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    SubjectRe: Negative "ios_in_flight" in the 2.4 kernel
    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
    > 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)
    if (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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