CPU utilization question

[zak100]

Hi,

I am trying to solve the following question:

 

What is the CPU utilization if there are 5 processes running at the same time, and on average the CPU spends 30% of its time waiting on I/O completion?

The formula is :

Quote

The formula for CPU utilization is 1−pn, in which n is number of process running in memory and p is the average percentage of time processes are waiting for I/O.

 

What is P in this formula?: 1-P ^n

 

I found a solution which finds the CPU utilization for each process separately.

Can we do it in the following way:

1- (0.3)^5

= 0.9975

Is the above answer correct?

Somebody please guide me.

 

Zulfi.

[Ghideon]

I do not know where formula comes from. In reality there are more factors to consider.

On 4/30/2020 at 6:50 AM, zak100 said:

What is P in this formula?: 1-P ^n

It seems like you provided the answer:

On 4/30/2020 at 6:50 AM, zak100 said:

p is the average percentage of time processes are waiting for I/O.

 

On 4/30/2020 at 6:50 AM, zak100 said:

found a solution which finds the CPU utilization for each process separately.

Can we do it in the following way:

1- (0.3)^5

= 0.9975

Is the above answer correct?

Some reasoning: If each process is waiting for I/O 30% of the time and during that time not consuming CPU* then the maximum CPU usage of a single process is 70%. When many processes are started they will not only wait for I/O, they will also have to wait for the CPU. Each process will utilise the CPU less than the maximum 70% of the time that the process could run. So 0.9975 is not correct.

0.3 (30%) is the probability of a process waiting for I/O. (0.3)^5 is the probability of having all five processes waiting for I/O. So 1-(0.3)^5 then is the probability having at least one process in a running state and not waiting, hence able to consume CPU.

Not sure how you wish to define CPU utilisation for a single process, maybe [math] \frac{1-p^{n}}{n} [/math]  ?

*) I assume a process waiting for I/O is not in a running state or performing some busy-wait

 

Edited May 2, 2020 by Ghideon

[zak100]

Hi,

Thanks. For the update otherwise I would be doing it wrongly. I found the formula which I wrote in my  (Question) post from google. But your formula is looking correct also. I have to discuss it with somebody. God bless you.

 

Zulfi.

 

[zak100]

Hi,

(1-P^n)/n is the average case.

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So 1-(0.3)^5 then is the probability having at least one process in a running state and not waiting, hence able to consume CPU.>

I think this is the correct way.

God bless you.

 

Thanks.

 

Zulfi.