Performance indicators of hardware

Time unit Conversion Relationship

 
Conversion Relationship1S =10^3MS =10^6US =10^9NS = 10 ^ 12 μs second millisecond microsecond nanosecond second
 

 

Cycle

To sum up, the relationship between them is,Instruction cycleConsists of several machine cycles,Bus CycleGenerally, there are fourClock cycle. 

Machine cycle and bus cycle ...... The machine cycle refers to the time when a basic operation is completed. This basic operation may sometimes contain bus reads and writes, and thus the bus cycle, but sometimes it may not be related to bus reads and writes, there is no explicit relationship between each other.

Command cycle: a key indicator of the CPU. It refers to the time when a command is taken and executed. Generally, a single command execution cycle and dual-command execution cycle are measured in the unit of machine cycle. Most of the current processor's commands (arm and DSP) use a single instruction execution cycle.

Machine cycle: the time unit for completing a basic operation, such as the period of the specified period and the period of the number.

CPU cycle:Also knownMachine cycleVarious operations in the machine can be attributed to the operations inside the CPU and the operations on the master memory. Because the internal operation speed of the CPU is fast, it takes a long time for the CPU to access the memory, therefore, readScriptThe baseline time is the CPU cycle (machine cycle ). OneInstruction cycleIt is usually composed of several CPU cycles.

Bus Cycle: the time used for an access memory or I/O port operation.

Clock cycle: the CPU Crystal OscillatorOperating frequency. (Fantaxy: time required for a crystal oscillator ),For example, for a 12 m crystal oscillator, the time period is 1/12 μs,If the 1 MHzClock frequency, The clock cycle is 1 μs; if the 4 MHz clock frequency is adopted, the clock cycle is 250ns.

Example: What is the period of the 22.1184mhz crystal oscillator?

Taking 51 as an example, the crystal oscillator is 22.1184 m, and the clock cycle (crystal oscillator cycle) is (1/22. 1184) μs. One machine cycle contains 12 clock cycles, and one machine cycle is 0.5425 μs. A machine cycle generally refers to the time it takes for an instruction, and some are instructions for two machine cycles. The djnz command is a dual-cycle instruction.

Cycle: the time when the task is completed.

Clock cycle: the English clock cycle; clock period; clock is used for timing, is a basic unit; in computer, the CPU crystal time is the most basic unit, therefore, the clock cycle is very basic, and other cycles are used for reference!

Context switching: The real overhead of context switching varies with the platform, but there is an empirical rule: In most General processors, context switching takes 5000 to 10000 clock cycles, or several microseconds.

 

Memory frequency

Memory clock speedSimilar to CPU clock speed, it is used to indicate the memory speed, which represents the maximum operation that the memory can achieveFrequency. The memory clock speed is measured in MHz (MHz. The higher the memory frequency, the faster the memory can reach. The memory clock speed determines the maximum operating frequency of the memory. Currently, the most popular memory frequencies are 333mhz and 400mhz.DDR memory, 667 MHz, MHz, and MHzDDR2 memory, 1066 MHz, 1333 MHz, MHzDdr3 memory.

MemoryIt does not have a crystal oscillator, so the clock signal during memory operation isMotherboard chipsetOr directly provided by the clock generator of the motherboard, that is, the memory cannot determine its ownOperating frequencyThe actual working frequency is determined by the motherboard.

 

Disk Disk performance indicators

Iops (input/output operations per second)

That is, the number of read/write (I/O) operations per second. It is often used in databases and other occasions to measure the performance of random access.StorageThe performance of iops is different from that of the host. iops refers to the number of accesses allowed by the host per second. Only one Io of the host can be completed after multiple accesses to the storage. For example, the host writes the smallestData BlockYou must also go through three steps, namely, sending a write request, writing data, and receiving a write confirmation.

Iops depends mainly on the arrayAlgorithm, Cache hit rate, and number of disks.

If an array has 120 blocks of 15 K rpmOptical fiber hard drive, The maximum iops it can support is 120*150 = 18000. This is the theoretical value of the hardware limitation. If it exceeds this value, the hard disk response may become very slow and cannot provide services normally.

 

Data throughput(Throughput)

The number of data records that can be successfully transferred per unit time. For applications with a large number of sequential reads and writes, such as VOD (video on demand), more attention is given to throughput indicators.

 

Disk Service Time

A traditional disk is essentially a type of mechanical device, such as FC, SAS, and SATA disks, with a speed of 5400/7200/10 k/15 K rpm. The key factor affecting the disk is the disk service time, that is, the time it takes for the disk to complete an I/O request. It consists of three parts: Seeking time, rotation delay, and data transmission time.

Seek timeTseek refers to the time required to move the read/write head to the correct track. The shorter the tracing time, the faster I/O operations are. Currently, the average tracing time of a disk is generally 3-15 ms.
Rotation DelayTrotation refers to the time required for disk rotation to move the sector where the request data is located to the bottom of the read/write head. The rotation delay depends on the disk speed, which is usually expressed by 1/2 of the time required for disk rotation for one week. For example, the average rotation latency of a 7200 RPM disk is about 60*1000/7200/2 = 4.17 ms, while the average rotation latency of a 15000 rpm disk is about 2 ms.
Data transmission timeTtransfer refers to the time required to complete the data requested for transmission. It depends on the data transmission rate, and its value is equal to the data size divided by the data transmission rate. At present, IDE/ATA can reach 133 Mb/s, and sata ii can reach the interface data transmission rate of 300 MB/S. The data transmission time is usually far earlier than the first two parts.

 

Theoretical calculation of maximum iops

Iops = 1000 MS/(tseek + troatation), ignore data transfer time. Assuming that the average physical tracing time of a disk is 3 ms, and the disk speed is, 10 K, and 15 K rpm, the theoretical maximum iops of the disk is,
Iops = 1000/(3 + 60000/7200/2) = 140
Iops = 1000/(3 + 60000/10000/2) = 167
Iops = 1000/(3 + 60000/15000/2) = 200

SSD is an electronic device that avoids the time spent on track finding and rotating traditional disks and greatly reduces the overhead of storage unit addressing. Therefore, the iops can be very high, it can reach tens of thousands or even hundreds of thousands.

 

Iops metrics

In general, iops can be subdivided into the following indicators:
Toatal iops, disk iops under mixed read/write and sequential random I/O load, which is the most consistent with the actual I/O situation, most applications focus on this indicator.
Random read iops: iops in the case of 100% random read loads.
Random write iops: iops under 100% random write loads.
Sequential read iops: iops in the case of 100% sequential load reads.
Sequential write iops: The iops under 100% sequential write loads.