Date: Wed, 2 Oct 1996 12:00:00 +0900 (KST) From: Heo Sung-Gwan <heo@cslsun10.sogang.ac.kr> To: freebsd-fs@FreeBSD.ORG Subject: nbuf in buffer cache Message-ID: <Pine.SUN.3.93.961002115902.4123A-100000@cslsun10>
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Hi,
I am curious about the number of buffers(= nbuf) in buffer cache.
The variable nbuf is determined in i386/i386/machdep.c as following:
#ifdef NBUF
int nbuf = NBUF;
#else
int nbuf = 0;
#endif
...
void
cpu_startup()
{
...
if (nbuf == 0) {
nbuf = 30;
if( physmem > 1024)
nbuf += min((physmem - 1024) / 12, 1024);
}
...
}
If NBUF is not defined and physical memory is less than 1024 pages(= 4Mbytes)
then nbuf becomes 30, and otherwise nbuf is 30 + min((physmem - 1024) / 12,
1024).
Why does the number of buffers is calculated in this fashion?
30 buffers, 1024 pages, and division by 12 have special meaning?
There is no comment on source code.
In addition, if there is no user application processes how many buffers
are enough to run the system without degrading the performance of the system?
Only 30 buffers? Or better as many as possible?
Please let me know.
--
Heo Sung-Gwan
Dept. of Computer Science, Sogang University, Seoul, Korea.
E-mail: heo@cslsun10.sogang.ac.kr
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