ISMv2, ch2: A Question about Disk Utilization

I am working through chapter 2 of ISM v2, and I am a little confused about how IOPs and disk utilization are connected.

[I will be using T(s) for T subscript s, etc...]

Starting on page 43, you get T(s) = T + L + X, where T(s) is your disk service time.  It's an average time.

Back on page 39, we see that utilization affects the average response time T(r) in a non-linear manner.  T(r) = T(s) / (1 - Utilization), "based on the fundamental laws of disk drive performance."  However, the specifics of controller utilization are left fuzzy.

• As an aside, can somebody point me to the fundamental laws they are referring to?

On page 40, it states, "it is common to utilize disks below their 70 percent of I/O serving capability."  Starting on page 44, the ISM starts to use 70% of total IOPs as the target utilization, therefore defining utilization in terms of IOPs.  100% utilization is the total IOPs available to the drive, 50% utilization is half the total IOPs available to the drive, etc...

So, here is the source of my confusion, using the example starting on page 43.

The disks are capable of 128 IOPs.  T(s) is 7.8ms.  The typical requirement is to go no further than 70% of the utilization, so we will use 128 x 0.7 = 88 IOPs in our calculations.

However, when you turn around and calculate T(r), it seems to contradict this.  As stated, 70% utilization is 88 IOPs.  At 70% utilization, T(r) = 7.8 / (1 - .70) = 26ms.  However, if my average response time T(r) is 0.026 seconds, then 1/0.026 is approximately 38 IOPs.

How does starting with 88 IOPs give a response time that only allows 38 IOPs?

Likewise, as IOPs approach capacity, average response time approaches infinity.  So, I can do 128 I/O operations per second, although each one never finishes?  It appears I don't understand a definition somewhere, and I'm guessing it's "utilization."  Can somebody please provide some clarity?

Thanks!