any given bus is not desirable when faced with performance concerns.
In the following iostat example, we use an I/O request size of 2K, which results in low service time and queue length but high disk utilization.
atlorca2:~ # dd if=/dev/raw/raw8 of=/dev/null bs=2k & atlorca2:~ # iostat -t -d -x 1 100 Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdj 0.00 0.00 5492.00 0.00 21968.00 0.00 10984.00 0.00 4.00 0.97 0.18 0.18 96.70 sdj 0.00 0.00 5467.00 0.00 21868.00 0.00 10934.00 0.00 4.00 0.95 0.17 0.17 94.80 sdj 0.00 0.00 5413.00 0.00 21652.00 0.00 10826.00 0.00 4.00 0.96 0.18 0.18 96.40 sdj 0.00 0.00 5453.00 0.00 21812.00 0.00 10906.00 0.00 4.00 0.98 0.18 0.18 97.80 sdj 0.00 0.00 5440.00 0.00 21760.00 0.00 10880.00 0.00 4.00 0.97 0.18 0.18 96.60
Notice how the %util remains high, while the request size falls to 4 sectors/(I/O), which equals our 2048-byte block size. In addition, the average queue size remains small, and wait time is negligible along with service time. Recall that wait time includes roundtrip time, as discussed previously. Now that we have low values for avgrq-sz, avgqu-sz, await, and svctm, we must decide whether we have a performance problem. In this example, the answer is both yes and no. Yes, the device is at its peak performance for a single thread data query, and no, the results for the fields typically focused on to find performance concerns are not high.
Now that we have covered the basics, let us address a multiple read request to a device.
In the following example, we proceed with the same block size, 2K, as discussed previously; however, we spawn a total of six read threads to the given device to illustrate how service time, queue length, and wait time differ. Let's run six dd commands at the same time.
atlorca2:~ # dd if=/dev/raw/raw8 of=/dev/null bs=2k & atlorca2:~ # dd if=/dev/raw/raw8 of=/dev/null bs=2k & atlorca2:~ # dd if=/dev/raw/raw8 of=/dev/null bs=2k & atlorca2:~ # dd if=/dev/raw/raw8 of=/dev/null bs=2k & atlorca2:~ # dd if=/dev/raw/raw8 of=/dev/null bs=2k & atlorca2:~ # dd if=/dev/raw/raw8 of=/dev/null bs=2k &
Note that the previous code can be performed in a simple for loop:
for I in 1 2 3 4 5 6 do dd if=/dev/raw/raw8 of=/dev/null bs=2k & done
Let's use iostat again to look at the dd performance.
atlorca2:~ # iostat -t -d -x 1 100 Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdj 0.00 0.00 5070.00 0.00 20280.00 0.00 10140.00 0.00 4.00 4.96 0.98 0.20 100.00 sdj 0.00 0.00 5097.00 0.00 20388.00 0.00 10194.00 0.00 4.00 4.97 0.98 0.20 100.00 sdj 0.00 0.00 5103.00 0.00 20412.00 0.00 10206.00 0.00 4.00 4.97 0.97 0.20 100.00
The queue length (avgqu-sz) is 4.97, while the max block request size holds constant. The service time for the device to act on the request remains at 0.20ms. Furthermore, the average wait time has increased to 0.98ms due to the device's response to multiple simultaneous I/O requests requiring a longer round-trip time. It is useful to keep the following example handy when working with a large multithreaded performance problem because the device may be strained, and striping at a volume manager level across multiple devices would help relieve this type of strain.
To illustrate the reduction of strain, let us create a VG and 4000MB lvol striped across two disks with a 16k stripe size.
atlorca2:/home/greg/sysstat-5.0.6 # pvcreate /dev/sdi
No physical volume label read from /dev/sdi
Physical volume "/dev/sdi" successfully created
atlorca2:/home/greg/sysstat-5.0.6 # pvcreate /dev/sdj
No physical volume label read from /dev/sdj
Physical volume "/dev/sdj" successfully created
atlorca2:/home/greg/sysstat-5.0.6 # vgcreate vg00 /dev/sdi /dev/sdj
Volume group "vg00" successfully created
atlorca2:/home/greg/sysstat-5.0.6 # lvcreate -L 4000m -i 2 -I 16 -n lvol1 vg00
Logical volume "lvol1" created
atlorca2:/home/greg/sysstat-5.0.6 # lvdisplay -v /dev/vg00/lvol1
Using logical volume(s) on command line
------ Logical volume ------
LV Name /dev/vg00/lvol1
VG Name vg00
LV UUID UQB5AO-dp8Z-N0ce-Dbd9-9ZEs-ccB5-zG7fsF
LV Write Access read/write
LV Status available
# open 0
LV Size 3.91 GB
Current LE 1000
Segments 1
Allocation next free (default)
Read ahead sectors 0
Block device 253:0
We again use sequential 2k reads with dd to measure the performance of the disks.
atlorca2:/home/greg/sysstat-5.0.6 # raw /dev/raw/raw9 /dev/vg00/lvol1 /dev/raw/raw9: bound to major 253, minor 0 atlorca2:/home/greg/sysstat-5.0.6 # dd if=/dev/raw/raw9 of=/dev/null bs=2k & atlorca2:/home/greg/sysstat-5.0.6 # dd if=/dev/raw/raw9 of=/dev/null bs=2k & atlorca2:/home/greg/sysstat-5.0.6 # dd if=/dev/raw/raw9 of=/dev/null bs=2k & atlorca2:/home/greg/sysstat-5.0.6 # dd if=/dev/raw/raw9 of=/dev/null bs=2k & atlorca2:/home/greg/sysstat-5.0.6 # dd if=/dev/raw/raw9 of=/dev/null bs=2k & atlorca2:/home/greg/sysstat-5.0.6 # dd if=/dev/raw/raw9 of=/dev/null bs=2k &
Note that the previous command can be performed in a simple for loop, as previously illustrated. Again we use iostat to measure disk throughput.
atlorca2:/home/greg # iostat -x 1 1000
avg-cpu: %user %nice %sys %iowait %idle
0.01 0.01 0.03 0.11 99.84
Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util
sdi 0.00 0.00 2387.00 0.00 9532.00 0.00 4766.00 0.00 3.99 3.04 1.28 0.42 100.00
sdj 0.00 0.00 2380.00 0.00 9536.00 0.00 4768.00 0.00 4.01 2.92 1.22 0.42 100.00
sdi 0.00 0.00 2318.00 0.00 9288.00 0.00 4644.00 0.00 4.01 3.14 1.35 0.43 99.70
sdj 0.00 0.00 2330.00 0.00 9304.00 0.00 4652.00 0.00 3.99 2.82 1.21 0.43 99.50
Notice that the average wait time per I/O and the service time have increased slightly in this example. However, the average queue has been cut almost in half, as well as the physical I/O demand on the device sdj. The result is similar to a seesaw effect: As one attribute drops, another rises. In the previous scenario, the LUN (sdj) is physically composed of multiple physical mechanisms in the array called (array group), which remains a hidden attribute to the OS. By using the LVM strategy, we reduce some of the contingency for one LUN or array group to handle the entire load needed by the device (lvol). With the previous demonstration, you can see the advantages of striping, as well as its weaknesses. It seems true here that, for every action, there is an equal and opposite reaction.
In the following example, we compare a striped raw lvol to a raw single disk. Our objective is to watch the wait time remain almost constant, while the queue size is cut almost in half when using a lvol stripe instead of a single disk.
First let's look at performance using the lvol. In this example, we start six dd commands that perform sequential reads with block size set to 512k. The dd commands run at the same time and read a raw device bound to the lvol (as illustrated previously) with a 16k stripe size. Remember, iostat shows two disk devices for our lvol test because the lvol is striped across two disks. At 512KB, iostat yields values as follows:
Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdi 0.00 0.00 152.00 0.00 156672.00 0.00 78336.00 0.00 1030.74 3.00 19.60 6.58 100.00 sdj 0.00 0.00 153.00 0.00 155648.00 0.00 77824.00 0.00 1017.31 2.99 19.69 6.54 100.00 sdi 0.00 0.00 154.00 0.00 157696.00 0.00 78848.00 0.00 1024.00 2.98 19.43 6.49 100.00 sdj 0.00 0.00 154.00 0.00 157696.00 0.00 78848.00 0.00 1024.00 3.01 19.42 6.49 100.00
Notice that the I/O queue length when reading lvol1 is much shorter than the following identical dd sequential read test on a raw disk sdj as shown next. Though the identical blocking size of 512k is used, the service time decreases. Here are the results of the test with a raw disk.
Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdj 0.00 0.00 311.00 0.00 318464.00 0.00 159232.00 0.00 1024.00 5.99 19.30 3.22 100.00 sdj 0.00 0.00 310.00 0.00 317440.00 0.00 158720.00 0.00 1024.00 5.99 19.31 3.23 100.00 sdj 0.00 0.00 311.00 0.00 318464.00 0.00 159232.00 0.00 1024.00 5.99 19.26 3.22 100.00
The raw device, sdj in the test using lvol1, reflects that the read requests per second (r/s) remain constant (152 on disk device sdi and 153 on disk device sdj), yielding a net result of 305 read requests per second. The lvol test also shows an improvement in the average wait time; however, we hurt the service time. The service time for the lvol test is about 6.5ms, whereas it is 3.2ms for the raw disk. Upon closer inspection, we notice that the service time is higher due to the I/O issued to the device. In the lvol example, we have in fact submitted 512KB every other time (because we are striping and blocking our I/O both on 512k) so that each total I/O submitted to the device is in a smaller queue, thereby reducing the wait time to be serviced. However, in the single device example, the queue wait time is high because we are waiting on the device to finish on the given I/O request, so with no overhead, the return is faster for the service. This example illustrates the seesaw effect discussed previously, in which a device (single LUN or lvol) is slammed, in which case the end user would need to address the application's need to perform such heavy I/O with a single device. In the previous example, tweaking the device or lvol buys no performance gain; it just moves the time wait status to another field.
