Re: Negative "ios_in_flight" in the 2.4 kernel

From: Marcelo Tosatti
Date: Wed Dec 22 2004 - 13:19:32 EST


On Wed, Dec 22, 2004 at 07:19:42AM -0800, M. Edward Borasky wrote:
> On Wed, 2004-12-22 at 12:16 +0100, Jens Axboe wrote:
> >
> > > Question: wouldn't a simple refusal to decrement ios_in_flight in
> > > "down_ios" if it's zero fix this, or am I missing something?
> >
> > That would paper over the real bug, but it will work for you.
> What is the "real bug", then? What will "work for me" is accurate disk
> usage tick counts. The intent of these statistics is something known as
> Operational Analysis of Queueing Networks.
>
> The "requirement" is that the operations on each device be accurately
> counted, and the "wall clock" time spent *waiting* for requests and the
> time spent *servicing* requests be accurately accumulated for each
> device. The sector count is a bonus.
>
> >From these raw counters, one can, and iostat does, compute throughput,
> utilization, average service time, average wait time and average queue
> length. An excellent and highly readable reference for the math involved
> can be found at
>
> http://www.cs.washington.edu/homes/lazowska/qsp/Images/Chap_03.pdf
>
> That is the intent behind these counters, and what will "work for me" is
> a kernel that captures the raw counters correctly. If forcing
> ios_in_flight to be non-negative is done at the expense of losing or
> gaining ticks in the wait or service time accumulators, then it will not
> work for me.

Well something is deaccounting uncorrectly (doh), probably the disk/partition
accounting logic is doing wrong in some condition, Jens?

void req_merged_io(struct request *req)
{
struct hd_struct *hd1, *hd2;

locate_hd_struct(req, &hd1, &hd2);
if (hd1)
down_ios(hd1);
if (hd2)
down_ios(hd2);
}

void req_finished_io(struct request *req)
{
struct hd_struct *hd1, *hd2;

locate_hd_struct(req, &hd1, &hd2);
if (hd1)
account_io_end(hd1, req);
if (hd2)
account_io_end(hd2, req);
}

We could eliminate that possibility if you ran your tests with a single
non-partitioned disk, but thats just a guess.

Jens has more of a clue than I certainly.

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