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Updates memory usage. The 24 was wrong, it was always 28.

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Andrea Mazzoleni 2013-09-27 00:44:51 +02:00
parent d387562df4
commit 63b7bce7a5
3 changed files with 9 additions and 9 deletions
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6
snapraid.1
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@ -791,14 +791,14 @@ memory to run SnapRAID.
As a rule of thumb, with 4 GiB or more memory use the default 256,
with 2 GiB use 512, and with 1 GiB use 1024.
.PP
In more details SnapRAID requires about TS*24/BS bytes
In more details SnapRAID requires about TS*28/BS bytes
of RAM memory to run. Where TS is the total size in bytes of
your disk array, and BS is the block size in bytes.
.PP
For example with 6 disk of 2 TiB and a block size of 256 KiB
For example with 4 disk of 3 TiB and a block size of 256 KiB
(1 KiB = 1024 Bytes) you have:
.PP
RAM = (6 * 2 * 2^40) * 24 / (256 * 2^10) = 1.1 GiB
RAM = (4 * 3 * 2^40) * 28 / (256 * 2^10) = 1.4 GiB
.PD 0
.PP
.PD

6
snapraid.d
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@ -623,14 +623,14 @@ Configuration
As a rule of thumb, with 4 GiB or more memory use the default 256,
with 2 GiB use 512, and with 1 GiB use 1024.
In more details SnapRAID requires about TS*24/BS bytes
In more details SnapRAID requires about TS*28/BS bytes
of RAM memory to run. Where TS is the total size in bytes of
your disk array, and BS is the block size in bytes.
For example with 6 disk of 2 TiB and a block size of 256 KiB
For example with 4 disk of 3 TiB and a block size of 256 KiB
(1 KiB = 1024 Bytes) you have:
:RAM = (6 * 2 * 2^40) * 24 / (256 * 2^10) = 1.1 GiB
:RAM = (4 * 3 * 2^40) * 28 / (256 * 2^10) = 1.4 GiB
You could instead decrease this value if you have a lot of
small files in the disk array. For each file, even if of few

6
snapraid.txt
View file

@ -683,14 +683,14 @@ memory to run SnapRAID.
As a rule of thumb, with 4 GiB or more memory use the default 256,
with 2 GiB use 512, and with 1 GiB use 1024.
In more details SnapRAID requires about TS*24/BS bytes
In more details SnapRAID requires about TS*28/BS bytes
of RAM memory to run. Where TS is the total size in bytes of
your disk array, and BS is the block size in bytes.
For example with 6 disk of 2 TiB and a block size of 256 KiB
For example with 4 disk of 3 TiB and a block size of 256 KiB
(1 KiB = 1024 Bytes) you have:
RAM = (6 * 2 * 2^40) * 24 / (256 * 2^10) = 1.1 GiB
RAM = (4 * 3 * 2^40) * 28 / (256 * 2^10) = 1.4 GiB
You could instead decrease this value if you have a lot of
small files in the disk array. For each file, even if of few