IO WAIT Information From IBM

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Date:         Thu, 8 Apr 2004 09:52:26 -0500
To: aix-l@Princeton.EDU

This is from a PDF and text file written by one of the IBM Performance Specialists. I don't think I can send attachments to the list so I have pasted everything here. It contains some tables so I have tried to format as best as possible in this email:

A great deal of controversy exists around the interpretation of
the I/O wait metric in AIX. This number shows up at the rightmost "wa"
column in vmstat output, the "% iowait" column in iostat, the %wio column
in the sar -P , and the ascii bar graph titled "wait" in topas. Confusion exists
when I/O wait is evaluated for performance or capacity planning as to
whether this number should be considered CPU cycles that are used or
cycles that should be added to the system idle time indicating unused
capacity. This paper will explain how this metric is captured and calculated
as well as provide a "case study" example to illustrate the effects.
A review of some of the basic AIX functions will assist in a better
understanding of how the I/O wait value is collected and calculated. The
AIX scheduler, the CPU "queues", the CPU states, and the idle or wait
process, will be discussed.
The scheduler is a part of the AIX kernel that is tasked with making sure the
individual CPUs have work to do and in the case where there are more
runnable jobs (threads) than CPUs, to make sure each one gets its fair share
of the CPU resource. The system contains a hardware timer which
generates 100 interrupts/second. This interrupt will then dispatch the kernel
scheduler process which runs at a fixed priority of 16. The scheduler will
first charge the running thread with the 10 millisecond time slice and then
dispatch another thread (context switch) of equal or higher priority on that
CPU assuming there are other runnable threads. This short term CPU usage
Demystifying I/O Wait
Harold Lee - ATS 12/11/2002 Page 1

Partial ps command output showing short term CPU usage in "C' column
#ps -aekl
F S UID PID PPID C PRI NI ADDR SZ WCHAN TTY TIME CMD
303 A 0 0 0 120 16 -- 12012 12 - 4:17 swapper
200003 A 0 1 0 0 60 20 c00c 732 - 0:22 init
303 A 0 516 0 120 127 -- 13013 8 - 31972:23 kproc
303 A 0 774 0 120 127 -- 14014 8 - 31322:34 kproc
303 A 0 1032 0 0 16 -- 17017 12 - 0:00 kproc
303 A 0 1290 0 0 36 -- 1e01e 16 - 0:32 kproc
303 A 0 1548 0 0 37 -- 1f01f 64 * - 5:09 kproc
303 A 0 1806 0 0 60 -- c02c 16 3127c558 - 0:04 kproc
240001 A 0 2638 1 0 60 20 12212 108 3127ab98 - 102:04 syncd

is reported in the "C" column when including a -l option with the ps
command.

One hundred times a second, the scheduler will take the process that is
currently running on each CPU, and increment the "C" value by one. It will
then recalculate that processes priority and rescan the process table looking
for the next process to dispatch. If there are no runnable processes the
scheduler will dispatch the "idle" kernel process. There are one of these
assigned to each CPU and are bound to that particular processor. The
following output shows a four way system with four wait processes each
bound to a CPU.
THREAD TABLE :
SLT ST TID PID CPUID POLICY PRI CPU EVENT PROCNAME
0 s 3 0 bound FIFO 10 78 swapper
flags: kthread
Demystifying I/O Wait
Harold Lee - ATS 12/11/2002 Page 2

