diff options
| author | Leonid Bloch <lbloch@janustech.com> | 2018-09-26 19:04:39 +0300 |
|---|---|---|
| committer | Kevin Wolf <kwolf@redhat.com> | 2018-10-01 12:51:12 +0200 |
| commit | 40fb215d483ce510e211b843352288894eb13285 (patch) | |
| tree | 7aa1b0a6bff65781222b16c4773f28fa3c70eb60 /docs | |
| parent | 543770bd2e8fbaac286a947dee91675adb3215c7 (diff) | |
qcow2: Options' documentation fixes
Signed-off-by: Leonid Bloch <lbloch@janustech.com>
Reviewed-by: Alberto Garcia <berto@igalia.com>
Reviewed-by: Kevin Wolf <kwolf@redhat.com>
Signed-off-by: Kevin Wolf <kwolf@redhat.com>
Diffstat (limited to 'docs')
| -rw-r--r-- | docs/qcow2-cache.txt | 21 |
1 files changed, 14 insertions, 7 deletions
diff --git a/docs/qcow2-cache.txt b/docs/qcow2-cache.txt index 8a09a5cc5f..7e28b41bd3 100644 --- a/docs/qcow2-cache.txt +++ b/docs/qcow2-cache.txt @@ -79,14 +79,14 @@ Choosing the right cache sizes In order to choose the cache sizes we need to know how they relate to the amount of allocated space. -The amount of virtual disk that can be mapped by the L2 and refcount +The part of the 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), that is +(16), this becomes: disk_size = l2_cache_size * 8192 disk_size = refcount_cache_size * 32768 @@ -97,12 +97,16 @@ 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 +For example, 1MB of L2 cache is needed to cover every 8 GB of the virtual +image size (given that the default cluster size is used): - 1048576 / 131072 = 8 GB of virtual disk covered by that cache - 262144 / 32768 = 8 GB + 8 GB / 8192 = 1 MB + +The refcount cache is 4 times the cluster size by default. With the default +cluster size of 64 KB, it is 256 KB (262144 bytes). This is sufficient for +8 GB of image size: + + 262144 * 32768 = 8 GB How to configure the cache sizes @@ -130,6 +134,9 @@ There are a few things that need to be taken into account: memory as possible to the L2 cache before increasing the refcount cache size. + - At most two of "l2-cache-size", "refcount-cache-size", and "cache-size" + can be set simultaneously. + Unlike L2 tables, refcount blocks are not used during normal I/O but only during allocations and internal snapshots. In most cases they are accessed sequentially (even during random guest I/O) so increasing the |