Take the 2-minute tour ×
Server Fault is a question and answer site for professional system and network administrators. It's 100% free, no registration required.

Possible Duplicate:
Meaning of the buffers/cache line in the output of free

I did free -tm on my system, and I got the output below.

Is the free buffers/cache part of the used memory? And therefore we can consider it as free memory?

             total       used       free     shared    buffers     cached
Mem:          5721       5689         32          0        137       4664
-/+ buffers/cache:        887       4834
Swap:         6000         13       5987
Total:       11722       5703       6019

Thanks

share|improve this question

marked as duplicate by sysadmin1138 Aug 24 '11 at 23:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4  
linuxatemyram.com –  Zoredache Mar 12 '10 at 6:20
add comment

2 Answers 2

up vote 1 down vote accepted

take a look at this post here on serverfault. it will explain the free output to you.

share|improve this answer
add comment

Whether you consider it "free memory" depends on what you really mean by "free memory". If you mean memory that was totally wasted because the system couldn't figure out any way to use it, then you should not consider it free memory. If you mean memory that's available for use should the system need memory for something, then you should consider it free memory.

Personally, I think the concept of "free memory" should simply be banished. The way modern operating systems manage physical memory just doesn't create a category that can sensibly be described as "free".

In particular, the following mindset is completely wrong: "If there's free physical memory, then I don't need more RAM. But if almost all of the physical memory is in use, then my system could benefit from more RAM."

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.