1 s 103 1 unbound other 3c 0 init
flags: local wakeonsig cdefer
unknown: 0x10000
2 r 205 204 0 FIFO ff 78 wait
flags: funnel kthread
3 r 307 306 1 FIFO ff 78 wait
flags: funnel kthread
4 r 409 408 2 FIFO ff 78 wait
flags: funnel kthread
5 r 50b 50a 3 FIFO ff 78 wait
flags: funnel kthread
6 s 60d 60c unbound RR 11 2b reaper
Also notice that the wait process priority is 0Xff. The MSB has been turned
off to give a priority range of 0-127 on AIX 5.1 and lower. In AIX 5.2 and
higher the range of priorities has been increased to 255 to allow more
granularity for control when using Workload Manager (WLM).
If there are no processes to dispatch, the scheduler will dispatch the "wait"
process which will run until any other process becomes runnable at which
time it will immediately be dispatched since it will always have a higher
priority. The wait processes only job is to increment the counters that report
if that particular processor is "idle" or "waiting for I/O". It is important to
remember that the "waiting for I/O" metric is incremented by the idle
process. The decision on whether the idle process decides to increment the
"idle" counter or the "waiting for I/O counter depends on whether there is a
process sitting in the blocked queue. Processes which are runnable but
waiting on data from a disk are placed on the blocked queue to wait for their
data. If no processes are sitting on that particular processors blocked queue,
then the wait process will charge the time to "idle". If there are one or more
processes on that particular processors blocked queue, then the system
Demystifying I/O Wait
Harold Lee - ATS 12/11/2002 Page 3

charges the time to "waiting for I/O". Waiting for I/O is considered to be a
special case of idle and therefore the percentage of time spent in waiting for
I/O is usable for process to perform work.
A case study will be presented to illustrate this concept. Consider a single
CPU system the has two tasks to perform. Task a is a CPU intensive
program and task B is an I/O intensive program. The effects of these
programs on the vmstat output will be considered separately and then
combined.
Task "A" which is CPU intensive is run on a single CPU system, the
majority of the CPU time will be spent in the "user" (us) mode. The vmstat
output below reflects the effects of a single process running.
$ vmstat 1
kthr memory page faults cpu
----- ----------- ------------------------ ------------ -----------
r b avm fre re pi po fr sr cy in sy cs us sy id wa
1 0 106067 164605 0 0 0 0 23 0 232 835 411 99 0 0 0
1 0 106072 164600 0 0 0 0 0 0 239 2543 413 99 1 0 0
1 0 106072 164600 0 0 0 0 0 0 234 2425 403 99 0 0 0
1 0 106072 164600 0 0 0 0 0 0 235 2426 405 98 2 0 0
1 0 106072 164600 0 0 0 0 0 0 241 2572 428 99 1 0 0
1 0 106072 164600 0 0 0 0 0 0 233 2490 475 99 0 0 0
Task "B" which is I/O intensive is run on a single CPU system, the majority
of the CPU time will be spent in the "waiting for I/O" (wa) mode. The vmstat
output below reflects the effects of a single process running.
$ vmstat 1
kthr memory page faults cpu
----- ----------- ------------------------ ------------ -----------
r b avm fre re pi po fr sr cy in sy cs us sy id wa
0 1 106067 164605 0 0 0 0 23 0 232 835 411 0 1 0 99
0 1 106072 164600 0 0 0 0 0 0 239 2543 413 0 1 0 99
Demystifying I/O Wait
Harold Lee - ATS 12/11/2002 Page 4

0 1 106072 164600 0 0 0 0 0 0 234 2425 403 0 1 0 99
1 1 106072 164600 0 0 0 0 0 0 235 2426 405 0 2 0 98
0 1 106072 164600 0 0 0 0 0 0 241 2572 428 0 1 0 99
0 1 106072 164600 0 0 0 0 0 0 233 2490 475 0 1 0 99
If while Task "B" which is I/O intensive is running, task "A" is started on a
single CPU system, the majority of the CPU time will be spent in the "user"
(us) mode. This shows that all of the CPU cycles spent in the "waiting for
I/O" mode have been recovered and are usable by other processes. The
vmstat output below reflects the effects of running a CPU intensive program
and an I/O intensive simultaneously.
$ vmstat 1
kthr memory page faults cpu
----- ----------- ------------------------ ------------ -----------
r b avm fre re pi po fr sr cy in sy cs us sy id wa
1 1 106067 164605 0 0 0 0 23 0 232 835 411 99 0 0 0
1 1 106072 164600 0 0 0 0 0 0 239 2543 413 99 2 0 0
1 1 106072 164600 0 0 0 0 0 0 234 2425 403 99 0 0 0
2 1 106072 164600 0 0 0 0 0 0 235 2426 405 98 1 0 0
1 1 106072 164600 0 0 0 0 0 0 241 2572 428 99 1 0 0
1 1 106072 164600 0 0 0 0 0 0 233 2490 475 99 0 0 0
One item to note from this example. I/O bound systems cannot always be
determined by looking at the "waiting for I/O" metrics only. A busy system
can mask the effects of I/O bottlenecks. To determine if an I/O bottleneck
exists, the blocked queue as well as the output from iostat must also be
considered.
Demystifying I/O Wait
Harold Lee - ATS 12/11/2002 Page 5

