The precise meaning of I/O wait time in Linux

这篇文章非常清楚介绍了top里面看到的iowait的含义，值得一读。本文系转发：Andreas Veithen的博客：http://veithen.github.io/2013/11/18/iowait-linux.html

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Some time ago I had a discussion with some systems guys about the exact meaning of the I/O wait time which is displayed by top as a percentage of total CPU time. Their answer was that it is the time spent by the CPU(s) while waiting for outstanding I/O operations to complete. Indeed, the man page for the top command defines this as the “time waiting for I/O completion”.

However, this definition is obviously not correct (or at least not complete), because a CPU never spends clock cycles waiting for an I/O operation to complete. Instead, if a task running on a given CPU blocks on a synchronous I/O operation, the kernel will suspend that task and allow other tasks to be scheduled on that CPU.

So what is the exact definition then? There is an interesting Server Fault question that discussed this. Somebody came up with the following definition that describes I/O wait time as a sub-category of idle time:

iowait is time that the processor/processors are waiting (i.e. is in an idle state and does nothing), during which there in fact was outstanding disk I/O requests.

That makes perfect sense for uniprocessor systems, but there is still a problem with that definition when applied to multiprocessor systems. In fact, “idle” is a state of a CPU, while “waiting for I/O completion” is a state of a task. However, as pointed out earlier, a task waiting for outstanding I/O operations is not running on any CPU. So how can the I/O wait time be accounted for on a per-CPU basis?

For example, let’s assume that on an otherwise idle system with 4 CPUs, a single, completely I/O bound task is running. Will the overall I/O wait time be 100% or 25%? I.e. will the I/O wait time be 100% on a single CPU (and 0% on the others), or on all 4 CPUs? This can be easily checked by doing a simple experiment. One can simulate an I/O bound process using the following command, which will simply read data from the hard disk as fast as it can:

dd if=/dev/sda of=/dev/null bs=1MB

Note that you need to execute this as root and if necessary change the input file to the appropriate block device for your hard disk.

Looking at the CPU stats in top (press 1 to get per-CPU statistics), you should see something like this:

%Cpu0  :  3,1 us, 10,7 sy,  0,0 ni,  3,5 id, 82,4 wa,  0,0 hi,  0,3 si,  0,0 st %Cpu1  :  3,6 us,  2,0 sy,  0,0 ni, 90,7 id,  3,3 wa,  0,0 hi,  0,3 si,  0,0 st %Cpu2  :  1,0 us,  0,3 sy,  0,0 ni, 96,3 id,  2,3 wa,  0,0 hi,  0,0 si,  0,0 st %Cpu3  :  3,0 us,  0,3 sy,  0,0 ni, 96,3 id,  0,3 wa,  0,0 hi,  0,0 si,  0,0 st

This output indicates that a single I/O bound task only increases the I/O wait time on a single CPU. Note that you may see that occasionally the task “switches” from one CPU to another. That is because the Linux kernel tries to schedule a task on the CPU it ran last (in order to improve CPU cache hit rates), but this is not always possible and the task is moved on another CPU. On some systems, this may occur so frequently that the I/O wait time appears to be distributed over multiple CPUs, as in the following example:

%Cpu0  :  5.7 us,  5.7 sy,  0.0 ni, 50.5 id, 34.8 wa,  3.3 hi,  0.0 si,  0.0 st %Cpu1  :  3.0 us,  3.3 sy,  0.0 ni, 72.5 id, 20.9 wa,  0.3 hi,  0.0 si,  0.0 st %Cpu2  :  7.0 us,  4.3 sy,  0.0 ni, 62.0 id, 26.7 wa,  0.0 hi,  0.0 si,  0.0 st %Cpu3  :  4.3 us,  2.3 sy,  0.0 ni, 89.6 id,  3.7 wa,  0.0 hi,  0.0 si,  0.0 st

Nevertheless, assuming that dd is the only task doing I/O on the system, there can be at most one CPU in state I/O wait at any given point in time. Indeed, 34.8+20.9+26.7+3.7=86.1 which is close to but lower than 100.

To make the experiment more reproducible, we can use the taskset command to “pin” a process to a given CPU (Note that the first command line argument is not the CPU number, but a mask):

taskset 1 dd if=/dev/sda of=/dev/null bs=1MB

Another interesting experiment is to run a CPU bound task at the same time on the same CPU:

taskset 1 sh -c "while true; do true; done"

The I/O wait time now drops to 0 on that CPU (and also remains 0 on the other CPUs), while user and system time account for 100% CPU usage:

%Cpu0  : 80,3 us, 15,5 sy,  0,0 ni,  0,0 id,  0,0 wa,  0,0 hi,  4,3 si,  0,0 st %Cpu1  :  4,7 us,  3,4 sy,  0,0 ni, 91,3 id,  0,0 wa,  0,0 hi,  0,7 si,  0,0 st %Cpu2  :  2,3 us,  0,3 sy,  0,0 ni, 97,3 id,  0,0 wa,  0,0 hi,  0,0 si,  0,0 st %Cpu3  :  2,7 us,  4,3 sy,  0,0 ni, 93,0 id,  0,0 wa,  0,0 hi,  0,0 si,  0,0 st

That is expected because I/O wait time is a sub-category of idle time, and the CPU to which we pinned both tasks is never idle.

These findings allow us to deduce the exact definition of I/O wait time:

For a given CPU, the I/O wait time is the time during which that CPU was idle (i.e. didn’t execute any tasks) and there was at least one outstanding disk I/O operation requested by a task scheduled on that CPU (at the time it generated that I/O request).

Note that the nuance is not innocent and has practical consequences. For example, on a system with many CPUs, even if there is a problem with I/O performance, the observed overall I/O wait time may still be small if the problem only affects a single task. It also means that while it is generally correct to say that faster CPUs tend to increase I/O wait time (simply because a faster CPU tends to be idle more often), that statement is no longer true if one replaces “faster” by “more”.