Hi Jlliagre,
Thank you so much for your reply. Really appreciate your guidance.
Quote:
Originally Posted by
jlliagre
No. The load would be 9+1 during the same 1 second when all processes compete then 0 during 9 seconds when all are idling so the average load would be 1. As you have only one core, the CPU utilization would be 10%.
Please pardon me for my ignorance, but I still did not quite get the full picture or the maths behind this.
Quote:
the load will be 9 (in queue) + 1 running during the 1st second, and 0 during the next 9 seconds.
q1) Does that means that the 10 threads/load are actually completed within the 1st second
(right before the 2nd second)
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so the average load would be 1.
q2) Do you mean that the load average is calculated as an average of 1 second for the past 10 seconds ?
Hence (9+1) =10 load in the past 10 seconds ? (10load/10sec)
so its essentially 1 load / per sec, for the rest of the 60 secs/1 minute ?
--but i thought you mentioned earlier that the load is sample every 10ms and not 10 sec?
Quote:
As you have only one core, the CPU utilization would be 10%
q3) My understanding is that I have 1 cpu/core.
It was utilized 100% on the 1st second of every 10 seconds.
In 1 minute, it would be utilized 5/60 second.
So the utilization for 1 minute is 5/60 * 100 = 8%.
How does it become 10% ?
Because the CPU is fully utilized for 1 sec in every 10 second = 1/10 *100 = 10% ?
Base on the above ->
Is both
a) the cpu utilization (1sec/10sec*100)
and
b) the cpu load ( (9+1)load / 10 sec) calculated per every 10second then ?
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Please do bear with me if i seems totally off.
Hope to hear your advice soon.
Regards,
Noob