What exactly is "iowait"?

To summarize it in one sentence, 'iowait' is the percentage
of time the CPU is idle AND there is at least one I/O
in progress.

Each CPU can be in one of four states: user, sys, idle, iowait.
Performance tools such as vmstat, iostat, sar, etc. print
out these four states as a percentage. The sar tool can
print out the states on a per CPU basis (-P flag) but most
other tools print out the average values across all the CPUs.
Since these are percentage values, the four state values
should add up to 100%.

The tools print out the statistics using counters that the
kernel updates periodically (on AIX, these CPU state counters
are incremented at every clock interrupt (these occur
at 10 millisecond intervals).
When the clock interrupt occurs on a CPU, the kernel
checks the CPU to see if it is idle or not. If it's not
idle, the kernel then determines if the instruction being
executed at that point is in user space or in kernel space.
If user, then it increments the 'user' counter by one. If
the instruction is in kernel space, then the 'sys' counter
is incremented by one.

If the CPU is idle, the kernel then determines if there is
at least one I/O currently in progress to either a local disk
or a remotely mounted disk (NFS) which had been initiated
from that CPU. If there is, then the 'iowait' counter is
incremented by one. If there is no I/O in progress that was
initiated from that CPU, the 'idle' counter is incremented
by one.

When a performance tool such as vmstat is invoked, it reads
the current values of these four counters. Then it sleeps
for the number of seconds the user specified as the interval
time and then reads the counters again. Then vmstat will
subtract the previous values from the current values to
get the delta value for this sampling period. Since vmstat
knows that the counters are incremented at each clock
tick (10ms), second, it then divides the delta value of
each counter by the number of clock ticks in the sampling
period. For example, if you run 'vmstat 2', this makes
vmstat sample the counters every 2 seconds. Since the
clock ticks at 10ms intervals, then there are 100 ticks
per second or 200 ticks per vmstat interval (if the interval
value is 2 seconds). The delta values of each counter
are divided by the total ticks in the interval and
multiplied by 100 to get the percentage value in that
interval.

iowait can in some cases be an indicator of a limiting factor
to transaction throughput whereas in other cases, iowait may
be completely meaningless.
Some examples here will help to explain this. The first
example is one where high iowait is a direct cause
of a performance issue.

Example 1:
Let's say that a program needs to perform transactions on behalf of
a batch job. For each transaction, the program will perform some
computations which takes 10 milliseconds and then does a synchronous
write of the results to disk. Since the file it is writing to was
opened synchronously, the write does not return until the I/O has
made it all the way to the disk. Let's say the disk subsystem does
not have a cache and that each physical write I/O takes 20ms.
This means that the program completes a transaction every 30ms.
Over a period of 1 second (1000ms), the program can do 33
transactions (33 tps). If this program is the only one running
on a 1-CPU system, then the CPU usage would be busy 1/3 of the
time and waiting on I/O the rest of the time - so 66% iowait
and 34% CPU busy.

If the I/O subsystem was improved (let's say a disk cache is
added) such that a write I/O takes only 1ms. This means that
it takes 11ms to complete a transaction, and the program can
now do around 90-91 transactions a second. Here the iowait time
would be around 8%. Notice that a lower iowait time directly
affects the throughput of the program.

