On my system with 16 G of ram, ram is full of cached data, at the end of the day. Most of the cached data is buffer_head type. AFAIK buffer_head is about mapping sectors of physical disk to memory.

- Is this status normal? Why kernel should not free cached memory?
- Will the situation have effects on the performance f the system. For example, swapping.
- How can I resolv the problem. Or should I accept as normal.

regards,


top:

Mem: 16633388k total, 16089760k used, 543628k free, 315296k buffers
Swap: 4399532k total, 96k used, 4399436k free, 14624576k cached

free:
total used free shared buffers cached
Mem: 16633388 16089704 543684 0 315308 14624592
-/+ buffers/cache: 1149804 15483584
Swap: 4399532 96 4399436


# df
Filesystem 1K-blocks Used Available Use% Mounted on
/dev/cciss/c0d1p1 9920592 2711904 6696620 29% /
/dev/cciss/c0d1p5 399696092 78576380 300488640 21% /var
/dev/cciss/c0d1p2 4956316 143108 4557376 4% /tmp
/dev/cciss/c0d0p1 194442 11194 173209 7% /boot
tmpfs 8316692 0 8316692 0% /dev/shm

# slabtop
Active / Total Objects (% used) : 1731551 / 1778509 (97.4%)
Active / Total Slabs (% used) : 37209 / 37331 (99.7%)
Active / Total Caches (% used) : 101 / 145 (69.7%)
Active / Total Size (% used) : 135210.79K / 142038.93K (95.2%)
Minimum / Average / Maximum Object : 0.01K / 0.08K / 128.00K

OBJS ACTIVE USE OBJ SIZE SLABS OBJ/SLAB CACHE SIZE NAME
1507248 1505309 99% 0.05K 20934 72 83736K buffer_head
85190 85167 99% 0.27K 6085 14 24340K radix_tree_node
51504 51155 99% 0.13K 1776 29 7104K dentry_cache
27676 20513 74% 0.09K 629 44 2516K vm_area_struct
17712 14315 80% 0.48K 2214 8 8856K ext3_inode_cache
11560 7654 66% 0.19K 578 20 2312K filp
9266 5449 58% 0.03K 82 113 328K size-32
9145 4420 48% 0.06K 155 59 620K size-64
6350 5708 89% 0.01K 25 254 100K anon_vma

# sysctl -a |grep dirty
vm.dirty_expire_centisecs = 2999
vm.dirty_writeback_centisecs = 499
vm.dirty_ratio = 8
vm.dirty_background_ratio = 2

# sysctl -a |grep commit
vm.overcommit_ratio = 50
vm.overcommit_memory = 0

Kernel is 2.6.18-162 from latest Centos EL5

AFAIK the kernel will freeing memory only when it need it, and keep as much cache as possible. That's why on a Linux system, RAM is always full.

1 members found this post helpful.

Yes

Why should it? Cached memory means that the system doesn't have to perform an IO to access the cached data. Cached data is read-only, and can be abandoned at will by the kernel without an IO action.

Yes, caching actually improves the performance of your system, otherwise they wouldn't have coded it into the kernel. It will not cause it to swap.

There is no problem.

1 members found this post helpful.

No it isn't - don't confuse slab usage (which is minute in the grand scheme of things) with page cache.
Despite the common use of the term "cache", each is caching something totally different. And of course there are hardware caches as well - (multiple) in the CPU as well as things like disk controllers. The term is used everywhere - describing function, not the specific storage usage.

As for your slabtop output, that by default sorts by object count - hit the "c" key whilst in slabtop to sort by usage. Note the sizes are basically irrelevant. And not included in the "cached" number from things like "free" anyway.

There have been some fixes for slab/slub allocators since that kernel level, but you really don't have a problem as the posters above indicated.

1 members found this post helpful.