Date: Mon, 2 Aug 2010 09:42:29 -0400 From: John Baldwin <jhb@freebsd.org> To: Alan Cox <alc@cs.rice.edu> Cc: alc@freebsd.org, freebsd-arch@freebsd.org Subject: Re: amd64: change VM_KMEM_SIZE_SCALE to 1? Message-ID: <201008020942.29654.jhb@freebsd.org> In-Reply-To: <4C54981B.9080209@cs.rice.edu> References: <4C4DB2B8.9080404@freebsd.org> <201007301614.40768.jhb@freebsd.org> <4C54981B.9080209@cs.rice.edu>
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On Saturday, July 31, 2010 5:39:39 pm Alan Cox wrote:
> John Baldwin wrote:
> > On Friday, July 30, 2010 2:49:59 pm Alan Cox wrote:
> >> With that in mind, the following patch slows the growth of "virt" from
> >> 2/5 of vm_kmem_size to 1/7. This has no effect on amd64. However, on
> >> i386. it allows desiredvnodes to grow slowly for machines with 1.5GB to
> >> about 2.5GB of RAM, ultimately exceeding the old desiredvnodes cap by
> >> about 17%. Once we exceed the old cap, we increase desiredvnodes at a
> >> marginal rate that is almost the same as your patch, about 1% of
> >> physical memory. It's just computed differently.
> >>
> >> Using 1/8 instead of 1/7, amd64 machines with less than about 1.5GB lose
> >> about 7% of their vnodes, but they catch up and pass the old limit by
> >> 1.625GB. Perhaps, more importantly, i386 machines only exceed the old
> >> cap by 3%.
> >>
> >> Thoughts?
> >
> > I think this is much better. My strawman was rather hackish in that it was
> > layering a hack on top of the existing calculations. I prefer your approach.
> > I do not think penalizing amd64 machines with less than 1.5GB is a big worry
> > as most x86 machines with a small amount of memory are probably running as
> > i386 anyway. Given that, I would probably lean towards 1/8 instead of 1/7,
> > but I would be happy with either one.
>
> I've looked a bit at an i386/PAE system with 8GB. I don't think that a
> default configuration, e.g., no changes to the mbuf limits, is at risk
> with 1/7.
Ok.
> > How is this value computed? I would prefer something like:
> >
> > '512 * 1024 * 1024 * 1024 / (sizeof(struct vnode) + sizeof(struct vm_object) / N'
> >
> > if that is how it is computed. A brief note about the magic number of 393216
> > would also be nice to have (and if it could be a constant with a similar
> > formula value that would be nice, too.).
> >
> >
>
> I've tried to explain this computation below.
Thanks, it looks good to me now.
> Index: kern/vfs_subr.c
> ===================================================================
> --- kern/vfs_subr.c (revision 210702)
> +++ kern/vfs_subr.c (working copy)
> @@ -282,23 +282,34 @@ SYSCTL_INT(_debug, OID_AUTO, vnlru_nowhere, CTLFLA
>
> /*
> * Initialize the vnode management data structures.
> + *
> + * Reevaluate the following cap on the number of vnodes after the physical
> + * memory size exceeds 512GB. In the limit, as the physical memory size
> + * grows, the ratio of physical pages to vnodes approaches sixteen to one.
> */
> #ifndef MAXVNODES_MAX
> -#define MAXVNODES_MAX 100000
> +#define MAXVNODES_MAX (512 * (1024 * 1024 * 1024 / PAGE_SIZE /
> 16))
> #endif
> static void
> vntblinit(void *dummy __unused)
> {
> + int physvnodes, virtvnodes;
>
> /*
> - * Desiredvnodes is a function of the physical memory size and
> - * the kernel's heap size. Specifically, desiredvnodes scales
> - * in proportion to the physical memory size until two fifths
> - * of the kernel's heap size is consumed by vnodes and vm
> - * objects.
> + * Desiredvnodes is a function of the physical memory size and the
> + * kernel's heap size. Generally speaking, it scales with the
> + * physical memory size. The ratio of desiredvnodes to physical
> pages
> + * is one to four until desiredvnodes exceeds 98,304.
> Thereafter, the
> + * marginal ratio of desiredvnodes to physical pages is one to
> + * sixteen. However, desiredvnodes is limited by the kernel's heap
> + * size. The memory required by desiredvnodes vnodes and vm objects
> + * may not exceed one seventh of the kernel's heap size.
> */
> - desiredvnodes = min(maxproc + cnt.v_page_count / 4, 2 *
> vm_kmem_size /
> - (5 * (sizeof(struct vm_object) + sizeof(struct vnode))));
> + physvnodes = maxproc + cnt.v_page_count / 16 + 3 * min(98304 * 4,
> + cnt.v_page_count) / 16;
> + virtvnodes = vm_kmem_size / (7 * (sizeof(struct vm_object) +
> + sizeof(struct vnode)));
> + desiredvnodes = min(physvnodes, virtvnodes);
> if (desiredvnodes > MAXVNODES_MAX) {
> if (bootverbose)
> printf("Reducing kern.maxvnodes %d -> %d\n",
>
>
--
John Baldwin
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