Why is the number of ticks inaccurate?

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I checked the numbers in proc/stat to the the number of clock ticks the CPU has had and I saw that the sum of the numbers in CPU1-4 is not equal to the numbers written in front of CPU as a whole. Why is that?







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    I checked the numbers in proc/stat to the the number of clock ticks the CPU has had and I saw that the sum of the numbers in CPU1-4 is not equal to the numbers written in front of CPU as a whole. Why is that?







    share|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I checked the numbers in proc/stat to the the number of clock ticks the CPU has had and I saw that the sum of the numbers in CPU1-4 is not equal to the numbers written in front of CPU as a whole. Why is that?







      share|improve this question














      I checked the numbers in proc/stat to the the number of clock ticks the CPU has had and I saw that the sum of the numbers in CPU1-4 is not equal to the numbers written in front of CPU as a whole. Why is that?









      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 10 '17 at 13:08









      Satō Katsura

      10.7k11533




      10.7k11533










      asked Nov 10 '17 at 8:06









      yukashima huksay

      429217




      429217




















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          The difference results from truncating the values after calculating the sums. /proc/stat calculates the sums of elapsed times in nanoseconds, and then converts the resulting values to USER_HZ using integer division which truncates.



          Thus if all four CPUs have individual values of 1.3, you’ll get a sum of 5.2, truncated to 5 for the first line, and 1 for each CPU line.






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            1 Answer
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            active

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            1 Answer
            1






            active

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            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            The difference results from truncating the values after calculating the sums. /proc/stat calculates the sums of elapsed times in nanoseconds, and then converts the resulting values to USER_HZ using integer division which truncates.



            Thus if all four CPUs have individual values of 1.3, you’ll get a sum of 5.2, truncated to 5 for the first line, and 1 for each CPU line.






            share|improve this answer
























              up vote
              3
              down vote



              accepted










              The difference results from truncating the values after calculating the sums. /proc/stat calculates the sums of elapsed times in nanoseconds, and then converts the resulting values to USER_HZ using integer division which truncates.



              Thus if all four CPUs have individual values of 1.3, you’ll get a sum of 5.2, truncated to 5 for the first line, and 1 for each CPU line.






              share|improve this answer






















                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                The difference results from truncating the values after calculating the sums. /proc/stat calculates the sums of elapsed times in nanoseconds, and then converts the resulting values to USER_HZ using integer division which truncates.



                Thus if all four CPUs have individual values of 1.3, you’ll get a sum of 5.2, truncated to 5 for the first line, and 1 for each CPU line.






                share|improve this answer












                The difference results from truncating the values after calculating the sums. /proc/stat calculates the sums of elapsed times in nanoseconds, and then converts the resulting values to USER_HZ using integer division which truncates.



                Thus if all four CPUs have individual values of 1.3, you’ll get a sum of 5.2, truncated to 5 for the first line, and 1 for each CPU line.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 10 '17 at 8:37









                Stephen Kitt

                143k22312377




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