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|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
|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
|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
|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.
|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.
|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.
|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.
|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.
|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.