Start a Conversation

This post is more than 5 years old


Go to Solution


May 20th, 2011 15:00

Information storage and management (ISM) Impact

Given the following exercise in the  course:



Consider an application that requires 1TB of storage

capacity and performs 4900 IOPS

Application I/O size is 4KB

As it is business critical application, response time must be within

acceptable range

Specification of available disk drive:

Drive capacity = 73 GB

15000 RPM

5 ms average seek time

40 MB/sec transfer rate

Calculate the number of disks required?

and its solution:

Calculate time required to perform one I/O

Seek time + (rotational delay)/speed in RPM + (block size/transfer rate)

Therefore, 5 ms + 0.5 /15000 + 4K/40MB = 7.1 msec

Calculate max. number of IOPS a disk can perform

1 / 7.1 ms = 140 IOPS

For acceptable response time disk controller utilization must be less

than 70%

Therefore, 140 X 0.7 = 98 IOPS

To meet application

Performance requirement we need 4900/98 i.e. 50 disk

Capacity requirement we need 1TB/ 73 GB i.e. 14 disk

Disk required = max (capacity, performance)

Can someone please explain how

5 ms + 0.5 /15000 + 4K/40MB = 7.1 ms

and how the 0.5 in the equation above was derived?

5 Practitioner


274.2K Posts

May 20th, 2011 19:00

When I was preparing for the exam last year, I was confused like h3ll just like you.

I tried to figure out everything but failed consistently. .

Finally, I found an explanation online which made a lot of sense. Turned out that the explanation was not helpful at all missing many important factors. Forget what you are reading there... here is what you need to understand

The formula for 1 I/O is:

Avg. Seek Time + Avg. disk Latency(or rotational delay) + (BlockSize/Transfer Rate)

The issue here is how to find the Avg. disk Latency(or rotational delay)

Understanding this will help immensly:

Avg. Rotational Delay is 1/2 of the time it takes for ONE rotation.

We know that the RPM (Rotation Per minute) is 15000


15000 Rotation in 60 seconds

=> 1  Rotation = 60/15000 seconds or (60/15000) x 1000 ms = 4 ms

so Avg. Rotational Delay is 1/2 of 4ms = 1/2 x 4 = 2 ms

Hope that helped

5 Practitioner


274.2K Posts

May 23rd, 2011 12:00

Thanks AQ.  I am very good at math and this problem threw me and got me mad and frustrated that I could not "reverse engineer" the numbers no matter how hard I tried.  So, am I to assume that ALL of the problems that have to do with calculations are also going to be wrong?

5 Practitioner


274.2K Posts

May 24th, 2011 20:00


I can't comment on the other problems involving calculation. I struggled with this one in particular hence remembered it so clearly...

Good luck with your first proven certification. I am sure you will do good.

No Events found!