Question

A computer has 4 GB of RAM of which the operating system occupies 512 MB. The processes are all 256 MB (for simplicity) and have the same characteristics. If the goal is 99% CPU utilization, what is the maximum I/O wait that can be tolerated?

Answer 1

Answer:

$1-p^{14} = 0.99$

$p^{14}= 0.01$

$p =(0.01)^{1/14}= 0.720$

So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %

Explanation:

Previous concepts

Input/output operations per second (IOPS, pronounced eye-ops) "is an input/output performance measurement used to characterize computer storage devices like hard disk drives (HDD)"

Solution to the problem

For this case since we have 4GB, but 512 MB are destinated to the operating system, we can begin finding the available RAM like this:

Available = 4096 MB - 512 MB = 3584 MB

Now we can find the maximum simultaneous process than can use with this:

$\frac{3584 MB}{256 MB/proc}= 14 processes$

And then we can find the maximum wait I/O that can be tolerated with the following formula:

$1- p^{14}= rate$

The expeonent for p = 14 since we got 14 simultaneous processes, and the rate for this case would be 99% or 0.99, if we solve for p we got:

$1-p^{14} = 0.99$

$p^{14}= 0.01$

$p =(0.01)^{1/14}= 0.720$

So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %