SQL Server 2012 : Storage Systems (part 5) - Measuring Performance - Storage Performance Counters, Disk Drive Performance

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The single most important performance metric is latency. Latency is a measure of system health and the availability of system resources. Latency is governed by queuing theory, a mathematical study of lines, or queues. An important contribution to queuing theory, now known as Little’s Law, was introduced in a proof submitted by John D. C. Little ( in 1961.

Put simply, Little’s Law states that given a steady-state system, as capacity reaches maximum performance, response time approaches infinity. To understand the power of Little’s Law, consider the typical grocery store. If the store only opens one cash register and ten people are waiting in line, then you are going to wait longer to pay than if the store opened five or ten cashiers.

Storage systems are directly analogous to a grocery store checkout line. Each component has a certain performance maximum. Driving the system toward maximum performance will increase latency. We have found that most users and application owners directly correlate latency with failure. For example, it won’t matter that a payroll system is online if it can’t process transactions fast enough to get everyone’s paycheck out on time.

You test I/O performance using several tools that are described later. You test storage using a logarithmic scale, starting with one I/O, moving to two, then four, then eight, and finally peaking at 256 I/Os that are all sent to storage in parallel (see Figure 1). As it turns out, this test perfectly demonstrates Little’s Law by defining how storage operates.

As you can see in Figure 11 the storage response time remains less than our goal of 10 milliseconds through eight outstanding I/Os. As we increase the workload to 16 outstanding I/Os, the latency increases to 20 milliseconds. We can determine from this test that our configuration is optimal when we issue between 8 and 16 I/Os. This is called the knee of the curve. The system is capable of a lot more work, but the latency is higher than our tolerance.



The goal of less than 10 milliseconds of latency is no accident. SQL Server best practices call for no more than 20 milliseconds of delay. If you implement synchronous data replication with AlwaysOn Availability Groups, you can’t tolerate more than 10 milliseconds of delay; and many applications are even more sensitive!

4.1 Storage Performance Counters

Windows Performance Monitor (perfmon) allows Windows Server users to capture storage performance metrics. For the purposes of storage monitoring, you utilize the LogicalDisk performance monitor object. Both logical and physical disk counters deliver storage performance metrics. The logical disk counters show the performance of a specific partition, while the physical disk counters cover the entire LUN (a Logical Unit Number is a term that describes a storage volume that is hosted by a storage controller). Table 1 shows a list of the available Windows storage performance counters.

TABLE 1: Windows Storage Performance Counters

Average Disk sec/Read
Average Disk sec/Write
Disk Reads/sec
Disk Writes/sec
Disk Read Bytes/sec
Disk Write Bytes/sec

The Average Disk Sec/Read and Write counters measure the time it takes for an input output (I/O) operation to be sent from the server to the storage system and back. This latency measure is the single biggest indicator of I/O system health. Reads and writes are treated separately because most storage systems perform one operation faster than the other. If you are using a storage array with a battery-backed cache, it will often write in just a few milliseconds, whereas a random read will take longer.

The latency counters are measured in milliseconds. A reading of 0.001 is one millisecond, 0.010 is 10 milliseconds, and .100 is 100 milliseconds. SQL Server best practices call for latency that is under 20 milliseconds. This is not a hard-and-fast rule, however, as many applications will not tolerate latency that exceeds several milliseconds.

It is important to understand the underlying hardware configuration, application, and workload. In some cases, such as a SQL Server standard backup, large I/O sizes will drastically increase latency. If you change the backup I/O size to 8MB, the latency will increase, but you can still achieve a lot of work.

If you are implementing a specialized system, such as a SQL Server Fast Track Data Warehouse, you will actively configure the data files so they issue sequential I/O. Be sure to test your specific configuration so you can properly interpret the results.

The Disk Reads and Writes per second counters list how many I/Os are generated each second (Storage administrators often refer to this as IOPS). Disk Read and Write Bytes per second demonstrate the throughput of your storage system. To calculate average I/O sizes, simply divide bytes per second by the number of operations per second.

Knowing the size of the I/O can reflect application behavior. When performing highly random I/O access, SQL Server will write 8K data pages and read 64KB data extents from the data files. Performing sequential operations, such as a table scan, will generate I/O that is dynamically sized from 8K to 512KB. Dynamic I/O sizing, also known as Read-Ahead, is one of the hidden gems of SQL Server. Increasing I/O size decreases the number of I/Os and increases efficiency.

4.2 Disk Drive Performance

A disk drive (see Figure 12) is made up of an external logic board and an internal hard drive assembly. The logic board provides connectivity between the disk and the host. Each drive interface supports one of many available communications protocols. Modern interfaces use a high-speed serial connection. The interfaces that are most commonly used for database applications are SATA (Serial Advanced Technology Attachment), FC (Fibre Channel), and SAS (Serial Attached SCSI).



The hard drive assembly is serviceable only inside a high-technology clean room. Opening the cover on a hard disk drive will void the warranty. The drive platter rotates around a spindle and is powered by a spindle motor. A drive is made up of several platters that are stacked. Each platter is double-sided and coated with a magnetic oxide.

Data is physically read and written by a hard drive head. Each drive platter has a dedicated drive head. An actuator arm houses all the drive heads, and a magnetic actuator moves the arm. You can think of a hard drive as a record player. The platter spins and the head reads and writes data. Unlike a record player, however, the disk drive head can move back and forth. The head actually rides just above the disk surface on a cushion of high-pressure air that is created when the platter spins at high speed.

SATA disk drives provide commodity storage. They offer much larger capacity than FC or SAS drives. At the time of this writing, SATA drives are available with a capacity of three terabytes (3TB). SATA drives spin at lower speeds — 5,400 to 7,200 RPM. They are sold for both the consumer and the enterprise markets. Enterprise drives are designed for more continuous use and higher reliability.

Both FC and SAS drives are considered enterprise-class disk drives. They are available in 7,200, 10,000, and 15,000 RPM models. With the exception of Nearline SAS (NL-SAS) drives, these disk drives are generally lower in capacity than SATA disk drives. SAS 6GB/s drives are displacing Fibre Channel 4GB/s drives in the marketplace.

Modern SAS drives are manufactured in a 2.52 form factor, unlike the traditional 3.52 form factor of Fibre Channel and SATA drives. This smaller drive enables more disk drives to be housed in a given space. NL-SAS drives offer a high-reliability enterprise SATA drive with a SAS interface.

The logic board governs how the disk operates. Each disk contains buffers, and some disk drives contain cache. For the proper operation of SQL Server write-ahead logging, volatile cache must be bypassed for write operations. Most array vendors will guarantee cached data with battery backing. When using disks directly in a JBOD, it is important to ensure that they meet SQL Server reliability requirements.

A disk drive is made of both electronic and mechanical components. When data is read sequentially, the drive can read it as fast as the drive spins. When the data needs to be accessed out of order, the head needs to move to the appropriate track. Head movement, or seeking, is not instantaneous.

A sequential read consists of a head movement followed by the sequential reading or writing of data. When data is not sequentially located, the disk drive executes a series of random operations. Random access patterns are much slower because the head needs to physically move between tracks.

Drive manufacturers provide a measurement called maximum seek time that reflects how long it will take for the drive head to move from the innermost tracks to the outermost tracks. The manufacturer also provides what is known as average seek time, the average time it will take to move the drive head to any location on the disk.

Disk Drive Latency

You can calculate the time it takes to move the head to a location on the disk mathematically. The number of times a disk can rotate in a millisecond limits the amount of data the drive can generate, a limitation called rotational latency. To calculate how many random I/Os a hard disk can perform, the following equation is used:


This equation works by normalizing all calculations to milliseconds. To find IOPS, you start by dividing 60,000 (because there are 60,000 milliseconds in a minute) by the hard disk rotations per minute. Dividing the revolutions per millisecond by 2 accounts for the fact that the head needs to exit the first track and enter the second track at specific points, requiring about two rotations. You add the revolutions result to the seek time and convert the result back to seconds by dividing 1,000 by this sum.

For example, consider a 10,000-RPM disk. This drive will rotate about 6 times per second. You account for the fact that it will take your drive two rotations to move between tracks and then add the seek time. This drive has a read seek time of 4 milliseconds. Dividing 1,000 by 7 results in 143 I/Os per second:


We have tested many drives over the years and this formula has proven reliable. Remember that each individual model of disk drive varies. Having said that, you can calculate IOPS for the most frequently used disk drives:


If you need more IOPS, then simply add more disks. If one disk will perform 150 IOPS, then two will perform 300. When you need 10,000 IOPS, you only need 54 physical disks. If you actually want to keep the data when one of the drives fails, then you need 108 disks. Those 108 disks will provide 10,000 IOPS when the database needs to read, but only 5,000 IOPS for writes. RAID causes overhead for both space and performance. RAID 1+0 is fairly easy to calculate. You will receive N number of reads and N divided by 2 writes. RAID 5 is much trickier to calculate.

For more information on how SQL Server actions translate into IOPS please visit

To generate the parity information, a RAID controller reads relevant data and performs an XOR calculation. Let’s take a small RAID set of four disks as an example. One write operation will generate two writes and two reads. We are assuming that there are two existing data chunks and we are writing the third. We need to write the resultant parity and the new data block.

RAID controllers vary greatly in design, but generally speaking, they utilize their internal cache to assist in the generation of parity information. Typically, Raid 5 enables N number of reads and N divided by 4 writes.

RAID 6 protects against double disk failure and therefore generates double the parity. An 8-disk RAID set consists of two parity chunks and six data chunks. You need to write the new data chunk and two parity chunks, so you know that you have three writes. You need to read the other five data chunks, so you are looking at eight operations to complete the RAID 6 write. Luckily, most RAID controllers can optimize this process into three reads and three writes.

Table 2 provides a guide to calculating common RAID overhead. Please remember that each system is different and your mileage may vary.

TABLE 2: RAID Overhead

0 N N
1+0 N N ÷ 2
5 N N ÷ 4
6 N N ÷ 6
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