Note
With a wider stripe, some performance would be gained in the previous sequential I/O example, but it is unrealistic in the real world. In addition to adding more disks for a wider stripe, you could add more paths to the storage for multipath I/O. However, multipath I/O comes with its own list of constraints.
Many administrators have heard about load balance drivers, which allow disk access through multiple paths. However, very few multipath I/O drivers provide load balance behavior to I/Os across multiple HBA paths as found in enterprise UNIX environments. For example, device drivers such as MD, Autopath, Secure Path (spmgr), and Qlogic's secure path are dedicated primarily to providing an alternate path for a given disk. Though HP's Secpath does offer a true load balance policy for EVA HSG storage on Linux, all the other drivers mentioned only offer failover at this time.
The one true load balancing driver for Linux (HP's Secure Path) provides a round robin (RR) load balance scheduling policy for storage devices on EVA and HSG arrays. Unfortunately, just because a driver that provides load balancing, such as the HP Secure Path driver, exists does not mean support is available for your system. Support for array types is limited. Review your vendor's storage requirements and device driver's hardware support list before making any decisions about which driver to purchase. Keeping in mind that restrictions always exist, let's review a typical RR policy and its advantages and disadvantages.
Though we want to discuss load balancing, the vast majority of Linux enterprise environments today use static (also known as "manual") load balancing or preferred path. With this in mind, we keep the discussion of RR to a minimum.
In the next example, we proceed with a new host and new array that will allow the RR scheduling policy.
The following example illustrates RR through Secure Path on Linux connected through Qlogic HBAs to an EVA storage array. Due to configuration layout, we use a different host for this example.
[root@linny5 swsp]# uname -a Linux linny5.cxo.hp.com 2.4.21-27.ELsmp #1 SMP Wed Dec 1 21:59:02 EST 2004 i686 i686 i386 GNU/Linux
Our host has two HBAs, /proc/scsi/qla2300/0 and /proc/scsi/qla2300/1, with Secure Path version 3.0cFullUpdate-4.0.SP, shown next.
[root@linny5 /]# cat /proc/scsi/qla2300/0
QLogic PCI to Fibre Channel Host Adapter for QLA2340:
Firmware version: 3.03.01, Driver version 7.01.01
Entry address = f88dc060
HBA: QLA2312 , Serial# G8762
[root@linny5 swsp]# cat /etc/redhat-release
Red Hat Enterprise Linux AS release 3 (Taroon Update 3)
Continuing with our raw device testing, we must bind the block device to the character device.
[root@linny5 swsp]# raw /dev/raw/raw8 /dev/spdev/spd
Use the Secure Path command spmgr to display the product's configuration.
[root@linny5 swsp]# spmgr display
Server: linny5.cxo.hp.com Report Created: Tue, May 10 19:10:04 2005
Command: spmgr display
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Storage: 5000-1FE1-5003-1280
Load Balance: Off Auto-restore: Off
Path Verify: On Verify Interval: 30
HBAs: 2300-0 2300-1
Controller: P66C5E1AAQ20AL, Operational
P66C5E1AAQ20AD, Operational
Devices: spa spb spc spd
To reduce space needed for this example, a large part of the spmgr display has been truncated, and we focus only on device spd, as per the following:
TGT/LUN Device WWLUN_ID #_Paths
0/ 3 spd 6005-08B4-0010-056A-0000-9000-0025-0000 4
Controller Path_Instance HBA Preferred? Path_Status
P66C5E1AAQ20AL YES
hsx_mod-0-0-0-4 2300-0 no Available
hsx_mod-1-0-2-4 2300-1 no Active
Controller Path_Instance HBA Preferred? Path_Status
P66C5E1AAQ20AD no
hsx_mod-0-0-1-4 2300-0 no Standby
hsx_mod-1-0-1-4 2300-1 no Standby
Notice that two HBAs and two controllers are displayed. In this case, the EVA storage controller P66C5E1AAQ20AL has been set to preferred active on this particular LUN, in which both of the fabric N_ports enable connection to the fabric. In this configuration, each N_Port connects to different fabrics, A and B, which are seen by Qlogic 0 and 1. In addition, each HBA also sees the alternate controller in case a failure occurs on the selected preferred controller.
We should also to mention that not all arrays are Active/Active on all paths for any LUN at any given time. In this case, the EVA storage array is an Active/Active array because both N_Ports on any given controller have the capability to service an I/O. However, any one LUN can only access a single N_Port at any moment, while another LUN can access the alternate port or alternate controller. Now that we have a background in EVA storage, we need to discuss how the worldwide name (WWN) of a given target device can be found. In the following illustration, we simply read the content of the device instance for the filter driver swsp.
[root@linny5 swsp]# cat /proc/scsi/swsp/2 swsp LUN information: Array WWID: 50001FE150031280
Next, we initiate load balancing and start our raw device test, which is identical to the test performed earlier in this chapter.
[root@linny5 swsp]# spmgr set -b on 50001FE150031280 [root@linny5 swsp]# dd if=/dev/raw/raw8 of=/dev/null bs=512k
While this simple test runs, we collect iostat-x-d1 100 measurements, and we collect a few time captures.
Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdd 0.00 0.00 5008.00 0.00 319488.00 0.00 159744.00 0.00 63.80 9.81 1.96 0.19 93.00 sdd 0.00 0.00 4992.00 0.00 320512.00 0.00 160256.00 0.00 64.21 9.96 1.96 0.18 91.00 sdd 0.00 0.00 4992.00 0.00 318464.00 0.00 159232.00 0.00 63.79 9.80 2.00 0.18 92.00 sdd 0.00 0.00 4992.00 0.00 319488.00 0.00 159744.00 0.00 64.00 9.89 1.98 0.19 95.00
Notice that the blocking factor for a given I/O has changed to 64 sectors per I/O, which equals 32k block size from the swsp module. To get a good comparison between a sequential read test with RR enabled and one with RR disabled, we must disable load balance and rerun the same test. We disable load balancing in the following example.
[root@linny5 swsp]# spmgr set -b off 50001FE150031280 [root@linny5 swsp]# dd if=/dev/raw/raw8 of=/dev/null bs=512k [root@linny5 swsp]# iostat -x 1 100 Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdd 0.00 0.00 4718.00 0.00 302080.00 0.00 151040.00 0.00 64.03 9.48 2.01 0.21 98.00 sdd 0.00 0.00 4710.00 0.00 302080.00 0.00 151040.00 0.00 64.14 9.61 2.02 0.20 94.00 sdd 0.00 0.00 4716.00 0.00 302080.00 0.00 151040.00 0.00 64.05 9.03 1.91 0.20 95.00 sdd 0.00 0.00 4710.00 0.00 301056.00 0.00 150528.00 0.00 63.92 8.23 1.76 0.20 96.00
Iostat reports that the block size remains constant and that the average wait time for a given I/O round trip is slightly higher. This makes sense now that all I/O is on a single path. Because more I/O is loaded on a single path, average wait time increases, as do service times. Now that we have drawn a quick comparison between spmgr being enabled and disabled on a sequential read, we need to recap the advantages seen thus far.
In the previous example, no performance gain was seen by enabling load balancing with regard to spmgr. As we can see, no obvious performance increase was seen when RR was enabled through the host measurements. However, though the host's overall benefit from enabling load balancing was insignificant, the SAN load was cut in half. Keep in mind that a simple modification can impact the entire environment, even outside the host. Something as minor as having a static load balance with a volume manager strip across multiple paths or having a filter driver automate the loading of paths can have a large impact on the overall scheme.
Finally, with respect to load balance drivers, it is important to watch for the max block size for any given transfer. As seen in the previous iostat examples, the Secure Path product reduces the block size to 32k, and if LVM were to be added on top of that, the block would go to 16K. This small block transfer is great for running small, block-heavy I/O traffic, but for large data pulls, it can become a bottleneck.
Application data access through a filesystem is much more common than access through raw storage. Filesystem meta structures are maintained by the filesystem's driver, removing the overhead from the application. For this reason, very few applications are written to perform raw I/O, except for a few database systems whose creators believe they can maintain the integrity or performance better than a standardized filesystem. This section addresses performance characteristics with regards to multiple filesystems and draws a comparison to the previous section on raw device access.
The configuration for this section is identical to the previous section. We use the IA64 host atlorca2. Filesystem types used for comparisons are xfs and ext3. To begin, we define a filesystem using the same disks from previous examples. By creating a filesystem, we simply add an additional layer to the data management overhead.
The following fdisk output shows device sdj with one primary partition of type 83 known as Linux native. The objective is to compare performance of a large sequential read. We compare performance with raw to performance with XFS, illustrating performance overhead.
The first step in comparing performance between raw and filesystem is to set a baseline. Here we set a performance baseline for a single threaded read on a raw device named /dev/sdj.
atlorca2:~ # fdisk -l /dev/sdj Disk /dev/sdj: 250.2 GB, 250219069440 bytes 255 heads, 63 sectors/track, 30420 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Device Boot Start End Blocks Id System /dev/sdj1 1 30420 244348618+ 83 Linux
The fdisk output shows that partition 1 is active, with 250GB of capacity.
