NetApp – Calculate Maximum Number of inodes per Volume

NetApp volumes allow for inodes to be dynamically allocated/increased on volumes which are provisioned on an array.  This begs the question, what is the maximum inode count supported by a volume and how is the maximum number calculated?

inodes = files

“The maximum number of inodes is limited to one inode per one block in the filesystem. (which is 1 inode per every 4KB).  It is generally recommended to NOT go that low.”

TB (Volume) GB MB KB
1.2 1,228.8 1,258,291 1,288,490,189

1,288,490,189KB / 4KB Blocks = 322,122,547 supported files / inodes per 1.2TB volume.

Credit where credit is due… https://communities.netapp.com/thread/2176

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Calculate Corrected Usable Hard Drive Size

Over the years there has been a  constant battle between storage engineers and  hard drive companies on whether or not advertised hard drive sizes  are misleading to consumers.  The advertised numbers are in fact misleading because they do not state the actual usable size of the hard drive being sold.  In order to find the corrected size of a hard drive you must account for the actual allocation size used by the hard drive manufacturer.

When a 300GB hard drive is sold the manufacturer is stating that the hard drive has a capacity which meets the following equation:

300GB = 300,000,000,000 bytes

With a 300GB hard drive the manufacturer would like you to think the above equation is true.  It in fact is not.  The reason why this value is false is because 1,000 bytes should be represented as 1,024 bytes.  So in order to find the true size of a  300GB hard drive you  must divide by 1024 three times to calculate its true size in Gigabytes.

300,000,000,000 bytes ÷ 1024 = 292,968,750 Kilobytes

292,968,750 kilobytes ÷ 1024 = 286,102.3 Megabytes

286,102.3 Megbytes ÷ 1024 = 279.3968 Gigabytes

The corrected size of an advertised 300GB drive is actually 279GB and change.

If hard drive manufacturers gave you the advertised capacity of 300GB the drive would have to be 322,122,547,200 Bytes in size. This can be calculated by multiplying 300GB x 1024 three times to find the required bytes.


2Gb Fibre Channel – Determine Maximum Throughput Speed

With storage area networks (SAN) performance issues sometimes arrise.  Knowing the maximum throughput speed of your network is key to remove it as a possible bottlekneck.  Bellow I will outline the required calculations for determining the maximum throughput speed of 2 Gigabit Fibre Channel.

Calculation Rules
Network Speeds = Always in Bits
Storage/Disk Values = Always in Bytes
Files = Always in Bytes

1 nybble = 4 bits
1 word = 16 bits
8 bits = 1 Byte
1024 Bytes = 1 Kilobyte
1024 KB = 1 Megabyte
1024 MB = 1 Gigabyte

Calculating Maximum Throughput for 2 Gigabit Fibre Channel in MegaBytes
2 Gigabits = 2/8bits = 0.25 GigaBytes
0.25 GigaBytes x 2 (2 HBAs) = 0.5 GigaBytes
0.5 x 1024 = 512 MegaBytes/sec throughput or 256 MB/sec per HBA.

Depending on the monitoring software used performance numbers may not be in MB format. Bellow I’ve broken down the values for each common throughput value.

256 MegaBytes/sec =
262,144 KiloBytes/sec
268,435,456 Bytes/sec
2,147,483,648.00 Bits/sec
33,554,432 Bits/sec

Since FC speeds have tripled since I wrote this post an update is in order. Bellow I’ve outlined speeds for 4Gb and 8Gb FC.
4Gb = 512 MB/s
8Gb = 1024 MB/s

Now that we have determined the maximum throughput of each HBA we need to rule out any bottlenecks found in the server itself or in the SAN fabric.  HBA throughput speed is directly impacted by the bus speed of each HBA as well as whether the switch ports utilized by either the storage array FA’s or the HBA itself are dedicated bandwidth or over subscribed ports.

…. brain dump in progress