On Mon, Mar 07, 2005 at 02:20:07PM -0000, sounak chakraborty wrote: > After reading the proc.txt file,i come to the following > conclusion: > > 1. I get the total clock speed of the cpu > from /proc/cpuinfo from cpu MHz entry. > 2. Now i add the values of all the colomns > in the first row of /proc/stat(the row which > starts with cpu).this gives the total cpu usage. > > Then i am dividing the above value with 1000 and the resulting answer > is again divided with the total clock speed to get the percentage of > cpu usage. > > Is my approach correct. No, the CPU MHz is not involved at all. Heck, your method will fail when the MHz changes (yes, this happens with cpufreq). total = user + nice + system + idle + iowait + irq + softirq user percentage = user / total * 100 nice percentage = nice / total * 100 etc. (Note that /proc/stat contains the total numbers. To calculate the difference, you need to look two times). > I am checking my results with system monitor but there is a variation. True, because you can't guarantee that your program and the system monitor sample at the same time. > (This is my best effort to write a mail) Thanks for your efforts. Now switch off posting in HTML and it's perfect. Only spammers have a reason to use HTML. Erik -- Erik Mouw J.A.K.Mouw@xxxxxxxxxxxxxx mouw@xxxxxxxxxxxx
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