Suppose that in a single processor system, an average of 6 process arrive each minute. Each process needs an average service time of 8 seconds. How would you calculate the fraction of time the CPU is kept busy ?
What I have done is -:
Since a 6 new processes arrive each minute and 8 seconds is required to complete one then in total 6×8=48 seconds would be required. So the CPU would be kept busy for 48 seconds out of 60 seconds.
Am I right ?
4
In the usual terminology, processes don’t arrive. Processes roughly corresponds to the programs running on a machine. What can arrive (and cause some processing to be done) are requests and/or events. I will assume you meant those when using “process” in the question.
Assuming that the processing of a request is CPU-bound (meaning that a single request keeps the CPU fully occupied for 8 seconds, with no spare time to do anything else), then your calculation of
6 requests/minute * 8 seconds processing time = 48/60 = 80% load
is a reasonable estimate for the average processor load.
If the processing of a single request isn’t CPU-bound, then the 8 second response time is not a correct figure for calculating CPU load, because you need to subtract the time that the process is waiting for external systems, like the disk drive or database.