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m Signing comment by 147.88.200.112 - "→Error in example of load average?: " |
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:i totaly Agree on the example being wrong. What if you have 1 highest-priority job, which will need 1minute of cpu time, and 3 lowest-priority jobs, which need 5seconds of cpu time. What if the lowest-priority jobs are only scheduled when the highest priority job is either blocked, sleeping oder done?<br>Then you would have a Load Average of 4 for the first minute. Remember, total CPU Time is 1 minute 15 seconds. So you'd only need a 25% faster CPU to do all the work in one minute. The example in the article is undoubtly wrong and missleading.<br> I'd say, you'd be able to calculate the percentage of cpu load if you knew the jobs that run in a given time and each of the jobs consumed cpu time, thus getting a total cpu time consumed over a given time period. cpu time consumed / time passed = percentage of cpu load. But i guess that's just not available on a Unix system, /proc/stat lists the average load in time since system start. So there should be no talking about a percentage in this part.
== Unix or windows or both ==
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