I was feeling Ubuntu was running a bit sluggish, and then I went to see the processes running on it, and found one which was using something like 50% of CPU called ksoftirqd/0
.
Does anyone know what this process does, why it is using so much CPU and how to make it less CPU-intensive?
Your computer communicates with the devices attached to it through IRQs (interrupt requests). When an interrupt comes from a device, the operating system pauses what it was doing and starts addressing that interrupt.
In some situations IRQs come very very fast one after the other and the operating system cannot finish servicing one before another one arrives. This can happen when a high speed network card receives a very large number of packets in a short time frame.
Because the operating system cannot handle IRQs as they arrive (because they arrive too fast one after the other), the operating system queues them for later processing by a special internal process named
ksoftirqd
.If
ksoftirqd
is taking more than a tiny percentage of CPU time, this indicates the machine is under heavy interrupt load.From the man page,
ksoftirqd
is a per-cpu kernel thread that runs when the machine is under heavy soft-interrupt load.You can tweak the settings a bit, by defining which cpu picks up a certain interrupt. You do this by changing the contents of
/proc/irq/$interrupt_number/smp_affinity
. You can get a list of interrupts and their meaning by doing:The number in
smp_affinity
is a bitmap of cpus, represented in hex code. The rightmost bit is the least significant. For instance, my system has 8 cores. If I wanted to use only cores 1, 3 and 4, I would set the smp_affinity to1a
:Personally, I set up any cpu to be able to pick up interrupt 29 (eth0 in my 8-core system) with:
ksoftirqd is a per-cpu kernel thread that runs when the machine is under heavy soft-interrupt load.So, it is not eating your cpu but rather reducing your IRQ load.