The XFS filesystem type offers the performance feature of creating a journal log on a disk separate from the filesystem data. Next we demonstrate this feature and test it. In this example, device sdk is the journal log device, and sdj is used for the metadata device.
atlorca2:~ # mkfs.xfs -f -l logdev=/dev/sdk1,size=10000b /dev/sdj1
meta-data=/dev/sdj1 isize=256 agcount=16, agsize=3817947 blks
= sectsz=512
data = bsize=4096 blocks=61087152, imaxpct=25
= sunit=0 swidth=0 blks, unwritten=1
naming =version 2 bsize=4096
log =/dev/sdk1 bsize=4096 blocks=10000, version=1
= sectsz=512 sunit=0 blks
realtime =none extsz=65536 blocks=0, rtextents=0
The following demonstrates a large block write and the performance boost that an external logger offers.
atlorca2:/xfs.test # dd if=/dev/zero of=/xfs.test/zero.out bs=512k atlorca2:~ # iostat -x 1 100|egrep "sdj|sdk" Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdj 0.00 32512.00 0.00 281.00 0.00 262144.00 0.00 131072.00 932.90 141.99 394.24 3.56 100.00 sdk 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 sdj 0.00 36576.00 0.00 277.00 0.00 294929.00 0.00 147464.50 1064.73 142.75 511.21 3.61 100.00 sdk 0.00 0.00 0.00 1.00 0.00 5.00 0.00 2.50 5.00 0.10 105.00 105.00 10.50 sdj 0.00 35560.00 0.00 279.00 0.00 286736.00 0.00 143368.00 1027.73 141.26 380.54 3.58 100.00 sdk 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 sdj 0.00 36576.00 0.00 277.00 0.00 294912.00 0.00 147456.00 1064.66 142.35 204.65 3.61 100.00 sdk 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
The journal log device provides little to no added performance. There is almost no I/O to disk sdk, which contains the log. Putting the log on a separate disk won't help performance because we are not diverting a meaningful amount of I/O. The journal log device provides no added benefit because the intent to modify/write is established at the beginning of the file access. From this point forward, the I/O is completed on the meta device, as depicted in the previous iostat.
As shown previously, the dd command bs option sets the block size to 512k for each I/O transaction. We can see this in iostat. To calculate this value, find the average request size (avgrq-sz) column from iostat. In this example, we find that avgrq-sz has a value of 1024 sectors. To calculate the block size, multiply avgrq-sz by the sector size (512 bytes). In this example:
1024 (sectors) x 512 (bytes/sector) = 524288 (bytes) / 1024 (KB/bytes) = 512KB
However, the same dd command using the XFS filesystem reveals that the largest avgrq-sz value set forth by a sequential read is equal to 256 (sectors) regardless of block size set by dd. Following the same calculations, we determine that an XFS sequential read has a block size set to 128KB. The block size of any I/O is an important item to understand because not all programs control the block size. A program can request a given block size; however, a lower-layer driver can require that the I/O request be broken into smaller requests as illustrated in the previous iostat output.
Thus we have demonstrated that using a remote journal provides no performance improvements for large data access on sequential reads and writes. However, when writing to our XFS filesystem with a large number of small files, the opposite becomes true: The remote journal does in fact help.
In the next test, we write 512KB files as fast as the system allows while watching the load on the journal device and the filesystem meta device. To run this test, we must write a short, simple program to control the number of files to create during our test phase. The program is as follows:
#!/bin/sh
count=1
total=0
while [ $total -ne $* ]
do
total='expr $total + $count'
touch $total
dd if=/xfs.test/zero.out of=/xfs.test/$total bs=512k > /dev/null 2>&1 #Using dd to control the BS.
done
This program uses the same dd command used throughout our testing. It is important to always keep the control in any test constant. By running the previous program, thousands of 512KB files are created, causing an increased load on the journal log device (sdk), as depicted in the following listing:
atlorca2:/xfs.test # ./count_greg.sh 10000 atlorca2:~ # iostat -x 1 100|egrep "sdj|sdk" Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdj 0.00 16219.00 0.00 181.00 0.00 131169.00 0.00 65584.50 724.69 9.31 54.90 2.77 50.20 sdk 0.00 0.00 0.00 53.00 0.00 3200.00 0.00 1600.00 60.38 1.43 28.96 9.40 49.80 sdj 0.00 20274.00 0.00 201.00 0.00 163936.00 0.00 81968.00 815.60 11.90 54.74 2.83 56.90 sdk 0.00 0.00 0.00 39.00 0.00 2752.00 0.00 1376.00 70.56 1.19 26.26 14.00 54.60 sdj 0.00 20273.00 0.00 198.00 0.00 163936.00 0.00 81968.00 827.96 10.96 54.34 2.77 54.80 sdk 0.00 0.00 0.00 50.00 0.00 3072.00 0.00 1536.00 61.44 1.43 30.48 10.90 54.50 sdj 0.00 16217.00 0.00 200.00 0.00 131138.00 0.00 65569.00 655.69 10.22 56.56 2.71 54.10 sdk 0.00 0.00 0.00 50.00 0.00 2982.00 0.00 1491.00 59.64 1.37 28.78 10.92 54.60
By creating thousands of files or modifying the same file thousands of times a minute, we can see the added load on the journal device sdk as well as the filesystem device sdj. By understanding the end goal, we can make better decisions about how to size and lay out a filesystem.
In addition, notice the block size of the I/O request submitted to the filesystem and how the filesystem responded. In the previous example on a sequential read, the block size is restricted to 128K. However, on a sequential write, the blocking structure on the XFS filesystem is that which is set forth by the command calling the SCSI write, setting the average request size to 512k as shown in the previous iostat illustration. However, will the same results be found using a completely different filesystem?
Let's repeat our test on ext3, also known as ext2 with journaling, to depict I/O latency and blocking factors.
atlorca2:/ext3.test # dd if=/ext3.test/usr.tar of=/dev/null bs=512k atlorca2:/ # iostat -x 1 100| grep sdj Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdj 60.00 0.00 898.00 0.00 229856.00 0.00 114928.00 0.00 255.96 0.98 1.10 1.09 98.20 sdj 58.00 0.00 939.00 0.00 240360.00 0.00 120180.00 0.00 255.97 0.98 1.04 1.04 97.50 sdj 62.00 1.00 918.00 3.00 234984.00 32.00 117492.00 16.00 255.17 0.97 1.06 1.05 97.00 sdj 62.00 0.00 913.00 0.00 233704.00 0.00 116852.00 0.00 255.97 0.98 1.07 1.07 97.80 sdj 58.00 0.00 948.00 0.00 242664.00 0.00 121332.00 0.00 255.97 0.96 1.01 1.01 96.20 sdj 62.00 0.00 933.00 0.00 238824.00 0.00 119412.00 0.00 255.97 0.97 1.04 1.04 97.30
The sequential read test holds the same blocking factor as previously seen on XFS, with a little savings on overhead with respect to average wait time and service time. To continue our example, let's proceed with the file create script discussed previously.
#!/bin/sh
count=1
total=0
while [ $total -ne $* ]
do
total='expr $total + $count'
touch $total
dd if=/ext3.test/zero.out of=/ext3.test/$total bs=512k > /dev/null 2>&1 #Using dd to control the BS.
done
atlorca2:/ext3.test # ./count_greg.sh 10000 atlorca2:/ # iostat -x 1 100| grep sdj Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s rkB/s wkB/s avgrq-sz avgqu-sz await svctm %util sdj 0.00 51687.00 0.00 268.00 0.00 416864.00 0.00 208432.00 1555.46 128.62 346.74 3.45 92.40 sdj 0.00 27164.00 1.00 264.00 8.00 219040.00 4.00 109520.00 826.60 139.42 521.48 3.77 100.00 sdj 0.00 656.00 1.00 113.00 8.00 5312.00 4.00 2656.00 46.67 21.95 516.42 3.85 43.90 sdj 0.00 0.00 1.00 0.00 8.00 0.00 4.00 0.00 8.00 0.00 1.00 1.00 0.10 sdj 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 sdj 0.00 52070.00 1.00 268.00 8.00 419936.00 4.00 209968.00 1561.13 128.44 346.09 3.43 92.30 sdj 0.00 27800.00 0.00 271.00 0.00 224160.00 0.00 112080.00 827.16 139.27 513.93 3.69 100.00 sdj 0.00 662.00 2.00 112.00 16.00 5368.00 8.00 2684.00 47.23 20.45 489.31 3.91 44.60 sdj 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 sdj 0.00 0.00 1.00 0.00 8.00 0.00 4.00 0.00 8.00 0.00 0.00 0.00 0.00 sdj 0.00 51176.00 0.00 273.00 0.00 412792.00 0.00 206396.00 1512.06 128.12 336.55 3.37 92.00 sdj 0.00 27927.00 0.00 274.00 0.00 225184.00 0.00 112592.00 821.84 138.00 510.66 3.65 100.00 sdj 0.00 658.00 0.00 105.00 0.00 5328.00 0.00 2664.00 50.74 17.44 492.92 3.57 37.50 sdj 0.00 0.00 1.00 128.00 8.00 1024.00 4.00 512.00 8.00 2.88 22.32 0.35 4.50
Notice how the filesystem buffers the outbound I/O, submitting them to the SCSI layer in a burst pattern. Though nothing is wrong with this I/O pattern, you must understand that the larger the burst, the larger the strain on the storage array. For example, exchange servers save up and de-stage out a burst of I/O operations, which can flood an array's write pending cache, so you should monitor for excessive write burst. However, the write block size maintains a 512k average block, which is similar to the XFS on writes with large block requests.
As we've seen, I/O block sizes, stripe size, and filesystem layouts have unique benefits that aid I/O performance. In addition to these items, most filesystems have unique characteristics that are designed to guess the next request, trying to save resources by anticipating requests. This is accomplished by read-ahead algorithms. Ext2, Ext3, JFS, and XFS all have the capability to perform read-ahead, as shown with XFS in the following XFS source code for mounting /usr/src/linux/fs/xfs/xfs_mount.c:
/*
* Set the number of readahead buffers to use based on
* physical memory size.
*/
if (xfs_physmem <= 4096) /* <= 16MB */
mp->m_nreadaheads = XFS_RW_NREADAHEAD_16MB;
else if (xfs_physmem <= 8192) /* <= 32MB */
mp->m_nreadaheads = XFS_RW_NREADAHEAD_32MB;
else
mp->m_nreadaheads = XFS_RW_NREADAHEAD_K32;
if (sbp->sb_blocklog > readio_log) {
mp->m_readio_log = sbp->sb_blocklog;
} else {
mp->m_readio_log = readio_log;
}
mp->m_readio_blocks = 1 << (mp->m_readio_log - sbp->sb_blocklog);
if (sbp->sb_blocklog > writeio_log) {
mp->m_writeio_log = sbp->sb_blocklog;
} else {
mp->m_writeio_log = writeio_log;
}
mp->m_writeio_blocks = 1 << (mp->m_writeio_log - sbp->sb_blocklog);
Although read-ahead is a powerful attribute, a concern exists. It is not fair to say that read-ahead causes these drawbacks, as a true increase in read performance can be seen on any filesystem that uses read-ahead functionality. When using read-ahead, filesystem block size is an important factor. For example, if filesystem block size is 8k and sequential read pattern exist where an application is reading 1K sequential blocks (index), readahead kicks in and pulls an extra predefined number of blocks, where each block is equal to the filesystem block size. To sum up the concern with read-ahead, one must be careful not to read in more data than is needed. Another performance boost can be found by utilizing buffer cache.
Filesystems such as XFS, Reiser, Ext2, and Ext3 use the buffer cache and reduce the amount of memory for an application to process data in the buffer cache, forcing more physical I/O (PIO). Later in this chapter we discuss the difference between raw, Oracle Cluster File System (OCFS), and XFS in an example with buffer cache and readahead. Before we jump too far ahead, though, we need to cover one last topic with respect to disk performance.
We have covered some in-depth I/O troubleshooting tactics for character and block devices. Now we need to address the rumor mill about disk geometry alignment. Geometry alignment, also known as sector alignment, is the new craze in Windows performance tweaking. Cylinders lie in a small band, like a ring on a platter. The cylinders are then divided into tracks (wedges), which contain sectors, which are described in great detail in Chapter 6, "Disk Partitions and File Systems."
Sector alignment provides little to no performance boost in Linux. To date, no issues exist with regards to how partitions and filesystems interact with sectors alignment for a given platter, regardless of whether the platter is logical or physical within Linux. However, for those who are interested, a performance boost has been documented with respect to DOS 6.X, Windows 2000, and greater. See http://www.microsoft.com/resources/documentation/Windows/2000/server/reskit/en-us/Default.asp?url=/resources/documentation/Windows/2000/server/reskit/en-us/prork/pree_exa_oori.asp and http://www.microsoft.com/resources/documentation/windows/2000/professional/reskit/en-us/part6/proch30.mspx for more information.
Now that we have covered some basics guidelines about I/O performance metrics, we need to revisit our primary goal. As already mentioned, our primary goal is to deliver methods for finding performance problems. In all circumstances, a good performance snapshot should be taken at every data center before and after changes to firmware roles, fabric changes, host changes, and so on.
The following is generalized performance data from a single LUN RAID 5 7d+1p, and it is provided to demonstrate the performance benchmark tool called bonnie. The following test does not depict the limit of the array used for this test. However, the following example enables a brief demonstration of a single LUN performance characteristic between three filesystems. The following bonnie++ benchmark reflects the results of the equipment used throughout this chapter with XFS and Ext3 filesystems.
atlorca2:/ext3.test # bonnie++ -u root:root -d /ext3.test/bonnie.scratch/ -s 8064m -n 16:262144:8:128
Version 1.03 ------Sequential Output------ --Sequential Input- --Random-
-Per Chr- --Block-- -Rewrite- -Per Chr- --Block-- --Seeks--
Machine Size K/sec %CP K/sec %CP K/sec %CP K/sec %CP K/sec %CP /sec %CP
atlorca2 8064M 15029 99 144685 44 52197 8 14819 99 124046 7 893.3 1
------Sequential Create------ --------Random Create--------
-Create-- --Read--- -Delete-- -Create-- --Read--- -
Delete--
files:max:min /sec %CP /sec %CP /sec %CP /sec %CP /sec %CP /sec %CP
16:262144:8/128 721 27 14745 99 4572 37 765 28 3885 28 6288 58
Testing with XFS and journal on the same device yields the following:
atlorca2:/ # mkfs.xfs -f -l size=10000b /dev/sdj1
meta-data=/dev/sdj1 isize=256 agcount=16, agsize=3817947 blks
= sectsz=512
data = bsize=4096 blocks=61087152, imaxpct=25
= sunit=0 swidth=0 blks, unwritten=1
naming =version 2 bsize=4096
log =internal log bsize=4096 blocks=10000, version=1
= sectsz=512 sunit=0 blks
realtime =none extsz=65536 blocks=0, rtextents=0
atlorca2:/ # mount -t xfs -o logbufs=8,logbsize=32768 /dev/sdj1 /xfs.test
atlorca2:/ # mkdir /xfs.test/bonnie.scratch/
atlorca2:/ # mount -t xfs -o logbufs=8,logbsize=32768 /dev/sdj1 /xfs.test
atlorca2:/xfs.test # bonnie++ -u root:root -d /xfs.test/bonnie.scratch/ -s 8064m -n 16:262144:8:128
Version 1.03 ------Sequential Output------ --Sequential Input- --
Random-
-Per Chr- --Block-- -Rewrite- -Per Chr- --Block-- --Seeks--
Machine Size K/sec %CP K/sec %CP K/sec %CP K/sec %CP K/sec %CP /sec %CP
atlorca2 8064M 15474 99 161153 21 56513 8 14836 99 125513 9 938.8 1
------Sequential Create------ -------- Random Create--
------
-Create-- --Read--- -Delete-- -Create-- --Read--- -
Delete--
files:max:min /sec %CP /sec %CP /sec %CP /sec %CP /sec %CP /sec %CP
16:262144:8/128 1151 24 12654 100 9705 89 1093 22 12327 99 6018 71
Testing with XFS without remote journal provides these results:
atlorca2:~ # mount -t xfs -o logbufs=8,logbsize=32768,logdev=/dev /sdk1/dev/sdj1 /xfs.test
atlorca2:/xfs.test # mkdir bonnie.scratch/
atlorca2:/xfs.test # bonnie++ -u root:root -d /xfs.test/bonnie.scratch/ -s 8064m -n 16:262144:8:128
Version 1.03 ------Sequential Output------ --Sequential Input- --
Random-
-Per Chr- --Block-- -Rewrite- -Per Chr- --Block-- --
Seeks--
Machine Size K/sec %CP K/sec %CP K/sec %CP K/sec %CP K/sec %CP /sec %CP
atlorca2 8064M 15385 99 146197 20 58263 8 14833 99 126001 9 924.6 1
------Sequential Create------ --------Random Create--------
-Create-- --Read--- -Delete-- -Create-- --Read--- -
Delete--
files:max:min /sec %CP /sec %CP /sec %CP /sec %CP /sec %CP /sec %CP
16:262144:8/128 1175 24 12785 100 10236 95 1097 22 12280 99 6060 72
Just by changing the filesystem and journal log location, we pick up some nice performance on sequential I/O access with respect to XFS. The point of the previous demonstration is to identify factors other than hardware that increase performance. As we have seen, simply changing the filesystem layout or type can increase performance greatly. One other performance tool we enjoy using for SCSI measurements is IOzone, found at www.iozone.org.
As with any performance problem, usually more than one factor exists. Our troubleshooting performance journey continues with coverage of application CPU usage and how to monitor it. In this section, CPU usage and application-specific topics are covered, focusing on process threads.
To begin, we want to demonstrate how a few lines of code can load a CPU to a 100% busy state. The C code "using SLES 9 with long integer" illustrates a simple count program, which stresses the CPU in user space. It is important to understand that our application is not in system space, also called kernel mode, because we are not focusing on any I/O as previously discussed in this chapter.
Example 1 goes like this:
#include <stdio.h>
#include <sched.h>
#include <pthread.h> /* POSIX threads */
#include <stdlib.h>
#define num_threads 1
void *print_func(void *);
int main ()
{
int x;
printf("main() process has PID= %d PPID= %d\n", getpid(),
getppid());
pthread_t tid[num_threads];
/* Now to create pthreads */
for (x=0; x <= num_threads;x++)
pthread_create(tid + x, NULL, print_func, NULL );
/*wait for termination of threads before main continues*/
for (x=0; x < num_threads;x++)
{
pthread_join(tid[x], NULL);
printf("Main() PID %d joined with thread %d\n", getpid(),
tid[x]);
}
}
void *print_func (void *arg)
{
long int y;
printf("PID %d PPID = %d TID = %d\n",getpid(), getppid(),
pthread_self());
/* creating a very large loop for the CPU to chew on :) */
/* Note, the following line may be changed to: for (y=1;y>0;y++) */
for (y=1; y<10000000000000;y++)
printf ("%d\n", y);
return 0;
}
Note that instead of just counting to a large integer, you may want to create an infinite loop. The C code in Example 2 (also shown in Chapter 8, "Linux Processes: Structures, Hangs, and Core Dumps") generates an infinite loop with a signal handler used to kill the threads.
#include <pthread.h> /* POSIX threads */
#include <signal.h>
#include <stdlib.h>
#include <linux/unistd.h>
#include <errno.h>
#define num_threads 8
void *print_func(void *);
void threadid(int);
void stop_thread(int sig);
_syscall0(pid_t,gettid)
int main ()
{
int x;
pid_t tid;
pthread_t threadid[num_threads];
(void) signal(SIGALRM,stop_thread); /*signal handler */
printf("Main process has PID= %d PPID= %d and TID= %d\n",
getpid(), getppid(), gettid());
/* Now to create pthreads */
for (x=1; x <= num_threads;++x)
pthread_create(&threadid[x], NULL, print_func, NULL );
sleep(60); /* Let the threads warm the cpus up!!! :) */
for (x=1; x < num_threads;++x)
pthread_kill(threadid[x], SIGALRM);
/*wait for termination of threads before main continues*/
for (x=1; x < num_threads;++x)
{
printf("%d\n",x);
pthread_join(threadid[x], NULL);
printf("Main() PID %d joined with thread %d\n", getpid(),
threadid[x]);
}
}
void *print_func (void *arg)
{
printf("PID %d PPID = %d Thread value of pthread_self = %d and
TID= %d\n",getpid(), getppid(), pthread_self(),gettid());
while(1); /* nothing but spinning */
}
void stop_thread(int sig) {
pthread_exit(NULL);
}
To continue with Example 1, compile the code and run the application as follows:
atlorca2:/home/greg # cc -o CPU_load_count CPU_Load_count.c -lpthread atlorca2:/home/greg # ./CPU_load_count | grep -i pid main() process has PID= 5970 PPID= 3791 PID 5970 PPID = 3791 TID = 36354240 PID 5970 PPID = 3791 TID = 69908672
When running top, shown next, we find that our CPUs have a load, yet the PID reports zero percent usage for the CPU_load_count program. So where is the load coming from? To answer this question, we must look at the threads spawned by the parent process.
atlorca2:~ # top top - 20:27:15 up 36 min, 3 users, load average: 1.24, 1.22, 1.14 Tasks: 79 total, 1 running, 78 sleeping, 0 stopped, 0 zombie Cpu0 : 0.1% us, 0.6% sy, 0.0% ni, 99.1% id, 0.2% wa, 0.0% hi, 0.0% si
Cpu1 : 1.7% us, 1.6% sy, 0.0% ni, 96.7% id, 0.0% wa, 0.0% hi, 0.0% si Cpu2 : 44.1% us, 19.0% sy, 0.0% ni, 36.9% id, 0.0% wa, 0.0% hi, 0.0% si Cpu3 : 43.3% us, 21.0% sy, 0.0% ni, 35.7% id, 0.0% wa, 0.0% hi, 0.0% si Mem: 4142992k total, 792624k used, 3350368k free, 176352k buffers Swap: 1049568k total, 0k used, 1049568k free, 308864k cached PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ PPID RUSER UID WCHAN COMMAND 5974 root 16 0 3632 2112 3184 R 0.7 0.1 0:00.13 3874 root 0 - top 5971 root 15 0 3168 1648 2848 S 3.3 0.0 0:02.16 3791 root 0 pipe_wait grep 5970 root 17 0 68448 1168 2560 S 0.0 0.0 0:00.00 3791 root 0 schedule_ CPU_load_count 5966 root 16 0 7376 4624 6128 S 0.0 0.1 0:00.01 5601 root 0 schedule_ vi 5601 root 15 0 5744 3920 4912 S 0.0 0.1 0:00.11 5598 root 0 wait4 bash 5598 root 16 0 15264 6288 13m S 0.0 0.2 0:00.04 3746 root 0 schedule_ sshd
You can see top shows that the CPU_Load_count program is running, yet it reflects zero load on the CPU. This is because top is not thread-aware in SLES 9 by default.
A simple way to determine a thread's impact on a CPU is by using ps with certain flags, as demonstrated in the following. This is because in Linux, threads are separate processes (tasks).
atlorca2:/proc # ps -elfm >/tmp/ps.elfm.out
vi the file, find the PID, and focus on the threads in running (R) state.
F S UID PID PPID C PRI NI ADDR SZ WCHAN STIME TTY TIME CMD 4 - root 5970 3791 0 - - - 4276 - 20:26 pts/0 00:00:00 ./CPU_load_count 4 S root - - 0 77 0 - - schedu 20:26 - 00:00:00 - 1 R root - - 64 77 0 - - schedu 20:26 - 00:02:41 - 1 S root - - 64 79 0 - - schedu 20:26 - 00:02:40 -
Though top provides a great cursory view of your system, other performance tools are sometimes required to find the smoking gun, such as the ps command in the preceding example. Other times, performance trends are required to isolate a problem area, and products such as HP Insight Manager with performance plug-ins may be more suited.
In the following example, we use Oracle's statistics package, called statspak, to focus on an Oracle performance concern.
DB Name DB Id Instance Inst Num Release Cluster Host
------------ ----------- ---------- -------- --------- ------- ----------
DB_NAME 1797322438 DB_NAME 3 9.2.0.4.0 YES Server_name
Snap Id Snap Time Sessions Curs/Sess Comment
------- ------------------ -------- --------- -------------------
Begin Snap: 23942 Date 11:39:12 144 3.0
End Snap: 23958 Date 11:49:17 148 3.1
Elapsed: 10.08 (mins)
Cache Sizes (end)
~~~~~~~~~~~~~~~~~
Buffer Cache: 6,144M Std Block Size: 8K
Shared Pool Size: 1,024M Log Buffer: 10,240K
Load Profile
~~~~~~~~~~~~ Per Second Per Transaction
--------------- ---------------
Redo size: 6,535,063.00 6,362.58
Logical reads: 30,501.36 29.70 <-Cache Reads LIO
Block changes: 15,479.36 15.07
Physical reads: 2,878.69 2.80 <-Disk reads PIO
Physical writes: 3,674.53 3.58 <-Disk Writes PIO
User calls: 5,760.67 5.61
Parses: 1.51 0.00
Hard parses: 0.01 0.00
Sorts: 492.41 0.48
Logons: 0.09 0.00
Executes: 2,680.27 2.61
Transactions: 1,027.11
% Blocks changed per Read: 50.75 Recursive Call %: 15.27
Rollback per transaction %: 0.01 Rows per Sort: 2.29
Instance Efficiency Percentages (Target 100%)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Buffer Nowait %: 99.98 Redo NoWait %: 100.00
Buffer Hit %: 90.57 In-memory Sort %: 100.00
Library Hit %: 100.00 Soft Parse %: 99.45
Execute to Parse %: 99.94 Latch Hit %: 98.37
Parse CPU to Parse Elapsed %: 96.00 % Non-Parse CPU: 99.99
Shared Pool Statistics Begin End
------ ------
Memory Usage %: 28.16 28.17
% SQL with executions>1: 82.91 81.88
% Memory for SQL w/exec>1: 90.94 90.49
Top 5 Timed Events
~~~~~~~~~~~~~~~~~~ % Total
Event Waits Time (s) Ela Time
---------------------------------------- ------------ ----------- --------
db file sequential read 1,735,573 17,690 51.88
log file sync 664,956 8,315 24.38
CPU time 3,172 9.32
global cache open x 1,556,450 1,136 3.36
log file sequential read 3,652 811 2.35
-> s - second
-> cs - centisecond - 100th of a second
-> ms - millisecond - 1000th of a second
-> us - microsecond - 1000000th of a second
Avg
Total Wait Wait Waits
Event Waits Timeouts Time (s) (ms) /txn
---------------------- ------------ ---------- ---------- ------ --------
db file sequential read 1,738,577 0 17,690 10 2.8
This statspak output is truncated to conserve space. Some of the key points of interest are highlighted in bold: Logical reads (LIO), Physical reads/writes (PIO), Latches, Buffered I/O, and db file sequential read. Please understand that many hundred parameters exist, and Oracle has published many performance documents and created many classes that we recommend reading and taking. However, this chapter's main goal is performance troubleshooting from a wide view, or in other words, a quick reference guide to performance troubleshooting. The bold areas in the previous listing are some critical places to focus, especially with LIO and PIO.
To elaborate on the bold elements in the previous listing, we begin with LIO and PIO. LIO is the access of a memory register, residing in the database buffer cache, and PIO is an I/O operation request from Oracle to the system for a data block fetch from spindle. In short, LIO is faster than PIO, but PIO is not always the bottleneck. A critical piece of the puzzle with regards to performance problems on any database connected to any storage device is understanding that a small percentage of PIOs (from Oracle's viewpoint) are actually a read from the host cache, storage cache, or both. Thus, a PIO from the database's perspective may in fact still be a read from cache, so having a high number of PIOs is not always a bad omen. In the previous example, the PIO reads were around 1.7 million with an average latency of 10ms, which, by the way, is not bad. However, although the I/O is fine, a memory control performance problem may still linger in the background.