Example 2:

Let's say that there is one program running on the system - let's assume
that this is the 'dd' program, and it is reading from the disk 4KB at
a time. Let's say that the subroutine in 'dd' is called main() and it
invokes read() to do a read. Both main() and read() are user space
subroutines. read() is a libc.a subroutine which will then invoke
the kread() system call at which point it enters kernel space.
kread() will then initiate a physical I/O to the device and the 'dd'
program is then put to sleep until the physical I/O completes.
The time to execute the code in main, read, and kread is very small -
probably around 50 microseconds at most. The time it takes for
the disk to complete the I/O request will probably be around 2-20
milliseconds depending on how far the disk arm had to seek. This
means that when the clock interrupt occurs, the chances are that
the 'dd' program is asleep and that the I/O is in progress. Therefore,
the 'iowait' counter is incremented. If the I/O completes in
2 milliseconds, then the 'dd' program runs again to do another read.
But since 50 microseconds is so small compared to 2ms (2000 microseconds),
the chances are that when the clock interrupt occurs, the CPU will
again be idle with a I/O in progress. So again, 'iowait' is
incremented. If 'sar -P <cpunumber>' is run to show the CPU
utilization for this CPU, it will most likely show 97-98% iowait.
If each I/O takes 20ms, then the iowait would be 99-100%.
Even though the I/O wait is extremely high in either case,
the throughput is 10 times better in one case.

Example 3:

Let's say that there are two programs running on a CPU. One is a 'dd'
program reading from the disk. The other is a program that does no
I/O but is spending 100% of its time doing computational work.
Now assume that there is a problem with the I/O subsystem and that
physical I/Os are taking over a second to complete. Whenever the
'dd' program is asleep while waiting for its I/Os to complete,
the other program is able to run on that CPU. When the clock
interrupt occurs, there will always be a program running in
either user mode or system mode. Therefore, the %idle and %iowait
values will be 0. Even though iowait is 0 now, that does not
mean there is NOT a I/O problem because there obviously is one
if physical I/Os are taking over a second to complete.

Example 4:

Let's say that there is a 4-CPU system where there are 6 programs
running. Let's assume that four of the programs spend 70% of their
time waiting on physical read I/Os and the 30% actually using CPU time.
Since these four programs do have to enter kernel space to execute the
kread system calls, it will spend a percentage of its time in
the kernel; let's assume that 25% of the time is in user mode,
and 5% of the time in kernel mode.
Let's also assume that the other two programs spend 100% of their
time in user code doing computations and no I/O so that two CPUs
will always be 100% busy. Since the other four programs are busy
only 30% of the time, they can share that are not busy.

If we run 'sar -P ALL 1 10' to run 'sar' at 1-second intervals
for 10 intervals, then we'd expect to see this for each interval:

         cpu %usr %sys %wio %idle
          0 50 10 40 0
          1 50 10 40 0
          2 100 0 0 0
          3 100 0 0 0
          - 75 5 20 0

Notice that the average CPU utilization will be 75% user, 5% sys,
and 20% iowait. The values one sees with 'vmstat' or 'iostat' or
most tools are the average across all CPUs.

Now let's say we take this exact same workload (same 6 programs
with same behavior) to another machine that has 6 CPUs (same
CPU speeds and same I/O subsytem). Now each program can be
running on its own CPU. Therefore, the CPU usage breakdown
would be as follows:

         cpu %usr %sys %wio %idle
          0 25 5 70 0
          1 25 5 70 0
          2 25 5 70 0
          3 25 5 70 0
          4 100 0 0 0
          5 100 0 0 0
          - 50 3 47 0

So now the average CPU utilization will be 50% user, 3% sy,
and 47% iowait. Notice that the same workload on another
machine has more than double the iowait value.

Conclusion:

The iowait statistic may or may not be a useful indicator of
I/O performance - but it does tell us that the system can
handle more computational work. Just because a CPU is in
iowait state does not mean that it can't run other threads
on that CPU; that is, iowait is simply a form of idle time.

Jason de la Fuente

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