Article from: http://www.cnblogs.com/renyb/p/3698664.html
Three factors for calculating disk IOPS: 1, read/write for RAID type with IOPS calculation formula for different RAID types:
RAID type |
Formula |
RAID5, RAID3 |
Drive iops=read ioPS + 4*write ioPS |
RAID6 |
Drive iops=read ioPS + 6*write ioPS |
RAID1, RAID10 |
Drive iops=read ioPS + 2*write ioPS |
2, hard disk type IOPS value different disk type IOPS:
Hard disk Type |
IOPS |
FC 15K RPM |
180 |
FC 10K RPM |
140 |
SAS 15K RPM |
180 |
SAS 10K RPM |
150 |
SATA 10K RPM |
290 |
SATA 7.2K RPM |
80 |
SATA 5.4K RPM |
40 |
Flash Drive |
2500 |
3, the specific business system of reading and writing than the second, case
1 Business requirements: 10TB FC 15K rpm storage space, to meet 6000 IOPS, calculate raid5,raid10 How many pieces of hard disk required respectively.
First you need to know the percentage of read and write operations in I/O. Assuming that the read/write ratio in 6000 ioPS is 2:1 different RAID types Drive hard disk actual IOPS load is as follows:
RAID10: (2/3) *6000+2* (1/3) *6000= 8000 IOPS
RAID5: (2/3) *6000+4* (1/3) *6000=12000 IOPS
Refer to the IOPS values for different hard disk types to calculate how many disks are required:
raid10:8000/180 = 45 block
raid5:12000/180 = 67 Block
2 A RAID5 is made up of 5 500G 10K rpm FC disks, which translates the maximum iops that RAID supports and the IOPS that can be provided to front-end applications.
First 10K rpm FC disk, the IOPS of the single disk is 140, 5 disk maximum IOPS value is 700.
Assuming the read/write ratio is 2:1, the IOPS that can be supplied to the front-end application are:
(2/3) *x+4* (1/3) *x = 2*x = x=350 The ioPS available to front-end applications is 350.