Thus, while I/O may in fact be fine, memory lock control also must be addressed for smooth, fast database operation. Note that in this example, the Latch hit percentage is around 98%, which raises a red flag. A latch is basically a protection agent for access control with regards to shared memory within the system's global area (SGA). In short, the goal is to keep memory available so that the latch-free count remains high, keeping the percentage of Latch hit percentage around 99.5%. Latch info can be viewed by reviewing both the willing-to-wait and not-willing-to-wait latches found in the immediate_gets and immediate_misses columns by using V$LATCH or by looking at Oracle's statspack. In addition to waiting on free memory segments with regards to latches, we need to touch on buffer waits.
When a database starts to show an increase in buffer waits, the objective is to focus on the two main issues. The first issue is that memory is running low, which is impacting the second issue, that of physical I/O read/writes. A buffer wait is logged when the database must flush a write to spindle to clear up some available cache for data processing. The quick solution to this problem is to run raw (or other proprietary filesystem such as OCFS) to bypass host buffer cache so that buffer cache is used for data processing only. However, the huge drawback to using a non-buffer cache filesystem is the loss of performance with respect to read-ahead as discussed earlier in this chapter.
Now that we have covered some I/O basics, both physical and logical, and memory control with respect to latches, we present an example of an application failing to initialize due to lack of shared memory space. We demonstrate a lack of shared memory without going into detail about system V message queues, semaphores, or shared memory. As with all applications, more memory equals faster performance, and without enough memory, system failure is imminent.
Our example shows a common error 28 example using a 64-bit kernel, with a 64-bit application failing to initialize due to a memory address problem. Our focus is on interprocess communication (IPCS), as with any large application that spawns multiple threads/processes. Using our 64-bit machine, we bring a 64-bit Oracle 10g instance online, which fails with the following error to demonstrate a failed IPCS.
ORA-27102: out of memory Linux-x86_64 Error: 28: No space left on device
System parameters are as follows:
# ipcs ------ Shared Memory Segments -------- key shmid owner perms bytes nattch status 0x00000000 1245184 gdm 600 393216 2 dest 0x852af124 127926273 oracle 640 8558477312 15 ------ Semaphore Arrays -------- key semid owner perms nsems 0x3fbfeb1c 1933312 oracle 640 154 ==== kernel parameters ====== # sysctl -a kernel.sem = 250 32000 100 128 kernel.msgmnb = 16384 kernel.msgmni = 16 kernel.msgmax = 8192 kernel.shmmni = 4096 kernel.shmall = 2097152 kernel.shmmax = 34359738368 ==== process ulimits (bash shell) $ ulimit -a core file size (blocks, -c) 0 data seg size (kbytes, -d) unlimited file size (blocks, -f) unlimited max locked memory (kbytes, -l) 4 max memory size (kbytes, -m) unlimited open files (-n) 65536 pipe size (512 bytes, -p) 8 stack size (kbytes, -s) 10240 cpu time (seconds, -t) unlimited max user processes (-u) 16384 virtual memory (kbytes, -v) unlimited
This failure is a result of the kernel not being able to fulfill the shared memory request. Not enough space is a condition explained in /usr/src/linux/ipc/shm.c, which reads:
if (shm_tot + numpages >= shm_ctlall)
return -ENOSPC;
The program we tried to start previously required more shared memory than we had allocated, which in turn caused the Oracle application to fail on initialization. The solution is to increase shared memory by the kernel parameter. In this example, we simply increase it to shmall=8388608.
As we conclude this chapter, we cover some uncommon tools that can be used to monitor performance characteristics and build charts in most cases. isag, RRDtool, Ganglia (which uses RRDtool to monitor grid computing and clustering), and Nagios are great performance tools. More monitoring tools exist, but for the most part, they are common tools used every day such as sar, iostat, top, and netstat. Due to space limitations, we only cover isag in this chapter. However, the other tools are easy to find and configure if one so desires. isag, found at http://www.volny.cz/linux_monitor/isag/index.html, provides a nice graphical front end to sar. After systat tools have been loaded, isag should be included, as depicted here:
atlorca2:/tmp # rpm -qf /usr/bin/isag sysstat-5.0.1-35.1
Most engineers who work with sar, iostat, and other performance tools will enjoy using isag, the GUI front end to sar. To give a quick demonstration of how the tool works, we must enable sar to collect some data. To achieve a real-world demonstration, we repeat our previous bonnie++ test while running sar -A to collect as much detail as possible and display it through isag.
To demonstrate, we mount an XFS filesystem with a local journal and use an NFS mount point to a HPUX server to demonstrate both disk and network load through bonnie++ while monitoring through sar and isag.
atlorca2:/var/log/sa # mount -t xfs -o logbufs=8,logbsize=32768 /dev/sdj1 /xfs.test
atlorca2:/var/log/sa # df
Filesystem 1K-blocks Used Available Use% Mounted on
hpuxos.atl.hp.com:/scratch/customers
284470272 50955984 218940848 19% /scratch/customers
/dev/sdj1 244308608 4608 244304000 1% /xfs.test
atlorca2:/var/log/sa # sar -A -o 1 100 #This will build a fine in /var/log/sa that isag will use.
atlorca2:/var/log/sa # isag
Performance troubleshooting relies upon comparison. To know that performance is bad, you must have a preexisting benchmark from when performance was good. Though we can continue giving hundreds of examples of performance-related issues, a better approach is to understand the tools and know what to focus on.
To solve storage-related performance problems, you must be familiar with common tools such as sar and iostat. For threaded applications, make sure your version of top shows threads. Otherwise, focus on the ps command with the m flag. Other performance monitoring tools exist, such as Oprofile, Prospect, q-tools (q-syscollect, q-view, and q-dot), Perfmon, and Caliper.
 |
 |
 |
The following steps are used as our quick tuning strategy:
1. Know your system.
2. Back up the system.
3. Monitor and analyze the system's performance.
4. Narrow down the bottleneck and find its cause.
5. Fix the bottleneck cause by trying only one single change at a time.
6. Go back to step 3 until you are satisfied with the performance of the system.
4.1.1 Gathering information
Mostly likely, the only first-hand information you will have access to will be statements such as "There is a problem with the server." It is crucial to use probing questions to clarify and document the problem. Here is a list of questions you should ask to help you get a better picture of the system.
Can you give me a complete description of the server in question?
Model
Age
Configuration
Peripheral equipment
Operating system version and update level
Can you tell me exactly what the problem is?
What are the symptoms?
Describe any error messages.
Some people will have problems answering this question, but any extra information the customer can give you might enable you to find the problem. For example, the customer might say "It is really slow when I copy large files to the server." This might indicate a network problem or a disk subsystem problem.
Who is experiencing the problem?
Is one person, one particular group of people, or the entire organization experiencing the problem? This helps determine whether the problem exists in one particular part of the network, whether it is application-dependent, and so on. If only one user experiences the problem, then the problem might be with the user's PC (or their imagination).
The perception clients have of the server is usually a key factor. From this point of view, performance problems may not be directly related to the server: the network path between the server and the clients can easily be the cause of the problem. This path includes network devices as well as services provided by other servers, such as domain controllers.
Can the problem be reproduced?
All reproducible problems can be solved. If you have sufficient knowledge of the system, you should be able to narrow the problem to its root and decide which actions should be taken.
Tip: You should document each step, especially the changes you make and their effect on performance.
Chapter 4. Analyzing performance bottlenecks 71
The fact that the problem can be reproduced enables you to see and understand it better.
Document the sequence of actions that are necessary to reproduce the problem:
What are the steps to reproduce the problem?
Knowing the steps may help you reproduce the same problem on a different machine under the same conditions. If this works, it gives you the opportunity to use a machine in a test environment and removes the chance of crashing the production server.
Is it an intermittent problem?
If the problem is intermittent, the first thing to do is to gather information and find a path to move the problem in the reproducible category. The goal here is to have a scenario to make the problem happen on command.
Does it occur at certain times of the day or certain days of the week?
This might help you determine what is causing the problem. It may occur when everyone arrives for work or returns from lunch. Look for ways to change the timing (that is, make it happen less or more often); if there are ways to do so, the problem becomes a reproducible one.
Is it unusual?
If the problem falls into the non-reproducible category, you may conclude that it is the result of extraordinary conditions and classify it as fixed. In real life, there is a high probability that it will happen again.
A good procedure to troubleshoot a hard-to-reproduce problem is to perform general maintenance on the server: reboot, or bring the machine up to date on drivers and patches.
When did the problem start? Was it gradual or did it occur very quickly?
If the performance issue appeared gradually, then it is likely to be a sizing issue; if it appeared overnight, then the problem could be caused by a change made to the server or peripherals.
Have any changes been made to the server (minor or major) or are there any changes in the way clients are using the server?
Did the customer alter something on the server or peripherals to cause the problem? Is there a log of all network changes available?
Demands could change based on business changes, which could affect demands on a servers and network systems.
Are there any other servers or hardware components involved?
Are any logs available?
What is the priority of the problem? When does it have to be fixed?
Does it have to be fixed in the next few minutes, or in days? You may have some time to fix it; or it may already be time to operate in panic mode.
How massive is the problem?
What is the related cost of that problem?
72 Tuning Red Hat Enterprise Linux on IBM Eserver xSeries Servers
4.1.2 Analyzing the server's performance
At this point, you should begin monitoring the server. The simplest way is to run monitoring tools from the server that is being analyzed. (See Chapter 2, "Monitoring tools" on page 15, for information.)
