How to determine which process produces disk i/o?

O

Oliver

Hi everyone,

on my xp machine i'm able to see how many cpu time is consumed by
every process through Task Manager, Sysinternals Process Explorer or
with the performance monitor. With this last tool i can also see the
usage of my physical and logical drives. But this usage is only a
global value (like my hdd led on my computer case) and it doesn't tell
my which process leads to this high disk usage (e.g. explorer while
copying a big file from one drive to another). At the task manager you
can enable some rows like I/O read or write, but they show only a
total number, not a current value about consumption. And here it seems
that not only disk i/o is added. There are all i/o uses of a process
added.

So is there any way to see which process is responisble for a high
usage of my hard drive??

Best regards,
Oliver
 
O

Oliver

Hi Detlev,

thanks for these infos, but that's not what i'm searching.

At diskmon you can see that someone is accessing the disk, but not
who.
At filemon you get a list of what is currently going on, but that's
not exactly what i like to have.

My whish would be a graph. At the x-axis you get the time, at the y-
axis you get the utilisation from 0 to 100 percent. And for every
process you'll get one line at the graph. At the performance monitor
you can generate such a graph for each drive (e.g. c:, d:, etc.) but
not for a specific process how it would access a hard drive.

Best regards,
Oliver
 
D

Detlev Dreyer

Oliver said:
My whish would be a graph. At the x-axis you get the time, at the y-
axis you get the utilisation from 0 to 100 percent. And for every
process you'll get one line at the graph. At the performance monitor
you can generate such a graph for each drive (e.g. c:, d:, etc.) but
not for a specific process how it would access a hard drive.

Well, I was often in need of special software and/or features beyond
the capabilities of the OS. 10-15 years ago, I was developing all of
the required software myself - today however, Google exists. ;)
 

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