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From: | Leonid Bloch |
Subject: | Re: [Qemu-devel] [PATCH v7 1/9] qcow2: Options' documentation fixes |
Date: | Sat, 11 Aug 2018 21:01:06 +0300 |
User-agent: | Mozilla/5.0 (X11; Linux x86_64; rv:60.0) Gecko/20100101 Thunderbird/60.0 |
On 8/10/18 2:50 PM, Alberto Garcia wrote:
On Fri 10 Aug 2018 08:26:39 AM CEST, Leonid Bloch wrote:Signed-off-by: Leonid Bloch <address@hidden> --- docs/qcow2-cache.txt | 16 +++++++++++----- qemu-options.hx | 9 ++++++--- 2 files changed, 17 insertions(+), 8 deletions(-) diff --git a/docs/qcow2-cache.txt b/docs/qcow2-cache.txt index 8a09a5cc5f..0f157d859a 100644 --- a/docs/qcow2-cache.txt +++ b/docs/qcow2-cache.txt @@ -97,12 +97,15 @@ need: l2_cache_size = disk_size_GB * 131072 refcount_cache_size = disk_size_GB * 32768-QEMU has a default L2 cache of 1MB (1048576 bytes) and a refcount-cache of 256KB (262144 bytes), so using the formulas we've just seen -we have +With the default cluster size, to cover each 8 GB of the virtual image +size, 1MB of L2 cache is needed:- 1048576 / 131072 = 8 GB of virtual disk covered by that cache- 262144 / 32768 = 8 GB + 65536 / 8 = 8192 = 8 GB / 1 MBWhere does this 65536 / 8 come from? The line that you changed follows directly from the formula immediately before that paragraph: l2_cache_size = disk_size_GB * 131072 that is l2_cache_size / 131072 = disk_size_GB 1048576 / 131072 = 8 GB Berto
How about the following (all the relevant section reproduced below, for a continuous readability):
"""""""""" Choosing the right cache sizes ------------------------------ In order to choose the cache sizes we need to know how they relate to the amount of the allocated space. The amount of virtual disk that can be mapped by the L2 and refcount caches (in bytes) is: disk_size = l2_cache_size * cluster_size / 8 disk_size = refcount_cache_size * cluster_size * 8 / refcount_bits With the default values for cluster_size (64KB) and refcount_bits (16), this becomes: disk_size = l2_cache_size * 8192 disk_size = refcount_cache_size * 32768 So in order to cover n GB of disk space with the default values we need: l2_cache_size = disk_size_GB * 131072 refcount_cache_size = disk_size_GB * 32768 For example, 1MB of L2 cache is needed to cover each 8 GB of the virtual image size (given that the default cluster size is used): 8 * 131072 = 1 MB A default refcount cache is 4 times the cluster size, which defaults to 256 KB (262144 bytes). This is sufficient for 8 GB of image size: 262144 / 32768 = 8 GB """"""""""
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