A performance log of the server should be created during its peak time of operation (for example, 9:00 a.m. to 5:00 p.m.); it will depend on what services are being provided and on who is using these services. When creating the log, if available, the following objects should be included:
Processor
System
Server work queues
Memory
Page file
Physical disk
Redirector
Network interface
Before you begin, remember that a methodical approach to performance tuning is important.
Our recommended process, which you can use for your xSeries server performance tuning process, is as follows:
1. Understand the factors affecting server performance. This Redpaper and the redbook
Tuning IBM Eserver xSeries Servers for Performance, SG24-5287 can help.
2. Measure the current performance to create a performance baseline to compare with your future measurements and to identify system bottlenecks.
3. Use the monitoring tools to identify a performance bottleneck. By following the instructions in the next sections, you should be able to narrow down the bottleneck to the subsystem level.
4. Work with the component that is causing the bottleneck by performing some actions to improve server performance in response to demands.
5. Measure the new performance. This helps you compare performance before and after the tuning steps. When attempting to fix a performance problem, remember the following:
Take measurements before you upgrade or modify anything so that you can tell whether the change had any effect. (That is, take baseline measurements.)
Examine the options that involve reconfiguring existing hardware, not just those that involve adding new hardware.
Important: Before taking any troubleshooting actions, back up all data and the configuration information to prevent a partial or complete loss.
Note: It is important to understand that the greatest gains are obtained by upgrading a component that has a bottleneck when the other components in the server have ample "power" left to sustain an elevated level of performance.
Chapter 4. Analyzing performance bottlenecks 73
4.2 CPU bottlenecks
For servers whose primary role is that of an application or database server, the CPU is a critical resource and can often be a source of performance bottlenecks. It is important to note that high CPU utilization does not always mean that a CPU is busy doing work; it may, in fact, be waiting on another subsystem. When performing proper analysis, it is very important that you look at the system as a whole and at all subsystems because there may be a cascade effect within the subsystems.
4.2.1 Finding CPU bottlenecks
Determining bottlenecks with the CPU can be accomplished in several ways. As discussed in
Chapter 2, "Monitoring tools" on page 15, Linux has a variety of tools to help determine this; the question is: which tools to use?
One such tool is uptime. By analyzing the output from uptime, we can get a rough idea of what has been happening in the system for the past 15 minutes. For a more detailed explanation of this tool, see 2.2, "uptime" on page 16.
Example 4-1 uptime output from a CPU strapped system
18:03:16 up 1 day, 2:46, 6 users, load average: 182.53, 92.02, 37.95
Using KDE System Guard and the CPU sensors lets you view the current CPU workload.
Using top, you can see both CPU utilization and what processes are the biggest contributors to the problem (Example 2-3 on page 18). If you have set up sar, you are collecting a lot of information, some of which is CPU utilization, over a period of time. Analyzing this information can be difficult, so use isag, which can use sar output to plot a graph. Otherwise, you may wish to parse the information through a script and use a spreadsheet to plot it to see any trends in CPU utilization. You can also use sar from the command line by issuing sar -u or
sar -U processornumber. To gain a broader perspective of the system and current utilization of more than just the CPU subsystem, a good tool is vmstat (2.6, "vmstat" on page 21).
4.2.2 SMP
SMP-based systems can present their own set of interesting problems that can be difficult to detect. In an SMP environment, there is the concept of CPU affinity, which implies that you bind a process to a CPU.
The main reason this is useful is CPU cache optimization, which is achieved by keeping the same process on one CPU rather than moving between processors. When a process moves between CPUs, the cache of the new CPU must be flushed. Therefore, a process that moves between processors causes many cache flushes to occur, which means that an individual process will take longer to finish. This scenario is very hard to detect because, when
Note: There is a common misconception that the CPU is the most important part of the server. This is not always the case, and servers are often overconfigured with CPU and underconfigured with disks, memory, and network subsystems. Only specific applications that are truly CPU-intensive can take advantage of today's high-end processors.
Tip: Be careful not to add to CPU problems by running too many tools at one time. You may find that using a lot of different monitoring tools at one time may be contributing to the high CPU load.
74 Tuning Red Hat Enterprise Linux on IBM Eserver xSeries Servers monitoring it, the CPU load will appear to be very balanced and not necessarily peaking on any CPU. Affinity is also useful in NUMA-based systems such as the xSeries 445 and xSeries 455, where it is important to keep memory, cache, and CPU access local to one another.
4.2.3 Performance tuning options
The first step is to ensure that the system performance problem is being caused by the CPU and not one of the other subsystems. If the processor is the server bottleneck, then a number of steps can be taken to improve performance. These include:
Ensure that no unnecessary programs are running in the background by using ps -ef. If you find such programs, stop them and use cron to schedule them to run at off-peak hours.
Identify non-critical, CPU-intensive processes by using top and modify their priority using
renice.
In an SMP-based machine, try using taskset to bind processes to CPUs to make sure that processes are not hopping between processors, causing cache flushes.
Based on the running application, it may be better to scale up (bigger CPUs) than scale out (more CPUs). This depends on whether your application was designed to effectively take advantage of more processors. For example, a single-threaded application would scale better with a faster CPU and not with more CPUs.
General options include making sure you are using the latest drivers and firmware, as this may affect the load they have on the CPU.
4.3 Memory bottlenecks
On a Linux system, many programs run at the same time; these programs support multiple users and some processes are more used than others. Some of these programs use a portion of memory while the rest are "sleeping." When an application accesses cache, the performance increases because an in-memory access retrieves data, thereby eliminating the need to access slower disks.
The OS uses an algorithm to control which programs will use physical memory and which are paged out. This is transparent to user programs. Page space is a file created by the OS on a disk partition to store user programs that are not currently in use. Typically, page sizes are 4 KB or 8 KB. In Linux, the page size is defined by using the variable EXEC_PAGESIZE in the include/asm-<architecture>/param.h kernel header file. The process used to page a process out to disk is called pageout.
4.3.1 Finding memory bottlenecks
Start your analysis by listing the applications that are running on the server. Determine how much physical memory and swap each application needs to run. Figure 4-1 on page 75 shows KDE System Guard monitoring memory usage.
Chapter 4. Analyzing performance bottlenecks 75
Figure 4-1 KDE System Guard memory monitoring
The indicators in Table 4-1 can also help you define a problem with memory.
Table 4-1 Indicator for memory analysis
Paging and swapping indicators
In Linux, as with all UNIX-based operating systems, there are differences between paging and swapping. Paging moves individual pages to swap space on the disk; swapping is a bigger operation that moves the entire address space of a process to swap space in one operation.
Swapping can have one of two causes:
A process enters sleep mode. This usually happens because the process depends on interactive action, as editors, shells, and data entry applications spend most of their time waiting for user input. During this time, they are inactive.
Memory indicator Analysis
Memory available This indicates how much physical memory is available for use. If, after you start your application, this value has decreased significantly, you may have a memory leak. Check the application that is causing it and make the necessary adjustments. Use free -l -t -o for additional information.
Page faults There are two types of page faults: soft page faults, when the page is found in memory, and hard page faults, when the page is not found in memory and must be fetched from disk. Accessing the disk will slow your application considerably. The sar -B command can provide useful information for analyzing page faults, specifically columns pgpgin/s and pgpgout/s.
File system cache This is the common memory space used by the file system cache. Use the free -l -t -o
command for additional information.
Private memory for process
This represents the memory used by each process running on the server. You can use the pmap
command to see how much memory is allocated to a specific process.
76 Tuning Red Hat Enterprise Linux on IBM Eserver xSeries Servers A process behaves poorly. Paging can be a serious performance problem when the amount of free memory pages falls below the minimum amount specified, because the paging mechanism is not able to handle the requests for physical memory pages and the swap mechanism is called to free more pages. This significantly increases I/O to disk and will quickly degrade a server's performance.
If your server is always paging to disk (a high page-out rate), consider adding more memory.
However, for systems with a low page-out rate, it may not affect performance.
4.3.2 Performance tuning options
It you believe there is a memory bottleneck, consider performing one or more of these actions:
Tune the swap space using bigpages, hugetlb, shared memory.
Increase or decrease the size of pages.
Improve the handling of active and inactive memory.
Adjust the page-out rate.
Limit the resources used for each user on the server.
Stop the services that are not needed, as discussed in 3.3, "Daemons" on page 38.
Add memory.
4.4 Disk bottlenecks
The disk subsystem is often the most important aspect of server performance and is usually the most common bottleneck. However, problems can be hidden by other factors, such as lack of memory. Applications are considered to be I/O-bound when CPU cycles are wasted simply waiting for I/O tasks to finish.
The most common disk bottleneck is having too few disks. Most disk configurations are based on capacity requirements, not performance. The least expensive solution is to purchase the smallest number of the largest-capacity disks possible. However, this places more user data on each disk, causing greater I/O rates to the physical disk and allowing disk bottlenecks to occur.
The second most common problem is having too many logical disks on the same array. This increases seek time and greatly lowers performance. The disk subsystem is discussed in 3.12, "Tuning the file system" on page 52. A recommendation is to apply the diskstats-2.4.patch to fix problems with disk statistics counters, which can occasionally report negative values.
4.4.1 Finding disk bottlenecks
A server exhibiting the following symptoms may be suffering from a disk bottleneck (or a hidden memory problem):
Slow disks will result in:
Memory buffers filling with write data (or waiting for read data), which will delay all
requests because free memory buffers are unavailable for write requests (or the response is waiting for read data in the disk queue)
Insufficient memory, as in the case of not enough memory buffers for network requests, will cause synchronous disk I/O
Chapter 4. Analyzing performance bottlenecks 77
Disk utilization, controller utilization, or both will typically be very high.
Most LAN transfers will happen only after disk I/O has completed, causing very long response times and low network utilization.
Disk I/O can take a relatively long time and disk queues will become full, so the CPUs will be idle or have low utilization because they wait long periods of time before processing the next request.
The disk subsystem is perhaps the most challenging subsystem to properly configure.
Besides looking at raw disk interface speed and disk capacity, it is key to also understand the workload: Is disk access random or sequential? Is there large I/O or small I/O? Answering these questions provides the necessary information to make sure the disk subsystem is adequately tuned.
Disk manufacturers tend to showcase the upper limits of their drive technology's throughput.
However, taking the time to understand the throughput of your workload will help you understand what true expectations to have of your underlying disk subsystem.
Table 4-2 Exercise showing true throughput for 8 KB I/Os for different drive speeds
Random read/write workloads usually require several disks to scale. The bus bandwidths of SCSI or Fibre Channel are of lesser concern. Larger databases with random access workload will benefit from having more disks. Larger SMP servers will scale better with more disks. Given the I/O profile of 70% reads and 30% writes of the average commercial workload, a RAID-10 implementation will perform 50% to 60% better than a RAID-5.
Sequential workloads tend to stress the bus bandwidth of disk subsystems. Pay special attention to the number of SCSI buses and Fibre Channel controllers when maximum throughput is desired. Given the same number of drives in an array, RAID-10, RAID-0, and RAID-5 all have similar streaming read and write throughput.
There are two ways to approach disk bottleneck analysis: real-time monitoring and tracing.
Real-time monitoring must be done while the problem is occurring. This may not be practical in cases where system workload is dynamic and the problem is not repeatable.
However, if the problem is repeatable, this method is flexible because of the ability to add objects and counters as the problem becomes well understood.
Tracing is the collecting of performance data over time to diagnose a problem. This is a good way to perform remote performance analysis. Some of the drawbacks include the potential for having to analyze large files when performance problems are not repeatable, and the potential for not having all key objects and parameters in the trace and having to wait for the next time the problem occurs for the additional data.
Disk speed Latency Seek time
Total random access timea
a. Assuming that the handling of the command + data transfer < 1 ms, total random access time = latency + seek time + 1 ms.
I/Os per second per diskb
b. Calculated as 1/total random access time.
Throughput given 8 KB I/O
15 000 RPM 2.0 ms 3.8 ms 6.8 ms 147 1.15 MBps
10 000 RPM 3.0 ms 4.9 ms 8.9 ms 112 900 KBps
7 200 RPM 4.2 ms 9 ms 13.2 ms 75 600 KBps
78 Tuning Red Hat Enterprise Linux on IBM Eserver xSeries Servers
vmstat command
One way to track disk usage on a Linux system is by using the vmstat tool. The columns of interest in vmstat with respect to I/O are the bi and bo fields. These fields monitor the movement of blocks in and out of the disk subsystem. Having a baseline is key to being able to identify any changes over time.
Example 4-2 vmstat output
[root@x232 root]# vmstat 2
r b swpd free buff cache si so bi bo in cs us sy id wa
2 1 0 9004 47196 1141672 0 0 0 950 149 74 87 13 0 0
0 2 0 9672 47224 1140924 0 0 12 42392 189 65 88 10 0 1
0 2 0 9276 47224 1141308 0 0 448 0 144 28 0 0 0 100
0 2 0 9160 47224 1141424 0 0 448 1764 149 66 0 1 0 99
0 2 0 9272 47224 1141280 0 0 448 60 155 46 0 1 0 99
0 2 0 9180 47228 1141360 0 0 6208 10730 425 413 0 3 0 97
1 0 0 9200 47228 1141340 0 0 11200 6 631 737 0 6 0 94
1 0 0 9756 47228 1140784 0 0 12224 3632 684 763 0 11 0 89
0 2 0 9448 47228 1141092 0 0 5824 25328 403 373 0 3 0 97
0 2 0 9740 47228 1140832 0 0 640 0 159 31 0 0 0 100
iostat command
Performance problems can be encountered when too many files are opened, being read and written to, then closed repeatedly. This could become apparent as seek times (the time it takes to move to the exact track where the data is stored) start to increase. Using the iostat tool, you can monitor the I/O device loading in real time. Different options enable you to drill down even farther to gather the necessary data.
Example 4-3 shows a potential I/O bottleneck on the device /dev/sdb1. This output shows average wait times (await) of about 2.7 seconds and service times (svctm) of 270 ms.
Example 4-3 Sample of an I/O bottleneck as shown with iostat 2 -x /dev/sdb1
[root@x232 root]# iostat 2 -x /dev/sdb1
avg-cpu: %user %nice %sys %idle
11.50 0.00 2.00 86.50
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/sdb1 441.00 3030.00 7.00 30.50 3584.00 24480.00 1792.00 12240.00 748.37
101.70 2717.33 266.67 100.00
avg-cpu: %user %nice %sys %idle
10.50 0.00 1.00 88.50
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/sdb1 441.00 3030.00 7.00 30.00 3584.00 24480.00 1792.00 12240.00 758.49
101.65 2739.19 270.27 100.00
avg-cpu: %user %nice %sys %idle
10.95 0.00 1.00 88.06
Chapter 4. Analyzing performance bottlenecks 79
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/sdb1 438.81 3165.67 6.97 30.35 3566.17 25576.12 1783.08 12788.06 781.01
101.69 2728.00 268.00 100.00
The iostat -x (for extended statistics) command provides low-level detail of the disk subsystem. Some things to point out:
%util Percentage of CPU consumed by I/O requests
svctm Average time required to complete a request, in milliseconds
await Average amount of time an I/O waited to be served, in milliseconds
avgqu-sz Average queue length
avgrq-sz Average size of request
rrqm/s Number of read requests merged per second that were issued to the device
wrqms Number of write requests merged per second that were issued to the device
For a more detailed explanation of the fields, see the man page for iostat(1).
Changes made to the elevator algorithm as described in "Tune the elevator algorithm in kernel 2.4" on page 55 will be seen in avgrq-sz (average size of request) and avgqu-sz (average queue length). As the latencies are lowered by manipulating the elevator settings, avgrq-sz will decrease. You can also monitor the rrqm/s and wrqm/s to see the effect on the number of merged reads and writes that the disk can manage.
4.4.2 Performance tuning options
After verifying that the disk subsystem is a system bottleneck, several solutions are possible.
These solutions include the following:
If the workload is of a sequential nature and it is stressing the controller bandwidth, the solution is to add a faster disk controller. However, if the workload is more random in nature, then the bottleneck is likely to involve the disk drives, and adding more drives will improve performance.
Add more disk drives in a RAID environment. This spreads the data across multiple physical disks and improves performance for both reads and writes. This will increase the number of I/Os per second. Also, use hardware RAID instead of the software implementation provided by Linux. If hardware RAID is being used, the RAID level is hidden from the OS.
Offload processing to another system in the network (users, applications, or services).
Add more RAM. Adding memory increases system memory disk cache, which in effect improves disk response times.
4.5 Network bottlenecks
A performance problem in the network subsystem can be the cause of many problems, such as a kernel panic. To analyze these anomalies to detect network bottlenecks, each Linux distribution includes traffic analyzers.
4.5.1 Finding network bottlenecks
We recommend KDE System Guard because of its graphical interface and ease of use. The tool, which is available on the distribution CDs, is discussed in detail in 2.10, "KDE System
Guard" on page 24. Figure 4-2 on page 80 shows it in action.
80 Tuning Red Hat Enterprise Linux on IBM Eserver xSeries Servers
Figure 4-2 KDE System Guard network monitoring
It is important to remember that there are many possible reasons for these performance problems and that sometimes problems occur simultaneously, making it even more difficult to pinpoint the origin. The indicators in Table 4-3 can help you determine the problem with your network.
Table 4-3 Indicators for network analysis
Network indicator Analysis
Packets received Packets sent
Shows the number of packets that are coming in and going out of the
specified network interface. Check both internal and external interfaces.
Collision packets Collisions occur when there are many systems on the same domain. The use of a hub may be the cause of many collisions.
Dropped packets Packets may be dropped for a variety of reasons, but the result may affect performance. For example, if the server network interface is configured to run at 100 Mbps full duplex, but the network switch is configured to run at 10 Mbps, the router may have an ACL filter that drops these packets. For example:
iptables -t filter -A FORWARD -p all -i eth2 -o eth1 -s 172.18.0.0/24
-j DROP
Errors Errors occur if the communications lines (for instance, the phone line) are of poor quality. In these situations, corrupted packets must be resent, thereby
decreasing network throughput.
Faulty adapters Network slowdowns often result from faulty network adapters. When this kind of hardware fails, it may begin to broadcast junk packets on the network.
Chapter 4. Analyzing performance bottlenecks 81
4.5.2 Performance tuning options
These steps illustrate what you should do to solve problems related to network bottlenecks:
Ensure that the network card configuration matches router and switch configurations (for example, frame size).
Modify how your subnets are organized.
Use faster network cards.
Tune the appropriate IPV4 TCP kernel parameters. (See Chapter 3, "Tuning the operating system" on page 35.) Some security-related parameters can also improve performance, as described in that chapter.
If possible, change network cards and recheck performance.
Add network cards and bind them together to form an adapter team, if possible.
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